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https://www.aimsciences.org/article/doi/10.3934/proc.2013.2013.781 | # American Institute of Mathematical Sciences
2013, 2013(special): 781-790. doi: 10.3934/proc.2013.2013.781
## Existence and uniqueness of entropy solutions to strongly degenerate parabolic equations with discontinuous coefficients
1 Department of General Education, Salesian Polytechnic, 4-6-8 Oyamagaoka, Machida-city, Tokyo, 194-0215
Received September 2012 Revised February 2013 Published November 2013
In this paper, we consider the initial value problem for strongly degenerate parabolic equations with discontinuous coefficients. This equation has the both properties of parabolic equation and hyperbolic equation. Therefore, we should choose entropy solutions as generalized solutions to the equation. Moreover, entropy solutions to the equation may not belong to $BV$ in our setting. These are difficult points for this type of equations.
In particular, we consider the case that coefficients are the functions of bounded variation with respect to the space variable $x$. Then, we prove the existence of Kružkov type entropy solutions. Moreover, we prove the uniqueness of the solution under additional conditions.
Citation: Hiroshi Watanabe. Existence and uniqueness of entropy solutions to strongly degenerate parabolic equations with discontinuous coefficients. Conference Publications, 2013, 2013 (special) : 781-790. doi: 10.3934/proc.2013.2013.781
##### References:
[1] J. Aleksić and D. Mitrovic, On the compactness for two dimensional scalar conservation law with discontinuous flux, Comm. Math. Science, 4 (2009), 963-971. Google Scholar [2] L. Ambrosio, N. Fusco and D. Pallara, "Functions of Bounded Variation and Free Discontinuity Problems", Oxford Science Publications, (2000). Google Scholar [3] J. Carrillo, Entropy solutions for nonlinear degenerate problems, Arch. Rational. Anal., 147 (1999), 269-361. Google Scholar [4] L. C. Evans and R. Gariepy, "Measure theory and fine properties of functions", Studies in Advanced Math., CRC Press, London, (1992). Google Scholar [5] K. H. Karlsen, M. Rascle and E. Tadmor, On the existence and compactness of a two-dimensional resonant system of conservation laws, Commun. Math. Sci. 5, (2007), 253-265. Google Scholar [6] K. H. Karlsen, N. H. Risebro, On the uniqueness and stability of entropy solutions of nonlinear degenerate parabolic equations with rough coefficients, Discrete Contin. Dyn., 9 (2003), 1081-1104. Google Scholar [7] K. H. Karlsen, N. H. Risebro and J. D. Towers, On a nonlinear degenerate parabolic transport-diffusion equation with a discontinuous coefficient, Electron. J. Differential Equations, 28 (2002), 1-23 (electronic). Google Scholar [8] K. H. Karlsen, N. H. Risebro and J. D. Towers, $L^{1}$ stability for entropy solutions of nonlinear degenerate parabolic convective-diffusion equations with discontinuous coefficients, Skr. K. Vidensk. Selsk., (3) (2003), 1-49. Google Scholar [9] S. N. Kružkov, First order quasilinear equations in several independent variables, Math. USSR Sbornik, 10 (1970), 217-243. Google Scholar [10] C. Mascia, A. Porretta and A. Terracina, Nonhomogeneous Dirichlet problems for degenerate parabolic-hyperbolic equations, Arch. Rational Mech. Anal., 163 (2002), 87-124. Google Scholar [11] E. Yu. Panov, Existence and strong pre-compactness properties for entropy solutions of a first-order quasilinear equation with discontinuous flux, Arch. Rational Mech. Anal., 195 (2010), 643-673. Google Scholar [12] L. Tartar, Compensated compactness and applications to partial differential equations, Nonlinear analysis and mechanics: Heriot-Watt Symposium, Vol. IV, Pitman, Boston, Mass. London, (1979), 136-212. Google Scholar [13] H. Watanabe, Initial value problem for strongly degenerate parabolic equations with discontinuous coefficients, Bulletin of Salesian Polytechnic 38 (2012), 13-20. Google Scholar [14] H. Watanabe, Solvability of boundary value problems for strongly degenerate parabolic equations with discontinuous coefficients, Discrete Contin. Dyn. Syst. Ser. S, 7(2014), no.1, 177-189. Google Scholar [15] H. Watanabe and S. Oharu, $BV$-entropy solutions to strongly degenerate parabolic equations, Adv. Differential Equations 15 (2010), 757-800. Google Scholar
show all references
##### References:
[1] J. Aleksić and D. Mitrovic, On the compactness for two dimensional scalar conservation law with discontinuous flux, Comm. Math. Science, 4 (2009), 963-971. Google Scholar [2] L. Ambrosio, N. Fusco and D. Pallara, "Functions of Bounded Variation and Free Discontinuity Problems", Oxford Science Publications, (2000). Google Scholar [3] J. Carrillo, Entropy solutions for nonlinear degenerate problems, Arch. Rational. Anal., 147 (1999), 269-361. Google Scholar [4] L. C. Evans and R. Gariepy, "Measure theory and fine properties of functions", Studies in Advanced Math., CRC Press, London, (1992). Google Scholar [5] K. H. Karlsen, M. Rascle and E. Tadmor, On the existence and compactness of a two-dimensional resonant system of conservation laws, Commun. Math. Sci. 5, (2007), 253-265. Google Scholar [6] K. H. Karlsen, N. H. Risebro, On the uniqueness and stability of entropy solutions of nonlinear degenerate parabolic equations with rough coefficients, Discrete Contin. Dyn., 9 (2003), 1081-1104. Google Scholar [7] K. H. Karlsen, N. H. Risebro and J. D. Towers, On a nonlinear degenerate parabolic transport-diffusion equation with a discontinuous coefficient, Electron. J. Differential Equations, 28 (2002), 1-23 (electronic). Google Scholar [8] K. H. Karlsen, N. H. Risebro and J. D. Towers, $L^{1}$ stability for entropy solutions of nonlinear degenerate parabolic convective-diffusion equations with discontinuous coefficients, Skr. K. Vidensk. Selsk., (3) (2003), 1-49. Google Scholar [9] S. N. Kružkov, First order quasilinear equations in several independent variables, Math. USSR Sbornik, 10 (1970), 217-243. Google Scholar [10] C. Mascia, A. Porretta and A. Terracina, Nonhomogeneous Dirichlet problems for degenerate parabolic-hyperbolic equations, Arch. Rational Mech. Anal., 163 (2002), 87-124. Google Scholar [11] E. Yu. Panov, Existence and strong pre-compactness properties for entropy solutions of a first-order quasilinear equation with discontinuous flux, Arch. Rational Mech. Anal., 195 (2010), 643-673. Google Scholar [12] L. Tartar, Compensated compactness and applications to partial differential equations, Nonlinear analysis and mechanics: Heriot-Watt Symposium, Vol. IV, Pitman, Boston, Mass. London, (1979), 136-212. Google Scholar [13] H. Watanabe, Initial value problem for strongly degenerate parabolic equations with discontinuous coefficients, Bulletin of Salesian Polytechnic 38 (2012), 13-20. Google Scholar [14] H. Watanabe, Solvability of boundary value problems for strongly degenerate parabolic equations with discontinuous coefficients, Discrete Contin. Dyn. Syst. Ser. S, 7(2014), no.1, 177-189. Google Scholar [15] H. Watanabe and S. Oharu, $BV$-entropy solutions to strongly degenerate parabolic equations, Adv. Differential Equations 15 (2010), 757-800. Google Scholar
[1] Hiroshi Watanabe. Solvability of boundary value problems for strongly degenerate parabolic equations with discontinuous coefficients. Discrete & Continuous Dynamical Systems - S, 2014, 7 (1) : 177-189. doi: 10.3934/dcdss.2014.7.177 [2] Sébastien Gouëzel. An interval map with a spectral gap on Lipschitz functions, but not on bounded variation functions. Discrete & Continuous Dynamical Systems, 2009, 24 (4) : 1205-1208. doi: 10.3934/dcds.2009.24.1205 [3] Yunho Kim, Luminita A. Vese. Image recovery using functions of bounded variation and Sobolev spaces of negative differentiability. Inverse Problems & Imaging, 2009, 3 (1) : 43-68. doi: 10.3934/ipi.2009.3.43 [4] Pierpaolo Soravia. Uniqueness results for fully nonlinear degenerate elliptic equations with discontinuous coefficients. Communications on Pure & Applied Analysis, 2006, 5 (1) : 213-240. doi: 10.3934/cpaa.2006.5.213 [5] Dung Le. Partial regularity of solutions to a class of strongly coupled degenerate parabolic systems. Conference Publications, 2005, 2005 (Special) : 576-586. doi: 10.3934/proc.2005.2005.576 [6] Kenneth Hvistendahl Karlsen, Nils Henrik Risebro. On the uniqueness and stability of entropy solutions of nonlinear degenerate parabolic equations with rough coefficients. Discrete & Continuous Dynamical Systems, 2003, 9 (5) : 1081-1104. doi: 10.3934/dcds.2003.9.1081 [7] Denis R. Akhmetov, Renato Spigler. $L^1$-estimates for the higher-order derivatives of solutions to parabolic equations subject to initial values of bounded total variation. Communications on Pure & Applied Analysis, 2007, 6 (4) : 1051-1074. doi: 10.3934/cpaa.2007.6.1051 [8] Gui-Qiang Chen, Kenneth Hvistendahl Karlsen. Quasilinear anisotropic degenerate parabolic equations with time-space dependent diffusion coefficients. Communications on Pure & Applied Analysis, 2005, 4 (2) : 241-266. doi: 10.3934/cpaa.2005.4.241 [9] Zhigang Wang, Lei Wang, Yachun Li. Renormalized entropy solutions for degenerate parabolic-hyperbolic equations with time-space dependent coefficients. Communications on Pure & Applied Analysis, 2013, 12 (3) : 1163-1182. doi: 10.3934/cpaa.2013.12.1163 [10] Franco Obersnel, Pierpaolo Omari. Multiple bounded variation solutions of a capillarity problem. Conference Publications, 2011, 2011 (Special) : 1129-1137. doi: 10.3934/proc.2011.2011.1129 [11] Roberto Alicandro, Andrea Braides, Marco Cicalese. $L^\infty$ jenergies on discontinuous functions. Discrete & Continuous Dynamical Systems, 2005, 12 (5) : 905-928. doi: 10.3934/dcds.2005.12.905 [12] Cai-Ping Liu. Some characterizations and applications on strongly $\alpha$-preinvex and strongly $\alpha$-invex functions. Journal of Industrial & Management Optimization, 2008, 4 (4) : 727-738. doi: 10.3934/jimo.2008.4.727 [13] Julien Jimenez. Scalar conservation law with discontinuous flux in a bounded domain. Conference Publications, 2007, 2007 (Special) : 520-530. doi: 10.3934/proc.2007.2007.520 [14] Jian Liu, Sihem Mesnager, Lusheng Chen. Variation on correlation immune Boolean and vectorial functions. Advances in Mathematics of Communications, 2016, 10 (4) : 895-919. doi: 10.3934/amc.2016048 [15] Renato C. Calleja, Alessandra Celletti, Rafael de la Llave. Construction of response functions in forced strongly dissipative systems. Discrete & Continuous Dynamical Systems, 2013, 33 (10) : 4411-4433. doi: 10.3934/dcds.2013.33.4411 [16] Ciprian Preda, Petre Preda, Adriana Petre. On the asymptotic behavior of an exponentially bounded, strongly continuous cocycle over a semiflow. Communications on Pure & Applied Analysis, 2009, 8 (5) : 1637-1645. doi: 10.3934/cpaa.2009.8.1637 [17] Karim Boulabiar, Gerard Buskes and Gleb Sirotkin. A strongly diagonal power of algebraic order bounded disjointness preserving operators. Electronic Research Announcements, 2003, 9: 94-98. [18] G. P. Trachanas, Nikolaos B. Zographopoulos. A strongly singular parabolic problem on an unbounded domain. Communications on Pure & Applied Analysis, 2014, 13 (2) : 789-809. doi: 10.3934/cpaa.2014.13.789 [19] Maria Alessandra Ragusa. Parabolic systems with non continuous coefficients. Conference Publications, 2003, 2003 (Special) : 727-733. doi: 10.3934/proc.2003.2003.727 [20] Teemu Lukkari, Mikko Parviainen. Stability of degenerate parabolic Cauchy problems. Communications on Pure & Applied Analysis, 2015, 14 (1) : 201-216. doi: 10.3934/cpaa.2015.14.201
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https://www.preprints.org/manuscript/201712.0191/v2 | Preprint Article Version 2 Preserved in Portico This version is not peer-reviewed
Lagrangian Function on the Finite State Space Statistical Bundle
Version 1 : Received: 26 December 2017 / Approved: 27 December 2017 / Online: 27 December 2017 (10:56:37 CET)
Version 2 : Received: 25 January 2018 / Approved: 26 January 2018 / Online: 26 January 2018 (04:25:24 CET)
A peer-reviewed article of this Preprint also exists.
Pistone, G. Lagrangian Function on the Finite State Space Statistical Bundle. Entropy 2018, 20, 139. Pistone, G. Lagrangian Function on the Finite State Space Statistical Bundle. Entropy 2018, 20, 139.
Journal reference: Entropy 2018, 20, 139
DOI: 10.3390/e20020139
Abstract
The statistical bundle is the set of couples $( Q , W )$ of a probability density Q and a random variable W such that $\mathbb{E}$Q [W] = 0. On a finite state space, we assume Q to be a probability density with respect to the uniform probability and give an affine atlas of charts such that the resulting manifold is a model for Information Geometry. Velocity and accelleration of a one-dimensional statistical model are computed in this set up. The Euler-Lagrange equations are derived from the Lagrange action integral. An example of Lagrangian using minus the entropy as potential energy is briefly discussed.
Subject Areas
information geometry; statistical bundle; lagrangian function
Views 0 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9364798069000244, "perplexity": 1495.6185528941508}, "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-2020-50/segments/1606141747774.97/warc/CC-MAIN-20201205104937-20201205134937-00059.warc.gz"} |
https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=38&t=16240 | ## Dipole Moment in a Lewis Structure
Jason Slater 3I
Posts: 16
Joined: Wed Sep 21, 2016 2:58 pm
### Dipole Moment in a Lewis Structure
Can someone explain why we draw an arrow to represent the dipole moment in a lewis structure? What does it tell us?
Chem_Mod
Posts: 19177
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 833 times
### Re: Dipole Moment in a Lewis Structure
The arrow points to the delta negative pole in the molecule, the tail of the arrow indicates the delta positive pole of the molecule.
Katherine_Zhuo_3B
Posts: 23
Joined: Wed Sep 21, 2016 2:57 pm
### Re: Dipole Moment in a Lewis Structure
With the arrow showing us the delta positive pole and delta negative pole, we can determine which atom in the molecule has a higher electronegativity than the other, and will pull the shared electrons closer when sharing.
William Shu 1G
Posts: 18
Joined: Wed Sep 21, 2016 2:56 pm
### Re: Dipole Moment in a Lewis Structure
Is it just a matter of determining which atom has a higher electronegativity, and then drawing the dipole moment from there?
Alexandra Watts 3L
Posts: 14
Joined: Wed Sep 21, 2016 2:56 pm
### Re: Dipole Moment in a Lewis Structure
How do we know which is negative and which is positive?
Noor_Chahal_3G
Posts: 18
Joined: Wed Sep 21, 2016 2:56 pm
### Re: Dipole Moment in a Lewis Structure
The delta negative pole is the atom that is higher in electronegativity.
The delta positive pole is the atom that is lower in electronegativity.
For example in H2O, oxygen is more electronegative than hydrogen, so oxygen would be delta negative. The 2 hydrogens have a lower electronegativity, so they would be delta positive. | {"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": 0.9254964590072632, "perplexity": 3863.871727990911}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610704799741.85/warc/CC-MAIN-20210126104721-20210126134721-00143.warc.gz"} |
http://swmath.org/software/8678 | # CSP
A tracking semantics for CSP. CSP is a powerful language for specifying complex concurrent systems. Due to the non-deterministic execution order of processes and to synchronizations, many analyses such as deadlock analysis, reliability analysis, and program slicing try to predict properties of the specification which can guarantee the quality of the final system. These analyses often rely on the use of CSP’s traces. In this work, we introduce the theoretical basis for tracking concurrent and explicitly synchronized computations in process algebras such as CSP. Tracking computations is a difficult task due to the subtleties of the underlying operational semantics which combines concurrency, non-determinism and non-termination. We define an instrumented operational semantics that generates as a side-effect an appropriate data structure (a track) which can be used to track computations. Formal definition of a tracking semantics improves the understanding of the tracking process, but also, it allows to formally prove the correctness of the computed tracks.
## References in zbMATH (referenced in 1 article )
Showing result 1 of 1.
Sorted by year (citations) | {"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": 0.9279512166976929, "perplexity": 1659.5881622754703}, "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/1600400189928.2/warc/CC-MAIN-20200919013135-20200919043135-00715.warc.gz"} |
https://www.glocktalk.com/threads/thunderbird-1-07-prob.483235/ | close
Privacy guaranteed - Your email is not shared with anyone.
Thunderbird 1.07 prob...
Discussion in 'Tech Talk' started by IndyGunFreak, Jan 7, 2006.
1. IndyGunFreak
Messages:
26,834
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Indiana
Ok, this isn't really a problem, I'm sure its just a setting. I've probably always had this problem, and just never knew, beccuase nobody on hotmail ever pointed it out to me.
When I send an email to a hotmail user, my sentences and paragraphs are broken up terribly. I always hit "enter" at the end of all my sentences, paragraph breaks, etc, but after they receive it, it is completely broken up. For what its worth, I forwarded an email to myself, that someone told me about this on, and it does indeed do this. I don't think its a hotmail setting.
For instance, in Thunderbird, it may will look like this...
bla bla bla bla bla bla bla bla bla bla[enter]
bla bla bla bla bla bla bla bla bla bla[enter]
Then they receive it, and it looks like this...
bla bla bla bla
bla bla bla bla bla bla bla bla[enter]
bla
bla bla bla bla bla
bla bla bla bla bla bla bla[enter]
I hope my bla's properly explain the problem.
2. IndyGunFreak
Messages:
26,834
2,198
Joined:
Jan 26, 2001
Location:
Indiana
OK, this is driving me crazy..
Someone's gotta know the answer to this.
I have a strange feeling it has to do with the "word wrap" setting, which is currently at 80 characters. Should it be lower? Should it be higher?
Could somoeone explain to me what word wrap is exactly?.. I assume that means it wraps the characters at the set amount and makes a new line, but if I'm hitting enter to start a new line, then it shouldn't be starting new lines for me...
Thanks
IGF | {"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": 0.9028708338737488, "perplexity": 1451.3748938300098}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934804518.38/warc/CC-MAIN-20171118021803-20171118041803-00032.warc.gz"} |
https://www.physicsforums.com/threads/finding-magnitudes-of-vectors-given-angles.426693/ | # Homework Help: Finding magnitudes of vectors given angles
1. Sep 5, 2010
### randomhobo
1. The problem statement, all variables and given/known data
Vector A points in the negative direction. Vector B points at an angle of 35.0 above the positive axis. Vector C has a magnitude of 16 and points in a direction 44.0 below the positive axis.
Given that A+B+C=0 , find the magnitudes of A and B .
2. Relevant equations
Ax cos θ
Ay sin θ
3. The attempt at a solution
I think I have the picture of the triangle graphed out, but since there are no 90 degree angles, how do I use the relevant equations to find the magnitudes?
Last edited: Sep 5, 2010
2. Sep 5, 2010
### CompuChip
That's not an equation, it's just a formula with some variables and a cosine.
Maybe you can look up precisely what formula you wanted to write there and explain to us what it means?
(This is not to be strict about the rules; but if you mean what I think you mean, then understanding what that formula says is the key to the solution)
Also, can you post the complete question? You just defined A, B and C, and you have to calculate the missing magnitudes, if I understand it correctly, but that is not enough info.
3. Sep 5, 2010
### brushman
Try using the law of sines (I believe this works).
4. Sep 21, 2010
### DrewBlay
_______________________________________
For angles games from the list of the subjects: http://www.free-training-tutorial.com/angles-games.html" [Broken]
http://www.free-training-tutorial.com/math-games.html" [Broken]
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https://search.datacite.org/works/10.4230/LIPICS.RTA.2011.139 | ### Modes of Convergence for Term Graph Rewriting
Patrick Bahr
Term graph rewriting provides a simple mechanism to finitely represent restricted forms of infinitary term rewriting. The correspondence between infinitary term rewriting and term graph rewriting has been studied to some extent. However, this endeavour is impaired by the lack of an appropriate counterpart of infinitary rewriting on the side of term graphs. We aim to fill this gap by devising two modes of convergence based on a partial order resp. a metric on term... | {"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": 0.9151432514190674, "perplexity": 836.0814050484067}, "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/1512948514051.18/warc/CC-MAIN-20171211203107-20171211223107-00074.warc.gz"} |
https://byjus.com/question-answer/to-reduce-a-rational-number-to-its-standard-form-we-divide-its-numerator-and-denominator/ | Question
To reduce a rational number to its standard form, we divide its numerator and denominator by their
A
LCM
B
HCM
C
product
D
multiple
Solution
The correct option is C HCMFor standard form, we have to divide numerator and denominator by their highest common factor(HCF).Mathematics
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View More | {"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": 0.9806967377662659, "perplexity": 2044.968808045246}, "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-2022-05/segments/1642320303868.98/warc/CC-MAIN-20220122164421-20220122194421-00338.warc.gz"} |
http://math.stackexchange.com/questions/30620/conditional-expectation?answertab=votes | # Conditional expectation
Consider two random variables $\theta$ and $x$ whose supports are $[0,1]$ with the joint distribution function $f(x,\theta).$ Consider a conditional expectation $E[\theta|x]$. Suppose I have some information about $\theta$ which is given as some measurable set $S$. I want to consider a condition that the rate of change of the conditional expectation with respect to $x$ is nonzero for almost every $x$ for each $S$. That is $\frac{\partial}{\partial x}E[\theta|x, S] \ne 0$ for every measurable set $S$. Does this condition makes sense? Can there be any sufficient condition in terms of $f(x, \theta)$?
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What is the utility of this condition? Also do you mean $E[x| \theta]$? – PEV Apr 3 '11 at 6:30
In the last line $f(x,\theta)$ should be $f(x|\theta)$. More importantly, it seems a condition such as the one you are after should involve the (absolute) distribution of $\theta$. But you only introduce the conditional distribution of $x$ conditionally on $\theta$. Any hypothesis on the distribution of $\theta$? – Did Apr 3 '11 at 7:52
Yes you can make assumption on the distribution of $\theta$. – Thales Apr 3 '11 at 11:09
It implies that, whatever information you obtain, when $x$ is different, its estimate of $\theta$ is different. – Thales Apr 3 '11 at 11:28
It means $E[\theta|x].$ – Thales Apr 3 '11 at 11:50 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.9803606867790222, "perplexity": 240.34744883272268}, "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/1394011210359/warc/CC-MAIN-20140305092010-00001-ip-10-183-142-35.ec2.internal.warc.gz"} |
http://www.centelcu.org/Minnesota/coefficient-of-error-stereology.html | Address 115 3rd Ave SW, Crosby, MN 56441 (218) 546-2717
# coefficient of error stereology Nisswa, Minnesota
Devel Neurobio. 2013 Jan;73(1):45-59. 47. • Perform a Pilot Study and check the CE• Understand the cellular distribution• Even distribution and/or high density: visit fewer sites per section• Uneven distribution and/or At least three sections must be sampled for this value to be computed. The Matheron transitive technique makes an important observation; the samples used in a typical stereological procedure are not independent. A step in the derivation for the formula replaces the curve with piecewise parabolic sections.
In order to design the studies in an optimal manner, with regard to the number of individuals, sections, probes, and to be able to critically evaluate the stereological studies made by Profile Counting: Size andOrientation Biasmbfbioscience.comC. Select another clipboard × Looks like you’ve clipped this slide to already. The Schmitz-Hof has such a large range as to be of no practical value.Last but not least think of it this way.
If less than three sections are displayed, these values are not shown in the results. There is another variance as well. This precision or reproducibility is often referred to as the coefficient of error of the estimate, which is a statistical expression for the size of the standard error of the mean You may consult with some statistician in your institution to see which one is more appropriate for your particular project.Richard Logged hokie SuperTech Posts: 43 which CE « Reply #4 on:
CE Scheaffer Estimated Mean Cell Count: Mean number of cells per counting frame across one or more sections. There is clearly a dependency here. Also – how many objects/cells do you expect to see? Hof.
Journal of Microscopy, Vol. 204, Pt 3, December2001, pp. 232±246. 42. Although this value depends to some degree upon the Variance Due to Noise, this dependency will normally be very small if the Variance Due to Noise is reasonably small. This means that instead of choosing every section at random, without regard to what has already been chosen, samples are chosen by choosing the starting one at random and then all NvVref The Fractionator Principle SRS Preferential Sections Independent Uniform Random (IUR) Sections Vertical Uniform Random (VUR) Sections Cycloid FAVER Sections Guide for Deciding on the Precision of Sampling Pitfalls and Solutions
It takes intoaccount:• Probe choice• Region of interest• Section thickness & histology• Object distributionThe Pilot Studymbfbioscience.com 49. CE Scheaffer Estimated Mean Cell Count: Mean number of cells per counting frame across one or more sections. There are several coefficients of error commonly used in Stereology. This value is the primary contributor to the final Coefficient Of Error for a given estimate over a series of sections.
Houllier F., J Microsc 1993;172:249 6. Scheaffer HL. The difference between sections has a variance that is evident even if exhaustive counting is used on each section. Ideally, this value will be both small and of roughly the same magnitude as the Variance Of Systematic Random Sampling. The fact that you are taking the time to learn about all of these different ideas suggests that you are going to do a fine job.Cheers Logged Do more, less well.
You want to publish. If this value is too high, more sampling needs to be performed on each section. Schmitz and P. Formula for the Optical FractionatorThe cell population is determined by sampling asubset or subfraction of tissue within the regionof interest.Population estimate, N, is equal to:Reciprocal of Volume Fraction X Sum of
With sampling, a given estimate of a population will vary from the true (and unknown) number. For quantifying volume with the Cavalieri estimator, various methods have been developed to predict the CE. Geometric Probes• Geometric probes usedfor the sampling• Points for volume• Lines for surface area• Planes for lengths• Volume for numbers• Geometric probes arerequired to report 3Ddatambfbioscience.comHoward CV, Reed MG: Unbiased Stereology. Demystifying Stereology 3.
That does not apply. While the Gundersen CE is the most commonly used, we recommend following the CE protocol established within your field, and reading the references on coefficients of errors to determine which CE Journal of Microscopy, Vol. 196, Pt 1, Oct1999, pp. 69±73. 24. The Scheaffer CE is not an alternative to the Matheron method.
Design-Based Stereology Model-Based Stereology What are you estimating? rumc Trainee Posts: 8 Which Coefficeint of Error value to use in Stereology « on: October 07, 2005, 06:47:52 PM » Hello,I was wondering if anyone has input on which coefficent Now customize the name of a clipboard to store your clips. Neuroscience Reviews Plaque Load in Alzheimer's Disease Placenta Reviews Pulmonary Reviews Destructive Index Estimating the Chord Length of Alveoli Quantifying Sweat Gland Innervation Reviews Renal Reviews Number of Glomeruli Vasculature Vasculature
Estimated Cell Population Count: (Contour Area or Counting Frame Area)*Estimated Mean Cell Count This is a simple linear, biased, approximation. This controversy has its origins by claims in early papers (for example Gundersen, 1986; Gundersen 1988) that it was necessary to count as few as 100 cells to accurately estimate a As mentioned earlier, all of the CE estimation methods are based on models and each method has conditions where the model is not met. Contour Area / Counting Frame Area: Inverse of the fraction of the total area actually sampled.
mbfbioscience.com• Thickness should be measured at every samplingsite• Assumptions pertaining to the post-processingthickness can lead to sampling bias and error• Processing of tissue results in shrinkage• With some techniques, tissue can It is unlikely that your work is ever independently random.If you read Scheaffer's book you see that the examples are things like sampling animals taken in traps. However, the objects being counted have to be randomly distributed and the ROI must be accurately traced for Scheaffer's CE to remain unbiased. The different CE methods and their associated formulas have been developed, based upon different assumptions and with different considerations taken into account, such as the shape of the region of interest,
It is difficult and time consuming to empirically derive the coefficient of error of estimates made of features observed in histological preparations. | {"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": 0.8235493302345276, "perplexity": 2512.6444693387775}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912202199.51/warc/CC-MAIN-20190320024206-20190320050206-00150.warc.gz"} |
http://tex.stackexchange.com/questions/42855/whats-behind-over/42856 | # What’s behind \over?
Typically a TeX command have arguments coming after it. But the command \over, which is used to produce fractions, can access the token before it. How exactly is it implemented and can I define a custom command like that?
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This is exactly the question that got me reading the TeXbook! My understanding is that this behaviour is compiled into TeX; it's always maintaining a tokslist, just in case it happens to hit an \over. Perhaps one could hook into this tokslist? – L Spice Jan 30 '12 at 21:31
I remember once I found a command that takes a command and prints its definition, with which I tested \over but got nothing. I too guessed it was built-in but still hoped there was a way to access the token list (although David’s answer seems to imply there’s no way to do that…). – JC Chu Jan 30 '12 at 21:45
@LSpice Actually TeX always maintains the "current math list" (in math mode) and when it finds \over it puts this list aside, starts another math list and when this one is over it builds the fraction from the two lists. – egreg Jan 30 '12 at 21:52
@egreg thanks! (Plus some characters.) – L Spice Jan 30 '12 at 22:12
It's a TeX primitive so no you can't define commands like it. It's also a pain in the neck and the cause of many of the problems in math mode, as it means that you can not be sure when you first encounter any math mode token what style things will end up in, hence the need for \mathchoice and various other horrors. If the primitive had had normal prefix syntax like LaTeX's \frac it wouldn't have been necessary.
It occurs to me that there is one other TeX primitive that can do this: \lastbox. It doesn't work quite the same; it just grabs the last box produced, if the last thing produced was indeed a box (and if you are not in the "main vertical mode" or math mode). It is somewhat limited, though, since the box is immutable once written, whereas \over switches its parts to "cramped" style, squeezing superscripts and subscripts. It is important to realize, however, that neither \over nor \lastbox can access previous tokens; once a token is expanded/executed, it is gone. They operate only on lists: math lists or vertical/horizontal lists.
As David Carlisle said, though, its syntax is a poor design decision. For user-visible commands you have the luxury of requiring the author to write the operation before its operands. For internal commands, \lastbox is useful for picking apart things that were already produced in order to process them using knowledge that's only available after the fact. | {"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": 0.9826219081878662, "perplexity": 1531.6843354041728}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-48/segments/1386164819343/warc/CC-MAIN-20131204134659-00005-ip-10-33-133-15.ec2.internal.warc.gz"} |
https://www.transtutors.com/questions/which-of-the-following-is-correct-concerning-pcaob-guidance-that-uses-the-term-ldquo-600334.htm | # Which of the following is correct concerning PCAOB guidance that uses the term “should”?
Which of the following is correct concerning PCAOB guidance that uses the term “should”?
a. The auditor must fulfill the responsibilities.
b. The auditor must comply with requirements unless s/he demonstrates that alternative actions were sufficient to achieve the objectives of the standard.
c. The auditor should consider performing the procedure; whether the auditor performs depends on the exercise of professional judgment in the circumstances.
d. The auditor has complete discretion as to whether to perform the procedure. | {"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": 0.8890071511268616, "perplexity": 2684.4385596863535}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875145533.1/warc/CC-MAIN-20200221142006-20200221172006-00431.warc.gz"} |
https://stats.stackexchange.com/questions/262277/why-would-you-predict-from-a-mixed-effect-model-without-including-random-effects | # Why would you predict from a mixed effect model without including random effects for the prediction?
This is more of a conceptual question, but as I use R I will refer to the packages in R. If the aim is to fit a linear model for the purposes of prediction, and then make predictions where the random effects might not be available, is there any benefit to using a mixed effects model, or should a fixed effect model be used instead?
For example, if I have data on weight vs. height with some other information, and build the following model using lme4, where subject is a factor with $n$ levels ($n=no.samples$):
mod1 <- lmer(weight ~ height + age + (1|subject), data=df, REML=F)
Then I want to be able to predict weight from the model using new height and age data. Obviously the by-subject variance in the original data is captured in the model, but is it possible to use this information in the prediction? Let's say I have some new height and age data, and want to predict weight, I can do so as follows:
predict(mod1,newdata=newdf) # newdf columns for height, age, subject
This will use predict.merMod, and I can either include a column for (new) subjects in newdf, or set re.form =~0. In the first instance, it is not clear what the model does with the 'new' subject factors, and in the second instance, will the by-subject variance captured in the model simply be ignored (averaged over) for the prediction?
In either case it would seem to me that a fixed effect linear model might be more appropriate. Indeed, if my understanding is correct, then a fixed effect model should predict the same values as the mixed model, if the random effect is not used in the prediction. Should this be the case? In R it is not, for example:
mod1 <- lmer(weight ~ height + age + (1|subject), data=df, REML=F)
predict(mod1,newdata=newdf, re.form=~0) # newdf columns for height, age, subject
yields different results to:
mod2 <- lm(weight ~ height + age, data=df)
predict(mod2,newdata=newdf) # newdf columns for height, age
• It could be yoy want to predict for a new group which was not included in the estimation Feb 16, 2017 at 14:19
• Yes, but in that instance why bother with a mixed effect model? What does it give you that a fixed effect model doesn't, if you ignore the random effects in the prediction? Feb 16, 2017 at 14:36
• Well, it might give better estimators, because you have a better (more correct) model of the error structure Feb 16, 2017 at 14:39
Simple thought experiment: You have measured weight and height of 5 infants after birth. And you measured it from the same babies again after two years. Meanwhile you measured weight and height of your baby daughter almost every week resulting in 100 value pairs for her. If you use a mixed effects model, there is no problem. If you use a fixed effects model you put undue weight on the measurements from your daughter, to a point where you would get almost the same model fit if you used only data from her. So, it's not only important for inference to model repeated measures or uncertainty structures correctly, but also for prediction. In general, you don't get the same predictions from a mixed effects model and from a fixed effects model (with violated assumptions).
and I can either include a column for (new) subjects in newdf
You can't predict for subjects which were not part of the original (training) data. Again a thought experiment: the new subject is obese. How could the model know that it is at the upper end of the random effects distribution?
will the by-subject variance captured in the model simply be ignored (averaged over) for the prediction
If I understand you correctly then yes. The model gives you an estimate of the expected value for the population (note that this estimate is still conditional on the original subjects).
• Thank you for the clear explanation and example, this all makes sense. However, where you state You can't predict for subjects which were not part of the original (training) data; isn't setting re.form=~0 and predicting from the population expected value allowing me to do just that? Granted, the model is not using any subject-specific information in the prediction, but is it fair to say that the estimate from a mixed effect model will still be more accurate than that from an equivalent fixed effect model where the subject-specific variation was ignored? Feb 20, 2017 at 14:05
• The fixed model is not applicable since its assumptions are violated. You must use a model that includes a dependence structure.re.form=~0 gives you the population-level prediction, which is the best you can do for new subjects. Feb 20, 2017 at 14:32
• I had the same question when using the glmmLasso package in R. The author of the package, Andreas Groll, stated the glmmLasso procedure uses just the fixed effects for making predictions for new subjects and fixed + random effects for existing subjects in the next time period. Apr 9, 2018 at 20:00 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.8484652042388916, "perplexity": 821.2153952696485}, "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/1652662545326.51/warc/CC-MAIN-20220522094818-20220522124818-00060.warc.gz"} |
https://www.physicsforums.com/threads/which-operator-is-suitable-to-define-a-qubit.269108/ | # Which operator is suitable to define a qubit?
1. Nov 3, 2008
### Lojzek
What requirements must an observable A meet to be usefull as a qubit for
quantum computers?
I think that the wave function must have time independent coefficients in the base
of A (when we are not applying quantum gates). This means that expected value <A> must be constant, so the necessary condition is [A,H]=0.
But is this enough? What if the eigenstates of A have different energies (eigenvalues of H)? Then the phase of the two states is changing with different frequencies. Is quantum computing possible anyway or must both eigenstates have the same energy?
2. Nov 3, 2008
### borgwal
I suppose you mean: for what A are two eigenstates of A useful as basis states for a qubit?
So yes, preferably you have [H,A]=0.
And indeed, that's not quite enough, as you observe, but almost: as long as you know the energy difference precisely enough, then it's okay. In practice, the energy difference (divided by \hbar) is often matched to the difference between two laser frequencies which are stabilized.
3. Nov 5, 2008
### Lojzek
Thanks for the reply. I suppose a precisely known energy difference would not be a big problem, since we could correct the phase with phase shifter gates. Or maybe we could even
use controled energy difference to create phase shifter gates? | {"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": 0.9635211229324341, "perplexity": 1317.5718933014446}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257644271.19/warc/CC-MAIN-20180317035630-20180317055630-00366.warc.gz"} |
https://www.ms.u-tokyo.ac.jp/seminar/2009/sem09-061.html | ## Lie群論・表現論セミナー
開催情報 火曜日 16:30～18:00 数理科学研究科棟(駒場) 126号室 小林俊行 https://www.ms.u-tokyo.ac.jp/~toshi/seminar/ut-seminar.html
### 2009年06月15日(月)
16:30-18:00 数理科学研究科棟(駒場) 126号室
The Schur-Szeg\\"o composition of the degree $n$ polynomials $P:=\\sum_{j=0}^na_jx^j$ and $Q:=\\sum_{j=0}^nb_jx^j$ is defined by the formula $P*Q:=\\sum_{j=0}^na_jb_jx^j/C_n^j$ where $C_n^j=n!/j!(n-j)!$. Every degree $n$ polynomial having one of its roots at $-1$ (i.e. $P=(x+1)(x^{n-1}+c_1x^{n-2}+\\cdots +c_{n-1})$) is representable as a Schur-Szeg\\"o composition of $n-1$ polynomials of the form $(x+1)^{n-1}(x+a_i)$ where the numbers $a_i$ are uniquely defined up to permutation. Denote the elementary symmetric polynomials of the numbers $a_i$ by $\\sigma_1$, $\\ldots$, $\\sigma_{n-1}$. The talk will focus on some properties of the affine mapping
$$(c_1,\\ldots ,c_{n-1})\\mapsto (\\sigma_1,\\ldots ,\\sigma_{n-1})$$ | {"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": 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": 0.9950844645500183, "perplexity": 334.77803555326057}, "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/1664030337731.82/warc/CC-MAIN-20221006061224-20221006091224-00643.warc.gz"} |
http://mathhelpforum.com/number-theory/127473-solve-congruence.html | # Math Help - Solve a Congruence
1. ## Solve a Congruence
49x is congruent to 4000 (mod 999).
I need to solve for x, and I can't figure out how to work it. Trying to use the Euclidean Algorithm.
Thanks all!
2. Originally Posted by meggnog
49x is congruent to 4000 (mod 999).
I need to solve for x, and I can't figure out how to work it. Trying to use the Euclidean Algorithm.
Thanks all!
To begin with one can notice the following:
4000 = 4*999 + 4 ≡ 4 (mod 999)
So, we've reduced the problem to 49x ≡ 4 (mod 999)
Now, by Euclid's algorithm
999 = 20*49 + 19
49 = 2*19 + 11
19 = 11 + 8
11 = 8 + 3
8 = 3*3 - 1
Going backwards in the algorithm you'll get the following
1 = 367*49 - 18*999 ≡ 367*49 (mod 999)
with other words: the inverse of 49 in Z_999 is 367
So,
49x ≡ 4 (mod 999)
367*49x ≡ 367*4 (mod 999)
x ≡ 367*4 = 1468 = 999 + 469 ≡ 469 (mod 999)
x ≡ 469 (mod 999)
and that's it! | {"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": 0.8292807936668396, "perplexity": 1296.6181900354743}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-26/segments/1466783396029.85/warc/CC-MAIN-20160624154956-00000-ip-10-164-35-72.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/287442/treatise-on-non-elementary-integrable-functions | # Treatise on non-elementary integrable functions
All of us mathematicians after some time (and trial-and-error, of course) we are able to guess with reasonable accuracy whether or not a given function is elementary integrable (test yourself: $$\int\frac1{x\sin\bigl(\frac1x\bigr)}\,dx\quad\text{vs}\quad\int\frac1{x^2\sin\bigl(\frac1x\bigr)}\,dx\ ;$$
surely the readers can give a lot more challenging and interesting examples).
I would like to know what is the most comprehensive work (survey, book, whatever) dealing with the theory of integration in elementary terms. I know about the pioneering work of Liouville, as well as the classic paper by Rosenlicht, but what else? what about allowing certain "VIP" non-elementary functions (erf function, for example)? Many thanks.
-
Manuel Bronstein, Symbolic Integration, published by Springer in 2004, may be of some use. – Gerry Myerson Jan 27 '13 at 7:19
Here is the list of references I've seen on the topic of elementary integration. It is eclectic, and not intended to be complete, but contains most of what seem to be the relevant benchmarks, and several expository accounts in varying degrees of detail. Sadly, I am not familiar with Liouville's or Ostrowski's original papers. (Perhaps I'll use this as an excuse to track them down.)
1. MR0223346 (36 #6394). Rosenlicht, Maxwell. Liouville's theorem on functions with elementary integrals. Pacific J. Math. 24 (1968), 153–161.
2. MR0237477 (38 #5759). Risch, Robert H. The problem of integration in finite terms. Trans. Amer. Math. Soc. 139 (1969), 167–189.
3. MR0269635 (42 #4530). Risch, Robert H. The solution of the problem of integration in finite terms. Bull. Amer. Math. Soc. 76, (1970), 605–608.
4. MR0321914 (48 #279). Rosenlicht, Maxwell. Integration in finite terms. Amer. Math. Monthly 79 (1972), 963–972.
5. MR0409427 (53 #13182). Risch, Robert H. Implicitly elementary integrals. Proc. Amer. Math. Soc. 57 (1), (1976), 1–7.
6. MR0536040 (81b:12029). Risch, Robert H. Algebraic properties of the elementary functions of analysis. Amer. J. Math. 101 (4), (1979), 743–759.
7. MR0815235 (87a:12009). Richtmyer, R. D. Integration in finite terms: a method for determining regular fields for the Risch algorithm. Lett. Math. Phys. 10 (2-3), (1985), 135–141.
8. Matthew P Wiener. Elementary integration and $x^x$. Sci.Math post. February 21, 1995. (The pdf version was typed by Apollo Hogan).
9. Manuel Bronstein. Symbolic Integration Tutorial. "Course notes of an ISSAC (International Symposium on Symbolic and Algebraic Computation) '98 tutorial."
10. MR1960772 (2004c:12010). Van der Put, Marius; Singer, Michael F. Galois theory of linear differential equations. Grundlehren der Mathematischen Wissenschaften [Fundamental Principles of Mathematical Sciences], 328. Springer-Verlag, Berlin, 2003. xviii+438 pp. ISBN: 3-540-44228-6.
11. MR2106657 (2005i:68092). Bronstein, Manuel. Symbolic integration. I. Transcendental functions. Second edition. With a foreword by B. F. Caviness. Algorithms and Computation in Mathematics, 1. Springer-Verlag, Berlin, 2005. xvi+325 pp. ISBN: 3-540-21493-3.
12. Brian Conrad. Integration in elementary terms. Unpublished note. (2011?).
13. Moshe Kamensky. Differential Galois theory. "An introduction to Galois theory of linear differential equations."
Bronstein's book in particular is highly recommended.
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– Andres Caicedo Apr 12 '13 at 2:50
– lhf May 13 '13 at 11:47
@lhf Many thanks for the link! – Andres Caicedo May 13 '13 at 12:56 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8298851847648621, "perplexity": 879.2026957845359}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-42/segments/1413507443438.42/warc/CC-MAIN-20141017005723-00242-ip-10-16-133-185.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/basic-definition-of-quadratic-fields.569359/ | # Basic definition of Quadratic Fields
1. Jan 20, 2012
### Math Amateur
I am reading Dummit and Foote Chapter 7.
D&F use a quadratic field as an example of a ring. I am trying to get a good understanding of this ring.
D&F define a quadratic field as follows:
Let D be a rational number that is not a perfect square in and define
$$\mathbb{Q} ( \sqrt D ) = \{ \ a + b \sqrt D \ | \ a,b \in \mathbb{Q} \ \}$$
as a subset of $\mathbb{C}$
In this example D&F write ... "... It is easy to show that the assumption that D is not a square implies that every element of $\mathbb{Q} ( \sqrt D )$ may be written uniquely in the form $a + b \sqrt D$."
How do you show this? Further, I am not sure why this assumption is needed?
Is it because we have both positive and negative roots of a square number like 4, but then only consider the principal root $+ \sqrt 3$ of 3? This seems slightly inconsistent!
Also how does the above fit with the idea that D must be not only not a perfect square but squarefree? Is the squarefree condition on D necessary? If so why?
Can someone please clarify this situation for me?
PeterMarshall
Newcomer
Posts: 8
Joined: Tue Jan 17, 2012 9:22 pm
Location: Tasmania, Australia
2. Jan 20, 2012
### Deveno
it's really pretty simple. if D is a square of a rational number, then Q(√D) = Q.
if D isn't square-free, say D = p2m, for some prime p, and integer m, then we get the same extension of Q with Q(√m) as we do for Q(√D) (in other words, we want "the totally irrational part" of the square root, with any rational factors (hence squares) already factored out).
for example, 12 isn't square free, so adjoining √12 = 2√3, gives us the same field as adjoining √3:
a+b√12 <=> a + 2b√3
c+d√3 <=> c + (d/2)√12
on the other hand, if D IS square-free, we get a UNIQUE quadratic extension (different D's, different extensions).
you can visualize such a ring (which is actually a field, since 1/√D = √D/D, and one can use "the conjugate trick" to find an inverse for a + b√D, as long as not both a and b are 0) as a lattice of rational points in the plane, that is, as a vector space it's isomorphic to Q x Q.
3. Jan 20, 2012
### Math Amateur
Thanks!
Thinking through this now!
4. Jan 20, 2012
### Math Amateur
Thanks again
I can see that if D is not a squarefree number then the field specified by $\mathbb{Q} ( \sqrt D )$ is not unique in the sense that $\mathbb{Q} ( \sqrt 12 )$ is the same field as$\mathbb{Q} ( \sqrt 3 )$.
But D&F's statement seemingly refers to consequences within the field if D is not squarefree.
They write:
"It is easy to show that the assumption that D is not a square implies that every element of $\mathbb{Q} ( \sqrt D )$ may be written uniquely in the form $a + b \sqrt D$."
So they are saying that there is a lack of uniqueness within the field.
Do you agree? Can you clarify?
Another thing that bothers me in D&F's statement is that they do not use the term "squarefree" but go for a lesser condition that D is not a square. Can you clarify this also?
5. Jan 20, 2012
### Deveno
if D is not a square, then Q(√D) is bigger than D. argue by contradiction:
suppose Q(√D) = Q. then √D is in Q, hence we have some rational number m (= √D) with m2 = D, so D is a square.
the uniqueness of the extension is what D being square-free entails. even if D is NOT square-free, there is still only one way to write an element of Q(√D):
suppose a+b√D = c+d√D, where √D is not in Q.
then a-c = (d-b)√D.
case 1) d≠ b:
then (a-c)/(d-b) = √D, and thus √D is in Q, contradiction.
case 2) d = b:
then a-c = 0, so that a = c, which shows uniqueness.
6. Jan 20, 2012
### Math Amateur
Thanks.
OK so the sqarefree part is only to get a unique description of the field
Yes, follow you arguments regarding $Q \sqrt D$ being bigger that Q.
I now regard D&F's statement - that got me going on this - as rather misleading!
7. Jan 20, 2012
### morphism
I fail to see what is misleading.
If D is not a square then $\sqrt D$ is irrational. So if $a+b\sqrt D = a'+b' \sqrt D$ with $a,b,a',b' \in \mathbb Q$, then from $a-a' = (b'-b) \sqrt D$ one easily sees that b'=b and hence a'=a (since otherwise $\sqrt D = (a-a')/(b'-b)$ would be rational). It follows that an element x of $\mathbb Q(\sqrt D)$ has a unique expression of the form $x=a+b\sqrt D$ where $a,b \in \mathbb Q$, in the sense that a and b are uniquely determined by x.
If D is a square, say D=E^2 with E rational, then it's easy to see that this fails. For example, we have $a+b\sqrt D = (a-|E|)+(b+1)\sqrt D$ for any $a,b \in \mathbb Q$.
8. Jan 21, 2012
### Math Amateur
Thanks for that help and guidance
Your last two sentences clarified the situation for me! | {"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": 0.9217544198036194, "perplexity": 712.2465200204158}, "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/1531676590295.61/warc/CC-MAIN-20180718154631-20180718174631-00475.warc.gz"} |
https://brilliant.org/discussions/thread/youtube-channel-continued/ | ×
As mentioned in my note, I uploaded about 50 videos on different topics on my YouTube channel.
I am planning to upload more videos. But, I don't know what topics to start with. Any Suggestions?
P.S. Let those topics be of moderate difficulty, topics which I know of.
Note by Anish Puthuraya
3 years, 4 months ago
Sort by:
How about something that will help in JEE as well as in brilliant??
- 3 years, 4 months ago
Sure..Ill include that
- 3 years, 4 months ago
Thank you very much Also remind everyone when you upload them ;)
- 3 years, 4 months ago
Sure
- 3 years, 4 months ago
Guys, I have uploaded a video on Electric Potential...I hope you guys like it...
Hoping for a like or two, so that Im motivated to do more such videos.
- 3 years, 4 months ago
- 3 years, 4 months ago
- 3 years, 4 months ago
Physics problems and solutions would be good ;)
- 3 years, 4 months ago
Perhaps you can make some based purely on magnetism?
- 3 years, 4 months ago
Problems or Theory?
- 3 years, 4 months ago
Both. Some videos may have theory whereas other videos on problems.
- 3 years, 4 months ago
Ok...
- 3 years, 4 months ago
Hi anish, your videos uploaded are awesome. I think now you should upload some good problem based on electricity and magnetism.
- 3 years, 4 months ago
Thank you...I will be doing that for sure..
- 3 years, 4 months ago
I think you must upload videos that teach how to solve problems that difficulty level of 4,5 on Brilliant?
- 3 years, 4 months ago
Do you mean that I show the solutions? I dont think thats allowed...
- 3 years, 4 months ago
No, I mean techniques involved in solving them............I can't solve a single one!!!!
- 3 years, 4 months ago
Different problems use different techniques...So, indirectly, I will be telling you the solutions if I do so...Perhaps, you could give me an example where showing the techniques does not reveal the solution to that problem...Or shall I just deal with the general techniques involved, rather than the techniques for specific problems...?
- 3 years, 4 months ago
I mean general techniques involved.BTW,how was your JEE MAINS Paper?
- 3 years, 4 months ago
It was horrible...Im getting just 212 marks and 86% in HSC Maharashtra Boards
- 3 years, 4 months ago
No , that's FINE performance.
- 3 years, 4 months ago
- 3 years, 4 months ago
I just have the basic knowledge about quantum mechanics...
- 3 years, 4 months ago
can you upload on relativit special and general
- 3 years, 4 months ago
I have very less knowledge and understanding about that as well...But, Ill try to do some videos on them afterwards...Lately, I havent had much time to do videos, since, I have been very busy with some work regarding my boards
- 3 years, 4 months ago
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http://slideplayer.com/slide/3191552/ | # Transport phenomena in electrochemical systems:
## Presentation on theme: "Transport phenomena in electrochemical systems:"— Presentation transcript:
Transport phenomena in electrochemical systems:
Charge and mass transport in electrochemical cells F. Lapicque, CNRS-ENSIC, Nancy, France Outline 1- Various phenomena in electrolyte solutions 2- Mass transport rates and current density 3- Flow fields in electrochemical cells (a brief introduction to) 4- Mass transfer rates to electrode surfaces
PhD in Chemical Engineering (Nafion Nanocomposite membranes for the Direct Methanol Fuel Cells) Currently working as a Postdoc for Francois Lapicque at CNRS ENSIC, Nancy France Originally from Australia (which is a long, long way from Serbia!)
1- Various physical phenomena in electrolyte solution
Metal ions (and also anions) are highly solvated. Solvatation Relaxation: caused by interactions between the cation and the ionic atmosphere This atmosphere is distorted by the motion of Men+ (It is a sphere for nil electric field) Electrophoretic effect : Force on the ionic atmosphere. acts as an increase in solvent viscosity Existing forces and hindrance to motion Ionic atmosphere (negative charge) Electric field Fionic atm. Men+ Fion NB: these effects are rarely accounted for in models
1- Transport phenomena: introduction to migration
Electrical force on ions (charge q) Velocity of the charged particle Absolute mobility of the ion Specific migration flux (Stokes’ law) Ion: very small particle mol m-3
2- Mass transport rates and current density
General equations of transport Consider a fluid in motion Species i Concentration Ci and velocity of ions vi Defining a barycentric molar velocity Convection flux Ci v Specific flux of species i Ci vi Flux for diffusion and migration Theory of irreversible processes µie: electrochemical potential Ion activity Elec. potential
2- Mass transport rates : the Nernst-Planck equation
From the expression for Ji and the relation between Ni and Ji: Assuming ideal solutions (ai = Ci) leads to the Nernst Planck equation (steady state): Convection : Overall motion of particles with barycentric velocity Migration : Motion of ions (zi) under the electric field Diffusion term (Fick) NB: This equation is not rigorous in most cases, however, it is often used because of its simplicity Other expressions available from the theory of irreversible processes (Stefan maxwell, Onsager …)
2- Mass transport rates : expression of the current density
Equations in electrochemical systems Current density Electroneutrality equation Without C gradients: Medium conductivity (low C) NB: The current density can be defined and calculated anywhere in the electrolytic medium
2- Mass transport rates: some more useful relations
* Relation between diffusivity and ion mobility For the expression of the migration flux and Nernst-Planck equation: only in dilute media which leads to the Stokes Einstein’s relation For more concentrated media, various laws…. Dµ0.7/T = Constant * Transference number: fraction of the current transported by species i
2- Mass transport rates: the trivial case of binary solutions
Binary solutions: one salt dissolved (one cation and one anion) Assuming total dissociation of the salt leads to (general transient expression): same for the anion Replacement of the electrical term, and algebraic rearrangement leads to The salt behaves like an non-dissociated species, with the overall diffusivity D being compromise between D+ and D- with Transference numbers Expression for the current density NB: Although extensively used, the relation is only valid for binary solutions
3- Flow fields in electrochemical cells (an introduction to)
Fluid in motion along a surface The stress applied to the fluid has two components - the normal component, corresponding to a pressure - the second one, along the plane, corresponds to viscous force The structure of the flow can be * Laminar, for which the fluid is divided into thin layers (« laminae » that slide one each another * Turbulent, where the fluid is divided into aggregates. The velocity of the aggregate possesses a random component, in addition to its steady component NB. For too short systems, with local changes in direction and cross-section, the flow is disturbed or non-established
3- Flow fields in electrochemical cells
Two dimensionless numbers allow the flow to be defined in the considered system Friction factor Reynolds number Tangent. stress/kinetic energy Inertia/viscous forces <u> Average velocity, d charactetistic dimension A few comments Laminar/turbulent transition: for Re = 2300? Only in pipes Very large systems are in turbulent flow… e.g. atmosphere, oceans Minimum length for the flow to be established Which characteristic length d? Gotta find the length of highest physical meaning Jf is used for estimation of the pressure drop
3- Flow fields in electrochemical cells: laminar and turbulent flows
Laminar flow (example of a pipe) Jf = 8/Re Parabolic velocity profile The pressure drop varies with <u> Turbulent flow (example of a pipe) More complex expression for the velocity, but the profiles are much flatter One example for the expression of the friction factor: Blasius’ relation Jf = Re for 104 < Re < (the pressure drop varies with <u> 1.8
4- Analogy between the transports of various variables
Specific flux = - Diffusivity x Gradient of the extensive variable Heat (J) Weight (kg) Example Dimensionless numbers: ratio of the diffusivities and orders of magnitude Sc = n/D Pr = n/a Le = a/D Gas Liquids
Did2Ci/dx2 = 0 4- Mass transfer to electrode surfaces
Mass balance (transient) in a fluid element near the electrode surface Whow! The Nernst-film model: a cool shortcut for approximate calculations of mass transfer rates Steady-state conditions Negligible migration Flux Vicinity of the electrode (low u) 1-D approach Did2Ci/dx2 = 0 NB: the velocity profile is Sc1/3 thicker than the concentration profile. i.e. 10 or so Linear profile of the concentration Only the diffusion term
neFkL 4- Mass transfer to electrode surfaces (C’td)
Expression for the current density Defining the mass transfer coefficient, kL neFkL Limiting current density: when CAS tends to 0 Miximum value for the current density iL : Fe can be equal to 1 maximum production rate
4- Mass transfer to electrode surfaces (C’td)
Two dimensionless numbers: Re and Sh (Sherwood) What’s the use of these dimensionless relations?? Hint: Possible change in velocity, dimensions and physicochemical properties (D, n…)
4- Mass transfer to electrode surfaces (C’td) Examples
L=1m dp=0.005 m n=10-6 m2/s D=10-9 m2/s kL = A <u>n Laminar flow 1/3 < n < 0.5 Turbulent flow < n < 0.8
4- Mass transfer to electrode surfaces How to determine them?
Poorly accurate!! Access to overall data, only * Measure pressure drops in the system and use energy correlations (bridge between the dissipated energy and the mass transfer rate) Find the most suitable correlation in your usual catalogue or in published works * Measure the limiting current at electrode surfaces Reliability of the data? Is your system so close? Access to local rates with microelectrodes Find the right electrochemical system (solution, electroactive species Do measurement with the academic system Deduce estimate for kL in the real case using dimensionless analysis | {"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": 0.9028856158256531, "perplexity": 4820.702682394153}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084891926.62/warc/CC-MAIN-20180123111826-20180123131826-00044.warc.gz"} |
https://quantumcomputing.stackexchange.com/questions/14786/what-are-the-possible-results-of-measuring-x-and-z-on-the-state-01-rangle | # What are the possible results of measuring $X$ and $Z$ on the state $|01\rangle+|10\rangle$?
When calculating the probability of getting +1 on X-basis on the first qubit of Bell's state $$|01\rangle+|10\rangle$$, the result is 1/2 with the state after measurement |++⟩ while the probability of measuring the second qubit with the collapse state is 1 and the state after measurement also $$|++\rangle$$.
When calculating the probability of getting +1 on Z-basis on the first qubit of Bell's state $$|01\rangle+|10\rangle$$, the result is 1/2 with the state after measurement |01⟩ while the probability of measuring the second qubit with the collapse state is 0 and the state after measurement also $$|01\rangle$$.
What is other possible result of measuring X and Z, for example, XX, XZ, ZX, ZZ ?
How to build a table to compile all of the possible results?
Starting with the state $$|\psi \rangle = \dfrac{|01\rangle + |10 \rangle }{\sqrt{2}} = \dfrac{1}{\sqrt{2}}\begin{pmatrix} 0 \\ 1 \\ 1 \\ 0 \end{pmatrix}$$.
If you want to find the probability of measuring $$+1$$ in observable $$X = \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}$$ for the first qubit, and $$+1$$ in the observable $$Z = \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}$$ for the second qubit then you can calculate it as $$\langle \psi| M | \psi \rangle = Tr(\rho M)$$ where $$\rho = |\psi \rangle \langle \psi |$$ and here $$M = |+\rangle\langle +| \otimes |0\rangle\langle 0 |$$ since $$|+\rangle = \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix}$$ is the eigenvector corresponding to the $$+1$$ eigenvalue of $$X$$ and $$|0\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$$ is the eigenvector corresponding to the $$+1$$ eigenvalue of $$Z$$. So explicitly
\begin{align} M = |+\rangle \langle+| \otimes |0\rangle\langle 0| &= \bigg[ \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix} \dfrac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \end{pmatrix} \bigg] \otimes \bigg[ \begin{pmatrix} 1 \\ 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \end{pmatrix} \bigg] \\ &= \dfrac{1}{2} \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \otimes \begin{pmatrix} 1 & 0 \\ 0 & 0\end{pmatrix} \\ &= \dfrac{1}{2}\begin{pmatrix} 1 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\\ 1 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\end{pmatrix} \end{align} Thus, $$\langle \psi| M | \psi \rangle = \dfrac{1}{\sqrt{2}}\begin{pmatrix} 0 & 1 & 1 & 0 \end{pmatrix} \dfrac{1}{2}\begin{pmatrix} 1 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\\ 1 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\end{pmatrix} \dfrac{1}{\sqrt{2}}\begin{pmatrix} 0 \\ 1 \\ 1 \\ 0 \end{pmatrix} = \dfrac{1}{4}$$
Also note that, the above is the same if we have done $$Tr(\rho M)$$ since $$Tr\bigg( \rho M \bigg) = Tr\bigg( \begin{pmatrix} 0 & 0 & 0 & 0\\ 0 & 1/2 & 1/2 & 0\\ 0 & 1/2 & 1/2 & 0\\ 0 & 0 & 0 & 0 \end{pmatrix} \dfrac{1}{2}\begin{pmatrix} 1 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\\ 1 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\end{pmatrix} \bigg) = \dfrac{1}{4}$$
You can extend this to other cases as well.
Update:
If you want to do sequential measurement, then you can find the the post measurement state $$|\psi\rangle_{post}$$ then follow the same procedure.
For instance, if we again start with $$|\psi \rangle = \dfrac{|01\rangle + |10 \rangle }{\sqrt{2}} = \dfrac{1}{\sqrt{2}}\begin{pmatrix} 0 \\ 1 \\ 1 \\ 0 \end{pmatrix}$$
and we want to find the probability of measuring $$+1$$ in observable $$X = \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}$$ for the first qubit. Then afterward, finding the probability of measuring $$+1$$ in observable $$X$$ for the first qubit, and $$+1$$ in the observable $$Z = \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}$$ for the second qubit on this collapsed state then we can do it as follow:
First Step: To find the probability of measuring $$+1$$ in observable $$X$$ we can construct $$M$$ as \begin{align} M = |+\rangle \langle+| \otimes I = \bigg[ \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix} \dfrac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \end{pmatrix} \bigg] \otimes \begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix} &= \dfrac{1}{2} \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \otimes \begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix} \\ &= \dfrac{1}{2}\begin{pmatrix} 1 & 0 & 1 & 0\\ 0 & 1 & 0 & 1\\ 1 & 0 & 1 & 0\\ 0 & 1 & 0 & 1\end{pmatrix} \end{align}
And therefore,
$$\langle \psi| M | \psi \rangle = \dfrac{1}{2}$$
and the state after measurement, $$|\psi_{post}\rangle$$, is going to be \begin{align} |\psi_{post}\rangle = \dfrac{ M |\psi \rangle }{ \sqrt{prob(+1)}} = \dfrac{ \dfrac{1}{2}\begin{pmatrix} 1 & 0 & 1 & 0\\ 0 & 1 & 0 & 1\\ 1 & 0 & 1 & 0\\ 0 & 1 & 0 & 1\end{pmatrix} \dfrac{1}{\sqrt{2}}\begin{pmatrix} 0 \\ 1 \\ 1 \\ 0 \end{pmatrix} }{ \sqrt{ 1/\sqrt{2} } } = \dfrac{1}{2} \begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix} \end{align}
Second Step: Now the probability of measuring $$+1$$ in observable $$X$$ for the first qubit, and $$+1$$ in the observable $$Z$$ for the second qubit on this collapsed state $$|\psi_{post} \rangle$$ can be calculated as $$\langle \psi_{post} | M | \psi_{post} \rangle$$ where again $$M = |+\rangle\langle +| \otimes |0\rangle\langle 0 |$$ (as indicated why on the top of this answer). Hence this probability is
$$\langle \psi_{post} | M | \psi_{post} \rangle = \dfrac{1}{2} \begin{pmatrix} 1 & 1 & 1 & 1 \end{pmatrix} \dfrac{1}{2}\begin{pmatrix} 1 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\\ 1 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\end{pmatrix} \dfrac{1}{2} \begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix} = \dfrac{1}{2}$$
Where the post state after this process, $$|\psi_{post 2} \rangle$$ is now in the state
$$|\psi_{post 2} \rangle = \dfrac{ \dfrac{1}{2}\begin{pmatrix} 1 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\\ 1 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\end{pmatrix} \dfrac{1}{2} \begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix} }{ \sqrt{1/2} } = \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 0 \\ 1 \\ 0 \end{pmatrix} = \dfrac{|00\rangle + |10\rangle }{\sqrt{2}}$$
• for the second qubit, suppose the measurement needs to use the collapse state of measuring the first qubit. can you show a measurement of observable Z for the first qubit and then measuring the second qubit using the collapse state? Nov 22 '20 at 19:42
• how to relate the measurement with the non-locality of quantum entanglement? how they are correlated? @KAJ226 Nov 22 '20 at 19:48
• can you explain why the M = |+><+| \otimes |0><0| not I\otimes |0><0| ? Nov 23 '20 at 0:02
• @EaraShahirah That is because I wanted to find the probability of measuring $+1$ in observable $X$ for the first qubit, and $+1$ in the observable $Z$ for the second qubit on this collapsed state... But if you want to find the probability of calculating $+1$ in the $Z$ observable for the second qubit then you would make $M = I \otimes |0\rangle \langle 0|$ like you wrote. Hope that helps. Nov 23 '20 at 0:51
• Thank you for your answer. I got it now. But, can you give me insight behind all this calculation? what actually this calculation is for? @KAJ226 Nov 23 '20 at 15:33
I would start by rewriting the same state in different bases:
• $$XX$$ basis: $$\frac{1}{\sqrt{2}}\left(|++\rangle-|--\rangle\right)$$
• $$XZ$$ basis: $$\frac{1}{2}\left(|+1\rangle+|-1\rangle+|+0\rangle-|-0\rangle\right)$$
• $$ZX$$ basis: $$\frac{1}{2}\left(|0+\rangle-|0-\rangle+|1+\rangle+|1-\rangle\right)$$
• $$ZZ$$ basis: $$\frac{1}{\sqrt{2}}\left(|01\rangle+|10\rangle\right)$$
Now we can look at all the possibilities where $$M_1M_2$$ indicates first measuring the first qubit in the $$M_1$$ basis and then measuring the second qubit in the $$M_2$$ basis:
$$X_1X_2$$:
$$P(X_1=1) = \frac{1}{2}, P(X_2=1|X_1=1) = 1, P(X_2=-1|X_1=1) = 0$$
$$P(X_1=-1) = \frac{1}{2}, P(X_2=1|X_1=-1) = 0, P(X_2=-1|X_1=-1) = 1$$
$$X_1Z_2$$:
$$P(X_1=1) = \frac{1}{2}, P(Z_2=1|X_1=1) = \frac{1}{2}, P(Z_2=-1|X_1=1) = \frac{1}{2}$$
$$P(X_1=-1) = \frac{1}{2}, P(Z_2=1|X_1=-1) = \frac{1}{2}, P(Z_2=-1|X_1=-1) = \frac{1}{2}$$
$$Z_1X_2$$:
$$P(Z_1=1) = \frac{1}{2}, P(X_2=1|Z_1=1) = \frac{1}{2}, P(X_2=-1|Z_1=1) = \frac{1}{2}$$
$$P(Z_1=-1) = \frac{1}{2}, P(X_2=1|Z_1=-1) = \frac{1}{2}, P(X_2=-1|Z_1=-1) = \frac{1}{2}$$
$$Z_1Z_2$$:
$$P(Z_1=1) = \frac{1}{2}, P(Z_2=1|Z_1=1) = 0, P(Z_2=-1|Z_1=1) = 1$$
$$P(Z_1=-1) = \frac{1}{2}, P(Z_2=1|Z_1=-1) = 1, P(Z_2=-1|Z_1=-1) = 0$$
In summary
1. if you're measuring in $$XX$$, it's $$\frac{1}{2}$$ for the first measurement and the second measurement is fully correlated to the first one,
2. if you're measuring in $$ZZ$$, it's also $$\frac{1}{2}$$ for the first measurement but the second measurement is fully anti-correlated to the first one,
3. For the other two combinations, it's all $$\frac{1}{2}$$ for both measurements regardless of the outcome of the first measurement (because the two measurements happen to be completely uncorrelated). | {"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": 3, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9997771382331848, "perplexity": 180.490906933912}, "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/1642320302740.94/warc/CC-MAIN-20220121071203-20220121101203-00419.warc.gz"} |
http://mathhelpforum.com/differential-equations/157874-critical-points-phase-portraits.html | # Math Help - Critical points and phase portraits
1. ## Critical points and phase portraits
Find the critical points and phase portrait of the given autonomous first-order differential equation: dx/dt=x^2 - x^4. Classify each critical point as asymptotically stable, semi-stable, or unstable. Sketch the graph of a typical solution x(t) where x0 has the given values: (a) x0>1 (b) 0<x0<1 (c) -1<x0<0 (d) x0<-1.
2. What did you get for $x(t)$ ?
3. I got (x∙ln(|x+1|)-x∙ln(|x-1|) - 2)/2x but couldn't figure out how to solve for x.
4. Originally Posted by kiddopop
I got (x∙ln(|x+1|)-x∙ln(|x-1|) - 2)/2x
what is this equal to?
5. (x∙ln(|x+1|)-x∙ln(|x-1|) - 2)/2x = t
6. I understand. I was going about the problem in the completely wrong way. | {"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": 0.883258581161499, "perplexity": 2831.708771787093}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-41/segments/1410657135777.13/warc/CC-MAIN-20140914011215-00145-ip-10-234-18-248.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/william-lowell-putnam-competition.48188/ | # William Lowell Putnam Competition
1. Oct 16, 2004
### vsage
Has anyone competed in or been part of an institution that participated in this competition? My school doesn't appear to have been involved in it past 2001 which is a shame because I really would like to pit myself against people from other schools or heck just challenge myself. Anyway I came across a problems archive and was doing a few practice problems so really this post is less about the competition and more about whether I am ready. This was question 1 on the 1995 test and I think I have a solution but I'm not sure it is "rigorous": (solution to problem A-1 located at http://www.unl.edu/amc/a-activities/a7-problems/putnam/-pdf/1995.pdf )
Let $$a, b, c \in T$$
$$abc \in T$$
$$(ab)c \in T$$
Let $$d, e, f \in U$$
$$def \in U$$
$$d(ef) \in U$$
Assume $$(ab) \in U$$
then $$(ab)ef \in U$$
$$ab(ef) \in U$$
For this to be true $$a, b \in U$$ but since U and T are disjoint this is a contradiction so $$ab \in T$$
$$Let g = ab \in T$$
$$gc \in T$$
Is it proven? Please pardon the bad LaTex I will edit this post if it doesn't come out right. Well come to think of it I don't need that g = ab part right?
Last edited by a moderator: Oct 17, 2004
2. Oct 17, 2004
### TenaliRaman
well u almost nailed it if i am right in interpreting your proof ....
If u like solving putnam problems i would recommend u to see ...
1> www.kalva.demon.co.uk
Also check rec.puzzles where many of the putnam problems get solved ...
-- AI
3. Oct 17, 2004
### zefram_c
Sorry, seems not proven to me.
First, a little unrelated matter: you denote the sets T, U as if they were finite. This is not possible, unless they only contain the elements 1 and/or 0. Otherwise, if there are 2 elements >1 (or <-1), multiply the two largest to get something larger than anything in the set; if there are 2 elements between -1 and 1, multiply the two of smallest absolute value to get a smaller still.
Now, if ab(ef) is in U and (ef) is in U, it does not follow that a,b are in U - or it needs more work to do so.
Finally, showing that if g=ab is in T, then for all c, gc is also in T, is not sufficient. You must show this for all g, or otherwise show that all elements in T can be written as a product of two other elements in T.
Here's a simpler proof:
Wolog T is not closed under multiplication, then there exist a,b in T s.t. ab is not in T. Since ab is in R but not T, ab is in U.
Assume that U is also not closed under multiplication. Then there exists c,d in U s.t. cd is in T (as before).
Now a,b and (cd) are in T, hence abcd is in T. But c,d and (ab) are in U, hence abcd is in U. This contradicts the fact that T,U are disjoint.
Heh. I can't believe I got this... someone prove me wrong, or I might regret not writing the Putnam!
4. Oct 17, 2004
### vsage
Yes I realized later my definitions of U and T were ultimately incorrect (and unncesessary for the proof). I'm still learning (1st yr college student) and haven't been able to take any set theory or logic classes so I appreciate your input on this matter. I'm rethinking my post right now so I can resubmit for criticism :). Edit: Bah but I really do see the flaw though I assumed for some reason that $$ef \in U$$ which is dumb because it's assuming what I'm trying to prove!
Last edited by a moderator: Oct 17, 2004
5. Oct 17, 2004
### vsage
OK I am ready to try my luck again! I am really tired now though so I am not sure if I am getting further away from the solution or not.
Suppose there are $$a, b, c \in T$$ and $$d, e, f \in U$$
so $$abc \in T$$ and $$def \in U$$
(The given)
Suppose $$ab not \in T$$ therefore $$ab \in U$$ for some $$a, b \in T$$ because $$T \bigcup U = S$$
Suppose $$ef not \in U$$ therefore $$ef \in T$$ for some $$e,f \in U$$ because $$T \bigcup U = S$$
then $$(ab)ef \in U$$ and $$ab(ef) \in T$$ from supposing the 1st and second lines, respectively.
This implies that $$(ab)(ef) \in U$$ AND $$(ab)(ef) \in T$$
which is a contradiction since U and T are disjoint.
Am I closer? :)
Last edited by a moderator: Oct 17, 2004
6. Oct 17, 2004
### zefram_c
If you highlight the seemingly empty space in my previous post, you'll see
7. Oct 17, 2004
### vsage
Very nice! We have nearly the same proof I see. Thank you so much (also thanks for the link to tenaliraman) | {"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": 0.9088802933692932, "perplexity": 850.4143948475445}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501173405.40/warc/CC-MAIN-20170219104613-00405-ip-10-171-10-108.ec2.internal.warc.gz"} |
http://mathhelpforum.com/algebra/33124-here-s-another-question-finding-x.html | # Math Help - Here's another question on finding X
1. ## Here's another question on finding X
Is there a trick to finding the value of X?
Example:
75% of x = 1050
90% of x = 1260
I know that x = 1400, but I'm wondering if there's an easy way to find it out quickly.
Thx!
2. Originally Posted by matoau
Is there a trick to finding the value of X?
Example:
75% of x = 1050
90% of x = 1260
I know that x = 1400, but I'm wondering if there's an easy way to find it out quickly.
Thx!
of = "Multiply" $75 \%=\frac{3}{4} \mbox{ and }90 \%=\frac{9}{10}$
so solving each equation gives.
$\frac{3}{4}x=1050 \iff x=\frac{4}{3}1050=4 \cdot 350=1400$
and the second
$\frac{9}{10}x=1260 \iff x =\frac{10}{9}1260=10 \cdot 140=1400$
Fractions can make calculations alot easier.
I hope this helps.
3. Yes Yes Yes!
4. actually, i'm confused (again) at this part:
$
x=\frac{4}{3}1050=4 \cdot 350=1400
$
how does this give us 4? and why is the 4 and 3 flipped? division?
$
x=\frac{4}{3}1050=4
$
also, where does this come from?
$
4 \cdot 350
$
yeah, so where does the the 350 and 140 come from? how is it derived?
5. $\frac{3}{4}x=1050$ so we divide both sides by 3/4
$\frac{\frac{3}{4}x}{\frac{3}{4}}=\frac{1050}{\frac {3}{4}}$
remember that 1050 can be written as a fraction $\frac{1050}{1}$
and if we have a fraction divided by a fraction it is the same as multiplying by the reciprocial.
So now we get...
$x=\frac{\frac{1050}{1}}{\frac{3}{4}}=\underbrace{\ frac{1050}{1}\cdot \frac{4}{3}}_{1050/3=350}=350 \cdot 4=1400$
I hope this clears it up
6. Yes, thank you...
7. ## hi
Just divide 1260 with 0,9.
1260/0.9 = 1400
8. Originally Posted by Twig
Just divide 1260 with 0,9.
1260/0.9 = 1400
Fractions > Decimals
Originally Posted by TheEmptySet
$\frac{3}{4}x=1050$ so we divide both sides by 3/4
$\frac{\frac{3}{4}x}{\frac{3}{4}}=\frac{1050}{\frac {3}{4}}$
remember that 1050 can be written as a fraction $\frac{1050}{1}$
and if we have a fraction divided by a fraction it is the same as multiplying by the reciprocial.
So now we get...
$x=\frac{\frac{1050}{1}}{\frac{3}{4}}=\underbrace{\ frac{1050}{1}\cdot \frac{4}{3}}_{1050/3=350}=350 \cdot 4=1400$
I hope this clears it up
Nice Latex | {"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": 14, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9563004970550537, "perplexity": 808.5898844789212}, "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/1422122086930.99/warc/CC-MAIN-20150124175446-00184-ip-10-180-212-252.ec2.internal.warc.gz"} |
https://www.catalyzex.com/paper/arxiv:2103.04546 | Get our free extension to see links to code for papers anywhere online!
# Bandit Linear Optimization for Sequential Decision Making and Extensive-Form Games
Mar 08, 2021
Gabriele Farina, Robin Schmucker, Tuomas Sandholm
Tree-form sequential decision making (TFSDM) extends classical one-shot decision making by modeling tree-form interactions between an agent and a potentially adversarial environment. It captures the online decision-making problems that each player faces in an extensive-form game, as well as Markov decision processes and partially-observable Markov decision processes where the agent conditions on observed history. Over the past decade, there has been considerable effort into designing online optimization methods for TFSDM. Virtually all of that work has been in the full-feedback setting, where the agent has access to counterfactuals, that is, information on what would have happened had the agent chosen a different action at any decision node. Little is known about the bandit setting, where that assumption is reversed (no counterfactual information is available), despite this latter setting being well understood for almost 20 years in one-shot decision making. In this paper, we give the first algorithm for the bandit linear optimization problem for TFSDM that offers both (i) linear-time iterations (in the size of the decision tree) and (ii) $O(\sqrt{T})$ cumulative regret in expectation compared to any fixed strategy, at all times $T$. This is made possible by new results that we derive, which may have independent uses as well: 1) geometry of the dilated entropy regularizer, 2) autocorrelation matrix of the natural sampling scheme for sequence-form strategies, 3) construction of an unbiased estimator for linear losses for sequence-form strategies, and 4) a refined regret analysis for mirror descent when using the dilated entropy regularizer. | {"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": 0.8232929110527039, "perplexity": 1002.9766860509684}, "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-2022-40/segments/1664030334974.57/warc/CC-MAIN-20220927002241-20220927032241-00318.warc.gz"} |
https://thoughtstreams.io/rrees/chains-of-satinav/6682/ | # Chains of Satinav
4 thoughts
last posted Nov. 9, 2014, 4:19 p.m.
1 later thought
0
I hit an untranslated piece of voice acting at the fairy gate, mildly amusing but I had no idea what I was meant to be doing so it was back to the hint sheet.
2 earlier thoughts | {"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": 0.9175734519958496, "perplexity": 4717.785179152644}, "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-2022-33/segments/1659882572198.93/warc/CC-MAIN-20220815175725-20220815205725-00650.warc.gz"} |
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## UBC Theses and Dissertations
### S-R-T division algorithms as dynamical systems McCann, Mark A. 2002
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S-R-T Division Algorithms As Dynamical Systems by Mark A . McCann B . S c , University of B r i t i s h Columbia, 1999 A THESIS S U B M I T T E D IN P A R T I A L F U L F I L L M E N T O F THE REQUIREMENTS FOR THE DEGREE OF Master of Science in T H E F A C U L T Y OF G R A D U A T E STUDIES (Department of Computer Science) We accept this thesis as conforming to the required standard The University of British Columbia A p r i l 2002 © M a r k A . M c C a n n , 2002 In presenting degree freely at this the thesis in partial fulfilment University of British Columbia, I agree that the available for copying of department publication this or of reference thesis by this his for and study. scholarly or thesis for her purposes of Compter Date DE-6 (2/88) It gain shall not Sciintt The University of British Columbia Vancouver, Canada may representatives. financial the requirements I further agree that permission. Department of be is Library permission granted by understood be for an advanced shall make for the that allowed without it extensive head of my copying or my written Abstract S - R - T division, as it was discovered in the late 1950s [4, 19, 23], represented an important improvement i n the speed of division algorithms for computers at the time. A variant of S - R - T division is still commonly implemented i n computers today. A l t h o u g h some bounds on the performance of the original S - R - T division method were obtained, a great many questions remained unanswered. In this thesis, S - R - T division is described as a d y n a m i c a l system. T h i s enables us to bring modern dynamical systems theory, a relatively new development i n mathematics, to bear on an older problem. In doing so, we are able to show that S - R - T division is ergodic, and is even Bernoulli, for a l l real divisors and dividends. ii Contents Abstract ii Contents iii List of Figures v Acknowledgements vi 1 Introduction and Background 1 1.1 Introduction to S - R - T Division 1 1.2 S - R - T D i v i s i o n as a D y n a m i c a l System 5 1.3 Shift Average for D € [|, 1) 12 2 Bernoulli Property 2.1 P r o o f of Bernoulliness 2.2 Entropy oi T 15 16 • D 3 Extensions to Multi-Divisor S-R-T Division 19 23 3.1 M u l t i - D i v i s o r S - R - T Division 23 3.2 P r o o f of Bernoulliness 27 3.3 Some Restrictions on a 32 3.4 Entropy of M u l t i - D i v i s o r S - R - T D i v i s i o n 37 iii 4 Future Work Bibliography L i s t of Figures 1.1 A pseudo state-machine for converting to binary 4 1.2 S - R - T division where p = 0 . 6 7 , and D = 0 . 7 5 5 1.3 Following partial remainder magnitudes graphically for D 0 0.75 = and p = 0 . 6 7 7 0 1.4 A p p l y i n g T / to x = j one hundred times 1.5 Applying T 1.6 A p p l y i n g P associated w i t h T / to f{x) = 1 six times 1.7 A p p l y i n g P associated w i t h T / to f(x) — ^~ f^ 3.1 S - R - T division where p = 0 . 6 7 , D = 0 . 7 5 , and a = ( 0 . 7 5 , 1 , 1 . 2 5 ) 2 7 3.2 C o m b i n e d plot of the regions where oti(e, D) < | and a (e, D) > 4 8 5 4 / 5 to x = f + 0 . 0 0 0 0 1 one hundred times 3 3 5 5 2 ' ^ 8 11 six times 12 0 2 1 for proof of Theorem 1 5 34 3.3 A n example of a non-ergodic system for T D 3.4 A n example of a non-ergodic system for To, 3.5 A n example of a non-ergodic system for T o ) Q a v ; a G 9Tt„>4 34 £ 9^3 35 G WI2 37 Acknowledgements I would like to thank my thesis supervisor, Nick Pippenger, for his patient guidance and for the inspiration for this thesis. I would like to thank Mark Greenstreet for the many excellent comments and suggestions that he provided as my second reader. I would like to thank my good friends Ellen Gethner, Wesley Wong, and Michael Forbes for many good conversations relating to my thesis, and for their numerous helpful suggestions. I would especially like to thank my wife Karen for being a constant support and encouragement to me throughout my Master's degree. Finally, I would like to thank my parents for all the support and encouragement they have given me through numerous years of schooling. M A R K The University April of British Columbia 2002 vi A . M C C A N N Chapter 1 Introduction and Background 1.1 Introduction to S-R-T Division " S - R - T division" roughly refers to a class of non-restoring, binary division algorithms that have been designed for floating-point computers [3, 5, 6, 7, 14, 22]. T h e term "non-restoring" refers to the fact that partial remainders are allowed to range freely through the interval (—1,1), rather than being restored to the positive realm before proceeding to the next step. T h i s feature reduces uses of the adder by about fifty percent. A n equally important feature of this algorithm is the " S - R - T " optimization from whence the algorithm gets its name. In the late 1950's, Sweeney [4], Robertson [19], and Tocher [23] independently made the observation that whenever a p a r t i a l remainder is i n the range (—|, | ) , there w i l l be one or more leading zeros that can be shifted through i n a very short amount of time (usually one cycle). T h e more leading zeros i n a given step, the more the algorithm can avoid costly uses of the adder. A further development of this original algorithm, which is still called S - R - T division, is the algorithm most often implemented in modern point units. floating- In modern S - R - T division, a fixed number of quotient digits are produced every cycle as opposed to a variable number [5, pp. 1 37-62]. A n example of modern S - R - T division i n use is Intel's first release of the P e n t i u m ™ C P U with its infamous "Pentium Bug," which was really just a small error in its S-R-T division implementation. This thesis will restrict its attention to the original version of S-R-T division. To present the simplest type of S-R-T division, we begin with a few definitions for an algorithm similar to that presented by Shively [22, pp. 3-4]: (a) n represents the number of iterations performed in the algorithm. (b) po is the dividend (or initial partial remainder) normalized so that po G [\, !)• (c) Pi G (—1,1), i G N, is the partial remainder after the ith step. (d) D is the divisor normalized to 1). (e) qi G {—1,0,1} (i G { 0 , . . . , n — 1}) is the quotient digit generated by the i t h step. 71-1 (f) Q = ^2 % is the "rounded off" quotient generated after n steps of the i=0 algorithm. n Given the above definitions, after n steps of the division algorithm, we would like it to be true that po = DQ n + s{n) where e(n) is a term that goes to zero as n goes to infinity. A recurrence relation for the S-R-T division algorithm can be stated as Pi+i = < 2pi \Pi\ < \ 2(pi - D) \pi\ > \ and pi > 0 2(pi + D) \pi\ > \ and pi < 0, 2 and : 0 Qi = < 1 P i : i \ < \Pi I : -1 \ > 5 a > 5 n d Pi > 0 and pi < 0 . B y observing that 2{ -{Q)D) \Pi\ < 5 \ 2 ( P i - (1)D) \Pi\ > \ ( Pi+i = Pi 2{ -{-l)D) and pi > 0 |pi| > 5 and pi < 0 , Pi we can rewrite the definition of pi+i as After n steps have been completed, we have = 2 p n n Po - 2qD n 0 - 2- D 2 q -lD.. n l l qi n and then after dividing by 2 and solving for Q we find that n P Pn , PO = 7T + qoD q\D 2° + +... + 71-1 Qn-lD Jn—1 Pn i=0 Now let e(n) = p /2 n n and let Q* = l i m ^ o o Q - Since | p | < 1, i n the limit as n n n goes to infinity Po = DQ*. The quotient bits being generated are not i n a standard binary representation, but it is a simple matter to convert the answer back to standard binary without using any expensive operations. Figure 1.1 shows a simple pseudo state-machine (really a push-down automaton) that converts positive floating-point numbers in the {—1,0,1} representation into binary. Figure 1.1: A pseudo state-machine for converting sequences of { — 1,0,1} into sequences of {0,1} (binary). We assume that the input sequence corresponds to a positive number. The letter ' Z ' is used to indicate that the end of the sequence has been reached, and the symbol e represents the null string. We represent a run of m zeros as 0 • • • 0 and a run of m ones as 1 • • • 1. Sequences of symbols should be read m from left to right. For example, the expression 1/10 ••• 0 means: if a 1 is encountered in the input sequence, write a 1 followed by m zeros. The above conversion automaton implies that conversion happens after the calculation is completed. In reality, the conversion from the generated quotient bits to standard binary is done in hardware on-the-fiy, using registers to convert runs of zeros into runs of zeros or ones in parallel, or by performing a single subtraction. Figure 1.2 shows an example of using the S-R-T division algorithm to divide 0.67 by 0.75. The steps that produce non-zero quotient bits have been shown. In this example, after six uses of the adder, the quotient (0.893) has been determined to four digits of precision. 4 0.67 Po = 0.67 -0.16 Pi = 2(0.67 - D) Pi = 2 ( 2 ( - 0 . 1 6 ) + £>) = 0.22 Pi = 2(2 (0.22) — D) = 0.26 P9 = 2(2 (0.26) - D) = -0.46 2 2 1 1 go = Qo = 1 Qz = - 1 qs = Qz = 0.875 1 Qe = 0.890625 1 Qs = 0.89453125 Pn = 2 ( 2 ( - 0 . 4 6 ) + D) = - 0 . 3 4 Qw = - 1 Qw = 0.8935546875 Pl3 = 2 ( 2 ( - 0 . 3 4 ) + D) = 912 -1 Ql2 = 0.8933105469 1 0.14 1 = Figure 1.2: A n example of S-R-T division when the dividend po = 0.67, and the divisor D = 0.75. The quotient Q* is 0.893. Now, with this simple system of division in hand, we might want to ask certain questions about its performance. For example, we could ask "How many bits of precision are generated per iteration of the algorithm on average?" To answer this question, we must look at the magnitude of \Q* — Q \ = \ /2 \. n n of bits of precision on the nth step is then n — log 2P n Pn The number . In the worst case, p is close n to 1, and therefore we get at least one bit of precision per iteration of the algorithm, regardless of the values of D or Q. Of course, a designer of actual floating-point P hardware probably wants to know the expected performance based on the expected values of p . To answer the many variants of this type of question, it is clear that we n must know something about the distribution of partial remainders over time. The remainder of this thesis is devoted to extending what is known about the answer to this type of question as it relates to S-R-T division and its variants. 1.2 S-R-T Division as a Dynamical System The example in figure 1.2 makes it clear that keeping track of the signs of successive partial remainders is irrelevant in determining how many times the adder will be 5 used for a p a r t i c u l a r calculation. F o r this reason, we only need to consider the magnitudes of successive p a r t i a l remainders. W e now give a reformulation of S - R - T d i v i s i o n that w i l l allow us to look at d i v i s i o n as a d y n a m i c a l system. Definition 1 (S-R-T Division Transformation). F o r D G [5,1), we define the function T D : [0,1) ->• [0,1) as TD(X) = { 2x 0 < x < \ 2(D - x) \ < x < D 2(x - D) D < x < 1 T h i s t r a n s f o r m a t i o n of the u n i t interval represents the successive p a r t i a l remainders that arise as S - R - T d i v i s i o n is carried out b y a divisor D o n a d i v i d e n d x. D is n o r m a l i z e d to 1). T h e d i v i d e n d x is normalized to the successive p a r t i a l remainders Tjj(x) 1) initially, while each of (n G N) subsequently ranges t h r o u g h [0,1). B y using the characteristic function for a set A defined as 1A(*) = we c a n rewrite T (x) D J 1 : i e A 0 : x g" A , as = 2x • l ^ i j ( 1 ) + 2(D - x) • l r i , 0 r ) + 2(x - D) • l o ) m ) (x) . (1.1) If we p l o t equation 1.1 o n t h e u n i t interval, we o b t a i n a very useful v i s u a l i z a t i o n of our transformation. F i g u r e 1.3 shows the plot of To.75(0;) c o m b i n e d w i t h a plot of the successive p a r t i a l remainders that arise while d i v i d i n g 0.67 b y 0.75. T h i s is the same system that was presented earlier i n figure 1.2. N o t i c e that a vertical line i n the interval [\,D) corresponds to a subsequent flip i n the sign of the next p a r t i a l remainder. 6 0 0.25 0.5 0.75 1 X F i g u r e 1.3: A n example of following p a r t i a l remainder magnitudes g r a p h i c a l l y for D — 0.75 a n d po = 0.67. T h e heavy solid lines represent the transf o r m a t i o n To.75, while the abscissa of the t h i n vertical lines represent successive p a r t i a l remainder magnitudes. F i g u r e 1.3 shows a n example of following the t r a j e c t o r y of a single p a r t i a l remainder for a p a r t i c u l a r divisor. A f t e r ten applications of the To.75, there is not any obvious regular pattern, although we expect to see one eventually since the quotient is r a t i o n a l i n this case. O f course, most numbers are not r a t i o n a l and we can deduce that for most numbers, the transformation w i l l never exhibit a repeating pattern. In figures 1.4 a n d 1.5, we see that a very s m a l l change i n the value of the i n i t i a l p a r t i a l remainder quickly produces large differences i n the observed behaviour of the subsequent p a r t i a l remainders. O u r system appears to be chaotic (it certainly has sensitive dependence o n i n i t i a l conditions a n d is topologically transitive), a n d , if this the case, we w i l l gain little understanding by s t u d y i n g the trajectories of 7 i n d i v i d u a l p a r t i a l remainders. T h e logical next step is to study the behaviour of d i s t r i b u t i o n s of points over the whole interval. l r 0.75 0.5 0.25 100 F i g u r e 1.4: T h e result of a p p l y i n g T / 4 5 to x = j one h u n d r e d times. 0.75 ST 0.5 0.25 0 J 20 40 F i g u r e 1.5: T h e result of a p p l y i n g T / 4 5 n 60 80 100 to x = j + 0.00001 one h u n d r e d times. T h e area of understanding the behaviour of ensembles of points under repeated transformation is the r e a l m of d y n a m i c a l systems theory. F o r the remainder of this thesis, we assume a certain amount of f a m i l i a r i t y w i t h the fundamentals of 8 d y n a m i c a l systems theory (or ergodic theory), w h i c h requires some basic unders t a n d i n g of measure theory. W e w i l l include a few helpful b a c k g r o u n d m a t e r i a l definitions as they are needed, b u t mostly we w i l l provide references. A very good i n t r o d u c t i o n to the study of chaotic systems is L a s o t a a n d M a c k e y ' s book Chaos, Fractals, and Noise [11]. For a more detailed i n t r o d u c t i o n to ergodic theory (along w i t h the necessary measure theory needed to understand this thesis), Peter W a l ters's book An Introduction to Ergodic Theory [24] a n d K a r l Petersen's book Ergodic Theory [18] are h i g h l y recommended. Definition 2 (Probability Space). If B is a cr-algebra on subsets of a set X a n d if m is a measure o n B where m(X) = 1, then the triple (X, B, m) is called a probability space. (See [24, p p . 3-9] a n d [11, pp. 19-31] for a good overview of basic measure theory a n d Lebesgue integration.) Definition 3 (Stationary Distribution). Let (X, B, m) be a p r o b a b i l i t y space, let P be the P e r r o n - F r o b e n i u s operator associated w i t h a non-singular t r a n s f o r m a t i o n T : X —>• X, a n d let L Pf = ft 1 denote the L space of (X,B,m)^. l If / € L 1 is such that then / is called a stationary distribution of T. Definition 4 (Perron-Frobenius operator). F o r a probability space the Perron-Frobenius (X,B,m), operator associated w i t h a non-singular t r a n s f o r m a t i o n T : X —)• X is defined b y f Pf(x) d mJT-^B) = [ f{x) d m , JB for B eB . For a piecewise C § transformation T w i t h n pieces, we c a n give a n explicit 2 f o r m u l a for the P e r r o n - F r o b e n i u s operator. Let A = {Ai,A ,... 2 ,A } n be the p a r t i - t i o n of X w h i c h separates T into n pieces. For i € { 1 , . . . , n } , let ti(x) represent the *For a probability space (X,B,m), the L space of (X,B,m) is the set of / : X -> E satisfying J \f(x) \ dm < co. *The o symbol will be used to indicate that a given relation holds except possibly on a set of measure zero. §C denotes the set of all functions with two continuous derivatives. 1 x 2 9 n a t u r a l extension of the ith. C 2 function T ( x ) | ^ . i T h e P e r r o n - P r o b e n i u s operator for T is t h e n tl {x) f(t-\x))-l {x) l dx u[Ai) I n p a r t i c u l a r , for T o (as i n equation 1.1), Pf{x) = \f{\x)-l {x) m + \f{D - \x) • l ,2D-i](aj) (0 + \f{D + \x) •l [ 0 l 2 -2D)(^ • (1-2) W i t h equation 1.2 we c a n show precisely what happens to a n i n i t i a l dist r i b u t i o n of points (described b y a n integrable function) after they are repeatedly t r a n s f o r m e d under Try. Figures 1.6 a n d 1.7 show w h a t happens to two different i n i t i a l d i s t r i b u t i o n of points after five applications of the P e r r o n - F r o b e n i u s operator associated w i t h T / ( z ) . B y the fifth a p p l i c a t i o n , the d i s t r i b u t i o n s look r e m a r k a b l y 3 5 similar. O n e might guess that they are b o t h approaching t h e same final d i s t r i b u t i o n . T h i s s i t u a t i o n is i n m a r k e d contrast to chaotic behaviour observed i n figures 1.4 a n d 1.5. 10 2 2 1.5 1.5 1 1 0.5 0. 5 0 0 0.25 0.5 0.75 0 1 0 0.25 2 2 1.5 1. 5 1 CO 1 0.75 1 0.75 1 1 0 0.25 0.5 0.75 0 1 0 0.25 2 2 1.5 1.5 1 lO 0. 0.5 0 0.25 0.5 a: 0.5 X X 0 0.75 0.5 0.5 0 0.5 X X 0.75 1 0.5 0 1 0 0.25 0.5 X u r e 1.6: T h e result of a p p l y i n g the P e r r o n - F r o b e n i u s operator P associated with T / 3 5 to f(x) = 1 six times. 11 2.5 2.5 2 ^ ^ 2 ^ 1.5 a, 1 i 0.5 0.5 0 0 0.25 0.5 X 0.75 0 1 0 0.25 0.5 X 0.75 1 0 0.25 0.5 0.75 1 0 0.25 0.5 0.75 1 2 2 ^ - 1. 5 1.5 1 CN 0.5 0.5 0 0 0.25 0.5 0.75 0 1 x 2 2 1.5 1.5 1 a, 0.5 0 1 lO 0.5 0 0.25 0.5 X 0.75 1 F i g u r e 1.7: T h e result of a p p l y i n g the P e r r o n - F r o b e n i u s operator P associated 1 f = -—/ !og2 J • • 1 w i t h T 3 / 5 to fix) 1 / 2 1.3 — six times. x Shift Average for D G [§, 1) A n exact equation for the stationary d i s t r i b u t i o n when D € [|, 1) was first given by F r e i m a n [6] a n d is restated by Shively [22] as 1 /(z) 1 = -Q\O,2D-\){?) + 2^1[2£>-i,i)(») • (1-3) T o verify that this is a stationary d i s t r i b u t i o n f u n c t i o n , we b e g i n by a p p l y i n g the P e r r o n - F r o b e n i u s operator as given i n equation 1.2 to equation 1.3 a n d verifying 12 that Pf(x) = f(x). So then, a p p l y i n g P to / we get Pf(x) = ^ ^ l [ o , 2 D - i ) ( H + 2^ [2i>-i,i)(5 )) [o,i)( ) 1 a; 1 + \ ^ l [ o , 2 i > - i ) ( - D - h ) + -fij [2D-\,\){D x L + \ ^l[0,2X?-l)(-D + \ ) x a; ~ k ) ) l(0,2£»-i](a;) + 2 ^ 1 [ 2 D - l , l ) P + \x)^j l[Q, -2D){x) 2 • A s s u m i n g that D € [5,1), a n d observing that a; € [0,1), Pf(x) = \ ^ l [ o , 4 D - 2 ) ( a j ) + ^ [ 4 D - 2 , i ) ( ) ) l[o,i)(a:) 1 a ; + £ ^ ( 2 - 2 D , l ) ( ) + ^ ( 0 , 2 - 2 £ > ] ( 2 ; ) ) l(0,2D-l](a;) 1 a ; 1 + £ ( ^ 1 [ 0 , 2 D - 1 ) ( ^ l[0,2-2O)(a;) • F i n a l l y , assuming that D € [|, 1), we have P f( ) x = 2^1[0,l)(a;) + 2 ^ ( 2 - 2 D , 2 i 3 - l ] ( ^ ) + ^ l ( o , 2 - 2 D ] ( a ; ) + 42jl[o,2-2D)(a:) = 3 1 ^ 1 [ 0 , 2 - 2 D ) ( ^ ) + ^ 1(0,2-20] (*) 1 + -^ [2-2D,2D-l){ ) l x + ^T5 (2-2D,2£)-l](a;) 1 + 2^1[22>-i,i)(a0 = ^l[o,2D-i)(a;) + 2^1[2D-i,i)(a:) = / ( » ) • O n e of the p r i m a r y uses of having a formula for t h e d i s t r i b u t i o n of p a r t i a l remainders is for calculating the shift average for a given divisor. T h e shift average is the average number uses of the shift register (single shift or m u l t i p l i c a t i o n by two) between uses of the adder. U n d e r the assumption that a register shift is a m u c h faster operation t h a n using the adder, the shift average gives a useful characterization of the expected performance of our a l g o r i t h m for a given divisor. W i t h equation 1.3, we know the fraction of bits that require the use of the adder. T o calculate the average number of zero bits generated between non-zero bits (bits requiring use of 13 the adder), we take the reciprocal of the fraction of bits that require the adder. W e calculate the shift average for a divisor D e [f, 1) to b e •<">-I4 * £ T = (L4) Since have not proven that the s t a t i o n a r y d i s t r i b u t i o n s f r o m S - R - T d i v i s i o n are unique, we have no way of k n o w i n g whether or not a shift average c a l c u l a t i o n i n equation 1.4 is correct. T o prove that a l l stationary d i s t r i b u t i o n s are unique, we need to show that TD is ergodic for a l l D G [^,1)- F r e i m a n [6] shows that T o is ergodic for r a t i o n a l D, b u t we extend this result for real D. I n the next section we show that a l l are B e r n o u l l i a n d it is k n o w n that h a v i n g the B e r n o u l l i property implies ergodicity. Before c o n c l u d i n g this chapter w i t h a definition for ergodicity, we w i l l briefly comment o n the derivation of stationary distributions for D G [|, f ) . F o r D G [|, | ) , the stationary d i s t r i b u t i o n functions have been derived, a n d their associated shift average functions have been shown to be constantly three [6, 22]. T h e layout of stationary d i s t r i b u t i o n functions i n the region D G [ j , § ) has several s u r p r i s i n g properties a n d is far f r o m being fully understood. W e discuss the c a l c u l a t i o n of shift averages as a n interesting area for future investigation i n C h a p t e r 4. Definition 5 (Ergodic [11]). L e t (X, B,m) be a p r o b a b i l i t y space a n d let a nonsingular t r a n s f o r m a t i o n T : X —> X b e given. T h e n T is ergodic i f for every set B G B such that T~ {B) L = B, either m(B) = 0 or m(X \B) 14 = 0. Chapter 2 Bernoulli Property In this chapter, we w i l l prove that the class of transformations of the interval that characterizes the S - R - T d i v i s i o n for a l l real divisors D has the property that each t r a n s f o r m a t i o n TD is B e r n o u l l i . A l t h o u g h the basic concept of a B e r n o u l l i shift (the things to w h i c h transformations having a B e r n o u l l i property are isomorphic to) is not difficult, a complete definition requires enough a u x i l i a r y concepts from measure theory (concepts not used anywhere else i n this thesis) that we chose to refer the interested reader to [17, 18, 21, 24] a n d other selections listed i n the B i b l i o g r a p h y . N e i t h e r a n understanding of B e r n o u l l i shifts, nor a f o r m a l definition of what it means to be B e r n o u l l i is required to follow the proofs i n this chapter. H a v i n g said this, we should mention informally the connection between B e r n o u l l i shifts and transformations h a v i n g the B e r n o u l l i property. T h e transformation TD is an non-invertible e n d o m o r p h i s m of the unit interval. T h i s means that f r o m a given p a r t i a l remainder we can predict a l l future p a r t i a l remainders, but we cannot uniquely predict past p a r t i a l remainders. T h e r e is a natu r a l way (called the n a t u r a l extension) to make our t r a n s f o r m a t i o n invertible (an a u t o m o r p h i s m ) on a larger space. Specifically, each non-invertible transformation TD h a v i n g the B e r n o u l l i property has a n extension to a n a u t o m o r p h i c transformat i o n , isomorphic to a two-sided B e r n o u l l i shift [18, p p . 13,276]. F r o m the way that 15 entropy for a transformation is denned, the entropy for a n a u t o m o r p h i c B e r n o u l l i t r a n s f o r m a t i o n associated w i t h a non-invertible B e r n o u l l i t r a n s f o r m a t i o n is the same as the entropy for the non-invertible B e r n o u l l i transformation. B y p r o v i n g that a l l transformations T p are B e r n o u l l i , a n d b y p r o v i n g that entropy of each To is the same, we w i l l be able to conclude that the n a t u r a l extensions of S - R - T d i v i s i o n algorithms are isomorphic to each other for a l l divisors. 2.1 Proof of Bernoulliness Definition 6 (of Bowen [1], Expanding). W e w i l l say that a t r a n s f o r m a t i o n T on a n interval is expanding i f it has the property that s u p u-(T U) = 1 for a l l open n n > 0 intervals U w i t h p{U) > 0, where / i is any n o r m a l i z e d measure that is absolutely continuous w i t h respect to Lebesgue measure. Definition 7 (Straddle). L e t U be a n interval of reals (either o p e n , closed, or half open) a n d let p G K . If p G U°J then we say that U straddles p. + Theorem 1. The S-R-T division transformation is expanding for all real divisors. Proof. L e t (X,B,m) be a p r o b a b i l i t y space where X = [0,1), B is the B o r e l o- algebra o n X a n d m is the Lebesgue measure o n H*. L e t T D : X ->• X be the S - R - T d i v i s i o n t r a n s f o r m a t i o n for a given normalized divisor D as defined i n equation 1.1. *The symbol o as the exponent of an interval denotes an open version of the interval. *For an interval [a, b], the Lebesgue measure is defined as m([a, b]) = b — a. 16 Let us define an infinite sequence of intervals U = {C/jjieN as and Ui = U C/°C[0,i) g [0, i ) and (7° g U° T (C7in[0,i)) D ort/?C[±,l) [i, 1) and m(C7i D [0, i ) ) > m ( < 7 [ i , l ) ) i n tf?g[0,±)andtf?g[±,l) n [ ± , i ) ) and n [ o , i ) ) Property 1. For all Proof. Ui such that \ U° and D g U°, m{U i) i+ TD i = 2m{Ui). [5,-D), or [D, 1), then we are in the first If a U° is a subset of either [0, case of the U definition and we apply <m((7 n[i,l)). directly. Since each of the three cases of the expand an interval by a factor of two, it is clear that m(T£>(Ui)) = m(Ui \) + = 2m(Ui). Property 2. Proof. For all Ui where D 0 Ui, m(Ui i) + > m(Ui). Assume that D $Ui. If j £ Ui, then according to Property 1, U{+\ doubles. Otherwise, \ 6 U{ and therefore, to find E/i+i, we must consider the second and third cases of the U sequence. In the worst case, m(Ui D [0, ^)) = m(Ui n [^,D)), and regardless of which half we choose, m(Ui By fl [0, ^)) = m(Ui fl [5, D)) = ^m(Ui). applying Trj to this truncated interval, we double what we halved so that m(Ui) = m{U i). i+ By way of contradiction, let us assume that there exists a sequence of U that never expands to fill X. Such a sequence can never include the point D and the following Property will hold: Property 3. (a) There exists N such m ( [ / ; n [ 0 , ^)),m(Uif][^, straddle that for all i > N 1)) > 0 (in other \), and 17 words, all subsequent intervals must (b) m{Ui PI [0, \)) < m(Ui n that the right Proof of Property half 1)) (in other words, of Ui is not discarded all subsequent by the definition Ui must be such ofU). 3(a) Property 1 says that the only way not to double is to straddle ^. Therefore, at a minimum, it must be the case that ^ is eventually included every time or else the interval will double a sufficient number of times to include D which would be a contradiction. Proof e,\ of Property 3(b) If m{Ui n [0, | ) ) + e') where e > e'. Now U i+1 > m(Ui = T {Ui) n [£, 1)), = T (\ D then we have Ui = {\ - - e, \) = (1 - 2e, 1). B u t , D since D is not i n C/j+i, \ cannot be in t/i+i and Property 3(a) fails, resulting in a contradiction. B y Property 3, we will eventually be in a situation where U{ = ( \ — e', \ +e), e' < e, and Property 3 will hold for every subsequent interval. So then Ui i+ =T {\-e',\+e) =T [\,\ D + e) = (2Z> - 1 - 2e, 2 D - 1] D by Property 3(b). But again by Property 3, U l+2 = T {2D D - 1 - 2e,2D - 1] = T [\,2D - 1] = [2 - 2D,2D D - 1]. It is now clear that ^ is at the midpoint of Ui+2 and that we must now pick the left half of the interval which contradicts Property 3(b). Therefore, D will eventually be included in an interval and the sequence will expand to fill all of X. • We can now prove that the S-R-T division process is weak-mixing, and therefore Bernoulli, by two theorems of Bowen [1]. Theorem 2 (of Bowen [1]). T-invariant system probability Let T be a piece-wise measure, (T, /j,) is weak-mixing, then C 2 map of [0,1], [i be a and A = infn< <i |/'0c)| > 1x the natural 18 extension U of (T, p) is smooth dynamical Bernoulli. W e mention here that the natural of (T, n) is the associated a u - extensions t o m o r p h i c transformation that we alluded to at the b e g i n n i n g of this chapter. See Petersen [18, p. 13] for a n exact definition. Theorem 3 (of Bowen [1]). With T and mixing ifT is / i as in Theorem 2, C Perron-Frobenius with T, then for any f operator (n Sfc=o f* has the property -f /)n^=i under ? s associated convergent that Pf* 2 function be a probability and let T : X —>• X be a piecewise invariant will be weak- expanding. Theorem 4 (of Lasota and Yorke [10]). Let (X,B,m) fc (T,n) such that i n f \T'\ > 1. If P is the in norm to a function = f* space f* € L\. and consequently, G L, the sequence 1 The limit function the measure d/Lt* = / * d m is T. H a v i n g established that TD is e x p a n d i n g , we now use the above three theorems to prove the central result of this thesis. Theorem 5. TD is Bernoulli. Proof. F r o m the definition of T o , we see that TD is C a n d that i n f o < < i |TD'(:E)| = 2 x 2 > 1 since \TD'(X)\ i n f n < < i \TD'(X)\ x = 2 for a l l x for w h i c h the derivative is defined. Since > 1, by T h e o r e m 4 there exists at least one \x such that fi is a s m o o t h T o - i n v a r i a n t p r o b a b i l i t y measure. B y T h e o r e m 1, we see that T h e o r e m 3 holds. Hence, (TD,U-) 2.2 is w e a k - m i x i n g a n d , by T h e o r e m 2 ( T o , ^ ) is B e r n o u l l i . • Entropy of T D K n o w i n g that a l l T o are B e r n o u l l i is a very useful property because we c a n use entropy as a complete invariant to show i s o m o r p h i s m amongst the two-sided B e r n o u l l i shifts associated w i t h T o that have the same entropy. T h i s comes f r o m the contrib u t i o n of O r n s t e i n to the K o l m o g o r o v - O r n s t e i n T h e o r e m . 19 Theorem 6 (of Kolmogorov [8, 9] and Ornstein [16]). Two Bernoulli shifts are isomorphic if and only if they have the same entropy. T h e purpose of this section is t o calculate the entropy o f T p . W e begin w i t h a m u l t i - p a r t definition of entropy along w i t h some s u p p o r t i n g definitions that follow the development presented by Walters [24, pp. 75-87]. Definition 8 (Partition). A partition of (X,B,m) is a disjoint collection of ele- ments of B whose u n i o n is X. Definition 9 (Join). Let V and Q be finite partitions of (X, B, m). T h e n ? V Q = {PnQ : P G V, a n d Q 6 Q } is called the join of V a n d Q. N o t e that V V Q is also a finite p a r t i t i o n of (X, B, m). Definition 10 (Entropy of a partition). L e t (X,B,m) a n d let V = {Pi,...,Pk} be a finite p a r t i t i o n of (X,B,m). be a p r o b a b i l i t y space T h e entropy of the partition is defined as H(V) k = -^miPi) log m(Pi). Definition 11 (Entropy of a transformation with respect to a partition). Suppose T : X —> X is a measure-preserving transformation of the p r o b a b i l i t y space (X, B, m). If V is a finite p a r t i t i o n of (X, B, m), then is called the entropy ofT with respect to partition V. Definition 12 (Entropy of a transformation). L e t T : X —> X be a measurepreserving t r a n s f o r m a t i o n of the p r o b a b i l i t y space (X, B,m) a n d suppose h(T) = s u p / i ( T , V), where the s u p r e m u m is taken over a l l finite partitions V of T h e n h(T) is called the entropy ofT. 20 (X,B,m). T h e following definitions a n d theorems involving C - m a p s a n d P C - m a p s are taken f r o m a paper o f L e d r a p p i e r [12] a n d have been streamlined for o u r argument. Definition 13 (of Ledrappier [12], C-map). A real f u n c t i o n / defined o n a n interval [a, b] is said to be a C-map if / is continuously differentiable a n d its derivative / ' has the following properties: (a) / ' satisfies a H o l d e r condition^ of order e > 0. (b) T h e r e are only a finite number of points x G [a, 6] where f'(x) t h e m by a < a\ < a •.. < a 2 n = 0. W e denote < b w i t h / ' ( a ; ) = 0 for 0 < i < n. (c) T h e r e exist positive numbers k~ (kf) such that log ,k |x—a| 1 -(+) is b o u n d e d i n a left (right) neighborhood of a ; . Definition 14 (of Ledrappier [12], PC-map). A m a p / : [0,1) —> [0,1) is called a PC-map if there exists a finite p a r t i t i o n 0 < b\ < b .. • < b 2 C - m a p f r o m [bj,bj i] + into [0,1), for any j. Theorem 7 (of Ledrappier [12]). Let f be a PC-map. lutely continuous < 1 such that / is a m invariant measure), If then Rohlin's formula is an a.c.i.m. \i (abso- [20] is true: Hf) = j log|/'|d/i. Theorem 8. The entropy h{T ), D ofT D for D e [^,1) is equal to / l o g | T o ' | d/z = log 2. Proof. W e begin by showing that T o is a P C - m a p . B y the definition of a P C - m a p , T p is a P C - m a p if each of the three functions Tu\^ ij ,T!D|[1 y Q D a n d To\y ^ is a D C-map. T r i v i a l l y , each Try restricted to any of the three domains [0, | ) , [D,l) or satisfies a H o l d e r c o n d i t i o n of order e = 1 because each piece of To is just a §A function f(x) defined on an interval [a, b] satisfies a Holder condition of order e E K if there exists c 6 R such that for any two points pi,P2 G [a,b], \f{pi) — f(p2)\ < c\pi - P2\ • + + e 21 line of slope two. T h u s c o n d i t i o n (a) of D e f i n i t i o n 13 is satisfied. C o n d i t i o n (b) is satisfied because there are no points for w h i c h the derivative is equal to zero w i t h i n a given line segment. Therefore, condition (c) is t r i v i a l l y satisfied. T h u s each of the three segments of TD are C - m a p s a n d by D e f i n i t i o n 14, T o is a P C - m a p . N o w , since each TD is B e r n o u l l i , there exists a unique a.c.i.m., call it /z, for each To- B y T h e o r e m 7, we can use R o h l i n ' s formula to calculate the entropy: • W i t h the proof of T h e o r e m 8 we have established i s o m o r p h i s m amongst the a u t o m o r p h i c transformations (or n a t u r a l extension) associated w i t h simple S - R - T d i v i s i o n transformations by a n a p p l i c a t i o n of the K o l m o g o r o v - O r n s t e i n T h e o r e m . T h e key to o b t a i n i n g this result was being able to show that TD has Bowen's exp a n d i n g property. In C h a p t e r 3, we extend the results of this chapter to a more general type of S - R - T d i v i s i o n . 22 Chapter 3 Extensions t o M u l t i - D i v i s o r S-R-T Division 3.1 Multi-Divisor S-R-T Division A c o m m o n o p t i m i z a t i o n to the S - R - T d i v i s i o n a l g o r i t h m is the i n c l u s i o n of a d d i t i o n a l divisors to increase the shift average. I n this section, we prove that a l l such d i v i s i o n algorithms w i t h reasonable assumptions o n the separation of the divisor multiples have the e x p a n d i n g property. It w i l l be useful to define precisely a class of m u l t i divisor S - R - T d i v i s i o n transformations. Definition 15. L e t a G W be such that 1 (a) 0 < ai < a < • • • < a , and (b) F o r a l l x,D e [|, 1), there exists i G { 1 , . . . , n) such that \c<iD — 2 n x\<\. W e define 2 l „ to be the set of a l l a G R , satisfying conditions (a) a n d (b). A l s o , n Definition 16 (Peaks and Valleys). G i v e n a n a between two lines f(x) G 2 l > 2 , the point of intersection N = 2(x - ctiD) a n d g(x) = 2(a iD i+ 23 - x) w i l l be called a peak a n d is denoted by ipi = (^D(cti i + ati), D(a{ i to the abscissa as ipf = ^D(oti i + a j ) , a n d to the ordinate as ipf = D{a.i \ + + + — a;)). For convenience, we w i l l refer — aj). + T h e point of intersection of the two lines f(x) = 2{aiD — x) a n d g(x) = 2(x — otiD) is (cxiD,0) a n d w i l l be called a valley. Definition For a D e 17. a n d Q 6 21, define the t r a n s f o r m a t i o n Tr), (x) a : [0,1) —* [0,1). F o r a S 2 t i , we get the familiar transformation 2x TDA ) 0 < \2(D-x)\ For X < i X < = X < 1. a € 2l , 2 Q<x<\ 2x \2{ D-x)\ i < x < ipf \2{a D-x)\ \ < x a n d ipf < x < 1 . ai 2 For a € 2 l n > 3 , 2x 0 <x < \ \2( D-x)\ \ < x < ipf \2( D-x)\ \<x \2{a D-x)\ \ < x a n d ipn-i < x < l . ai ai n Definition 1 8 . Define Wl = {T n UneN^n- We c a n * n es e Di(X t °^ we c a l l iVd the set of multi-divisor a n a n d ipf < x < tpf +1 : D <E ( ± , 1 ] , a € 2 t } a n d define iW = n -divisor n S-R-T S-R-T division division transformations and transformations. C o n d i t i o n (b) i n D e f i n i t i o n 15 guarantees that the d i v i s i o n a l g o r i t h m generates a new quotient bit every step. A l t h o u g h the c o n d i t i o n makes intuitive sense, it is not i m m e d i a t e l y obvious i f a n a satisfies the c o n d i t i o n just by inspection. L e m m a 10 below provides a n easier way t o check. 24 Lemma 9. If a = ( a i ) , then condition (b) of Definition 15 is satisfied if and only if ai = 1. Proof. If a\ = 1, then maxjr> e[i/2,i) \ \D « i ^ 1 and e G M + — x\ < | . N o w consider the cases when a ]X . If ct\ = 1 + e, then when D = a n d x = \ , \OL\D — x\ = 1 — i = i ^ i . O n the other h a n d , if ct\ = 1 — e , then when D = \ a n d a; = 1 — | , |ai£> - x | = 1 - | - (1 - e ) ± = i ^ \ . Lemma 10. A n condition (i) G 2l a{ € ( 0 , 7 ? ] and (Sketch). if for some otj € a^c? satisfies n (b) if and only ( M J a» € [^,1] Proof a • [1,1 + a*], condition i,j (a) of Definition G { 1 , . . . ,n} (possibly 15 also i = j), satisfies either or € [l,3«i]. L e m m a 9 has shown that a single component a of a w i t h a = 1 is sufficient to ensure that the range of f(x) = 2 \aD — x\ is equal to [0,1) as x a n d D range over [5,1). It is easy, to see based on the proof of L e m m a 9 that if there does not exist i 6 { 1 , . . . ,n} such that ai = 1, then there must exist i,j 6 { 1 , . . . ,n} (i < j) where OJJ < 1 a n d a > 1. 3 Let us assume that i is the largest value where CKJ < 1, a n d let us assume that j is the smallest value where ctj > 1 (therefore j = i + 1). W e make this a s s u m p t i o n because no other scalars of D w i l l have an influence on whether or not c o n d i t i o n (b) is satisfied. C o n s i d e r the case where c*j G (0, ^]. In this case where ai 6 (0, | ] , when D is close enough to 1, some of the line f(x) = 2(x — aiD) appears i n the region (denoted R) where \ < x < 1, 0 < T (x) a < 1. W h e n a p o r t i o n of the line f(x) appears i n region R, we must put restrictions on aj i n terms of ai so that the peak ?/>! is always i n R. ipf is greatest when D = 1. W e find the m a x i m u m allowable value of aj by setting D = 1 a n d solving ipf = 1 for aj: ipf = I Therefore, if a ; G (0, => D(aj then - ai) = 1 G [1,1 + aj]. 25 aj = a ; + 1. In the case where a j € [^, 1], for large enough values of D, the line f(x) 2(x — Dai) = crosses the line x = 1 i n the range [0,1). Because of this, we must loosen the restriction that ctj G [1,1 + ctj]. It is straightforward to calculate that begins to cross the line x = 1 i n the range [0,1) w h e n D = f(x) W e c a n ensure that as D becomes smaller, the peak ipi w i l l always be i n region R by solving tpf = 1 for ctj when D = tp\ = 1 D(aj - on) = 1 => 7f—((Xj - = 1 on) =4> ctj = 3a . t Therefore, if ai € [^, 1], then ctj £ [1, 3aj]. • Definition 19 (Separation). F o r a € 2 l , we define the separation i n a as n separation(a) = max i€{l,...,n-l} 1 + 1 . ttj L i m i t i n g the separation is a convenient way to restrict the subset of 21 being considered. If separation(a) = r, we say that "the divisor multiples i n a are separated by at most a factor of r." F i g u r e 3.1 shows a n example of m u l t i - d i v i s o r S - R - T d i v i s i o n . T h i s example is performing the same calculation as i n figure 1.2, but it has c o m p u t e d the d i v i d e n d w i t h twice as m a n y digits of precision w i t h the same effective n u m b e r of uses of the adders. W e say "effective" because i n m u l t i - d i v i s o r S - R - T d i v i s i o n , there are several adders w o r k i n g i n parallel. In a real i m p l e m e n t a t i o n of m u l t i - d i v i s o r S - R - T d i v i s i o n , the values for a must be carefully chosen so that not too m u c h overhead is required to select a good p a r t i a l remainder. T h e r e is also a tradeoff between the amount of overhead i n choosing a good p a r t i a l remainder a n d the precision to w i t h w h i c h a g o o d p a r t i a l remainder is selected. 26 0.67 Po = 0.67 Pi = 2(0.67 - -0.16 a D) 2 P4 = 2(2 (-0.16) + anD) 2 P7 = -0.155 = 2(2 (-0.155) + a D) = -0.115 2 x Pll = 2(2 (-0.115) + a D) 3 3 = Qo = 1 Qo = 0.035 Q3 = -a Qz = 0.90625 <76 = Q = 0.89453125 9io = - « 3 Qw = 0.8933105469 x 6 = 2(2 (0.035) - a D) = -0.005 Qis = Ql5 = 0.8933334351 P24= 2(2 (0.005) + a D) = -0.155 923 = Q2S = 0.8933333456 Pl6 4 x 7 x Figure 3.1: An example of S-R-T division where three multiples of the divisor are used. In this example the dividend po — 0.67, and the divisor D = 0.75 with divisor multiples a = (0.75,1,1.25). The quotient Q* is 0.893. 3.2 Proof of Bernoulliness In this section, we will show that all multi-divisor S-R-T division transformations are Bernoulli, given a necessary restriction on the multiples of the divisor. As in the case for a single divisor, it will be useful to define a sequence of intervals that are subsets of the sequence of sets that would arise from repeatedly applying TD,Q. to an initial open interval. Unless otherwise noted, assume that the function m denotes the Lebesgue measure. Definition 20. Given an initial open interval U C [0,1) and To^a G 9DT, we define 27 the infinite sequence of intervals U = {Ui}^ as and Ui = U C/°C[0,i) T , (Ui) D a T , (Ui D a t / ° £ [0, orC7?C[i,l) n [0, i ) ) a n d t/? g [ i , 1) a n d m(f/<h[0,i)) T ^ a ^ i a n d C7? g [ ± , l ) U°%%\) n [ i , i ) ) >m{Uif\{\,\)) and m(f7 n[0,i)) i Definition 21 (Critical Points). For a given Tu, a C = { Ci n : i 6 {1, • • •, m } , a E B U {0, ±, 1}} n C2 < . . . < 1. C is called the set of critical Lemma 11 (Doubling). GivenTr),a points for T o £ m be the set of critical j G {1,... , r a - 1}, then m(Ui+i) Proof. Since Ui C [CJ,CJ \] + = ) a • • •, V>n-i}} a n d c i < . ^ ^ sequence of intervals U be defined e 20 and let Ui be some interval C = {ci,... ,c } ) i where a € 2 l , define the set where 5 = {& : i < & < 1 and b G { a i - D , • • •, a - D } U O f , as in Definition <m(f/ n[i l)). in the sequence. points for If Ui C T D , « . Furthermore, [ C J , C J + I ] for let some 2m(Ui). for some j £ { l , . . . , m - l } , because we are i n the first case of the definition of U, either U? C [0, \) or U° C [ i , 1). B y simple inspection of the i n d i v i d u a l cases that define Trj , i(X the points Cj and Cj+i, it is apparent that a l l of Ui, except possibly fall w i t h i n the same case of Tr), a Therefore, the resulting interval Ui+i w i l l be double the length of Ui. Definition 22 (Active Valleys). G i v e n T • Dt(X € 93T , define n V = {ctiD : i 6 { 1 , . . . , n) and \ < a D t V is called the set of active valleys for To, a 28 < 1} . Definition 23 (Active Peaks). Given Tn, G Tl , define a n P = {if>f : t e { l , . . . , n - l } andi<^f<l}. P is called the set of active peaks for To, a Lemma 12 (Non-shrinking). the sequence of intervals {Ui}i^ active valleys for T r j , . For any interval a m(U i) i+ > m(Ui) or m(U ) > i+2 Proof, separation(a) with separation(a) Given TD,OC G < | , let be defined as above and let V denote the set of Ui G U such that V f l Ui = 0, either m(Ui). < | implies that cti \ < + For a given separation, the value of ip\ is maximized when ipf = 1. This implies that aj = We calculate the value of ipf with the assumption that ipf = 1 to get a bound on ipf for D < 1: iPf < D(l ai - ai) = D(l ) ai = £>(§&) = 5 • Case 1: Consider when Ui C [0, \ ] . In this case, m(Ui \) — 2m(Ui). + Case 2: Consider when Ui C [^,1). The interval Ui can span at most one peak. Therefore, m(£/j+i) > m(Ui). m{U ) i+2 = A further observation is that since Ui+i C [0, 2m{Ui). Case 3: Consider when Ui £ [0, \] and Ui % 1). In this case, Ui straddles \ . From the definition of U, we see that in the worst case we might throw away up to half of Ui. Call the part not thrown away Ui and observe that m(Ui) > \m(Ui). Now, either Ui' C [0, \) or Ui' C [i, 1). If [// C [0, ±], then m(C/i+i) = 2m(<7;') > m{Ui). If t/i' C [ i , 1), then m ( C / i ) = 2m(£//) > m(c/;). Lemma 13. A multi-divisor when separation(a) • +2 S-R-T division transformation Tr>, G 9JI is expanding a < |. Proof. Let V be the set of active valleys (as defined in Definition 22) for a Trj,a- Let P be the set of active peaks (as defined in Definition 23) for a To,a- Let U = be the sequence of intervals associated with a To, and an initial interval U. a 29 {Ui\i^n B y way of c o n t r a d i c t i o n , assume that a T£> >a is not e x p a n d i n g . T h i s means that for some T r j , , there does not exist a n interval Ui where any of the points i n V a are contained i n Ui. T h i s is true because i f any of the valley points are i n Ui, then Ui i + = [0, e) or C / j + 1 = [0,e], a n d after a finite number of steps, Ui w i l l have grown to include a l l of [0,1). If there is a sequence U that avoids a l l points i n V, t h e n b y L e m m a 12 it must be true that the intervals i n the sequence can only double a finite number of times. L e t i e N b e the first index for w h i c h there is no j > i such that m(Uj) > 2m(Ui). It now follows that Ui straddles ^ . T h e proof for L e m m a 12 reveals that this is the only s i t u a t i o n where it is not necessarily the case that either m{Ui \) + or rn(Ui+2) = 2m(Ui). I n fact, Ui must straddle b o t h \ a n d m i n P . not s t r a d d l e d a n d m(Ui f l or m(Ui z) + m(Ui n [0, > 2m(Ui). [0, ^)) < m(Ui f l [|,1)), then either m(Ui+2) = 2m{Ui) If m i n P is > 2m([/j) I n the other possibility where m i n P is not straddled a n d ^)) > m{Ui n [1,1)), we find that m(U ) i+2 > 2m{Ui). A s s u m i n g that Ui straddles b o t h ^ a n d m i n P , we also observe that there can be no j > i such that m(Uj D [0, 5)) > m(Uj f l [5,1)) because this q u i c k l y leads to d o u b l i n g . I n other words, the right side must be larger t h a n the left side whenever we straddle | . Therefore, we must be i n the s i t u a t i o n where t/i = ( i - e ' , i + e), e'<£ Ui+i = (min{2(i - aiD),2(a D- U l + 2 = (2min{2(i u i+3 = (min{2(i - aiD),2{a Ui+4 = (2min{2(i => U i + 5 i+l - aiD),2(a D i+1 D- i+1 -aiD),2{a D i+1 = (min{2(i - aiD),2{a D- i+1 _(i+ ))}, 2^?) e 2vf)},^r) -2ltf)},2#) 2 ^ ) } , ^ ) = l7 3 i+ It is apparent that the interval represented by [/j+4 w i l l re-occur every other interval ad infinitum. W e now use this interval to show that i n fact such a sequence of n o n - e x p a n d i n g intervals is not possible. 30 Since straddles \ , we can compare the length of the left a n d right sides of J7j+4. L e t R = [^,2i/>f) denote the right side a n d let L = (4(^ — o^-D), \ ) a n d V = (4(cti+iD — 2ipy), ^) denote the two possibilities for the left side. T h e length of the right side is = 2Vf - m(R) i, while the length of the left side is the larger of two possible lengths m(L) = \ - 4(i - D) ai and m(L') = i-4(a i + i£>-2^) . W e then compare the differences between the right side a n d each of the two possible left sides. T h e first possibility is m(R) - m(L) = 2$ a D)) - I - (I - 4(i - = 2D(a - i+l = 2a D a i t ) - 1 + 2- 4a D { - GctiD + 1 , i+1 while the second possibility is m(R) - m(L') = 2^f - ± - (± - 4 ( a = 2D{a l+x = -2a D i+i l + 1 2\$)) - - ai) - 1 + 4{a D l+1 + 6aiD - - 2D{a l+1 - a )) t 1. It is now clear that m(R) - m(L) = - (m{R) - m(L')) . B u t this means that the length of the left side is always greater t h a n or equal to the length of the right side, w h i c h contradicts our a s s u m p t i o n t h a t the right side must be bigger t h a n the left side whenever the interval straddles ^. • 31 Theorem 14. To,a G %Jl is Bernoulli Proo/. Let T = T o i a . when separation(cx) < |. F r o m the definition of T , we see that T o is C a n d that 2 > a i n f o < < i |T'(a;)| = 2 > 1 since | T ' ( x ) | = 2 for a l l x for w h i c h the derivative is defined. x Since i n f o < < i |T"(a;)| > 1, by T h e o r e m 4, there exists at least one /z such that zz is x a s m o o t h T - i n v a r i a n t p r o b a b i l i t y measure. B y L e m m a 13 we see that T h e o r e m 3 holds w h e n separation(a) < §. Hence, (T, /z) is w e a k - m i x i n g a n d b y T h e o r e m 2, (T,/z) is B e r n o u l l i w h e n separation(a) 3.3 < |. • Some Restrictions on ex. In section 3.2, we showed that i f a l l T D ) Q G 9Jt, i f separation(a) < §, t h e n TD >OC is B e r n o u l l i . I n this section, we construct examples of T G %Jl , for every n, that fail n to be B e r n o u l l i w h e n the restriction that separation(a) Theorem 15. For TD,O. G i/iere ea;zsr; uncountably if separation(a) ffin>4, many a. for which Tu < § is relaxed. [|,1), > §, then for each D G is not ergodic. t0l Proof. W e b e g i n this proof by considering T G 9 J l = 4 n A s s u m e that we relax the restrictions o n a b y e > 0. T h i s means that separation(a) < | + e a n d that no peak can be above the line f(x) = this r e l a x a t i o n , we can define a = (c*i, a ,0:3,0:4) 2 a subset of [0,1) is n o n - e x p a n d i n g . 40C+2 4 Pe 9 £ 1 a n ( ^ a 4 = 40P+24Pe • ^ o r o u W e let ai = r With w i t h respect t o a given D so that g+48D > 2 = a 8 0 £ 8 or>+48£>e' 0 : 3 = constructed a to be v a l i d , we must be careful that conditions (a) a n d (b) of D e f i n i t i o n 15 h o l d . C o n d i t i o n (a) requires that the components of a r e m a i n i n ascending order. T h i s is satisfied w h e n e G (0, ^ ] . ordering is m a i n t a i n e d , separation(a) Since < 3, a n d minrj [i/2,i),ee(o,2/i5] 4 = 1-2 > 1, a g to verify that c o n d i t i o n (b) of D e f i n i t i o n 15 holds, it is sufficient to show (by L e m m a 10) that for a l l values of D a n d e, either ot\, a , or a G [5,1]- B y m a x i m i z i n g a n d 2 3 m i n i m i z i n g over e a n d D, we find that a\ G [0.375,0.7] a n d a G [0.625,1.3]. F i g u r e 2 32 3.2 provides a v i s u a l proof that as e is varied over [0, ^ ] a n d D is varied over [5,1], it is never the case that b o t h a\ < \ a n d a 2 that either a\ or a 2 > 1. Therefore, it is always the case G [5, !]• H a v i n g verified that our defined a satisfies D e f i n i t i o n 15, we calculate that peak ^ = ( i ± i f , l{j±if§) a n d peak </> 3 while r e m a i n i n g above the line g(x) below f(x) = (f^, ftl||). — ^, a n d the point ip W i t h this definition for 1 w i l l always be slightly while r e m a i n i n g above the line g(x) = \ . A l l of the definitions have been chosen so that we are i n a s i t u a t i o n where 1 — ip% = ^ 3 ~~ \ — 2(V'i ~~ \) = 2{ip\ A n o t h e r i m p o r t a n t feature i n this construction is the interval between a D 2 Since ip 2 and |)- — a^D. is not used i n our construction, it is possible to insert a n a r b i t r a r y number of divisor multiples between a D 2 a n d a D . T h u s , the results i n this proof a p p l y 3 to T G 9Jt„ for a r b i t r a r i l y large n. F i g u r e 3.3 illustrates the type of t r a n s f o r m a t i o n that we have constructed. We are now i n a p o s i t i o n to show that there exists a set of points A w i t h 0 < m(A) < 1, for w h i c h TD,OL{A) = A. T h i s is the definition of a t r a n s f o r m a t i o n 5). i + w - A* = [h- m -M+m - h)i 4. = [i - 2(1 - 1]. being non-ergodic [11, p. 59]. Define A = A\ U A It can be shown by calculation that T o ( A i ) ] a TD,OL{M) = A. 2 Therefore, TD,CI(A) U A 2 = A, 2 3 where A\ and = [\ - (ipf - = A\ U A 3 , and T^A^) = A, a n d by definition, To, a is non-ergodic or • non-expanding. 33 Figure 3.2: Combined plot of the regions where a\(e,D) < \ and ct2{e,D) > 1. Over the domain e 6 [0, ^ ] and D € [\, 1], it is never true that both ct\ < \ and «2 > 1- X Figure 3.3: A n example of a non-ergodic system for Trj a ample n = 4, D = a = (||, 1, | | ) , | + j^-. The thick lines represent T r j . ]a € 9Jt„>4- In this exand separation(a) = The coarse dashed line represents the necessary separation restriction on a to guarantee that T D A = ) Q §] is ergodic. u tl> I] u In this case, partial remainders in the set ( i t . !) a r e mapped back to A by T , . This D a means that T D . is not ergodic, and therefore not Bernoulli. ) Q 34 Theorem 16. For To, G 9Jt a 3) if separation(a) > | , then for each D £ [5,1), there exists an a for which T o i a is not ergodic. Proof. T h e proof for this theorem comes as a special case from the proof for T h e orem 15. C o n s i d e r a separation{a) = (oti, 0:2, 0^3,0^4) as defined i n the proof for 15. = | = | + we are i n the special situation where a 2 When = 0:3. Since a l l of the results for the proof of T h e o r e m 15 s t i l l h o l d , we now have a n example transformation T w i t h only three unique multiples of D a n d this T has been proven to be non-ergodic. F i g u r e 3.4 gives a n example of a non-ergodic transformation for D = l. • 2 0.25 0 0.5 0.75 1 X F i g u r e 3.4: A n example of a non-ergodic system for T o , D = a n d « = ( § > f> i f ) - T n e t m c ^ u n e s a £ 9^3- In this example, r e P r e s e n t I'D,a- The coarse dashed line represents the necessary separation restriction o n a to guarantee that To,a is ergodic. In this case, p a r t i a l remainders i n the set A = [±, ^ ] U [ ^ , i § M y § , 1) are m a p p e d back to A by T h i s means that Trj, a T,. is not ergodic, a n d therefore not B e r n o u l l i . 35 D a Theorem 17. For Trj, a there exist uncountably € 9#2, if separation(a) many a for which TD is not ergodic. >(X Proof. A s s u m e that separation(a) ai = jjj > 3, then for some D G (^,1), < 3 + e a n d D G {\, ^ p ) . F i r s t , we choose so that a i D = | a n d a = 1 + OL\. O u r restriction o n D i n terms of e has 2 been chosen so that a /a\ 2 < 3 + e w h e n 0:2 = 1 + « i - Since 0 2 > (a) of D e f i n i t i o n 15 is satisfied. Since a\ G (\, 5), a n d a condition G (1,1 + a\], b y L e m m a 2 10, c o n d i t i o n (b) of D e f i n i t i o n 15 is satisfied. T h u s , o u r defined a is always v a l i d . Define A = [±, D]. W e now a p p l y T = T D>a to A: T [ i , D] = [ m i n { 2 ( i - D),2(a D ai - 2 £>)},^] = [ m i n { 2 ( ± - 4%), 2(D + \ - D)},D(a 2 =[min{I,i},D(l + 4 )] ai ^- ^)] 4 N o w , since \ < D < 1, 0 < m ( A ) < 1 a n d T A Dt(x ergodic. - = A, b y D e f i n i t i o n T D ] a is not • 36 X F i g u r e 3.5: A n example of a non-ergodic system for Trj, a D = | and S 93?2. In this example, a = ( ^ , ^ ) . T h e thick lines represent Tr>,a- T h e coarse dashed line represents the necessary separation restriction o n a guarantee that T a to is ergodic. In this case, p a r t i a l remainders w i t h i n the interval [|, |] m a p back to |] a n d w h i c h means T o , a is not ergodic, a n d therefore not B e r n o u l l i . 3.4 Entropy of Multi-Divisor S-R-T Division T h e c a l c u l a t i o n for entropy i n m u l t i - d i v i s o r S - R - T d i v i s i o n follows the same m e t h o d used for single divisor S - R - T d i v i s i o n . W e begin by showing that To,a is a P C - m a p . Lemma 18. Proof. Trj tOC eiVl is a PC-map B y i n s p e c t i o n , each T o , a (as defined in Definition 14)- is a finite collection of line segments each w i t h slope 2. E a c h of these line segments is a C - m a p by the same argument used i n the proof for T h e o r e m 8. Therefore, by definition, each To,a is a P C - m a p . Theorem 19. The entropy of any TD >(X S 9 ^ with separation(a) 37 • < | is log 2. Proof. B y L e m m a 18, a l l To, a when separation(ot) G 9Jt are P C - m a p s . B y T h e o r e m 14, TD,<X is B e r n o u l l i < | and hence there exists a unique a.c.i.m. /J. T h e o r e m 7 says that R o h l i n ' s f o r m u l a for the entropy is true and therefore: h{T , ) D a = J log | T b , ' | d i i = log 2 ^ dfi = log 2. a • 38 Chapter 4 Future Work T h e o r i g i n a l question that inspired this thesis was "Is simple S - R - T d i v i s i o n ergodic for a l l real d i v i s o r s ? " In p u r s u i n g the answer to this p r o b l e m , we discovered that not o n l y is simple S - R - T d i v i s i o n ergodic for a l l divisors, but it is also B e r n o u l l i . H a v i n g established a B e r n o u l l i property, a n d having calculated the entropy for our transformations, we were able to use the K o l m o g o r o v - O r n s t e i n theorem to conclude that our transformations are isomorphic to each other. In p r o v i n g these i m p o r t a n t properties for simple S - R - T d i v i s i o n , we made extensive use of more general results f r o m d y n a m i c a l systems theory. Consequently, our results were shown to be easily extensible to more general d i v i s i o n systems. In general, it is difficult to prove that a p a r t i c u l a r class of transformations are ergodic or B e r n o u l l i . O u r results have p r o v i d e d a n effective means of p r o v i n g b o t h of these properties for a large class of S - R - T - l i k e d i v i s i o n algorithms. P r o m the standpoint of understanding a n algorithm's expected performance, it is necessary to know that when a stationary d i s t r i b u t i o n is f o u n d , it is unique. H a v i n g established the uniqueness of stationary distributions, the next step is to find the a c t u a l stationary d i s t r i b u t i o n for as wide a class of transformations as possible. In section 1.3, we verified a k n o w n expression for the stationary d i s t r i b u t i o n f u n c t i o n for TD where D e [|, 1). In a d d i t i o n , m a n y of the stationary d i s t r i b u t i o n functions 39 have been classified by Shively and F r e i m a n for D G [§, §], a l t h o u g h the derivations are not nearly as simple as for D G [§, 1). It turns out that things become very c o m p l i c a t e d w h e n D G [^, §]. In his thesis [22], Shively shows m a n y interesting properties for the stationary d i s t r i b u t i o n functions i n this region. F o r example, he shows that there are many different intervals of D where there are a n infinite number of different stationary d i s t r i b u t i o n equations. A s such, the g r a p h of the shift average for D G [5, |] is far f r o m complete a n d appears to have a°complex p a t t e r n (from the few points that have been plotted i n this region). T h i s is s u r p r i s i n g considering the s i m p l i c i t y of the u n d e r l y i n g transformation. A better u n d e r s t a n d i n g of this final region of simple S - R - T d i v i s i o n would be an interesting goal to pursue. In the work of F r e i m a n [6], it was first shown that the shift average for D G [|, |] is constantly 3, w h i c h can be easily shown to be the m a x i m u m possible shift average. T h i s property was then used by M e t z e [15] to o b t a i n a version of S - R - T d i v i s i o n that has a n expected shift average of 3 for a l l divisors. A n o t h e r area to pursue w o u l d be to explore shift averages for m u l t i - d i v i s o r S - R - T d i v i s i o n a n d , if other plateaus are f o u n d , they could possibly be used to o b t a i n higher expected shift averages for a l l possible divisors. U n d o u b t e d l y , o b t a i n i n g a complete u n d e r s t a n d i n g of the stationary d i s t r i b u t i o n functions for m u l t i - d i v i s o r d i v i s i o n w o u l d be even more difficult t h a n it is for simple S - R - T d i v i s i o n . It is possible that such results i n this area could lead to improvements i n m o d e r n S - R - T d i v i s i o n . R e l a t e d to this, it w o u l d be interesting to attempt to extend the results of this thesis to m o d e r n S - R - T division. 40 Bibliography [1] R u f u s B o w e n . B e r n o u l l i maps of the interval. Israel J. Math., 28(1-2):161-168, 1977. [2] A b r a h a m B o y a r s k y and Pawel G o r a . Laws of chaos. B i r k h a u s e r B o s t o n Inc., B o s t o n , M A , 1997. Invariant measures a n d d y n a m i c a l systems i n one dimension. [3] Digital Joseph F. Cavanagh. tion. Computer Arithmetic: Design and Implementa- M c G r a w - H i l l , N e w Y o r k , N Y , 1984. [4] J . Cocke a n d D. W . Sweeney. H i g h speed arithmetic i n a parallel device. T e c h n i c a l report, I B M , F e b r u a r y 1957. [5] M i l o s D. Ercegovac a n d Tomas L a n g . Division algorithms and implementations. and square root: digit-recurrence Kluwer Academic Publishers Group, Norwell, M A , U S A , a n d Dordrecht, T h e Netherlands, 1994. [6] C . V . F r e i m a n . S t a t i s t i c a l analysis of certain b i n a r y d i v i s i o n algorithms. IRE, 49:91-103, 1961. [7] D. H a r r i s , S. O b e r m a n , and M . H o r o w i t z . S R T d i v i s i o n architectures i m p l e m e n t a t i o n s . In 13th IEEE Symposium on Computer ings, California, USA, v o l u m e 13 of Symposium July Computer 6-9, 1997, Asilomar, Arithmetic, [8] A . N . K o l m o g o r o v . Akad. Proc. Nauk SSSR, Arithmetic: and proceedon pages 18-25. I E E E C o m p u t e r Society Press, 1997. A new invariant for transitive d y n a m i c a l systems. 119:861-864, 1958. 41 Dokl. [9] A . N . K o l m o g o r o v . p h i s m s . Dokl. E n t r o p y per unit time as a metric invariant of automor- Akad. Nauk SSSR, 124:754-755, 1959. [10] A . L a s o t a a n d James A . Yorke. O n the existence of invariant measures for Trans. piecewise m o n o t o n i c transformations. Amer. Math. Soc, 186:481-488 (1974), 1973. [11] A n d r z e j L a s o t a a n d M i c h a e l C . Mackey. Chaos, fractals, and noise. Springer- V e r l a g , N e w Y o r k , second edition, 1994. Stochastic aspects of d y n a m i c s . [12] Frangois L e d r a p p i e r . Some properties of absolutely continuous invariant measures on a n i n t e r v a l . Ergodic Theory Dynamical Systems, l ( l ) : 7 7 - 9 3 , 1981. [13] T i e n Y i e n L i a n d James A . Yorke. E r g o d i c transformations f r o m a n interval into itself. Trans. Amer. Math. Soc., 235:183-192, 1978. [14] O . L. M a c S o r l e y . High-speed arithmetic i n b i n a r y computers. Proc. IRE, 4 9 : 6 7 91, 1961. [15] G e m o t M e t z e . A class of b i n a r y divisions y i e l d i n g m i n i m a l l y represented quotients. IRE Trans, on Electronic [16] D o n a l d S. O r n s t e i n . Advances [17] Computers, EC-ll:761-764, D e c . 1961. B e r n o u l l i shifts w i t h the same entropy are isomorphic. in Math.,.4:337-352, D o n a l d S. O r n s t e i n . Ergodic 1970. theory, randomness, and dynamical systems. Yale U n i v e r s i t y Press, N e w H a v e n , C o n n . , 1974. James K . W h i t t e m o r e Lectures i n M a t h e m a t i c s given at Y a l e University, Y a l e M a t h e m a t i c a l M o n o g r a p h s , N o . 5. [18] K a r l Petersen. Ergodic theory. C a m b r i d g e U n i v e r s i t y Press, C a m b r i d g e , 1983. [19] J . E . R o b e r t s o n . A new class of d i g i t a l d i v i s i o n methods. IRE Trans. Computers, E C - 7 : 2 1 8 - 2 2 2 , Sept. 1958. 42 Electronic [20] V . A . R o h l i n . E x a c t endomorphisms of a Lesbesgue space. Amer. Transl., Math. Soc. 39, 1964. [21] P a u l Shields. The theory of Bernoulli shifts. T h e U n i v e r s i t y of C h i c a g o Press, C h i c a g o , I l l . - L o n d o n , 1973. C h i c a g o Lectures i n M a t h e m a t i c s . [22] R o b e r t Shively. Stationary division. distribution of partial remainders in S-R- T digital P h D thesis, U n i v e r s i t y of Illinois, 1963. [23] K . D . Tocher. Techniques of m u l t i p l i c a t i o n a n d d i v i s i o n for a u t o m a t i c computers. Quart. J. Mech. [24] Peter Walters. An introduction Appl. Math., to ergodic 1982. 43 binary 11:364-384, J u l y - S e p t e m b e r 1958. theory. Springer-Verlag, N e w Y o r k ,
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http://mathhelpforum.com/advanced-algebra/227663-subspace-proof-finding-basis-dimension.html | Math Help - Subspace Proof, finding a basis and dimension
1. Subspace Proof, finding a basis and dimension
Hey guys, I have a problem that I have to do, I am having a little trouble with it.
Consider the set of matrices
Show that W is a subspace ofM22.
Find a basis for W, and hence find dim(W).
(Make sure to demonstrate that your basis is linearly independent and spans W.)
2. Re: Subspace Proof, finding a basis and dimension
You have the set of all 2 by 2 matrices of the form $\begin{bmatrix}a & b- d \\ b+ a & d\end{bmatrix}$ and you want to
1) Show that this is a subspace of the space of all 2 by 2 matrices.
You must show it is closed under matrix addition and scalar multiplication. The sum of two such matrices can be written
$\begin{bmatrix}u & v- w \\ v+ u & w\end{bmatrix}+\begin{bmatrix}x & y- z \\ y+ x & z\end{bmatrix}= \begin{bmatrix}u+ x & v- w+ y- z \\ v+ u+ y+ x & w+ z\end{bmatrix}$
Can that be written in the form $\begin{bmatrix}a & b- d \\ b+ a & d\end{bmatrix}$? What must a, b, and d be?
The product of the number "k" and the matrix $\begin{bmatrix}x & y- z\\ y+ x & z\end{bmatrix}$ is $\begin{bmatrix}kx & k(y- z) \\ k(y+ x) & kz\end{bmatrix}$.
Can that be written in the form $\begin{bmatrix}a & b- d \\ b+ a & d\end{bmatrix}$? What must a, b, and d be?
The third condition, that the subset be non-empty, can be done by showing that the 0 matrix is in the set. Take a= b= d= 0.
2) Find a basis for this subspace.
Well, $\begin{bmatrix}a & b- d \\ b+ a & d\end{bmatrix}$ $= \begin{bmatrix}a & 0 \\ a & 0\end{bmatrix}+ \begin{bmatrix}0 & b \\ b & 0 \end{bmatrix}+ \begin{bmatrix}0 & -d \\ 0 & d\end{bmatrix}$ $= a\begin{bmatrix}1 & 0 \\ 1 & 0 \end{bmatrix}+ b\begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}+ d\begin{bmatrix}0 & -1 \\ 0 & 1\end{bmatrix}$. Does that give you any ideas?
3) Find the dimension.
After (2), this should be trivial. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 9, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9229753613471985, "perplexity": 313.7269313580096}, "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-2016-18/segments/1461860116929.30/warc/CC-MAIN-20160428161516-00114-ip-10-239-7-51.ec2.internal.warc.gz"} |
https://dsweb.siam.org/Media-Gallery/natural-invariant-measures | # Natural invariant measures
### Fractals and Choas
Print
Shown are approximations to natural invariant measures ("SRB-measures") in the Lorenz system and in Chua's circuit. The underlying computational technique consists of (1) computing a covering of the underlying invariant set via a multilevel-subdivision algorithm and (2) discretization of the corresponding transfer-operator (Perron-Frobenius operator) using a Galerkin-approach. An invariant vector of the discretized operator yields an approximate invariant measure.
Natural invariant measure in Chua's circuit for the parameter values $$\alpha=18$$, $$\beta=33$$, $$m_0=-0.2$$ and $$m_1=0.01$$.
Natural invariant measures in the Lorenz system for the parameter values $$\beta = 0.4, 0.8, 1.2$$ and $$8/3$$ (from left to right, top to bottom). The other parameter values were fixed to $$\sigma = 10$$ and $$\rho = 28$$.
Author Institutional Affiliation University of Paderborn, Germany Author Email Author Postal Mail Institute for Mathematics, University of Paderborn, Warburger Str. 100, 33098 Paderborn, Germany Notes The pictures have been rendered in collaboration with Martin Rumpf ([email protected]) using the software platform GRAPE. Keywords natural invariant measure, SRB-measure, transfer operator, set oriented method
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http://www.sequencesandseries.com/limit-example/ | # Limit example 1
$$\def \C {\mathbb{C}}\def \R {\mathbb{R}}\def \N {\mathbb{N}}\def \Z {\mathbb{Z}}\def \Q {\mathbb{Q}}\def\defn#1{{\bf{#1}}}$$
Example
Claim: The sequence $a_n=(4n+3)/n$ has limit $4$.
Note that this intuitively obvious since $a_n=4+3/n$ and $3/n\to 0$ is reasonable.
For a rigorous proof we need to use the definition (applying a definition to find something is often called using first principles).
Given an $\varepsilon$ we will need to find an $N$ which depends on that $\varepsilon$.
First we simplify $|a_n-a|$ as it plays a central role in the definition. This simplification follows from a straightforward chain of equalities:
$|a_n-a|=|a_n-4| = \left| \frac{4n+3}{n} -4 \right| =\left| \frac{4n+3-4n}{n} \right| = \left| \frac{3}{n} \right| = \frac{3}{n} .$
Note that the last equality occurs because $n>0$ and hence $3/n>0$.
Now we can investigate $|a_n-a|\lt \varepsilon$:
\begin{align*}
|a_n-4| \lt \varepsilon &\iff \frac{3}{n} \lt \varepsilon \\
&\iff \frac{3}{\varepsilon } \lt n .
\end{align*}
Reading from bottom to top, we have the condition
$n>\frac{3}{\varepsilon } \Longrightarrow |a_n-4| \lt \varepsilon .$
This is almost what we want. We want a $N$ such that when $n>N$ we get $|a_n-4| \lt \varepsilon$.
Hence take $N$ to be a natural number greater than $\frac{3}{\varepsilon }$. Then if $n>N$ we obviously have $n>\frac{3}{\varepsilon }$.
Let’s recap what we have proved:
1. Let $N$ be a natural number greater than $\frac{3}{\varepsilon }$.
2. If $n>N$, then $n > \frac{3}{\varepsilon }$.
3. If $n>\frac{3}{\varepsilon }$, then $|a_n-4| \lt \varepsilon$.
This shows that $a_n$ has limit $4$ but we do not write the solution in this way. We rewrite it without showing where we got $N$ from. We just show that our $N$ gives us what we want. (No doubt you will have had teachers who say that you should show your working. Well, here we are not showing all our working, just the bit that proves the result!)
Here is the polished version that proves that $(4n+3)/n\to 4$. Note that we use the working from above but rearrange it into a different format.
Polished version: The sequence $a_n=(4n+3)/n$ has limit $4$.
Given $\varepsilon >0$, let $N$ be a natural number greater than $\frac{3}{\varepsilon }$. If $n>N$, then $\frac{3}{n} \lt \varepsilon$ and we have
\begin{align*}
|a_n-4| &= \left| \frac{4n+3}{n} -4 \right| \\
&= \left| \frac{4n+3-4n}{n} \right| \\
&= \left| \frac{3}{n} \right| \\
&= \frac{3}{n} \\
&\lt \varepsilon .
\end{align*}
Notice that in our working we had a string of equivalences marked by $\iff$ and that these involved $\varepsilon$. In the write up, we did not use $\varepsilon$ and the implication $\Longrightarrow$ was used but not explicitly. It is used implicitly in the logic of our argument. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 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": 0.99257892370224, "perplexity": 220.18324127247172}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627998959.46/warc/CC-MAIN-20190619103826-20190619125826-00399.warc.gz"} |
https://socratic.org/questions/how-do-you-write-the-equation-of-the-line-parallel-to-y-3x-4-and-passing-through | Algebra
Topics
# How do you write the equation of the line parallel to y = 3x - 4 and passing through the point (-2, 5)?
Aug 25, 2016
$y = 3 x + 11$
#### Explanation:
Parallel lines have the same slopes.
The slope of the new line will have the same slope as the given line:
$y = \textcolor{red}{3} x - 4 \Rightarrow m = \textcolor{red}{3}$
One point on the line is given, $\textcolor{b l u e}{\left(\left(- 2 , 5\right)\right)} . \text{This is } \textcolor{b l u e}{\left(\left({x}_{1} , {y}_{1}\right)\right)}$
The formula for slope is $m = \left(\text{change in y values"/"change in x-values}\right) = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$
If you have the slope and one point , substitute them into a formula which is based on the formula for slope given above.
$y - \textcolor{b l u e}{{y}_{1}} = \textcolor{red}{m} \left(x - \textcolor{b l u e}{{x}_{1}}\right)$
$y - \textcolor{b l u e}{5} = \textcolor{red}{3} \left(x - \textcolor{b l u e}{\left(- 2\right)}\right)$
$y - 5 = 3 \left(x + 2\right)$
Simplify to get the required equation of the line.
$y = 3 x + 6 + 5$
$y = 3 x + 11$
##### Impact of this question
7156 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": 9, "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": 0.8625078201293945, "perplexity": 351.2233261019791}, "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/1570986647517.11/warc/CC-MAIN-20191013195541-20191013222541-00544.warc.gz"} |
http://tex.stackexchange.com/questions/17940/how-does-one-extract-specific-attributes-of-a-label-e-g-just-the-section-number | # How does one extract specific attributes of a label e.g. just the section number?
I'm using hyperref and calling \autoref to refer to tables, sections, chapters, etc.. However, sometimes I just want to refer to the section number of a table (e.g. 7.2 as opposed to Table 7.2).
How do we do this?
More generally, how do we extract the (5?) attributes of a label. I know \pageref gets the page, but what about the others?
Also, some commands are issued in .aux like:
\@writefile{lot}{\contentsline
{figure}{\numberline
{5.2}{\ignorespaces Galois
Correspondence}}{20}{figure.5.2}}
How does extract, say, the 2.2nd attribute ({\ignorespaces Galois Correspondence})?
-
In addition to @lockstep's answer, plain LaTeX also has the \pageref command. – Marc van Dongen Mar 21 '13 at 9:34
Use the \ref command (core LaTeX) and the \nameref command (provided by the nameref package, which is loaded by hyperref).
\documentclass{article}
\usepackage{hyperref}
\begin{document}
\begin{table}
\centering
(Table content)
\caption{foo}
\label{tab:foo}
\end{table}
This is a reference to \autoref{tab:foo}.
One could also refer to the number \ref{tab:foo} or to the name \nameref{tab:foo}.
\end{document}
-
Thanks lockstep, that answers an important part of my question. I have found the package "cleveref" (yes there's no double ‘r’), which sort of does some of the other things I was asking about (e.g. it is able to spit out the type of section which has been labelled — chapter, section, table, etc.). The question about the contents line thing still remains. – R17 May 13 '11 at 3:08 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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": 0.9310083985328674, "perplexity": 3240.125967196876}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-35/segments/1408500830746.39/warc/CC-MAIN-20140820021350-00197-ip-10-180-136-8.ec2.internal.warc.gz"} |
http://mmonoplayer.com/standard-error/when-to-use-standard-deviation-vs-standard-error.html | Home > Standard Error > When To Use Standard Deviation Vs Standard Error
# When To Use Standard Deviation Vs Standard Error
## Contents
Notice that s x ¯ = s n {\displaystyle {\text{s}}_{\bar {x}}\ ={\frac {s}{\sqrt {n}}}} is only an estimate of the true standard error, σ x ¯ = σ n The standard error estimated using the sample standard deviation is 2.56. Recent popular posts Extracting Tables from PDFs in R using the Tabulizer Package Writing Good R Code and Writing Well How to send bulk email to your students using R Efficiently If σ is known, the standard error is calculated using the formula σ x ¯ = σ n {\displaystyle \sigma _{\bar {x}}\ ={\frac {\sigma }{\sqrt {n}}}} where σ is the http://mmonoplayer.com/standard-error/standard-error-and-standard-deviation-difference.html
pp. 249–255.6. The problem is that when conducting a study we have one sample (with multiple observations), eg, s1 with mean m1 and standard deviation sd1, but we do not have or sdm. In an article by Copay et al, SEM (standard error of measurement) was quoted as one of the many approaches in evaluating the MCID. This is usually the case even with finite populations, because most of the time, people are primarily interested in managing the processes that created the existing finite population; this is called
## When To Use Standard Deviation Vs Standard Error
doi: 10.1136/bmj.331.7521.903PMCID: PMC1255808Statistics NotesStandard deviations and standard errorsDouglas G Altman, professor of statistics in medicine1 and J Martin Bland, professor of health statistics21 Cancer Research UK/NHS Centre for Statistics in Medicine, Seven samples (3, 11, 29, 39, 54, 59, and 96) have a 95% confidence interval ...Fig. 2The cascade from the distribution of the parameter in the population, to the sampling distribution of Can a free radical be created by chemical reaction of non-radical species? The standard deviation of the age for the 16 runners is 10.23.
Standard deviations and standard errors. Should a country name in a country selection list be the country's local name? The ages in one such sample are 23, 27, 28, 29, 31, 31, 32, 33, 34, 38, 40, 40, 48, 53, 54, and 55. Standard Error In Excel Standard Error of Mean vs.
Of the 2000 voters, 1040 (52%) state that they will vote for candidate A. Standard Error And Standard Deviation Difference In the example of 100 samples of tumor size, seven samples (3, 11, 29, 39, 54, 59, and 96) have a confidence interval that does not include the true population mean Clark-Carter D. The sample proportion of 52% is an estimate of the true proportion who will vote for candidate A in the actual election.
Compare the true standard error of the mean to the standard error estimated using this sample. Standard Error Of Estimate The standard deviation of the means of those samples is the standard error. The proportion or the mean is calculated using the sample. If people are interested in managing an existing finite population that will not change over time, then it is necessary to adjust for the population size; this is called an enumerative
• JSTOR2682923. ^ Sokal and Rohlf (1981) Biometry: Principles and Practice of Statistics in Biological Research , 2nd ed.
• In this scenario, the 2000 voters are a sample from all the actual voters.
• Think about the following situation.
• The graph shows the ages for the 16 runners in the sample, plotted on the distribution of ages for all 9,732 runners.
• Published online 2011 May 10.
• In: Everitt BS, Howell D (eds).
• As an example of the use of the relative standard error, consider two surveys of household income that both result in a sample mean of \$50,000.
## Standard Error And Standard Deviation Difference
Bence (1995) Analysis of short time series: Correcting for autocorrelation. If you take a sample of 10 you're going to get some estimate of the mean. When To Use Standard Deviation Vs Standard Error Privacy policy About Wikipedia Disclaimers Contact Wikipedia Developers Cookie statement Mobile view GraphPad Statistics Guide The SD and SEM are not the same The SD and SEM are not the same Standard Error Vs Standard Deviation Example ISBN 0-7167-1254-7 , p 53 ^ Barde, M. (2012). "What to use to express the variability of data: Standard deviation or standard error of mean?".
All Rights Reserved. http://mmonoplayer.com/standard-error/standard-error-example.html With n = 2 the underestimate is about 25%, but for n = 6 the underestimate is only 5%. Deng at 8:17 AM Email ThisBlogThis!Share to TwitterShare to FacebookShare to Pinterest 2 comments: Nick Barrowman said... The mean age for the 16 runners in this particular sample is 37.25. Standard Error In R
The SD does not change predictably as you acquire more data. pp. 1891–1892.5. The next graph shows the sampling distribution of the mean (the distribution of the 20,000 sample means) superimposed on the distribution of ages for the 9,732 women. Source The tops of the marshalled row form a flowing curve of invariable proportion; and each element, as it is sorted in place, finds, as it were, a pre-ordained niche, accurately adapted
ISBN 0-7167-1254-7 , p 53 ^ Barde, M. (2012). "What to use to express the variability of data: Standard deviation or standard error of mean?". Standard Error Of Mean Word that includes "food, alcoholic drinks, and non-alcoholic drinks"? It seems from your question that was what you were thinking about.
## Compare the true standard error of the mean to the standard error estimated using this sample.
JSTOR2682923. ^ Sokal and Rohlf (1981) Biometry: Principles and Practice of Statistics in Biological Research , 2nd ed. Of course deriving confidence intervals around your data (using standard deviation) or the mean (using standard error) requires your data to be normally distributed. Be careful that you do not confuse the two terms (or misinterpret the values). Standard Error Calculator Using a sample to estimate the standard error In the examples so far, the population standard deviation σ was assumed to be known. | {"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": 0.8347147107124329, "perplexity": 1354.8828481545431}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-39/segments/1505818688158.27/warc/CC-MAIN-20170922022225-20170922042225-00285.warc.gz"} |
https://math.stackexchange.com/questions/3624664/distance-of-point-from-a-particular-line | # Distance of point from a particular line
The number of points on the line $$3x + 4y = 5$$, which are at a distance of $$sec^2\theta+2cosec^2\theta$$ ,$$\theta \in \mathbb{R}$$ from the point $$(1, 3)$$, is
(1) 1
(2) 2
(3) 3
(4) infinite
My approach is as follow the least distance of $$3x+4y=5$$ from the point $$(1,3)$$ is $$2$$ which is perpendicular distance. The point from $$3x+4y=5$$ from the point $$(1,3)$$ may not be perpendicular. The distance $$sec^2\theta+2cosec^2\theta$$ is always greater than $$2$$ so we need to find the number of points valid for $$sec^2\theta+2cosec^2\theta$$ which I am not able to find.
• Is the answer supposed to depend on $\theta$? – Parcly Taxel Apr 14 at 7:16
The function $$\sec^2\theta+2\csc^2\theta$$ is unbounded in the positives, so there are infinitely many points that can satisfy the condition.
By C-S $$\frac{1}{\cos^2\theta}+\frac{2}{\sin^2\theta}=(\cos^2\theta+\sin^2\theta)\left(\frac{1}{\cos^2\theta}+\frac{2}{\sin^2\theta}\right)\geq(1+\sqrt2)^2,$$ which gives a range of $$\frac{1}{\cos^2\theta}+\frac{2}{\sin^2\theta}$$: $$\left[(1+\sqrt2)^2,+\infty\right)$$ and there are infinitely many such points.
• if the minimum is $(1+\sqrt{2})^2$, the range starts with it, does not it? – farruhota Apr 14 at 9:49
• should not the range be: $[(1+\sqrt{2})^2,+\infty)$? – farruhota Apr 14 at 10:26 | {"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": 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": 16, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9089899063110352, "perplexity": 170.05794387264234}, "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/1600400188841.7/warc/CC-MAIN-20200918190514-20200918220514-00001.warc.gz"} |
http://mathoverflow.net/questions/59633/maximum-flow-with-negative-capacities/59785 | # Maximum flow with negative capacities?
I'm trying to compute an (s-t) maximum flow through a network which includes a number of arc pairs ((u,v), (v,u)) that have equal, negative capacities (weights). I'm not aware of any efficient algorithms that solve this problem directly, so I am trying to think of a way to transform the problem so that it can be passed to standard max-flow algorithms that assume all arc capacities are non-negative.
The only hint I could find online was a question in a 1999 homework assignment which asks students to "Show how to reduce the problem of maximum flow with possibly negative capacities to two maximum flow problems both with nonnegative capacities." (The original can be seen here: http://www.cs.cmu.edu/afs/cs/academic/class/15750-s99/www/homeworks/hw4.ps )
How can this be done?
Edit #1: I should explain the origin of these ((u,v),(v,u)) pairs. What I am really trying to do is solve an s-t maximum flow in a graph that is essentially undirected. By "essentially undirected" I mean a graph that is undirected except for arcs with source s and target in the undirected graph, and arcs with source in the undirected graph and target t. Of course, this kind of "partially directed" graph must be translated into a normal directed graph for any s-t max flow algorithm. To do this, I am applying the standard transformation of replacing every undirected edge (u<->v) (with weight c) with a pair of directed arcs (u->v) and (v->u) with weight c. Since the original graph sometimes has negative edge weights, these ((u,v),(v,u)) pairs with equal negative weights arise.
-
Can you clarify the problem? I assume that an arc with capacity -7 needs to have a flow of -7 or less, right? But can an arc with positive capacity get a negative flow? – Stefan Geschke Mar 26 '11 at 8:17
Ok, I have just read the problem in the link. Still not so clear. If you have an arc pair $\{(u,v),(v,u)\}$, both with the same negative capacity, what would be an admissible flow on these arcs? Or is one capacity positive and the other negative, of the same absolute value, so that you essentially specify your flow at this arc? – Stefan Geschke Mar 26 '11 at 8:21
If arcs ${(u,v),(v,u)}$ both have capacities of -7, isn't this the same as saying that both arcs must have positive flows of at least 7 units? – Brian Borchers Mar 26 '11 at 14:20
The correspondence between max flow with lower capacities and max flow with negative capacities as described in the linked homework problem is based on the convention that $f(u,v)=-f(v,u)$ for all arc pairs $\{(u,v),(v,u)\}$. http://computingscience.nl/docs/vakken/an/an-maxflow.ppt contains a reduction of max flow with lower capacities to the solution of two standard max flow problems, thus solving the homework question. If I understand your problem correctly, it is a bit more complicated: A weight of $-7$ on an (undirected) edge $\{u,v\}$ means that you want to send at least 7 units of flow either from $u$ to $v$ or from $v$ to $u$. As far as I can see, for this problem the standard transformation of replacing the edge by an arc pair does not work. In the standard case only one of the arcs $(u,v)$ and $(v,u)$ is used in an optimal solution, and its flow value can be used as the value for the undirected edge. But how would you transfer $f(u,v)=-9$, $f(v,u)=-12$ back to the undirected model?
Practically, the first thing I would try is to write the original problem as a mixed integer program (I guess you have to model the choices for the arc directions with binary variables) and throw it into a general purpose solver (Cplex, Gurobi, http://zibopt.zib.de, http://www.coin-or.org/ ).
-
Thanks everyone for your comments. Unfortunately, I now realize that I was trying to do the impossible. As Brian's comment suggests, I can't have a net flow of +7 in both directions between a pair of nodes. As Kali points out, a choice must be made at every such arc-pair that leads to a combinatorial explosion.
I came to the realization yesterday: by the duality of max-flow and min-cut, if I could solve a max-flow problem with negative edge weights then I could solve a min-cut problem with negative edge weights. However, since I can translate any (NP-complete) max-cut problem into a min-cut problem by negating all the edge weights, I can't expect to solve such problems in P-time.
I see now (thanks to Kali) that the homework question makes the (standard) assumption of what I think is called "skew symmetry". The problem I have is incompatible with this and therefore isn't merely a network flow problem with negative edge weights.
-
If you are just interested in the solution to your problem (and not in obtaining it by a provably polynomial algorithm) I still recommend to simply try solving the MIP. – Thomas Kalinowski Mar 28 '11 at 1:33
Yes, I will probably do this. Since the negative capacity edge pairs are static and reasonably limited in number, an MIP solver is probably going to do a reasonable job. I can also make use of suboptimal solutions, so I agree it is a good way forward. – Fumiyo Eda Mar 28 '11 at 6:26
@Fumiyo, What did you end up doing? I have been working on this problem to come to the same conclusion you did regarding the incompatibility of negative undirected weights with the standard "skew symmetry" assumption. I suspected from the beginning this problem would be NP-hard. Did you find a good approximation? – SchighSchagh May 3 '14 at 6:21
The problem you describe sounds like an energy minimization on a Markov Random Field (MRF).
Min E = sum[D(p)] + sum[V(p,q)]
where p, q is a binary labelling = {sink=0, source=1}.
The first term of E sums over all the nodes in the undirected graph. D(p=0) is the weight on the source-to-node edge and D(p=1) the weight on the node-to-sink edge.
The second term of E sums over all edges in the undirected graph, where p is the label on one end of the edge and q the label on the other end.
The standard min-cut with positive weights correponds to: V(0,0)=V(1,1)=0 and V(0,1)=V(1,0)=weight. More general, if V(p,q) satisfies the submodular condition on all edges, then minimizing E is the same as an s-t max-flow\min-cut.
Submodular: V(0,1) + V(1,0) >= V(0,0) + V(1,1)
If there are (some) negative weighted edges, these do not satisfy submodularity. Instead they are supermodular.
Supermodular: V(0,1) + V(1,0) <= V(0,0) + V(1,1)
If all edges are supermodular, I guess there is an equivalence with max-cut.
However, the problem you describe coontains a mixture of submodular (positive weighted) and supermodular (negative weighted) edges. This is indeed NP-hard. But how to solve it? You might consider looking into Quadratic Pseudo Boolean Optimization (QPBO): Wikipedia
- | {"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": 0.8422703742980957, "perplexity": 373.9414170146415}, "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/1429246661733.69/warc/CC-MAIN-20150417045741-00303-ip-10-235-10-82.ec2.internal.warc.gz"} |
https://de.maplesoft.com/support/help/view.aspx?path=Database/Result/GetType&L=G | GetType - Maple Help
Result
GetType
get the SQL type of a column
Calling Sequence result:-GetType( index )
Parameters
result - Result module index - integer or string; the column for which the type is returned
Description
• GetType returns the SQL type of the column indicated by index.
• If index is an integer, it is used to index into the current Result module's table. If index is a string, it is the name of the column.
• The type returned by GetType can be used in an explicit cast to make Maple perform a particular type conversion. For more information, see conversions.
Examples
Create a Result.
> $\mathrm{driver}≔\mathrm{Database}\left[\mathrm{LoadDriver}\right]\left(\right):$
> $\mathrm{conn}≔\mathrm{driver}:-\mathrm{OpenConnection}\left(\mathrm{url},\mathrm{name},\mathrm{pass}\right):$$\mathrm{res}≔\mathrm{conn}:-\mathrm{ExecuteQuery}\left("SELECT name FROM animals"\right):$$\mathrm{res}:-\mathrm{GetType}\left(1\right)$
${\mathrm{INTEGER}}$ (1)
> $\mathrm{res}:-\mathrm{GetType}\left(2\right)$
${\mathrm{VARCHAR}}$ (2)
> $\mathrm{res}:-\mathrm{GetType}\left(3\right)$
${\mathrm{INTEGER}}$ (3)
> $\mathrm{res}:-\mathrm{GetType}\left(4\right)$
${\mathrm{DOUBLE}}$ (4)
> $\mathrm{res}:-\mathrm{GetType}\left("id"\right)$
${\mathrm{INTEGER}}$ (5)
> $\mathrm{res}:-\mathrm{GetType}\left("name"\right)$
${\mathrm{VARCHAR}}$ (6)
> $\mathrm{res}:-\mathrm{GetType}\left("number"\right)$
${\mathrm{INTEGER}}$ (7)
> $\mathrm{res}:-\mathrm{GetType}\left("mass"\right)$
${\mathrm{DOUBLE}}$ (8) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 19, "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": 0.8232582211494446, "perplexity": 2721.3020079771663}, "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/1674764500154.33/warc/CC-MAIN-20230204205328-20230204235328-00698.warc.gz"} |
http://mathhelpforum.com/algebra/47434-integer-help.html | # Math Help - Integer help!
1. ## Integer help!
Thank you taking the time.
I need to write the expression in words that described the interval of numbers below.
(4,21/2)
3. Given that A = {x|x > 3} and B = {x|-3< x < 2}, express the following in interval notation. (The < has a line under it so it means equal to. I can't make it equal to.)
A U B interval ____________
A n B interval ____________
2. Originally Posted by Pangu
Thank you taking the time.
I need to write the expression in words that described the interval of numbers below.
(4,21/2)
3. Given that A = {x|x > 3} and B = {x|-3< x < 2}, express the following in interval notation. (The < has a line under it so it means equal to. I can't make it equal to.)
A U B interval ____________
A n B interval ____________
i can try this 3 u mentioned
A = {3,4,5,6....} - u mentioned > 3
B = {-3,-2,-1,1,2.3}
A u B ={-3,-2,-1,0,1,2,3,,4,5,6....}
A n B ={3} | {"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": 0.9510456323623657, "perplexity": 2079.0343973909344}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-41/segments/1410657129229.10/warc/CC-MAIN-20140914011209-00273-ip-10-196-40-205.us-west-1.compute.internal.warc.gz"} |
https://www.unige.ch/math/tggroup/doku.php?id=workshop | workshop
# Workshop "Real geometry in the footsteps of Gabriel Cramer"
Geneva, 2019 , October 28th - November 1st, Villa Battelle
Monday, October 28, 16:30, Battelle, Ilia Itenberg, “Planes in four-dimensional cubics”.
Tuesday, October 29, 11:45, Battelle, Nikita Kalinin, “Symplectic packing problem”.
Tuesday, October 29, 14:45, Battelle, Kristin Shaw, “Poincaré duality for tropical manifolds”.
Wednesday, October 30, 10:30, Battelle, Kristin Shaw, “Poincaré duality for tropical manifolds”.
Thursday, October 31, 14:30,Battelle, Kristin Shaw, “Poincaré duality for tropical manifolds”.
Thursday, October 31, 16:00, Battelle, Nikita Kalinin,“Symplectic packing problem and Nagata's conjecture”.
Friday, November 1, 15:00, Battelle, Nikita Kalinin, “Symplectic packing problem and Nagata's conjecture”.
Abstracts:
Ilia Itenberg (Sorbonne University)
Planes in four-dimensional cubics
We discuss possible numbers of 2-planes in a smooth cubic hypersurface in the 5-dimensional projective space. We show that, in the complex case, the maximal number of planes is 405, the maximum being realized by the Fermat cubic. In the real case, the maximal number of planes is 357.
The proofs deal with the period spaces of cubic hypersurfaces in the 5-dimensional complex projective space and are based on the global Torelli theorem and the surjectivity of the period map for these hypersurfaces, as well as on Nikulin's theory of discriminant forms.
Joint work with Alex Degtyarev and John Christian Ottem.
Kristin Shaw (University of Oslo)
Minicourse: Poincaré duality for tropical manifolds
The series of lectures will focus on the different formulations and approaches to Poincaré duality for the tropical homology group of tropical manifolds. Tropical homology is the homology of certain sheaves on polyhedral spaces and was introduced by Itenberg, Katzarkov, Mikhalkin, and Zharkov. The first formulation of tropical Poincaré duality was in terms of a non-degenerate pairing between compactly supported and usual tropical cohomology by Jell, Shaw, and Smacka. This pairing was formulated via the integration of superforms in the sense of Lagerberg. Tropical manifolds also satisfy a version of Poincaré duality for tropical (co)homology with integral coefficients by Jell, Rau, and Shaw. This version is formulated in terms of the cap product with the fundamental class. Another recent approach by Gross and Shokreih is via Verdier duality in the derived category and removes the assumption on the existence of a suitable covering of the tropical manifolds required in the two formulations above.
The proof of these duality statements in all three cases boils down to a local version of Poincaré duality for the tropical (co)homology of matroidal fans which is proved by using the deletion and contraction operations. The matroid property of a fan is sufficient but not necessary for tropical Poincaré duality. I will point out some partial results of Edvard Aksnes on necessary and sufficient conditions for tropical Poincaré duality for fans, and also highlight some consequences of Poincaré duality in the global case.
Nikita Kalinin (HSE University)
Talk 1: Symplectic packing problem. This will be an introductory talk. I will mention several instances of symplectic packing problems and present simple geometric methods for tackling them. Based on (unpublished) survey of Felix Schlink.
Talk 2: Symplectic packing problem and Nagata’s conjecture. Curiously, the question of what is the maximal R such that we can embed k<10 symplectic balls of radius R in CP^2 is related to the question of what is the minimal degree d of an algebraic curve in CP^2 passing through given generic points with given multiplicities. Nagata’s conjecture (still open for all n>10 except squares) states that d>m\sqrt n if we draw a curve through n points of multiplicity m. I will highlight the connections between symplectic packing and Nagata’s conjecture (based on works of McDuff, Polterovich, Biran, among others).
workshop.txt · Dernière modification: 2019/11/03 21:09 de weronika | {"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": 0.8505528569221497, "perplexity": 1552.5219168297167}, "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/1652662522270.37/warc/CC-MAIN-20220518115411-20220518145411-00752.warc.gz"} |
https://www.dspe.nl/knowledge/thermomechanics/chapter-2-in-depth/2-4-thermal-radiation/2-4-3-gebhard-method/near-black-surfaces/ | Chapter 2 - In depth
# 2.4.3.1 Near black surfaces
The Gebhart method describes the radiative heat transfer in terms of heat coming from one surface and absorbed by another, including all reflections. The Gebhart matrix traces exactly the radiative transfer contribution of each surface to the others.
For two surfaces with high emissivities (close to 1) the Gebhart factor can be simplified by:
Reflections are then not taken into account, but because of the low reflectivity its fraction is relatively small and therefore also the error you make. In that case the relation becomes:
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http://math.stackexchange.com/questions/67773/how-do-i-integrate-e-sqrtx-using-the-substitution-rule | # ¿How do I integrate $e^\sqrt{x}$, using the substitution rule?
I've been trying to integrate that function, but it looks like I'm missing something, so can any one please show me how to integrate $e^\sqrt{x}$, in order to correct my procedure. Thanks
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If $x=u^2$, then $\mathrm dx=\dots$ – Guess who it is. Sep 26 '11 at 21:53
Hint: substitute $u = \sqrt{x}$. – Robert Israel Sep 26 '11 at 21:54
No is not a homework I'd tried to do the same substitution that is shown down in the first answer, but then I don't know what to do then, since there is a product in the integrand. – Davynch0 Sep 26 '11 at 22:13
You're going to want to start with a $u$-substitution $u=\sqrt{x}$. Notice then that $u^2=x$, so $2udu=dx$. Then $$\int e^\sqrt{x}dx=2\int e^u u\ du.$$
You can then proceed by integration by parts.
To get started, you can set $w=u$ and $dv=e^u\ du$, and use the fact that $\int w\ dv= wv-\int v\ dw$. Lastly, don't forget to substitute back to get the integral in terms of $x$.
Continuing, $dw=du$ and $v=e^u$. Then $$\int w\ dv= wv-\int v\ dw = e^u u-\int e^u\ du=e^u u-e^u.$$ So $$\int e^\sqrt{x}dx=2\int e^u u\ du=2e^uu-2e^u+C=2e^\sqrt{x}(\sqrt{x}-1)+C$$ after back substituting $u=\sqrt{x}$ and adding a possible constant term.
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I don't know how to integrate by parts, can't it be solved by the substitution rule at all? – Davynch0 Sep 26 '11 at 22:07
@Dav Since you say it is not homework, I've tried to add a more complete solution. Hopefully it's clearer. – yunone Sep 26 '11 at 22:20
thanks a lot @yunone – Davynch0 Sep 26 '11 at 22:21
Not homework, and you don't know integration by parts? Maybe that is why it is not (yet) homework. Maybe it WILL be homework later, after you do learn integration by parts! – GEdgar Sep 27 '11 at 0:12
@Davyncho: It really is (after the natural substitution) a standard integration by parts. But if we insist on substitution alone, we can (I am joking, but it is true) use $u=(\sqrt{x}-1)e^\sqrt{x}$. – André Nicolas Sep 27 '11 at 1:37 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9552062749862671, "perplexity": 403.76888809223664}, "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/1438042987127.36/warc/CC-MAIN-20150728002307-00162-ip-10-236-191-2.ec2.internal.warc.gz"} |
https://www.zora.uzh.ch/id/eprint/48254/ | # Implications of high-resolution simulations on indirect dark matter searches
Pieri, L; Lavalle, J; Bertone, G; Branchini, E (2011). Implications of high-resolution simulations on indirect dark matter searches. Physical Review D, 83(2):023518.
## Abstract
We study the prospects for detecting the annihilation products of dark matter (DM) in the framework of the two highest-resolution numerical simulations currently available, i.e., Via Lactea II and Aquarius. We propose a strategy to determine the shape and size of the region around the Galactic center that maximizes the probability of observing a DM signal, and we show that, although the predicted flux can differ by a factor of 10 for a given DM candidate in the two simulation setups, the search strategy remains actually unchanged, since it relies on the angular profile of the annihilation flux, not on its normalization. We present mock γ-ray maps that keep into account the diffuse emissions produced by unresolved halos in the Galaxy, and we show that, in an optimistic DM scenario, a few individual clumps can be resolved above the background with the Fermi-LAT. Finally, we calculate the energy-dependent boost factors for positrons and antiprotons and show that they are always of O(1), and, therefore, they cannot lead to the large enhancements of the antimatter fluxes required to explain the recent PAMELA, ATIC, Fermi, and HESS data. Still, we show that the annihilation of 100 GeV weakly interacting massive particles into charged lepton pairs may contribute significantly to the positron budget.
## Abstract
We study the prospects for detecting the annihilation products of dark matter (DM) in the framework of the two highest-resolution numerical simulations currently available, i.e., Via Lactea II and Aquarius. We propose a strategy to determine the shape and size of the region around the Galactic center that maximizes the probability of observing a DM signal, and we show that, although the predicted flux can differ by a factor of 10 for a given DM candidate in the two simulation setups, the search strategy remains actually unchanged, since it relies on the angular profile of the annihilation flux, not on its normalization. We present mock γ-ray maps that keep into account the diffuse emissions produced by unresolved halos in the Galaxy, and we show that, in an optimistic DM scenario, a few individual clumps can be resolved above the background with the Fermi-LAT. Finally, we calculate the energy-dependent boost factors for positrons and antiprotons and show that they are always of O(1), and, therefore, they cannot lead to the large enhancements of the antimatter fluxes required to explain the recent PAMELA, ATIC, Fermi, and HESS data. Still, we show that the annihilation of 100 GeV weakly interacting massive particles into charged lepton pairs may contribute significantly to the positron budget.
## Statistics
### Citations
Dimensions.ai Metrics
115 citations in Web of Science®
123 citations in Scopus®
### Altmetrics
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http://www.math.colostate.edu/graduate.shtml | Education is not the filling of a pail, it is the lighting of a fire. - W. B. Yeats
### NEWS in the Department of Mathematics Graduate Announcements
Eric Hanson - PhD defense
Date: Tuesday, June 23, 2015
Time: 10:00 a.m.
Place: Weber 201
Title: Algorithms in Numerical Algebraic Geometry and Applications
Committee:
Dr. Chris Peterson
Dr. Renzo Cavalieri
Dr. Anthony Maciejewski
Abstract: The topics in this talk, while independent, are unified under the field of numerical algebraic geometry. With ties to some of the oldest areas in mathematics, numerical algebraic geometry is relatively young as a field of study in its own right. The field is concerned with the numerical approximation of the solution sets to systems of polynomial equations and the manipulation of the these sets. Given a system of polynomial equations, the methods of numerical algebraic geometry produce numerical approximations of the isolated solutions, as well as points on any positive-dimensional components of the solution set. In a short time, the work done in numerical algebraic geometry has significantly pushed the boundary of what is computable. This talk will provide an introduction to numerical algebraic geometry and then discuss three algorithms/applications: perturbed homotopies, exceptional sets and fiber products, and a numerical approach to finding unit distance embeddings of finite simple graphs. A short description of each topic appears below:
One of the most recent advances in numerical algebraic geometry is regeneration, an equation-by-equation homotopy method that is often more efficient then other approaches. However, the basic form of regeneration will not necessarily find all isolated singular solutions of a polynomial system without the additional cost of using deflation. We present an alternative to deflation in the form of perturbed homotopies for solving polynomial systems. In particular, we propose first solving a perturbed version of the polynomial system, followed by a parameter homotopy to remove the perturbation. The aim of this is two-fold. First, such perturbed homotopies are sometimes more efficient than regular homotopies, though they can also be less efficient. Second, a useful consequence is that the application of this perturbation to regeneration will yield all isolated solutions, including all singular isolated solutions. This version of regeneration -- {\em perturbed regeneration} -- can decrease the efficiency of regeneration but increases its applicability.
The second topic of this talk considers families of polynomial systems which depend on parameters. There is a typical dimension for the variety defined by a system in the family; however, this dimension may jump for parameters in algebraic subsets of the parameter space. Sommese and Wampler exploited fiber products to give a numerical method for identifying these special parameter values. We propose a refined numerical approach to fiber products, which uses recent advancements in numerical algebraic geometry, such as regeneration extension. We show that this method is sometimes more efficient then known techniques. In part, this gain in efficiency is due to the fact that regeneration extension allows the construction of the fiber product to be restricted to specified irreducible components. This work is motivated by applications in Kinematics - the study of mechanisms. As such we use an algebraic model of a two link arm to illustrate the algorithms we develop.
The final topic is the identification of unit distance embeddings of finite simple graphs. Given a graph $G(V,E)$, a unit distance embedding is a map $\phi$ from the vertex set $V$ into a metric space $M$ such that if $\{v_i,v_j\}\in E$ then the distance between $\phi(v_i)$ and $\phi(v_j)$ in $M$ is one. Given $G$ we cast the question of the existence of a unit distance embedding in $\mathbb{R}^n$ as the question of the existence of a real solution to a system of polynomial equations. As a consequence, we are able to develop theoretic algorithms for determining the existence of a unit distance embedding and for determining the smallest dimension of $\mathbb{R}^n$ for which a unit distance embedding of $G$ exists (that is, we determine the minimal embedding dimension of $G$). We put these algorithms into practice using the methods of numerical algebraic geometry. In particular, we consider unit distance embeddings of the Heawood Graph. This is the smallest example of a point-line incidence graph of a finite projective plan. In 1972, Chv\'{a}tal conjectured that point-line incidence graphs of finite projective planes do not have unit-distance embeddings into $\mathbb{R}^2$. In other words, Chv\'{a}tal conjectured that the minimal embedding dimension of any point-line incidence graph of a finite projective plane is at least $3$. We disprove this conjecture, adding hundreds of counterexamples to the $11$ known counterexamples found by Gerbracht.
Steve Ihde - PhD defense
Date: Wednesday, June 24, 2015
Time: 2:00 p.m.
Place: Weber 201
Title: Preconditioning Polynomial Systems Using Macaulay Dual Spaces
Committee
Dr. Chris Peterson
Dr. Alexander Hulpke
Dr. Peter Young
Abstract:
Numerical Algebraic Geometry is concerned with using numerical methods to solve polynomial systems. The main tool from this area is the method of Homotopy Continuation. Sometimes, this method is not efficient due to a number of factors. This project concerns the preconditioning of polynomial systems in order to be better suited for homotopy continuation.
In 2010, Hauenstein showed that computations could be done on the basis of a polynomial ideal by working in the Macaulay dual space. We will show that these computations can be used to find an H-basis, first introduced in 1916 by Macaulay. This H-basis will create a preconditioned set of homogeneous polynomials from a zero-dimensional system. We then improve upon these methods using the Closedness Subspace, first introduced by Zeng in 2009.
Background on homotopy continuation, h-bases, and the closedness subspace will be presented. This will lead to algorithms and examples of preconditioning for homotopy continuation.
Sofya Chepushtanova - PhD defense
Date: Thursday, June 25, 2015
Place: Weber, 201
Time: 10:00 a.m.
Title: Algorithms for Feature Selection and Pattern Recognition on Grassmann Manifolds
Committee:
Dr. Chris Peterson
Dr. Dan Bates
Dr. Asa Ben-Hur
Abstact: Three distinct application-driven research projects, inspired by ideas and topics from geometric data analysis, machine learning, and computational topology, are presented. We first consider feature selection based on sparse support vector machines (SSVMs) applied to hyperspectral band selection problem. A supervised embedded approach is proposed using the property of SSVMs to exhibit a model structure that includes a clearly identifiable gap between zero and non-zero feature vector weights that permits important spectral bands to be definitively selected in conjunction with the classification problem. Second, we describe an approach for performing set-to-set pattern recognition, via classification of data on embedded Grassmannians. A set of points from a given class characterizes the variability of the class information, so we propose organizing sets of data as subspaces on a Grassmann manifold. Multidimensional scaling finds Euclidean embeddings of the manifold, preserving or approximating the Grassmannian geometry based on a chosen distance measure. SSVMs are trained in the Euclidean space for classification and identification of optimal dimensions of embedded subspaces. Lastly, we consider an application of persistent homology to hyperspectral data analysis on the Grassmann manifolds. Persistent homology (PH) is a method of topological data analysis (TDA) that can be applied to a data set to capture the persistence of topological structure. Using the Grassmannian framework affords a form of data compression while retaining pertinent data structure. In this setting it becomes feasible to analyze large volumes of hyperspectral data as the high computational cost of PH applied to the original data space is greatly reduced.
Mike Mikucki - PhD Defense
Date: Friday, June 26, 2015
Time: 1:00 p.m.
Place: Weber 201
Title: Electromechanical and curvature driven molecular flows for lipid membranes | {"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": 0.8581714630126953, "perplexity": 777.1517836601398}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-27/segments/1435375097475.46/warc/CC-MAIN-20150627031817-00208-ip-10-179-60-89.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/dark-energy-is-space.34743/ | # Dark energy is space
1. Jul 12, 2004
### kurious
If I double the size of a classical,non-quantum mechanical vacuum,
I double the volume of space and its density stays constant at zero
kg/ m ^3.As the amount of dark energy in the universe increases the volume of space increases and the density stays constant.Is dark energy space?
2. Jul 16, 2004
### force5
Hi kurious;
I don't think all space is dark energy. But, I think all dark energy is space.
Just my thoughts.
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| ̃y[W̃gbv | | {"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": 0.9981544017791748, "perplexity": 28.26564531334152}, "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-2017-39/segments/1505818696677.93/warc/CC-MAIN-20170926175208-20170926195208-00215.warc.gz"} |
http://www.ni.com/documentation/en/vision-development-module/19.0/ni-vision-algorithms-node-ref/convert-roi-to-rectangle/ | # Convert from ROI (Convert ROI to Rectangle) (G Dataflow)
Converts an ROI descriptor to a rectangle or rotated rectangle.
The contour of the ROI Descriptor is always of the type rotated rectangle.
## ROI descriptor
Descriptor that defines the rectangle.
### global rectangle
Minimum rectangle required to contain all of the contours in the ROI.
### Contours
Individual shapes that define an ROI.
#### ID
Object specifying if contour is the external or internal edge of an ROI.
#### Type
Shape type of the contour.
#### Coordinates
Relative position of the contour.
## error in
Error conditions that occur before this node runs.
The node responds to this input according to standard error behavior.
Standard Error Behavior
Many nodes provide an error in input and an error out output so that the node can respond to and communicate errors that occur while code is running. The value of error in specifies whether an error occurred before the node runs. Most nodes respond to values of error in in a standard, predictable way.
error in does not contain an error error in contains an error
If no error occurred before the node runs, the node begins execution normally.
If no error occurs while the node runs, it returns no error. If an error does occur while the node runs, it returns that error information as error out.
If an error occurred before the node runs, the node does not execute. Instead, it returns the error in value as error out.
Default: No error
## rectangle
Rectangular region.
The left, top, right, and bottom coordinates describe a non-rotated rectangle. The rotation parameter specifies the rotation angle of the rectangle around its center.
### left
X-coordinate of the upper left corner of the rectangle.
### top
Y-coordinate of the upper left corner of the rectangle.
### right
X-coordinate of the bottom right corner of the rectangle.
### bottom
Y-coordinate of the bottom right corner of the rectangle.
### rotation
Rotation angle in degrees of the rectangle with its center as point of rotation.
If the rotation angle does not equal zero, the left, top, right, and bottom coordinates are not the actual coordinates of the upper left and bottom right corner of the rectangle, but their position if the rotation angle equals zero.
## error out
Error information.
The node produces this output according to standard error behavior.
Standard Error Behavior
Many nodes provide an error in input and an error out output so that the node can respond to and communicate errors that occur while code is running. The value of error in specifies whether an error occurred before the node runs. Most nodes respond to values of error in in a standard, predictable way.
error in does not contain an error error in contains an error
If no error occurred before the node runs, the node begins execution normally.
If no error occurs while the node runs, it returns no error. If an error does occur while the node runs, it returns that error information as error out.
If an error occurred before the node runs, the node does not execute. Instead, it returns the error in value as error out.
Where This Node Can Run:
Desktop OS: Windows
FPGA: Not supported
Web Server: Not supported in VIs that run in a web 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": 0.8140066862106323, "perplexity": 2176.8621321337714}, "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/1573496670987.78/warc/CC-MAIN-20191121204227-20191121232227-00085.warc.gz"} |
https://mathoverflow.net/questions/118505/are-maximal-connected-semisimple-subgroups-automatically-closed | # Are maximal connected semisimple subgroups automatically closed?
(Yet another question in a series demonstrating my rather embarrassing ignorance of standard Lie theory... I hope this is not too basic for MO!)
To be a little more precise: let $G$ be a real connected Lie group, not necessarily simply connected, and let $R$ be its solvable radical. Then (see the EOM entry, for instance) $G$ can be written as $R\cdot S$ for some maximal semisimple, connected Lie subgroup $S$. Having looked online, I get the impression that $S$ need not be closed. Is this impression correct?
(As mentioned in the link above, if $G$ is simply connected then a Levi--Malcev decomposition of its Lie algebra exponentiates to give $G=R\rtimes S$ and $S$ is closed.)
The motivation is in some joint work, where we have a property of locally compact groups that is known to pass to closed subgroups, and which we have shown is not satisfied by any connected semisimple Lie group. It would be nice if we could deduce from this that a connected Lie group with this property has to be solvable...
• If by a Lie subgroup you mean that $S$ has the induced topology, then every Lie subgroup is closed, since the complement of $S$ in its closure is a union of orbits, and all orbits are submanifolds of the same dimension. Jan 10, 2013 at 8:52
• @Angelo: perhaps my terminology is incorrect or ambiguous. I meant that $S$ is an immersed submanifold of $G$ (as in Alain Valette's answer). Jan 10, 2013 at 22:56
Your impression is correct. Let $H$ be the universal cover of $SL_2(\mathbb{R})$, it has infinite cyclic center, let $z$ be a generator of the center. Consider the product $H\times U(1)$ (where $U(1)$ is the group of complex numbers of modulus 1), and mod out by the (discrete, central) subgroup generated by $(z,e^{2\pi i\theta})$, where $\theta$ is irrational. You get a reductive group with radical $S^1$, in which the Levi factor $H$ is dense. | {"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": 0.928867757320404, "perplexity": 105.53399702817549}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103619185.32/warc/CC-MAIN-20220628233925-20220629023925-00288.warc.gz"} |
https://www.statstutor.net/downloads/question-2000009-descriptive-statistics/ | # Question #2000009: Descriptive Statistics
Question: A sample of 36 eggs is taken by the owner of chicken farm. He measures a mean weight of 30 grams with a standard deviation of 3 grams. The standard deviation of the mean weight is…??
Solution: The solution consists of 72 words (1 page)
Deliverables: Word Document
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": 0.8775147795677185, "perplexity": 1664.1379012711811}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550247526282.78/warc/CC-MAIN-20190222200334-20190222222334-00312.warc.gz"} |
http://physics.stackexchange.com/questions/13259/evidence-on-the-equation-of-state-for-dark-energy | # evidence on the equation of state for dark energy?
If dark energy contributes mass-energy density $\rho$ and pressure $p$ to the stress-energy tensor, then you can define $w=p/\rho$, where $w=-1$ gives a cosmological constant, $w<-1$ gives a big rip, and $w<-1/3$ if we want to use dark energy to explain cosmological acceleration. The WP "Big Rip" article cites a paper that dates back to 2003 http://arxiv.org/abs/astro-ph/0302506 , which states that the empirical evidence was at that time only good enough to give $-2 \lesssim w \lesssim -.5$.
Have observations constrained $w$ any more tightly since 2003?
I've heard many people express the opinion that $w<-1$ is silly or poorly motivated, or that "no one believes it." What are the reasons for this? Other than mathematical simplicity, I don't know of any reason to prefer $w=-1$ over $w\ne -1$. Considering that attempts to calculate $\Lambda$ using QFT are off by 120 orders of magnitude, is it reasonable to depend on theory to give any input into what values of $w$ are believable?
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Combining with previous measurements of the acoustic scale, we obtain a value of $w_0$ = -1.03 +/- 0.16 for the equation of state parameter of the dark energy
cited here : Cosmology today–A brief review (25 pages on Theory and data, 2007/Jul)
You can find a new model of the Universe without Dark Energy here:
A self-similar model of the Universe unveils the nature of dark energy (21 pages Jul/2011, not peer-reviewed, it uses only Newton and Coulomb laws). I do not know how this novel viewpoint can be discarded.
Argument:
From $$F=m\cdot a,\, F=G\cdot\frac{m_{1}\cdot m_{2}}{d^{2}},\, F=\frac{1}{4\pi\varepsilon_{0}}\cdot\frac{q_{1}\cdot q_{2}}{d^{2}},\, c=\frac{1}{\sqrt{\varepsilon_{0}\mu_{0}}}.$$ obtain the dimensional equations $$\left[G\right]=M^{-1}L^{3}T^{-2},\left[\varepsilon\right]=M^{-1}Q^{2}L^{-3}T^{2},\left[c\right]=LT^{-1}$$ because the the sum of the exponents is zero, a cute coincidence ;), we know that $[M],[Q],[L],[T]$ can scale in the same way and $[c],[G],[\varepsilon]$ are constants. If the universe can scale thru time, keeping always the same basic physical laws, it will scale.
The scaling law $\alpha(t_{S})=e^{-H_{0}\cdot t_{S}}$ is derived from the observational data, where $t_S$ is considered from the viewpoint of a comoving invariant referential, and Dark Energy is absent from it.
The accelerated expansion is an artifact of the standard model
The statement that space expansion is accelerating is not the result of some direct measurement more or less independent of the cosmological model but, on the contrary, it is a consequence of the theoretical framework of the standard model. The deceleration parameter at the present moment, $q_{0}$, in the $\Lambda$CDM model, for flat space and $\Omega_{R}=0$ , is given by $q_{0}=\frac{1}{2}\left(\Omega_{M}-2\Omega_{\Lambda}\right)$ therefore, for $\Omega_{M}+\Omega_{\Lambda}=1$ , the value of $q_{0}$ is negative for $\Omega_{\Lambda}>1/3$ ; a value of $\Omega_{\Lambda}$ lower than 1/3 leads to a comoving distance largely in disagreement with observations, hence, in the framework of $\Lambda$CDM model it has to be $\Omega_{\Lambda}>1/3$ and, so, $q_{0}<0$ .
(Needless to say: this is my preferred viewpoint because it has no free lunches: growing space, growing dark energy, and apply to all scales, even to Solar system)
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Thanks for the reference to the Carnero paper -- that's exactly what I was looking for! If your answer had stopped there, I would have accepted and upvoted. But the Oliveira paper is pure crackpottery. – Ben Crowell Aug 8 '11 at 16:09
I would still be interested in posts as to why Big Rip scenarios seem not to be taken seriously by so many people. – Ben Crowell Aug 8 '11 at 16:29
@Ben Crowell I used a few lines to present the 'argument' that a scaling model is inscribed in the gravitation and electrostatic laws. Somewhere you argued that $M_e/M_p$ is constant. D'accord it is as constant as the ratio of the mass of my hand versus the one of my body. But my hand is not invariant. There is no evidence that the mass of the proton is an 'absolut' invariant (irt itself). From you I expect more that nasty word : Be the first to present evidence of the invariance or to advance a physical counter-argument to the scaling model. To you it should be a piece of cake. – Helder Velez Aug 8 '11 at 18:05 | {"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": 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": 0.8145462274551392, "perplexity": 438.4582622470277}, "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/1461861831994.45/warc/CC-MAIN-20160428164351-00066-ip-10-239-7-51.ec2.internal.warc.gz"} |
http://blekko.com/wiki/Integration_by_parts?source=672620ff | # Integration by parts
From Wikipedia, the free encyclopedia - View original article
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In calculus, and more generally in mathematical analysis, integration by parts is a theorem that relates the integral of a product of functions to the integral of their derivative and antiderivative. It is frequently used to transform the antiderivative of a product of functions into an antiderivative for which a solution can be more easily found. The rule can be derived in one line simply by integrating the product rule of differentiation.
If u = u(x), v = v(x), and the differentials du = u '(xdx and dv = v'(xdx, then integration by parts states that
$\int u(x) v'(x) \, dx = u(x) v(x) - \int u'(x) v(x) \, dx$
or more compactly:
$\int u \, dv=uv-\int v \, du.\!$
More general formulations of integration by parts exist for the Riemann–Stieltjes integral and Lebesgue–Stieltjes integral. The discrete analogue for sequences is called summation by parts.
## Theorem
### Product of two functions
The theorem can be derived as follows. Suppose u(x) and v(x) are two continuously differentiable functions. The product rule states (in Leibniz’ notation):
$\frac{d}{dx}\left(u(x)v(x)\right) = v(x) \frac{d}{dx}\left(u(x)\right) + u(x) \frac{d}{dx}\left(v(x)\right).\!$
Integrating both sides with respect to x, over an interval axb:
$\int_a^b \frac{d}{dx}\left(u(x)v(x)\right)\,dx = \int_a^b u'(x)v(x)\,dx + \int_a^b u(x)v'(x)\,dx$
then applying the fundamental theorem of calculus,
$\int_a^b \frac{d}{dx}\left(u(x)v(x)\right)\,dx = \left[u(x)v(x)\right]_a^b$
gives the formula for integration by parts:
$\left[u(x)v(x)\right]_a^b = \int_a^b u'(x)v(x)\,dx + \int_a^b u(x)v'(x)\,dx.$
Since du and dv are differentials of a function of one variable x,
$du=u'(x)dx, \quad dv=v'(x)dx.$
The original integral ∫uv′ dx contains v′ (derivative of v); in order to apply the theorem, v (antiderivative of v′) must be found, and then the resulting integral ∫vu′ dx must be evaluated.
### Product of many functions
Integrating the product rule for three multiplied functions, u(x), v(x), w(x), gives a similar result:
$\int_a^b u v \, dw = [u v w]^{b}_{a} - \int_a^b u w \, dv - \int_a^b v w \, du.$
In general for n factors
$\frac{d}{dx} \left(\prod_{i=1}^n u_i(x) \right)= \sum_{j=1}^n \prod_{i\neq j}^n u_i(x) \frac{du_j(x)}{dx},$
which leads to
$\Bigl[ \prod_{i=1}^n u_i(x) \Bigr]_a^b = \sum_{j=1}^n \int_a^b \prod_{i\neq j}^n u_i(x) \, du_j(x),$
where the product is of all functions except for the one differentiated in the same term.
## Visualization
Graphical interpretation of the theorem. The pictured curve is parametrized by the variable t.
Define a parametric curve by (x, y) = (f(t), g(t)). Assuming that the curve is locally one-to-one, we can define
$x(y) = f(g^{-1}(y))$
$y(x) = g(f^{-1}(x))$
The area of the blue region is
$A_1=\int_{y_1}^{y_2}x(y)dy$
Similarly, the area of the red region is
$A_2=\int_{x_1}^{x_2}y(x)dx$
The total area A1 + A2 is equal to the area of the bigger rectangle, x2y2, minus the area of the smaller one, x1y1:
$\overbrace{\int_{y_1}^{y_2}x(y)dy}^{A_1}+\overbrace{\int_{x_1}^{x_2}y(x)dx}^{A_2}=\biggl.x_iy_i\biggl|_{i=1}^{i=2}$
Assuming the curve is smooth within a neighborhood, this generalizes to indefinite integrals:
$\int xdy + \int y dx = xy$
Rearranging:
$\int xdy = xy - \int y dx$
Thus integration by parts may be thought of as deriving the area of the blue region from the total area and that of the red region.
This visualisation also explains why integration by parts may help find the integral of an inverse function f−1(x) when the integral of the function f(xv) is known. Indeed, the functions x(y) and y(x) are inverses, and the integral ∫x dy may be calculated as above from knowing the integral ∫y dx.
## Application to find antiderivatives
### Strategy
Integration by parts is a heuristic rather than a purely mechanical process for solving integrals; given a single function to integrate, the typical strategy is to carefully separate it into a product of two functions u(x)v(x) such that the integral produced by the integration by parts formula is easier to evaluate than the original one. The following form is useful in illustrating the best strategy to take:
$\int uv \,dx = u \int v \,dx - \int \left ( u' \int v \,dx \right )\,dx.\!$
Note that on the right-hand side, u is differentiated and v is integrated; consequently it is useful to choose u as a function that simplifies when differentiated, and/or to choose v as a function that simplifies when integrated. As a simple example, consider:
$\int \frac{\ln x}{x^2}\,dx.\!$
Since the derivative of ln x is 1/x, we make (ln x) part of u; since the antiderivative of 1/x2 is −1/x, we make (1/x2) part of v. The formula now yields:
$\int \frac{\ln x}{x^2}\,dx = -\frac{\ln x}{x} - \int \biggl( \frac{1}{x} \biggr) \biggl( -\frac{1}{x} \biggr) \, dx.\!$
The antiderivative of −1/x2 can be found with the power rule and is 1/x.
Alternatively, we may choose u and v such that the product u' (∫v dx) simplifies due to cancellation. For example, suppose we wish to integrate:
$\int\sec^2x\ln|\sin x|dx.$
If we choose u(x) = ln |sin x| and v(x) = sec2x, then u differentiates to 1/ tan x using the chain rule and v integrates to tan x; so the formula gives:
$\int\sec^2x\ln|\sin x|dx=\tan x\ln|\sin x|-\int\tan x\frac{1}{\tan x}dx.$
The integrand simplifies to 1, so the antiderivative is x. Finding a simplifying combination frequently involves experimentation.
In some applications, it may not be necessary to ensure that the integral produced by integration by parts has a simple form; for example, in numerical analysis, it may suffice that it has small magnitude and so contributes only a small error term. Some other special techniques are demonstrated in the examples below.
Polynomials and trigonometric functions
In order to calculate
$I=\int x\cos (x) \,dx\,,$
let:
$u = x \Rightarrow d u = dx$
$dv = \cos(x)\,dx \Rightarrow v = \int\cos(x)\,dx = \sin(x)$
then:
\begin{align} \int x\cos (x) \,dx & = \int u \, dv \\ & = uv - \int v \, du \\ & = x\sin (x) - \int \sin (x) \,dx \\ & = x\sin (x) + \cos (x) + C, \end{align} \!
where C is an arbitrary constant of integration.
For higher powers of x in the form
$\int x^n e^x \,dx,\,\int x^n\sin (x) \,dx,\,\int x^n\cos (x) \,dx\,,$
repeatedly using integration by parts can evaluate integrals such as these; each application of the theorem lowers the power of x by one.
Exponentials and trigonometric functions
An example commonly used to examine the workings of integration by parts is
$I=\int e^{x} \cos (x) \,dx.$
Here, integration by parts is performed twice. First let
$u = \cos(x) \Rightarrow du = -\sin(x)\,dx$
$dv = e^x \, dx \Rightarrow v = \int e^x \,dx = e^x$
then:
$\int e^{x} \cos (x) \,dx = e^{x} \cos (x) + \int e^{x} \sin (x) \,dx.\!$
Now, to evaluate the remaining integral, we use integration by parts again, with:
$u = \sin(x) \Rightarrow du = \cos(x)\, dx$
$dv = e^x \,dx \Rightarrow v = \int e^x \,dx = e^x.$
Then:
$\int e^x \sin (x) \,dx = e^x \sin (x) - \int e^x \cos (x) \,dx.$
Putting these together,
$\int e^x \cos (x) \,dx = e^x \cos (x) + e^x \sin (x) - \int e^x \cos (x) \,dx.$
The same integral shows up on both sides of this equation. The integral can simply be added to both sides to get
$2 \int e^{x} \cos (x) \,dx = e^{x} ( \sin (x) + \cos (x) ) + C\!$
which rearranges to:
$\int e^x \cos (x) \,dx = {e^x ( \sin (x) + \cos (x) ) \over 2} + C'\!$
where again C (and C' = C/2) is an arbitrary constant of integration.
A similar method is used to find the integral of secant cubed.
Functions multiplied by unity
Two other well-known examples are when integration by parts is applied to a function expressed as a product of 1 and itself. This works if the derivative of the function is known, and the integral of this derivative times x is also known.
The first example is ∫ ln(x) dx. We write this as:
$I=\int \ln (x) \cdot 1 \,dx.\!$
Let:
$u = \ln(x) \Rightarrow du = \frac{dx}{x}$
$dv = dx \Rightarrow v = x\,$
then:
\begin{align} \int \ln (x) \,dx & = x \ln (x) - \int \frac{x}{x} \,dx \\ & = x \ln (x) - \int 1 \,dx \\ & = x \ln (x) - x + C \end{align}
where C is the constant of integration.
The second example is the inverse tangent function arctan(x):
$I=\int \arctan (x) \cdot 1 \,dx.$
Rewrite this as
$\int \arctan (x) \cdot 1 \,dx.$
Now let:
$u = \arctan(x) \Rightarrow du = \frac{dx}{1 + x^2}$
$dv =dx \Rightarrow v = x$
then
\begin{align} \int \arctan (x) \,dx & = x \arctan (x) - \int \frac{x}{1 + x^2} \,dx \\[8pt] & = x \arctan (x) - {1 \over 2} \ln \left( 1 + x^2 \right) + C \end{align}
using a combination of the inverse chain rule method and the natural logarithm integral condition.
### LIATE rule
A rule of thumb proposed by Herbert Kasube of Bradley University advises that whichever function comes first in the following list should be u:[1]
L - Logarithmic functions: ln x, logb x, etc.
I - Inverse trigonometric functions: arctan x, arcsec x, etc.
A - Algebraic functions: x2, 3x50, etc.
T - Trigonometric functions: sin x, tan x, etc.
E - Exponential functions: ex, 19x, etc.
The function which is to be dv is whichever comes last in the list: functions lower on the list have easier antiderivatives than the functions above them. The rule is sometimes written as "DETAIL" where D stands for dv.
To demonstrate the LIATE rule, consider the integral
$\int x\cos x \, dx.\!$
Following the LIATE rule, u = x and dv = cos x dx, hence du = dx and v = sin x, which makes the integral become
$x\sin x - \int 1\sin x \, dx\!$
which equals
$x\sin x + \cos x+C.\!$
In general, one tries to choose u and dv such that du is simpler than u and dv is easy to integrate. If instead cos x was chosen as u and x as dv, we would have the integral
$\frac{x^2}2\cos x + \int \frac{x^2}2\sin x\, dx,$
which, after recursive application of the integration by parts formula, would clearly result in an infinite recursion and lead nowhere.
Although a useful rule of thumb, there are exceptions to the LIATE rule. A common alternative is to consider the rules in the "ILATE" order instead. Also, in some cases, polynomial terms need to be split in non-trivial ways. For example, to integrate
$\int x^3e^{x^2}\, dx,$
one would set
$u=x^2, \quad dv=xe^{x^2}\, dx,$
so that
$du = 2x\,dx, \quad v = \frac12 e^{x^2}.$
Then
$\int x^3e^{x^2}\, dx = \int \left(x^2\right) \left( xe^{x^2} \right) \, dx = \int u \, dv = uv - \int v\,du = \frac12 x^2 e^{x^2} - \int xe^{x^2}\,dx.$
Finally, this results in
$\int x^3e^{x^2}\, dx=\frac{1}{2}e^{x^2}(x^2-1)+C.$
## Applications in pure mathematics
Integration by parts is often used as a tool to prove theorems in mathematical analysis. This section gives a few of examples.
### Use in special functions
The gamma function is an example of a special function, defined as an improper integral. Integration by parts illustrates it to be an extension of the factorial:
\begin{align} \Gamma(z) & = \int_0^\infty d\lambda e^{-\lambda} \lambda^{z-1} \\ & = - \int_0^\infty d\left(e^{-\lambda}\right) \lambda^{z-1} \\ & = - \left[e^{-\lambda}\lambda^{z-1}\right]_0^\infty + \int_0^\infty d\left(\lambda^{z-1}\right) e^{-\lambda} \\ & = 0 + \int_0^\infty d\lambda\left(z-1\right) \lambda^{z-2} e^{-\lambda} \\ & = (z-1)\Gamma(z-1) \\ \end{align}
yielding the famous identity
$\Gamma(z) = (z-1)\Gamma(z-1)\,.$
For integer z, applying this formula repeatedly gives the factorial (denoted by the !):
$\Gamma(z+1) = z!$
### Use in harmonic analysis
Integration by parts is often used in harmonic analysis, particularly Fourier analysis, to show that quickly oscillating integrals with sufficiently smooth integrands decay quickly. The most common example of this is its use in showing that the decay of function's Fourier transform depends on the smoothness of that function, as described below.
Fourier transform of derivative
If f is a k-times continuously differentiable function and all derivatives up to the kth one decay to zero at infinity, then its Fourier transform satisfies
$(\mathcal{F}f^{(k)})(\xi) = (2\pi i\xi)^k \mathcal{F}f(\xi),$
where f(k) is the kth derivative of f. (The exact constant on the right depends on the convention of the Fourier transform used.) This is proved by noting that
$\frac{d}{dy} e^{-2\pi iy\xi} = -2\pi i\xi e^{-2\pi iy\xi},$
so using integration by parts on the Fourier transform of the derivative we get
\begin{align} (\mathcal{F}f')(\xi) &= \int_{-\infty}^\infty e^{-2\pi iy\xi} f'(y)\,dy \\ &=\left[e^{-2\pi iy\xi} f(y)\right]_{-\infty}^\infty - \int_{-\infty}^\infty (-2\pi i\xi e^{-2\pi iy\xi}) f(y)\,dy \\ &=2\pi i\xi \int_{-\infty}^\infty e^{-2\pi iy\xi} f(y)\,dy \\ &=2\pi i\xi \mathcal{F}f(\xi). \end{align}
Applying this inductively gives the result for general k. A similar method can be used to find the Laplace transform of a derivative of a function.
Decay of Fourier transform
The above result tells us about the decay of the Fourier transform, since it follows that if f and f(k) are integrable then
$\vert\mathcal{F}f(\xi)\vert \leq \frac{I(f)}{1+\vert 2\pi\xi\vert^k}$, where $I(f)=\int_{-\infty}^\infty\Bigl(\vert f(y)\vert + \vert f^{(k)}(y)\vert\Bigr) dy$.
In other words, if f satisfies these conditions then its Fourier transform decays at infinity at least as quickly as 1/|ξ|k. In particular, if k ≥ 2 then the Fourier transform is integrable.
The proof uses the fact, which is immediate from the definition of the Fourier transform, that
$\vert\mathcal{F}f(\xi)\vert \leq \int_{-\infty}^\infty \vert f(y) \vert \,dy.$
Using the same idea on the equality stated at the start of this subsection gives
$\vert(2\pi i\xi)^k \mathcal{F}f(\xi)\vert \leq \int_{-\infty}^\infty \vert f^{(k)}(y) \vert \,dy.$
Summing these two inequalities and then dividing by 1 + |2πξk| gives the stated inequality.
### Use in operator theory
One use of integration by parts in operator theory is that it shows that the -∆ (where ∆ is the Laplace operator) is a positive operator on L2 (see Lp space). If f is smooth and compactly supported then, using integration by parts, we have
\begin{align} \langle -\Delta f, f \rangle_{L^2} &= -\int_{-\infty}^\infty f''(x)\overline{f(x)}\,dx \\ &=-\left[f'(x)\overline{f(x)}\right]_{-\infty}^\infty + \int_{-\infty}^\infty f'(x)\overline{f'(x)}\,dx \\ &=\int_{-\infty}^\infty \vert f'(x)\vert^2\,dx \geq 0. \end{align}
## Recursive integration by parts
Integration by parts can often be applied recursively on the ∫ v du term to provide the following formula
$\int uv = u v_1 - u' v_2 + u'' v_3 - \cdots + (-1)^{n-1}\ u^{(n-1)} \ v_{n} + (-1)^n \int{u^{(n)}v_{n}}.\!$
Here, u′ is the first derivative of u and u′′ is the second derivative. Further, u(n) is a notation to describe its nth derivative with respect to the independent variable. Another notation approved in the calculus theory has been adopted:
$v_{n+1}(x)=\int\! \int\ \cdots \int v \ (dx)^{n+1}.\!$
There are n + 1 integrals.
Note that the integrand above (uv) differs from the previous equation. The dv factor has been written as v purely for convenience.
The above mentioned form is convenient because it can be evaluated by differentiating the first term and integrating the second (with a sign reversal each time), starting out with uv1. It is very useful especially in cases when u(k + 1) becomes zero for some k + 1. Hence, the integral evaluation can stop once the u(k) term has been reached.
### Tabular integration by parts
While the aforementioned recursive definition is correct, it is often tedious to remember and implement. A much easier visual representation of this process is often taught to students and is dubbed either "the tabular method",[2] "the Stand and Deliver method",[3] "rapid repeated integration" or "the tic-tac-toe method". This method works best when one of the two functions in the product is a polynomial, that is, after differentiating it several times one obtains zero. It may also be extended to work for functions that will repeat themselves.
For example, consider the integral
$\int x^3 \cos x \, dx.\!$
Let u = x3. Begin with this function and list in a column all the subsequent derivatives until zero is reached. Secondly, begin with the function v (in this case cos(x)) and list each integral of v until the size of the column is the same as that of u. The result should appear as follows.
Derivatives of u (Column A)Integrals of v (Column B)
$x^3 \,$$\cos x \,$
$3x^2 \,$$\sin x \,$
$6x \,$$-\cos x \,$
$6 \,$$-\sin x \,$
$0 \,$$\cos x \,$
Now simply pair the 1st entry of column A with the 2nd entry of column B, the 2nd entry of column A with the 3rd entry of column B, etc... with alternating signs (beginning with the positive sign). Do so until further pairing leads to sums of zeros. The result is the following (notice the alternating signs in each term):
$(+)(x^3)(\sin x) - (3x^2)(-\cos x) + (6x)(-\sin x) - (6)(\cos x) + C \,.$
Which, with simplification, leads to the result
$x^3\sin x + 3x^2\cos x - 6x\sin x - 6\cos x + C. \,$
With proper understanding of the tabular method, it can be extended. Consider
$\int e^x \cos x \,dx.$
Derivatives of u (Column A)Integrals of v (Column B)
$e^x \,$$\cos x \,$
$e^x \,$$\sin x \,$
$e^x \,$$-\cos x \,$
In this case in the last step it is necessary to integrate the product of the two bottom cells obtaining:
$\int e^x \cos x \,dx = e^x\sin x + e^x\cos x - \int e^x \cos x \,dx,$
which leads to
$2 \, \int e^x \cos x \,dx = e^x\sin x + e^x\cos x,$
and yields the result:
$\int e^x \cos x \,dx = {e^x ( \sin x + \cos x ) \over 2} + C.\!$
## Higher dimensions
The formula for integration by parts can be extended to functions of several variables. Instead of an interval one needs to integrate over an n-dimensional set. Also, one replaces the derivative with a partial derivative.
More specifically, suppose Ω is an open bounded subset of ℝn with a piecewise smooth boundary Γ. If u and v are two continuously differentiable functions on the closure of Ω, then the formula for integration by parts is
$\int_{\Omega} \frac{\partial u}{\partial x_i} v \,d\Omega = \int_{\Gamma} u v \, \nu_i \,d\Gamma - \int_{\Omega} u \frac{\partial v}{\partial x_i} \, d\Omega,$
where $\hat{\mathbf{\nu}}$ is the outward unit surface normal to Γ, νi is its i-th component, and i ranges from 1 to n.
By replacing v in the above formula with vi and summing over i gives the vector formula
$\int_{\Omega} \nabla u \cdot \mathbf{v}\, d\Omega = \int_{\Gamma} u (\mathbf{v}\cdot \hat{\nu})\, d\Gamma - \int_\Omega u\, \nabla\cdot\mathbf{v}\, d\Omega,$
where v is a vector-valued function with components v1, ..., vn.
Setting u equal to the constant function 1 in the above formula gives the divergence theorem
$\int_{\Gamma} \mathbf{v} \cdot \hat{\nu}\, d\Gamma = \int_\Omega \nabla\cdot\mathbf{v}\, d\Omega.$
For $\mathbf{v}=\nabla v$ where $v\in C^2(\bar{\Omega})$, one gets
$\int_{\Omega} \nabla u \cdot \nabla v\, d\Omega = \int_{\Gamma} u\, \nabla v\cdot\hat{\nu}\, d\Gamma - \int_\Omega u\, \nabla^2 v\, d\Omega,$
which is the first Green's identity.
The regularity requirements of the theorem can be relaxed. For instance, the boundary Γ need only be Lipschitz continuous. In the first formula above, only u, vH1(Ω) is necessary (where H1 is a Sobolev space); the other formulas have similarly relaxed requirements.
## Notes
1. ^ Kasube, Herbert E. (1983). "A Technique for Integration by Parts". The American Mathematical Monthly 90 (3): 210–211. doi:10.2307/2975556. JSTOR 2975556.
2. ^ Khattri, Sanjay K. (2008). "FOURIER SERIES AND LAPLACE TRANSFORM THROUGH TABULAR INTEGRATION". The Teaching of Mathematics XI (2): 97–103.
3. ^ Horowitz, David (1990). "Tabular Integration by Parts". The College Mathematics Journal 21 (4): 307–311. doi:10.2307/2686368. JSTOR 2686368. | {"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": 98, "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": 0.9976322054862976, "perplexity": 714.9712813938104}, "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-2014-49/segments/1416400376197.4/warc/CC-MAIN-20141119123256-00225-ip-10-235-23-156.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/potentials-and-de-broglie-wavelength.150543/ | # Homework Help: Potentials and de Broglie wavelength
1. Jan 8, 2007
### Brewer
1. The problem statement, all variables and given/known data
2. Relevant equations
E=hc/λ?
and the Time Independent Schrodinger Equation.
3. The attempt at a solution
Now, would I be right in thinking for the first section that the energy for the E=hc/λ bit is just the energy of the particle given in the question (10eV).
Then for the other sections is the relevant energy just 10eV minus the potential?
As the last section will give a negative KE of the particle does that mean that it doesn't have a de Broglie wavelength? Its just a negatively decaying exponential (i.e. tending to 0)?
Does that sound like I'm on the right lines?
2. Jan 9, 2007
### Brewer
Anyone have any hints about this one? The way that I described it above is the way that I attempted it, and after using the wrong values from my data sheet (first I used k(why I picked this I don't know - maybe I just can't read!) and then $$\hbar$$ before reading the sheet properly for the value of h!) I think I have values that give decent answers.
3. Jan 9, 2007
### Dick
It sounds fine to me. Decaying exponential in the last region, yes. Is there still a question?
4. Jan 9, 2007
### Gokul43201
Staff Emeritus
Are we talking about photons or electrons? What is the general de Broglie relation for any particle? | {"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": 0.9111815690994263, "perplexity": 1004.9098164329657}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-34/segments/1534221212910.25/warc/CC-MAIN-20180817202237-20180817222237-00371.warc.gz"} |
http://math.stackexchange.com/questions/41821/upper-bound-on-probability-of-not-choosing-certain-subsets | # Upper bound on probability of not choosing certain subsets
Given $m \geq 2$ subsets of $\{1,...,n\}$, say $A_1,...,A_m$, each of size $r$, we pick $B \subset \{1,...,n\}$ such that for any $i \in [n]$, $\displaystyle Pr\left[i \in B\right] = m^{-\frac{1}{r+1}}$, independently of the other choices. I want to show that there is a constant $\gamma_r$ (dependent only on $r$) such that
$$Pr\left[A_1\not\subseteq B \wedge ... \wedge A_m \not\subseteq B\right] \leq \gamma_r \left(1-m^{-\frac{r}{r+1}}\right)^m$$
It seems that Janson's inequality should be used here, however I can't get that to give me a constant which depends only on r. I'd appreciate any idea, and if this is at all correct.
Thanks.
Edit: I had mistakenly written that I used Azuma's inequality, however I meant Janson's.
-
You seem to ask for the probability of the event $C=[A\not\subseteq B]$, where $A$ is the union of $m$ given sets $A_k$, and $B$ is random. This is the union of the events $[i\notin B]$ for every $i\in A$. These are independent and $P(i\in B)=p$ does not depend on $i$ hence $P(C)=1-p^a$ where $a$ is the size of $A$.
In your context, $p=m^{-1/(r+1)}$ and, unless some hypothesis is missing, $a$ can be anything between $r$ and $rm$, hence the only upper bound available for every family $(A_k)$ like in your post is $$P(C)\le1-m^{-mr/(r+1)}.$$ When $m\to\infty$, the RHS converges to $1$ and the RHS of the inequality to prove converges to $0$ hence the former cannot be bounded by the latter, for any value of $\gamma_r$.
Or maybe you are asking for the probability of the event that, for every $k$, $A_k\not\subseteq B$...
Didier: Sorry, indeed the way I wrote it was misleading. I'm looking to bound $Pr\left[A_1 \not\subseteq B \wedge A_2 \not\subseteq B \wedge ... \wedge A_m \not\subseteq B\right]$. I'll fix my post. – shay May 28 '11 at 16:07 | {"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": 0.9899120330810547, "perplexity": 88.537930966954}, "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-2015-40/segments/1443737913039.71/warc/CC-MAIN-20151001221833-00002-ip-10-137-6-227.ec2.internal.warc.gz"} |
https://cstheory.stackexchange.com/questions/42506/is-there-a-universal-gate-set-for-classical-probabilistic-computing | Is there a universal gate set for classical probabilistic computing?
We know that NAND gates are universal for deterministic classical circuits, Toffoli gates are universal for reversible deterministic classical circuits, and Clifford+T is universal for quantum circuits. However, what about classical probabilistic circuits?
It seems like if you had a "coin-flip" gate (takes no input/any input and produces 0 or 1, each with probability 1/2) and NAND, you could build any probabilistic circuit you want (up to some level of precision) with a rejection sampling procedure. The desired probabilistic circuit can be seen as a set of deterministic circuits, chosen according to some distribution; use rejection sampling to simulate that distribution and then apply the required deterministic circuit.
(by the same argument, any gate that produces a fixed, non-trivial probability distribution, plus NAND, would be universal).
Is this a question that has been looked into? Any results that I can cite?
• Not sure what you mean by a "coin-flip gate ". Could you give us the truth table for such a gate ? – William Hird Mar 11 at 15:16
• I don't know how to represent a probabilistic gate with a truth table. A "coin-flip gate" would be a 1/2 probability of an identity gate, and a 1/2 probability of a NOT gate – Sam Jaques Mar 11 at 16:15
• Well Sam, here is your chance to become famous, invent a whole new form of truth table for the gate, it will be named the " Jaques Gate" LOL – William Hird Mar 17 at 19:28 | {"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": 0.9255308508872986, "perplexity": 1020.8330670148338}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232256980.46/warc/CC-MAIN-20190522223411-20190523005411-00333.warc.gz"} |
http://aas.org/archives/BAAS/v31n4/dps99/259.htm | 31st Annual Meeting of the DPS, October 1999
Session 23. Edgeworth-Kuiper Belt I
Contributed Oral Parallel Session, Tuesday, October 12, 1999, 9:10-10:00am, Sala Pietro d'Abano
## [23.02] Stability of Volatiles in the Kuiper Belt and in Extra-Solar Dust Disks.
J. Stansberry, R.H. Brown, J. Lunine, D. Trilling, W. Grundy (U of Arizona)
Recent observations of a near-IR absorption feature near 2.3\mum suggest the presence of methane or higher hydrocarbons on grains (~10\mum diameter) in the circumstellar disk of 55 Cancri (Trilling and Brown (1998) Nature 395, 775). Other observations (Brown et al. (1998) Science 276, 937) suggest the presence of methane or higher hydrocarbons on the surface of the large Kuiper Belt Object 1993SC. Several processes set timescales for the persistence of volatiles in the outer reaches of a dust disk or in the Kuiper Belt: Jeans escape, energy limited escape, ballistic escape, photolysis/radiolysis, and cold-trapping. Likewise there are processes which set lifetimes for dust and surface processing under relevant conditions: radiation pressure, impacts, and fragmentation. We examine the timescales for the above processes in the context of our Kuiper Belt and the circumstellar environment of 55 Cancri and in the presence of methane and/or more carbon-rich molecules derived from it. We also examine the optical properties of grains composed of H2O, CH4, and a neutral absorber to place constraints on the abundance of methane (if that is in fact the absorber responsible for the 2.3\mum feature) in the 55 Cancri disk. Devolatilization of grains smaller than 1mm is very efficient even in the cold environment of the 55 Cancri disk. The presence of methane or its photolytic products may require sequestration in grain interiors. Conversely, on large bodies such as 1993SC cold trapping of volatiles offers ample opportunity for photolytic/radiolytic processing of methane, and probably can even stabilize methane itself over timescales comparable to the age of the Solar System. We discuss implications of these processes for nitrogen as well, although there is very little hope of detecting it in such environments on human timescales unless the Pluto--Kuiper Express mission becomes a reality. The potential for addressing issues such as these via observations with SIRTF will also be addressed. | {"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": 0.8111760020256042, "perplexity": 4917.331402191045}, "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-2015-40/segments/1443737896527.58/warc/CC-MAIN-20151001221816-00127-ip-10-137-6-227.ec2.internal.warc.gz"} |
https://authorea.com/users/299657/articles/429144-multi-dimensional-legendre-wavelet-matrix-approach-for-hyperbolic-telegraph-equation-with-dirichlet-boundary-condition | Multi-dimensional Legendre wavelet matrix approach for hyperbolic telegraph equation with dirichlet boundary condition
• Vijay Patel,
• Dhirendra Bahuguna
Vijay Patel
Indian Institute of Technology Kanpur
Author Profile
Dhirendra Bahuguna
Indian Institute of Technology Kanpur
Author Profile
## Abstract
The present article is devoted to developing the Legendre wavelet operational matrix method (LWOMM) to find the numerical solution of two-dimensional hyperbolic telegraph equations (HTE) with appropriate initial time boundary space conditions. The Legendre wavelets series with unknown coefficients has used for approximating the solution in both of the spatial and temporal variables. The basic idea for discretizing two-dimensional HTE is based on differentiation and integration of operational matrices. By implementing LWOMM on HTE, HTE is transformed into algebraic generalized Sylvester equation. Numerical experiments are provided to illustrate the accuracy and efficiency of the presented numerical scheme. Comparisons of numerical results associated with the proposed method with some of the existing numerical methods confirm that the method is easy, accurate and fast experimentally. Moreover, we have investigated the convergence analysis of multidimensional Legendre wavelet approximation. | {"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": 0.9065399169921875, "perplexity": 1036.9669963132733}, "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/1606141191511.46/warc/CC-MAIN-20201127073750-20201127103750-00119.warc.gz"} |
https://infoscience.epfl.ch/record/219911 | Infoscience
Conference paper
# Optimal Computational Split-state Non-malleable Codes
Non-malleable codes are a generalization of classical error-correcting codes where the act of "corrupting" a codeword is replaced by a "tampering" adversary. Non-malleable codes guarantee that the message contained in the tampered codeword is either the original message m, or a completely unrelated one. In the common split-state model, the codeword consists of multiple blocks (or states) and each block is tampered with independently. The central goal in the split-state model is to construct high rate non-malleable codes against all functions with only two states (which are necessary). Following a series of long and impressive line of work, constant rate, two-state, non-malleable codes against all functions were recently achieved by Aggarwal et al. [2]. Though constant, the rate of all known constructions in the split state model is very far from optimal (even with more than two states). In this work, we consider the question of improving the rate of splitstate non-malleable codes. In the "information theoretic" setting, it is not possible to go beyond rate 1/2. We therefore focus on the standard computational setting. In this setting, each tampering function is required to be efficiently computable, and the message in the tampered codeword is required to be either the original message m or a "computationally" independent one. In this setting, assuming only the existence of one-way functions, we present a compiler which converts any poor rate, two-state, (sufficiently strong) non-malleable code into a rate-1, two-state, computational non-malleable code. These parameters are asymptotically optimal. Furthermore, for the qualitative optimality of our result, we generalize the result of Cheraghchi and Guruswami [10] to show that the existence of one-way functions is necessary to achieve rate > 1/2 for such codes. Our compiler requires a stronger form of non-malleability, called augmented non-malleability. This notion requires a stronger simulation guarantee for non-malleable codes and simplifies their modular usage in cryptographic settings where composition occurs. Unfortunately, this form of non-malleability is neither straightforward nor generally guaranteed by known results. Nevertheless, we prove this stronger form of non-malleability for the two-state construction of Aggarwal et al. [3]. This result is of independent interest. | {"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": 0.9442670941352844, "perplexity": 842.4537639549754}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084890947.53/warc/CC-MAIN-20180122014544-20180122034544-00008.warc.gz"} |
http://mathoverflow.net/questions/17959/effect-on-connectivity-when-partitioning-a-graph/17995 | Effect on connectivity when partitioning a graph
I have a connected graph $G=(V,E)$, $V$ being the vertex set and $E$ being the edge set. I partition the graph into components $C=\{C_1,\dots,C_n\}$ such that all $C_i$ are pairwise disjoint.
Take two vertices $s,t \in V$ such that $s,t$ are connected by a path. Is there an $O(|V|+|E|)$ algorithm to find out the list of all $C_i \in C$ such that if we remove the vertices in $C_i$ from the graph, then $s$ will become disconnected from $t$.
I know there is the $O(|C|(|V|+|E|))$ algorithm to do so by removing vertices in $C_i \in C$ from the graph for all $1\leq i\leq n$ and then checking if $s$ and $t$ are connected.
This can be somewhat improved if we take all edge weights as 1 and compute the shortest path and then consider only components whose vertices are present in the shortest path but this still has worst case complexity $O(|C|(|V|+|E|))$.
-
Here is a thought: consider the color-adjacency matrix A_i, where (u,v)=1 iff u and v have a path using vertices only from at most i different partition classes, i running from 1 to n. Perhaps s and t have a path involving log(n) colors, and then the worst case for your shortest path algorithm would involve log(n) instead of n. You may get lucky and find an algorithm to compute, say, A_(2^j) quickly. Gerhard "Ask Me About System Design" Paseman, 2010.03.12 – Gerhard Paseman Mar 12 '10 at 9:04
Would A_i take $O(n^2)$ time to setup though? – Opt Mar 12 '10 at 9:09
It might. Even computing the row for vertex s might take some time. However, computing A_i for all pairs s,t might be useful in other applications. Gerhard "Ask Me About System Design" Paseman, 2010.03.12 – Gerhard Paseman Mar 13 '10 at 7:55
I am not sure if I get the point. Are there any restrictions on the components C_i? If there isn't any, then why not take all neighbors of s? – Sangxia Huang Mar 22 '10 at 12:37
You want a list of all the C_i, that's why just taking the neighbors does not suffice. – Opt Mar 22 '10 at 16:30
Consider the problem in which every "component" is a single vertex. Then the problem is to find every $s$-$t$ cutvertex in $O(|E|)$ time, which is fairly easy (each lies on every $s$-$t$ path).
If your components are not necessarily connected, it might be more difficult. You would have to solve the case in which each $C_i$ is an independent set. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8206678628921509, "perplexity": 329.65123650157443}, "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-2014-41/segments/1410657104131.95/warc/CC-MAIN-20140914011144-00118-ip-10-196-40-205.us-west-1.compute.internal.warc.gz"} |
https://www.jiskha.com/questions/13314/a-4-kg-block-is-pushed-along-the-ceiling-with-a-constant-applied-force-of-85-n-that-acts | # PHYSICS
A 4 kg block is pushed along the ceiling with a constant applied force of 85 N that acts at an angle of 55 degrees with the horizontal. The block accelerates to the right at 6 m/s^2. determine the coefficient of kinetic friction between the block and the ceiling.
The applied force in the direction of motion is 85 cos 55 = 48.75 N
The friction force that acts opposite to the direction of motion is
(F sin55 - M g)* (mu,k)
To solve for (mu,k), the coefficient of kinetic friction, solve the Newton's Second Law equation
Fnet = F cos55 -F sin55 + M g(mu,k]
= M a
You already know F, M, g and a.
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1. 1.14
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3. 4.00 kg block is pushed along
the ceiling with a constant
applied force of 85.0 N that acts
at an angle of 55.0° with the
horizontal, as in the figure. The block accelerates to the right at 6.00 m/s2. Determine the coefficient of kinetic friction between the block and the ceiling.
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4. 4.00 kg block is pushed along
the ceiling with a constant
applied force of 85.0 N that acts
at an angle of 55.0° with the
horizontal, as in the figure. The block accelerates to the right at 6.00 m/s2. Determine the coefficient of kinetic friction between the block and the ceiling.
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http://volokh.com/posts/1142373636.shtml | pageok
pageok
pageok [Puzzleblogger Kevan Choset, March 15, 2006 at 11:16am] Trackbacks The West Wing & Math: This dialogue is from the West Wing episode, "Evidence of Things Not Seen." C.J. There's a spot on the earth where the temperature is exactly the same as it would be if you drilled through the earth to the other side. WILL No, there isn't. How about six dollars if you do it with a face card? C.J. Yes, there is, and it's called the antipode. And if that's true, then why can't it be that you could stand an egg on end at the equinox? Now, the problem. Prove that at any given moment, what C.J. is saying is true: there are two points on the earth exactly opposite each other which have the exact same temperature. You can assume that temperature is continuous (i.e., there's no spot on the earth where the temperature just "jumps" from 0 degrees at one spot to 100 degrees at a spot infinitesimally close to it). (link) Aaron Bergman (mail): Ah, the joys of the intermediate value theorem. 3.15.2006 12:18pm (link) Very Anonymous Coward: Suppose not; then construct the obvious retraction; Brouwer fixed point theorem; contradiction. A related result is that there is at least one place on the Earth where there is no wind. 3.15.2006 12:44pm (link) Freder Frederson (mail): With what precision are you measuring the temperature. If you are measuring the temperature with sufficient precision, I would imagine the statement is never true. If you are measuring it within a degree or two then I would imagine it is often true if you are looking at the tropics over the oceans and in temperate zones during spring and fall. 3.15.2006 12:47pm (link) Anoymous Koward: There are probably serveral points on the equator where that's true. So what is her point? 3.15.2006 12:49pm (link) Anoymous Koward: Except that when it's daytime on one side, it's night on the other. So maybe not. 3.15.2006 12:52pm (link) Kevan Choset (mail): Very Anonymous Coward -- that all sounds fine, but would you care to explicate? Freder Frederson -- no matter how precisely you measure, it will always still be true. Anonymous Koward -- maybe it's true on the equator and maybe it's not. But unfortunately that doesn't a proof make... 3.15.2006 12:59pm (link) Kevan Choset (mail): (And for those of you turned off by the math, I promise you don't need anything complicated. As for the Intermediate Value Theorem, you don't need anything more sophisticated than the fact that if it's 0 degrees in NY and 10 degrees in Boston then it must be 5 degrees somewhere between them. (And 4 degrees somewhere between them, and 9.99 degrees somewhere between them, and pi degrees somewhere between them, and 0.[TheNumbersFromLost] degrees somewhere between them, etc., etc.) 3.15.2006 1:01pm (link) HeScreams (mail): I haven't wrapped my head around the proof yet, but I'm certian that this spot has never been in my kitchen. Koward: on the equator, it's always dusk somewhere, and it's dawn on the other side. At those spots it's at least possible that they have the same temp; but the problem is actually mathematical. Freder: actually, you have it backwards. A lack of precision would allow the statement to be false; with sufficient precision the statement is always true (assuming it's ever true). 3.15.2006 1:03pm (link) Robert Cote (mail) (www): Temperature is not a discrete number. All you need is two antipodes with daily highs and lows where one place has a higher high and lower low and sometime during that day their temeperatures passed each other. Duplicate except hold temperature is held constant and location is varied. Is this some kind of test? What did I win? 3.15.2006 1:12pm (link) A. Zarkov (mail): Very Anonymous has got it. The key assumption is continuity of the temperature field. But we need to prove it and my topology is rusty. 3.15.2006 1:20pm (link) JLR (mail) (www): This is technically not a "proof," but there aren't too many places where land-to-land antipodes are even possible. For example, I believe that anywhere in the lower 48 states, if you "drilled a hole through to the other side of the earth," you'd hit the Indian Ocean. About four-fifths or so of Earth's land is antipodal to water. However, I believe parts of Argentina and Chile are antipodal to China. Also, I believe parts of Norway are antipodal to Antarctica. Furthermore, if memory serves me, parts of Spain are antipodal to parts of New Zealand as well. 3.15.2006 1:22pm (link) Aaron Bergman (mail): I'm not sure what the Brouwer fixed point theorem has to do with this. Just pick a diameter of globe. Pick some plane through the center of the earth to rotate it around. Let f(theta) be the difference in the tempreatures between the two antipodes. Then it's clear that f(theta + pi) = -f(theta). So, the intermediate value theorem tells us that there's a value of theta between zero and pi such that f(theta) = 0. QED. 3.15.2006 1:26pm (link) Dan Levine (mail): Kevin, it has to be true around the Equator. Just as it has to be true that there is a place on the Arctic Circle where directly across the Arctic Circle (not going through the center of the Earth) the temperature is the same. 3.15.2006 1:27pm (link) James Grimmelmann (mail) (www): Consider just the equator. For each point on the equator, consider the difference in temperature between that point and the point directly opposite it (more precisely, the temperature at that point minus the temperature directly opposite). If this difference-function is zero for every point on the equator, then we're done -- any two opposite points have the same temperature. If the difference-function is nonzero at some point (call it A), then the difference-function is the negative of that nonzero value at the exact antipode of A. Consider the segment of the equator from A to A's antipode. The difference-function goes from some positive value to the negative of that value as you go along the segment. At some point along the segment, the difference-function must be zero. That point and the point directly opposite it have the same temperature. 3.15.2006 1:31pm (link) Dan Levine (mail): I like James's proof. Here's a graphic way to think about it. Let the x-axis refer to degrees around a circle. Let the y-axis refer to temperature. Draw any continuous graph that crosses the x-axis at (0,0) and (360,0)--the idea being that when you make it all the way around the circle and wind up at the same point the temperature is the same. Now draw the same-shaped graph shifted over 180 degrees to the right---i.e., the same shape, but crossing the x-axis at (180,0) and (540,0). At any point where the two graphs cross, there will be a point on the first graph, 180 "degrees" to the left, that has the same y-value (i.e., the same temperature). 3.15.2006 1:48pm (link) SLS 1L: This is slightly less obvious to me, because I don't remember the relevant bits of calculus very well, so I'm going to have a go at it by intuition. If it's true in the following one-dimensional case, then it's true on the earth as well: Consider two points A and B at opposite ends of the earth and any "equator" defined by them. Let t(theta) be the temperature at the point theta radians from A along that circle (so t(0) = t(2pi) = the temperature at A and t(pi) = the temperature at B). The claim in this one-dimensional case is that there is a point theta0 such that t(theta0) = t(theta0 +- pi). If t is constant, then the claim is true and we are done. If t is not constant, then it has at least one maximum or minimum. Let theta0 denote any such maximum or minimum; since t(0) = t(2pi), t(theta) must take on every value between t(0) and t(theta0) at least twice. Here's the non-rigorous graphical intuition element of the proof. Graph f(theta) on a standard cartesian graph (with theta on the "x-axis" and f(theta) on the "y-axis") from 0 to 2pi. Imagine a line segment of length pi parallel to the theta-axis "moving" along the function's curve starting at theta = 0 and and ending at theta = pi (i.e. the line segment runs from (theta, f(theta)) to (theta + pi, f(theta))). By graphical intuition, you can see that at some point theta' in this process both ends of the line segment must intersect the curve of f. Since the line segment is pi long, this means that f(theta') = f(theta' + pi). Since we have proved that the theorem holds along every equator of the earth, it is true on the earth itself. Q.E.D. 3.15.2006 1:53pm (link) FXKLM: I think CJ is wrong about the egg thing though. And even if s/he is right, is it in any way related to the antipode? 3.15.2006 1:54pm (link) SLS 1L: I see several people have posted the same intuitive idea in a less complicated way while I was writing mine. I like Dan's best. 3.15.2006 1:55pm (link) Gavin Peters (mail): Ok, so any great circle has two antipodal points at the same temperature on it. That's boring first year analysis stuff. Here's the real question: demonstrate that on the earth there are two antipodal points with the same temperature _and_ pressure! 3.15.2006 2:10pm (link) LTEC (mail) (www): Coward -- I think you are making this too difficult. As the above arguments show, this works on any circle, that is, in two dimensions. Your argument about (essentially) not being able to comb a hairy coconut only holds in three dimensions, since we can comb a hairy circle. 3.15.2006 2:14pm (link) Philip S: Consider just the equator. Let the temperature fcn. at time t be defined as f ranging from 0 to 2pi. If f is constant the result is trivial. If f is not constant define f over the whole real line periodically in intervals of 2pi. Now define g(x) = f(x) - f(x+pi). f is continuous by assumption, and since g is the difference of two continuous fcns. it too is continuous. Since f is continuous but not constant, therefore g must have both pos. and neg. values. Since g is also continuous, therefore there exists x s.t. g(x) = 0. QED 3.15.2006 2:16pm (link) Philip S: whoops. Instead of "Since f is continuous but not constant" I should have written "Since f is continuous, periodic, and not constant." I think that should do it. 3.15.2006 2:21pm (link) Siona Sthrunch (mail): In fact, there are antipodal points on the earth that have the same temperature and barometric pressure. Let T be a temperature distribution on the earth and let U(T) be the set of points whose temperature is the same as their antipodal points. We know from the comments above that U(T) intersects each great circle on the earth. Of course U(T) is symmetrical. In general, what properties are sufficient for a set S on the earth's surface to equal U(T) for some T? 3.15.2006 2:23pm (link) rbj: Perhaps the Arctic after 6 months of winter and the Antarctic after 6 months of summer? Or a spot on the equator during sunrise and sunset? I have absolutely no way of proving these. 3.15.2006 2:25pm (link) Jim T: I don't think that you are reading the quote correctly. Her first sentence is: "There's a spot on the earth where the temperature is exactly the same as it would be if you drilled through the earth to the other side." You could obviously say that she (or the writers) mis-spoke, but what she is claiming has absolutely nothing to do with two spots on either side of the Earth. She is saying there is some point on the Earth's surface, let's call it 'X', at which the temperature would be completely uneffected if a hole were to be drilled through the Earth all the way to the other side. It seems to me that such a hypothetical hole would have to have SOME effect, though I suppose that would depend on it's radius. I also don't see how such a thing could be proven or disproven. 3.15.2006 2:45pm (link) Kevan Choset (mail): Dan, Yes, it's definitely true on the Equator. (It's definitely true on any circle, in fact.) But I believe the earlier suggestion about the Equator was trying to rely on some sort of special temperature-related properties that are true on the Equator, which I was trying to discourage. 3.15.2006 2:49pm (link) Kevan Choset (mail): James's solution was the one I had in mind. You can actually view the whole earth instead of just the equator, but use his trick of looking at the difference between the temperature of every point on earth and the point completely opposite it. This is, of course, still a continuous function. It can't be positive everywhere (because if it's positive in NY then it's negative in the point directly opposite NY), and similarly can't be negative everywhere. So it's positive in some places and negative in others. This means that it must be 0 somewhere. Wherever it's 0, the temperature is the same at two opposite points. 3.15.2006 2:54pm (link) Aaron Bergman (mail): There might be an easier way to do the pressure/temperature thing, but I came up with this (which is essentially an obstruction theory argument). For p a point on the sphere and -p its antipode, define the function f(p) = (T(p) - T(-p), P(p) - P(-p)) This is a map from S^2 into R^2. Assume that it never hits the origin. Then we have a map into R^2 \ {0}. We also have that if (x,y) is in the image of f, then so is (-x,-y). Since the image of f is also connected, it contains a (non-contractible) circle that goes around the origin. Take the inverse image of this circle. This gives a region on S^2 that contains a circle which maps to the circle around the origin of R^2. However, any circle on the sphere can be "filled in" (ie, the sphere is simply connected). Apply f to this disc. It must map to a contractible region bounded by the unit circle in R^2 \ {0}. No such region exists, so we have a contradiction. Thus the image of f must include the origin. 3.15.2006 3:03pm (link) Tennessean (mail): Is there any sense to the assumption that temperature is continuous? Maybe there is; certainly, it makes sense to assume that there is no place where the temperature is 0 immediately next to a place where the temperature is 100. But if you assume, as I think we are, that the notion of place is not necessarily continuous (i.e., temperature is a measure of a discrete place or field), then why would temperature be continuous? 3.15.2006 3:07pm (link) Juan Notwithstanding the Volokh: You can assume that temperature is continuous (i.e., there's no spot on the earth where the temperature just "jumps" from 0 degrees at one spot to 100 degrees at a spot infinitesimally close to it). But this is not a safe assumption. Take, for instance, a lava flow or an iceberg. It is possible that the temperature may well jump or drop by 100 degrees from one point to the next. More mundanely, it depends on how you measure "on" the earth. A cliff may have a particular temperature at it's edge, while the next "spot on the earth", which is "infinitesimally close" for the purposes of determining an antipode could be the ocean below, with a significant rise or drop in temperature due solely to the change in altitude. Maybe you live on a perfect sphere, but my planet is more interesting. 3.15.2006 3:20pm (link) John (mail): There are an infinite number of such points. Connect any two points that are at the two ends of a diameter of the earth. Assume they are of different temperatures. Now rotate the diameter line until the ends are reversed. As each point was traversing the temperature range from what it started at to what the other end started at, the two temperatures must, at some point be the same. 3.15.2006 3:27pm (link) exfizz (mail): Let C be a great circle (e.g. the equator) on the sphere S. Let L=[0,360] be a parameter that functions as a generalized longitude on C. Let T(L) be the temperature (or any continuous scalar field) on S. Let d(L) = T(L)-T(180-L) be the difference between temperatures at L and its antipode L'. Since d(L)=-d(L'), the intermediate value theorem says there must exist at least one Lo for which d(Lo)=0, i.e. for which T(Lo)=T(Lo'). The choice of equator was arbitrary; the proof holds for any great circle, or indeed any circle or any closed loop on the surface of S. It's just that the tidy notion of "antipodes" becomes less tidy in that case. (For me, the fixed point theorem comes into play when I visualize the problem: a closed loop with a rising and falling line above it. Rotate the loop 180o, i.e. map it onto itself. There must be at least one spot where the two lines cross.) 3.15.2006 3:51pm (link) lpdbw: I have personally balanced an egg on end during an equinox. Being a scientist, I repeated the experiment every day for 10 days. It's a matter of eggs and steady hands, and the equinox is irrelevant. I don't watch that TV show, and the comment by "Will" makes no sense to me. There must be some context missing... 3.15.2006 3:52pm (link) exfizz: Juan Notwithstanding the Volokh: [it] is not a safe assumption [that temperature is continuous]. Take, for instance, a lava flow or an iceberg. It is possible that the temperature may well jump or drop by 100 degrees from one point to the next. Below the distance scale of the mean free path (~0.1 μm in room air), it is true that the continuity of the T field gets a little dicey, but (1) I think you can still define it using the KE of an individual molecule, and (2) outside of a physics lab you will not find a temperature measuring device capable of resolving at a distance scale and a time scale where this matters. Besides, even a lava/air system or ice/air system has a transition layer where temperatures match up. Maybe you live on a perfect sphere, but my planet is more interesting. I realize you're joking, but (I think) the result holds for any closed loop on any surface equivalent to a sphere, including our beloved lumpy home. And far from reducing reality to a bland abstraction, math and science IMHO reveal aspects of its richness in a way that other ways of seeing do not. 3.15.2006 4:18pm (link) eng: The trouble is the phrase "through the Earth" which can quite reasonably be taken as through the center of the Earth. If you make such a restriction that there must be a line passing through the center of the earth, then it is false that there must be a pair of points on the perimeter intersecting such a line that are of equal temperature. It is only once the statement is relaxed to mean "any straight tunnel" that we can be assured of finding a solution. 3.15.2006 4:19pm (link) SLS 1L: eng:It is only once the statement is relaxed to mean "any straight tunnel" that we can be assured of finding a solution.I don't quite follow. If you consider e.g. my "proof" above, I think it's clear that any great circle has two points of equal temperatuer separated by pi radians - i.e. on opposite sides of the earth. I am hardly 100% confident of my reasoning or anyone else's here, but can you explain? 3.15.2006 4:28pm (link) james tierney (mail): Eng-- CJ's quote doesn't require that it hold true for every (any?) point. Rather, she says that there is "a spot on the earth" for which this will be true, given that the line in question is through the center of the earth. If the spot is on the equator, runs through the center of the earth, and hits the equator on the other side, there is probably a solution (as other posters have pointed out). 3.15.2006 4:31pm (link) Siona Sthrunch (mail): For p a point on the sphere and -p its antipode, define the function Bergmann - The first error in your proof is that it is not the case that a symmetric connected set in R^2 not containing the origin necessarily contains a circle. Bergmann wrote: Since the image of f is also connected, it contains a (non-contractible) circle that goes around the origin. 3.15.2006 4:32pm (link) DK: The temperature _is_ continuous, although this gets into physics, not math, and you have to measure very precisely. In any medium allowing heat diffusion, diffusion instantly smooths any transient heat discontinuity into a continuous but initially steep shape. Thus, if you have a magic wand that can create a temporary temperature discontinuity, the discontinuity will vanish right way. Moving objects and the differences between materials complicate things, but to simplify, just replace "the earth" with "the atmosphere X meters above all buildings, ground, moving objects, and near-ground heat sources". The same argument applies to that layer of atmosphere. 3.15.2006 4:39pm (link) Jam (mail): http://en.wikipedia.org/wiki/Antipodes In geography, the antipodes (from Greek anti- "opposed" and pous "foot") of any place on Earth is its antipodal point; that is, the region on the Earth's surface which is diametrically opposite to it. Two points which are antipodal to one another are connected by a straight line through the centre of the Earth. At any given moment half of the Earth surface is absorbing energy (daytime) and the other half is releasing energy (nightime). The temperature will be the same at the point of equilibrium; where a spot has gained energy and it's antipode has released energy. Antipodal equilibrium will happen: 1) At noon when the rate of energy absorption and release are the same. 2) Towards sunset if the rate of energy absorption is faster than the rate of energy release. 3) Towards sunrise if the rate of energy absorption is slower than the rate of energy release. 3.15.2006 4:49pm (link) Tennessean (mail): As an uneducated remembrance, it was my impression that temperature and pressure were closely related, and that pressure was a sampled statistic of a point _oer time_ used to "represent" an area. Also, although there is diffusion, it is not clear that this is an instant process. In light of these three assertions, is temperature necessarily continuous? (Not that this in any impugns the solutions offered to the actual problem posed - just an aside that seems interesting from here.) 3.15.2006 4:59pm (link) Juan Notwithstanding the Volokh: exfizz: But how would you account for cliffs and the like where an altitude difference might result in a significant temperature difference between two points that are adjacent for purposes of determining their antipode? JNOV 3.15.2006 5:34pm (link) Philistine (mail): Maybe I'm just confused, but isn't it pretty trivial to find counterexamples to the contention that there would mathematically always be an antipode with equal temperatures? Take the equator example. Picture the equator as a clockface--from 12 to 2 on the clockface (i.e. for 60 degrees of a circle)it is 100 degrees fahrenheit. At 2, it changes to 105. At 6 (180 degrees) it changes to 115 and at 7 is 110, and remains there until 11 when it goes back down to 100, returning to 100 at 12. It makes more sense is you draw it on a piece of scrap paper. :) There do not seem to be two antipodal points with the same temperature. 3.15.2006 6:09pm (link) SenatorX (mail): I think John is right and there are an infinite number of points. Basically that if the condition to refute is " there are two points on the earth exactly opposite each other" it infers that there are two points that share a common midpoint. This midpoint defines two equal lines. Now the condition "on the Earth" is ambiguous. But it clearly infers that any place "on the Earth" is fair game to satisfy the condition. Since the Earth is "sphere like" and further no condition AGAINST rotating the lines around the common midpoint exist... We define the possible range of probabilities as the area of a perfect sphere. Further though is "on the Earth" which contrast to "under the earth" or "above the earth" infers an area as infinitely as close to being "on the Earth" as possible. Let us decide as a tightest possible condition this area. People think of the equator because the greatest chance of similar variables would occur to produce the same temperatures. Temperature is the final condition and so this is a valid source of points but not limited to as the asymmetrical nature of "the Earth" is a factor. We have now defined the area that the matching temperature points could exist by refuting the any possible area where they could not exist. We do this at an infinite level because the statement to prove allows this. The places where the points can occur have extreme conditions BUT they are infinite because the movement of two points around a common point allows for infinite possibilities of success when MEASURING is the judge of success. Measuring can be as accurate as necessary, infinitely. So measuring temperature can in finely succeed at refinement(or grossness). Basically because of "on the Earth", "two points exactly opposite", and a measurement(temperature) the condition to prove contains many infinites and so an infinite number of successful conditions can occur if only one condition of success can occur. 3.15.2006 6:21pm (link) Jadagul (mail) (www): Juan notwithstanding the Volokh: since no cliff will be perfectly verical, there will always be some slope and thus a smooth gradient. I guess you have to assume that the earth's surface is relatively smooth (sorry, I'm sure there's a technical term for this and I'm sure I'll know it by the end of the semester when I finish my topology course). Philistine: we assumed that the temperature was distributed continuously. You have four points of discontinuity there, which is why it doesn't work. But physically, you can't have a discontinuous temperature field for a non-infinitesimal duration. 3.15.2006 6:29pm (link) Kevan Choset (mail): Philistine- Nope, it's not pretty trivial to find a counterexample, since there are no counterexamples! In the case you describe, the antipodes will be between 2 and 6 and between 7 and 11. Between 2 and 6 the temperature changes from 100 to 115 and between 7 and 11 the temperature changes from 100 to 110. Somewhere in there, the points opposite will have the same temperature as each other. 3.15.2006 6:34pm (link) Philistine (mail): Kevan Choset In the case you describe, the antipodes will be between 2 and 6 and between 7 and 11. Between 2 and 6 the temperature changes from 100 to 115 and between 7 and 11 the temperature changes from 100 to 110. Somewhere in there, the points opposite will have the same temperature as each other. I still don't see it. In my hypothetical, there could be equivalent antipodal temperatures--but only coincidentally. The vast majority of situations would not have them. For instance--keep my hypothetical, but refine it. All temperature changes are not "instantaneous" but occur in the 1 minute (6 degrees of the circle) before they are given. (Thus from 100 to 105 occurs just before 2, 105 to 115 just before 6, 115 to 110 just before 7 and 110 to 105 just before 11. If this is the case, I don't see antipodal equivalents. Am I missing something? --Philistine 3.15.2006 7:05pm (link) Kevan Choset (mail): Philistine-- In your refined example, it is 105 degrees at both 11:00 and 5:00, which are antipodal. You can change your setup to avoid this problem, but it will create another problem. The beauty of a mathematical proof (like James's, above) is that we don't have to worry about trying lots of different examples, since his logic holds for all possible examples. 3.15.2006 7:28pm (link) Dr. T (mail): I do not believe that any of the given math proofs work. They make too many assumptions about the physics of temperature distributions that are not true. They assume that because one spot is temperature A and its antipode is temperature B, the difference A-B can be brought to zero by rotating around a plane that passes through the center of the earth. As others have noted, this assumes no temperature discontinuities. The only way that we can guarantee no temperature discontinuities is if we assume that the spots at which temperatures are measured are infinitely small, a physical impossibility (because temperature requires moving molecules, and molecules have size). Another false assumption is that the A-B difference as we rotate around the bisecting plane is linear. Why should it be? If the gradient is non-linear, then there may never be a place where the antipodal temperature difference is zero. 3.15.2006 7:58pm (link) SLS 1L: Dr. T - if I understand the physics correctly, temperature becomes undefined, not discontinuous, when you try to consider it on a small enough scale. But continuity is a reasonable physical approximation. As for your claim about the A-B difference, t(theta) (the temperature at angle theta along a given great circle as we rotate around the circle) is a continuous function. So is t(antipode(theta)). Since both are continuous, t(theta) - t(antipode(theta)) is also continuous. If t is nonconstant, then the difference is positive at some point and negative at some other point, and by the intermediate value theorem it must take on the value zero at some point. No assumptions about linearity are required. 3.15.2006 8:13pm (link) Tennessean (mail): SLS - Concedering arguendo that temperature becomes undefined when you try to consider it on a small enough scale (i.e., for a point as opposed to for a volume), why would you assume that "continuity is a reasonable physical approximation"? (Again, this is not to detract from the elegance of the proof adduced above by many.) 3.15.2006 8:21pm (link) The Real Bill (mail): I think uniformity is required as well, not just continuity. 3.15.2006 8:25pm (link) The Real Bill (mail): Nope, uniformity is not required. 3.15.2006 8:49pm (link) SLS 1L: Tennessean - few, if any, physical quantities that we treat as continuous without running into problems are actually continuous. Charge is quantized, but that doesn't mean "charge is continuous" is a bad physical approximation. Basically the same point holds for temperature. 3.15.2006 9:02pm (link) Tennessean (mail): SLS: The normative judgment "bad physical approximation" is appropriate or inappropriate depending on the context, obviously, and, for many purposes, the assumption of continuity renders feasible many calculations and understandings with few errors or other problems. Here, however, the proofs all depend on the fact of continuity. So, it is a reasonable to question to ask. Just saying that it is a good assumption does not make it so (especially where we've already discussed that temperature is a statistic, not an attribute, as far as the temperature of a particular point at a particular time). 3.15.2006 9:21pm (link) Dick King: Some cliffs are overhung [so when you climb them you are holding yourself on with your hands]. In that case, there is a particular latitude and longitude that has three or more points. -dk 3.15.2006 9:56pm (link) Aaron Bergman (mail): Siona: By connected, I meant path connected. Sorry. Let U be a path connected subset of R^2 \ {0} s.t. if p \in U, then -p \in U. Pick a point p. Then, by assumption, there exists a path q(t) in U s.t. q(0) = p and q(1) = -p. Define the following path r(x) = q(2x) for x \in [0,1/2] r(x) = -q(2x - 1) for x in [1/2,1] By assumption again, r(x) is in U. It's easy to see that r is continuous and that r(0) = r(1), so it is a circle. It can be seen to be noncontractible in R^2 \ {0} by looking at polar coordinates, for example. 3.15.2006 9:57pm (link) Siona Sthrunch (mail): Bergman writes: It's easy to see that r is continuous and that r(0) = r(1), so it is a circle It is not true that r must be a circle. For example, r could be an ellipse, correct? Or any closed antipodally symmetric loop. 3.15.2006 10:06pm (link) SenatorX (mail): From the "problem" :"You can assume that temperature is continuous (i.e., there's no spot on the earth where the temperature just "jumps" from 0 degrees at one spot to 100 degrees at a spot infinitesimally close to it)." So let’s leave out of it the "real" world for the sake of the conditions in the problem to solve. I don't have a problem with the mathematical proof given so far for opposite points from an equal midpoint. Dr.T adds some excellent points I think. The thing is (and I wish someone could put this in mathematical terminology for me) that proving that is not enough. The problem calls for equal temperatures as well as antipodes. The problem calls for a cross correlation of "same area" between the math proof given for the equidistant points AND the ranges where the temperatures are the same at any given instant. It is worth noting that it doesn’t say the antipodes(as far as space-time are concerned)must remain in any sort of continuity. They could exist in any of possible area that the conditions are able to be satisfied in at any given instant. I think the key to the problem is given in the final statement about the temperature being continuous. It then even defines continuous for us by saying you cannot find a place to measure temperature, "at no spot on the earth", where a "spot infinitesimally close to it" has a non-continuous jump of value. What does this mean if not that two points(or the range of area defined by the points in the surface of a sphere intersecting with the surface of the earth) are connected by a continuum of temperature? That you can not MEASURE any two spots "infinitesimally close", or more to the point, any two points NOT that close(range of min/max variable in earth topology) without finding a gradient of values between the two points. It doesn't just jump from 0 to 100, 25 to 75, or anything to any other thing without being able to MEASURE a gradient between the two points in space. I am sure there is a mathematical way to formulate this. To me it means since there are an infinite number of gradients between two points equidistant from each other then measuring for value in those gradients(temperature) can give infinite results. Somewhere in infinity will be a matching condition at any given instant? I also think you would never be able to prove this type of "problem" by observation of nature. It's mostly mental masturbation. 3.15.2006 10:07pm (link) Aaron Bergman (mail): Sorry. I was speaking loosely. Read "loop" where ever you see "circle". 3.15.2006 10:37pm (link) Lev: Antipodes, islands, New Zealand - rocky uninhabited islands, 24 sq mi (62 sq km), South Pacific, c.550 mi (885 km) SE of New Zealand, to which they belong. Explored by British seamen in 1800, the Antipodes are so named because they are diametrically opposite Greenwich, England. Prove to me that Greenwich and The Antipodes never have the same temperature. 3.15.2006 11:19pm (link) exfizz: Lev: "Prove to me that Greenwich and The Antipodes never have the same temperature." That's a fun example, Lev, and contra SenatorX, easy to investigate. Considering that air temperature is a roughly periodic function (strong components at 1d-1 and 1y-1) varying widely about the mean, and that antipodal points share the same solar warming (just 180o out of phase), I'd wager that TG(t)=TA(t) many times each year. (Of course, applied to the proxy data below the IVT guarantees at least twice/year.) It won't happen tomorrow. Using proxies for the Antipodes, tomorrow Auckland will range from 52-71 and Wellington from 54-68, vs. London only from 37-42; thus the ranges are disjoint. But April-June and again September-October the monthly average temperature ranges overlap significantly and we might expect to see TG(t)=TA(t) more often. A simple "bot" could scrape the weather pages repeatedly and compare temps during those months. Good project for a smart HS kid. Appendix: Monthly avg temp for Wellington, NZ / London, UK Jan 58-67 / 33-44 F Feb 57-67 / 33-45 Mar 56-65 / 35-50 Apr 53-61 / 38-55 May 49-56 / 43-61 Jun 46-53 / 49-67 Jul 45-51 / 52-71 Aug 45-52 / 52-71 Sep 47-55 / 48-66 Oct 49-57 / 44-59 Nov 51-60 / 37-50 Dec 55-64 / 34-46 3.16.2006 2:47am (link) Daryl Herbert (www): I agree with Kevan that James' explanation was best, and I think Kevan's illustration is best. "If the difference-function is nonzero at some point (call it A), then the difference-function is the negative of that nonzero value at the exact antipode of A." I missed that insight, which makes his solution much more elegant than mine. I proved it by showing that: 1 if your difference function ever returns 0, then the points match 2 if it's always negative, that means temp always decreases as you go around the circle, but that's no good because if you go around and around multiple times it keeps getting colder--but at 0* it must be the same temperature as at x* when x approaches 360. 3 if it's always positive, that means that means the temp always increases as you go around, with the same problem. 4 if it's both positive and negative at times, then it must pass through the origin (be equal to zero at some place), because it is continuous. (this is where the Mean Value Theorum actually comes into play) But my proof is better, because it can be expanded to show to show that for ANY continuous cyclical path with continuous temperatures, for ANY distance between the two points (not necessarily "halfway" but 10%, or 10 miles, etc.) you will be able to find some two points that satisfy the distance requirement AND are at equal temperature. So there :-) 3.16.2006 4:21am (link) BobGo: So much nonsense is inspired by asking for a mathematical proof. Only Philistine seemed to appreciate that day/night and winter/summer real-world variations make it improbable, though not impossible, for temperatures at antipodes to be equal. On the equator, early morning and early evening could have equal temperatures, but only if heating and cooling were exactly symmetrical and local weather didn't interfere. In the tropical and temperate zones it would be possible that a winter daytime temperature would equal a summer nighttime temperature, but given the scarcity of antipodal land areas, variations of elevation, and local weather conditions, a rare coincidence. Let's try the polar regions: thick icecaps provide good insulation, probably even from extreme seasonal variations, though no guarantee against geothermal variations. Jim T's interpretation of CJ's original assertion is the cleanest answer: the temperature at a spot on the Earth would be unaffected by drilling (figuratively) through to the other side. 3.16.2006 6:33am (link) AppSocRes (mail): By definition the "antipode" of a point on a sphere is the point defined by the intersection of a line through the original point and the center of the sphere and the surface of the sphere and that is not the original point. All arguments using the Intermediate Value Theorem, Mean Value Theorem, Rolle's Theorem or any variation just show that at least two points on a circumference of a sphere have the same value for any continuous function mapping the surface of the sphere to the real number line. None of these proofs demonstrates that these two points are antipodes except in very special cases. The property everyone is attempting to prove is a global property of the sphere and dependent on the sphere's topology. What everyone's trying to prove must be a corollary of the Brouwer Fixed Point Theorem: probably a proof by contradiction involving a constructed pull-back function. I don't have time to work out the details. By the way, there is an elementary introduction to category theory -- a mathematically inclined high school student could handle it -- that contains a proof of the fixed point theorem using only category theory. Very neat stuff. I didn't see a proof of Brouwer's theorem till my junior year at university. 3.16.2006 8:25am (link) Aaron Bergman (mail): None of these proofs demonstrates that these two points are antipodes except in very special cases. I'm too tired to check if all of the proofs demonstrate this, but the majority of them definitely show that the antipodal temperatures are equal. 3.16.2006 10:59am (link) AppSocRes (mail): I was wrong. Aaron Bergman has convinced me the Mean Value proofs work. 3.16.2006 2:03pm (link) Robert Ayers: Aaron Bergman et all have the fine proof that involves picking a pair of points A and A*, seeing that they do not satisfy T(A) == T(A*), then rotating the diameter until A goes to A* and A* goes to A. It is amusing that the same proof applies to this puzzle: Given a true chair and a warped floor (continuous and with some requirements on max error) show that you can positon the chair so it won't wobble. Bob Ayers 3.16.2006 2:33pm (link) Jam (mail): The Earth is tilted, so that the poles are in complete darkness or complete light at certain times of the year. The only times that the poles may have the same temperatures is when the Earth's tilt is tangental to the Sun -- this happens twice a year. 3.16.2006 9:24pm (link) Siona Sthrunch (mail): Bergmann wrote: Sorry. I was speaking loosely. Read "loop" where ever you see "circle". The current version of your proof (of Borsuk-Ulam) is still either malformed, substantively incorrect, or both. It's malformed because you wrote in an earlier draft "unit circle" and there is no "unit loop." Frankly, the logic of your proof is still not written clearly, although it does use many standard mathematical terms. It's substantively incorrect, to the extent that I can analyze its nonstandard terminology, because there is a continuous map from S^2 onto the unit circle in R^2, which sends an equator onto the unit cirle, and the unit circle is of course antipodally symmetric. I believe the point of your proof may have been that there is no such map, since your construction of the map f:S^2->R^2 from the vector field on S^2 has only the named properties, and yet is in fact existent, contradicting your analysis. 3.16.2006 11:13pm (link) Siona Sthrunch (mail): Also, Bergmann, I am only picking on you because you seem to be one of only about 5% of the posters on this thread who do not clearly have no idea whatsoever what mathematics or a mathematical proof even is; the remaining 95% of the posts cannot even be discussed. 3.16.2006 11:17pm (link) Siona Sthrunch (mail): AppSocRes wrote: By the way, there is an elementary introduction to category theory -- a mathematically inclined high school student could handle it -- that contains a proof of the fixed point theorem using only category theory. I would be curious to see this since the fixed point theorem depends on the topology of R^n. Brouwer does usually come from the homology or homotopy chain functors (assigning sequences of groups to embeddings of topological spaces); I don't see how this can be replicated at a high-school level using category theory. 3.16.2006 11:21pm (link) Oren (mail): I had no idea so many other math people read the VC! /feels a little better now 3.17.2006 4:13am (link) BobGo: Exfizz has nicely shown, with his temperature range comparisons for NZ and UK, that there are overlaps in late spring and early fall, with the likelihood of equal temperatures twice daily during those seasons. Not very likely during the rest of the year, however. And even during the overlap seasons, not at just any moment; you have to wait for the varying times when the temperature lines cross. There are a few other antipodal places on Earth where such seasonal temperature range overlaps could occur, but antipodal land areas are quite sparse. I am assuming we are limited to land areas because “antipodes” means “feet growing out of heads” (as necessary for walking down under), and feet are for walking on land, aren’t they? So you’d have to look pretty hard for antipodal equal temperatures during much of the year, except maybe in the polar regions. Is the proposed problem intended to apply to any pairs of opposite points on the globe, whether land, water, mountains, or icecaps? It would take a lot more investigation to look for daily temperature range overlaps between water/water opposite points and water/land opposites points. Opposite-point equal-temperature moments might be quite common in some seasons, But do they occur somewhere “at any given moment”? I can’t say for sure, and there is little in the comments that is even remotely relevant. It’s amazing to me how little the math discussion in these comments has to do with understanding of seasons and weather conditions on a real Earth. 3.17.2006 5:56am (link) Very Anonymous Coward: Siona Sthrunch: Perhaps it's just for the 2-dimensional case? There are a number of cute ways of proving it that don't require very much machinery at all: Sperner's lemma, the game of Hex, some easy covering space arguments to get the fundamental group of the circle, etc. 3.17.2006 6:51am (link) Very Anonymous Coward: AppSocRes: Yes, that was my first impression as well. But I agree now; the IVT is all you really need. 3.17.2006 6:53am (link) Some Guy: BobGo-- I believe it is correct that there are not only at least two such points, but there are an infintite number of points (in fact, I think that these points must form a continuous path, that crosses the equator at least twice--and so it also crosses any circumference--but I could be wrong about that. In fact, there may be more than one such path, but there is at least one.) The trick to thinking about it is, the points on this path need not all be the same temperature. We're only talking about equivalent temperatures--but any temperature can have an equivalent. We're also assuming that a temperature scale is smooth, so you can slice it up into infinitely fine degrees. This also means that it's not necessarily true for every single temperature on the scale--in other words, it might only be true at any given time for the range between 50F and 51F, for example. It's mathematically true, but in this case the mathematics can reflect the real world. 3.17.2006 2:31pm (link) PT (mail): What's "IVT" ? 3.17.2006 2:57pm (link) John Bouvier (mail): Since this does require continuity of both the temperature and the surface of the earth, wouldn't a discontinuity allow for the conjecture to fail? Continuity of temperature requires a careful definition. It's true that at the sub-molecular level temperature loses it meaning. So we can reasonbly say that the temperature of a point in space is defined as the average for a sphere centered on that point. (Pick any size sphere, as long as it contains enough vibrating molecules to make the concept of temperature meaningful.) But ... In the real world there are truly discontinuities in the surface. The obvious example is an OVERHANG, noted by Dick above. If your great circle passes over an overhang, you truly get a discontinuity in elevation. And therefore you truly get a discontinuity in temperature. Say the overhang is 10 feet high and in the sun. Say you step over the edge to the bottom, which is in the shade. The temperature change may be continuous over this 10-foot drop, but that 10-foot path is not part of the great circle. So does this allow the conjecture to possibly fail?? 3.17.2006 6:00pm (link) Oren (mail): PT - http://en.wikipedia.org/wiki/IVT (jeez, minimum of effort). Alright, here's another good one for you guys to flame over. Prove (minimally!) that given the existence of some wind somewhere, there must be a cyclone somewhere on earth. You may assume that the earth is a perfect sphere with a 2D (flat) atmosphere (ie: wind has no radial component). 3.18.2006 11:55am pageok pageok | {"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": 0.8256565928459167, "perplexity": 816.8753511895064}, "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-2015-14/segments/1427131299054.80/warc/CC-MAIN-20150323172139-00118-ip-10-168-14-71.ec2.internal.warc.gz"} |
https://mathoverflow.net/questions/384909/showing-the-positivity-of-the-determinant-of-mathfrakspn-without-making-u | # Showing the positivity of the determinant of $\mathfrak{sp}(n)$ without making use of diagonalization
Let $$\mathfrak{sp}(n)$$ be the lie algebra of compact symplectic group $$\mathrm{SP}(n)$$, regarded as a compact form of $$\mathfrak{sp}(2n,\mathbb{C})$$, so we can talk about its (complex) determinant.
Let $$M\in \mathfrak{sp}(n)$$, then $$M$$ has purely imaginary eigenvalues $$(ix_1,ix_2,\dots,ix_n,-ix_1,-ix_2,\dots,-ix_n)$$, so $$\det(M)=(x_1x_2\cdots x_n)^2\geq 0.$$
My question is
Is there a coordinate independent way to show that every element of $$\mathfrak{sp}(n)$$ has nonnegative determinant?
I would want an argument without using eigenvalues, nor anything that cannot be expressed as a function of the matrix entries.
1. I will like to know if there is some geometric arguments.
I also want to understand the algebra behind. For $$n=2$$, I have tried expanding $$\det(M)$$, but I cannot find a way to express it as a sum of non-negative terms.
A useful way to show the positivity of an algebraic expression is to write it as a sum of terms, and each term is either a norm square, or can be shown to be non-negative by a direct application of the Cauchy-Schwartz inequality. For example, we know $$\mathrm{tr}(A^4)\geq 0$$ because $$\mathrm{tr}(A^4)=||A^2||^2$$.
Of course we have $$\det(M)=(x_1x_2\cdots x_n)^2$$, but the problem $$(x_1x_2\cdots x_n)$$ is not expressible by $$M$$.
1. Can $$\det(M)$$ be expressed a sum of such non-negative terms? If yes, what are they? If not, what are the extra ingredients we need to show the positivity apart from Cauchy Schwartz or completing squares?
• Because some now-deleted answers were confused about this, might be worth recording that the matrices $M \in \mathfrak{sp}(n)$ are those which satisfy $\Omega M = -M^T\Omega$ where $\Omega$ is some fixed non-singular skew-symmetric matrix, e.g. $\Omega = \begin{pmatrix} 0 & I \\ -I & 0 \end{pmatrix}$. Feb 25 at 17:05
Here is a proof of $$\det M\geq 0$$ for $$M\in \mathfrak{sp}(n)$$ based on the lemma that every complex matrix is consimilar to a real matrix.
Acknowledgment: In what follows I was helped by feedback I received at MSE.
By construction, the $$2n\times 2n$$ complex matrix $$M\in \mathfrak{sp}(n)$$ is skew-Hermitian and Hamiltonian, which means that it has the $$n\times n$$ block decomposition $$M=\begin{pmatrix} A&B\\ C&-A^T\end{pmatrix},\;\;\text{with}\;\;A=-A^\ast,\;\;B=B^T=-C^\ast=-\bar{C}.$$ Here $$M^T$$ denotes the transpose, $$\bar{M}$$ the complex conjugate, and $$M^\ast$$ the conjugate transpose.
By continuity of the determinant it is sufficient to consider $$\det A\neq 0$$. Then Schur's determinant identity gives
$$\det M=\det(-AA^T-ACA^{-1}B)=\det(A\bar{A}+A\bar{B}A^{-1}B)$$ $$\qquad=\det(A\bar{A})\det(1+\bar{A}^{-1}\bar{B}A^{-1}B)$$ $$\qquad=|\det A|^2\det(1+\bar{X}X),\;\;\text{with}\;\;X=A^{-1}B.$$
Now I apply the consimilarity lemma, to write $$X=SR\bar{S}^{-1}$$ with $$R$$ a real matrix. This gives $$\det M=|\det A|^2\det(1+\bar{S}R^2\bar{S}^{-1})=|\det A|^2\det(1+R^2)$$ $$\qquad=|\det A|^2\det(1+iR)\det(1-iR)$$ $$\quad=|\det A|^2|\det(1+iR)|^2\geq 0.$$
• Ultimately this seems to use eigenvalues in a way that's pretty similar to what the OP did. Feb 26 at 1:22
• For $X=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$, $|b|^4\det(1+X\overline{X})=||b|^2-\overline{b}^2 \det(X)|^2+|b|^2|a\overline{b}+b\overline{d}|^2$. So it is possible to write the determinant as a sum of norm square in the case $n=2$. Feb 26 at 3:47
• @SamHopkins --- I worked a bit more on the proof, I think I have now removed any reliance on eigenvalues. Feb 26 at 12:43
• @CarloBeenakker Can $R$ be expressed using $X$? I think you proof is using the fact that $X\overline{X}$ is conjugate to the square of a real matrix, which is not so much different from a proof using eigenvalues Feb 27 at 1:05
• a proof along these lines using eigenvalues would rely on the fact that the negative eigenvalues of $X\bar{X}$ have multiplicity two, which is not what I am using here (although indeed this property can be derived from the consimilarity of $X$ with a real matrix). Feb 27 at 7:42 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9841428995132446, "perplexity": 118.84116489704134}, "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/1637964362619.23/warc/CC-MAIN-20211203091120-20211203121120-00049.warc.gz"} |
https://community.freefem.org/t/which-mesh-coordinate-is-used-to-compute-dx-du-after-moving-mesh/788 | # Which mesh/coordinate is used to compute dx(du) after moving mesh
Hi,
I am a bit confused with the following dxd(u) – based upon which mesh the derivative is calculated?
Tho=Th;
fespace Vh(Th,P2);
fespace Vho(Tho,P2);
Vh u;
Vho uo;
Tho=Th;
uo=u;
Th=movemesh (Th, …);
dxd(u); //first
u=u; // Is this just an interpolation?
dxd(u); // second
u[ ] = uo[ ]; // is this a Larangian update?
dxd(u); //third
Can I say the first dxd(u) is computed using the old mesh, and the second dxd(u) uses the new mesh, but the derivatives are the same? both of which are different to the third dxd(u)?
Best,
Yongxing.
dx(u) is the derivative of Finite element function u def on mesh Th.
if u is almost == x then du(u) is almost == 1 and this do not depend of the mesh.
The problem is when you say u==x, this is still a Eulerian view, isn’t it? Once you define u=x, for example the density function, on a mesh, then you move (may twist or squeeze) the mesh and let the assiciated u follow the mesh, i.e. a Lagrangian movement, the u cannot be x on the new mesh anymore.
Anyway, can I confirm that in FreeFEM++, u[ ]=uold[ ] is just a Lagrangian update? | {"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": 0.8302003145217896, "perplexity": 2860.1603438764196}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103033925.2/warc/CC-MAIN-20220625004242-20220625034242-00369.warc.gz"} |
https://www.physicsforums.com/threads/maximum-acceleration-of-a-dot-on-a-slinky.365909/ | # Maximum acceleration of a dot on a slinky?
1. Dec 28, 2009
### uberifrit
A dot (representing vibration) on a slinky exhibits simple harmonic motion as the longitudinal wave passes. The wave has an amplitude of 5,4 * 10^-3 m and a frquency of 4,0 Hz. Find the maximum acceleration of the dot.
Please could you explain what equations to use and how to answer in detail. This would be much appreciated as I am just starting this subject
Last edited: Dec 28, 2009
2. Dec 28, 2009
### pgardn
Since you are studying simple harmonic motion, could you give the equation that describes the position of your dot along a straight line through time? The one that has sine in it?
Similar Discussions: Maximum acceleration of a dot on a slinky? | {"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": 0.9546424150466919, "perplexity": 1005.3781129715967}, "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-2018-09/segments/1518891814249.56/warc/CC-MAIN-20180222180516-20180222200516-00065.warc.gz"} |
http://simbad.cds.unistra.fr/simbad/sim-ref?bibcode=2006ApJS..166..128Z | other querymodes : Identifierquery Coordinatequery Criteriaquery Referencequery Basicquery Scriptsubmission TAP Outputoptions Help
2006ApJS..166..128Z - Astrophys. J., Suppl. Ser., 166, 128-153 (2006/September-0)
A comprehensive study of 2000 narrow line Seyfert 1 galaxies from the Sloan Digital Sky Survey. I. The sample.
ZHOU H., WANG T., YUAN W., LU H., DONG X., WANG J. and LU Y.
Abstract (from CDS):
This is the first paper in a series dedicated to the study of the emission-line and continuum properties of narrow line Seyfert 1 galaxies (NLS1s). We carried out a systematic search for NLS1s from objects assigned as QSOs'' or galaxies'' in the spectroscopic sample of the Sloan Digital Sky Survey Data Release 3 (SDSS DR3) by a careful modeling of their emission lines and continua. The result is a uniform sample comprising ∼2000 NLS1s. This sample dramatically increases the number of known NLS1s by a factor of ∼10 over previous compilations. This paper presents the parameters of the prominent emission lines and continua, which were measured accurately with typical uncertainties <10%. Taking advantage of such an unprecedented large and uniform sample with accurately measured spectral parameters, we carried out various statistical analyses, some of which were only possible for the first time. The main results found are as follows. (1) Within the overall Seyfert 1 population, the incidence of NLS1s is strongly dependent on the optical, X-ray, and radio luminosities as well as the radio loudness. The fraction of NLS1s peaks around SDSS g-band absolute magnitude Mg~-22 mag in the optical and ∼1043.2 ergs/s in the soft X-ray band, and decreases quickly as the radio loudness increases. (2) On average the relative Fe II emission, R4570=Fe II λλ4434-4684/Hβ, in NLS1s is about twice that in normal active galactic nuclei (AGNs) and is anticorrelated with the broad component width of the Balmer emission lines. (3) The well-known anticorrelation between the width of broad low-ionization lines and the soft X-ray spectral slope for broad line AGNs extends down to FWHM∼1000 km/s in NLS1s, but the trend appears to reverse at still smaller line widths. (4) The equivalent width of Hβ and Fe II emission lines are strongly correlated with the Hβ and continuum luminosities. (5) We do not find any difference between NLS1s and normal AGNs in regard to the narrow line region. (6) We have examined the black hole mass versus stellar velocity dispersion (MBH*) relation for a subsample of 308 NLS1s for which σ*could be measured directly from fitting the starlight in the SDSS spectra with our stellar spectral templates. A significant correlation between MBHand σ*is found, but with the bulk of black hole masses falling below the values expected from the MBH* relation for normal galaxies and normal AGNs. This result indicates that NLS1s are underage AGNs, where the growth of the SMBH lags behind the formation of the galactic bulge. (7) We also find that the FWHM of [N II] line is well correlated with σ*in 206 NLS1s, for which both parameters could be derived with reasonable accuracy. The [N II] width can predict the stellar velocity dispersion to an accuracy of ∼30%. A similar MBH*relation could be found for a larger sample of 613 NLS1s on making use of the more reliable measurements of FWHM[N II].
Journal keyword(s): Galaxies: Active - Galaxies: Seyfert
VizieR on-line data: <Available at CDS (J/ApJS/166/128): table1.dat> | {"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": 0.9223390221595764, "perplexity": 3871.9567167379255}, "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/1664030337595.1/warc/CC-MAIN-20221005073953-20221005103953-00170.warc.gz"} |
https://calculla.com/fraction_inverse | Fraction inverse (reciprocal) calculator
Calculator finds multiplicative inverse of given fraction or number.
# Fraction, which you're going to invert#
Enter fraction (number)
# Result - fraction inverse step by step#
1 3 4
Step I: Remove whole parts (if exists)
1 · 4 + 3 4
=
7 4
Step II: Invert fraction
4 7
Step III: Convert improper fraction to mixed number (if needed)
4 7
# Some facts#
Remember:
• "Multiplicative inverse" and "reciprocal" is the same thing
• To invert a number X you have to divide 1 by that number. So, multiplicative inverse of X is 1/X.
• To invert fraction you simply need to swap its nominator with denominator. For example inverse of 3/4 is improper fraction 4/3.
• Inverse of number X is Y, when X × Y is equal to one:
$X \times Y = 1$
ⓘ Example: The inverse of number 2 is fraction 1/2 because:
$2 \times \dfrac{1}{2} = 1$
So, whatever number X times inverted X equals one !
• There is no inverse of zero. This also means: any fraction with zero in nominator has no inverse too - no matter what is denominator. Do not be fooled!
• When you invert number lower than one, you get number greater than one and vice versa.
• The inverse of number 1 is 1.
• If you invert some number twice, then you will get the same number. In other words, if you invert the number X and then invert the result again - you get X !
# How to use this tool#
Enter fraction (number), which you want to invert and calculla will present you the step-by-step inversion. To find fraction (number) inverse you need to do below steps:
• I. Remove wholes part. If you want to invert mixed number (or integer number), then you need conver it to improper fraction. For example: 2 1/2 should be converted to 5/2. In other case i.e. when there is no wholes part part, nothing to do in this step. After this step your number will be ready to invert.
• II. Invert fraction i.e. swap its numerator and denominator.
• III. Pull out wholes part. If inverted fraction is improper (numerator greater than denominator) convert it to mixed number.
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https://www.physicsforums.com/threads/simple-rocket-problem.572502/ | # Homework Help: Simple Rocket Problem
1. Jan 30, 2012
### fisselt
I know I've done this problem a few times before but I'm just a fumbling idiot at the moment..
1. The problem statement, all variables and given/known data
Rocket has an acceleration of 12m/s^2 and at an altitude of 1km the motor shuts off. What is the maximum height?
2. Relevant equations
V^2=V_i+2a(x_f-X_i)
x_f=x_i+V_i(t)+1/2at^2
3. The attempt at a solution
V^2=0+2(12)(1000)=154.92m/s
0=154.92t-9.8t^2, t=17.213s
x_f=1000+154.95(17.213)-1/2(9.8)(17.213)^2=2214.82meters
I feel like I'm doing it wrong. 12m/s^2 is the acceleration of the motor, shouldnt their be some force from gravity on the rocket while going towards 1000m?
Thanks for the help.
2. Jan 30, 2012
### PeterO
The acceleration of the motor/rocket combination is the change noticed under the combined influence of the force of gravity, and the thrust force of the rocket motor, and presumably any frictional forces from the air. You are probably supposed to ignore the air resistance - a common approximation used with this sort of problem.
3. Jan 30, 2012
### fisselt
thanks for the 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": 0.8036049008369446, "perplexity": 1343.9867402336515}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376830479.82/warc/CC-MAIN-20181219025453-20181219051453-00331.warc.gz"} |
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• # Artykuł - szczegóły
## Colloquium Mathematicum
2006 | 105 | 2 | 283-295
## On tame dynamical systems
EN
### Abstrakty
EN
A dynamical version of the Bourgain-Fremlin-Talagrand dichotomy shows that the enveloping semigroup of a dynamical system is either very large and contains a topological copy of β𝓝, or it is a "tame" topological space whose topology is determined by the convergence of sequences. In the latter case we say that the dynamical system is tame. We show that (i) a metric distal minimal system is tame iff it is equicontinuous, (ii) for an abelian acting group a tame metric minimal system is PI (hence a weakly mixing minimal system is never tame), and (iii) a tame minimal cascade has zero topological entropy. We also show that for minimal distal-but-not-equicontinuous systems the canonical map from the enveloping operator semigroup onto the Ellis semigroup is never an isomorphism. This answers a long standing open question. We give a complete characterization of minimal systems whose enveloping semigroup is metrizable. In particular it follows that for an abelian acting group such a system is equicontinuous.
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• Department of Mathematics, Tel Aviv University, Tel Aviv, Israel | {"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": 0.9455052018165588, "perplexity": 1421.5963623298476}, "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/1652662619221.81/warc/CC-MAIN-20220526162749-20220526192749-00673.warc.gz"} |
https://math.stackexchange.com/questions/1802813/prove-upper-bound-for-recurrence | # Prove upper bound for recurrence
I am working on problem set 8 problem 3 from MIT's Fall 2010 OCW class 6.042J. This is covered in chapter 10 which is about recurrences.
Here is the problem:
$$A_0 = 2$$ $$A_{n+1} = A_n/2 + 1/A_n, \forall n \ge 1$$
Prove
$$A_n \le \sqrt2 + 1/2^n, \forall n \ge 0$$
I have graphed the recurrence and the upper bound and they seem to both converge on $\sqrt2$.
Also, if you ignore the boundary condition $A_0 = 2$ then you find that $\sqrt2$ is a solution to the main part of the recurrence. i.e. $\sqrt2 = \sqrt2/2 + 1/\sqrt2$.
The chapter and videos on recurrences have a lot to say about a kind of cookbook solution to divide and conquer recurrences which they call the Akra-Bazzi Theorem. But this recurrence does not seem to be in the right form for that theorem. If it were in the form $A_{n+1} = A_n/2 + g(n)$ then the theorem would give you an asymptotic bound. But $1/A_n$ is not a simple function of $n$ like a polynomial. Instead it is part of the recurrence.
Also, the chapter has a variety of things to say about how to guess the right solution and plug it into an inductive proof, but I haven't had much success. I have tried possible solutions of various forms like $a_n = \sqrt2+a/b^n$ and tried solving for the constants $a$ and $b$ but to no avail.
So, if someone can point me in the right direction that would be great. I always assume that the problem sets are based on something taught in the videos and in the text of the book but I am having trouble tracking this one down.
Bobby
We show by induction that $$\sqrt{2}\lt A_n\le \sqrt{2}+\frac{1}{2^n}.\tag{1}$$ Suppose the result holds at $n=k$. We show the result holds at $n=k+1$.
For the inequality on the right of (1), we need to show that $$\frac{A_k}{2}+\frac{1}{A_k}\le \sqrt{2}+\frac{1}{2^{k+1}}.$$ By the induction hypothesis, we have $$\frac{A_k}{2}+\frac{1}{A_k}\le \frac{\sqrt{2}}{2}+\frac{1}{2^{k+1}}+\frac{1}{\sqrt{2}}=\sqrt{2}+\frac{1}{2^{k+1}},$$ which takes care of the induction step for the inequality on the right of (1).
We still need to show that $\sqrt{2}\lt A_{k+1}$. Let $A_k=\sqrt{2}+\epsilon$, where $\epsilon$ is positive. Then $$A_{k+1}=\frac{\sqrt{2}+\epsilon}{2}+\frac{1}{\sqrt{2}+\epsilon}=\frac{4+2\sqrt{2}\epsilon+\epsilon^2}{2(\sqrt{2}+\epsilon)}\gt \frac{4+2\sqrt{2}\epsilon}{2(\sqrt{2}+\epsilon)}=\sqrt{2}.$$ This completes the induction step for the inequality on the left of (1).
Remark: The inequality (1) and squeezing show that $A_n$ indeed has limit $\sqrt{2}$.
• It didn't occur to me to prove that $A_n \ge \sqrt2$ and then substitute $\sqrt2$ in the $1/A_n$ term. Much easier that way. Thanks. – Bobby Durrett May 28 '16 at 2:36
• @BobbyDurrett: You are welcome. The inequality $A_n\gt \sqrt{2}$ is not mentioned explicitly in what you are asked to show, but when one tries to push the induction through, it becomes clear that it is necessary for the proof. – André Nicolas May 28 '16 at 2:40
• Nice induction for the $>\sqrt{2}$ part. I didn't manage it immediately, that's why I showed it studying the map $x\mapsto \frac{x}2+\frac{1}{x}$. – Daniel Robert-Nicoud May 28 '16 at 11:09
Let $$B_k = \frac{A_k}{\sqrt{2}}$$ Then $B_0 = \sqrt{2}$ and $$B_{n+1} = \frac12\left(B_n+\frac{1}{B_n} \right)$$ Thus $b_n$ is the $n$-th guess if you perform Newton's algorithm to try to find $\sqrt{1}$ starting with a guess of $\sqrt{2}$.
This recursion can be solved in closed form using the formula for the hyperbolic tangent of $2x$ in terms of $\tanh x$: $$\tanh(2x) = \frac{2 \tanh x}{1+\tanh^2{x}}$$ . The result looks something like $$B_n = \tanh\left( 2^n \theta\right)$$ where $\theta = \tanh^{-1} B_0$; the answer I have given is off in that every other term needs to be the reciprocal of what I wrote, but the general idea will work.
Once you have that, you can know exactly what $B_n$ is and prove the relation; or better yet, use the error analysis for Newton's method to get an estimate of how close to 1 you would be.
We want to apply induction, but we need also a lower bound on $A_n$ for the $\frac{1}{n}$ term. We can show that $A_n\ge\sqrt{2}$ as follows:
Let $f(x) = \frac{x}{2} + \frac{1}{x}$. Then $f'(x) = \frac{1}{2} - \frac{1}{x^2}$ which has a zero at $x = \sqrt{2}$ where $f(\sqrt{2}) = \sqrt{2}$, and is positive for $x>\sqrt{2}$. This shows that $f(x)\ge\sqrt{2}$ whenever $x\ge\sqrt{2}$, and as $A_{n+1} = f(A_n)$, we have that $A_n\ge\sqrt{2}$ for all $n\ge0$.
Then to conclude:
The case $n=0$ is trivially true.
For $n\ge0$ we have that $$A_{n+1} = \frac{A_n}{2} + \frac{1}{A_n}.$$ By induction hypothesis and what we have shown above, this has as upper bound $$A_{n+1}\le \frac{\sqrt{2} + 2^{-n-1}}{2} + \frac{1}{\sqrt{2}} = \sqrt{2} + 2^{-n-1}.$$
Three steps: Use the definition of $A_n$ and algebra to establish that $$A_{n+1}-\sqrt 2 = {(A_n-\sqrt 2)^2\over 2A_n}.\tag1$$ Next, use (1) to prove by induction that $$A_n\ge\sqrt 2\ \text{for every n.}\tag2$$ Finally, use (1) and (2) to prove by induction that $$A_n-\sqrt2 \le \frac1{2^{n}}\ \text{for every n.}\tag2$$
Your sequence is $$A_{n+1} = \frac{1}{2} A_n + \frac{1}{A_n}$$ where the last term features a division which reminds of the Newton-Raphson iteration.
Newton Raphson iteration takes the root of the tangent as next step for estimating a root of $f$: $$0 = T'(x_{n+1}) = f(x_n) + f'(x_n) (x_{n+1} - x_n) \iff \\ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$ by comparison we have $$-\frac{1}{2}A_n + \frac{1}{A_n} = - \frac{A_n^2-2}{2A_n} = - \frac{f(A_n)}{f'(A_n)}$$ and see $f(x) = x^2 - 2$, the function used to iterate against the root $\sqrt{2}$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9608221650123596, "perplexity": 119.9180937019314}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560628000367.74/warc/CC-MAIN-20190626154459-20190626180459-00518.warc.gz"} |
https://www.physicsforums.com/threads/area-under-the-curve-problem-but-theres-a-twist.41402/ | # Area under the curve problem, but there's a twist
1. Sep 1, 2004
### relativitydude
I have a unique area under a curve problem. Say we have a function y = x^2. Ok, so, the area under that curve is just the integral of x^2 from 0 to some value of x. That's easy. Lets call this area A.
Now normally if we wanted to find the area under a height, we would simply mulitply height by the value of x and subtract the integrated form to find the are above the curve to that height. Lets call this area B.
I cant use this method. I cant simply say h*x minus the integral of whatever to find area B. Is there another way to find area B while knowing the function but not involving area A in any way?
2. Sep 1, 2004
### mathwonk
try making this confusing question a little more precise. an example could help.
3. Sep 1, 2004
### relativitydude
I included a diagram. I need to find area A and area B separately. I cant find area B by subtracting A out of it and conversely I cant find area A by subtracting B out of it.
File size:
5.5 KB
Views:
118
4. Sep 1, 2004
### Hurkyl
Staff Emeritus
Why can't you?
5. Sep 1, 2004
I just cant.
6. Sep 1, 2004
### Ethereal
Why not? Try it for some value of x and h. What's the problem? On the other hand, maybe you could try expressing the f(x) in terms of y and integrating wrt y instead to find B directly.
PS. What did you use to create the picture? It looks rather neat.
7. Sep 1, 2004
### relativitydude
I cant do it the normal way because my problem that this assists boils down to 1=1, a true yet uninspiring finding. I will try it with your suggestions, Ethereal.
That chart? Eh, that's nothing. I made it in Adobe Illustrator. You can draw ANYTHING in that program.
8. Sep 1, 2004
### Hurkyl
Staff Emeritus
If you found one of them via the integral, couldn't you find the other by subtracting?
9. Sep 1, 2004
### relativitydude
I said I cant use the subtraction method.
10. Sep 2, 2004
### Galileo
Do you mean you don't know how? Or aren't you allowed to use it for solving the problem?
If you're looking for another way to find area B, here's a hint:
11. Sep 2, 2004
### relativitydude
I am not allowed.
Thanks Gilleo, will do.
12. Sep 2, 2004
### JonF
wouldn't the inverse function of x^2 have the same area if you integrated along the x-axis and used the height as the upper bound limit?
13. Sep 3, 2004
### SpatialVacancy
Here's how
Find $f^{-1}(x)$. In this case, you are using $f(x) = x^2 [/tex]. The inverse of this function is [itex] f(x) = \sqrt{x}$. You can now integrate, but make sure to adjust your limits of integration. If you wanted from 0 to 3 of the original function, that would translate to 0 to 9 of the inverse. Just plug in the original limits to the inverse (y) to obtain your new limits.
Hope this helps!
14. Sep 3, 2004
### Galileo
I don't know if inversion will help. Inversion mirrors the function in (through?) the line y=x.
My idea was to mirror the function in the x-axis, then add the height h to the function and then integrate. (or alternatively subtract the height from the function and then mirror it).
But the calculation would be exactly the same if you were calculating the area of the rectangle and subtract area A. So it seems silly not to use this method and do use the above method.
Subtract height: f(x)-h
Mirror: h-f(x)
Integrate: $h\Delta x - \int f(x)dx$
It's the same :)
Last edited: Sep 3, 2004
15. Sep 3, 2004
### JonF
Well basically it is the same exact thing as what you planed on doing. The area between the function and the y-axis is the same as its inverse function and the x-axis.
16. Sep 8, 2004
### Hessam
ok i'm not sure if anyone mentioned htis but....
lets say the equation was y = x^2.... you wanted to find area B
could you not find the integral from 0 -> x by changing the equation ... by flipping it upside down (reflect upon x-axis) and shift it up h units?
thus you get
y = -x^2 + h
and find integral from 0-->x now?
correct me if i'm wrong | {"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": 0.8810521960258484, "perplexity": 1208.3940466733015}, "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/1480698543316.16/warc/CC-MAIN-20161202170903-00046-ip-10-31-129-80.ec2.internal.warc.gz"} |
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Results: 1 to 30 of 40 found Go to page: 1 2 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9504256844520569, "perplexity": 1728.127652430574}, "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-2014-35/segments/1408500834663.62/warc/CC-MAIN-20140820021354-00069-ip-10-180-136-8.ec2.internal.warc.gz"} |
http://stfi.ru/en/issues/2020/02/STFI_2020_02_MorozovEA.html | 2020, no. 2
### ON THE USE OF BIVECTOR FORMALIZM IN HAMILTONIAN MECHANICS
#### Morozov E. A., Morozova A.R., Morozova L.E
A bivector formalism of Hamiltonian mechanics is constructed. The extended affine space of impulses, coordinates, and time is determined based on the determinism principle. The space attached to it is considered as a direct sum of the covariant space of pulses and the contravariant space of coordinates and time, after which the bivector space of pulses, coordinates and time is determined. The resulting point-bivector correspondence allows us to determine the corresponding extended phase space and flow. It turns out that the bivector analog of the dynamic Hamilton equations has the form of the dynamic Newton equation for the potential field. We consider a bivector variant of canonical transformations that define the geometry of a bivector phase space. The use of covariant and contravariant vector spaces, as well as basic tensor operations, makes it possible to significantly simplify the transformation algebra in proofs.
Keywords: Hamilton mechanics, Hamilton equations, phase space, canonical transformations.
UDC: 531.011
PACS: 04.50.Kd
DOI: 10.17238/issn2226-8812.2020.2.64-70 | {"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": 0.9584131836891174, "perplexity": 920.7491154502652}, "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/1664030335365.63/warc/CC-MAIN-20220929194230-20220929224230-00468.warc.gz"} |
http://mathoverflow.net/questions/4125/if-spec-z-is-like-a-riemann-surface-whats-the-analogue-of-integration-along-a-c/4141 | ## If Spec Z is like a Riemann surface, what’s the analogue of integration along a contour?
Rings of functions on a nonsingular algebraic curve (which, over C, are holomorphic functions on a compact Riemann surface) and rings of integers in number fields are both examples of Dedekind domains, and I've been trying to understand the classical analogy between the two. As I understand it, should be thought of as the curve / Riemann surface itself. Is there a good notion of integration in this setting? Is there any hope of recovering an analogue of the Cauchy integral formula?
(If I am misunderstanding the point of the analogy or stretching it too far, please let me know.)
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For a variety $X$ over a finite field, I guess one can take $\ell$-adic sheaves to replace differential forms. Then the local integral around a closed point $x$ (like integral over a little loop around that point) is the trace of the local Frobenius $Frob_x$ on the stalk of sheaf, the so-called naive local term. Note that $Frob_x$ can be regarded as an element (or conjugacy class) in $\pi_1(X)$, "a loop around $x$". The global integral would be the global trace map $$H^{2d}_c(X,\mathbf{Q}_{\ell})\to\mathbf{Q}_{\ell}(-d),$$ and the Tate twist is responsible for the Hodge structure in Betti cohomology (or the $(2\pi i)^d$ one has to divide by). The Lefschetz trace formula might be the analog of the residue theorem in complex analysis on Riemann surfaces.
For the case of number fields, each closed point $v$ in $Spec\ O_k$ still defines a "loop" $Frob_v$ in $\pi_1(Spec\ k)$ (let's allow ramified covers. One can take the image of $Frob_v$ under $\pi_1(Spec\ k)\to\pi_1(Spec\ O_k)$, but the target group doesn't seem to be big enough). For global integral, there's the Artin-Verdier trace map $H^3(Spec\ O_k,\mathbb G_m)\to\mathbb{Q/Z}$ and a "Poincar\'e duality" in this setting, but I don't know if there is a trace formula. The fact that 3 is odd always makes me excited and confused.
So basically I think of trace maps (both local and global) as counterpart of integrals. Correct me if I was wrong.
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This is interesting, but a little over my head. If you don't mind, could you explain how one thinks about the contour integral in this abstract setting, i.e. in terms of the fundamental group? – Qiaochu Yuan Nov 4 2009 at 19:58
A loop in this setting is a map from a scheme with a "cyclic" fundamental group. Finite fields have this property, so their spectra can be viewed as circles. For Spec Z, the only interesting loops we see are the canonical maps from spectra of finite fields. One has an analogy between integration of differentials and parallel transport along connections, so we are determining how a vector bundle (our l-adic sheaf) is transformed as we follow flat sections around a "circle". – S. Carnahan Nov 5 2009 at 2:37
"The fact that 3 is odd always makes me excited and confused." Fantastic. – Cam McLeman May 8 2011 at 18:56
I think that you have understood the analogy correctly, and you have pinpointed one of its weaknesses. Although number fields are like one dimensional functional fields in many ways, one of the differences is that the vector space of Kahler differentials for a number field has dimension 0, not 1. Here Kahler differentials are the vector space generated by symbols dx, subject to the relations d(x+y) = dx + dy and d(xy) = x dy + y dx.
Therefore, there is nothing like differentials, and nothing we can integrate.
But it is possible that there is some more sophisticated way to solve this problem. (Maybe using Arakelov geometry?) I'm looking forward to reading the other answers.
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This certainly does have a bit of the Arakelov smell about it. Maybe if you include the infinite places there's some different notion of Kahler differential? – Ben Webster Nov 4 2009 at 18:40
That makes sense; it's related to the fact that the abc conjecture is much harder than the Mason-Stothers theorem. I think a lot of people would want to see a notion of differential in the number field setting for this reason. – Qiaochu Yuan Nov 4 2009 at 18:45
I think David's module of differentials on R should include the rule that dc = 0 when c is a constant: here the ring of constants is some specified subring R_0 of R. (So e.g. when R = C[t] one takes R_0 = C; otherwise, the differentials are way bigger than you want.) So we see the problem: David's module is Omega_{Z/Z}, and the relative dimension is 0, while for Omega_{C[t]/C} the relative dimension is 1. In order to have a good notion of differentials you'd need a "field of constants" "inside" Spec Z, which would have... oh, I dunno.... one element? – JSE Nov 5 2009 at 1:31
Yup! JSE's comment is exactly right, and gives a great way of seeing why we want a field with one element. – David Speyer Nov 5 2009 at 1:38
Lichtenbaum talked about that and interesting applications last December: institut.math.jussieu.fr/projets/tn/STN/11/… – Thomas Riepe May 8 2011 at 15:05
For a number f the local-global formula written as
where is the inverse power of p that is equal to f in the local field of p-adic numbers and is f should provide a reasonable analogue (in other words, integration over all holes gives 0 on a closed surface).
(An example for f = 75: 1 * 1/3 * 1/25 * 1 * ...* 75 = 1)
A better formulation would involve adeles: the adele ring is a semi-restricted product of p-adic rationals and rationals themselves. There is a map and every element in the image has the property that the product over all places gives 1.
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I think we have to be careful about the two senses of the word "residue" here. Either way my hope was that one could recover the residue as part of a more general computation. – Qiaochu Yuan Nov 4 2009 at 18:43
I'm rewriting the answer and making it better. I remember things one by one, but there should definitely be a good answer! – Ilya Nikokoshev Nov 4 2009 at 18:54
I take the product formula as an analog of the fact that a principal divisor on a curve has degree 0, so it is a (at least special case of the) residue theorem: the local integral at all points add up to the integral over the boundary, and there's no boundary if we also include \infty! – shenghao Nov 4 2009 at 19:59
That's absolutely what I mean, yes. – Ilya Nikokoshev Nov 4 2009 at 20:56 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 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": 0.9362817406654358, "perplexity": 393.2906895211812}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368706082529/warc/CC-MAIN-20130516120802-00029-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://forum.bebac.at/forum_entry.php?id=17962 | ## critical values [Nonparametrics]
Hi Václav and GM,
see also this rather old post with tables of the positions of the lower critical values and exact probabilities for m=n=3–32. Generally the normal approximation should not be used if m <8 or n <8.
Dif-tor heh smusma 🖖🏼 Довге життя Україна!
Helmut Schütz
The quality of responses received is directly proportional to the quality of the question asked. 🚮
Science Quotes | {"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": 0.9327317476272583, "perplexity": 3393.9426409945886}, "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-2023-14/segments/1679296949107.48/warc/CC-MAIN-20230330070451-20230330100451-00512.warc.gz"} |
http://math.stackexchange.com/questions/238659/continuous-function-proof-help | Continuous function proof help
Let $f : [a, b]\to R$ be a continuous function such that $[a,b] \subset [f(a), f(b)]$. Prove that there exists $x\in [a,b]$ such that $f(x) = x$.
My attempt: I said let there be a $\delta > 0$and defined $c$ and $d$ to be $x + \delta$ and $x-\delta$ respectively. From here since $f$ is continuous $[f(c), f(d)]\subset [f(a), f(b)]$. Then I assumed by definition $[c,d]$ is also a subset of $[f(c), f(d)]$. Then I claimed $\delta$ can be arbitrarily small so that $f(c) = f(d) = f(x)$.
Is this correct or is there a better approach?
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If $[a,b]\cap [f(a),f(b)]=\emptyset$ then for any function $f:[a,b]\to \mathbb{R}$ (with this property) there is no such $x^*$. – vesszabo Nov 16 '12 at 13:23
It's not necessarily true that $[f(c),f(d)]$ is a subset of $[f(a),f(b)]$. Consider a function where $f(0) = 0$, $f(1) = (1)$ where $f(x)$ starts decreasing at $0$. – Jason DeVito Nov 16 '12 at 13:28
Your approach is not correct, since you are assuming what you are supposed to show. You cannot define $c$ and $d$ to be something which depends on $x^*$ before you have shown that there is such a number as $x^*$.
A better approach would be to consider the function $g(x)=f(x)-x$. Argue that $g$ is continuous, that $g(a)\leq 0$, and that $g(b)\geq 0$. Then there should be a result available that you can use.
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I see. I'm trying to show $g(a)\leq 0$ but I'm having a hard time seeing how knowing $[a,b] \subset [f(a), f(b)]$ will help. I only know it is a subset and can't say much about their values. – Vincent Nov 16 '12 at 13:56
Try to draw the two intervals along a real line. What must the order of the four points be? What does that tell you about the size of $a$ and $f(a)$? – Per Manne Nov 16 '12 at 14:00
I know that $a < b$ and that $f(a) < f(b)$. But isn't it still possible for either $a$ to be greater than $f(a)$ or vice versa? – Vincent Nov 16 '12 at 14:08
Did you draw the two intervals with one being contained in the other? Try to find a numerical example where $[a,b]\subset [f(a),f(b)]$ and $a<f(a)$. (Hint: It can't be done.) – Per Manne Nov 16 '12 at 14:32
In $\mathbb R^2$ draw the straight line $y=x$ and consider any interval at the positive part of $x$-axis, you'll get some intuition. Your problem is saying that under some conditions the graph of your function is going to intersept the straight line $y=x$ in some point. Think about the conditions for it to happen..
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http://mathoverflow.net/questions/102453/map-with-prescribed-jacobian | # Map with prescribed Jacobian
Recently I came up with the following problem.
Suppose $U$ is an open subset of $\mathbb{R}^n$ and we are given a continuous map $M:U\to GL(n;\mathbb{R})$. Does anybody know if there are conditions for the existence of a $C^1$ map $f:U\to \mathbb{R}^n$ such that $$Jf(x)=M(x)\quad \forall x \in U$$ (here $Jf(x)$ is the Jacobian matrix of $f$ at $x$)?
I know this question is somehow a little bit general; I will really appreciate even a reference for something related to this problem. :)
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Suppose we can find such an $f$. If $f$ is continuously differentiable, then the i'th row and j'th column of the Jacobian (in the standard basis) is the j'th partial derivative of the the i'th component of $f$. Indeed, the components of $f$ are continuously differentiable if and only if $f$ is. So it suffices to consider each of the components of $f$ along with its corresponding row in M(x) seperately - i.e. we want $\nabla f_{i}=M_{i}$ where $M_{i}$ is the i'th row of $M$. So the ith row of $M$ is given by a scalar potential - e.g. if $M$ is (continuously) differentiable $\omega_{i}=\sum_{j}m_{i,j}dx^{j}$ is exact, where $M(x)=(m_{i,j}(x))$. A neccesary condition for this to be true is $\partial_{k}m_{i,j}=\partial_{j}m_{i,k}$ for all $i,j,k\in\{1,..n\}$. The sufficiency of this condition depends on the topology of $U$. See http://en.wikipedia.org/wiki/Closed_and_exact_differential_forms.
This is a nice answer, though somehow tautological. Also notice that we are not allowed to take derivatives of the matrix $M$ since it is assumed to depend only continuously on $x.$ – A. Lerario Jul 17 '12 at 15:52
Why is this "somehow tautological"? Also, the derivatives of $M$ do exist in the weak or distributional sense and the answer remains valid using that. – Deane Yang Jul 17 '12 at 16:07
Maybe another thing to notice (maybe its obvious) is that $M$ is invertible on all of $U$. So if such an f exists, the inverse function theorem guarantees that it is a local $C^{1}$ diffeomorphism. – Brian Trundy Jul 17 '12 at 16:07
So if such an $f$ exists, there is (at least locally) a $C^{1}$ map $g$ satifying $Dg(x)=M^{-1}(f(x))$. This suggests that there should be a further restriction on the components of $M$ (Using Cramer's Rule to get the components of $M^{-1}$ in terms of the components of $M$) in terms of weak/distributional derivatives. – Brian Trundy Jul 17 '12 at 16:56 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9542188048362732, "perplexity": 103.95599218122273}, "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/1408500834663.62/warc/CC-MAIN-20140820021354-00088-ip-10-180-136-8.ec2.internal.warc.gz"} |
https://export.arxiv.org/abs/2110.11627 | cs.IT
(what is this?)
# Title: On the largest singular values of certain large random matrices with application to the estimation of the minimal dimension of the state-space representations of high-dimensional time series
Authors: Daria Tieplova (LIGM), Philippe Loubaton (LIGM)
Abstract: This paper is devoted to the estimation of the minimal dimension P of the state-space realizations of a high-dimensional time series y, defined as a noisy version (the noise is white and Gaussian) of a useful signal with low rank rational spectral density, in the high-dimensional asymptotic regime where the number of available samples N and the dimension of the time series M converge towards infinity at the same rate. In the classical low-dimensional regime, P is estimated as the number of significant singular values of the empirical autocovariance matrix between the past and the future of y, or as the number of significant estimated canonical correlation coefficients between the past and the future of y. Generalizing large random matrix methods developed in the past to analyze classical spiked models, the behaviour of the above singular values and canonical correlation coefficients is studied in the high-dimensional regime. It is proved that they are smaller than certain thresholds depending on the statistics of the noise, except a finite number of outliers that are due to the useful signal. The number of singular values of the sample autocovariance matrix above the threshold is evaluated, is shown to be almost independent from P in general, and cannot therefore be used to estimate P accurately. In contrast, the number s of canonical correlation coefficients larger than the corresponding threshold is shown to be less than or equal to P, and explicit conditions under which it is equal to P are provided. Under the corresponding assumptions, s is thus a consistent estimate of P in the high-dimensional regime. The core of the paper is the development of the necessary large random matrix tools.
Subjects: Information Theory (cs.IT); Probability (math.PR) Cite as: arXiv:2110.11627 [cs.IT] (or arXiv:2110.11627v1 [cs.IT] for this version)
## Submission history
From: Philippe Loubaton [view email]
[v1] Fri, 22 Oct 2021 07:21:34 GMT (151kb,D)
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https://www.physicsforums.com/threads/increment-or-decrement-a-parameter-in-latex.354839/ | # Increment (or decrement) a parameter in LaTeX
1. Nov 14, 2009
### robbins
How can I increment or decrement a parameter in LaTeX?
For example, suppose I create an environment foo as follows:
\newenvironment{foo}[2]{\begin{tabular}{*{#1}c*{#2}r}}{\end{tabular}}
I can write
\begin{foo}{4}{3}
<tabular data in seven columns>
\end{foo}
But in my application, if the first parameter is n, then second is always n-1. Is there a way to write this that passes only one parameter (in this case n), but which will produce the table with n + (n - 1) columns as in the environment foo?
Decrementing the parameter #1 inside the newenvironment -- if it is possible -- would surely work, but I don't know how to do it.
Note: The snippet
\let\temp#1
\advance\temp by -1
doesn't work because if I pass the parameter 4, \temp takes the value 'the character 4', which can't follow the \advance command.
I'm stuck. Any help?
2. Nov 16, 2009
### robbins
This solution was posted by Joseph Wright on LaTeX Community:
"You don't want \let, you want to do things with numbers proper. The thing is that they then [need] to be assigned to TeX counters. You seem to want something like
\newcount\mycount
\newenvironment{foo}[1]
{%
\mycount #1\relax
\advance\mycount -1\relax
\begin{tabular}{*{#1}c*{\the\mycount}r}%
}
{\end{tabular}}
Here, I'm using a TeX count register to do the maths. You can do the same with a LaTeX counter, but I find this route a bit easier for going downward. TeX assigns count registers locally, so the above should work in most cases."
Thanks, Joseph.
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https://www.physicsforums.com/threads/discrete-metric-and-continuity-equivalence.710512/ | # Discrete metric and continuity equivalence
• Start date
• #1
mahler1
222
0
Homework Statement .
Prove that a metric space X is discrete if and only if every function from X to an arbitrary metric space is continuous.
The attempt at a solution.
I didn't have problems to prove the implication discrete metric implies continuity. Let f:(X,δ)→(Y,d) where (Y,d) is an arbitrary metric space. Let V be an open set in (Y,d). Every set in X is open so, in particular, f^-1(V) is an open set. This proves f is continuous.
I don't know how to prove the other statement: if every function from the metric space X to some arbitrary metric space Y is a continuous function, then X is discrete.
• #2
Homework Helper
2,273
877
Homework Statement .
Prove that a metric space X is discrete if and only if every function from X to an arbitrary metric space is continuous.
The attempt at a solution.
I didn't have problems to prove the implication discrete metric implies continuity. Let f:(X,δ)→(Y,d) where (Y,d) is an arbitrary metric space. Let V be an open set in (Y,d). Every set in X is open so, in particular, f^-1(V) is an open set. This proves f is continuous.
I don't know how to prove the other statement: if every function from the metric space X to some arbitrary metric space Y is a continuous function, then X is discrete.
Proof by contrapositive: Suppose X is not discrete. Then there exists a function from X to {0,1} which is not continuous.
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• #3
mahler1
222
0
Proof by contrapositive: Suppose X is not discrete. Then there exists a function from X to {0,1} which is not continuous.
Thanks for the advice. I've come up with this idea but I want to check if it is ok:
Suppose X is not discrete. Take (Y,d') to be the Y={0,1} and d'=δ distance. By definition of discrete space, if X is not discrete then there exists some x for which for all ε>0, the ball B(x,ε) contains other point y different from x. Define f to be f(x)=0 and f(y)=1 for all y in X with y≠x. If f is continuous, in particular, is continuous at the point x. Take ε'=1/2. Then, for all ε, there is some y such that d(x,y)<ε but δ(f(x),f(y))=δ(1,0)=1>1/2=ε'. But this is absurd since f was continuous. Then X must be discrete.
I have a question regarding the concept of discrete space. If the derived set of some metric space is empty, is this equivalent to say that X is discrete? I think it is but I'm not 100% sure. If this equivalence holds, I have a similar proof to the original problem:
Suppose X is not discrete, then there exists x in X'. Take (Y,d')=({0,1},δ) and define f(x)=0 and f(y)=1 for all y in X-{x}. By definition of derived set, there exists a sequence {x_n} in X such that x_n→x when n→∞. The function f is continuous, so f(x_n)→f(x) when n→∞. But f(x_n)=1≠0=f(x), which is absurd. It follows that X must be discrete.
• #4
Homework Helper
26,263
619
Thanks for the advice. I've come up with this idea but I want to check if it is ok:
Suppose X is not discrete. Take (Y,d') to be the Y={0,1} and d'=δ distance. By definition of discrete space, if X is not discrete then there exists some x for which for all ε>0, the ball B(x,ε) contains other point y different from x. Define f to be f(x)=0 and f(y)=1 for all y in X with y≠x. If f is continuous, in particular, is continuous at the point x. Take ε'=1/2. Then, for all ε, there is some y such that d(x,y)<ε but δ(f(x),f(y))=δ(1,0)=1>1/2=ε'. But this is absurd since f was continuous. Then X must be discrete.
I have a question regarding the concept of discrete space. If the derived set of some metric space is empty, is this equivalent to say that X is discrete? I think it is but I'm not 100% sure. If this equivalence holds, I have a similar proof to the original problem:
Suppose X is not discrete, then there exists x in X'. Take (Y,d')=({0,1},δ) and define f(x)=0 and f(y)=1 for all y in X-{x}. By definition of derived set, there exists a sequence {x_n} in X such that x_n→x when n→∞. The function f is continuous, so f(x_n)→f(x) when n→∞. But f(x_n)=1≠0=f(x), which is absurd. It follows that X must be discrete.
That sounds fine. Both of those proofs are really the same. Why are you not 100% sure that the set of all limit points of a metric space is empty if and only if the space is discrete? You should be able to prove that. And for your first proof, since you have 'if and only if' you still have to prove any function on a discrete space is continuous. But that should be pretty easy. Actually, I see you've already gotten that part. Nevermind.
Last edited:
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• #5
Homework Helper
2,273
877
Thanks for the advice. I've come up with this idea but I want to check if it is ok:
Suppose X is not discrete. Take (Y,d') to be the Y={0,1} and d'=δ distance. By definition of discrete space, if X is not discrete then there exists some x for which for all ε>0, the ball B(x,ε) contains other point y different from x. Define f to be f(x)=0 and f(y)=1 for all y in X with y≠x. If f is continuous
You don't need to derive a contradiction; you can show directly that f as defined is not continuous at x:
If $\epsilon < d'(0,1)$ then for every $\delta > 0$ there exists $y \in B(x,\delta)$ with $d'(f(x),d(y)) > \epsilon$ and thus $f$ is not continuous at $x$.
That's sufficient to complete the proof: the object is to show that if X is not discrete then there exists a metric space Y such that there exists a function from X to Y which is not continuous, which is equivalent to showing that if every function from X to an arbitrary metric space is continuous then X is discrete.
A direct proof would probably start with the observations that if every function from X to an arbitrary metric space is continuous then every function from X to {0,1} is continuous, and that every such function can be written as
$$f : x \mapsto \left\{\begin{array}{rc} 0 & \mbox{ if } x \in U \\ 1 & \mbox{ otherwise} \end{array} \right.$$
for some $U \subset X$.
• #6
mahler1
222
0
You're right. I've messed up with logic. We've shown that the contrapositive is true, so the statement is true.
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989 | {"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": 0.9774895310401917, "perplexity": 339.6053596601184}, "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/1659882573399.40/warc/CC-MAIN-20220818185216-20220818215216-00255.warc.gz"} |
http://mathhelpforum.com/algebra/46872-average.html | # Math Help - Average
1. ## Average
The average of 5 results is 46 and that of first 4 is 45.The fifth result is
(a)1
(b)10
(c)12.5
(d)50
2. Hello,
Originally Posted by devi
The average of 5 results is 46 and that of first 4 is 45.The fifth result is
(a)1
(b)10
(c)12.5
(d)50
The average of 5 results is given this way :
$A=\frac{a+b+c+d+e}{5}$, where a,b,c,d,e are the results. In general, you divide the sum of the results by the number of results.
We know that the sum of the first 4 is 45, let's say a+b+c+d=45 and that the average is 46.
$46=\frac{45+e}{5}$
Solve for e...
e+45=46
46-45=e
e=1 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9259002208709717, "perplexity": 1404.6241130245521}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-22/segments/1464053379198.78/warc/CC-MAIN-20160524012939-00089-ip-10-185-217-139.ec2.internal.warc.gz"} |
http://www.cfd-online.com/Forums/main/16437-pressure-enigma-print.html | CFD Online Discussion Forums (http://www.cfd-online.com/Forums/)
- Main CFD Forum (http://www.cfd-online.com/Forums/main/)
- - Pressure enigma. (http://www.cfd-online.com/Forums/main/16437-pressure-enigma.html)
Sara March 6, 2009 05:57
Pressure enigma.
Hey,
1. What is the physical interpretation of the pressure in an incompressible fluid?
2. It seems to me that if we are solving the momentum equations explicitly in an incompressible fluid, then we use the pressure-velocity coupling to ensure that we actually are solving the problem implicitly (which sounds correct). Do we actually need the pressure to make this implicit coupling of the velocities in the domain?
otd March 6, 2009 14:18
Re: Pressure enigma.
There's always the option of using the sream function / vorticity formulation (for 1 or 2-d problems with constant viscosity). An extension of that with further manipulation gives a biharmonic equation for the stream function alone (with the same restrictions).
In either case the pressure is eliminated from the calculation - although a pressure field can be retrieved from the stream function solution.
kenn k.q. zhang March 6, 2009 16:30
Re: Pressure enigma.
the disavantages of stream function vorticity include, but not limited to:
6 components in 3d, while 4 components in 3d for velocity-pressure;
vorticty is the derivative of velocity, so it demands more smoothness
kenn k.q. zhang March 6, 2009 17:11
Re: Pressure enigma.
physical interpretation of the pressure in incompressible flows?
pressure is just the pressure, the normal-direction force per area acting on the surfaces you are interested in. nothing fancy for the pressure in incompressible flows, in terms of concept.
however, the propagation of the pressure in the incompressible flows is at an infinite speed. when you want to consider the hydroacoutsics, this is not the case (the speed of sound in water is more than 1000m/s, i don't remember exactly), but otherwise it is. in air, the propagation of pressure is the speed of the sound in air.
that "the pressure progagate at infinite speed" equates a tiny chnage in the density can have a consequence of big change in pressure field. this is called stiffness.
i don't understand your second question, so her i just say something i know.
the role of pressure in incompressible flows is a lagrangian multiplier, which means the pressure demands the velocity field you obtained (with the moving the pressure to the right-hand-side of momentum equations) satify the conservation of mass.
with pressure on the right hand side, where the pressure can be obtained by the approximation from the previous time step, you can get a velocity field. but you can a lot of, as a matter of fact infinite number of, velocity fields. however, such a field must satisfy the continuity equation.
in penalty formulation, you can incorporate the continuity equation into the momentum equation & kind of limintate the pressure. so, you can work out the velocity field alone. penalty formulation reflects the epcicenter of "lagrangian multiplier" literatelly, but it is a bad approach and i surely suggest you not go for it.
in compressible flows, the navier-stokes flows are solved in a mixed formulation, which simply means the most natural approach. unfornulately, at low sonic speed, the stiffness issue occurs and the mixed approaches does very poorly. you may use artificail compressibility or pressure stabilization methods, to try to extend the "mixed formulation"-which is excellent for compressibel flows & compressible elasticity to the incompressible limit. unfortunately, the extension is not successful, either due to implemental complexibity or performance/speed or both.
so, the best category of handling incompressible flows is the splitting. and there are 3 categories for this business:
1. continuous splitting:
you can derive a pressure poisson equation from the momentum and continuity, in a continuous sense; and impose boundary a neumann boundary condition for the pressure poisson; the boundary condition can be derived from application of momentum equation on the normal direction of the boundary. this approach is a little pain at the beginning, due to the boundary condtion; but once you have then ready, it's easy to proceed;
2. discrete splitting:
you can continue the "mixed formulation" or the natural formulation, and get all matrices; then you do some manipulations to get a sub systems for the velocity and a pressure-poisson-like subsystem for the pressure. this apporach share a lot with the mixed formulation, so that it's easy to communicate with traditional compressible slovers. unfortunately, you still need add additonal boundary conditions onto matrices you just got. the reason, after you convert the matrices you just got into subsytems for the velocity and pressure, some information may get inactive, even the conversion is a one-on-one conversion.
there are other types of discrete splitting, such as the one presented in the famous paper published in 1965 on "physics of fluids", in which this technqiue, finite volume method, staggered grid, and MAC free-surface capturing technique can be found (in one single paper -- can we find this kind of highly original papers now?)
3. semi-discrete approach
this is simply the famous "fractional step" or "projection" by Chorin.
I don't the "velocity correction" approach implemented in "Fluent" belongs to which category, such as SIMPLE, SIMPLER, and so on. but the procedures being used by commercial software must be good, which is not necessary a case for those published on journa papers.
kenn k.q. zhang March 6, 2009 17:20
Re: Pressure enigma.
a little calrification:
suppose we onvert system A to system B, in a one-on-one manner.
we might think: since these two systems are equivalent to each other, solving B is the same as soliving A.
unfortunately, this is a wrong idea.
system B indeed equates system A, they contain the same information; however, some information may hide inside the system and simply can't be utilized. in other words, some information stored in system A may become intrinsic in B. as a consequence, we can't solve B and get the correct results.
Sara March 6, 2009 19:31
Re: Pressure enigma.
"pressure is just the pressure, the normal-direction force per area acting on the surfaces you are interested in. nothing fancy for the pressure in incompressible flows, in terms of concept. "
I'm a bit confused here, let me reformulate my first question:
Do we have some pressure acting in the incompressible system other than the dynamic pressure?
ag March 6, 2009 19:46
Re: Pressure enigma.
You have the static pressure, but in terms of the solution of the equations all that is important is the pressure gradient. You can add or subtract a gauge pressure from the flo field and get the same velocity field.
Sara March 6, 2009 19:56
Re: Pressure enigma.
Is it possible to eliminate the pressure term (by say assuming p = 0.5*rho*u²) and do a least-squares fit to the resulting two equations (x-momentum and continuity) in a 1d problem?
Sara March 6, 2009 20:06
Re: Pressure enigma.
Yes. So what is the pressure gradient in an incompressible fluid? I understand that it resolves from the solution of the equations but is there a physical meaning behind it?
Tom March 6, 2009 20:13
Re: Pressure enigma.
In the incompressible equations the pressure (usually referred to as the dynamic pressure to distinguish it from the thermodynamic one) is the internal force required maintain mass conservation (which also prevents cavitation). Basically it describes the coupling between the momentum equation and the continuity equation;i.e. it's the force required to ensure that the velocity evolves in a manner that conserves mass (strictly speaking it's volume in the incompressible equations). If you know about classical mechanics (Lagrange & Hamilton) then in an inviscid fluid the pressure (p) is the "Lagrange multiply" that multiplies the mass-conservation constraint when added to the Lagrangian -variations with respect to p gives the mass conservation equation while variation of the other parameters yields the momentum equations.
Jonas Holdeman March 7, 2009 11:32
Re: Pressure enigma.
The interpretation of the pressure in incompressible flow as a Lagrange multiplier is the result of a particular mathematical analysis of NSE. If the equations are rewritten in the stream function-vorticity form or the stream function form, there is no pressure in the problem. The incompressible velocity field exists and can be computed without regard to the pressure. The incompressibility/mass conservation is being maintained without reference to a pressure. Where is a Lagrange multiplier in this case? If one expands the velocity field in terms of a divergence-free basis, the expansion will always be divergence-free no matter what expansion coefficients are chosen. If the expansion coefficients are chosen to best approximate the solution to the NSE, this solution will be divergence-free without regard to the pressure. The incompressibility condition acts like a conservation law. The "pressure" derived from the velocity field (say by the pressure Poisson equation) is that pressure-like quantity consistent with the flow field. How can we say that the pressure drives incompressible flow if there is no pressure appearing in the governing equation? Incompressibility drives the flow and pressure responds!
Tom March 7, 2009 12:16
Re: Pressure enigma.
"The interpretation of the pressure in incompressible flow as a Lagrange multiplier is the result of a particular mathematical analysis of NSE."
No - it follows from the derivation of the inviscid equations using a Lagrangian. This is also true for compressible flows!
The streamfunction is the mathematical device. It's just a way of hiding the pressure, just like using the vorticity form in the 3D problem.
"The incompressibility condition acts like a conservation law."
No - it behaves as a constraint on the flow. The pressure is gradient is the force required to enforce the constraint. The constraint actually corresponds to the conservation of volume not mass (unless density is constant).
Sara, If you're OK with the (thermodynamics) pressure in the compressible equations then one way of understanding the corresponding term in the incompressible equations is to nondimensionalize the compressible equations and perform a low Mach number asymptotic expansion of the equations. There are many examples of this procedure in papers within the Journal of Fluid mechanics.
Sara March 7, 2009 18:46
Re: Pressure enigma.
Excellent answer Jonas, thank you, this was the essence of my confusion.
Ahmed March 10, 2009 17:03
Re: Pressure enigma.
No No No The confusion,if there is any, is the result of the definition of pressure in the mechanical sense, i.e., the normal component of force ..etc. The Kinetic theory of Gases gives the definition of pressure as the momentum exchange between the molecules or between the molecules and solid boundaries. in other words, it is another thermodynamic property of matter. If you adopt this definition then there will be a pressure whether we treat the flow as incompressible or compressible or whatever. It is a property of matter. Good Luck.
otd March 10, 2009 19:51
Re: Pressure enigma.
But the original question was about INCOMPRESSIBLE flow.
kenn k.q. zhang March 11, 2009 21:22
Re: Pressure enigma.
i suggest you not to try these kinds of approaches.
in fundamenatal and long-standing things, it's better to use something popular; once you reach to the front of the research field, try hard something unconventional
jwm March 16, 2009 09:29
The original question was about the PHYSICAL interpretation of pressure, so forget Lagrange multipliers.
Suppose you have a fluid completely at rest. The molecules are moving at high speed with random thermal motion. The molecules collide either with a real surface, or with the molecules on the other side of an imaginary surface. The average of these forces of collision give a normal force (the tangential forces cancel statistically) called the pressure.
Now, if there is a velocity gradient the tangential forces do not quite cancel and a normal and a shear force are exerted on the surface. The normal force is called pressure and the tangential force is called viscosity.
Incompressible or compressible - same thing
Jonas Holdeman March 16, 2009 17:13
No, John M, the question was
"1. What is the physical interpretation of the pressure in an incompressible fluid?"
The key word here is "incompressible." We must distinguish between "incompressible flow" (a property of real fluids) and an "incompressible fluid" (which is an idealized concept). The incompressible fluid provides a model for incompressible flow. Being a concept, we do not have access to an incompressible fluid to experiment with and experimentally infer its properties. The closest thing in nature to an incompressible fluid might be the core of a neutron star, but we don't have access to that. So we must be careful of applying intuition, and examine the fluid from its mathematical description.
There are (at least) two time (and spacial) scales associated with a real fluid, one associated with the speed of the fluid and one associated with the speed of sound c(pressure variations) in the fluid. We may charactize the relation between these two by the Mach number . For very small Mach number the nonstationary equations become stiff and computations become difficult. Taking the limit as eliminates one of these scales giving the incompressible NSE and better-conditioned computations, but at a price.
If we expand the density , pressure and velocity in powers of and collect terms with the same power, we find that to lowest order, the pressure appearing in the momentum equation (which is now the incompressible NSE) is the coefficient of , two orders higher than the density or velocity. Thus an important part of what was the physical pressure is missing in the INSE momentum equation. So we should not be surprised that the pressure in the momentum equation does not behave quite like the physical pressure. Since leading pieces (which lead via energy conservation to the continuity equation) are missing, this cannot be a thermodynamic pressure.
So what is the interpretation of this pressure? I can say a lot about it, but I am not entirely comfortable saying what it is in physical terms.
Ford Prefect March 17, 2009 06:11
Quote:
Originally Posted by Jonas Holdeman (Post 209664) No, John M, the question was "1. What is the physical interpretation of the pressure in an incompressible fluid?" The key word here is "incompressible." We must distinguish between "incompressible flow" (a property of real fluids) and an "incompressible fluid" (which is an idealized concept). The incompressible fluid provides a model for incompressible flow. Being a concept, we do not have access to an incompressible fluid to experiment with and experimentally infer its properties. The closest thing in nature to an incompressible fluid might be the core of a neutron star, but we don't have access to that. So we must be careful of applying intuition, and examine the fluid from its mathematical description. There are (at least) two time (and spacial) scales associated with a real fluid, one associated with the speed of the fluid and one associated with the speed of sound c(pressure variations) in the fluid. We may charactize the relation between these two by the Mach number . For very small Mach number the nonstationary equations become stiff and computations become difficult. Taking the limit as eliminates one of these scales giving the incompressible NSE and better-conditioned computations, but at a price. If we expand the density , pressure and velocity in powers of and collect terms with the same power, we find that to lowest order, the pressure appearing in the momentum equation (which is now the incompressible NSE) is the coefficient of , two orders higher than the density or velocity. Thus an important part of what was the physical pressure is missing in the INSE momentum equation. So we should not be surprised that the pressure in the momentum equation does not behave quite like the physical pressure. Since leading pieces (which lead via energy conservation to the continuity equation) are missing, this cannot be a thermodynamic pressure. So what is the interpretation of this pressure? I can say a lot about it, but I am not entirely comfortable saying what it is in physical terms.
Interesting discussion. So where does this put compressible flows, seeing that we use the thermodynamic pressure in the cNSE, which magnitude depends on effects we do not model? Is it one part "physical reasoning" and one part numerical?
Ahmed March 17, 2009 22:49
Quote:
Originally Posted by Jonas Holdeman (Post 209664) No, John M, the question was "1. What is the physical interpretation of the pressure in an incompressible fluid?" The key word here is "incompressible." We must distinguish between "incompressible flow" (a property of real fluids) and an "incompressible fluid" (which is an idealized concept). The incompressible fluid provides a model for incompressible flow. Being a concept, we do not have access to an incompressible fluid to experiment with and experimentally infer its properties. The closest thing in nature to an incompressible fluid might be the core of a neutron star, but we don't have access to that. So we must be careful of applying intuition, and examine the fluid from its mathematical description. There are (at least) two time (and spacial) scales associated with a real fluid, one associated with the speed of the fluid and one associated with the speed of sound c(pressure variations) in the fluid. We may charactize the relation between these two by the Mach number . For very small Mach number the nonstationary equations become stiff and computations become difficult. Taking the limit as eliminates one of these scales giving the incompressible NSE and better-conditioned computations, but at a price. If we expand the density , pressure and velocity in powers of and collect terms with the same power, we find that to lowest order, the pressure appearing in the momentum equation (which is now the incompressible NSE) is the coefficient of , two orders higher than the density or velocity. Thus an important part of what was the physical pressure is missing in the INSE momentum equation. So we should not be surprised that the pressure in the momentum equation does not behave quite like the physical pressure. Since leading pieces (which lead via energy conservation to the continuity equation) are missing, this cannot be a thermodynamic pressure. So what is the interpretation of this pressure? I can say a lot about it, but I am not entirely comfortable saying what it is in physical terms.
This is a very old discussion that dates back to the 19th centuray and when the Navier Stokes equation was written.
But it would be of benfit if we recall the origin of the incompressible flow definition, so here it is, Engineers have found that in a process where changes in the density of a fluid are less than 3% can be neglected and hence we can treat the flow as a constant density fluid/flow. The result of this engineering criterion is that the divergence is Zero and hence the simplified Navier Stokes equation.
In short, when that criterion is adopted, we say that density changes are neglected and nothing more than that. (bear in mind that density changes may be the result of pressure or temperature changes or both) but for engineering design purposes we neglect that change in the magnitude of the density but not in the causes of that change.
Hope that will help
All times are GMT -4. The time now is 15:36. | {"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": 0.9175254702568054, "perplexity": 595.8356518787601}, "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/1461862047707.47/warc/CC-MAIN-20160428164727-00124-ip-10-239-7-51.ec2.internal.warc.gz"} |
https://datascience.stackexchange.com/questions/23690/fractions-or-probabilities-as-training-labels | # Fractions or probabilities as training labels
This it a problem that has come on my path a few times now and I don't have a satisfying solution yet. The goal is to predict probabilities or fractions based on some $x$ where our training $y$ has these probabilties or fractions and thus is in the domain $[0,1]$ as opposed to $\{0,1\}$. My question is with regards to my loss function. In the case of fractions, if the error between 0.4 and 0.5 and the error between 0.89 and 0.99 is the same I can just use MSE if I want to predict the expected value.
In case of probabilities where we want to approach it similarly as classification problems, where the difference between 0.89 and 0.99 is much bigger than that of 0.4 and 0.5, we want to put this in our loss function. Does cross entropy still work properly if I feed it fractions in $y$?
$\mathcal{L}(y,\hat{y})=-y\log(\hat{y}) - (1-y)\log(1-\hat{y})$
Let's say our $y=0.5$ and our current prediction is $\hat{y}=0.6$ we would get:
$\mathcal{L}(0.5,0.6)=-0.5\log(0.6) - 0.5\log(0.4)$
I don't really see why this would go wrong? The function is still convex. Everywhere it says that the target should be in $\{0, 1\}$ however. Maybe my math is lacking or I'm missing something obvious, why is this a bad idea?
Cross-entropy loss still works with probabilities in $[0,1]$ as well as $\{0,1\}$. Most importantly, $\hat{y} = y$ is still a stationary point (although it will not equal $0$).
It is also the case that the possible improvement in loss (and immediate gradient) is larger for $\hat{y} = 0.99, y = 0.89$ than for $\hat{y} = 0.4, y = 0.5$. If you use a sigmoid output, then the gradient at the logit $\hat{y} - y$ still applies - the larger gradient of the loss function scales inversely to the lower gradient of the sigmoid at that point.
So, in short, yes use binary cross-entropy loss for single-class probabilities, even when they are not strictly in $\{0,1\}$. | {"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": 0.941820502281189, "perplexity": 211.53655240629803}, "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/1664030336978.73/warc/CC-MAIN-20221001230322-20221002020322-00613.warc.gz"} |
https://en.wiktionary.org/wiki/logarithm | # logarithm
## English
Wikipedia has an article on:
Wikipedia
### Etymology
From New Latin logarithmus, term coined by Scot mathematician John Napier from Ancient Greek λόγος (lógos, word, reason) and ἀριθμός (arithmós, number); compare rational number, from analogous Latin.
### Pronunciation
• (US) IPA(key): /ˈlɑ.ɡə.ɹɪ.ð(ə)m/, /ˈlɑɡəɹ.ɹɪ.ðəm/, /ˈlɑɡ.ə.ɹɪðm/, /ˈlɑɡ.əɹ.ɹɪðm/
• Hyphenation: log‧a‧ri‧thm
### Noun
logarithm (plural logarithms)
1. (mathematics) For a number ${\displaystyle x}$, the power to which a given base number must be raised in order to obtain ${\displaystyle x}$. Written ${\displaystyle \log _{b}x}$. For example, ${\displaystyle \log _{10}1000=3}$ because ${\displaystyle 10^{3}=1000}$ and ${\displaystyle \log _{2}16=4}$ because ${\displaystyle 2^{4}=16}$.
For a currency which uses denominations of 1, 2, 5, 10, 20, 50, 100, 200, 500, 1000, etc., each jump in the base-10 logarithm from one denomination to the next higher is either 0.3010 or 0.3979. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 7, "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": 0.8890751600265503, "perplexity": 4105.97882387396}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891814538.66/warc/CC-MAIN-20180223075134-20180223095134-00271.warc.gz"} |
http://physics.stackexchange.com/questions/31474/higgs-field-requires-a-large-cosmological-constant-does-the-zero-point-field/31495 | # Higgs field requires a large cosmological constant — does the Zero Point Field balance it?
I just read Wolfram's blog post on the Higgs discovery.
Still, there’s another problem. To get the observed particle masses, the background Higgs field that exists throughout the universe has to have an incredibly high density of energy and mass. Which one might expect would have a huge gravitational effect—in fact, enough of an effect to cause the universe to roll up into a tiny ball. Well, to avoid this, one has to assume that there’s a parameter (a “cosmological constant”) built right into the fundamental equations of gravity that cancels to incredibly high precision the effects of the energy and mass density associated with the background Higgs field.
Then I recalled that one of the great unsolved problems in physics is why the zero-point energy of the vacuum predicts a very large cosmological constant which is not observed.
The language used to describe these two effects confuses me, but as far as I can tell, Higgs->contraction and ZPF->expansion
Any chance these two effects are in balance?
- | {"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": 0.886253297328949, "perplexity": 142.02915030714217}, "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/1429246651873.94/warc/CC-MAIN-20150417045731-00260-ip-10-235-10-82.ec2.internal.warc.gz"} |
https://projecteuclid.org/euclid.aoms/1177706898 | The Annals of Mathematical Statistics
Estimates of the Mean and Standard Deviation of a Normal Population
W. J. Dixon
Abstract
Several simple estimates of the mean and standard deviation of a normal population are discussed. The efficiencies of these estimates are compared to the sample mean and sample standard deviation and to the best linear unbiased estimates. Little efficiency is lost when simple rather than optimum weights are used. Since moments of the order statistics are now available for samples of sizes $N \leqq 20$ from normal populations [3] it is a simple matter to find the variances of linear combinations of order statistics. The sample values are denoted $X_1 \leqq X_2 \leqq X_3 \leqq \cdots \leqq X_N$.
Article information
Source
Ann. Math. Statist., Volume 28, Number 3 (1957), 806-809.
Dates
First available in Project Euclid: 27 April 2007
https://projecteuclid.org/euclid.aoms/1177706898
Digital Object Identifier
doi:10.1214/aoms/1177706898
Mathematical Reviews number (MathSciNet)
MR91590
Zentralblatt MATH identifier
0082.13605
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": 0.8460453748703003, "perplexity": 522.9407840435074}, "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-2019-09/segments/1550247481624.10/warc/CC-MAIN-20190217051250-20190217073250-00119.warc.gz"} |
https://math.eretrandre.org/tetrationforum/showthread.php?tid=424&pid=7869 | • 1 Vote(s) - 5 Average
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using sinh(x) ? tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 11/28/2012, 08:07 PM (This post was last modified: 11/28/2012, 08:11 PM by tommy1729.) Another thing I would like to talk about is the following property. I will start with an example. Let x be a real. Consider $f(n,x)=sin(n^2 x)/n$. If n goes to oo the limit gives $f(oo,x)=f(x)=0$. Now intuitively one would expect about the derivative with respect to x that $D f(x) dx =$ lim n-> oo $D f(n,x) dx = 0.$ Clearly $D f(x) dx = D 0 dx = 0.$ However $D sin(n^2 x)/n = n cos(n^2 x)$ so lim n-> oo $D f(n,x) dx$ =/= 0 ! This is an important and classic lesson in math. So many of our iterations used here require formal and carefull analysis. If you are not convinced notice F(n,x) = g^[n](Q(n,x)) IS something that occurs in the majority of limits related to tetration including e.g. tommysexp , basechange , interpolation methods , matrix methods , iterations to improve on riemann mappings , ... ! I do not know if $D^m F(n,x) dx = D^m F(x) dx$ has been proven here for any method ? Also note that I took x to be a real. So this is about Coo functions defined over the reals but for analytic functions defined over the complex number the issue is even bigger. Also note that in my example both f(n,x) and f(x) were Coo , in fact even analytic. So being analytic or Coo does not solve this issue. I think we need to work on this. --- For those clueness on how to such things , I point out that putting boundaries on values or signs and using the intermediate value theorem frequently does miracles. tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 12/17/2012, 04:34 PM I was wondering about the following conjectured uniqueness criterion for real-analytic half iterates ( easily generalized to all iterates btw) : (based upon my sinh method) for all real x: (D is the derivative operator) D exp^[1/2](x) > 0 D^2 exp^[1/2](x) > 0 exp^[1/2](x) - ln(ln(2sinh^[1/2](exp(exp(x))))) > 0. It seems intuitive , especially considering how the sinh method works. However tetration is sometimes (or often) counterintuitive. A stronger conjecture would be : D exp^[1/2](x) > 0 D^2 exp^[1/2](x) > 0 ln(ln(2cosh^[1/2](exp(exp(x))))) > exp^[1/2](x) > ln(ln(2sinh^[1/2](exp(exp(x))))) Although this requires some concept of cosh^[1/2]. ( I once considered a method based upon cosh but made a mistake ) Im not even sure about the uniqueness of these uniqueness criterions. For instance how this might be related to D^n exp^[1/2](x) > 0 for integer n with n>1 and/or curvature and/or length conditions. Intuitively it seems like f(x) = exp^[1/2](x) computed by the sinh method gives the shortest path/length between f(a) and f(b) for sufficiently large real a and b but im not even sure if that is a selfconsistant statement for ANY exp^[1/2](x). Also clearly shortest length and smallest curvature relate alot. Recently I posted a new thread that involved the point w := f ' (w) = 1 which might relate to uniqueness. Clearly there is still work. And I must admit its not all clear to me yet. Tetration is calculus on drugs regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 02/10/2015, 12:02 AM To extend the idea to smaller bases see this link : http://math.eretrandre.org/tetrationforu...21#pid7621 regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 04/22/2015, 12:27 PM Numerical test Use the Taylor expansions at 3,4,5 and see if they agree on the values of f(3),f(4),f(5) for say f= semi-exp. Regards Tommy1729 tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 04/23/2015, 04:49 PM Recently i started 2 threads about analytic like 2sinh type functions. Regards Tommy1729 tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 10/26/2015, 12:56 AM The 2sinh method is for bases larger than exp(1/2). There are easy analogues for bases > eta. For instance replace 2sinh with t(x) = exp(x) - 1/2 [ exp(-x) + exp( (3 - 2e) x) ]. Notice t(x) is close to 2sinh , is bijective from R to R and has Taylor t(x) = 0 + e x + O(x^2). Regards Tommy1729 tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 11/26/2015, 11:56 PM There are many ways to solve f(f(x)) = exp(x) or g(g(x)) = 2sinh(x). If g < f then g is uniquely so. Also using the 2sinh method on g gives us f. NOTICE it is impossible to directly solve for g in the 2sinh method. So we must use g < f. This implies a bijection between g and f , by the 2sinh or equivalently the inequality. Since there exist both analytic half iterates for both 2sinh and exp , for Every f there is a g and vice versa , we get the next BIG conclusion ; There MUST exist analytic solution of type 2sinh method. --- That is nice. But one wonders if f and g related as above must 1) both be analytic or both not. ? In other words is the bijection actually univalent ? I kinda repeat myself , but I assumed it belongs here better , and it was not understood by most. Regards tommy1729 sheldonison Long Time Fellow Posts: 626 Threads: 22 Joined: Oct 2008 11/27/2015, 03:41 PM (This post was last modified: 11/27/2015, 07:34 PM by sheldonison.) (11/26/2015, 11:56 PM)tommy1729 Wrote: There are many ways to solve f(f(x)) = exp(x) or g(g(x)) = 2sinh(x). If g < f then g is uniquely so. Also using the 2sinh method on g gives us f. .... But one wonders if f and g related as above must 1) both be analytic or both not. If g=2sinh(z), and f=exp(z), One can start with the analytic Abel function $\alpha_g(z)\;\;$ and the superfunction for g. $\alpha^{-1}_g(z)$. For example, the abel and superfunction for 2sinh(z) can be developed using Koenig's method. Now this gives us $g^{[0.5]}(z) = \alpha_g^{-1}(\alpha_g(z)+0.5)$ We wish to generate the superfunction for f(z) from $\alpha_g\;\alpha_g^{-1}$, with the equation below. In the case at hand this limit for the superfunction of f converges at the real axis, but it is not analytic; it does not converge in the complex plane. So far, this seems always be the case for arbitrary exponentially growing functions, f and g. $\alpha^{-1}_f(x)\; = \; \lim_{n \to \infty} \; f^{[-n]}( \alpha^{-1}_g(x+n))\;\;\;$the superfunction for f(z) So the resulting half iterate of f developed in this way is not analytic either, even thought it converges at the real axis $f^{[0.5]}(z) = \alpha_f^{-1}(\alpha_f(z)+0.5)$ - Sheldon tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 11/28/2015, 01:21 PM I think you are confused. Sorry. But it seems you are not agreeing with what I said ... For dubious reasons. Again the 2sinh starts at real x > 0. Then we take the Limit of the continuations of the n th step iteration. NOT the continuation of the Limit of the n th step iteration. Afterwards we can take analytic continuation of THAT limit. Depending on the super/Abel you use for 2sinh this works infinitely AND fails infinitely often. AND .. Not or. Regards Tommy1729 sheldonison Long Time Fellow Posts: 626 Threads: 22 Joined: Oct 2008 11/29/2015, 03:03 PM (This post was last modified: 11/29/2015, 03:03 PM by sheldonison.) Given half iterates, f(f(x))=exp(x), and g(g(x))=2sinh(x), which uniquely satisfy the Op's relationship, then the obvious conjecture from my post would be that only one of the two (f,g) can be analytic. If one of them is analytic, then the other is infinitely differentiable and nowhere analytic, so that the other does not converge to its Taylor series anywhere. This is kind of an interesting conjecture; that Tommy might be inclined to agree with based on earlier posts in this thread (see Tommy's post#61 for example). As far as "dubious reasons" go, I was merely answering Tommy's question, "both be analytic or both not.", as well as searching for cool mathematical truths. - Sheldon « Next Oldest | Next Newest »
Possibly Related Threads... Thread Author Replies Views Last Post exp^[3/2](x) > sinh^[1/2](exp(x)) ? tommy1729 7 5,252 10/26/2015, 01:07 AM Last Post: tommy1729 2*sinh(3^h*asinh(x/2)) is the superfunction of (...) ? Gottfried 5 4,952 09/11/2013, 08:32 PM Last Post: Gottfried zeta and sinh tommy1729 0 1,809 05/30/2011, 12:07 PM Last Post: tommy1729
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https://tex.stackexchange.com/questions/51757/how-can-i-use-tikz-to-make-standalone-svg-graphics/71648 | # How can I use TikZ to make standalone (SVG) graphics?
I would like to use TikZ in other settings than TeX and I'd especially like to use the drawings on websites where the text should be searchable and selectable. However, I'm unsure of the best way to convert my drawings to SVG.
How do I best take some TikZ code, render the drawing and turn the output into SVG?
You can use the standalone class to produce tight PDF files for one or multiple TikZ pictures. I originally wrote it to simplify the creation of the many pictures of my thesis. Since v1.0 it includes a convert option which can convert the produced PDF into a graphics file automatically (using external software, which requires the -shell-escape compiler option).
This is very similar to Compile a LaTeX document into a PNG image that's as short as possible, but SVG needs some extra care.
You can write your TikZ pictures the following way:
\documentclass[tikz,convert={outfile=\jobname.svg}]{standalone}
\begin{document}
\begin{tikzpicture}% Example:
\draw (0,0) -- (10,10); % ...
\draw (10,0) -- (0,10); % ...
\node at (5,5) {Lorem ipsum at domine standalonus};
\end{tikzpicture}
\end{document}
Then either you compile the file as usual with pdflatex or another latex and convert the PDF to a SVG manually or compile it with the -shell-escape option and let standalone convert it for you.
Manual conversion can be done with a number of tools. It is simpler under Linux, because these tools are easily available there, but should be possible under Windows as well. (The convert options isn't really tested under Windows, btw.) By default standalone uses Image Magick's convert, which can do PDF to SVG but will not always give you good results. The pdf2svg tool seems to be better suited, but isn't supported out-of-the-box by standalone yet. It can of course be used manually as shown in Exporting all equations from a document as individual svg files.
You can configure standalone to use pdf2svg directly by using the command key of the convert option. Unfortunately, there is a small bug in standalone preventing it. I just fixed that and will upload the new version today.
With this you can write:
\documentclass[crop,tikz,convert={outext=.svg,command=\unexpanded{pdf2svg \infile\space\outfile}},multi=false]{standalone}[2012/04/13]
\makeatletter
\begin{document}
\begin{tikzpicture}
\draw (0,0) -- (10,10); % ...
\draw (10,0) -- (0,10); % ...
\draw (5,0) -- (0,10); % ...
\node at (5,5) {Lorem ipsum at domine standalonus};
\end{tikzpicture}
\end{document}
The \unexpanded is required because LaTeX expands class options. You can also add \noexpand before every macro instead.
If you need this more often you can also use a standalone.cfg file which enables this for all (local) standalone files. Simply create this file as follows in the same directory:
% Local standalone.cfg file
\standaloneconfig{convert={command={pdf2svg \infile\space\outfile}}}
I might add a special pdf2svg key in the next version as well, so you only need to write the following then:
\documentclass[crop,tikz,convert=pdf2svg]{standalone}[2012/04/13]
% ...
• The new version v1.0b is now in CTAN. Mirrors might need up to 24 hours to sync. TeXLive should have it inside the next few days. – Martin Scharrer Apr 13 '12 at 13:01
• Escaping some special characters in the command key seems to be too complicated. That is why I prefer to invoke the conversion outside the standalone. – kiss my armpit Mar 9 '13 at 21:33
• I am sorry, but ImageMagick produces raster images only, see Avoid using ImageMagick for 'Vector Image' to 'Vector Image' conversions EG: converting between formats like: PDF, PS, SVG imagemagick.org/Usage/formats/#vector pdf2svg is the only way to go – escalator Feb 18 '15 at 7:35
• @escalator: Thanks for the link. I actually didn't know that. Note that standalone uses ghostscript on Linux by default (ImageMagick by defauly only on Windows), which should be able to create true vector graphics. That is also the reason I didn't noted the issue with IM, I'm a Linux user ;-) – Martin Scharrer Feb 18 '15 at 20:48
• I tried to compile your latex file with pdflatex -shell-escape filename.tex. It generated an svg image. But the size was horrible: 17 MB. pdf was 13 KB. So, this method of conversion seems useless. Or I did something wrong? – user4035 Sep 3 '15 at 12:38
I use tex4ht and set the PGF output format to SVG. This solution comes from page 110 of the pgfmanual http://mirrors.ctan.org/graphics/pgf/base/doc/generic/pgf/pgfmanual.pdf.
I've only used it to create SVG of a TikZ picture in an otherwise empty \documentclass{article} but it looks like you could use this to make html with SVG graphics of a large document.
The advantage of this approach is that the SVG is produced by PGF and you know you're getting vector graphics. Also you get the result in a single step. It won't do functional shading or matricies and text in the pictures can be a problem but there's more on fixing that in the pgfmanual.
• In the TeX or LaTeX document preamble before you load the TikZ package, e.g. with \usepackage{tikz} in LaTeX type:
\def\pgfsysdriver{pgfsys-tex4ht.def}
• Then process the TeX or LaTeX with httex or htlatex as appropriate. You may need to add tex4ht to your tex installation; it's at http://www.tug.org/applications/tex4ht/. For example to process my file called logoname.tex I do
htlatex logoname.tex
• The following output files are created in the current directory
logoname.html
logoname.css
logoname-1.svg
If you have more TikZ pictures in the document I assume they would become logoname-2.svg and so on.
I'm able to look at the SVG output in firefox and inkscape so it seems to produce good results.
• This things works for me (with $htlatex file.tex, not with pdflatex)! (Cf. the same solution suggested in this answer and a not on using htlatex) As I had some problems with installing pdf2svg (I am on OS X), this answer solves my problem whereas the most upvoted/accepted - does not. – Piotr Migdal Apr 22 '14 at 15:55 • This generally works, but one needs to be aware of a bug in the tex4ht driver of pgf 3.0.0. See sourceforge.net/p/pgf/bugs/327 There's also a thread on tex.SE about this, but I can't remember where it is. – Fizz Aug 12 '14 at 19:56 On linux, you can use pdf2svg (an opensource tool). All glyphs are converted to paths, thus you can't edit your text. But it is the only tool that seems to give good results for images mixing drawings and texts. In your MWE, to get smooth result, I add the smooth option to plot. Here is a snapshot of the svg file rendered by Firefox (click on the image to download the SVG file): Here the two commands used to convert TEX to SVG: pdflatex file.tex pdf2svg file.pdf file.svg Your MWE with my changes: \documentclass{standalone} \usepackage{tikz} \usepackage{lmodern} \usepackage[T1]{fontenc} \begin{document} \begin{tikzpicture}[domain=-2:2,samples=100,scale=1.0,>=latex] \tikzset{bgrid/.style={help lines,color=blue!10,very thin}} \draw[bgrid] (-1.5,-3.5) grid (7.5,3.5); \draw[<->, color=black] (-1.5,0) -- (7.5,0) node[right] {$x$}; \draw[<->, color=black] (0,-3.5) -- (0,3.5) node[above] {$y$}; \foreach \x/\xtext in {-1,1,2,3,4,5,6,7} \draw (\x cm,1pt) -- (\x cm,-1pt) node[anchor=north] {$\xtext$}; \foreach \y/\ytext in {-3,-2,-1,1,2,3} \draw (1pt,\y cm) -- (-1pt,\y cm) node[anchor=east] {$\ytext$}; \draw[thick,color=black,domain=0:7.5,smooth] plot (\x,{sqrt(\x)}) node[anchor=south] {$y = \sqrt{x}$}; \draw[dashed,color=black,domain=0:7.5,smooth] plot (\x,{(-1)*(sqrt(\x))}) node[anchor=north] {$y = -\sqrt{x}$}; \draw[thick,color=black,domain=-1.5:5.5,samples=3] plot (\x,{(\x)-2}) node[anchor=south] {$y = x - 2$}; \filldraw[black] (4,2) circle(2pt) node[anchor=south east] {$(4, 2)$}; \filldraw[red] (1,-1) circle(2pt); \draw[red] (1.5,-1) node[anchor=west] {$(1, -1)\$};
\end{tikzpicture}
\end{document}
• I have tried to implement your suggestion, but this doesn't work for me in MiKTeX in Windows. I can produce a pdf file as usual, but when I use the command pdf2svg file.pdf file.svg I get the following error message: 'pdf2svg' is not recognized as an internal or external command, operable program or batch file. Using the commands htlatex or mzlatex produce html files that do not have the graphics in them, and do not produce svg files. – Santo D'Agostino Sep 16 '12 at 19:29
• @SantoD'Agostino: To my knowledge, opensource pdf2svg exists only for linux. – Paul Gaborit Sep 16 '12 at 21:27
• Sorry to hear this, Paul, but thanks very much for your time anyway. I now face the question of whether I should transfer my latex operation from my Windows machine to my Linux machine; if I do this (which seems inevitable, the only question being when), then your answer will be relevant, and I will store it for use at that time. Thanks again! – Santo D'Agostino Sep 16 '12 at 22:51
• @SantoD'Agostino here is a cross-compiled version, that works under windows for me. Just download, unzip and add the location to the Path system environment variable. – Carsten Aug 24 '18 at 10:42
You can use dvisvgm, which is a built-in tool of some TeX distributions (e.g. MiKTeX).
The basic usage:
latex testsvg.tex
dvisvgm testsvg.dvi
Then the SVG file will be saved in the current directory.
If you converted a simple math formula but get garbage when displaying the SVG, you can run dvisvgm with option --no-fonts to replace the fonts with path elements.
dvisvgm --no-fonts testsvg.dvi
As of version 2.0, you can call dvisvgm with option --font-format=woff, you should get SVG files that render correctly in almost all recent web browsers.
dvisvgm --font-format=woff testsvg.dvi
If you use TeXstudio, you can edit the texstudio.ini file (located in AppData\Roaming\TeXstudio for Windows) and add the following line below [texmaker].
Tools\Commands\dvi2svg="txs:///latex | dvisvgm --font-format=woff %.dvi"
Restart TeXstudio and you can find your custom command at Tools->User->dvi2svg.
• The --font-format=woffis a great hint which is missing from the dvisvgm FAQ. It enabled me to get sensible output in Chrome, Firefox, IE and Edge, while keeping all text as text. – DaveP Apr 27 '18 at 1:38
If I run dvisvgm with option -n and the evaluation of PostScript specials is enabled, I get the expected result:
Since MiKTeX doesn't provide a dvisvgm binary through its repositories, you have to install it manually. Recent builds for MiKTeX are available from the dvisvgm website. Simply extract dvisvgm.exe to the MiKTeX subfolder miktex\bin, or even better, install it in a local texmf tree.
• This doesn't work for me in MiKTeX (Windows). I get a one-line mash up of symbols, and no graph. I placed the dvisvgm.exe in the miktex\bin folder (I'm using MiKTeX 2.8). How do I enable the evaluation of PostScript specials? – Santo D'Agostino Sep 16 '12 at 19:39
• In order to enable the processing of PS specials, you have to install a recent version of Ghostscript, and add the directory containing gsdll32.dll to the PATH environment variable. You can check the availability of PS processing with dvisvgm -l. If the ps entry is present, everything should work as expected. – Martin Sep 17 '12 at 6:09
• This works! Thank-you very much, Martin! All the best! – Santo D'Agostino Sep 17 '12 at 13:18
• You're welcome. I'm glad to hear that you've got it working now. – Martin Sep 17 '12 at 13:34
• Hi Martin, I think I spoke a little too soon. The program works if I use the example file, but if I try to use standalone with the following first line <br /><br /> \documentclass[tikz,crop=true,border=0.5cm]{standalone} <br /><br /> then I get an error message <br /><br /> DVI error: no font selected <br /><br /> and no svg file is produced. Any advice? (Using dvisvgm -l shows that the ps entry is present, so that is not the problem.) – Santo D'Agostino Sep 17 '12 at 19:14
You might want to check out a tool I just wrote: tikz2svg.
Given in.tikz (or stdin):
> cat in.tikz
\begin{tikzpicture}
\fill[red!90!black] ( 90:.6) circle (1);
\fill[green!80!black] (210:.6) circle (1);
\fill[blue!90!black] (330:.6) circle (1);
\end{tikzpicture}
It outputs:
> cat tikz2svg < in.tikz
<?xml version="1.0" encoding="UTF-8"?> | {"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": 0.9123871326446533, "perplexity": 3476.337085414384}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575540542644.69/warc/CC-MAIN-20191212074623-20191212102623-00359.warc.gz"} |
https://www.physicsforums.com/threads/what-energy-density-expected-at-lhc-point-of-collision.171170/ | # What energy density expected at LHC point of collision?
1. May 21, 2007
### marcus
I'm more used to seeing colliders rated by the kinetic energy of the colliding particles, but just recently I've been seeing estimates of the ENERGY DENSITY to be expected right at the point of collision.
This raises doubts about exactly what is meant. what is the nominal area of collision? what nominal volume is the kinetic energy spread out in?
but there should be some way to give a plausible rough figure for the energy density. experimental physicists are good at that kind of guesstimation.
So what do our experts in this department say?
Would you say it is closer to 10 GeV per cubic fermi
or to, say, 1016 GeV per cubic fermi?
Can you offer guidance or do you also need help?
Draft saved Draft deleted
Similar Discussions: What energy density expected at LHC point of collision? | {"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": 0.8971245884895325, "perplexity": 1292.7029108648924}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891815560.92/warc/CC-MAIN-20180224112708-20180224132708-00503.warc.gz"} |
https://www.physicsforums.com/threads/coefficient-of-restitution-problem.907173/ | # Homework Help: Coefficient of Restitution problem
1. Mar 10, 2017
### CCC1212
Hello, I have this simple COR problem that I seem to be a bit stuck on.
1. The problem statement, all variables and given/known data
A 1.25 kg squash racquet swung at 37 m/s contacts a ball moving at 42 m/s in the opposite direction. Immediately after impact the racquet has a velocity of 25 m/s and the ball has a velocity of 50 m/s. What is the coefficient of restitution associated with the impact?
2. Relevant equations
I've been using e= (V1-V2)/(U1-U2)
3. The attempt at a solution
When I use this equation I get 5, but obviously that's wrong since it shouldn't be greater than 1.
Can anyone point me in the right direction? Thanks so much
2. Mar 10, 2017
### PeroK
The operative word is direction!
3. Mar 10, 2017
### CCC1212
I sort of had a feeling this may be the case. Should I be using "U1+U2" instead of "U2-U1". Or am I way off?
4. Mar 10, 2017
### PeroK
Then depends on the direction of $U_1$ and $U_2$. Are you using speeds or velocities?
5. Mar 10, 2017
### CCC1212
Velocities since it says 'in the opposite direction' I'm assuming.
6. Mar 10, 2017
### PeroK
So, what are the initial velocities of the racket and ball?
7. Mar 10, 2017
### CCC1212
They are 37 m/s (racket) and 42 m/s (ball)
Ahhh I think I see what you may be getting at. Because it says the ball is moving in the opposite direction, should I be writing the equation like...
50-25/37- -42
8. Mar 10, 2017
### PeroK
No, for velocities that can't be correct if they are in opposite directions.
Only if you know why!
9. Mar 10, 2017
### CCC1212
Could you explain that statement. I thought the difference between speed and velocity is that velocity considers direction?
10. Mar 10, 2017
### PeroK
When you have a problem in 1D it's up to you to decide what direction is positive and what direction is negative. In this case I might decide that the racket is moving in the positive direction, which means that the velocity of the racket is $37m/s$.
The ball is moving in the opposite direction, so its velocity must be $-42 m/s$.
11. Mar 10, 2017
### CCC1212
Ahhh of course. Thanks so much! | {"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": 0.9139354228973389, "perplexity": 855.9292222578765}, "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/1531676590046.11/warc/CC-MAIN-20180718021906-20180718041906-00194.warc.gz"} |
https://byjus.com/question-answer/reaction-a-2b-c-rightarrow-product-follows-the-rate-law-displaystyle-frac-d-left-c/ | Question
Reaction $$\:A+2B+C\rightarrow product$$, follows the rate law$$\displaystyle-\frac{d\left[C\right]}{d\,t}\,=\,k\left[A\right]^2$$False statement regarding the above reaction is :
A
On doubling the conc. of B and C the rate of the reaction remains unaffected
B
Reducing the conc. of A to half, the rate becomes one-fourth
C
Half life period of the reaction depends upon the conc. of B
D
Half life period of the reaction is inversely proportional to the first power conc. of A
Solution
The correct option is C Half life period of the reaction depends upon the conc. of BHalf life period of the reaction depends upon the concentration of A and is independent of the concentration of B. ChemistryNCERTStandard XII
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View More | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9640202522277832, "perplexity": 3443.1654899529676}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320300573.3/warc/CC-MAIN-20220129062503-20220129092503-00697.warc.gz"} |
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http://mathhelpforum.com/geometry/139148-vectors-question.html | Math Help - Vectors question
1. Vectors question
I can't finish this attached vectors question.
Attached Thumbnails
2. Originally Posted by Stuck Man
I can't finish this attached vectors question.
Dear Stuck Man,
There is a calculation error in your result for $\overline{PQ}$. It must be, $\overline{PQ}=\frac{1}{2}\underline{b}-\frac{3}{10}\underline{a}$.
After you correct this you can equate the two results you obtained for $\overline{QR}$. Then considering the coefficients of $\underline{a}~and~\underline{b}$ you can find k and n.
3. I thought that might be wrong earlier. I still don't know how to find n and k.
4. Originally Posted by Stuck Man
I thought that might be wrong earlier. I still don't know how to find n and k.
Dear Stuck Man,
$\overline{QR}=\frac{n}{10}(5\underline{b}-3\underline{a})=\frac{1}{2}(1+2k)\underline{b}-\frac{1}{2}\underline{a}$
Now, you can equate the coefficients of $\underline{a}~and~\underline{b}$.
5. I have done it now. My mistake earlier did not help. I originally made an equation from the two expressions fo r the vector QR. Thanks. | {"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": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.948655366897583, "perplexity": 877.7692922426554}, "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-2015-48/segments/1448398449793.41/warc/CC-MAIN-20151124205409-00008-ip-10-71-132-137.ec2.internal.warc.gz"} |
https://mathoverflow.net/questions/243490/how-do-these-primes-jump | # How do these primes jump?
Update 2017.08.28: I am still looking for references. I have posted a request to https://cs.stackexchange.com/q/79971 which includes some literature references I found which are of interest but still miss the mark for this question. End Update.
Edit 2018.08.08 This answer https://mathoverflow.net/a/307881 will be updated to give recent information about S, especially a forthcoming preprint. End Edit 2018.08.08
I have several questions regarding the analysis, behaviour, and expression of a simple sieving algorithm which uses associative arrays. The pseudocode below assumes integer addition, string concatenation, checking that an index (key) exists in an array (so at the beginning, (n in c) is false for all n, but when c[m]=p is carried out, then (m in c) returns true), and sufficient memory. (Or set LIM to 100.)
for n = 2 to LIM
p = n
if (n in c) p = c[n]
m = n + p
while (m in c) m = m + p
c[m] = p
t[p] = t[p] "," m
(Compare a similar algorithm which in part computes the list of distinct prime factors for each n, found here: https://mathoverflow.net/a/50691 .) When this loop is run, at the end c[m] contains a prime factor p of m for each composite number m at most LIM, as well as for some composite m greater than LIM. t contains for each p a comma-delimited string of indices m for which c[m] gets p. It is a nice exercise to show that (at the end of the loop body) p is a prime, that c has only composite indices m, that c[m] when defined is always a prime dividing m, and that t[p] encodes an increasing sequence of some multiples of p. When I simulate this by hand, I imagine the primes p jumping over occupied slots in c until they find an empty slot c[m] and land there.
Question 1: Does this sieve implementation (perhaps slightly modified) exist in the computer science or number theory literature?
I am aware of wheel sieving which is faster, but this algorithm is not too shabby, even if you have to write your own data structure to perform (n in c).
Question 2: What does t[2] look like? Calling the $j$th member $m_j$, how good an approximation can we get to $m_j$ in terms of $j$?
Here are the first few terms of t[2] and of t[3]:
4 8 12 16 24 30 32 40 50 64 78 90 104 108 128 140 156 176 192 208
6 9 15 18 27 36 48 54 63 72 81 96 105 126 144 162 180 189 210 231
I will make a different post with other questions related to this algorithm and variants. If you want to play with it, here is an (untested) version which saves memory by removing c[n] after it is used, and by not storing p values larger than LIM:
BEGIN{ LIM = 100; LIM2 = LIM*LIM
for ( n=1; LIM2 > n++; ) {
if ((m=n) in c) { m += (p=c[n]); delete c[n]
while ( m in c ) m += p }
else m += (p=n)
if (p < LIM ) { c[m] = p; t[p] = t[p] "," m }
}
for ( p=1; p < LIM; p++ ) if (p in t) print t[p]
}
Question 3: How much slower does t[2] grow ( as a function of LIM ) than the sequence above where all primes are "jumped"?
Gerhard "Is Awed By Elegant Simplicity" Paseman, 2016.07.01.
• One key aspect which should be mentioned is that every prime $p$ jumping from $n$ to $n+kp$ must pick $k$ so that the greatest prime factor of $n+kp$ is less than $kp$. This along with good bounds on $kp$ gives a degree of smoothness to the trajectory, and may help with other aspects related to distribution of primes. Gerhard "Could Be Something Really Wide" Paseman, 2016.08.24. – Gerhard Paseman Aug 24 '16 at 18:52
• Yes, those agree with my output as well as the test that the next entry is "jump-length" smooth: The largest prime factor of the next entry is no bigger than the length of the jump from the previous entry. Gerhard "Hopes To Use At Length" Paseman, 2017.03.13. – Gerhard Paseman Mar 14 '17 at 1:07
• I've looked you up on LinkedIn. – user114263 Mar 6 at 16:55
• @user Thanks! I need to finish my current write-up and post a link to it from my profiles. Gerhard "Many Paths To Jumping Primes" Paseman, 2020.03.06. – Gerhard Paseman Mar 6 at 19:07
Just to confirm your list of $t[2]$ and extend it a bit, here is what those terms look like:
$t[2]=$ $$4,8,12,16,24,30,32,40,50,64,78,90,104,108,128,140,156,176,192,208,216,234,250,256,280,304,320,338,350,374,392,420,440,468,486,500,512,540,570,598,630,648,676,704,726,750,768,800,832,858,882,910,950,972,1008,1024,1056,1088,1122,1150,1176,1210,1248,1280,1296,1344,1372,1426,1458,1500,1536,1568,1584,1600,1632,1672,1694,1728,1760,1792,1824,1856,1904,1936,1976,2016 \;.$$
• This looks good! It suggests a subquadratic growth, and in fact I hope to prove that going from t[j] to t[j+1] involves a jump proportional to $\pi(\sqrt{}$t[j] $)$, or about proportional to the number of primes less than the square root of the origin of the jump. Gerhard "Will Square Off With This" Paseman, 2017.03.13. – Gerhard Paseman Mar 14 '17 at 1:16
Here is some information on how the primes jump. It seems the trajectories are somewhat smooth. Skipping numbers seems to be a topic which is a tempting if not low-hanging fruit.
As a prime traverses through its multiples according to the algorithm, it will skip over some multiples because a different prime got there first. For a prime $p$ I define skip($p$) to be the smallest positive multiple of $p$ not in the trajectory of $p$. I will use aliases of $sk(p)$ or $sk$ when $p$ is understood. One can simulate the first few steps to note skip($2$) is $6$, skip($3$) is $12$, skip($5$) is $15$, and skip($7$) is $35$.
Tracking the value of skip($p$) for many primes $p$, I note that $p^2 \lt 3sk(p) \lt 39p^2/11$ for $11 \lt p \lt 4000$. How close can we get to proving such equalities for all primes $p$?
One approach is to show for a fixed integer $k$ which primes $p$ that have $kp$ in their trajectory also have $(k+1)p$ in that trajectory, or which $p$ jump from $k$ to $k+1$. For $k=1$, there are two simple proofs: a) note that $2$ jumps from $1$ to $2$, and note for an odd prime $p$ that there are more even numbers between $p$ and $2p$ than odd primes less than $p$, so there is an even number between $p$ and $2p$ to be part of the trajectory of $2$, giving $p$ a chance to jump to $2p$ before $2$ does, and b) between $p$ and $2p$ is a power of $2$, so $2$ must land somewhere above $p$ and less than or equal to that power of $2$, again leaving $2p$ available for odd $p$.
For $k=2$, jumping from $2$ to $3$ is straightforward if there is a power of $3$ between $2p$ and $3p$. When there isn't, one can use $3$-smooth numbers (numbers whose distinct prime factors are at most $3$) for sufficiently large $p$. In particular, if $r$ is the largest $3$-smooth number less than $2p$ and $r$ is a multiple of $6$, then $2p \lt 4r/3 \lt 3r/2 \lt 3p$ and so there is a place for both $2$ and $3$ to land strictly below $3p$. If $r$ is a power of $2$ at least $8$, then one can use $9r/8$ and $3r/2$ in a similar way, and if $r$ is a power of $3$ at least $27$ then one can use $32r/27$ and $4r/3$. So for all $p$ with $2p \gt 12$, we get a pair of places (which are $3$-smooth numbers) less than $3p$ for both $2$ and $3$ to land. This takes care of $p \gt 7$, while $p=3$ can be settled by hand, leaving $p=2$ and $5$ missing out on jumping from $2$ to $3$, which they do.
In general, jumping from $k$ to $k+1$ involves the set of distinct prime divisors of $k+1$, and one may have to throw in a few more small primes to pull the argument through. Namely, find a nice set $Q$ of primes such that $k+1$ is a $Q$-smooth number (all the prime divisors come from $Q$), and so that if $Q$ has $t$ distinct primes, then find those $p$ such that the interval $(kp, (k+1)p)$ contains at least $t$ $Q$-smooth numbers. Then when $kp$ is processed, all the primes in $Q$ are guaranteed a place to land above $kp$ and below $(k+1)p$, leaving the latter available for $p$.
A problem with this approach is that I don't know an estimate for $Q$-smooth numbers in an interval that has small enough error. A rough estimate, given that there are $t$ many primes $q$ in $Q$ , and involving logarithms to different prime bases $q$, is $$\big[\prod_{q \in Q} (1 + \lfloor\log_q((k+1)p)\rfloor) - \prod_{q \in Q} (1 + \lfloor\log_q(kp) \rfloor)\big]/t!,$$ but I think this is a bad approximation when $k$ and $p$ are close to the same size.
A different approach is to consider what primes would jump to a number between $kp$ and $(k+1)p$. Using the statistics above as a guide, this interval has labels from all the primes in the interval $(kp/2,(k+1)p/2)$, and in general from $(kp/i,(k+1)p/i)$ for $i$ up to some integer hopefully close to $k$. Again, I am unsure that the error in such an estimate is easy to control.
Suppose we are able to determine $G(k)$, a non-decreasing function so that for all $p \geq G(k)$ that $p$ jumps from $k$ to $k+1$. Then define $F(p)$ to be the largest $k$ such that $p \geq G(k)$, and we now have that $F(p)=k$ means the multiples $2p$ up to $(k+1)p$, or skip($p$) $\gt F(p)$. Now the goal is to show something like $G(k)=3k+1$.
In any case, we can give a rough description now of how the labels p are arranged in c. When $n$ is processed with label $p$ dividing $n$ (so c[n] has the value p), we have all the primes $q$ from $n/2$ up to $n$ have jumped to $2q$, and for many small $t$ each prime $q$ from $n/(t+1)$ up to $n/t$ has landed on $(t+1)q$, up to about $t \in O(\sqrt{n})$. This just leaves about $O(\sqrt{n})$ many primes whose location above $n$ is unknown, but are likely below $n + K{\sqrt{n}}^{1+\epsilon}$ for some given value of $\epsilon$ and some small value of $K$ depending only on $\epsilon$. This suggests that the trajectory for a prime grows irregularly but somewhat quadratically. (Since there are at most $O(n/\log(n))$ obstructions for a given prime, we should expect strictly subquadratic growth.) Also suggested is that if $n$ has label $p$ and $p$ is not the largest prime factor of $n$, then the largest prime factor of $n$ is not much bigger than $\sqrt{3n}$, which I suggest as a conjecture.
Additional question: does it make sense to define skip($n$) for composite $n$, and will that help in the trajectory analysis? (I can see the definition depending on the prime assigned to $n$ as well as depending on the set of distinct prime factors.)
Gerhard "More Answers Gives More Questions" Paseman, 2016.08.22.
• We have skip(11)=143 and skip(89)=8277, and many instances where skip(p) is larger than p^2, but not by much. There are fewer instance where skip(p) is less than p^2, with the sample showing that for most remaining primes skip(p) is between (p^2)/2 and 2(p^2)/3. There is probably a combinatorial explanation for this, which I would like to see. Gerhard "Analytic Number Theorems Needn't Apply" Paseman, 2016.08.22. – Gerhard Paseman Aug 22 '16 at 20:28
I've made some headway on Question 3. It is related to my studies of Jacobsthal's function. Although a theory for general square free numbers $Q$ can be developed, I will take $Q=P_n$, the product of the first $n$ primes, for this post. Of course, $2$ can only make a jump of size at most $2n$, but it will rarely make a jump of that size for large $n$.
If I restrict the algorithm to primes dividing $Q$, the union of the trajectories are those numbers not coprime to $Q$, and as the system can be modeled as a finite state automaton, the trajectory of each prime will be periodic with a period related to a multiple of $Q$. More precisely, for a given prime $p$ dividing $Q$, one has that a positive number $m$ is in the trajectory if and only if $m+Q$ is in the trajectory. (Temporarily, I include $p$ in its own trajectory, contrary to the post above.)
To see this, note that $\gcd(Q,Q+p)=p$ for every prime divisor $p$ (and also every divisor) of $Q$. Thus, modulo $Q$, there is only one way to arrange the $n$ primes at the start ( the pattern of primes assigned to the indices in $[mQ+2, mQ+p_n]$ is independent of $m$), and thus the pattern of primes assigned to c repeats modulo $Q$ as well.
It is tempting to conjecture that c[$m$]$=$c[$Q-m$] for positive $m$ less than and not coprime to $Q$, but when $Q=30$ one has c[$12$]$=2$ and c[$18$]$=3$ (and similar examples exist for some larger $Q$). It is also tempting to conjecture that each prime $p$ does slightly less than $Q/p$ jumps in a range of length $Q$, with the difference related to inclusion-exclusion. Simulations show much smaller numbers than this prediction. Note that the longest jump is bounded above by the sum of the $n$ primes, and is likely to be much less. For small values of $n$ $(n\leq 9)$ I see the largest jump as less than $3p_n$. I would not be surprised if the largest jump were of the same order as $g(Q)$, the Jacobsthal function applied to $Q$, which would be conjectured at or near $p_n(\log p_n)^2$ but I would settle for strictly less than $O(n^2)$. In any case, it is apparent that the set of primes stay in a rather tight cluster not much larger than the largest prime as they jump according to the restricted version of the algorithm. How tight is hopefully less hard than various open questions regarding the distribution of prime numbers.
Gerhard "More Jumping Means More Excitement" Paseman, 2016.09.11.
I've decided to bust through my writer's block and post a partial result.
Of course when a prime $p$ jumps, it skims over multiples of $p$ occupied by other primes, and one question is how far a prime can jump from its location at $n$. The brief answer is that $p \cdot f_0(n/p)$ is a strict upper bound, and I am trying to tighten that to $f_0(n/p)$, maybe multiplied by a fractional power of $p$. Here $f_0(n)$ relates to the number $w$ of distinct prime factors of the product of integers in the interval $[n+1,n+k]$, and is the largest integer $k$ for which $w\geq k$. $f_0()$ (and also $f_2()$ which relates to Grimm's conjecture in another post) comes from a 1971 paper of Erdos and Selfridge.
The proof of this upper bound is clear: for each of the multiples of $p$ above $n$ which have a different prime divisor occupying that multiple, we run out of (a non repetitive pattern of) possible prime divisors when we hit $f_0(n/p)$ many such multiples. (Indeed we run out before encountering $1+f_2(n/p)$ many such multiples.) However, some of the primes $q$ involved may actually not be on a multiple of $p$, but occur on a smaller multiple of $q$ above $n$, and it is reasonable (but not well supported) to assume that about $1/p$ of these primes $q$ reside on multiples of $p$.
The nice thing is that the 1971 paper has a proof that $f_0(n)$ is $O(\sqrt{n/\log n})$, and $f_2(n)$ seems not much smaller. I hope to make a proof with explicit constants, and add it to a write-up. I also hope to get an inspiring idea for a lower bound on the length of a jump of $p$ from a large multiple of $p$. However, I find it interesting already that a trajectory of any prime is eventually smoother than somewhat smooth, and hope to use this in further analysis. Also, intriguing is how small the ratio skip(p)/p gets as p gets large.
Gerhard "Going For Block Busting Results" Paseman, 2017.07.05. | {"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": 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": 0.895319938659668, "perplexity": 257.5656906280442}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655886121.45/warc/CC-MAIN-20200704104352-20200704134352-00445.warc.gz"} |
https://www.physicsforums.com/threads/difficult-polynomial-and-determinant-formula.273000/ | # Difficult polynomial and determinant formula
1. Nov 18, 2008
### jostpuur
Anyone having any ideas about how this formula could be proved?
$$\underset{k<l}{\prod_{k,l=1}^n} (x_k - x_l) = (-1)^{n(n-1)/2} \left|\begin{array}{ccc} 1 & \cdots & 1 \\ x_1 & \cdots & x_n \\ \vdots & & \vdots \\ x_1^{n-1} & \cdots & x_n^{n-1} \\ \end{array}\right|$$
I tried induction without success. The induction step seems to only require other difficult formulas.
2. Nov 19, 2008
### morphism
Look up the Vandermonde determinant.
3. Dec 10, 2008
### jostpuur
I have now seen this Vandermonde determinant been mentioned in several places, but I've still not encountered the proof. It seems that most authors prefer mentioning the result instead of proving it.
4. Dec 10, 2008
### Anthony
What can you say about the ith and jth columns if $$x_i=x_j$$? Can you see how to construct a proof from there?
5. Dec 10, 2008
### jostpuur
I can see that in the equation I wrote, both sides are zero precisely when $x_i=x_j$ for some $i\neq j$, but I don't see how this remark would help with a proof yet.
6. Dec 10, 2008
### Anthony
Do you agree that the determinant is a polynomial in the variables $$x_i$$?
7. Dec 10, 2008
### jostpuur
Yes.
8. Dec 11, 2008
### Anthony
What order will the polynomial be in each of the $$x_i$$?
9. Dec 11, 2008
### jostpuur
n-1.
10. Dec 11, 2008
### epenguin
Subtract x1 (or if you prefer xn) times each row from the next row. You'll then find you get a factor in each column which you can take out and is (x2-x1)(x3-x1)...(xn-x1) and the other factor is the next smallest Vandermonde determinant on x2, x3,... xn. You can also formulate the above as matrix multiplication.
The fiddly may be is getting the expression for sign right. Actually I am not sure you have.
11. Dec 11, 2008
### jostpuur
This seems to imply that there exists some constant $C\in\mathbb{R}$ so that
$$\underset{k<l}{\prod_{k,l=1}^n} (x_k-x_l) = C \left|\begin{array}{ccc} 1 & \cdots & 1 \\ x_1 & \cdots & x_n \\ \vdots & & \vdots \\ x_1^{n-1} & \cdots & x_n^{n-1} \\ \end{array}\right|$$
I'll continue thinking.
12. Dec 11, 2008
### jostpuur
By using the formula
$$x_k^m - x_l^m = (x_k - x_l)(x_k^{m-1} + x_k^{m-2}x_l + \cdots + x_k x_l^{m-2} + x_l^{m-1})$$
I managed to do this:
$$\left|\begin{array}{ccc} 1 & \cdots & 1 \\ x_1 & \cdots & x_n \\ \vdots & & \vdots \\ x_1^{n-1} & \cdots & x_n^{n-1} \\ \end{array}\right| = (x_2 - x_1)\left|\begin{array}{ccccc} 1 & 0 & 1 & \cdots & 1 \\ x_1 & 1 & x_3 & \cdots & x_n \\ x_1^2 & x_2+x_1 & x_3^2 & \cdots & x_n^2 \\ \vdots & \vdots & \vdots & & \vdots \\ x_1^{n-1} & x_2^{n-2} + x_2^{n-3}x_1 + \cdots + x_1^{n-2} & x_3^{n-1} & \cdots & x_n^{n-1} \\ \end{array}\right| = \cdots$$
$$\cdots = (-1)^{n-2}\Big(\prod_{k=2}^n (x_1 - x_k)\Big)\left|\begin{array}{cccc} 1 & 0 & \cdots & 0 \\ x_1 & 1 & \cdots & 1 \\ x_1^2 & x_2 + x_1 & \cdots & x_n + x_1 \\ \vdots & \vdots & & \vdots \\ x_1^{n-1} & x_2^{n-2} + \cdots + x_1^{n-2} & \cdots & x_n^{n-2} + \cdots + x_1^{n-2} \\ \end{array}\right|$$
It looks like difficult to proceed. If I subtract the second column from the third column, the three first rows will become this
$$0 - 0 = 0$$
$$1 - 1 = 0$$
$$(x_3+x_1) - (x_2+x_1) = x_3 - x_2$$
but what about the rows below these? It becomes an algebraic mess.
13. Dec 11, 2008
### Anthony
Hurrah! So all you need do is find the constant $$C$$. Can you do that?
14. Dec 11, 2008
### jostpuur
The fourth row will be
$$(x_3^2 + x_3x_1 + x_1^2) - (x_2^2 + x_2x_1 + x_1^2) = (x_3-x_2)(x_3 + x_2 + x_1).$$
This looks good, and most obviously $x_3-x_2$ will become a common factor, but it is a mystery to me how one proves this. When one proceeds like this, the algebraic expressions at some arbitrary step seem to be complicated.
15. Dec 11, 2008
### jostpuur
Maybe. I'll try induction with choice $x_n=0$, and see what happens.
16. Dec 11, 2008
### jostpuur
Yes, it seems that a recursion formula
$$C_n = (-1)^{n+1} C_{n-1}$$
holds, and by checking the case $n=2$ first, the constants
$$C_n = (-1)^{n(n-1)/2}$$
follow.
Good job Anthony
17. Dec 11, 2008
### Anthony
Glad to be of service.
18. Dec 11, 2008
### jostpuur
I must admit I suspected that you had underestimated the problem at the time of your first post. You made it sound too simple.
19. Dec 11, 2008
### jostpuur
Actually this is not all clear to me yet. There was one result that I considered trivial earlier, but which turned out to be more difficult. How does on prove that if the vectors
$$\left(\begin{array}{c} 1 \\ x_1 \\ \vdots \\ x_1^{n-1} \\ \end{array}\right), \left(\begin{array}{c} 1 \\ x_2 \\ \vdots \\ x_2^{n-1} \\ \end{array}\right),\cdots, \left(\begin{array}{c} 1 \\ x_n \\ \vdots \\ x_n^{n-1} \\ \end{array}\right)$$
are linearly dependent, then $x_i=x_j$ with some $i\neq j$?
20. Dec 11, 2008
### epenguin
I can't do tex determinants, but if you subtract x1 X (penultimate row) from last row,..., x1 X first row from second row, then don't you get for first column {1, 0, 0, ...}, for second column {1, (x2-x1), x2(x2-x1), x22(x2-x1), ... ?
Similar Discussions: Difficult polynomial and determinant formula | {"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": 0.8978968858718872, "perplexity": 1408.753716080415}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-22/segments/1495463607120.76/warc/CC-MAIN-20170522211031-20170522231031-00492.warc.gz"} |
http://www.phy.duke.edu/~rgb/Class/phy51/phy51/node23.html | Next: Damped Oscillation Up: The Pendulum Previous: The Pendulum Contents
### The Physical Pendulum
In the treatment of the ordinary pendulum above, we just used Newton's Second Law directly to get the equation of motion. This was possible only because we could neglect the mass of the string and because we could treat the mass like a point mass at its end.
However, real grandfather clocks often have a large, massive pendulum like the one above - a long massive rod (of length and mass ) with a large round disk (of radius and mass ) at the end. The round weight rotates through an angle of in each oscillation, so it has angular momemtum. Newton's Law for forces no longer suffices. We must use torque and the moment of inertia to obtain the frequency of the oscillator.
To do this we go through the same steps (more or less) that we did for the regular pendulum. First we compute the net gravitational torque on the system at an arbitrary (small) angle :
(76)
(The - sign is there because the torque opposes the angular displacement from equilibrium.)
Next we set this equal to , where is the total moment of inertia for the system about the pivot of the pendulum and simplify:
(77)
(78)
and make the small angle approximation to get:
(79)
Note that for this problem:
(80)
(the moment of inertia of the rod plus the moment of inertial of the disk rotating about a parallel axis a distance away from its center of mass). From this we can read off the angular frequency:
(81)
With in hand, we know everything. For example:
(82)
gives us the angular trajectory. We can easily solve for the period , the frequency , the spatial or angular velocity, or whatever we like.
Note that the energy of this sort of pendulum can be tricky. Obviously its potential energy is easy enough - it depends on the elevation of the center of masses of the rod and the disk. The kinetic energy, however, is:
(83)
where I do not write as usual because it confuses (the angular frequency of the oscillator, roughly contant) and (the angular velocity of the pendulum bob, which varies between 0 and some maximum value in every cycle).
Next: Damped Oscillation Up: The Pendulum Previous: The Pendulum Contents
Robert G. Brown 2004-04-12 | {"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": 0.9781979322433472, "perplexity": 436.1781005458815}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-42/segments/1414637900248.22/warc/CC-MAIN-20141030025820-00102-ip-10-16-133-185.ec2.internal.warc.gz"} |
https://www.gradesaver.com/textbooks/math/algebra/intermediate-algebra-12th-edition/chapter-7-section-7-2-rational-exponents-7-2-exercises-page-449/67 | ## Intermediate Algebra (12th Edition)
$x^{5/12}$
$\bf{\text{Solution Outline:}}$ Use the laws of exponents to simplify the given expression, $x^{2/3}\cdot x^{-1/4} .$ $\bf{\text{Solution Details:}}$ Using the Product Rule of the laws of exponents which is given by $x^m\cdot x^n=x^{m+n},$ the expression above is equivalent to \begin{array}{l}\require{cancel} x^{\frac{2}{3}+\left(-\frac{1}{4}\right)} \\\\= x^{\frac{2}{3}-\frac{1}{4}} .\end{array} To simplify the expression, $\dfrac{2}{3}-\dfrac{1}{4} ,$ find the $LCD$ of the denominators $\left\{ 3,4 \right\}.$ The $LCD$ is $12$ since it is the lowest number that can be exactly divided by the denominators. Multiplying both the numerator and denominator of each terms by the constant that will make the denominators equal to the $LCD$ results to \begin{array}{l}\require{cancel} x^{\frac{2}{3}\cdot\frac{4}{4}-\frac{1}{4}\cdot\frac{3}{3}} \\\\= x^{\frac{8}{12}-\frac{3}{12}} \\\\= x^{\frac{5}{12}} \\\\= x^{5/12} .\end{array} | {"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": 0.9988040924072266, "perplexity": 505.63744563334683}, "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/1539583508988.18/warc/CC-MAIN-20181015080248-20181015101748-00254.warc.gz"} |
http://www.denizyuret.com/2015/05/what-is-wrong-with-p-values.html | ## May 27, 2015
### What is wrong with p-values?
Earlier this year, editors of the journal Basic and Applied Social Psychology announced that the journal would no longer publish papers containing p-values. The latest American Psychological Association Publication Manual states that researchers should "wherever possible, base discussion and interpretation of results on point and interval estimates," i.e. not p-values. FDA has been encouraging Bayesian analysis. What is wrong with p-values?
What is a p-value? In the classical statistical procedure known as "significance testing", we have a default hypothesis, usually called the null hypothesis and denoted by H0, and we wish to determine whether or not to reject H0 based on some observations X. We choose a statistic S=f(X) (a scalar function of X) that summarizes our data. The p-value is the probability of observing a value at least as extreme as S under H0. We reject H0 if the p-value is below some specified small threshold like α=0.05 and we say something like "H0 is rejected at 0.05 significance level." This threshold or significance level (α) upper bounds the probability of false rejection, i.e. rejecting H0 when it is correct.
Example: We toss a coin 1000 times and observe 532 heads, 468 tails. Is this a fair coin? In this example the null hypothesis H0 is that the coin is fair, observation X is the sequence of heads and tails, and the statistic S is the number of heads. The p-value, or probability of S ∉ [469,531] under H0, can be calculated as: $1 - \sum_{k=469}^{531} {1000 \choose k} \left(\frac{1}{2}\right)^{1000} = 0.04629$ We can reject the null hypothesis at 0.05 significance level and decide the coin is biased. But should we?
Objection 1: (MacKay 2003, pp.63) What we would actually like to know is the probability of H0 given that we observed 532 heads. Unfortunately the p-value 0.04629 is not that probability (although this is a common confusion). We can't calculate a probability for H0 unless we specify some alternatives. Come to think of it, how can we reject a hypothesis if we don't look at what the alternatives are? What if the alternatives are worse? So let's specify a "biased coin" alternative (H1) which assumes that the head probability of the coin (θ) is distributed uniformly between 0 and 1 (other ways of specifying H1 are possible and do not effect the conclusion). We have: $P(S=532 \mid H_0) = {1000 \choose 532} \left(\frac{1}{2}\right)^{1000} = 0.003256$ $P(S=532 \mid H_1) = \int_0^1 {1000 \choose 532} \theta^{532} (1-\theta)^{468} d\theta = 0.001$ So H0 makes our data 3.2 times more likely than H1! And here the p-value almost made us think the data was 1:20 in favor of the "biased" hypothesis.
Objection 2: (Berger 1982, pp.13) Well, now that we understand p-value is not the probability of H0, does it tell us anything useful? According to the definition it limits the false rejection rate, i.e. if we always use significance tests with a p-value threshold α=0.01, we can be assured of incorrectly rejecting only 1% of correct hypotheses in the long run. So does that mean when I reject a null hypothesis I am only mistaken 1% of the time? Of course not! P(reject|correct) is 1%, P(correct|reject) can be anything! Here is an example:
X=1 X=2 H0 .01 .99 H1 .01001 .98999
The table gives the probabilities the two hypotheses H0 and H1 assign to different outcomes X=1 or X=2. Say we observe X=1. We reject H0 at α=0.01 significance level. But there is very little evidence against H0, the likelihood ratio P(X|H1)/P(X|H0) is very close to 1, so the chance of being in error is about 1/2 (assuming H0 and H1 are a-priori equally likely). Thus α=0.01 is providing a very misleading and false sense of security when rejection actually occurs.
Objection 3: (Murphy 2013, pp.213) Consider two experiments. In the first one we toss a coin 1000 times and observe 474 tails. Using T=474 as our statistic the one sided p-value is P(T≤474|H0): $\sum_{k=0}^{474} {1000 \choose k} \left(\frac{1}{2}\right)^{1000} = 0.05337$ So at a significance level of α=0.05 we do not reject the null hypothesis of an unbiased coin.
In the second experiment we toss the coin until we observe 474 tails, and it happens to take us 1000 trials. Different intention, same data. This time N=1000 is the natural test statistic and the one sided p-value is P(N≥1000|H0): $\sum_{n=1000}^\infty {n-1 \choose 473} \left(\frac{1}{2}\right)^n = 0.04994$ Suddenly we are below the magical α=0.05 threshold and we can reject the null hypothesis. The observed data, thus the likelihoods of any hypotheses for this data have not changed. The p-value is based not just on what actually happened, but what could have happened. This is clearly absurd.
Objection 4: (Cumming 2012) If we base the fate of our hypotheses on p-values computed from experiments, at the very least we should expect the p-values (thus our critical decisions) to change very little when we replicate the experiments. Unfortunately p-values do not even give us stability, as this wonderful video "Dance of the p values" by Geoff Cumming illustrates:
Conclusion: (Jaynes 2003, pp.524) expressed the absurdity of significance testing best:
In order to argue for an hypothesis H1 that some effect exists, one does it indirectly: invent a "null hypothesis" H0 that denies any such effect, then argue against H0 in a way that makes no reference to H1 at all (that is, using only probabilities conditional on H0). To see how far this procedure takes us from elementary logic, suppose we decide that the effect exists; that is, we reject H0. Surely, we must also reject probabilities conditional on H0; but then what was the logical justification for the decision? Orthodox logic saws off its own limb.
Harold Jeffreys (1939, p. 316) expressed his astonishment at such limb-sawing reasoning by looking at a different side of it: "An hypothesis that may be true is rejected because it has failed to predict observable results that have not occurred. This seems a remarkable procedure. On the face of it, the evidence might more reasonably be taken as evidence for the hypothesis, not against it." | {"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": 0.8699479699134827, "perplexity": 921.9959729496144}, "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/1627046154175.76/warc/CC-MAIN-20210801092716-20210801122716-00327.warc.gz"} |
http://mathhelpforum.com/advanced-algebra/95818-algebraic-geometry-elliptic-quartic-curve.html | ## Algebraic Geometry: Elliptic Quartic Curve
This is a problem about an elliptic quartic in $P^{3}$
$Y$ is defined by
$x^{2} - xz - yw = 0$ and
$yz - xw -zw = 0$
Let $P$ be the point $(0,0,0,1)$
Let $f$ be the projection from $P$ to the plane $w=0$.
I need to show that $f$ induces an isomorphism of $Y - P$ with the plane cubic curve $K$ defined by:
$zy^{2} - x^{3} + xz^{2}=0$
minus the point $Q:=(1,0,-1)$
The point of the exercise is to prove that $Y$ is irreducible and nonsingular. This follows after showing that $f$ is an isomorphism since then we can show that $K$ is irreducible and nonsingular which is straightforward.
I have been trying to show that $f(Y)$ is contained in $K$ and vice versa.
Am I correct to right the $f(Y)$ as the equations that define $Y$ with $w==0$?
if so $f(Y)= x^{2} - xz = 0 = yz$ (excluding the point $(0,0,0)$)
But the algebra is not working out for me. Is there another approach? | {"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": 24, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9831420183181763, "perplexity": 90.60384691816633}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "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-2015-11/segments/1424936462762.87/warc/CC-MAIN-20150226074102-00234-ip-10-28-5-156.ec2.internal.warc.gz"} |
http://mathhelpforum.com/calculus/22650-integral-problem.html | 1. ## Integral problem
Find f(x) if f(1) = -1 and the tangent line at (x, f(x)) has slope 2e^x + 1
This is a problem that was thrown into my indefinite integral homework, and I'm kinda stuck.
any help appreciated.
2. Originally Posted by leviathanwave
Find f(x) if f(1) = -1 and the tangent line at (x, f(x)) has slope 2e^x + 1
This is a problem that was thrown into my indefinite integral homework, and I'm kinda stuck.
any help appreciated.
the derivative gives the slope. thus if the slope at any $x$ is $2e^x + 1$ it means that:
$f'(x) = 2e^x + 1$
$\Rightarrow f(x) = \int f'(x)~dx$
and use the fact that $f(1) = -1$ to solve for the arbitrary constant of integration | {"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": 0.9943254590034485, "perplexity": 534.0216032262972}, "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-2013-20/segments/1368696382360/warc/CC-MAIN-20130516092622-00071-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://crypto.stackexchange.com/questions/75777/rsa-exercises-example/75779 | # RSA exercises example
Consider the following textbook RSA example. Let be p = 7, q = 11 and e = 3. Give a general algorithm for calculating d and run such algorithm with the above inputs. What is the max integer that can be encrypted? Is there any changes in the answers, if we swap the values of p and q?
I tried to apply RSA in this way:
$$p=7$$ and $$q = 11$$ so $$n=(pq)$$ and $$n=77$$, therefore $$\phi(pq) = (p-1)(q-1) = (7-1)(11-1)= 60$$
Pick $$e$$ that is $$<60$$ and isn't coprime (which means I can't use $$2,3$$ and $$5$$)
So after that, I'm blocked, and my solution is to choose another coprime, for example $$e=7$$ in order to have public-key: $$(n,e)= (77,7)$$
But I'm wrong somewhere, because if I use $$7$$ (or another coprime like $$11$$) I can't compute $$d$$. In fact, for all numbers that I used, for example with $$e=7$$, if I pick $$d=9$$ I have $$63$$ that if I divided for $$60$$ I have $$1$$ with rest $$3$$.
So my questions are these :
• Am I wronging? (where?)
• As a request , can I pick 3, even if it isn't coprime? (because as I saw on theory, I need at least a coprime).
• pick $$e= p$$ or $$q$$ seams wrong.right?
• You find the inverse of $e$ according to $\phi(n)$ not according to $n$. The inverse can be calculated with ext-GCD. – kelalaka Nov 15 '19 at 18:50
• $7^{-1} = 43 \bmod 60$ see at Wolfram. You cannot choose 3 since the setup will not be a permutation. An example here – kelalaka Nov 15 '19 at 19:16
• is this homework? – kelalaka Nov 16 '19 at 8:18
• I'm studying computer security, I have an exam in a few months and I want to understand these topics before the exam (so I'm studying now). This question has already been asked in an old exam task (in 2015). @kelalaka – theantomc Nov 16 '19 at 9:01
Am I wrong? (where?)
Yes, there is a small mistake in the way you are computing $$d$$: you need to compute $$d$$ as being the inverse of $$e$$ modulo $$\phi(n) = (p-1)(q-1) = 60$$. So, if you pick $$e= 7$$ (since you cannot pick $$e = 3$$ because it would be coprime with $$\phi(n)$$), you need to compute its inverse modulo (which is typically done using Euclid's algorithm). As said in the comment, the modular inverse of $$7 \bmod{60}$$ is $$7^{-1} = 43 \bmod{60}$$.
As a request , can I pick 3, even if it isn't coprime? (because as I saw on theory, I need at least a coprime).
No, picking a value $$e$$ that is not coprime with $$\phi(n)$$ does not allow to guarantee unique decryption of ciphertexts.
Since this is not desirable, it is required that $$e$$ is coprime with $$\phi(n)$$, which in turn implies that $$\gcd(e,p-1)=1=\gcd(e, q-1)$$ when using RSA with $$n=pq$$ for $$p,q$$ two primes.
Picking $$e=p$$ or $$q$$ seems wrong, right?
Yes, because then you are literally giving away your private key, since anybody can see that $$n \bmod e \equiv 0$$, which means that $$e$$ divides $$n$$! And, let's say you set $$e=p$$, then it is easy to recover $$q$$ as well by computing $$\frac{n}{e}=q$$ and so anybody knowing your public key $$(n,e)$$ would be able to recover your private key $$(p,q,d)$$.
• My problem continues to be that calculation of d ... I'm trying to understand The Euclidean Algorithm – theantomc Nov 16 '19 at 9:09
• @theantomc: you want to understand the Extended Euclidean Algorithm. See there for a more efficient and easier to implement variant. – fgrieu Nov 16 '19 at 9:19
• @fgrieu Ups, indeed, missing a "not". – Lery Nov 18 '19 at 15:14
if I use $$e=7$$ (or another coprime like $$11$$) I can't compute $$d$$
You can use $$e=7$$. When $$n$$ is squarefree, a private exponent $$d$$ will work if (not: only if) $$e\;d\equiv1\pmod{\phi(n)}$$, that is by definition when $$e\;d-1$$ is divisible by $$\phi(n)$$. There are solutions to that if and only if $$e$$ is coprime with $$\phi(n)$$. The textbook systematic way to find such $$d$$ is the Extended Euclidean Algorithm. See there for a more efficient and easier to implement variant; or there for a "binary" variant.
Note: When $$n$$ is squarefree, the necessary and sufficient condition for $$d$$ to work in RSA is: $$e\;d\equiv1\pmod{\lambda(n)}$$ (where $$\lambda$$ is the Carmichael function). That simplifies computation of $$d$$, and typically leads to a smaller one. $$d=e^{-1}\bmod\lambda(n)$$ is required by some RSA standards including FIPS 186-4. When $$n$$ is the product of distinct primes $$p$$ and $$q$$, $$\lambda(n)$$ can be computed as \begin{align}\lambda(n)&=\operatorname{lcm}(p-1,q-1)\\&=\frac{(p-1)(q-1)}{\gcd(p-1,q-1)}\end{align}
Can I pick $$e=3$$, even if it isn't coprime with $$\phi(n)$$?
No. It is required that $$e$$ is coprime with $$\phi(n)$$ [equivalently: that $$\gcd(e,p-1)=1=\gcd(e,q-1)$$ ] in order to insure unique decryption of ciphertexts. Otherwise, there will be multiple plaintexts $$m\in[0,n)$$ leading to the same ciphertext $$m^e\bmod n$$. In your case $$(n,e)=(77,3)$$, for example, $$m=4$$ and $$m=15$$ would lead to the same ciphertext $$64$$.
Picking $$e=p$$ or $$q$$ seams wrong
For large $$n$$, it would be bad to choose $$e$$ equal to a factor of $$n$$ (or with any other approximate relation between $$e$$ and a factor of $$n$$) since that would allow factoring $$n$$. But when one deliberately illustrates RSA with a toy $$n$$ such as $$n=77$$ which is trivial to factor, choosing $$e$$ equal to one of the factors is a non-issue. Still, one could use $$e=13$$ to avoid that special case.
What is the largest integer that can be encrypted?
In textbook RSA, plaintext and ciphertext space is the integer interval $$[0,n)$$. The largest integer that can be encrypted (and decrypts correctly) is thus $$n-1$$. Notice that it is always encrypted to itself, thus trivial to decipher. More generally, textbook RSA is insecure when directly used to encipher data. It is conjectured secure when $$p$$ and $$q$$ are large random secret primes, and a random $$x$$ in the plaintext space $$[0,n)$$ is enciphered.
• What is the max integer that can be encrypted? is m<n (in my case 77)? – theantomc Nov 16 '19 at 13:25 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 100, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9117564558982849, "perplexity": 309.8969217746375}, "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/1600402131986.91/warc/CC-MAIN-20201001174918-20201001204918-00529.warc.gz"} |
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1. CMB Online first
Cruz-Uribe, David; Rodney, Scott; Rosta, Emily
Poincaré Inequalities and Neumann Problems for the $p$-Laplacian We prove an equivalence between weighted Poincaré inequalities and the existence of weak solutions to a Neumann problem related to a degenerate $p$-Laplacian. The Poincaré inequalities are formulated in the context of degenerate Sobolev spaces defined in terms of a quadratic form, and the associated matrix is the source of the degeneracy in the $p$-Laplacian. Keywords:degenerate Sobolev space, $p$-Laplacian, Poincaré inequalitiesCategories:30C65, 35B65, 35J70, 42B35, 42B37, 46E35
2. CMB Online first
Loring, Terry A.; Schulz-Baldes, Hermann
Spectral flow argument localizing an odd index pairing An odd Fredholm module for a given invertible operator on a Hilbert space is specified by an unbounded so-called Dirac operator with compact resolvent and bounded commutator with the given invertible. Associated to this is an index pairing in terms of a Fredholm operator with Noether index. Here it is shown by a spectral flow argument how this index can be calculated as the signature of a finite dimensional matrix called the spectral localizer. Keywords:index pairing, spectral flow, topological materialsCategories:19K56, 46L80
3. CMB Online first
Banica, Teodor
Tannakian duality for affine homogeneous spaces Associated to any closed quantum subgroup $G\subset U_N^+$ and any index set $I\subset\{1,\dots,N\}$ is a certain homogeneous space $X_{G,I}\subset S^{N-1}_{\mathbb C,+}$, called affine homogeneous space. We discuss here the abstract axiomatization of the algebraic manifolds $X\subset S^{N-1}_{\mathbb C,+}$ which can appear in this way, by using Tannakian duality methods. Keywords:quantum isometry, noncommutative manifoldCategories:46L65, 46L89
4. CMB Online first
Schmidt, Simon; Weber, Moritz
Quantum symmetries of graph $C^*$-algebras The study of graph $C^*$-algebras has a long history in operator algebras. Surprisingly, their quantum symmetries have never been computed so far. We close this gap by proving that the quantum automorphism group of a finite, directed graph without multiple edges acts maximally on the corresponding graph $C^*$-algebra. This shows that the quantum symmetry of a graph coincides with the quantum symmetry of the graph $C^*$-algebra. In our result, we use the definition of quantum automorphism groups of graphs as given by Banica in 2005. Note that Bichon gave a different definition in 2003; our action is inspired from his work. We review and compare these two definitions and we give a complete table of quantum automorphism groups (with respect to either of the two definitions) for undirected graphs on four vertices. Keywords:finite graph, graph automorphism, automorphism group, quantum automorphism, graph C*-algebra, quantum group, quantum symmetryCategories:46LXX, 05CXX, 20B25
5. CMB 2018 (vol 61 pp. 236)
Boutonnet, Remi; Roydor, Jean
A Note on Uniformly Bounded Cocycles into Finite von Neumann Algebras We give a short proof of a result of T. Bates and T. Giordano stating that any uniformly bounded Borel cocycle into a finite von Neumann algebra is cohomologous to a unitary cocycle. We also point out a separability issue in their proof. Our approach is based on the existence of a non-positive curvature metric on the positive cone of a finite von Neumann algebra. Keywords:Borel cocycle, von Neumann algebraCategories:46L55, 46L40, 22D40
6. CMB Online first
Bénéteau, Catherine Anne; Fleeman, Matthew C.; Khavinson, Dmitry S.; Seco, Daniel; Sola, Alan
Remarks on inner functions and optimal approximants We discuss the concept of inner function in reproducing kernel Hilbert spaces with an orthogonal basis of monomials and examine connections between inner functions and optimal polynomial approximants to $1/f$, where $f$ is a function in the space. We revisit some classical examples from this perspective, and show how a construction of Shapiro and Shields can be modified to produce inner functions. Keywords:inner function, reproducing Kernel Hilbert Space, operator-theoretic function theoryCategories:46E22, 30J05
7. CMB Online first
Abrahamsen, Trond A.; Hájek, Petr; Nygaard, Olav; Troyanski, Stanimir L.
Strongly extreme points and approximation properties We show that if $x$ is a strongly extreme point of a bounded closed convex subset of a Banach space and the identity has a geometrically and topologically good enough local approximation at $x$, then $x$ is already a denting point. It turns out that such an approximation of the identity exists at any strongly extreme point of the unit ball of a Banach space with the unconditional compact approximation property. We also prove that every Banach space with a Schauder basis can be equivalently renormed to satisfy the sufficient conditions mentioned. Keywords:denting point, strongly extreme point, unconditional compact approximation propertyCategories:46B20, 46B04
8. CMB 2017 (vol 60 pp. 855)
Suárez de la Fuente, Jesús
The Kottman Constant for $\alpha$-Hölder Maps We investigate the role of the Kottman constant of a Banach space $X$ in the extension of $\alpha$-Hölder continuous maps for every $\alpha\in (0,1]$. Keywords:Kottman constant, $\alpha$-Hölder mapCategories:46B60, 46B80
9. CMB 2017 (vol 60 pp. 673)
Abtahi, Fatemeh; Azizi, Mohsen; Rejali, Ali
Character Amenability of the Intersection of Lipschitz Algebras Let $(X,d)$ be a metric space and $J\subseteq [0,\infty)$ be nonempty. We study the structure of the arbitrary intersections of Lipschitz algebras, and define a special Banach subalgebra of $\bigcap_{\gamma\in J}\operatorname{Lip}_\gamma X$, denoted by $\operatorname{ILip}_J X$. Mainly, we investigate $C$-character amenability of $\operatorname{ILip}_J X$, in particular Lipschitz algebras. We address a gap in the proof of a recent result in this field. Then we remove this gap, and obtain a necessary and sufficient condition for $C$-character amenability of $\operatorname{ILip}_J X$, specially Lipschitz algebras, under an additional assumption. Keywords:amenability, character amenability, Lipschitz algebra, metric spaceCategories:46H05, 46J10, 11J83
10. CMB 2017 (vol 61 pp. 225)
Bichon, Julien; Kyed, David; Raum, Sven
Higher $\ell^2$-Betti Numbers of Universal Quantum Groups We calculate all $\ell^2$-Betti numbers of the universal discrete Kac quantum groups $\hat{\mathrm U}^+_n$ as well as their half-liberated counterparts $\hat{\mathrm U}^*_n$. Keywords:$\ell^2$-Betti number, free unitary quantum group, half-liberated unitary quantum group, free product formula, extensionCategories:16T05, 46L65, 20G42
11. CMB Online first
Figiel, Tadeusz; Johnson, William
Quotients of Essentially Euclidean Spaces A precise quantitative version of the following qualitative statement is proved: If a finite dimensional normed space contains approximately Euclidean subspaces of all proportional dimensions, then every proportional dimensional quotient space has the same property. Keywords:essentially euclidean spaceCategories:46B20, 46B07, 46B99
12. CMB 2017 (vol 60 pp. 402)
Shravan Kumar, N.
Invariant Means on a Class of von Neumann Algebras Related to Ultraspherical Hypergroups II Let $K$ be an ultraspherical hypergroup associated to a locally compact group $G$ and a spherical projector $\pi$ and let $VN(K)$ denote the dual of the Fourier algebra $A(K)$ corresponding to $K.$ In this note, we show that the set of invariant means on $VN(K)$ is singleton if and only if $K$ is discrete. Here $K$ need not be second countable. We also study invariant means on the dual of the Fourier algebra $A_0(K),$ the closure of $A(K)$ in the $cb$-multiplier norm. Finally, we consider generalized translations and generalized invariant means. Keywords:ultraspherical hypergroup, Fourier algebra, Fourier-Stieltjes algebra, invariant mean, generalized translation, generalized invariant meanCategories:43A62, 46J10, 43A30, 20N20
13. CMB 2017 (vol 61 pp. 301)
Józiak, Paweł
Remarks on Hopf Images and Quantum Permutation Groups $S_n^+$ Motivated by a question of A. Skalski and P.M. SoÅtan (2016) about inner faithfulness of the S. Curran's map of extending a quantum increasing sequence to a quantum permutation, we revisit the results and techniques of T. Banica and J. Bichon (2009) and study some group-theoretic properties of the quantum permutation group on $4$ points. This enables us not only to answer the aforementioned question in positive in case $n=4, k=2$, but also to classify the automorphisms of $S_4^+$, describe all the embeddings $O_{-1}(2)\subset S_4^+$ and show that all the copies of $O_{-1}(2)$ inside $S_4^+$ are conjugate. We then use these results to show that the converse to the criterion we applied to answer the aforementioned question is not valid. Keywords:Hopf image, quantum permutation group, compact quantum groupCategories:20G42, 81R50, 46L89, 16W35
14. CMB 2017 (vol 61 pp. 114)
Haralampidou, Marina; Oudadess, Mohamed; Palacios, Lourdes; Signoret, Carlos
A characterization of $C^{\ast}$-normed algebras via positive functionals We give a characterization of $C^{\ast}$-normed algebras, among certain involutive normed ones. This is done through the existence of enough specific positive functionals. The same question is also examined in some non normed (topological) algebras. Keywords:$C^{\ast}$-normed algebra, $C^*$-algebra, (pre-)locally $C^*$-algebra, pre-$C^*$-bornological algebra, positive functional, locally uniformly $A$-convex algebra, perfect locally $m$-convex algebra, $C^*$-(resp. $^*$-) subnormable algebraCategories:46H05, 46K05
15. CMB 2017 (vol 60 pp. 449)
Alaghmandan, Mahmood; Crann, Jason
Character Density in Central Subalgebras of Compact Quantum Groups We investigate quantum group generalizations of various density results from Fourier analysis on compact groups. In particular, we establish the density of characters in the space of fixed points of the conjugation action on $L^2(\mathbb{G})$, and use this result to show the weak* density and norm density of characters in $ZL^\infty(\mathbb{G})$ and $ZC(\mathbb{G})$, respectively. As a corollary, we partially answer an open question of Woronowicz. At the level of $L^1(\mathbb{G})$, we show that the center $\mathcal{Z}(L^1(\mathbb{G}))$ is precisely the closed linear span of the quantum characters for a large class of compact quantum groups, including arbitrary compact Kac algebras. In the latter setting, we show, in addition, that $\mathcal{Z}(L^1(\mathbb{G}))$ is a completely complemented $\mathcal{Z}(L^1(\mathbb{G}))$-submodule of $L^1(\mathbb{G})$. Keywords:compact quantum group, irreducible characterCategories:43A20, 43A40, 46J40
16. CMB 2017 (vol 60 pp. 791)
Jiang, Chunlan
Reduction to Dimension Two of Local Spectrum for $AH$ Algebra with Ideal Property A $C^{*}$-algebra $A$ has the ideal property if any ideal $I$ of $A$ is generated as a closed two sided ideal by the projections inside the ideal. Suppose that the limit $C^{*}$-algebra $A$ of inductive limit of direct sums of matrix algebras over spaces with uniformly bounded dimension has ideal property. In this paper we will prove that $A$ can be written as an inductive limit of certain very special subhomogeneous algebras, namely, direct sum of dimension drop interval algebras and matrix algebras over 2-dimensional spaces with torsion $H^{2}$ groups. Keywords:AH algebra, reduction, local spectrum, ideal propertyCategory:46L35
17. CMB 2017 (vol 60 pp. 690)
Bao, Guanlong; Göğüş, Nıhat Gökhan; Pouliasis, Stamatis
$\mathcal{Q}_p$ Spaces and Dirichlet Type Spaces In this paper, we show that the Möbius invariant function space $\mathcal {Q}_p$ can be generated by variant Dirichlet type spaces $\mathcal{D}_{\mu, p}$ induced by finite positive Borel measures $\mu$ on the open unit disk. A criterion for the equality between the space $\mathcal{D}_{\mu, p}$ and the usual Dirichlet type space $\mathcal {D}_p$ is given. We obtain a sufficient condition to construct different $\mathcal{D}_{\mu, p}$ spaces and we provide examples. We establish decomposition theorems for $\mathcal{D}_{\mu, p}$ spaces, and prove that the non-Hilbert space $\mathcal {Q}_p$ is equal to the intersection of Hilbert spaces $\mathcal{D}_{\mu, p}$. As an application of the relation between $\mathcal {Q}_p$ and $\mathcal{D}_{\mu, p}$ spaces, we also obtain that there exist different $\mathcal{D}_{\mu, p}$ spaces; this is a trick to prove the existence without constructing examples. Keywords:$\mathcal {Q}_p$ space, Dirichlet type space, Möbius invariant function spaceCategories:30H25, 31C25, 46E15
18. CMB 2017 (vol 60 pp. 816)
Moslehian, Mohammad Sal; Zamani, Ali
Characterizations of Operator Birkhoff--James Orthogonality In this paper, we obtain some characterizations of the (strong) Birkhoff--James orthogonality for elements of Hilbert $C^*$-modules and certain elements of $\mathbb{B}(\mathscr{H})$. Moreover, we obtain a kind of Pythagorean relation for bounded linear operators. In addition, for $T\in \mathbb{B}(\mathscr{H})$ we prove that if the norm attaining set $\mathbb{M}_T$ is a unit sphere of some finite dimensional subspace $\mathscr{H}_0$ of $\mathscr{H}$ and $\|T\|_{{{\mathscr{H}}_0}^\perp} \lt \|T\|$, then for every $S\in\mathbb{B}(\mathscr{H})$, $T$ is the strong Birkhoff--James orthogonal to $S$ if and only if there exists a unit vector $\xi\in {\mathscr{H}}_0$ such that $\|T\|\xi = |T|\xi$ and $S^*T\xi = 0$. Finally, we introduce a new type of approximate orthogonality and investigate this notion in the setting of inner product $C^*$-modules. Keywords:Hilbert $C^*$-module, Birkhoff--James orthogonality, strong Birkhoff--James orthogonality, approximate orthogonalityCategories:46L05, 46L08, 46B20
19. CMB 2016 (vol 60 pp. 655)
Zhuo, Ciqiang; Sickel, Winfried; Yang, Dachun; Yuan, Wen
Characterizations of Besov-Type and Triebel-Lizorkin-Type Spaces via Averages on Balls Let $\ell\in\mathbb N$ and $\alpha\in (0,2\ell)$. In this article, the authors establish equivalent characterizations of Besov-type spaces, Triebel-Lizorkin-type spaces and Besov-Morrey spaces via the sequence $\{f-B_{\ell,2^{-k}}f\}_{k}$ consisting of the difference between $f$ and the ball average $B_{\ell,2^{-k}}f$. These results give a way to introduce Besov-type spaces, Triebel-Lizorkin-type spaces and Besov-Morrey spaces with any smoothness order on metric measure spaces. As special cases, the authors obtain a new characterization of Morrey-Sobolev spaces and $Q_\alpha$ spaces with $\alpha\in(0,1)$, which are of independent interest. Keywords:Besov space, Triebel-Lizorkin space, ball average, Calderón reproducing formulaCategories:42B25, 46E35, 42B35
20. CMB 2016 (vol 60 pp. 350)
Ma, Yumei
Isometry on Linear $n$-G-quasi Normed Spaces This paper generalizes the Aleksandrov problem: the Mazur-Ulam theorem on $n$-G-quasi normed spaces. It proves that a one-$n$-distance preserving mapping is an $n$-isometry if and only if it has the zero-$n$-G-quasi preserving property, and two kinds of $n$-isometries on $n$-G-quasi normed space are equivalent; we generalize the Benz theorem to n-normed spaces with no restrictions on the dimension of spaces. Keywords:$n$-G-quasi norm, Mazur-Ulam theorem, Aleksandrov problem, $n$-isometry, $n$-0-distanceCategories:46B20, 46B04, 51K05
21. CMB 2016 (vol 60 pp. 217)
Wang, Yuanyi
Condition $C'_{\wedge}$ of Operator Spaces In this paper, we study condition $C'_{\wedge}$ which is a projective tensor product analogue of condition $C'$. We show that the finite-dimensional OLLP operator spaces have condition $C'_{\wedge}$ and $M_{n}$ $(n\gt 2)$ does not have that property. Keywords:operator space, local theory, tensor productCategory:46L07
22. CMB 2016 (vol 60 pp. 104)
Diestel, Geoff
An Extension of Nikishin's Factorization Theorem A Nikishin-Maurey characterization is given for bounded subsets of weak-type Lebesgue spaces. New factorizations for linear and multilinear operators are shown to follow. Keywords:factorization, type, cotype, Banach spacesCategories:46E30, 28A25
23. CMB 2016 (vol 60 pp. 122)
Ghanei, Mohammad Reza; Nasr-Isfahani, Rasoul; Nemati, Mehdi
A Homological Property and Arens Regularity of Locally Compact Quantum Groups We characterize two important notions of amenability and compactness of a locally compact quantum group ${\mathbb G}$ in terms of certain homological properties. For this, we show that ${\mathbb G}$ is character amenable if and only if it is both amenable and co-amenable. We finally apply our results to Arens regularity problems of the quantum group algebra $L^1({\mathbb G})$; in particular, we improve an interesting result by Hu, Neufang and Ruan. Keywords:amenability, Arens regularity, co-amenability, locally compact quantum group, homological propertyCategories:46L89, 43A07, 46H20, 46M10, 58B32
24. CMB 2016 (vol 60 pp. 173)
Oubbi, Lahbib
On Ulam Stability of a Functional Equation in Banach Modules Let $X$ and $Y$ be Banach spaces and $f : X \to Y$ an odd mapping. For any rational number $r \ne 2$, C. Baak, D. H. Boo, and Th. M. Rassias have proved the Hyers-Ulam stability of the following functional equation: \begin{align*} r f \left(\frac{\sum_{j=1}^d x_j}{r} \right) & + \sum_{\substack{i(j) \in \{0,1\} \\ \sum_{j=1}^d i(j)=\ell}} r f \left( \frac{\sum_{j=1}^d (-1)^{i(j)}x_j}{r} \right) = (C^\ell_{d-1} - C^{\ell -1}_{d-1} + 1) \sum_{j=1}^d f(x_j) \end{align*} where $d$ and $\ell$ are positive integers so that $1 \lt \ell \lt \frac{d}{2}$, and $C^p_q := \frac{q!}{(q-p)!p!}$, $p, q \in \mathbb{N}$ with $p \le q$. In this note we solve this equation for arbitrary nonzero scalar $r$ and show that it is actually Hyers-Ulam stable. We thus extend and generalize Baak et al.'s result. Different questions concerning the *-homomorphisms and the multipliers between C*-algebras are also considered. Keywords:linear functional equation, Hyers-Ulam stability, Banach modules, C*-algebra homomorphisms.Categories:39A30, 39B10, 39A06, 46Hxx
25. CMB 2016 (vol 60 pp. 586)
Liu, Feng; Wu, Huoxiong
Endpoint Regularity of Multisublinear Fractional Maximal Functions In this paper we investigate the endpoint regularity properties of the multisublinear fractional maximal operators, which include the multisublinear Hardy-Littlewood maximal operator. We obtain some new bounds for the derivative of the one-dimensional multisublinear fractional maximal operators acting on vector-valued function $\vec{f}=(f_1,\dots,f_m)$ with all $f_j$ being $BV$-functions. Keywords:multisublinear fractional maximal operators, Sobolev spaces, bounded variationCategories:42B25, 46E35
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© Canadian Mathematical Society, 2018 : https://cms.math.ca/ | {"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": 0.9677823185920715, "perplexity": 994.3632319268808}, "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-2018-17/segments/1524125945459.17/warc/CC-MAIN-20180421223015-20180422003015-00267.warc.gz"} |
http://www.physicsforums.com/showthread.php?p=3855005 | # Inviscid flow around cylinder in presence of wall
by Armin1986
Tags: cylinder, flow, inviscid, presence, wall
P: 1 Hi there, Is there someone who would know how to solve the following potential flow theory problem: How to find the flow field and resulting force on a cylinder moving at a constant velocity through a stagnant fluid in the presence of a wall. The motion of the cylinder is parallel to the wall. Without the wall it can be found that there is no force due to the inviscid nature of the flow. However, due to the wall, "ground effect" occurs and therefore there will be a lift force. I have two questions: - How to find the flow field (e.g. streamlines or vector field) - How to find the force as a function of the distance from the wall I would be very thankfull if someone could help me with this problem.
Related Discussions General Physics 4 Mechanical Engineering 7 Engineering, Comp Sci, & Technology Homework 2 Engineering, Comp Sci, & Technology Homework 1 Mechanical Engineering 2 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8826754689216614, "perplexity": 319.6879277705075}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1397609539447.23/warc/CC-MAIN-20140416005219-00160-ip-10-147-4-33.ec2.internal.warc.gz"} |
http://stats.stackexchange.com/questions/13301/assuming-u-sim-n0-sigma2-when-y-is-highly-skewed | # Assuming $u\sim N(0,\sigma^2)$ when y is highly skewed
does it make sense to assume $u\sim N(0,\sigma^2)$ when I know from a histogram that $y$ is highly skewed. Because from the assumption $u\sim N(0,\sigma^2)$ it follows that $y\sim N(x\beta,\sigma^2)$ and I'm absoluteley not sure if the assumption $u\sim N(0,\sigma^2)$ makes sense in a case where I know that the distribution of $y$ is not bell shaped. The alternative would be just to make OLS without any assumption about the error term, but in this case I can't analyze outliers and leverages (what I'd really like to do, because otherwise I can't present much more than a line which minimizes the sqaured sum of the residuals). [addendum: I can't make an outlier analysis because I can not define "outlier" in a context where I don't assume a normal distribution, because there is no outlying without a distribution] Besides your answers I'd really like to have a recommendation for a good book, where I can find some thoughts about what assumptions should we make when y is obviously not normal distributed.
Regards
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There is Taleb's book, although it will merely point the risk of this assumption. – wok Jul 20 '11 at 15:41
You need to put some more background into this. In the linear model $y_i = \alpha + \beta x_i + \epsilon_i$, it is not unusual to assume the $\epsilon_i$ are iid with a normal distribution of mean $0$, but there is no requirement for the $x_i$ or $y_i$ to be normally distributed. – Henry Jul 20 '11 at 15:41
Well, just because $y|x \sim N(x \beta, \sigma^2)$ doesn't mean that a histogram of the marginal distribution of $y$ will look bell shaped; I believe that will only happen if $x$ is also normally distributed. – Macro Jul 20 '11 at 16:37
@Henry For background you could read the almost identical series of questions here, here, and here. "What we've got here is failure to communicate." – whuber Jul 20 '11 at 17:17
@Mark Pick any $\beta$ and $\sigma$, generate 10 iid draws $\epsilon_1$, ..., $\epsilon_{10}$ from a normal(0, $\sigma$) distribution, and create the dataset $((2^i, \beta 2^i + \epsilon_i), i=1,\ldots,10)$. When $|\beta|$ and $\sigma$ are near $1$, the y's will be highly positively skewed but the data are perfectly linear with beautifully normal errors. In short, the skewness of the y's comes from the skewness of the x's but (of itself) reveals nothing at all about the distribution of the residuals. You check distributional assumptions by studying the residuals, not the y's. – whuber Jul 20 '11 at 19:44
If u is the residual from a regression of y on some other variable x, then I think this is a variant of an earlier question. The residuals of u can be Gaussian even if the distribution of y is highly skewed as it may simply be that the distribution of x is highly skewed.
Consider an example of estimating temperature (y) as a function of lattitude (x); here u represents the measurment error of the thermometer (and is Gaussian). The distribution of y values in our sample will depend on where we choose to site out weather stations. If we place them all either at the poles or the equator, then we will have a bimodal distribution. If we place them on a regular equal area grid, we will get a unimodal distribution of y values, even though the physics of climate is the same for both samples and the measurement uncertainty u is normal.
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Thanks for the answer! Do you have some good references in this context for me? A good book or a paper. – MarkDollar Jul 21 '11 at 5:41 | {"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": 0.951729953289032, "perplexity": 349.9692632532123}, "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-2013-48/segments/1386163052912/warc/CC-MAIN-20131204131732-00024-ip-10-33-133-15.ec2.internal.warc.gz"} |
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