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https://eo4society.esa.int/2021/06/14/carbon-dioxide-emission-plumes-from-a-large-power-station-detected-from-space/	# Carbon dioxide emission plumes from a large power station detected from space Researchers at the Finnish Meteorological Institute developed a new methodology to derive source-specific NOₓ-to-CO₂ emission ratios using satellite observations. The method was applied to Matimba power station in South-Africa. The results can be used to estimate carbon dioxide emissions. Missions to observe CO2 plumes Since the Paris agreement was adopted in 2015, the role of satellite observations in understanding anthropogenic CO2 emissions has become increasingly important. Currently, the NASA’s CO2 instrument Orbiting Carbon Observatory-2 (OCO-2), launched in 2014, provides CO2 observations with the best coverage and resolution. However, the observations are obtained on a narrow swath (less than 10 km), which does allow the detection of the cross-sections of the emission plumes, but not the plumes in their entirety. Satellite observations of co-emitted species, such as NO2, facilitate the detection of the CO2 emission plumes. The European Commission is currently planning a new CO2 monitoring mission CO2M via the Copernicus Programme, which will observe both CO2 and NO2 over a larger swath (about 400 km). NOx-to-CO2 emission ratio case study: the Matimba power station Estimating CO2 emissions from individual sources using satellite data can be challenging due to the large background levels, while it is easier for short-lived gases like NO2. In a recently published study, a new methodology to calculate source-specific NOₓ-to-CO₂ emission ratio from satellite observations is developed. This ratio provides information on how clean the employed technology is and can be used to convert NOₓ emission into CO2 emission. The method was tested for the Matimba power station in South Africa, which is an optimal case study as it is a large emission source with several satellite overpasses, and it is also well isolated from other sources. The results are based on the CO2 observations from the NASA’s OCO-2 satellite and the NO2 retrievals from the European TROPOMI (TROPOspheric Monitoring Instrument), operating onboard the Sentinel 5 Precursor satellite since late 2017. During the 2018–2020 period, 14 collocations over Matimba enabled the simultaneous detection of the CO2 and NO2 plumes. The mean NOx-to-CO2 emission ratio was estimated as (2.6 ± 0.6) × 10⁻³ and the CO2 emission as 60 kton/day. The obtained CO₂ emission estimates are similar to those reported in existing inventories such as ODIAC. The research was carried on in the DACES project, which focuses on detecting anthropogenic CO₂ emissions sources by exploiting the synergy between satellite-based observations of short-lived polluting gases (such as NO₂) and greenhouse gases. The full publication by Hakkarainen and co-authors can be found at the following link: https://doi.org/10.1016/j.aeaoa.2021.100110 Featured image : Wind-rotated TROPOMI monthly and annual NO2 means for the year 2019 (see the publication for the full set). The data are oversampled to 0.01°x0.01° resolution and the local background (median) is removed. SHARE	{"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": 0.8833394050598145, "perplexity": 3657.5442596039575}, "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/1664030335396.92/warc/CC-MAIN-20220929225326-20220930015326-00679.warc.gz"}
https://ftp.aimsciences.org/article/doi/10.3934/ipi.2010.4.721	# American Institute of Mathematical Sciences November 2010, 4(4): 721-734. doi: 10.3934/ipi.2010.4.721 ## Local Sobolev estimates of a function by means of its Radon transform 1 Department of Mathematics, Stockholm University, 10691 Stockholm, Sweden 2 Department of Mathematics, Tufts University, Medford, MA 02155, United States Received September 2008 Revised June 2009 Published September 2010 In this article, we will define local and microlocal Sobolev seminorms and prove local and microlocal inverse continuity estimates for the Radon hyperplane transform in these seminorms. The relation between the Sobolev wavefront set of a function $f$ and of its Radon transform is well-known [18]. However, Sobolev wavefront is qualitative and therefore the relation in [18] is qualitative. Our results will make the relation between singularities of a function and those of its Radon transform quantitative. This could be important for practical applications, such as tomography, in which the data are smooth but can have large derivatives. Citation: Hans Rullgård, Eric Todd Quinto. Local Sobolev estimates of a function by means of its Radon transform. Inverse Problems & Imaging, 2010, 4 (4) : 721-734. doi: 10.3934/ipi.2010.4.721 ##### References: [1] M. A. Anastasio, Y. Zou, E. Y. Sidky and X. Pan, Local cone-beam tomography image reconstruction on chords, Journal of the Optical Society of America A, 24 (2007), 1569-1579. doi: doi:10.1364/JOSAA.24.001569. Google Scholar [2] E. Candès and L. Demanet, Curvelets and Fourier Integral Operators, C. R. Math. Acad. Sci. Paris. Serie I, 336 (2003), 395-398. Google Scholar [3] E. J. Candès and D. L. Donoho, Curvelets and Reconstruction of Images from Noisy Radon Data, in "Wavelet Applications in Signal and Image Processing VIII'' (eds. M. A. U. A. Aldroubi, A. F. Laine), Proc. SPIE. 4119 (2000). Google Scholar [4] D. V. Finch, I.-R. Lan and G. Uhlmann, Microlocal Analysis of the restricted X-ray transform with sources on a curve, in "Inside Out, Inverse Problems and Applications,'' (ed. G. Uhlmann), MSRI Publications, Cambridge University Press, 47 (2003), 193-218. Google Scholar [5] A. Greenleaf and G. Uhlmann, Non-local inversion formulas for the X-ray transform, Duke Math. J., 58 (1989), 205-240. doi: doi:10.1215/S0012-7094-89-05811-0. Google Scholar [6] A. Greenleaf and G. Uhlmann, Estimates for singular Radon transforms and pseudodifferential operators with singular symbols, J. Funct. Anal., 89 (1990), 202-232. doi: doi:10.1016/0022-1236(90)90011-9. Google Scholar [7] A. Greenleaf and G. Uhlmann, Microlocal techniques in integral geometry, Contemp. Math., 113 (1990), 121-136. Google Scholar [8] V. Guillemin and D. Schaeffer, Fourier integral operators from the Radon transform point of view, Proc. Sympos. Pure Math., 27 (1975), 297-300. Google Scholar [9] V. Guillemin and S. Sternberg, "Geometric Asymptotics,'' American Mathematical Society, Providence, RI, 1977. Google Scholar [10] M. G. Hahn and E. T. Quinto, Distances between measures from 1-dimensional projections as implied by continuity of the inverse Radon transform, Zeit. Wahr., 70 (1985), 361-380. doi: doi:10.1007/BF00534869. Google Scholar [11] A. Hertle, Continuity of the Radon transform and its inverse on Euclidean space, Math. Z., 184 (1983), 165-192. doi: doi:10.1007/BF01252856. Google Scholar [12] A. Katsevich, Improved cone beam local tomography, Inverse Problems, 22 (2006), 627-643. doi: doi:10.1088/0266-5611/22/2/015. Google Scholar [13] A. I. Katsevich, Cone beam local tomography, SIAM J. Appl. Math., 59 (1999), 2224-2246. doi: doi:10.1137/S0036139998336043. Google Scholar [14] A. K. Louis, "Analytische Methoden in der Computer Tomographie," Habilitationsschrift, Universität Münster, 1981. Google Scholar [15] F. Natterer, The mathematics of computerized tomography, in "Classics in Mathematics," Society for Industrial and Applied Mathematics, New York, 2001. Google Scholar [16] F. Natterer and F. Wübbeling, Mathematical methods in image reconstruction, in "Monographs on Mathematical Modeling and Computation," Society for Industrial and Applied Mathematics, New York, 2001. Google Scholar [17] B. Petersen, "Introduction to the Fourier Transform and Pseudo-Differential Operators," Pittman, Boston, 1983. Google Scholar [18] E. T. Quinto, Singularities of the X-ray transform and limited data tomography in $R^2$ and $R^3$, SIAM J. Math. Anal., 24 (1993), 1215-1225. doi: doi:10.1137/0524069. Google Scholar [19] E. T. Quinto, T. Bakhos and S. Chung, A local algorithm for Slant Hole SPECT, in "Mathematical Methods in Biomedical Imaging and Intensity-Modulated Radiation Therapy (IMRT),'' 321-348, CRM Series, 7, Ed. Norm., Pisa, 2008. Centro De Georgi. Google Scholar [20] E. T. Quinto and O. Öktem, Local tomography in electron microscopy, SIAM J. Appl. Math., 68 (2008), 1282-1303. doi: doi:10.1137/07068326X. Google Scholar [21] A. G. Ramm and A. I. Zaslavsky, Singularities of the Radon transform, Bull. Amer. Math. Soc., 25 (1993), 109-115. doi: doi:10.1090/S0273-0979-1993-00350-1. Google Scholar [22] M. Beals and M. Reed, Propagation of singularities for hyperbolic pseudodifferential operators with nonsmooth coefficients, Comm. Pure Appl. Math., 35 (1982), 169-184. doi: doi:10.1002/cpa.3160350203. Google Scholar [23] M. Beals and M. Reed, Microlocal regularity theorems for nonsmooth pseudodifferential operators and applications to nonlinear problems, Trans. Amer. Math. Soc. 285 (1984), 159-184. Google Scholar [24] H. Triebel, "Interpolation Theory, Function Spaces, Differential Operators,'' Second edition. Johann Ambrosius Barth, Heidelberg, 1995. Google Scholar [25] Y. Ye, H. Yu and G. Wang, Cone beam pseudo-lambda tomography, Inverse Problems, 23 (2007), 203-215. doi: doi:10.1088/0266-5611/23/1/010. Google Scholar show all references ##### References: [1] M. A. Anastasio, Y. Zou, E. Y. Sidky and X. Pan, Local cone-beam tomography image reconstruction on chords, Journal of the Optical Society of America A, 24 (2007), 1569-1579. doi: doi:10.1364/JOSAA.24.001569. Google Scholar [2] E. Candès and L. Demanet, Curvelets and Fourier Integral Operators, C. R. Math. Acad. Sci. Paris. Serie I, 336 (2003), 395-398. Google Scholar [3] E. J. Candès and D. L. Donoho, Curvelets and Reconstruction of Images from Noisy Radon Data, in "Wavelet Applications in Signal and Image Processing VIII'' (eds. M. A. U. A. Aldroubi, A. F. Laine), Proc. SPIE. 4119 (2000). Google Scholar [4] D. V. Finch, I.-R. Lan and G. Uhlmann, Microlocal Analysis of the restricted X-ray transform with sources on a curve, in "Inside Out, Inverse Problems and Applications,'' (ed. G. Uhlmann), MSRI Publications, Cambridge University Press, 47 (2003), 193-218. Google Scholar [5] A. Greenleaf and G. Uhlmann, Non-local inversion formulas for the X-ray transform, Duke Math. J., 58 (1989), 205-240. doi: doi:10.1215/S0012-7094-89-05811-0. Google Scholar [6] A. Greenleaf and G. Uhlmann, Estimates for singular Radon transforms and pseudodifferential operators with singular symbols, J. Funct. Anal., 89 (1990), 202-232. doi: doi:10.1016/0022-1236(90)90011-9. Google Scholar [7] A. Greenleaf and G. Uhlmann, Microlocal techniques in integral geometry, Contemp. Math., 113 (1990), 121-136. Google Scholar [8] V. Guillemin and D. Schaeffer, Fourier integral operators from the Radon transform point of view, Proc. Sympos. Pure Math., 27 (1975), 297-300. Google Scholar [9] V. Guillemin and S. Sternberg, "Geometric Asymptotics,'' American Mathematical Society, Providence, RI, 1977. Google Scholar [10] M. G. Hahn and E. T. Quinto, Distances between measures from 1-dimensional projections as implied by continuity of the inverse Radon transform, Zeit. Wahr., 70 (1985), 361-380. doi: doi:10.1007/BF00534869. Google Scholar [11] A. Hertle, Continuity of the Radon transform and its inverse on Euclidean space, Math. Z., 184 (1983), 165-192. doi: doi:10.1007/BF01252856. Google Scholar [12] A. Katsevich, Improved cone beam local tomography, Inverse Problems, 22 (2006), 627-643. doi: doi:10.1088/0266-5611/22/2/015. Google Scholar [13] A. I. Katsevich, Cone beam local tomography, SIAM J. Appl. Math., 59 (1999), 2224-2246. doi: doi:10.1137/S0036139998336043. Google Scholar [14] A. K. Louis, "Analytische Methoden in der Computer Tomographie," Habilitationsschrift, Universität Münster, 1981. Google Scholar [15] F. Natterer, The mathematics of computerized tomography, in "Classics in Mathematics," Society for Industrial and Applied Mathematics, New York, 2001. Google Scholar [16] F. Natterer and F. Wübbeling, Mathematical methods in image reconstruction, in "Monographs on Mathematical Modeling and Computation," Society for Industrial and Applied Mathematics, New York, 2001. Google Scholar [17] B. Petersen, "Introduction to the Fourier Transform and Pseudo-Differential Operators," Pittman, Boston, 1983. Google Scholar [18] E. T. Quinto, Singularities of the X-ray transform and limited data tomography in $R^2$ and $R^3$, SIAM J. Math. Anal., 24 (1993), 1215-1225. doi: doi:10.1137/0524069. Google Scholar [19] E. T. Quinto, T. Bakhos and S. Chung, A local algorithm for Slant Hole SPECT, in "Mathematical Methods in Biomedical Imaging and Intensity-Modulated Radiation Therapy (IMRT),'' 321-348, CRM Series, 7, Ed. Norm., Pisa, 2008. Centro De Georgi. Google Scholar [20] E. T. Quinto and O. Öktem, Local tomography in electron microscopy, SIAM J. Appl. Math., 68 (2008), 1282-1303. doi: doi:10.1137/07068326X. Google Scholar [21] A. G. Ramm and A. I. Zaslavsky, Singularities of the Radon transform, Bull. Amer. Math. Soc., 25 (1993), 109-115. doi: doi:10.1090/S0273-0979-1993-00350-1. Google Scholar [22] M. Beals and M. Reed, Propagation of singularities for hyperbolic pseudodifferential operators with nonsmooth coefficients, Comm. Pure Appl. Math., 35 (1982), 169-184. doi: doi:10.1002/cpa.3160350203. Google Scholar [23] M. Beals and M. Reed, Microlocal regularity theorems for nonsmooth pseudodifferential operators and applications to nonlinear problems, Trans. Amer. Math. Soc. 285 (1984), 159-184. Google Scholar [24] H. Triebel, "Interpolation Theory, Function Spaces, Differential Operators,'' Second edition. Johann Ambrosius Barth, Heidelberg, 1995. Google Scholar [25] Y. Ye, H. Yu and G. Wang, Cone beam pseudo-lambda tomography, Inverse Problems, 23 (2007), 203-215. doi: doi:10.1088/0266-5611/23/1/010. Google Scholar 2020 Impact Factor: 1.639	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.8920266032218933, "perplexity": 3817.999524863383}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585201.94/warc/CC-MAIN-20211018093606-20211018123606-00091.warc.gz"}
https://mathoverflow.net/questions/46986/countable-connected-hausdorff-space	# Countable connected Hausdorff space Let me start by reminding two constructions of topological spaces with such exotic combination of properties: 1) The elements are non-zero integers; base of topology are (infinite) arithmetic progressions with coprime first term and difference. 2) Take $\mathbb{R}^{\infty}\setminus \{0\}$ with product-topology and factorize by the relation $x\sim y \Leftrightarrow x=ty$ for some $t>0$ (infinite-dimensional sphere). Then consider only points with rational coordinates, all but finitely of them vanishing. The first question is whether are these two examples homeomorphic or somehow related. The second is an historical one. I've heard that the first example of such space belongs to P. S. Urysohn. What was his example? • Just to be sure I get the example (2): would it be the same starting from $\mathbb{Q}^\infty$ with the product topology; then taking the subspace of sequences with all but finitely many vanishing coord.; then the set of all rays in it with the quotient topology? – Pietro Majer Nov 23 '10 at 12:17 • @Pietro: yes, it is the same (except isolated point 0 maybe) – Fedor Petrov Nov 23 '10 at 12:56 • Sorry for the late comment. Connected metrizable spaces cannot be countable. This means that infinite dimensional sphere with rational coordinates is not metrizable. How is that possible? Any two points lie on some $S^n$ and we can define the distance between them. Why isn't it a distance function for this topology? – Mihail Oct 22 '17 at 13:00 • @Mihail this defines another topology! – Fedor Petrov Oct 22 '17 at 13:42 First let us fix the terminology. The space (1) is known in General Topology as the Golomb space. More precisely, the Golomb space $\mathbb G$ is the set $\mathbb N$ of positive integers, endowed with the topology generated by the base consisting of arithmetic progressions $a+b\mathbb N_0$ where $a,b$ are relatively prime natural numbers and $\mathbb N_0=\{0\}\cup\mathbb N$. Let us call the space (2) the rational projective space and denote it by $\mathbb QP^\infty$. Both spaces $\mathbb G$ and $\mathbb QP^\infty$ are countable, connected and Hausdorff but they are not homeomorphic. A topological property distinguishing these spaces will be called the oo-regularity. Definition. A topological space $X$ is called oo-regular if for any non-empty disjoint open sets $U,V\subset X$ the subspace $X\setminus(\bar U\cap\bar V)$ of $X$ is regular. Theorem. 1. The rational projective space $\mathbb QP^\infty$ is oo-regular. 2. The Golomb space $\mathbb G$ is not oo-regular. Proof. The statement 1 is relatively easy, so is left to the interested reader. The proof of 2. In the Golomb space $\mathbb G$ consider two basic open sets $U=1+5\mathbb N_0$ and $V=2+5\mathbb N_0$. It can be shown that $\bar U=U\cup 5\mathbb N$ and $\bar V=V\cup 5\mathbb N$, so $\bar U\cap\bar V=5\mathbb N$. We claim that the subspace $X=\mathbb N\setminus (\bar U\cap\bar V)=\mathbb N\setminus 5\mathbb N$ of the Golomb space is not regular. Consider the point $x=1$ and its neighborhood $O_x=(1+4\mathbb N)\cap X$ in $X$. Assuming that $X$ is regular, we can find a neighborhood $U_x$ of $x$ in $X$ such that $\bar U_x\cap X\subset O_x$. We can assume that $U_x$ is of basic form $U_x=1+2^i5^jb\mathbb N_0$ for some $i\ge 2$, $j\ge 1$ and $b\in\mathbb N\setminus(2\mathbb N_0\cup 5\mathbb N_0)$. Since the numbers $4$, $5^j$, and $b$ are relatively prime, by the Chinese remainder Theorem, the intersection $(1+5^j\mathbb N_0)\cap (2+4\mathbb N_0)\cap b\mathbb N_0$ contains some point $y$. It is clear that $y\in X\setminus O_x$. We claim that $y$ belongs to the closure of $U_x$ in $X$. We need to check that each basic neighborhood $O_y:=y+c\mathbb N_0$ of $y$ intersects the set $U_x$. Replacing $c$ by $5^jc$, we can assume that $c$ is divisible by $5^j$ and hence $c=5^jc'$ for some $c'\in\mathbb N_0$. Observe that $O_y\cap U_x=(y+c\mathbb N_0)\cap(1+4^i5^jb\mathbb N_0)\ne\emptyset$ if and only if $y-1\in 4^i5^jb\mathbb N_0-5^jc'\mathbb N_0=5^j(4^ib\mathbb N_0-c'\mathbb N_0)$. The choice of $y\in 1+5^j\mathbb N_0$ guarantees that $y-1=5^jy'$. Since $y\in 2\mathbb N_0\cap b\mathbb N_0$ and $c$ is relatively prime with $y$, the number $c'=c/5^j$ is relatively prime with $4^ib$. So, by the Euclidean Algorithm, there are numbers $u,v\in\mathbb N_0$ such that $y'=4^ibu-c'v$. Then $y-1=5^jy'=5^j(4^ibu-c'v)$ and hence $1+4^i5^ju=y+5^jc'v\in (1+4^i5^jb\mathbb N_0)\cap(y+c\mathbb N_0)=U_x\cap U_y\ne\emptyset$. So, $y\in\bar U_x\setminus O_x$, which contradicts the choice of $U_x$. Remark. Another well-known example of a countable connected space is the Bing space $\mathbb B$. This is the rational half-plane $\mathbb B=\{(x,y)\in\mathbb Q\times \mathbb Q:y\ge 0\}$ endowed with the topology generated by the base consisting of sets $$U_{\varepsilon}(a,b)= \{(a,b)\}\cup\{(x,0)\in\mathbb B:\|x-(a-\sqrt{2}b)\|<\varepsilon\}\cup \{(x,0)\in\mathbb B:\|x-(a+\sqrt{2}b)\|<\varepsilon\}$$ where $(a,b)\in\mathbb B$ and $\varepsilon>0$. It is easy to see that the Bing space $\mathbb B$ is not oo-regular, so it is not homeomorphic to the rational projective space $\mathbb QP^\infty$. Problem 1. Is the Bing space homeomorphic to the Golomb space? Remark. It is clear that the Bing space has many homeomorphisms, distinct from the identity. So, the answer to Problem 1 would be negative if the answer to the following problem is affirmative. Problem 2. Is the Golomb space $\mathbb G$ topologically rigid? Problem 3. Is the Bing space topologically homogeneous? Since the last two problems are quite interesting I will ask them as separate questions on MathOverFlow. Added in an edit. Problem 1 has negative solution. The Golomb space and the Bing space are not homeomorphic since 1) For any non-empty open sets $U_1,\dots,U_n$ in the Golomb space (or in the rational projective space) the intersection $\bigcap_{i=1}^n\bar U_i$ is not empty. 2) The Bing space contain three non-empty open sets $U_1,U_2,U_3$ such that $\bigcap_{i=1}^3\bar U_i$ is empty. Added in a next edit. Problem 2 has a partial affirmative solution: 1 is a fixed point of any homeomorphism of $\mathbb G$. This implies that $\mathbb G$ is not homeomorphic to the Bing space or the rational projective space (which do not have such a fixed point). • I was about to reflexively fix your non-TeX (oo → $\infty$) before I realised that I was missing the point. :-) – LSpice Nov 8 '17 at 20:53 • @LSpice I am also hesitating about this new term oo-regularity; oo is chosen to symbolize two disjoint open sets. I also considered something like "rim-regular", "corim-regular", but "rim-regular" should mean having a base of the topology consisting of sets with topologically regular boundaries; corim-regular is too long. So, no better idea than oo-regularity :( By the way, the Bing space is an example of a non-regular rim-regular space. – Taras Banakh Nov 8 '17 at 20:58 • For the record, I wasn't criticising it; I think it's on the right side of cuteness. (The example of "fascist functor" in place of "cofree functor" probably shows that there is also a wrong side of cuteness.) I just misread it at first, and fortunately caught myself before 'correcting' it. – LSpice Nov 8 '17 at 21:26 Urysohn's example of a countable connected Hausdorff space with a countable base was published in his last paper «Über die Mächtigkeit der zusammenhängenden Mengen», Math Annalen 94 (1925), 262—295. Urysohn's original description of the space occupies about 4 pages so I would rather refrain from reproducing it here. This is probably the most complicated construction of its kind. Simpler examples were later obtained by Bing, Hewitt, Stone and others («Countable connected spaces» by Miller contains many relevant references). There is also a Russian translation of Urysohn's paper in his collected works • Урысон П.С. Труды по топологии и другим областям математики [Том 1], ГИТТЛ, 1951, стр. 177-214. • In example (2) the space is a quotient of a products of real lines, then the phrase "only points with rational coordinates" isnt well definited. May be you considering the point of the product with all but a finite vanish, and have distance $1$ from ${0}$? (i.e . the point on the 1-sphere by rational coordinates) Anyway here there is a (very nice) example of Bing: ams.org/journals/proc/1953-004-03/S0002-9939-1953-0060806-9/… – Buschi Sergio Nov 22 '10 at 20:37 • @Sergio: well, we may consider say "points with rational ratios of coordinates", it is maybe more rigorous, but actually does not matter. Thanks for the link! – Fedor Petrov Nov 22 '10 at 20:58 • by the way, how may translation of 34-pages paper consist of 5 pages? – Fedor Petrov Nov 22 '10 at 21:04 • @Fedor: I have edited the answer. The example itself is described in Section 3 of the paper, pp. 189-193 in the Russian translation. – Andrey Rekalo Nov 22 '10 at 21:24	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9209873676300049, "perplexity": 354.58517050126204}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550247489282.7/warc/CC-MAIN-20190219000551-20190219022551-00511.warc.gz"}
http://math.stackexchange.com/questions/19575/reasoning-about-lie-theory-and-the-exponential-map	# Reasoning about Lie theory and the Exponential Map I'm having a little difficulty wrapping my head around Lie theory (I'm a computer scientist, so perhaps that's to be expected). Specifically, considering the following definition from Wikipedia for the exponential map. What is the significance of the identity element? I think that this could be to do with the derivative of exp being equal to the function itself, but I'm not quite sure if I'm on the right track here. Any nudges in the right direction would be highly appreciated. "Let G be a Lie group and $\mathfrak{g}$ be its Lie algebra (thought of as the tangent space to the identity element of G). The exponential map is a map $\exp\colon \mathfrak g \to G$ which can be defined in several different ways." - ## 3 Answers The answers given are both really good; I just want to add an additional reason why the identity element is preferred in consideration of the exponential map. The reason is because the exponential map gives a one to one correspondence between tangent vectors at the identity and one parameter subgroups of $G$. A one parameter subgroup of $G$ is a homomorphism $\mathbb R \to G$, or equivalently a curve $\gamma(t)\in G$ such that $\gamma(t+s) = \gamma(t)\gamma(s)$. Since $\gamma$ is a group homomorphism, we need $\gamma(0)$ to be the identity $e\in G$ (so here we are using the special property of $e$) and any such one parameter subgroup is determined by $\gamma'(0) \in T_e G$. Conversely, given $X \in T_e G$, the curve $t\mapsto \exp(tX)$ is a one-parameter subgroup with derivative $X$ at time 0. - The elements of the Lie algebra can be thought of as "infinitesimal elements," so it's natural to think of them as being infinitesimally close to the identity; if we denote by $\epsilon$ an infinitesimal that squares to zero, then you can informally think of the Lie algebra as the set of elements of the form $I + \epsilon X$ where $I$ is the identity. For example, the elements of the orthogonal group satisfy $A \cdot A^T = I$, so the infinitesimal elements of the orthogonal group satisfy $(I + \epsilon X)(I + \epsilon X^T) = I + \epsilon (X + X^T) = 0$, or $X + X^T = 0$. Hence the Lie algebra of the orthogonal group is precisely the space of skew-symmetric matrices. (This argument can be made completely formal if we work with algebraic groups instead of Lie groups.) There is a precise sense in which the Lie algebra consists of "infinitesimal elements": whenever a Lie group $G$ acts on a manifold $M$ by symmetries, the Lie algebra $\mathfrak{g}$ acts by differential operators on the smooth functions $M \to \mathbb{R}$. So they "infinitesimally generate" symmetries of $M$ (and the precise sense in which this is true is the exponential map). If you prefer, there is an equivalent definition which does not privilege the identity element: the elements of the Lie algebra are the left-invariant vector fields on the Lie group. (Since a Lie group acts transitively on itself, a left-invariant vector field is determined by its value at any group element, and again, it's natural to look at the identity element.) - I am novice in Lie theory myself, but I'll try to answer your question about the significance of identity element. There is, in fact, no significance. The identity element is as good as any other element in the connected component of Lie group, and the tangent to the identity element can be moved anywhere, as long as we stay in the connected component. However, it is easier to perform calculations involving the identity element than those that involve other elements. This is why the identity element is used. -	{"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": 0.9395377039909363, "perplexity": 73.0100812044602}, "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-2016-26/segments/1466783403825.35/warc/CC-MAIN-20160624155003-00180-ip-10-164-35-72.ec2.internal.warc.gz"}
https://infoscience.epfl.ch/record/166539	Infoscience Conference paper # Practical low complexity linear equalization for MIMO-CDMA systems This article first reviews recently proposed techniques for adaptive and direct linear MIMO equalization in the context of MIMO-CDMA systems and in particular with application to a MIMO-extended UMTS-FDD downlink. The focus is thereby mainly on the complexity of the algorithms. The second part of the paper proposes frequency domain (FD) MIMO equalization using the overlap/add FFT method in conjunction with two different low-complexity FD-deconvolution techniques to obtain the equalizer coefficients based on explicit channel impulse response estimates. The effects of imperfect channel estimation are discussed. An architecture for the VLSI implementation of the proposed method is suggested and an estimate of the complexity of the proposed circuit is given in the conclusions.	{"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": 0.8172260522842407, "perplexity": 1026.3128041317198}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501171834.68/warc/CC-MAIN-20170219104611-00295-ip-10-171-10-108.ec2.internal.warc.gz"}
http://www.cfd-online.com/Forums/openfoam-programming-development/93745-one-question-about-turbulence-ueqn-print.html	CFD Online Discussion Forums (http://www.cfd-online.com/Forums/) - OpenFOAM Programming & Development (http://www.cfd-online.com/Forums/openfoam-programming-development/) - - one question about turbulence + Ueqn ? (http://www.cfd-online.com/Forums/openfoam-programming-development/93745-one-question-about-turbulence-ueqn.html) lfgmarc October 25, 2011 10:57 HsEqn ReactingFoam Hypothesis? Hi, I'm a little confuse with the assumptions under HsEqn implemented in reactingFoam, the energy equation in terms of the sensible enthalpy, can be written as: My confusion is with two terms: 1. the term associated with the species diffusion: 2, the term associated with the energy dissipation by viscous effects: The implementation of this equation in reactingFoam is: Code: fvm::ddt(rho, hs) + mvConvection->fvmDiv(phi, hs) - fvm::laplacian(turbulence->alphaEff(), hs) // - fvm::laplacian(turbulence->muEff(), hs) // unit lewis no. == DpDt + chemistrySh In this implementation I see that is neglected the viscous dissipation, but I don't understand very well why? i think that may be under the hypothesis of an Brinkman number<<1? On the other hand it is clear that the species diffusion term was neglected, but reviewed bibliography I see that this term ussually is neglected if : a) if mixture contains only one specie b) if all species have the same sensible enthalpy. c) sometimes is set to zero because it is usually negligible compared with The first assumption in this case is not suitable, I'm confused with the second one, it is applicable in this case or simply is considered the third hypothesis? if someone could shed some light on this I'll be very grateful	{"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": 0.879414439201355, "perplexity": 3212.137537102154}, "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-2014-10/segments/1394021767060/warc/CC-MAIN-20140305121607-00049-ip-10-183-142-35.ec2.internal.warc.gz"}
http://mathhelpforum.com/differential-geometry/183901-prove-application-closed.html	# Thread: Prove that the application is closed 1. ## Prove that the application is closed Hi, everyone, I resolved the following exercise. I need help reviewing it and giving the correct answer (if mine isn't correct of course). Let's have: $I_1=\{(x,y) \in R^2\|y=0, 0\leq x\leq 1\}$ $I_2=\{(x,y)\in R^2\|x=1, 0\leq y\leq 1\}$ $I_3=\{(x,y)\in R^2\|y=x; x,y \in [0 \;1]\}$ $Q=I_1 \cup I_2 \cup I_3$ where the following equivalence relation is defined $\A (x,y), (x',y') \in Q$, $(x,y)\sim (x',y')$ if $(x,y)=(x',y')$ or $(x, y), (x', y')\in I_1$, $(x,y),(x',y')\in I_2$ Prove that $\pi : Q\rightarrow Q/\sim$ is closed. Solution: $Q$ is connected and compact so even $Q/\sim$ must be connected and compacted. At this point we have to prove that $Q/\sim$ is $T_2$ (Hausdorff). That is to prove that the set $F= \{ (x;y),(u;v): \pi(x;y)=\pi(u;v) \}=\{ (x;y),(u;v): (x;y)\sim(u;v) \}$ is closed on $Q\times Q$. $C_1=\{ (x;y),(x;y)\in Q\times Q\| (x;y)\in Q}$, $C_2=\{ (x;0),(x;0)\in Q\times Q\| x\in[0\;1]\}$ $C_3=\{ (1;y),(1;y)\in Q\times Q\| y\in[0\;1]\}$. $C_1$ is closed because $Q$ is $T_2$. $C_2$, $C_3$ are closed being subspaces of space $T_2$ and immages of the compact $[0\;1]$ on $R$ through continuous applications. Because $F=C_1 \cup C_2 \cup C_3$ then even $F$ will be closed, this means that $Q/\sim$ is $T_2$. I conclude that $\pi$ is closed. 2. ## Re: Prove that the application is closed There are things I don't understant in your explanation. F, as defined, seems to be en element of Q/~ X Q/~ while C1,C2 and C3 are in Q X Q 3. ## Re: Prove that the application is closed Sorry, but F is defined in $Q \times Q$, then I use the theorem that say: If F is closed in $Q\times Q$ then $Q/\sim$ is $T_2$ and this means that $\pi$ is closed. 4. ## Re: Prove that the application is closed I think that $C_2\subset C_1$ and $C_3\subset C_1$ and threre are elements of F that are not in C1	{"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": 39, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9391905069351196, "perplexity": 514.8809898136789}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560280587.1/warc/CC-MAIN-20170116095120-00220-ip-10-171-10-70.ec2.internal.warc.gz"}
http://mathhelpforum.com/discrete-math/218389-numerical-differentiation.html	# Math Help - Numerical differentiation 1. ## Numerical differentiation What is the basic principle of numerical differentiation? 2. ## Re: Numerical differentiation What kind of derivative do you want to calculate? Do you want to for example, take some signal and find a smooth function that approximates that and take the derivative of that function? Can you give an example of what you want to do? 3. ## Re: Numerical differentiation This question was set in a University exam. 4. ## Re: Numerical differentiation What topic? As it stands the question is too broad and not clear enough to know what the questioner wants. 5. ## Re: Numerical differentiation I can't imagine what kind of answer was expected to that! My first thought was that you cannot just "do the obvious", convert the limit of $\frac{f(a+h)- f(a)}{h}$ to the fraction itself, with small h, because both numerator and denominator are so small "round off error" will be too large. What most numerical algorithms do is approximate f(x) by some specific kind of function, a polynomial or exponential, and take the derivative of that function. But I don't think I would call that the "basic principle". 6. ## Re: Numerical differentiation Hi, I would disagree a little with the previous response. Sometimes you can get away with using the difference quotient with carefully chosen h. A short discussion is found in Numerical Recipes by Press et. al. I don't have the 3rd edition, but the 2nd edition discussion starts on page 186. 7. ## Re: Numerical differentiation It was asked in the context of Numerical analysis. The syllabus of that paper includes Newtons forward and backward interpolation formula and differentiation based on these. The question carries 2 marks. 8. ## Re: Numerical differentiation In that case, find an interpolating function (polynomial) and get its derivative.	{"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": 0.944983720779419, "perplexity": 786.4966795318455}, "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/1408500804220.17/warc/CC-MAIN-20140820021324-00315-ip-10-180-136-8.ec2.internal.warc.gz"}
https://www.maplesoft.com/support/help/errors/view.aspx?path=geometry/sides	sides - Maple Help geometry sides compute the sides of a given triangle or a given square Calling Sequence sides(g) Parameters g - triangle or square Description • The routine computes the sides of triangle g or square g. • In case g is a triangle, the output is a list of three element denoting the three sides of g. • The command with(geometry,sides) allows the use of the abbreviated form of this command. Examples > $\mathrm{with}\left(\mathrm{geometry}\right):$ > $\mathrm{triangle}\left(\mathrm{ABC},\left[\mathrm{point}\left(A,0,0\right),\mathrm{point}\left(B,3,0\right),\mathrm{point}\left(C,0,4\right)\right]\right):$ > $\mathrm{sides}\left(\mathrm{ABC}\right)$ $\left[\sqrt{{9}}{,}\sqrt{{25}}{,}\sqrt{{16}}\right]$ (1) > $\mathrm{point}\left(C,0,3\right),\mathrm{point}\left(E,3,3\right):$ > $\mathrm{square}\left(\mathrm{Sq},\left[A,B,E,C\right]\right):$ > $\mathrm{sides}\left(\mathrm{Sq}\right)$ $\frac{\sqrt{{18}}{}\sqrt{{2}}}{{2}}$ (2)	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 8, "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": 0.9860460758209229, "perplexity": 875.8010017383718}, "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-2022-49/segments/1669446710712.51/warc/CC-MAIN-20221129232448-20221130022448-00753.warc.gz"}
https://gateoverflow.in/321846/isi2015-mma-31	16 views Consider the sets defined by the real solutions of the inequalities $$A = \{(x,y):x^2+y^4 \leq 1 \} \:\:\:\:\:\:\:\: B = \{ (x,y):x^4+y^6 \leq 1\}$$ Then 1. $B \subseteq A$ 2. $A \subseteq B$ 3. Each of the sets $A – B, \: B – A$ and $A \cap B$ is non-empty 4. none of the above edited \| 16 views 1 +1 vote 2 +1 vote 3	{"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": 0.9910415410995483, "perplexity": 2086.7705408422516}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250604397.40/warc/CC-MAIN-20200121132900-20200121161900-00404.warc.gz"}
https://www.isa-afp.org/entries/MFOTL_Monitor.html	# Formalization of a Monitoring Algorithm for Metric First-Order Temporal Logic Title: Formalization of a Monitoring Algorithm for Metric First-Order Temporal Logic Authors: Joshua Schneider and Dmitriy Traytel Submission date: 2019-07-04 Abstract: A monitor is a runtime verification tool that solves the following problem: Given a stream of time-stamped events and a policy formulated in a specification language, decide whether the policy is satisfied at every point in the stream. We verify the correctness of an executable monitor for specifications given as formulas in metric first-order temporal logic (MFOTL), an expressive extension of linear temporal logic with real-time constraints and first-order quantification. The verified monitor implements a simplified variant of the algorithm used in the efficient MonPoly monitoring tool. The formalization is presented in a RV 2019 paper, which also compares the output of the verified monitor to that of other monitoring tools on randomly generated inputs. This case study revealed several errors in the optimized but unverified tools. Change history: [2020-08-13]: added the formalization of the abstract slicing framework and joint data slicer (revision b1639ed541b7) BibTeX: @article{MFOTL_Monitor-AFP, author = {Joshua Schneider and Dmitriy Traytel}, title = {Formalization of a Monitoring Algorithm for Metric First-Order Temporal Logic}, journal = {Archive of Formal Proofs}, month = jul, year = 2019, note = {\url{https://isa-afp.org/entries/MFOTL_Monitor.html}, Formal proof development}, ISSN = {2150-914x}, } License: BSD License Depends on: Containers Used by: Generic_Join, MFODL_Monitor_Optimized	{"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": 0.8601087331771851, "perplexity": 3156.2884909167365}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323583408.93/warc/CC-MAIN-20211016013436-20211016043436-00714.warc.gz"}
https://math.stackexchange.com/questions/2873722/alternative-definition-of-sheaf	# Alternative definition of “sheaf” Let $$(X,\tau)$$ denote a topological space and $$\mathcal{O}$$ denote a presheaf on this space with codomain $$\mathbf{Set}$$. We can take the category of elements of $$\mathcal{O}$$, which consists of a poset $$\mathrm{el}(\mathcal{O}) = \{(U,f) : U \in \tau, f \in \mathcal{O}(U)\}$$ together with a forgetful map $$\pi : \mathrm{el}(\mathcal{O}) \rightarrow \tau$$ satisfying certain properties. If $$\cal O$$ happens to be a sheaf, this should be reflected in the structure of $$(\mathrm{el}(\mathcal{O}),\pi).$$ There should consequently be a definition of sheaf like so: Let $$(X,\tau)$$ denote a topological space. Then a sheaf on $$X$$ consists of a poset $$P$$ togther with a monotone map $$\pi : P \rightarrow \tau$$ such that the following axioms are satisfied: (a) (b) (c) (whatever)... I'm a bit unsure what these conditions should be (even for a presheaf). We want to be able to restrict elements of $$P$$ to arbitrary opens, which makes me think we should view $$P$$ as a "$$\tau$$-module", by which I mean that for all opens $$U \in X$$ and all $$f \in P$$, we can form the restriction $$U \cap f$$ which would normally be denoted $$f \restriction_U$$. The usual axioms of an action hold, e.g $$X \cap f = f, \qquad U \cap (V \cap f) = (U \cap V) \cap f.$$ I'm not quite sure whether this module structure should be viewed as extra data, or whether it can be recovered from the map $$\pi$$. Note that we have $$\pi(U \cap f) = U \cap \pi(f)$$, for example. Ideas, anyone? Addendum. I just learned that local homeomorphisms into a space $$X$$ are in bijective correspondence with sheaves on $$X$$. This doesn't actually answer the question, but it's related. • For the "$\tau$"-module structure, you just want $\pi$ to be a discrete Grothendieck fibration. – Pece Aug 6 '18 at 9:27 • Why should $P$ be a poset ? I mean how do you order $(U,f)$ and $(U,g)$ for $f,g\in \mathcal{O}(U)$ ? Shouldn't $P$ be a category ? – Max Aug 6 '18 at 11:31 • @Max, the order relation should be $$(U,f) \leq (V,g) \iff V \supseteq U \wedge f\restriction_U = g$$ if I'm not mistaken. – goblin Aug 6 '18 at 11:35 • @goblin : ah indeed, my bad ! – Max Aug 6 '18 at 11:45 • It seems to me that with this ordering there may be a way to recover some stuff with $\pi$ and the notions of lower bounds : $s,t\in P$ are compatible if and only if they have a lower bound $r$ such that $\pi(r) = \pi(s)\cap \pi(t)$; and so you can express the gluing axiom (it seems) – Max Aug 6 '18 at 11:52	{"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": 24, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9520228505134583, "perplexity": 172.79165505341976}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514574182.31/warc/CC-MAIN-20190921022342-20190921044342-00022.warc.gz"}
https://forum.math.toronto.edu/index.php?PHPSESSID=vhlf7me7j3nf33mm7i366pnm57&action=printpage;topic=2547.0	# Toronto Math Forum ## MAT244--2020F => MAT244--Lectures & Home Assignments => Chapter 4 => Topic started by: Sophie Liao on November 18, 2020, 12:26:03 PM Title: Variation of Parameter for higher order eqautions Post by: Sophie Liao on November 18, 2020, 12:26:03 PM Since we have learned the variation of parameter method for second and third order equation, I just want to confirm whether my idea is correct or not. For fourth order eqatuion or higher, should we compute the det of 4 by 4 matrix or even higher to use this method? Is there any general theorem for this method? Title: Re: Variation of Parameter for higher order eqautions Post by: shiyuancao on November 18, 2020, 09:34:38 PM I think your idea is correct and it's unmistakable to apply the same principle when solving higher order equations	{"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": 0.8199793696403503, "perplexity": 1418.038703345813}, "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/1637964363520.30/warc/CC-MAIN-20211208175210-20211208205210-00149.warc.gz"}
http://www.maa.org/press/maa-reviews/how-to-solve-applied-mathematics-problems	# How to Solve Applied Mathematics Problems ###### B. L. Moiseiwitsch Publisher: Dover Publications Publication Date: 2011 Number of Pages: 324 Format: Paperback Price: 19.95 ISBN: 9780486479279 Category: Textbook [Reviewed by Tom Schulte , on 04/14/2012 ] This original Dover publication boasts a title suggesting that it is a blueprint for mathematical modeling. On cracking open the text I expected to find out what gamut of problems are approached and at what level the mathematics is. I was disappointed to find contents more like that REA's Problem Solvers series: bluntly stated problems immediately following by a complete solution. There is no discussion of notation, theory, or building up a model from a real world situation. What it lacks in REA’s emphatic redundancy, this collection makes up in breadth. Engineering and physics are the focus of the presented problems. Topics covered include vector algebra, kinematics, fluid dynamics, electricity, magnetism, Fourier series, Laplace transforms, wave motion, heat conduction, tensor analysis, relativity, quantum theory, and more. The result is a survey of the main tools of applied mathematics in what is a potpourri of examples that is illustrative although not educational. Rather than an enlightening text, this book is more appropriate for students seeking an adjunct to existing texts or educators that need a question bank. The cover promises that the book "bridges the gap between lectures and practical applications, offering students of mathematics, engineering, and physics the chance to practice solving problems from a wide variety of fields." I think anyone expecting exactly what that implies will be disappointed. I do not see this work as useful to anyone that has not already grasped the relevant lectures or has other tools to do so. If the idea is to practice solving problems, then this work would have multiple problems of each type and solutions would be separated from the problems. The general principles laid out at the end of the preface are sage advice, but curiously are not followed by the author. It is indeed very good advice to “introduce a good notation” when solving applied mathematical problems, yet this work assumes the reader already knows and appreciates the presented notation, which is taken as established and without need of explanation. It is of course key to “try to understand what you want to prove” but here the breathless pace does not allow for context nor gives the reader a chance to make any inference. Finally, it is a surprise to see so many problems from classical mechanics, wave motion, etc. without a single illustration. Tom Schulte, a teacher of mathematics, enjoys many Dover titles in his home in Waterford, Michigan. Preface Introduction 1. Vector Algebra 2. Kinematics 3. Dynamics of a Particle 4. Vector Field Theory 5. Newtonian Gravitation 6. Electricity and Magnetism 7. Fluid Dynamics 8. Classical Dynamics 9. Fourier Series, Fourier and Laplace Transforms 10. Integral Equations 11. Wave Motion 12. Heat Conduction 13. Tensor Analysis 14. Theory of Relativity 15. Quantum Theory 16. Variational Principles Bibliography Index	{"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": 0.8293270468711853, "perplexity": 1302.3757691517549}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-07/segments/1454701152959.66/warc/CC-MAIN-20160205193912-00167-ip-10-236-182-209.ec2.internal.warc.gz"}
https://www.parabola.unsw.edu.au/1990-1999/volume-34-1998/issue-2/article/how-construct-regular-7-sided-polygons-part-2	# How to Construct Regular 7-sided Polygons - Part 2 In Part 1 (Parabola Vol. 34, No 1) we introduced you to a basic construction whereby we folded down m times at the top of a tape and folded up n times at the bottom of the tape (see Figure 1).	{"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": 0.9113622307777405, "perplexity": 852.3242763008711}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780057131.88/warc/CC-MAIN-20210921011047-20210921041047-00297.warc.gz"}
http://physics.stackexchange.com/questions/71428/intensive-radiative-heat-transfer-in-very-hot-gas-5000k-gas-core-nuclear-rea	# Intensive radiative heat transfer in very hot gas ( >5000K, gas core nuclear reactor related ) The question: At temperatures above ~5000K are not stable any solid or liquid materials or even more complex molecules (such as fullerenes and Polycyclic aromatic hydrocarbons) which emit/absorb wide-spectral black-body radiation effectively. Simple molecules and atoms which survive such temperatures have very narrow absorption/emission lines and small absorption/emission cross-section. => They are transparent for thermal radiation. Is there anything which would emit/absorb broadband thermal radiation at temperatures >10,000 K ? What about plasma? I guess plasma must be both highly ionized (~hot) and at the same time dense in order to have high opacity and absorption for thermal light (Am I right?). This is a bit contradictory. Density is inversely proportional to temperature if we are limited by pressure let's say 100MPa, while ionization is increasing with temperature. Is it possible to increase opacity/absorption of gas/plasma at ~10,000-100,000 K by seeding with alkali metal which release electron very easily ? background: Gas core nuclear reactor, and derived gas core nuclear rocket is a fission nuclear reactor theoretically able to achieve temperatures above 5000K (some proposals talk about 40,000-100,000 Kelvin ) which is important for high energy efficiency of electricity production (using MHD generators ) and high specific impulse of nuclear rocket. There is a concept of space propulsion called nuclear lightbulb rocket which should achieve high specific impulse exhausting moleculer/atomic hydrogen propellant at velocity up to 20-40 km/s which means temperature ~25,000-90,000 Kelvin. For effective function of such engine it is necessary that all the heat from gaseous nuclear core is transferred to propellant by thermal radiation. However, hydrogen propellant itself is almost transparent for thermal radiation. This would cause that the heat is transfered to the walls of rocket nozzle instead, which would melt the walls and destroy the reactor. In order to make propellant opaque it was proposed that tiny dust particles of tungsten or hafnium-tantalum carbide particles would be dispersed in propellant gas. These particles would however evaporate above ~5000K making propellant gas transparent again. -	{"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": 0.9263139963150024, "perplexity": 3143.2014673063377}, "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-2015-18/segments/1429246656887.93/warc/CC-MAIN-20150417045736-00090-ip-10-235-10-82.ec2.internal.warc.gz"}
http://preppedandpolished.com/tag/isee-middle-level/	## New SSAT Changes Alexis Avila Founder/President of Prepped & Polished, LLC in South Natick, Massachusetts lists the five most important changes to the SSAT test. 1. The SSAT created a new Elementary Level Test for 3rd and 4th graders. 2. The SSAT Lower Level test is now called the SSAT Middle Level Test. 3. Teachers will write the SSAT questions. 4. There is now an experimental section. 5. The SSAT Writing prompts have changed. Transcript (PDF) Full Word-for-Word Transcription Hey everyone, Alexis Avila, Prepped & Polished, LLC here at South Natick, Massachusetts, and Happy New Year. It’s 2013. There have been some changes to the SSAT, the private school admissions test as of late. I’m going to let you know about five of these changes.The first change is the SSAT test now has the elementary level test for students currently in third and fourth grade. Basically, the elementary level test is an abbreviated version of the SSAT test. It has all the classic SSAT sections. You have the quantitative math section, verbal section, which consists of synonyms and analogies, the reading comprehension section, which is basically 7 passages each with 4 questions, and then you’ll have a writing section, which is basically a 15 minute section of a student is shown a picture and then asked to tell a story with a beginning, a middle, and end about what happened in the picture. The writing section, of course, is not officially scored, but sent right to private schools.Change number two is SSAT has officially renamed what was previously known as the Lower Level Test to the SSAT Middle Level Test. If you are currently in grades 5th, 6th, and 7th grade, you will take the SSAT Middle Level Test. If you are currently in grades 8th through 11th grade, you will take the SSAT Upper Level Test as always.Change number 3 is now 100% of the test questions on the SSAT will be written by independent and private school teachers rather than the corporate test writing service to write the questions. The SSAT basically wants these questions to adequately depict the material found in independent and private schools.Change number four to the SSAT is now the SSAT will incorporate an experimental section, but it does not count towards the student score. It’s kind of similar to the SAT test that you find in high school, but in this case, you’ll get 16 extra questions that the SSAT will analyze to determine if they’re relevant for future tests.Change number five to the SSAT is in the writing section for Middle Level and Upper Level test takers. So if you’re taking the Middle Level SSAT test, you will be presented with two creative prompts, and you chose to write one. If you are going to take the Upper Level SSAT test, you will be presented with a creative prompt and an essay prompt and choose one. If you need some examples on types of creative prompts that they might present you, I highly encourage you to go to the SSAT.org website and order the official SSAT Study Guide. Just look around that website to see if you can get some free information.If you have further questions, feel free to email me at [email protected]. I wish you good luck on the SSAT. I will talk to you soon. Are you taking the SSAT? What questions do you have about the SSAT changes? Subscribe to our Blog Feed January 30th, 2013 Tagged , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , Posted in Featured, SSAT ## Mistake to avoid in the SSAT Math Section Alexis Avila Founder/President of Prepped & Polished, LLC in South Natick, Massachusetts shows you on his whiteboard one crucial mistake you do not want to make on the SSAT Math section. After you do your math steps, make sure you go back to the question and answer exactly what the question is asking. Transcript (PDF) Full Word-for-Word Transcription Hey everyone, Alexis Avila, founder of Prepped & Polished LLC, here in South Natick, Massachusetts. If you want to do well on the SSAT math section, you have to wipe out those careless errors, and do steady, careful work on the SSAT math.Time and time again, whether you’re a lower level or upper level student, I see the same student make the same careless mistake on this particular problem. So let’s go to the board. I’m going to show you this problem. This is a relatively easy problem, but almost 75% of students get this one wrong.”A \$15 shirt is on sale for 20% off. What is the sale price of the shirt?” Students get really happy and excited, because they think to themselves that this is an easy and manageable problem. So, what they do is they’ll take \$15, get the 20% discount, so they’ll multiply it by .2, and then they will get \$3. Knowing this is a time-pressure test, they’re going to instantly go to choice ‘A’, and circle \$3. However, you don’t want to go with ‘A’. You don’t want to go with \$3. You went for the trap answer.You have to re-read the question every time. After doing the math, go back to the question and make sure you’re answering what the question is asking. They want what the sale price of the shirt is. You got the discount. Now you have to subtract 3 from 15. The new price of the shirt is \$12. You go with choice ‘C’ and you move on to the next question on the test. So just remember, you could be a really good math student, but not do well on the SSAT math if you keep making careless mistakes. Avoid careless mistakes and you’ll do well on the SSAT math section. I’ll talk to you soon. Good luck. How do you avoid making careless mistakes on the SSAT Math Section? Have you fallen trap to this type of question before? Subscribe to our Blog Feed December 4th, 2012 Tagged , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , Posted in Featured, SSAT ## Common SSAT Math Mistake Alexis Avila Founder/President of Prepped & Polished, LLC demonstrates on his whiteboard what not to do when solving an SSAT math problem. Make sure that you pay attention to the units of measurement on a math question, and when necessary, make sure you convert the units of measurement. Transcript (PDF) Full Word-for-Word Transcription Hey, everyone. Alexis Avila, founder of Prepped and Polished, LLC, here in South Natick, Massachusetts. Now, half the battle to doing well on the SSAT math section, pace yourself well and to avoid making careless mistakes. Now, I often see students make the same careless mistake over and over again with this particular problem, so I’m going to show you this problem, so you don’t make the same mistake. Let’s go to the board.”Nick buys a piece of licorice 150 inches long. If he plans to give away all of the licorice by giving each of his 5 friends an equal piece, how long should he cut each piece?” So the math is really easy with this problem. We’re simply going to take 150, which is the total length of the piece of licorice, and divide it by the 5 friends that he shares it with. That is going to give us 30. Now this is where students get the problem wrong. They’re going to say, “Oh, well, I solved the problem. I got 30.” They’re going to go to answer choice ‘B’ and circle it, which says 30 feet. But they didn’t convert their units of measurement correctly.They forgot that they have to take 30 inches, because 150 inches divided by five gives each friend 30 inches of licorice, and now we got to convert 30 inches into feet. So don’t forget to do your units of measurement. 30 inches, we’re going to divide it by 12 inches, and that’s going to give us how many feet is 30 inches. 2.5 feet. The answer is ‘E’, 2.5 feet, not the trap answer ‘B’. Just make sure you notice the units of measurement on math problems and convert them when appropriate. And overall, don’t make careless mistakes on the SSAT math section, and you’ll do fine. Do you make careless mistakes on the SSAT math section? How do you avoid making careless errors on this section? Subscribe to our Blog Feed October 25th, 2012 Tagged , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , Posted in Featured, SSAT ## SSAT and ISEE: Matching Grade with Level Alexis Avila Founder/President of Prepped & Polished, LLC uses his online whiteboard to show you which specific level SSAT or ISEE test you need to take. If your student is currently in 5th, 6th, or 7th grade he or she will take the SSAT Lower Level Test. SSAT Elementary Level is for students currently in 3rd or 4th grade, the SSAT Middle Level Test is for students currently in 5th, 6th, or 7th grade, the SSAT Upper Level is for students currently in 8th, 9th, 10th, or 11th grade. The ISEE Lower Level test is for students currently in 4th or 5th grade. The ISEE Middle Level test is for student currently in 6th or 7th grade. The ISEE Upper Level test is for students currently in 8th, 9th, 10th, or 11th grade. Transcript (PDF) Full Word-for-Word Transcription Hi, everyone. Alexis Avila, founder of Prepped and Polished LLC, here in South Natick, Massachusetts. Now, parents come to me all the time and ask me which SSAT or ISEE level test does my student need to take? So let’s get this out of the way once and for all. So let’s look on the board here. If your student is currently in 5th, 6th, or 7th grade, he or she takes the SSAT lower level test. (Update: Please note for students currently in 5th, 6th, and 7th grade, the SSAT recently renamed the test “SSAT Middle Level” and the new “SSAT Elementary Level Test” is for students currently in 3rd or 4th grade.)The ISEE lower level test is for students currently in 4th or 5th grade.Now, there is no SSAT middle level test, so we don’t have to worry about it. (Update!! There is now indeed an SSAT Middle Level Test -formerly it was called the “SSAT Lower Level Test “- and this is for students currently in 5th, 6th, and 7th grade). But there is an ISEE middle level test, and that’s for students currently in 6th or 7th grade. And finally, both the SSAT and ISEE upper level tests are for students currently in 8th through 11th grades. So this is how you’d match your student’s grade with the appropriate level. I hope this helps and I’ll talk to you soon.Updated Recap: SSAT Elementary Level is for students currently in 3rd or 4th grade, the SSAT Middle Level Test is for students currently in 5th, 6th, or 7th grade, the SSAT Upper Level is for students currently in 8th, 9th, 10th, or 11th grade. The ISEE Lower Level test is for students currently in 4th or 5th grade. The ISEE Middle Level test is for student currently in 6th or 7th grade. The ISEE Upper Level test is for students currently in 8th, 9th, 10th, or 11th grade. Which level SSAT and ISEE test will your student take? 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https://www.unav.edu/web/investigacion/nuestros-investigadores/detalle-investigadores-cv?investigadorId=58119&investigador=Maza+Ozcoidi%2C+Diego+Martin+	## Diego Maza Ozcoidi Departamento Líneas de investigación Medios Granulares y Materia Blanda, Cáos y Dinámica No Lineal, Fluidos complejos Índice H 24, (WoS, 20/06/2018) ## Publicaciones científicas más recientes (desde 2010) Autores: Gella, D.; Maza, D; Zuriguel, Iker; Revista: PHYSICAL REVIEW E ISSN 2470-0045 Vol. 95 Nº 5 2017 págs. 052904 We experimentally analyze the effect that particle size has on the mass flow rate of a quasi two-dimensional silo discharged by gravity. In a previous work, Janda et al. [Phys. Rev. Lett. 108, 248001 (2012)PRLTAO0031-900710.1103/PhysRevLett.108.248001] introduced a new expression for the mass flow rate based on a detailed experimental analysis of the flow for 1-mm diameter beads. Here, we aim to extend these results by using particles of larger sizes and a variable that was not explicitly included in the proposed expression. We show that the velocity and density profiles at the outlet are self-similar and scale with the outlet size with the same functionalities as in the case of 1-mm particles. Nevertheless, some discrepancies are evidenced in the values of the fitting parameters. In particular, we observe that larger particles lead to higher velocities and lower packing fractions at the orifice. Intriguingly, both magnitudes seem to compensate giving rise to very similar flow rates. In order to shed light on the origin of this behavior we have computed fields of a solid fraction, velocity, and a kinetic-stress like variable in the region above the orifice. Autores: Gella, D.; Maza, D; Zuriguel, Iker; et al. Revista: PHYSICAL REVIEW FLUIDS ISSN 2469-990X Vol. 2 Nº 8 2017 págs. 084304 We demonstrate experimentally that clogging in a silo correlates with some features of the particle velocities in the outlet proximities. This finding, that links the formation of clogs with a kinematic property of the system, is obtained by looking at the effect that the position of the lateral walls of the silo has on the flow and clogging behavior. Surprisingly, the avalanche size depends nonmonotonically on the distance of the outlet from the lateral walls. Apart from evidencing the relevance of a parameter that has been traditionally overlooked in bottleneck flow, this nonmonotonicity supposes a benchmark with which to explore the correlation of clogging probability with different variables within the system. Among these, we find that the velocity of the particles above the outlet and their fluctuations seem to be behind the nonmonotonicity in the avalanche size versus wall distance curve. Autores: Maza, D; Cruz, Raúl, (Autor de correspondencia) Revista: COMPUTATIONAL PARTICLE MECHANICS ISSN 2196-4378 Vol. 4 Nº 4 2017 págs. 419 - 427 Very recently, we have examined experimentally and numerically the micro-mechanical details of monodisperse particle flows through an orifice placed at the bottom of a silo (Rubio-Largo et al. in Phys Rev Lett 114:238002, 2015). Our findings disentangled the paradoxical ideas associated to the free-fall arch concept, which has historically served to justify the dependence of the flow rate on the outlet size. In this work, we generalize those findings examining large-scale polydisperse particle flows in silos. In the range of studied apertures, both velocity and density profiles at the aperture are self-similar, and the obtained scaling functions confirm that the relevant scale of the problem is the size of the aperture. Moreover, we find that the contact stress monotonically decreases when the particles approach the exit and vanish at the outlet. The behavior of this magnitude is practically independent of the size of the orifice. However, the total and partial kinetic stress profiles suggest that the outlet size controls the propagation of the velocity fluctuations inside the silo. Examining this magnitude, we conclusively argue that indeed there is a well-defined transition region where the particle flow changes its nature. The general trend of the partial kinetic pressure profiles and the location of the transition region results the same for all particle types. We find that the partial kinetic stress is larger for bigger particles. However, the small particles carry a higher fraction of kinetic stress respect to their concentration, which suggest that the small particles have larger velocity fluctuations than the large ones and showing lower strength of correlation with the global flow. Our outcomes explain why the free-fall arch picture has served to describe the polydisperse flow rate in the discharge of silos. Revista: PHYSICAL REVIEW E ISSN 2470-0045 Vol. 96 Nº 2 2017 págs. 022904 We present numerical and experimental results on the mass flow rate during the discharge of three-dimensional silos filled with a bidisperse mixture of grains of different sizes. We analyzed the influence of the ratio between coarse and fine particles on the profile of volume fraction and velocity across the orifice. By using numerical simulations, we have shown that the velocity profile has the same shape as that in the monodisperse case and is insensitive to the composition of the mixture. On the contrary, the volume fraction profile is strongly affected by the composition of the mixture. Assuming that an effective particle size can be introduced to characterize the mixture, we have shown that previous expression for the mass flow rate of monodisperse particles can be used for binary mixtures. A comparison with Beverloo's correlation is also presented. Autores: Gago, P. A.; Maza, D; Pugnaloni, L. A.; Revista: PAPERS IN PHYSICS ISSN 1852-4249 Vol. 8 2016 págs. 080001 Static granular packs have been studied in the last three decades in the frame of a modified equilibrium statistical mechanics that assumes ergodicity as a basic postulate. The canonical example on which this framework is tested consists in the series of static configurations visited by a granular column subjected to taps. By analyzing the response of a realistic model of grains, we demonstrate that volume and stress variables visit different regions of the phase space at low tap intensities in different realizations of the experiment. We show that the tap intensity beyond which sampling by tapping becomes ergodic coincides with the forcing necessary to break all particle-particle contacts during each tap. These results imply that the well-known "reversible" branch of tapped granular columns is only valid at relatively high tap intensities. Autores: Gago, P. A. ; Maza, D; Pugnaloni, L. A. ; Revista: PHYSICAL REVIEW E ISSN 1539-3755 Vol. 91 Nº 3 2015 págs. 032207 We investigate the steady-state packing fraction phi and force moment tensor Sigma of quasi-two-dimensional granular columns subjected to tapping. Systems of different height h and width L are considered. We find that phi and Sigma which describe the macroscopic state of the system, are insensitive to L for L > 50d (with d the grain diameter). However, results for granular columns of different heights cannot be conciliated. This suggests that comparison between results of different laboratories on this type of experiments can be done only for systems of same height. We show that a parameter epsilon = 1 + (A omega)(2)/(2gh), with A and omega the amplitude and frequency of the tap and g the acceleration of gravity, can be defined to characterize the tap intensity. This parameter is based on the effective flight of the granular bed, which takes into account the h dependency. When phi is plotted as a function of epsilon, the data collapses for systems of different h. However, this parameter alone is unable to determine the steady state to be reached since different Sigma can be observed for a given epsilon if different column heights are considered. Autores: Zuriguel, Iker; Garcimartín, Ángel; et al. Revista: GRANULAR MATTER ISSN 1434-5021 Vol. 17 Nº 5 2015 págs. 545 - 551 We report experimental results for pipe flow of granular materials discharged through vertical narrow tubes by means of a conveyor belt placed at the bottom. When the diameter of the tube is not much larger than the particle size, the system clogs due to the development of hanging arches that are able to support the weight of the grains above them. We find that the time it takes to develop a stable clog decays exponentially, which is compatible with a clogging probability that remains constant during the discharge. From this, and making an analogy with the discharge of silos, we introduce the avalanche size, measured in terms of the number of discharged tubes before the system clogs. The mean avalanche size is found to increase as the tube diameter is enlarged, the velocity of the conveyor belt grows, and the tube tilt deviates from the vertical. © 2015, Springer-Verlag Berlin Heidelberg. Autores: Lind, P. G. ; Maza, D; et al. Revista: COMPUTATIONAL PARTICLE MECHANICS ISSN 2196-4378 Vol. 2 Nº 2 2015 págs. 127 - 138 We present a hybrid GPU¿CPU implementation of an accurate discrete element model for a system of ellipsoids. The ellipsoids have three translational degrees of freedom, their rotational motion being described through quaternions and the contact interaction between two ellipsoids is described by a force which accounts for the elastic and dissipative interactions. Further we combine the exact derivation of contact points between ellipsoids (Wang et al. in Computing 72(1¿2):235¿246, 2004) with the advantages of the GPU-NVIDIA parallelization strategy (Owens et al. in Comput Graph Forum 26:80¿113, 2007). This novelty makes the analytical algorithm computationally feasible when dealing with several thousands of particles. As a benchmark, we simulate a granular gas of frictionless ellipsoids identifying a classical homogeneous cooling state for ellipsoids. For low dissipative systems, the behavior of the granular temperature indicates that the cooling dynamics is governed by the elongation of the ellipsoids and the restitution coefficient. Our outcomes comply with the statistical mechanical laws and the results are in agreement with previous findings for hard ellipsoids (Bereolos et al. in J Chem Phys 99:6087, 1993; Villemot and Talbot in Granul Matter 14:91¿97, 2012). Additionally, new insight is provided namely suggesting that the mean field description of the cooling dynamics of elongated particles is conditioned by the particle shape and the degree of energy equipartition. Autores: Janda, A. ; Maza, D; et al. Revista: PHYSICAL REVIEW LETTERS ISSN 0031-9007 Vol. 114 Nº 23 2015 págs. 238002 Several theoretical predictions of the mass flow rate of granular media discharged from a silo are based on the spontaneous development of a free-fall arch region, the existence of which is still controversial. In this Letter, we study experimentally and numerically the particle flow through an orifice placed at the bottom of 2D and 3D silos. The implementation of a coarse-grained technique allows a thorough description of all the kinetic and micromechanical properties of the particle flow in the outlet proximities. Though the free-fall arch does not exist as traditionally understood¿a region above which particles have negligible velocity and below which particles fall solely under gravity action¿we discover that the kinetic pressure displays a well-defined transition in a position that scales with the outlet size. This universal scaling explains why the free-fall arch picture has served as an approximation to describe the flow rate in the discharge of silos. Autores: Pastor, José Martín; Garcimartín, Ángel; Gago, P. A. ; et al. Revista: PHYSICAL REVIEW E ISSN 2470-0045 Vol. 92 Nº 6 2015 págs. 062817 The ¿faster-is-slower¿ (FIS) effect was first predicted by computer simulations of the egress of pedestrians through a narrow exit [D. Helbing, I. J. Farkas, and T. Vicsek, Nature (London) 407, 487 (2000)]. FIS refers to the finding that, under certain conditions, an excess of the individuals' vigor in the attempt to exit causes a decrease in the flow rate. In general, this effect is identified by the appearance of a minimum when plotting the total evacuation time of a crowd as a function of the pedestrian desired velocity. Here, we experimentally show that the FIS effect indeed occurs in three different systems of discrete particles flowing through a constriction: (a) humans evacuating a room, (b) a herd of sheep entering a barn, and (c) grains flowing out a 2D hopper over a vibrated incline. This finding suggests that FIS is a universal phenomenon for active matter passing through a narrowing. Autores: Ardanza-Trevijano, Sergio; Zuriguel, Iker; Arévalo, R.; et al. Revista: PHYSICAL REVIEW E ISSN 1539-3755 Vol. 89 2014 págs. 052212 We use the first Betti number of a complex to analyze the morphological structure of granular samples in mechanical equilibrium. We investigate two-dimensional granular packings after a tapping process by means of both simulations and experiments. States with equal packing fraction obtained with different tapping intensities are distinguished after the introduction of a filtration parameter which determines the particles (nodes in the network) that are joined by an edge. This is accomplished by just using the position of the particles obtained experimentally and no other information about the possible contacts, or magnitude of forces. Autores: Acevedo, M. ; Zuriguel, Iker; Maza, D; et al. Revista: GRANULAR MATTER ISSN 1434-5021 Vol. 16 Nº 4 2014 págs. 411 - 420 We present experimental and numerical results for particle alignment and stress distribution in packings of faceted particles deposited in a small-scale bi-dimensional silo. First, we experimentally characterize the deposits' morphology in terms of the particles' aspect ratio and feeding rate. Then we use the experimental results to validate our discrete element method (DEM) based on spheropolygons. After achieving excellent agreement, we use contact forces and fabric provided by the simulations to calculate the coarse-grained stress tensor. For low feeding rates, square particles display a strong tendency to align downwards, i.e., with a diagonal parallel to gravity. This morphology leads to stress transmission towards the walls, implying a quick development of pressure saturation, in agreement with the Janssen effect. When the feed rate is increased, both the disorder and the number of horizontal squares in the silo increase, hindering the Janssen effect. Conversely, for elongated particles the feed rate has a weak effect on the final deposit properties. Indeed, we always observe highly ordered structures of horizontal rods where the stress is transmitted mainly in the vertical direction. Autores: Arévalo, R.; Zuriguel, Iker; Maza, D; et al. Revista: PHYSICAL REVIEW E ISSN 1539-3755 Vol. 89 Nº 4 2014 págs. 042205 We present numerical results of the effect that the driving force has on the clogging probability of inert particles passing through a bottleneck. When the driving force is increased by four orders of magnitude, the mean avalanche size remains almost unaltered (increases 1.6 times) while the flow rate and the avalanche duration display strong dependence on this magnitude. This indicates that in order to characterize the ability of a system to clog, the right variable to consider is the number of particles that pass through the outlet. The weak dependence of this magnitude on the driving force is explained in terms of the average kinetic energy of the flowing grains that has to be dissipated in order to get an arch stabilized. Autores: Zuriguel, Iker; Parisi, D. R.; Cruz, Raúl; et al. Revista: SCIENTIFIC REPORTS ISSN 2045-2322 Vol. 4 2014 págs. 7324 When a large set of discrete bodies passes through a bottleneck, the flow may become intermittent due to the development of clogs that obstruct the constriction. Clogging is observed, for instance, in colloidal suspensions, granular materials and crowd swarming, where consequences may be dramatic. Despite its ubiquity, a general framework embracing research in such a wide variety of scenarios is still lacking. We show that in systems of very different nature and scale -including sheep herds, pedestrian crowds, assemblies of grains, and colloids- the probability distribution of time lapses between the passages of consecutive bodies exhibits a power-law tail with an exponent that depends on the system condition. Consequently, we identify the transition to clogging in terms of the divergence of the average time lapse. Such a unified description allows us to put forward a qualitative clogging state diagram whose most conspicuous feature is the presence of a length scale qualitatively related to the presence of a finite size orifice. This approach helps to understand paradoxical phenomena, such as the faster-is-slower effect predicted for pedestrians evacuating a room and might become a starting point for researchers working in a wide variety of situations where clogging represents a hindrance. Autores: Arévalo, R.; Pugnaloni, L. A.; Maza, D; et al. Revista: PHYSICAL REVIEW E ISSN 1539-3755 Vol. 87 Nº 2 2013 págs. 022203 We analyze the contact network of simulated two-dimensional granular packings in different states of mechanical equilibrium obtained by tapping. We show that topological descriptors of the contact network allow one to distinguish steady states of the same mean density obtained with different tap intensities. These equal-density states were recently proven to be distinguishable through the mean force moment tensor. In contrast, geometrical descriptors, such as radial distribution functions, bond order parameters, and Voronoi cell distributions, can hardly discriminate among these states. We find that small-order loops of contacts-the polygons of the network-are especially sensitive probes for the contact structure. DOI: 10.1103/PhysRevE.87.022203 Autores: Arévalo, R.; Pugnaloni, L. A.; Maza, D; et al. Revista: PHILOSOPHICAL MAGAZINE ISSN 1478-6435 Vol. 93 Nº 31 - 33 2013 págs. 4078 - 4089 We characterize the structure of simulated two-dimensional granular packings using concepts from complex networks theory. The packings are generated by a simulated tapping protocol, which allows us to obtain states in mechanical equilibrium in a wide range of densities. We show that our characterization method is able to discriminate non-equivalent states that have the same density. We do this by examining differences in the topological structure of the contact network of the packings. In particular, we find that the polygons of the network are specially sensitive probes for the contact structure. Additionally, we compare the network properties obtained in two different scenarios: the tapped and a compressed system. Autores: Acevedo, Manuel Francisco; Cruz, Raúl; Zuriguel, Iker; et al. Revista: PHYSICAL REVIEW E ISSN 1539-3755 Vol. 87 Nº 1 2013 págs. 012202 In a previous paper [Hidalgo et al., Phys. Rev. Lett. 103, 118001 (2009)] it was shown that square particles deposited in a silo tend to align with a diagonal parallel to the gravity, giving rise to a deposit with very particular properties. Here we explore, both experimentally and numerically, the effect on these properties of the filling mechanism. In particular, we modify the volume fraction of the initial configuration from which the grains are deposited. Starting from a very dilute case, increasing the volume fraction results in an enhancement of the disorder in the final deposit characterized by a decrease of the final packing fraction and a reduction of the number of particles oriented with their diagonal in the direction of gravity. However, for very high initial volume fractions, the final packing fraction increases again. This result implies that two deposits with the same final packing fraction can be obtained from very different initial conditions. The structural properties of such deposits are analyzed, revealing that, although the final volume fraction is the same, their micromechanical properties notably differ. Autores: Maza, D; et al. Revista: REVISTA DE EDIFICACION ISSN 0213-8948 Nº 41 - 42 2013 págs. 102 - 107 Los medios granulados ¿materia inerte compuesta por sólidos divididos¿ al pasar por una abertura que sea sólo un poco mayor al tamaño de las partículas se pueden atascar. Se ha estudiado como disminuye la probabilidad de que este sistema se atasque gracias a la colocación de un obstáculo delante de la salida. El resultado obtenido demuestra que se disminuye la probabilidad de atasco dependiendo dónde esté colocado el obstáculo, siendo una posición óptima donde la distancia entre el obstáculo y la salida es similar al tamaño de la abertura. Estos resultados pueden ser interesantes para el diseño óptimo de una salida de evacuación. Autores: Altshuler, E.; Pastor, José Martín; Garcimartín, Ángel; et al. Revista: PLOS ONE ISSN 1932-6203 Vol. 8 Nº 8 2013 págs. e67838 While ¿vibrational noise¿ induced by rotating components of machinery is a common problem constantly faced by engineers, the controlled conversion of translational into rotational motion or vice-versa is a desirable goal in many scenarios ranging from internal combustion engines to ultrasonic motors. In this work, we describe the underlying physics after isolating a single degree of freedom, focusing on devices that convert a vibration along the vertical axis into a rotation around this axis. A typical Vibrot (as we label these devices) consists of a rigid body with three or more cantilevered elastic legs attached to its bottom at an angle. We show that these legs are capable of transforming vibration into rotation by a ¿ratchet effect¿, which is caused by the anisotropic stick-slip-flight motion of the leg tips against the ground. Drawing an analogy with the Froude number used to classify the locomotion dynamics of legged animals, we discuss the walking regime of these robots. We are able to control the rotation frequency of the Vibrot by manipulating the shaking amplitude, frequency or waveform. Furthermore, we have been able to excite Vibrots with acoustic waves, which allows speculating about the possibility of reducing the size of the devices so they can perform tasks into the human body, excited by ultrasound waves from the outside. Autores: Arévalo, R.; Pugnaloni, L. A.; Maza, D; et al. Revista: PHILOSOPHICAL MAGAZINE ISSN 1478-6435 Vol. 93 Nº 31 - 33 2013 págs. 4078 - 4089 We characterize the structure of simulated two-dimensional granular packings using concepts from complex networks theory. The packings are generated by a simulated tapping protocol, which allows us to obtain states in mechanical equilibrium in a wide range of densities. We show that our characterization method is able to discriminate non-equivalent states that have the same density. We do this by examining differences in the topological structure of the contact network of the packings. In particular, we find that the polygons of the network are specially sensitive probes for the contact structure. Additionally, we compare the network properties obtained in two different scenarios: the tapped and a compressed system. Autores: Zuriguel, Iker; Maza, D; Revista: PHYSICAL REVIEW LETTERS ISSN 0031-9007 Vol. 108 Nº 24 2012 págs. 248001 "Beverloo's law" is considered as the standard expression to estimate the flow rate of particles through apertures. This relation was obtained by simple dimensional analysis and includes empirical parameters whose physical meaning is poorly justified. In this Letter, we study the density and velocity profiles in the flow of particles through an aperture. We find that, for the whole range of apertures studied, both profiles are self-similar. Hence, by means of the functionality obtained for them the mass flow rate is calculated. The comparison of this expression with the Beverloo's one reveals some differences which are crucial to understanding the mechanism that governs the flow of particles through orifices. Autores: Garcimartín, Ángel; et al. Revista: PHYSICAL REVIEW E ISSN 1539-3755 Vol. 86 Nº 3 2012 págs. 031306 In a recent paper [Zuriguel et al., Phys. Rev. Lett. 107, 278001 (2011)] it has been shown that the presence of an obstacle above the outlet can significatively reduce the clogging probability of granular matter pouring from a silo. The amount of this reduction strongly depends on the obstacle position. In this work, we present new measurements to analyze different outlet sizes, extending foregoing results and revealing that the effect of the obstacle is enhanced as the outlet size is increased. In addition, the effect of the obstacle position on the flow rate properties and in the geometrical features of arches is studied. These results reinforce previous evidence of the pressure reduction induced by the obstacle. In addition, it is shown how the mean avalanche size and the average flow rate are not necessarily linked. On the other hand, a close relationship is suggested between the mean avalanche size and the flow rate fluctuations. Autores: Pugnaloni, L. A. ; Damas, J. ; Zuriguel, Iker; et al. Revista: PAPERS IN PHYSICS ISSN 1852-4249 Vol. 3 Nº 0 2011 págs. 030004 We prepare static granular beds under gravity in different stationary states by tapping the system with pulsed excitations of controlled amplitude and duration. The macroscopic state---defined by the ensemble of static configurations explored by the system tap after tap---for a given tap intensity and duration is studied in terms of volume, V, and force moment tensor, \Sigma. In a previous paper [Pugnaloni et al., Phys. Rev. E 82, 050301(R) (2010)], we reported evidence supporting that such macroscopic states cannot be fully described by using only V or \Sigma, apart from the number of particles N. In this work, we present an analysis of the fluctuations of these variables that indicates that V and \Sigma may be sufficient to define the macroscopic states. Moreover, we show that only one of the invariants of \Sigma is necessary, since each component of \Sigma falls onto a master curve when plotted as a function of \rm{Tr}(\Sigma). This implies that these granular assemblies have a common shape for the stress tensor, even though it does not correspond to the hydrostatic type. Although most results are obtained by molecular dynamics simulations, we present supporting experimental results. Autores: Garcimartín, Ángel; Zuriguel, Iker; et al. Revista: PHYSICAL REVIEW E ISSN 1539-3755 Vol. 84 Nº 3 2011 págs. 1 - 8 We present experimental data corresponding to a two-dimensional dense granular flow, namely, the gravity-driven discharge of grains from a small opening in a silo. We study the local velocity field at the scale of single grains at different places with the help of particle-tracking techniques. From these data, the velocity profiles can be obtained and the validity of some long-standing approaches can be assessed. Moreover, the fluctuations of the velocities are taken into consideration to characterize the features of the advective motion (due to the gravity force) and the diffusive motion, which shows nontrivial behavior. Autores: Kanzaki, T. ; Acevedo, Manuel Francisco; Zuriguel, Iker; et al. Revista: EUROPEAN PHYSICAL JOURNAL E ISSN 1292-8941 Vol. 34 Nº 12 2011 págs. 133 We present experimental and numerical results of the effect that a partial discharge has on the morphological and micro-mechanical properties of non-spherical, convex particles in a silo. The comparison of the particle orientation after filling the silo and its subsequent partial discharge reveals important shear-induced orientation, which affects stress propagation. For elongated particles, the flow induces an increase in the packing disorder which leads to a reduction of the vertical stress propagation developed during the deposit generated prior to the partial discharge. For square particles, the flow favors particle alignment with the lateral walls promoting a behavior opposite to the one of the elongated particles: vertical force transmission, parallel to gravity, is induced. Hence, for elongated particles the flow developed during the partial discharge of the silo leads to force saturation with depth whereas for squares the flow induces hindering of the force saturation observed during the silo filling. Autores: Zuriguel, Iker; Garcimartín, Ángel; et al. Revista: PHYSICAL REVIEW LETTERS ISSN 0031-9007 Vol. 107 Nº 27 2011 págs. 278001 Autores: Zuriguel, Iker; Maza, D; Revista: PHYSICAL REVIEW E ISSN 1539-3755 Vol. 81 Nº 4 2010 págs. 41302 The jamming transition of an isotropically compressed granular packing is studied by means of molecular dynamics simulations. The system is shown to undergo a critical transition which is analyzed by looking at the topological structure of the force network. At the critical packing fraction there is a sudden growth of the number of polygons in the network. Above the critical packing fraction the number of triangles keeps growing while the number of the rest of polygons is weakly reduced. Then, we prove that in the jammed regime, there is a linear relationship between the number of triangles and the coordination number. Furthermore, the presence of these minimal structures is revealed to be connected with the evolution of some important topological properties, suggesting its importance to understand the physical properties of the packing and the onset of rigidity during the compression. Autores: Kanzaki, T; Hidalgo, RC; Maza, D; et al. Revista: JOURNAL OF STATISTICAL MECHANICS-THEORY AND EXPERIMENT ISSN 1742-5468 2010 págs. P06020 Autores: Pugnaloni, Luis A.; Sánchez, Iván; Gago, Paula A.; et al. Revista: PHYSICAL REVIEW E ISSN 1539-3755 Vol. 82 Nº 5 2010 págs. 050301 We analyze, experimentally and numerically, the steady states, obtained by tapping, of a two-dimensional granular layer. Contrary to the usual assumption, we show that the reversible (steady state branch) of the density-acceleration curve is nonmonotonous. Accordingly, steady states with the same mean volume can be reached by tapping the system with very different intensities. Simulations of dissipative frictional disks show that equal volume steady states have different values of the force moment tensor. Additionally, we find that steady states of equal stress can be obtained by changing the duration of the taps; however, these states present distinct mean volumes. These results confirm previous speculations that the volume and the force moment tensor are both needed to describe univocally equilibrium states in static granular assemblies. Autores: Hidalgo, RC; Zuriguel, Iker; Maza, D; et al. Revista: JOURNAL OF STATISTICAL MECHANICS-THEORY AND EXPERIMENT ISSN 1742-5468 2010 págs. P06025 Autores: Zuriguel, Iker; Ardanza-Trevijano, Sergio; et al. Revista: International Journal of Bifurcation and Chaos ISSN 0218-1274 Vol. 20 Nº 3 2010 págs. 897 - 903 The existence of small order loops of contacts is presented as an intrinsic characteristic of force granular networks. Based on molecular dynamics simulations, it is proposed that the presence of these small order loops - and in particular third order loops of contacts - is important to understand the transition from fluid-like to solid-like behavior of granular packings. In addition, we show a close relationship between the development of third order loops and the small forces of the granular packing in the sense that almost all third order loops allocate a force component smaller than the average. Autores: Sánchez, I; Gutiérrez, G; Zuriguel, Iker; et al. Revista: PHYSICAL REVIEW E ISSN 1539-3755 Vol. 81 Nº 6 2010 págs. 062301 We present an experimental study of the displacement of a light intruder immersed in a vibrated granular bed. Using high speed video we resolve the motion, during one cycle of oscillation, of a cylindrical object inside a Plexiglas box partially filled with grains. We report experimental evidence that, in the absence of convection, at least two forces are behind the intruder's motion: an air drag force-due to the airflow through the granular bed-and a buoyancy force produced by an air-mediated granular fluid. Autores: Janda, A. ; Zuriguel, Iker; et al. Libro: Particle-based methods V : fundamentals and applications 2015 págs. 71 - 80 Autores: Cruz, Raúl; Acevedo, Manuel Francisco; Zuriguel, Iker; et al. Libro: Powders and grains 2013 : Proceedings of the 7th International Conference on Micromechanics of Granular Media 2013 págs. 895 - 898 The effect of the filling mechanism on the packing of faceted particles with different aspect ratios has been examined. We have experimentally measured the particle angular distribution and the packing fraction of ensembles of faceted particles deposited in a bidimensional box. The granular system has been numerically simulated using a two-dimensional model of faceted particles. We found that increasing the feeding rate results in an enhancement of the disorder in the final deposit and, consequently, in a reduction of the number of particles oriented in their most stable configuration. In this regime, the final packing fraction monotonically decreases as the feeding rate increases. The correlations between the final packing morphology and the stress transmission were examined by describing the micromechanical properties of the deposits. For the case of elongated particles, increasing the feeding rate leads to an enhancement of the stress transmission towards the sides of the box. On the contrary, for the case of square particles, increasing the feeding rate promotes vertical transmission of the stress. Autores: Zuriguel, Iker; Bienzobas, J.; et al. Libro: Powders and grains 2013 : Proceedings of the 7th International Conference on Micromechanics of Granular Media 2013 págs. 710 - 713 Transport of material through pipes or channels in mines or gravel quarries seems to be a simple and economic form of conveying blasted ore between different levels. Despite the apparent advantages of moving the material by means of the gravity force, there exists an important problem that makes the applicability of this method more difficult: the election of the pipe diameter to prevent clogging of the stones. It was R. Kvapil in the sixties who extended the ideas of granular flows in silos to underground mining. Nevertheless, after his pioneering works there are only a few manuscripts focused on this topic, and many questions remain unsolved. In this work, we present experimental results about the flow of particles (gravel) driven by gravity through tilted tubes. The amount of material discharged between clogs shows that the probability of clogging can be estimated by the same procedures introduced for silos. Finally, by changing the ratio between the tube diameter and the typical particle size, we discuss about the existence or not of a critical size beyond which clogging is not possible. Autores: Zuriguel, Iker; et al. Libro: Powders and grains 2013 : Proceedings of the 7th International Conference on Micromechanics of Granular Media 2013 págs. 698 - 701 We present an experimental study of the effect that an obstacle above the outlet of a silo has on the clogging probability. Both, the size of the orice and the obstacle position are varied for a chosen obstacle size and shape. If the position of the obstacle is properly selected the clogging probability can be importantly reduced. Indeed, as the outlet size is increased ¿ and we approach the critical size above which there is not clogging ¿ the obstacle effect is enhanced. For the largest outlet size studied, the clogging probability is reduced by a factor of more than one hundred. We will show, using numerical simulations, that the physical parameter behind the reduction of the silo clogging seems to be the decrease of the vertical pressure at the outlet proximities. Autores: Maza, D; Rubio-Largo, S. M.; et al. Libro: Powders and grains 2013 : Proceedings of the 7th International Conference on Micromechanics of Granular Media 2013 págs. 674 - 677 The role of density and velocity profiles in the flow of particles through apertures has been recently put on evidence in a two-dimensional experiment (Phys. Rev. Lett. 108, 248001). For the whole range of apertures studied, both velocity and density profiles are selfsimilar and the obtained scaling functions allow to derive the relevant scales of the problem. Indeed, by means of the functionality obtained for these profiles, an exact expression for the mass flow rate was proposed. Such expression showed a perfect agreement with the experiential data. In this work, we generalize this study to the three dimensional case.We perform numerical simulations of a 3D silo in which the velocity and volume fraction profiles are determined. Both profiles shows that the scaling obtained for 2D can be generalized to the 3D case. Finally, the scaling of the mass flow rate with the outlet radius is discussed.	{"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": 0.8568910360336304, "perplexity": 1639.4031449817483}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676591596.64/warc/CC-MAIN-20180720115631-20180720135631-00175.warc.gz"}
http://jmlr.org/papers/v18/16-079.html	## A Robust-Equitable Measure for Feature Ranking and Selection A. Adam Ding, Jennifer G. Dy, Yi Li, Yale Chang; 18(71):1−46, 2017. ### Abstract In many applications, not all the features used to represent data samples are important. Often only a few features are relevant for the prediction task. The choice of dependence measures often affect the final result of many feature selection methods. To select features that have complex nonlinear relationships with the response variable, the dependence measure should be equitable, a concept proposed by Reshef et al. (2011); that is, the dependence measure treats linear and nonlinear relationships equally. Recently, Kinney and Atwal (2014) gave a mathematical definition of self- equitability. In this paper, we introduce a new concept of robust-equitability and identify a robust- equitable copula dependence measure, the robust copula dependence (RCD) measure. RCD is based on the $L_1$-distance of the copula density from uniform and we show that it is equitable under both equitability definitions. We also prove theoretically that RCD is much easier to estimate than mutual information. Because of these theoretical properties, the RCD measure has the following advantages compared to existing dependence measures: it is robust to different relationship forms and robust to unequal sample sizes of different features. Experiments on both synthetic and real-world data sets confirm the theoretical analysis, and illustrate the advantage of using the dependence measure RCD for feature selection. [abs][pdf][bib]	{"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": 0.9570485949516296, "perplexity": 1255.7770937667838}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376828697.80/warc/CC-MAIN-20181217161704-20181217183704-00404.warc.gz"}
https://www.physicsforums.com/threads/effect-of-temperature-on-entropy.517807/	# Effect of temperature on Entropy 1. Jul 29, 2011 ### weng cheong i'm a pre-U student, and i came across this when i study the topic on entropy dS = dQ / T i understand that as temperature increases, entropy increases as well, as there are more quanta of energy and more thermal states(energy levels) available. however according to this equation, it seems to indicate that with a lower temperature, we can get a greater entropy? 2. Jul 29, 2011 ### Bloodthunder Re: how temperature affect entropy S is entropy. dS is change in entropy. So, for a given temperature T, the change in entropy is equal to the change in heat energy divided by whatever the temperature that system is currently in. 3. Jul 29, 2011 ### Mapes The equation you wrote describes the infinitesimal increase in entropy of a system at constant temperature when infinitesimal energy is added reversibly by heating. That's not the same as the change in entropy of a system with increasing temperature, which is $\partial S/\partial T$. This quantity is, indeed, always positive. You can integrate your equation to get $\Delta S=Q\ln(T_2/T_1)$, which confirms that entropy increases with temperature when a system is heated. Does this make sense? 4. Jul 29, 2011 ### Mike H You get a larger change in entropy, that's correct. ETA - I figured that constant temperature and reversibility were givens, considering the Clausius formalism used by the OP. (Although that dQ should be dQrev, which is how I recall seeing it written in textbooks.) ETA 2 - One analogy that might be useful - if you scream at a sports game (add a tiny bit of energy to a high-temperature system), most likely you will not make that much of a dent in the general soundscape (a small change in entropy). If you yell in the middle of a wedding (adding a tiny bit of energy to a low-temperature system), it would be far more dramatic (a large change in entropy). Last edited: Jul 29, 2011 5. Jul 29, 2011 ### Studiot Re: how temperature affect entropy Careful! The temperature of the the system at the point (moment) of heat exchange, currently is a bit vague and could mean after the exchange. Note also it is a differential relationship so may be integrated from point to point. 6. Jul 30, 2011 ### thebiggerbang Imagine this situation. You provide X joules of heat to a system at 1)say 100 K and 2)say 300K So, in the first case, what you will observe is the there will be a larger change in the randomness than in the second case, as due to the pre-existing higher temperature, there will be already a large amount of disorder in the system! It's just like shuffling a deck of kinda properly arranged cards (analogous to a lower T) will give you a larger change in disorder than shuffling a pack of cards that are already random in order (higher temperature!). Am I correct? 7. Jul 30, 2011 ### Studiot The entropy change in the system, in both cases, depends upon the conditions of energy input, which have not been specified. 8. Jul 30, 2011 ### thebiggerbang 9. Jul 30, 2011 ### Studiot If both additions of heat are reversible then the entropy change of the second case input is 1/3 of the first. The addition of heat may become the work of expansion, which allows greater entropy in the form of freedom of space to occupy, or It may be taken up in a phase (state) change so the particles become more disordered. Similar Discussions: Effect of temperature on Entropy	{"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": 0.8700895309448242, "perplexity": 955.1624596221427}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934804518.38/warc/CC-MAIN-20171118021803-20171118041803-00739.warc.gz"}
http://mathoverflow.net/questions/107192/dehn-function-for-undistorted-subgroups-of-a-product-of-free-groups?sort=newest	# Dehn function for undistorted subgroups of a product of free groups Let $G$ be a finitely generated subgroup of a product of two finite rank free groups $F_m \times F_n$. If there is a Lipschitz retraction $F_m \times F_n \to G$ with respect to word metrics, then $G$ is undistorted in $F_m \times F_n$, and the Dehn function of $G$ has a quadratic upper bound. Suppose now that we require only that $G$ is undistorted in $F_m \times F_n$. Is it still true that the Dehn function of $G$ has a quadratic upper bound? - Lee: All subgroups G are either free or direct products of free groups or are not finitely-presented. Among latter, I do not know any examples of undistorted ones, do you? In fact, the only undistorted subgroups of semi simple Lie groups that I know are retracts, non-uniform lattices and Anosov (in the sense of Guichard and Wienhard) hyperbolic groups (the latter could in the end turn out to be retracts, this is unknown). – Misha Sep 14 '12 at 17:32 ...I forgot one more class of undistorted subgroups: Some polycyclic groups which are peripheral subgroups of non-uniform lattices. However, one should regard them as lattices in solvable Lie groups. – Misha Sep 14 '12 at 17:41 The Bieri-Stallings subgroup of $F_n\times F_n$ is undistorted and finitely generated, but not finitely-presented, so in some sense it has an infinite Dehn function. It's the kernel of the map $F_n\times F_n\to \mathbb{Z}$ which sends each generator to 1, and it's generated by elements of the form $g_ih_j^{-1}$ where $g_i$ and $h_j$ are generators of the two different factors. - Welcome to MO Robert! – Andy Putman Sep 14 '12 at 18:23 How does one know, or find a reference for, the statement that this subgroup is undistorted? – Lee Mosher Sep 14 '12 at 18:32 There's a proof for the theorem that Yves mentions in Olshanskii, Sapir, "Length and area functions on groups and quasi-isometric Higman embeddings" (Theorem 2). For this subgroup, you can construct words explicitly -- if $$w=w_1(g_1,\dots, g_n)w_2(h_1,\dots,h_n)$$ and $w$ lies in the kernel, we can rewrite $w$ as $$w=w_1(g_1 h_1^{-1},\dots, g_n h_1^{-1})(h_1 g_1^{-1})^k w_2(h_1g_1^{-1},\dots,h_n g_1^{-1})$$ where $k$ is the sum of the exponents in $w_1$. This is a product of generators of the subgroup and its length increases by at most a constant. – Robert Young Sep 14 '12 at 19:37 That's pretty. – Lee Mosher Sep 14 '12 at 21:18 The answer is no: actually by Baumslag-Roseblade (JLMS 1984) either $G$ is commensurable to a product of free groups (hence has linear or quadratic Dehn function), or is not finitely presented (so the Dehn function is infinite, or not defined, as you wish). The latter case occurs if $H$ is an infinite word hyperbolic group and $f:F_m\to H$ is a non-bijective surjection and $H$ is the fibre product $[(g,h)\in F_m\times F_m:f(g)=f(h)]$. - Sorry, I can't type braces so I put brackets. – YCor Sep 14 '12 at 17:44 @Yves, I have the same question here as for Robert Young's answer: how does one obtain nondistortion in these examples? – Lee Mosher Sep 14 '12 at 18:36 In general the Dehn function of $H$ is equivalent to the distortion of $G$. This is not completely formal (there's a little Van Kampen diagram cuisine) but in the case of $H=\mathbf{Z}$ it's simple to verify by hand. I saw the general result written somewhere but I can't remember right now. – YCor Sep 14 '12 at 18:42	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.810000479221344, "perplexity": 272.3824302431649}, "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-2015-48/segments/1448398452385.31/warc/CC-MAIN-20151124205412-00329-ip-10-71-132-137.ec2.internal.warc.gz"}
http://math.stackexchange.com/questions/154692/lim-limits-n-to-inftyx-n-x-n1-0-x-n-k-converges-but-x-n-does-not	# $\lim\limits_{n \to\infty}x_n-x_{n+1}=0$, $x_{n_k}$ converges but $x_n$ does not converge This is a counter example homework question that I can't seem to solve. I need to find a sequence of real numbers $(x_n)_{n=1}^{\infty}$, and a monotonic increasing sequence of natural numbers $(n_k)_{k=1}^{\infty}$such that: $(i)\lim\limits_{n\to\infty}(x_n-x_{n+1})=0$, $(ii)(x_{n_k})_{k=1}^{\infty}$ converges, but $(iii)(x_n)_{n=1}^{\infty}$ do not converges. Only thing I know so far is that for all $k$, $[n_{k+1}-n_k]$ cannot be bounded. - it converges to 0. – Amihai Zivan Jun 6 '12 at 14:43 I changed the term "sequence" to "series". Pardon my English. – Amihai Zivan Jun 6 '12 at 14:45 @AmihaiZivan: I think "sequence" is what you meant. The word "series" implies that you're summing the terms, but that is not what your formulas say -- they speak about the raw elements of the sequence converging. – Henning Makholm Jun 6 '12 at 14:47 I know this is calculus, but I found this to be an enlightening book on series and sequences. amazon.com/Counterexamples-Analysis-Dover-Books-Mathematics/dp/… – rckrd Jun 6 '12 at 14:54 @Henning Thanks. I'll change that once again.. :-) – Amihai Zivan Jun 6 '12 at 15:10 It's sometimes easiest to go for the most blatant counterexample you can get away with. So decide that you will have $x_{n_k}=0$ for all $k$ and $x_n=1$ for some $n$ between each set of $x_{n_k}$ and $x_{n_{k+1}}$. This calls for a sequence that climbs from 0 to 1 and back down to 0 and up again infinitely many times. It must eventually do this slower and slower due to the condition on the successive difference, but you can decide how slow or fast the differences go to 0. For example, decide that $\|x_n-x_{n+1}\|$ must be $1/k$ whenever $n$ is between $n_k$ and $n_{k+1}$... - Hint: consider the sequence $0\,,{1\over2}\,,1\,,{2\over3}\,,{1\over3}\,,0,\,{1\over4}\,,{2\over4}\,,{3\over4}\,,1\,,{4\over5}\,,{3\over5}\,,{2\over5}\,,{1\over5}\,,0\,,{1\over6}\,,\ldots$ . - +1 This is a really neat example. – Derek Allums Jun 6 '12 at 14:49 Excellent! Thanks David Mitra. – Amihai Zivan Jun 6 '12 at 15:13	{"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": 0.9288472533226013, "perplexity": 627.5956704555109}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-22/segments/1432207932182.13/warc/CC-MAIN-20150521113212-00222-ip-10-180-206-219.ec2.internal.warc.gz"}
https://www.afterecon.com/economics-and-finance/statistical-reasoning-uncertainty/	# Statistical Reasoning with Uncertainty Risk and uncertainty are fundamentally different notions. Risk means there is a probability some event will occur, and it usually implies that the probability is known or at least estimated with some level of confidence. For example, betting on a coin toss, betting on a lottery ticket, or insuring a person’s health. Uncertainty, on the other hand, is not quantifiable. In economics we generally conceive of uncertainty as knightian uncertainty, and it exists in two forms. The first form is a known event that occurs with an unknown probability. The second form is an unknown event. Interestingly, it might be the case that unknown events occur with relative predictability. This seems to be the case with the economics of innovation. We don’t know what the inventions will be, but we know the rate at which innovation is happening. That is a bit of a rabbit trail. My point in this article is that the sort of a priori statistical reasoning we have developed can be used to estimate uncertainty in a remarkably straightforward way. Ceteris paribus, the probability of an uncertain event should be taken as equal to the probability of a risky event. Moreover, again assuming ceteris paribus, the expected cost of uncertainty is equal to the expected risk cost. Why should this be so? An uncertain event will occur in one of three probability ranges. It might occur with less probability than a comparative risk event, equal probability to a comparative risk event, or greater probability than a comparative risk event. Given the a priori statistical reasoning we spoke of earlier, the rational estimates for the probability of the uncertain event are: • Case 1, p(U).5 = p(R) • Case 2, p(U) = p(R) • Case 3, p(U)2 = p(R) If we are genuinely uncertain then we have no reason to think any probability case is more likely than another, so a rational estimate would average the three cases. The average yields the expectation that p(U) = p(R). Let’s bring this to an application. You are considering investing in a security. The security has a known past value history from which we can create a forecast of expected future values. These future forecasts include estimated risk, but not uncertainty. We can now include a rational estimate of uncertainty by simply doubling the expected risk premium (in both directions, of course). • • • •	{"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": 0.963141143321991, "perplexity": 641.3060753460478}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178362741.28/warc/CC-MAIN-20210301151825-20210301181825-00262.warc.gz"}
https://www.gradesaver.com/textbooks/math/calculus/calculus-3rd-edition/chapter-10-introduction-to-differential-equations-10-1-solving-differential-equations-preliminary-questions-page-504/4	## Calculus (3rd Edition) Yes it can be, for example $$y'= \frac{x}{y}\Longrightarrow y'=\frac{1}{y} x=f(y)g(x).$$	{"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": 0.907272219657898, "perplexity": 3032.3111156419627}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575540490743.16/warc/CC-MAIN-20191206173152-20191206201152-00276.warc.gz"}
https://materialsmodeling.org/publications/2010-Pressure-induced-phase-transition-in-the-electronic-structure-of-palladium-nitride/	# Pressure-induced phase transition in the electronic structure of palladium nitride D. Åberg, P. Erhart, J. Crowhurst, J. M. Zaug, A. F. Goncharov, and B. Sadigh Physical Review B 82, 104116 (2010) arXiv:1009.0041 doi: 10.1103/PhysRevB.82.104116 We present a combined theoretical and experimental study of the electronic structure and equation of state (EOS) of crystalline PdN2. The compound forms above 58 GPa in the pyrite structure and is metastable down to 11 GPa. We show that the EOS cannot be accurately described within either the local density or generalized gradient approximations. The Heyd-Scuseria-Ernzerhof exchange-correlation functional (HSE06), however, provides very good agreement with experimental data. We explain the strong pressure dependence of the Raman intensities in terms of a similar dependence of the calculated band gap, which closes just below 11 GPa. At this pressure, the HSE06 functional predicts a first-order isostructural transition accompanied by a pronounced elastic instability of the longitudinal-acoustic branches that provides the mechanism for the experimentally observed decomposition. Using an extensive Wannier function analysis, we show that the structural transformation is driven by a phase transition of the electronic structure, which is manifested by a discontinuous change in the hybridization between d and p electrons as well as a conversion from single to triple bonded nitrogen dimers. We argue for the possible existence of a critical point for the isostructural transition, at which massive fluctuations in both the electronic as well as the structural degrees of freedom are expected. Tags: Updated:	{"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": 0.8995080590248108, "perplexity": 1207.7993292304307}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400192783.34/warc/CC-MAIN-20200919173334-20200919203334-00356.warc.gz"}
https://astarmathsandphysics.com/university-maths-notes/matrices-and-linear-algebra/4316-radiactive-decay-with-alternative-decay-routes-in-matrix-form.html	## Radiactive Decay With Alternative Decay Routes in Matrix Form We can represent a radioactive decay series via two possible routes. There is initially a number of atoms $N$ of nuclide 1. Suppose the two possible decay routes are~ $1 \rightarrow 2 \rightarrow 4$ or $1 \rightarrow 3 \rightarrow 4$ Th initial stage nuclei 1 decay according to the equation $\frac{dN_1}{dt} =-\lambda_{12} N_1 -\lambda_{13} N_1$ Where the subscript $N_{ik}$ denotes decay from state i to state k. Th intermediate stage nuclei 2 are created according to the equation $\frac{dN_2}{dt} = \lambda_{12} N_1 -\lambda_{24} N_2$ Th intermediate stage nuclei 3 are created according to the equation $\frac{dN_3}{dt} = -\lambda_{13} N_1 - \lambda_{34} N_2$ The end stage nuclei 4 are created according to the equation $\frac{dN_4}{dt} = \lambda_{24} N_2 + \lambda_{34} N_3$ We can write this in matrix form as $\begin{pmatrix}dN_1\\dN_2\\dN_3\\dN_4 \end{pmatrix} = \left( \begin{array}{cccc} - \lambda_{12} - \lambda_{13} & 0 & 0 & 0 \\ \lambda_{12} & - \lambda_{24} & 0 & 0 \\ 0 & \lambda_{13} & - \lambda_{34} & 0 \\ 0 & \lambda_{24} & \lambda_{34} & 0 \end{array} \right) \begin{pmatrix}N_1\\N_2\\N_3\\N_4\end{pmatrix}$	{"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": 0.8577893972396851, "perplexity": 1981.9482564265243}, "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/1518891812306.16/warc/CC-MAIN-20180219012716-20180219032716-00282.warc.gz"}
https://research.nsu.ru/ru/publications/cp-properties-of-higgs-boson-interactions-with-top-quarks-in-the-	CP Properties of Higgs Boson Interactions with Top Quarks in the t t ¯ H and tH Processes Using H →γγ with the ATLAS Detector The ATLAS collaboration Результат исследования: Научные публикации в периодических изданияхстатьярецензирование 2 Цитирования (Scopus) Аннотация A study of the charge conjugation and parity (CP) properties of the interaction between the Higgs boson and top quarks is presented. Higgs bosons are identified via the diphoton decay channel (H→γγ), and their production in association with a top quark pair (tt¯H) or single top quark (tH) is studied. The analysis uses 139 fb-1 of proton-proton collision data recorded at a center-of-mass energy of s=13 TeV with the ATLAS detector at the Large Hadron Collider. Assuming a CP-even coupling, the tt¯H process is observed with a significance of 5.2 standard deviations. The measured cross section times H→γγ branching ratio is 1.64-0.36+0.38(stat)-0.14+0.17(sys) fb, and the measured rate for tt¯H is 1.43-0.31+0.33(stat)-0.15+0.21(sys) times the Standard Model expectation. The tH production process is not observed and an upper limit on its rate of 12 times the Standard Model expectation is set. A CP-mixing angle greater (less) than 43 (-43)° is excluded at 95% confidence level. Язык оригинала английский 061802 21 Physical Review Letters 125 6 https://doi.org/10.1103/PhysRevLett.125.061802 Опубликовано - 7 авг 2020	{"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": 0.9942873120307922, "perplexity": 3093.0415513449743}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703519923.26/warc/CC-MAIN-20210120054203-20210120084203-00031.warc.gz"}
https://stacks.math.columbia.edu/tag/0CSM	## 37.64 Descent finiteness properties of complexes This section is the continuation of Derived Categories of Schemes, Section 36.12. Lemma 37.64.1. Let $X \to S$ be locally of finite type. Let $\{ f_ i : X_ i \to X\}$ be an fppf covering of schemes. Let $E \in D_\mathit{QCoh}(\mathcal{O}_ X)$. Let $m \in \mathbf{Z}$. Then $E$ is $m$-pseudo-coherent relative to $S$ if and only if each $Lf_ i^E$ is $m$-pseudo-coherent relative to $S$. Proof. Assume $E$ is $m$-pseudo-coherent relative to $S$. The morphisms $f_ i$ are pseudo-coherent by Lemma 37.54.6. Hence $Lf_ i^E$ is $m$-pseudo-coherent relative to $S$ by Lemma 37.53.16. Conversely, assume that $Lf_ i^E$ is $m$-pseudo-coherent relative to $S$ for each $i$. Pick $S = \bigcup U_ j$, $W_ j \to U_ j$, $W_ j = \bigcup W_{j, k}$, $T_{j, k} \to W_{j, k}$, and morphisms $\alpha _{j, k} : T_{j, k} \to X_{i(j, k)}$ over $S$ as in Lemma 37.44.2. Since the morphism $T_{j, K} \to S$ is flat and of finite presentation, we see that $\alpha _{j, k}$ is pseudo-coherent by Lemma 37.54.7. Hence $L\alpha _{j, k}^Lf_{i(j, k)}^E = L(T_{i, k} \to S)^E$ is $m$-pseudo-coherent relative to $S$ by Lemma 37.53.16. Now we want to descend this property through the coverings $\{ T_{j, k} \to W_{j, k}\}$, $W_ j = \bigcup W_{j, k}$, $\{ W_ j \to U_ j\}$, and $S = \bigcup U_ j$. Since for Zariski coverings the result is true (by the definition of $m$-pseudo-coherence relative to $S$), this means we may assume we have a single surjective finite locally free morphism $\pi : Y \to X$ such that $L\pi ^E$ is pseudo-coherent relative to $S$. In this case $R\pi _L\pi ^E$ is pseudo-coherent relative to $S$ by Lemma 37.53.9 (this is the first time we use that $E$ has quasi-coherent cohomology sheaves). We have $R\pi _L\pi ^E = E \otimes ^\mathbf {L}_{\mathcal{O}_ X} \pi _\mathcal{O}_ Y$ for example by Derived Categories of Schemes, Lemma 36.22.1 and locally on $X$ the map $\mathcal{O}_ X \to \pi _\mathcal{O}_ Y$ is the inclusion of a direct summand. Hence we conclude by Lemma 37.53.12. $\square$ Lemma 37.64.2. Let $X \to T \to S$ be morphisms of schemes. Assume $T \to S$ is flat and locally of finite presentation and $X \to T$ locally of finite type. Let $E \in D(\mathcal{O}_ X)$. Let $m \in \mathbf{Z}$. Then $E$ is $m$-pseudo-coherent relative to $S$ if and only if $E$ is $m$-pseudo-coherent relative to $T$. Proof. Locally on $X$ we can choose a closed immersion $i : X \to \mathbf{A}^ n_ T$. Then $\mathbf{A}^ n_ T \to S$ is flat and locally of finite presentation. Thus we may apply Lemma 37.53.17 to see the equivalence holds. $\square$ Lemma 37.64.3. Let $f : X \to S$ be locally of finite type. Let $\{ S_ i \to S\}$ be an fppf covering of schemes. Denote $f_ i : X_ i \to S_ i$ the base change of $f$ and $g_ i : X_ i \to X$ the projection. Let $E \in D_\mathit{QCoh}(\mathcal{O}_ X)$. Let $m \in \mathbf{Z}$. Then $E$ is $m$-pseudo-coherent relative to $S$ if and only if each $Lg_ i^E$ is $m$-pseudo-coherent relative to $S_ i$. Proof. This follows formally from Lemmas 37.64.1 and 37.64.2. Namely, if $E$ is $m$-pseudo-coherent relative to $S$, then $Lg_ i^E$ is $m$-pseudo-coherent relative to $S$ (by the first lemma), hence $Lg_ i^E$ is $m$-pseudo-coherent relative to $S_ i$ (by the second). Conversely, if $Lg_ i^E$ is $m$-pseudo-coherent relative to $S_ i$, then $Lg_ i^E$ is $m$-pseudo-coherent relative to $S$ (by the second lemma), hence $E$ is $m$-pseudo-coherent relative to $S$ (by the first lemma). $\square$ In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 2, "x-ck12": 0, "texerror": 0, "math_score": 0.9996730089187622, "perplexity": 183.69422383394456}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780057580.39/warc/CC-MAIN-20210924201616-20210924231616-00100.warc.gz"}
http://mathhelpforum.com/math-topics/1124-exponential-decay.html	# Math Help - exponential decay 1. ## exponential decay Hi! I don't understand the following: F = 1/tf * V F... vector quantity, represents the friction tf ... friction timescale V ... velocity of an air parcel V= (u,v,w), also called wind velocity In the Article "Ray tracing volume densities" Kajiya and Von Herzen (1984) writes: "Frictional effects are approximated by a simple relation yielding an exponential decay of wind velocities with time" (the wind velocities are described by a simplified equation of motion and also change at every time step) Why is this a exponential decay ?! hope you can help me ldenk 2. Originally Posted by ldenk Hi! I don't understand the following: F = 1/tf * V F... vector quantity, represents the friction tf ... friction timescale V ... velocity of an air parcel V= (u,v,w), also called wind velocity In the Article "Ray tracing volume densities" Kajiya and Von Herzen (1984) writes: "Frictional effects are approximated by a simple relation yielding an exponential decay of wind velocities with time" (the wind velocities are described by a simplified equation of motion and also change at every time step) Why is this a exponential decay ?! hope you can help me ldenk The friction on the parcel should be in the opposite direction to the velocity so you should have: $F = -V/t_f$ Then Newton tells us that: $\frac{dV}{dt}\ =\ m.F$ which is: $\frac{dV}{dt}\ =\ -m.V/t_f$ and as both $m$ and $t_f$ are positive the solution of the differential equation has the form of exponential decay. (I would also assume that at some point things have been normalised so that F is the frictional force per unit mass, and so the value of $m$ that appears here should be $1$.) RonL	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 7, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9698894023895264, "perplexity": 2727.8534639257637}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1404776435471.2/warc/CC-MAIN-20140707234035-00047-ip-10-180-212-248.ec2.internal.warc.gz"}
https://www.arxiv-vanity.com/papers/1207.6300/	# On the decomposition of the Foulkes module Eugenio Giannelli ###### Abstract The Foulkes module is the permutation module for the symmetric group given by the action of on the collection of set partitions of a set of size into sets each of size . The main result of this paper is a sufficient condition for a simple -module to have zero multiplicity in . A special case of this result implies that no Specht module labelled by a hook partition with appears in . ## 1 Introduction For and natural numbers, let denote the collection of all set partitions of into sets each of size . Let denote the corresponding -permutation module, known as the Foulkes module. Let be the permutation character of afforded by . At the end of Section 1 of [4], Foulkes made a conjecture which can be stated as follows. ###### Conjecture (Foulkes’ Conjecture). For all and natural numbers such that , there exists an injective -homomorphism from to . The conjecture has been proved to be true only when by Thrall (see [9]), when by Dent (see [3, Main Theorem]), when by McKay (see [7, Theorem 1.2]) and when is very large compared to by Brion (see [2, Corollary 1.3]). Foulkes’ original statement of the conjecture was as an inequality between multiplicities, namely that, for all and natural numbers such that and for all partitions of , ⟨ϕ(ab),χλ⟩≤⟨ϕ(ba),χλ⟩, where is the irreducible character of canonically labelled by . From this point of view the decomposition of the Foulkes module as a direct sum of simple modules becomes central. Except in the case when or (see [9, Chapter 2] and [8]) and when (see [3, Theorem 4.1]), little is known about the multiplicities of simple modules in this decomposition. In [5, Theorem 5.4.34] an explicit fomula is given for the specific case of simple modules labelled by two-row partitions: in this case Foulkes’ Conjecture holds with equality. We give a short alternative proof of this result in Corollary 2.12 below. In [10] Paget and Wildon gave a combinatorial description of the minimal partitions that label simple modules appearing as summands of Foulkes modules. The aim of this paper is to prove a number of new results on when these multiplicities vanish. We start by giving some standard notation and definitions in Section 2.1. In Section 2.2 we discuss some basic properties of the Foulkes module and we describe its restriction to the subgroups of . In Section 3 we prove the following result which shows that no Specht module labelled by a hook partition is a direct summand of the Foulkes module . ###### Theorem 1.1. If and are natural numbers such that , then ⟨ϕ(ab),χ(ab−r,1r)⟩=0. In Section 4 we extend this result, by giving a sufficient condition on a partition of for to equal zero. We need the following notation: let be a partition of , let be such that and . Define to be the partition (ab−k−m,α1+1,…,αt+1,1k−t) of . (The value of will be always clear from the context.) It is obvious that every partition of can be expressed uniquely in the form . We will call the inside-partition of . The main result of this paper is as follows. ###### Theorem 1.2. Let and be natural numbers and let be a partition of with and . Let . Suppose that and . Then ⟨ϕ(ab),χ[k:α]⟩=0. Notice that for every simple -module labelled by , a partition of satisfying the hypothesis of Theorem 1.2, Foulkes’ Conjecture holds with equality. Indeed for all we have ⟨ϕ(ab),χλ⟩=0=⟨ϕ(ba),χλ⟩, since there is not any restriction on and in the statement of the theorem. By Proposition 2.7 below, if then has at most parts. When we consider only characters labelled by such partitions, it occurs that a significant proportion of the characters appearing with zero multiplicity in satisfy the hypotheses of Theorem 1.2. For example, computations using the computer algebra package magma [1] show that there are partitions of with at most parts such that ; of these satisfy the hypotheses of Theorem 1.2. For an important subclass of partitions to which Theorem 1.2 applies we refer the reader to Corollary 4.4. ## 2 Preliminaries ### 2.1 Notation and definitions A partition of is a non-increasing finite sequence of positive integers λ=(λ1,λ2,…,λs) such that . We write to denote that is a partition of . The number of parts of a partition will be denoted by . Denote by the conjugate partition of , as defined by for such that , and notice that . We may also denote a partition by (λm11,…,λmrr) to underline that has exactly parts equal to for all . It will often be useful to think of a partition as a -Young diagram, as explained by James in [6, Chapter 3]. ###### Definition 2.1. Let be a partition of and be a partition of . We say that is a subpartition of , and write , if , for all such that . In particular is a subpartition of if and only if the Young diagram of is contained in the Young diagram of . ###### Definition 2.2. A hook partition is a partition of the form λ=(n−k,1k) where . The number is called the leg length of the hook partition . We shall also need the dominance order on the set of partitions of a fixed natural number . Given , we say dominates , and write , if j∑i=1λi≥j∑i=1μi for all such that . Following the definitions and notation of [6] we denote by the Young permutation -module linearly spanned by the -tabloids and by the Specht module linearly spanned by the -polytabloids; let and respectively denote the associated characters. From [6], we have the following fundamental results. ###### Theorem 2.3. Let and be two partitions of . If is a direct summand of then dominates . ###### Theorem 2.4 (Branching Theorem). Let be a partition of . Let be the set of all the partitions of corresponding to the Young diagrams obtained by adding a box to the Young diagram of . Then the induced module decomposes as follows: Sμ↑⏐Sn+1=⨁λ∈ΛSλ. The following theorems are straightforward corollaries of the Littlewood-Richardson rule, as stated in [6, Chapter 16]. ###### Theorem 2.5. Let be a natural number such that and let be a partition of . If is the set of all the partitions of corresponding to the Young diagrams obtained by adding boxes, no two in the same column, to the Young diagram of , then (χλ×1Sk)↑⏐SnSn−k×Sk=∑μ∈Lχμ. ###### Theorem 2.6. Let be a natural number such that , let be a partition of , let be a partition of and let be a partition of . If ⟨(χλ×χμ)↑⏐Sn,χν⟩≠0 then , and . ### 2.2 The Foulkes module Here we present some properties of the Foulkes module that will be needed to prove the two main theorems. ###### Proposition 2.7. Let be a partition of such that . Then ⟨ϕ(ab),χλ⟩=0. ###### Proof. It is easily seen that there is a injective map from to . The proposition now follows from Theorem 2.3. ∎ ###### Definition 2.8. Let and be natural numbers. We define to be the set of all partitions of with at most parts and first part of size at most . An element of can be denoted by , where for each , is a subset of of size and for all such that it holds . ###### Definition 2.9. Let be a natural number such that and let be in . We will say that an element {A1,…,Ab}∈Ω(ab) is linked to if the composition of whose parts are ∣∣{1,2,…,r}∩Ai∣∣for 1≤i≤b. has underlying partition . ###### Definition 2.10. Let be a natural number, such that and let be in . We denote by the set of all the set partitions in linked to and by the transitive permutation module for linearly spanned by the elements of . In the following proposition we show how the restriction of the Foulkes module decomposes into the direct sum of transitive permutation modules. Such decompositions will be used in all the proofs of the main theorems of this paper. ###### Proposition 2.11. Let be a natural number such that . Then H(ab)⏐↓Sr×Sab−r=⨁λ∈P(r)baVλ. ###### Proof. Let . The restriction of to decomposes as a direct sum of transitive permutation modules, one for each orbit of on . Observe that two set partitions , are in the same orbit of on if and only if and are linked to the same partition . The result follows. ∎ An immediate corollary of Proposition 2.11 is the following result about the multiplicity of characters labelled by two-row partitions, proved by a different argument in [5, Theorem 5.4.34]. ###### Corollary 2.12. Let and be natural numbers. Then ###### Proof. It is well known that π(ab−r,r)=1Sab−r×Sr↑⏐Sab. Therefore, by Frobenius reciprocity and Proposition 2.11 we have ⟨ϕ(ab),π(ab−r,r)⟩= ⟨ϕ(ab)⏐↓Sab−r×Sr,1Sab−r×Sr⟩ = ∑λ∈P(r)ba⟨χVλ,1Sab−r×Sr⟩ = ∑λ∈P(r)ba1=∣∣P(r)ba∣∣. To complete the proof of part (i), it suffices to observe that the conjugation of partitions induces a one to one map between and , and so ∣∣P(r)ba∣∣=∣∣P(r)ab∣∣ for all , and natural numbers. Part (ii) follows immediately from (i) since . ∎ We conclude this section with the definition and a description of a generalized Foulkes module that will be used in the proof of Theorem 1.2. ###### Definition 2.13. Let be a partition of , where , and let . We define Hη=(H(ab11)⊗H(ab22)⊗⋯⊗H(abrr))↑⏐SnG We denote by the character of the generalized Foulkes module . ###### Definition 2.14. Let be a partition of . Define to be the collection of all the set partitions of into sets of sizes . The proof of the following proposition is left to the reader. ###### Proposition 2.15. Let be a partition of . Then ## 3 The multiplicities of hook characters are zero In this section we will prove that no Specht module labelled by a hook partition appears in the Foulkes module . ###### Definition 3.1. Let be a -module with character . For all we denote by the exterior power of , and by the corresponding character. Let be the sign character of the symmetric group for any natural number . We leave to the reader the proofs of the following two well known results. ###### Lemma 3.2. Let and be natural numbers, then Akπ(n−1,1)=(ϵk×1n−k)↑⏐Sn. ###### Lemma 3.3. Let and be natural numbers such that . Then χ(n−k,1k)=Akχ(n−1,1) and Akπ(n−1,1)=χ(n−k,1k)+χ(n−(k−1),1k−1). In the following proposition we will calculate the inner product between the Foulkes character and the character . This is a fundamental step in the proof of Theorem 1.1. ###### Proposition 3.4. Let and be natural numbers and let . Then ⟨ϕ(ab),Akψ⟩={0if k≥21if k=0,1. ###### Proof. Firstly consider the case . Let . By Lemma 3.2 ⟨ϕ(ab),Akψ⟩=⟨ϕ(ab),(ϵk×1ab−k)↑Sab⟩=⟨ϕ(ab)↓K,ϵk×1ab−k⟩. The final inner product above is equal to the number of -submodules in whose associated character is . By Proposition 2.11 it suffices to show that if then has no submodule with character . Suppose that spans such a submodule. Let , where the sum is over all set partitions . Choose such that . If then there exist such that and appear in the same set in . Hence whereas , a contradiction. Therefore . If then Q={{1,x12,…,x1a},{2,x22,…,x2a},…,{k,xk2,…,xka},…} for some . Taking , we obtain a contradiction again, since but . Hence there are no -submodules of having character . The two cases and are left to the reader. ∎ We are now ready to prove Theorem 1.1. This theorem follows at once from Proposition 3.4, since, from Lemma 3.3 we have that χ(ab−r,1r)=Arχ(ab−1,1)=(−1)rr∑k=0(−1)kAkψ. We end this section with a corollary of Theorem 1.1 that will be needed in the proof of Theorem 1.2. Recall that is the character of the generalized Foulkes module , as defined in Definition 2.13. ###### Corollary 3.5. Let be a partition of , where . If then ⟨ψη,χ(n−r,1r)⟩=0. ###### Proof. From the definition of generalized Foulkes module, we can write as a character induced from ϕ(ab11)×⋯×ϕ(abtt). It follows from Theorems 2.6 and 1.1 that in order to obtain as an irreducible constituent of the induced character, we have to take the trivial character in each factor. Therefore ⟨ψη,χ(n−r,1r)⟩=⟨(1Sa1m1×⋯×1Satmt)↑⏐Sn,χ(n−r,1r)⟩. Observe that the right-hand side is the multiplicity of in the Young permutation character . By Theorem 2.3, the constituents of are labelled by partitions with at most parts, so we need to get a non-zero multiplicity. ∎ ## 4 A sufficient condition for zero multiplicity In this section we will prove Theorem 1.2 by an inductive argument. Part of the section will be devoted to the proof of the base step of such induction. Firstly we need to state two technical lemmas. Let be a natural number. Denote by the subgroup . Let be in and let and be as in Definition 2.10. Then by a standard result on orbit sums we have the following lemma. ###### Lemma 4.1. The largest -submodule of on which acts trivially is U:=⟨∑σ∈SβPσ \| P∈O(λ)⟩C. With the next lemma we will understand precisely the structure of this particular module . ###### Lemma 4.2. Let . If then U≅CSβ⊗Hη where and is a generalized -Foulkes module. ###### Proof. By Proposition 2.15 it suffices to show that the set W:={∑σ∈SβPσ \| P∈O(λ)} is isomorphic as a -set to the set of all -partitions of . Let . We define a map by Pfλ={A1∩X,A2∩X,…,Ab∩X} where . It is easy to see that is well defined since by definition of . The map is surjective, and for all and in we have that if and only if and are in the same -orbit of . It is easy to see that is an -map and that for all we have that , since fixes the numbers greater than . To conclude the proof we define ~fλ:W⟶Ωη by (∑σ∈SβPσ)~fλ=Pfλ for all . The map is well defined and the surjectivity of follows directly from the surjectivity of . The map is also injective since (∑σ∈SβPσ)~fλ=(∑σ∈SβQσ)~fλ⟺Pfλ=Qfλ⟺P=Qτ⟺∑σ∈SβPσ=∑σ∈SβQσ for some . Finally is an -map since is an -map and for all and . Therefore is the desired isomorphism. ∎ In the following proposition we use the notation as defined in the introduction. In particular we consider partitions of with trivial inside-partition (one row). The proposition is actually the base step of the inductive proof of Theorem 1.2. ###### Proposition 4.3. Let and be natural numbers. For all we have ⟨ϕ(ab),χ[k:(β)]⟩=0. ###### Proof. By Theorem 2.5 and Frobenius reciprocity, we have ⟨ϕ(ab),χ[k:(β)]⟩ ≤ ⟨ϕ(ab),(1Sβ×χ(ab−(k+β),1k))↑⏐SabSβ×Sab−β⟩ = ⟨ϕ(ab)⏐↓Sβ×Sab−β,1Sβ×χ(ab−(k+β),1k)⟩. Let . By Proposition 2.11 we have: H(ab)⏐↓K=⨁λ∈P(β)baVλ. Fix . Let be the number of parts of . We are now interested in submodules such that acts trivially on . By Lemmas 4.1 and 4.2, the largest submodule of is isomorphic to , where . From now on we will denote . Note that U≅Hη⊗CSβ≅(H(ab−r)⊗Hζ)↑⏐Sab−β⊗CSβ. Hence ⟨χVλ,χ(ab−(k+β),1k)×1Sβ⟩= ⟨χU,χ(ab−(k+β),1k)×1Sβ⟩ = ⟨(ϕ(a(b−r))×ψζ)↑⏐Sab−β×1Sβ,χ(ab−(k+β),1k)×1Sβ⟩ = ⟨(ϕ(a(b−r))×ψζ)↑⏐Sab−β,χ(ab−(k+β),1k)⟩ = ∑ν,μdμν⟨(χν×χμ)↑⏐Sab−β,χ(ab−(k+β),1k)⟩ where is an irreducible character of with non zero multiplicity in , is an irreducible character of having non zero multiplicity in , and is the multiplicity of their tensor product in the decomposition of . Notice that the last sum is not equal to zero if and only if there exist and such that contains a character of having leg length equal to in its decomposition. By Theorem 2.6, we have that both and must be subpartitions of . This means that and are hooks or trivial partitions. In particular we deduce from Theorem 1.1 that . So we need to be a hook with leg length at least to have ⟨(χν×χμ)↑⏐Sab−β,χ(ab−(k+β),1k)⟩≠0. On the other hand ψζ=(ϕ((a−λ1)m1)×⋯×ϕ((a−λs)ms))↑⏐Sar−β So by Corollary 3.5 we have that the hooks that have non-zero multiplicity in the decomposition of have at most parts, where is the number of different parts of . We observe that the smallest number having a partition with different parts is , with . So under our hypothesis we obtain that cannot be a hook character with leg length at least . Hence for all we have that ⟨χVλ,χ(ab−(k+β),1k)×1Sβ⟩=0. We are now ready to prove Theorem 1.2. ###### Proof of Theorem 1.2. We proceed by induction on , the number of parts of the inside-partition . If then ⟨ϕ(ab),χ[k:(α1)]⟩=0 by Proposition 4.3. Suppose now that and the theorem holds when the inside-partition has less then parts. Denote . By Theorem 2.5, Lemma 4.2 and Frobenius reciprocity we have that ⟨ϕ(ab),χ[k:α]⟩≤ ⟨ϕ(ab),(χ[k:ν]×1Sαt)↑⏐Sab⟩ = ⟨ϕ(ab)⏐↓Sab−αt×Sαt,χ[k:ν]×1Sαt⟩ = ∑λ∈P(αt)ba⟨χVλ,χ[k:ν]×1Sαt⟩ = ∑λ∈P(αt)ba⟨χUλ,χ[k:ν]×1Sαt⟩ = ∑λ∈P(αt)ba⟨(ϕ(a(b−p(λ)))×ψ(a−λp(λ),…,a−λ1))↑⏐Sab−αt,χ[k:ν]⟩ = where, for each , is the largest submodule of on which acts trivially and is the decomposition into irreducible characters of the character . For every , observe that every simple summand of is a simple summand of the Young permutation module . Hence by Theorem 2.3 we have that the partition has at most parts; in particular it has at most parts. It follows that, by Theorem 2.6, we need to have at least parts, and to be a subpartition of in order to have ⟨(χζ×χμ)↑⏐Sab−αt,χ[k:ν]⟩≠0. Therefore must be of the form [kζ:β]⊢a(b−p(λ)) with • , and • . We conclude proving that such a cannot label any irreducible summand of the Foulkes character . Define ; if then let . We observe that such a partition has inside-partition having parts and it satisfies the initial hypothesis, since • , and Hence has zero multiplicity in by induction. Therefore ⟨ϕ(ab),χ[k:α]⟩≤ The theorem is then proved. ∎ As mentioned in the introduction, and as we will prove in the following corollary, a consequence of our main theorem is that every Specht module labelled by a partition having leg length equal to and at most boxes inside the hook has zero multiplicity, except when the boxes are column-shaped (i.e. the inside-partition is ). In that particular case we are able to prove that the multiplicity equals 1, for all the values of . The proof is similar to that of Proposition 4.3 and is omitted. In [3, Lemma 3.3] Dent proved the same result in the specific case . ###### Corollary 4.4. Let and be natural numbers. Let and be a partition of not equal to . Then ⟨ϕ(ab),χ[k:α]⟩=0 ###### Proof. Let be an arbitrary partition of not equal to . Then α=(α1,α2,…,αt). Write . We will show that satisfies the hypothesis of Theorem 1.2 that: • , and • . The first condition is trivial since k≥m=α1+n. To prove the second condition proceed by contradiction: suppose that α1≥(k−n)(k−n+1)2. Then k−n≥(k−n)(k−n+1)2. This implies or . The first situation is impossible because . The second is also impossible because implies and with . ∎	{"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": 0.9918069839477539, "perplexity": 426.31501997992154}, "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/1642320306181.43/warc/CC-MAIN-20220129122405-20220129152405-00099.warc.gz"}
https://www.physicsforums.com/threads/linearizing-an-explicit-differentiation-scheme.591518/	# Linearizing an explicit differentiation scheme 1. Mar 29, 2012 ### Pietair 1. The problem statement, all variables and given/known data Consider the following implicit scheme: $$y_{n+1}=y_{n}+\frac{\Delta t}{2}\left [f(y_{n+1})+f(y_{n})]$$ By linearization one can obtain an explicit scheme which is an approximation to this - with approximation error $$O(\Delta t^{3})$$ 2. Relevant equations The solution is: $$y_{n+1}=y_{n}+\Delta t\left [1-\frac{1}{2}\Delta t f'(y_{n}) \right ]^{-1}f(y_{n})$$ And the notation: $$y'(x)=f(x,y(x))$$ 3. The attempt at a solution I think I have to take the Taylor expansion of $$f(y_{n+1})$$ I get: $$f(y_{n+1})=f(y_{n})+\Delta tf'(y_{n})+O(\Delta t^{2})$$ Substituting in the main scheme: $$y_{n+1}=y_{n}+\frac{\Delta t}{2}\left [f(y_{n})+\Delta tf'(y_{n})+O(\Delta t^{2})+f(y_{n})]f(y_{n+1})=y_{n}+\frac{\Delta t}{2}\left [f(y_{n})+\Delta ty'(y_{n})+O(\Delta t^{2})+f(y_{n})]f(y_{n+1})$$ This expression is, as far as I can see, not equal to the expression of the solution. Any help would be appreciated! 2. Mar 29, 2012 ### SammyS Staff Emeritus You are missing the "\right" code with several of your "]" symbols. (Corrected in Quote.) 3. Mar 29, 2012 ### hunt_mat This is what I did, it may help you it may not. $$\begin{array}{rcl} f(y_{n+1}) & = & f\left( y_{n}+\frac{\delta t}{2}X\right) \\ & = & f(y_{n}+\frac{X\delta t}{2}f'(y_{n}) \\ & = & f(y_{n})+ \frac{\delta t}{2}(f(y_{n+1})+f(y_{n}))f'(y_{n}) \end{array}$$ where $X=f(y_{n+1})+f(y_{n})$ and we can solve for $f(y_{n+1})$, to obtain: $$f(y_{n+1})=\left( 1-\frac{\delta t}{2}f'(y_{n})\right)^{-1}\left( 2f(y_{n})+\frac{\delta t}{2}f'(y_{n})\right)$$ Last edited: Mar 29, 2012 Similar Discussions: Linearizing an explicit differentiation scheme	{"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": 0.9815182089805603, "perplexity": 1665.3207366020893}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886105927.27/warc/CC-MAIN-20170819220657-20170820000657-00345.warc.gz"}
http://www.uah.edu/science/departments/math/research/dissertations/290-main/science/science-mathematical-science/3827-jianchao-zhu	# Jianchao Zhu ## On Linear Differential Equations with Functionally Commutative Coefficient Matrices Let C and R be the fields of complex and real numbers, respectively. Let Mn(C) be the space of matrices of order n over C. Let I ⊆ R and D ⊆ C. A matrix-valued function FI → Mn(C) is said to be functionally commutative on I if F(t)F(τ) = F(τ)F(t), for all t, τ in I, and F is said of the proper if F(t) = f(tA) where A in Mn(C) and fI × C → C is a scalar function. The system of linear differential equations x' = PxPI → Mn(C), will be called semiproper if P is functionally commutative on I. It is known that a semiproper system has a closed form fundamental solution matrix where the matrix exponential function is defined by the power series Therefore the problem of solving a semiproper system amounts to that of finding a finite form for the matrix exponential. in this thesis, systematic approaches are developed for finding finite form analytical solutions for semiproper systems. Stability criteria for semiproper systems are also presented. To this end, functionally commutative matrix functions are investigated in terms of proper matrix functions.	{"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": 0.9464857578277588, "perplexity": 351.0831582283447}, "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/1368701638778/warc/CC-MAIN-20130516105358-00022-ip-10-60-113-184.ec2.internal.warc.gz"}
https://appliedmachinelearning.wordpress.com/2017/03/09/understanding-support-vector-machines-a-primer/	One of the most frequently encountered task in many ML applications is classification. From the plethora of classifiers at our disposal, Support Vector Machines (SVM) are arguably the most widely used one. Yet, the intuition behind their working and the key concepts are seldom understood and their understanding is reduced to just learning a few buzzwords such as hyperplanes, kernels, support vectors, etc. This post is an attempt to unroll the mystery that SVMs tend to become for beginners in ML. ### Introduction – Hyperplane and Margin An SVM performs two class classification by building a classifier based on a training set, making it a supervised algorithm. The training set points have features and their class label. An SVM aims to find a separating boundary from which the distances of all the training points are as large as possible. This boundary is what is called the optimal hyperplane, while the distances give an idea about the margin. Notice the word optimal used with hyperplane – its easy to imagine that there will be a lot of boundaries that can separate the data points of the 2 classes, but the fact that SVM finds the one located as far as possible from the data points of either class makes it optimal. The figure below illustrates this important concept about SVMs. The points plotted in the figure are feature vectors for 40 speech files belonging to 2 emotion classes – happy and sad (20 each). The feature vectors have been projected to a 2D space from their high dimensional space by applying principal component analysis (PCA) for visualization purposes. You may be having many questions after reading this – from where these data points are coming, why PCA has been used, what are the 2 axes, etc. Don’t worry too much. Just understand we have a dataset of points from 2 classes (happy and sad) which we want to separate. In the figure above, while the pink, cyan and brown lines are also separating the data points of these 2 classes, its the blue line which is the optimal separating hyperplane. We move ahead by explaining the concept of margin. Margin is the distance between the optimal hyperplane and the training data point closest to the hyperplane. This distance, when taken on both the sides of the hyperplane, forms a region where no data point will ever lie. SVM intends to make this region as wide as possible and the selection of the optimal hyperplane is such that it aids in fulfilling this target. If it were to choose a hyperplane which is quite close to any of the data point, the margin would have been small. Thus, an SVM builds a classifier by searching for a separating hyperplane which is optimal and maximizes the margin. Its time to delve into the mathematics behind the above keywords. Firstly, lets clear the air about what a hyperplane is. In the figures above, we saw our separating hyperplane to be a line. Then why did we call it a plane at all, moreover a hyperplane? In essence, a hyperplane is just a fancy word for a plane. It becomes a point for 1 dimensional data points, a line when data points are of 2 dimensions, a plane in 3 dimensions and is called a hyperplane if dimensionality exceeds 3. The equation of a hyperplane is $\textbf{w}^T\textbf{x} + b = 0$ (1) where $\mathbf{w}$ and $\mathbf{x}$ are both vectors. $\textbf{w}^T\textbf{x}$ is actually the dot product of $\mathbf{w}$ and $\mathbf{x}$thus $\textbf{w}^T\textbf{x}$ and $\textbf{w}.\textbf{x}$ are interchangeable. Considering the case of 2 dimensional data points, implying every data point $\mathbf{x}$ is of the form $\begin {bmatrix}x\\y \end{bmatrix}$ , putting $\mathbf{w}$ = $\begin {bmatrix}-1\\m \end{bmatrix}$ gives us $\textbf{w.x} = -mx + y$ $\textbf{w.x} = y - mx$ The right hand side of the above equation is the equation of a line with slope ‘m’. The above formulation is to convince the reader that in 2 dimensions, a hyperplane is essentially a line. Adding ‘b’ to both sides of the equation $\textbf{w.x} + b = y - mx + b$ Again, the RHS of the equation is the equation of a line with slope ‘m’ and intercept ‘b’. The equation of a hyperplane when written like in (1) (vectorized form) should provide you the intuition that points with more than 2 dimensions can also be easily dealt with. The vector $\mathbf{w}$ is an important one. From the definition of a hyperplane, $\mathbf{w}$ is the normal to the hyperplane. Keeping this in mind, we will now determine the margin $\mathbf{m}$ of a hyperplane (not to be confused with slope ‘m’ written earlier). That will give us a very nice understanding of the optimization that an SVM performs. ### Problem Formulation We consider a dataset containing $\mathbf{n}$ points, denoted as $\mathbf{x_i}$ and each $\mathbf{x_i}$ has a label $\mathbf{y_i}$, which is either +1 or -1. The label can be anything, but lets consider +1 and -1 for understanding the concept. Consider 3 separating hyperplanes HP0, HP1 and HPas follows $\textbf{w.x} + b = 0$ $\textbf{w.x} + b = -1$ $\textbf{w.x} + b = +1$ HP1 and HP2 are no ordinary hyperplanes; as per our objective of having a margin devoid of any data points lying inside it, HP1 and HP2 are selected such that for each data point $\mathbf{x_i}$ if $y_i = +1$, then $\mathbf{w}\cdot\mathbf{x_i} + b \geq +1$ or if $y_i = -1$, then $\mathbf{w}\cdot\mathbf{x_i} + b \geq -1$ By introducing above constraints, we have assured that the region between HP1 and HP2 will have no data points in it. These 2 constraints can be concisely written as one as follows $y_i(\mathbf{w}\cdot\mathbf{x_i} + b ) \geq 1$ Having selected the above constraint obeying hyerplanes, next in line is the task of maximizing the distance $\mathbf{m}$ (i.e.,the margin) between HP1 and HP2. Look at the figure shown below. Lets consider a point $\mathbf{x_1}$ lying on the plane HP1. Notice that HP0 is equidistant from HP1 and HP2. If we had a point $\mathbf{z_1}$ that lies on HP2 ,we could simply calculate the distance between $\mathbf{x_1}$ and $\mathbf{z_1}$ and that would have been all. You might think that $\mathbf{z_1}$ can be obtained by simply adding $\mathbf{m}$ to $\mathbf{x_1}$, but remember, that while $\mathbf{x_1}$ is a vector, $\mathbf{m}$ is a scalar, and thus the two can’t be added. By looking at figure below, lets think about what we need in order to get $\mathbf{z_1}$. Point $\mathbf{x_1}$ can also be seen as a vector from the origin to the $\mathbf{x_1}$ (as shown in figure below). Similarly, we can have a vector beginning from the origin to $\mathbf{z_1}$. From the discussion above, we know this much that we require a new vector with magnitude $\mathbf{m}$, the distance between HP1 and HP2. Lets call this vector $\mathbf{k}$. Being a vector, $\mathbf{k}$ also requires a direction and we make $\mathbf{k}$ have the same direction as $\mathbf{w}$, the normal to HP2. Then $\mathbf{k}$ on addition to $\mathbf{x}$ will give us a point that lies on HPas shown below (triangle law of vector addition). Direction of $\mathbf{w} = \mathbf{ \frac{w}{\|\|w\|\|}}$. Thus $\mathbf{k} = m \mathbf{\frac{w}{\|\|w\|\|}}$ Now, $\mathbf{z_1} = \mathbf{x_1} + \mathbf{k}$. We have obtained a point that lies on HP2. Now since $\mathbf{z_1}$ lies on HP2, it satisfies the equation of HP($\mathbf{w}\cdot\mathbf{x} + b = 1$). Thus $w.z_1 + b = 1$ $w.(x_1 + k) + b = 1$ $w.(x_1 + m\frac{w}{\|\|w\|\|}) + b = 1$ $w.x_1 + m\frac{w.w}{\|\|w\|\|}) + b = 1$ $w.x_1 + m\frac{\|\|w\|\|^2}{\|\|w\|\|}) + b = 1$ $w.x_1 + m\|\|w\|\|+ b = 1$ $w.x_1 + b = 1 - m\|\|w\|\|$ Since $\mathbf{x_1}$ lies on HP1, thus $\mathbf{w}\cdot\mathbf{x_1} + b = -1$. $- 1 = -1 - m \|\|w\|\|$ $m = \frac{2}{\|\|w\|\|}$ You must realize we have landed at a very important result here. Maximizing margin $\mathbf{m}$ is the same as minimizing \|\|w\|\|, the norm of the vector $\mathbf{w}$. Thus, we now have a formulation for the constrained optimization problem that an SVM solves. minimize $\|\|w\|\|^2$ subject to $y_i(\mathbf{w}\cdot\mathbf{x_i} + b ) \geq 1$ for all i ### Primal and Dual The above formulation is known as the primal optimization problem for an SVM. What is actually solved when determining the maximum margin classifier is known as the dual of the primal problem. The dual is solved by using the Lagrange multipliers method. The motivation behind solving the dual in place of the primal is that by solving the dual, the algorithm can be easily written in terms of the dot products between the points in the input data set, a property which becomes very useful for the kernel trick (introduced later). Lets see what the dual formulation looks like. $max_\alpha W(\alpha) = \sum_{i=1}^n \alpha_i - \frac{1}{2} \sum_{i,j=1}^n y^{(i)} y^{(j)} \alpha_i \alpha_j (x^{(i)},x^{(j)})$ subject to $\alpha_i \geq 0, i = 1,2,..., n$ $\sum_{i=1}^n \alpha_i y^{(i)} = 0$ Seems overwhelming? Don’t bother about its complexity and just take away the crucial points explained next. Here, $\mathbf{\alpha_i}$ are the Lagrange multipliers. Notice the term $(x^{(i)},x^{(j)})$, which is the dot product of 2 input data points, $x^{(i)}$ and $x^{(j)}$ . Also notice that while the primal was a minimization problem, the dual is a maximization one. Thus we have obtained a maximization problem in which the parameters are the $\mathbf{\alpha_i}$ and we are dealing completely with the dot products between the data points. The benefit in this comes from the fact that dot products are very fast to evaluate. This dot product formulation comes in handy further too when kernels (introduced later) are used. Another interesting point is that the constraint $\mathbf{\alpha_i > 0}$ will be true only for a small number of points, the ones which lie closest to the decision boundary. These points are known as the support vectors, and for making any future predictions for a new test data point $\mathbf{X}$, we will require only the inner products between $\mathbf{X}$ and these few support vectors rather than the entire training set for determining which class the point $\mathbf{X}$ belongs to. The points surrounded by a blue rectangle (nearest to the decision boundary) are the support vectors in our 2 class case. ### Linearly Non-Separable Data – Soft Margin & Kernels Till now, we have very conveniently assumed our training set to be linearly separable. However, if the 2 classes are not linearly separable, or are somehow linearly separable but the resulting margin is very narrow, we have the concept of soft margin in SVMs. It often happens that by violating the constraints for a few points, we get a better margin for the separating hyperplane rather than by trying to fit a strictly separating hyperplane, as in figure shown below. If the constraints are strictly followed for all the points in this case, then we can see that the resulting separating hyperplane has a very small margin. Thus, sometimes by allowing a few mistakes while partitioning the training data, we can get a better margin. At the same time, the SVM applies a penalty $\mathbf{\zeta}$ on the points for which the constraint are being violated. $\zeta_i = max(0, 1 - y_i (w.x_i + b))$ Looking at the equation, we can see that $\mathbf{\zeta_i}$ will be 0 for data points on correct side of the margin, while it will be a positive number for data points on the wrong side. This alters our minimization problem from before to minimize $\|\|w\|\|^2 + \lambda\sum_{i=1}^n\zeta_i$ subject to $y_i(\mathbf{w}\cdot\mathbf{x_i} + b) \geq 1 - \zeta_i, \zeta_i > 0$ for all i The variable $\mathbf{\lambda}$ is the regularization parameter and is actually determining the trade off between maximizing the margin and ensuring that all training data points lie on the correct side of the margin. If $\mathbf{\lambda}$ is small, then we are effectually making the total regularization term $\lambda\sum_{i=1}^n\zeta_i$ small in the optimization function and we get a large margin. The opposite becomes true if $\mathbf{\lambda}$ chosen is large. Another way of SVMs to deal with the case of non-linearly separable data points is the kernel trick. The idea stems from the fact that if the data cannot be partitioned by a linear boundary in its current dimensionality, then projecting the data into a higher dimensional space may make it linearly separable, such as in figure shown below (taken from the internet). When such is the case, there will be a function $\mathbf{\phi(x)}$ that will project the data points from their current space to the higher dimensional space, and $\phi(x)$ will replace $\mathbf{x}$ in all above equations. In the figure above, we can see that when the 1 dimensional points were projected to 2 dimensional space (higher dimensional space), they became linearly separable. Like we mentioned before, we get the benefit of solving the optimization problem in terms of the dot products only here too, as the kernel function $\mathbf{k(x,z)}$ can then simply be written as the dot product of the transformed points $\mathbf{x}$ and $\mathbf{z}$. $k(x,z) = \phi(x).\phi(z)$ Now whats interesting to note is that often, the kernel $\mathbf{k(x,z)}$ is rather easier to calculate than first calculating $\mathbf{\phi(x)}$ and $\mathbf{\phi(x)}$ and then their dot product, since the vectors have now become high dimensional vectors after their projection. This is why the method has been called kernel trick, since our effective implementation of the kernel can allow us to learn in the higher dimensional space even without explicitly performing any projection and determining the vectors $\mathbf{\phi(x)}$ and $\mathbf{\phi(x)}$ . ### Conclusion This blog was an effort to give a comprehensive idea behind the working of SVMs, right from introducing the key concepts, developing intuition about the keywords to the actual problem formulation, albeit with much of the mathematics intentionally skipped. We aimed to cover as many keywords related to SVMs as possible, nevertheless its expected that some important content was missed. Readers are encouraged to delve more into the mathematical formulation and to discuss it with us, and to also let us know about any more SVM concepts they would like to be elaborated upon in the comments. It will motivate us to cover those concepts in future posts. ### Acknowledgement 1. Alexandre KOWALCZYK’s blog – “SVM Tutorial”. 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https://space.stackexchange.com/questions/8002/how-much-sulphuric-acid-is-on-venus	# How much sulphuric acid is on Venus? According to Wikipedia there are just 150 ppm of sulphur dioxide and 20 ppm of water in Venus' atmosphere. At the same time it is known that there is a considerable amount of sulphuric acid in the clouds. Venus' clouds are made of concentrated sulphuric acid, if I understand it correctly. This would suggest that the amount of sulphuric acid on Venus should be higher than the amount of water (otherwise sulphuric acid would absorb the water since it is strongly hygroscopic). Also, if data cited on Wikipedia are from some spectrographic measurements, this may be affected by some bias: above the clouds there is probably a much lower concentration of sulphuric acid vapors than under the clouds. I ask especially concerning possible colonization and local resource utilization on Venus. Two of the limiting substances are water and hydrogen, and a lot of $H_2SO_4$ would be helpful as a source of both hydrogen and water. To be more specific with the question. 20 ppm of water means $9.6 \times 10^{15} \text{ kg } H_2O$ in total (Venus' atmosphere's mass is $4.8 \times 10^{20} \text{ kg}$). Is there less or more sulphuric acid? Just for scale ... the amount of various sources of carbon on Earth follows: CO2 in Earth atmosphere 0.810e+15 kg Earth coal reserves 0.9e+15 kg (proven recoverable) Earth biosphere 1.9e+15 kg CO2 in Earth sea water 36e+15kg First of all, thanks to @Robert Walker, who informed us about this dissertation of Yeon Joo Lee in his comment on the other answer. In this article on page 14, we can see figure 1.3 with the vertical structure of Venus' cloud layers with mass loading profiles. From this figure we can detect a cloud layer at about 48 to 50 km height with roughly a mass loading of 100 mg/m$^3$. At 50 to 54 km height there is a mass loading of about 10 mg/m$^3$ and from 54-66 km i guess a mean mass loading of roughly 5 mg/m$^3$ would be suitable. All these mass loadings from the 3 mentioned layers added up equals a roughly 3 km thick layer with a mass loading of about 100 mg/m$^3$. To calculate the volume of this assumed layer we use Venus' surface area which is 4.6.10$^1$$^4 m^2 and enlarge this area to 5.10^1$$^4$ m$^2$ for the layer. Accordingly the total weight of the droplets in this layer is 5.10 $^1$$^4 m^2 x 3.10^3 m x 10^-$$^4$ kg/m$^3$ = 15.10 $^1$$^3 kg. Assuming 80% of H_2SO_4 in these droplets gives a total of 12.10 ^1$$^3$ kg H$_2$SO$_4$. According to the question the total amount of water is: 9.6.10 $^1$$^5$ kg H$_2$O So there is much less sulfuric acid than water ! OK, so I found something which more or less answers the question. Presentation Aerosols and Clouds on Earth and Venus from Owen Brian Toon of Department of Atmospheric and Oceanic Sciences and the abstract of BEYOND SULPHURIC ACID – WHAT ELSE IS IN THE CLOUDS OF VENUS? In short the density of the clouds is about 1-50 $mg / m^3$ of liquid phase. Clouds are formed from 75-90% $H_2SO_4$. $6 \times 10^{13} \text{ kg }H_2SO_4$ is produced each year by photochemical processes. But still it does not answer exactly what is the total amount. • Doesn't answer your original question, sorry, but just to say that in the clouds it's much denser. Assuming the droplets that form the cloud layer at 60 km are made of sulfiuric acid then it's an estimated 0.1 grams per cubic meter for the densest layer at 48.5 km, which compares favourably with fog levels on Earth of .05 g/m3 and is at the lower end of the density of cumulonimbus (thunderstorm clouds) at 0.1 - 0.3 g/m3. See page 14 of this paper: mps.mpg.de/phd/theses/… – Robert Walker Sep 25 '16 at 12:28 • @RobertWalker Thanks to you i could answer the question ! – Conelisinspace May 26 '18 at 9:58	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.877607524394989, "perplexity": 1699.887525105679}, "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/1570987751039.81/warc/CC-MAIN-20191021020335-20191021043835-00447.warc.gz"}
https://courses.lumenlearning.com/boundless-statistics/chapter/what-are-the-chances/	## Fundamentals of Probability Probability is the branch of mathematics that deals with the likelihood that certain outcomes will occur. There are five basic rules, or axioms, that one must understand while studying the fundamentals of probability. ### Learning Objectives Explain the most basic and most important rules in determining the probability of an event ### Key Takeaways #### KEY POINTS • Probability is a number that can be assigned to outcomes and events. It always is greater than or equal to zero, and less than or equal to one. • The sum of the probabilities of all outcomes must equal $1$. • If two events have no outcomes in common, the probability that one or the other occurs is the sum of their individual probabilities. • The probability that an event does not occur is $1$ minus the probability that the event does occur. • Two events $\text{A}$ and $\text{B}$ are independent if knowing that one occurs does not change the probability that the other occurs. #### KEY TERMS • experiment: Something that is done that produces measurable results, called outcomes. • outcome: One of the individual results that can occur in an experiment. • event: A subset of the sample space. • sample space: The set of all outcomes of an experiment. In discrete probability, we assume a well-defined experiment, such as flipping a coin or rolling a die. Each individual result which could occur is called an outcome. The set of all outcomes is called the sample space, and any subset of the sample space is called an event. For example, consider the experiment of flipping a coin two times. There are four individual outcomes, namely $\text{HH},\text{HT},\text{TH},\text{TT}$. The sample space is thus $\{\text{HH},\text{HT},\text{TH},\text{TT}\}$. The event “at least one heads occurs” would be the set $\{\text{HH},\text{HT},\text{TH}\}$. If the coin were a normal coin, we would assign the probability of 1/4 to each outcome. In probability theory, the probability $\text{P}$ of some event $\text{E}$, denoted $\text{P}\left(\text{E}\right)$, is usually defined in such a way that $\text{P}$ satisfies a number of axioms, or rules. The most basic and most important rules are listed below. ### Probability Rules Probability is a number. It is always greater than or equal to zero, and less than or equal to one. This can be written as $0\leq{\text{P}}\left(\text{A}\right)\geq{1}$. An impossible event, or an event that never occurs, has a probability of $0$. An event that always occurs has a probability of $1$. An event with a probability of $0.5$ will occur half of the time. The sum of the probabilities of all possibilities must equal $1$. Some outcome must occur on every trial, and the sum of all probabilities is 100%, or in this case, $1$. This can be written as $\text{P}\left(\text{S}\right)=1$, where $\text{S}$ represents the entire sample space. If two events have no outcomes in common, the probability that one or the other occurs is the sum of their individual probabilities. If one event occurs in 30% of the trials, a different event occurs in 20% of the trials, and the two cannot occur together (if they are disjoint), then the probability that one or the other occurs is 30%+20%=50%. This is sometimes referred to as the addition rule, and can be simplified with the following: $\text{P}\left({\text{A}} \text{ or} {\text{ B}}\right)=\text{P}\left(\text{A}\right)+\text{P}\left(\text{B}\right)$. The word “or” means the same thing in mathematics as the union, which uses the following symbol: $\cup$. Thus when $\text{A}$ and $\text{B}$ are disjoint, we have $\text{P}\left(\text{A}\cup{\text{B}}\right)=\text{P}\left(\text{A}\right)+\text{P}\left(\text{B}\right)$. The probability that an event does not occur is $1$ minus the probability that the event does occur. If an event occurs in 60% of all trials, it fails to occur in the other 40%, because 100%−60%=40%. The probability that an event occurs and the probability that it does not occur always add up to 100%, or $1$. These events are called complementary events, and this rule is sometimes called the complement rule. It can be simplified with $\text{P}\left(\text{A}^\text{c}\right)=1−\text{P}\left(\text{A}\right)$, where $\text{A}^\text{c}$ is the complement of $\text{A}$. Two events $\text{A}$ and $\text{B}$ are independent if knowing that one occurs does not change the probability that the other occurs. This is often called the multiplication rule. If $\text{A}$ and $\text{B}$ are independent, then $\text{P}\left(\text{A} \text{ and} \text{ B}\right)=\text{P}\left(\text{A}\right)\text{P}\left(\text{B}\right)$. The word “and” in mathematics means the same thing in mathematics as the intersection, which uses the following symbol: $\cap$. Therefore when $\text{A}$ and $\text{B}$ are independent, we have $\text{P}\left(\text{A}\cap{\text{B}}\right)=\text{P}\left(\text{A}\right)\text{P}\left(\text{B}\right)$. #### Extension of the Example Elaborating on our example above of flipping two coins, assign the probability $1/4$ to each of the $4$ outcomes. We consider each of the five rules above in the context of this example. 1. Note that each probability is $1/4$, which is between $0$ and $1$. 2. Note that the sum of all the probabilities is $1$, since $\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=1$. 3. Suppose $\text{A}$ is the event exactly one head occurs, and B is the event exactly two tails occur. Then $\text{A}=\{\text{HT},\text{TH}\}$ and $\text{B}=\{\text{TT}\}$ are disjoint. Also, $\text{P}\left(\text{A}\cup{\text{B}}\right)=\frac{3}{4}=\frac{2}{4}+\frac{1}{4}=\text{P}\left(\text{A}\right)+\text{P}\left(\text{B}\right)$. 4. The probability that no heads occurs is $1/4$, which is equal to $1−3/4$. So if $\text{A}=\{\text{HT},\text{TH},\text{HH}\}$ is the event that a head occurs, we have $\text{P}\left(\text{A}^\text{c}\right)=\frac{1}{4}=1−\frac{3}{4}=1−\text{P}\left(\text{A}\right)$. 5. If $\text{A}$ is the event that the first flip is a heads and $\text{B}$ is the event that the second flip is a heads, then $\text{A}$ and$\text{B}$ are independent. We have $\text{A}=\{\text{HT},\text{HH}\}$ and $\text{B}={\text{TH},\text{HH}}$ and $\text{A}\cap{\text{B}}={\text{HH}}$. Note that $\text{P}\left(\text{A}\cap{\text{B}}\right)=\frac{1}{4}=\frac{1}{2}\cdot{\frac{1}{2}}=\text{P}\left(\text{A}\right)\text{P}\left(\text{B}\right)$. ## Conditional Probability The conditional probability of an event is the probability that an event will occur given that another event has occurred. ### Learning Objectives Explain the significance of Bayes’ theorem in manipulating conditional probabilities ### Key Takeaways #### KEY POINTS • The conditional probability $\text{P}\left(\text{B}\mid{\text{A}}\right)$ of an event $\text{B}$, given an event $\text{A}$, is defined by: $\text{P}\left(\text{B}\mid{\text{A}}\right)=\frac{\text{P}\left(\text{A}\cap{\text{B}}\right)}{\text{P}\left(\text{A}\right)}$, when $\text{P}\left(\text{A}\right)>0$. • If the knowledge that event $\text{A}$ occurs does not change the probability that event $\text{B}$ occurs, then $\text{A}$ and$\text{B}$ are independent events, and thus, $\text{P}\left(\text{B}\mid{\text{A}}\right)=\text{P}\left(\text{B}\right)$. • Mathematically, Bayes’ theorem gives the relationship between the probabilities of $\text{A}$ and $\text{B}, \text{P}\left(\text{A}\right)$ and $\text{P}\left(\text{B}\right)$, and the conditional probabilities of $\text{A}$ given $\text{B}$ and $\text{B}$ given $\text{A}, \text{P}\left(\text{A}\cap{\text{B}}\right)$ and $\text{P}\left(\text{B}\cap{\text{A}}\right)$. In its most common form, it is: $\text{P}\left(\text{A}\cap{\text{B}}\right)=\frac{\text{P}\left(\text{B}\mid{\text{A}}\right)\text{P}\left(\text{A}\right)}{\text{P}\left(\text{B}\right)}$. #### KEY TERMS • conditional probability: The probability that an event will take place given the restrictive assumption that another event has taken place, or that a combination of other events has taken place • independent: Not dependent; not contingent or depending on something else; free. ### Probability of B Given That A Has Occurred Our estimation of the likelihood of an event can change if we know that some other event has occurred. For example, the probability that a rolled die shows a $2$ is $1/6$ without any other information, but if someone looks at the die and tells you that is is an even number, the probability is now $1/3$ that it is a $2$. The notation $\text{P}\left(\text{B}\mid{\text{A}}\right)$ indicates a conditional probability, meaning it indicates the probability of one event under the condition that we know another event has happened. The bar “$\mid$” can be read as “given”, so that $\text{P}\left(\text{B}\mid{\text{A}}\right)$ is read as “the probability of $\text{B}$ given that $\text{A}$ has occurred”. The conditional probability $\text{P}\left(\text{B}\mid{\text{A}}\right)$ of an event $\text{B}$, given an event $\text{A}$, is defined by: $\text{P}\left(\text{B}\mid{\text{A}}\right)=\frac{\text{P}\left(\text{A}\cap{\text{B}}\right)}{\text{P}\left(\text{A}\right)}$ When $\text{P}\left(\text{A}\right)>0$. Be sure to remember the distinct roles of $\text{B}$ and $\text{A}$ in this formula. The set after the bar is the one we are assuming has occurred, and its probability occurs in the denominator of the formula. ### Example Suppose that a coin is flipped 3 times giving the sample space: $\text{S}=\{\text{HHH},\text{HHT},\text{HTH},\text{THH},\text{TTH},\text{THT},\text{HTT},\text{TTT}\}$ Each individual outcome has probability $1/8$. Suppose that $\text{B}$ is the event that at least one heads occurs and $\text{A}$ is the event that all 3 coins are the same. Then the probability of $\text{B}$ given $\text{A}$ is $1/2$, since $\text{A}\cap{\text{B}}=\{\text{HHH}\}$ which has probability $1/8$ and $\text{A}=\{\text{HHH},\text{TTT}\}$ which has probability $2/8$, and $\frac{1/8}{2/8}=\frac{1}{2}$. ### Independence The conditional probability $\text{P}\left(\text{B}\mid{\text{A}}\right)$ is not always equal to the unconditional probability $\text{P}\left(\text{B}\right)$. The reason behind this is that the occurrence of event $\text{A}$ may provide extra information that can change the probability that event $\text{B}$ occurs. If the knowledge that event $\text{A}$ occurs does not change the probability that event $\text{B}$ occurs, then $\text{A}$ and $\text{B}$ are independent events, and thus, $\text{P}\left(\text{B}\mid{\text{A}}\right)=\text{P}\left(\text{B}\right)$. ### Bayes’ Theorem In probability theory and statistics, Bayes’ theorem (alternatively Bayes’ law or Bayes’ rule) is a result that is of importance in the mathematical manipulation of conditional probabilities. It can be derived from the basic axioms of probability. Mathematically, Bayes’ theorem gives the relationship between the probabilities of $\text{A}$ and $\text{B}$, $\text{P}\left(\text{A}\right)$ and $\text{P}\left(\text{B}\right)$, and the conditional probabilities of $\text{A}$ given $\text{B}$ and $\text{B}$ given $\text{A}$. In its most common form, it is: $\text{P}\left(\text{A}\mid{\text{B}}\right)=\frac{\text{P}\left(\text{B}\mid{\text{A}}\right)\text{P}\left(\text{A}\right)}{\text{P}\left(\text{B}\right)}$ This may be easier to remember in this alternate symmetric form: $\frac{\text{P}\left(\text{A}\mid{\text{B}}\right)}{\text{P}\left(\text{B}\mid{\text{A}}\right)}=\frac{\text{P}\left(\text{A}\right)}{\text{P}\left(\text{B}\right)}$ ### Example Suppose someone told you they had a nice conversation with someone on the train. Not knowing anything else about this conversation, the probability that they were speaking to a woman is 50%. Now suppose they also told you that this person had long hair. It is now more likely they were speaking to a woman, since women in in this city are more likely to have long hair than men. Bayes’s theorem can be used to calculate the probability that the person is a woman. To see how this is done, let $\text{W}$ represent the event that the conversation was held with a woman, and $\text{L}$ denote the event that the conversation was held with a long-haired person. It can be assumed that women constitute half the population for this example. So, not knowing anything else, the probability that $\text{W}$ occurs is $\text{P}\left(\text{W}\right)=0.5$. Suppose it is also known that 75% of women in this city have long hair, which we denote as $\text{P}\left(\text{L}\mid{\text{W}}\right)=0.75$. Likewise, suppose it is known that 25% of men in this city have long hair, or $\text{P}\left(\text{L}\mid{\text{M}}\right)=0.25$, where $\text{M}$ is the complementary event of $\text{W}$, i.e., the event that the conversation was held with a man (assuming that every human is either a man or a woman). Our goal is to calculate the probability that the conversation was held with a woman, given the fact that the person had long hair, or, in our notation, $\text{P}(\text{W}\mid{\text{L}})$. Using the formula for Bayes’s theorem, we have: $\text{P}\left(\text{W}\mid{\text{L}}\right)=\frac{\text{P}\left(\text{L}\mid{\text{W}}\right)\text{P}\left(\text{W}\right)}{\text{P}\left(\text{L}\right)}=\frac{\text{P}\left(\text{L}\mid{\text{W}}\right)\text{P}\left(\text{W}\right)}{\text{P}\left(\text{L}\mid{\text{W}}\right)\text{P}\left(\text{W}\right)+\text{P}\left(\text{L}\mid{\text{M}}\right)\text{P}\left(\text{M}\right)}=\frac{0.75\cdot{0.5}}{0.75\cdot{0.5}+0.25\cdot{0.5}}=0.75$ ## Unions and Intersections Union and intersection are two key concepts in set theory and probability. ### Learning Objectives Give examples of the intersection and the union of two or more sets ### Key Takeaways #### KEY POINTS • The union of two or more sets is the set that contains all the elements of the two or more sets. Union is denoted by the symbol $\cup$. • The general probability addition rule for the union of two events states that $\text{P}\left(\text{A}\cup{\text{B}}\right)=\text{P}\left(\text{A}\right)+\text{P}\left(\text{B}\right)−\text{P}\left(\text{A}\cap{\text{B}}\right)$, where $\text{A}\cap{\text{B}}$ is the intersection of the two sets. • The addition rule can be shortened if the sets are disjoint: $\text{P}\left(\text{A}\cup{\text{B}}\right)=\text{P}\left(\text{A}\right)+\text{P}\left(\text{B}\right)$. This can even be extended to more sets if they are all disjoint: $\text{P}\left(\text{A}\cup{\text{B}}\cup{\text{C}}\right)=\text{P}\left(\text{A}\right)+\text{P}\left(\text{B}\right)+\text{P}\left(\text{C}\right)$. • The intersection of two or more sets is the set of elements that are common to every set. The symbol $\cap$ is used to denote the intersection. • When events are independent, we can use the multiplication rule for independent events, which states that $\text{P}\left(\text{A}\cap{\text{B}}\right)=\text{P}\left(\text{A}\right)\text{P}\left(\text{B}\right)$. #### KEY TERMS • independent: Not contingent or dependent on something else. • disjoint: Having no members in common; having an intersection equal to the empty set. Probability uses the mathematical ideas of sets, as we have seen in the definition of both the sample space of an experiment and in the definition of an event. In order to perform basic probability calculations, we need to review the ideas from set theory related to the set operations of union, intersection, and complement. ### Union The union of two or more sets is the set that contains all the elements of each of the sets; an element is in the union if it belongs to at least one of the sets. The symbol for union is $\cup$, and is associated with the word “or”, because $\text{A}\cup{\text{B}}$ is the set of all elements that are in $\text{A}$ or $\text{B}$ (or both.) To find the union of two sets, list the elements that are in either (or both) sets. In terms of a Venn Diagram, the union of sets $\text{A}$ and $\text{B}$ can be shown as two completely shaded interlocking circles. Union of Two Sets: The shaded Venn Diagram shows the union of set $\text{A}$ (the circle on left) with set $\text{B}$ (the circle on the right). It can be written shorthand as $\text{A}\cup{\text{B}}$ In symbols, since the union of $\text{A}$ and $\text{B}$ contains all the points that are in $\text{A}$ or $\text{B}$ or both, the definition of the union is: $\text{A}\cup{\text{B}}=\{\text{x}:\text{x}\in{\text{A}} \text{ or } \text{x}\in{\text{B}}\}$ For example, if $\text{A}=\{1,3,5,7\}$ and $\text{B}=\{1,2,4,6\}$ , then $\text{A}\cup{\text{B}}=\{1,2,3,4,5,6,7\}$. Notice that the element 1 is not listed twice in the union, even though it appears in both sets $\text{A}$ and $\text{B}$. This leads us to the general addition rule for the union of two events: $\text{P}\left(\text{A}\cup{\text{B}}\right)=\text{P}\left(\text{A}\right)+\text{P}\left(\text{B}\right)−\text{P}\left(\text{A}\cap{\text{B}}\right)$ Where $\text{P}\left(\text{A}\cap{\text{B}}\right)$ is the intersection of the two sets. We must subtract this out to avoid double counting of the inclusion of an element. If sets $\text{A}$ and $\text{B}$ are disjoint, however, the event $\text{A}\cap{\text{B}}$ has no outcomes in it, and is an empty set denoted as ∅, which has a probability of zero. So, the above rule can be shortened for disjoint sets only: $\text{P}\left(\text{A}\cup{\text{B}}\right)=\text{P}\left(\text{A}\right)+\text{P}\left(\text{B}\right)$ This can even be extended to more sets if they are all disjoint: $\text{P}\left(\text{A}\cup{\text{B}}\cup{\text{C}}\right)=\text{P}\left(\text{A}\right)+\text{P}\left(\text{B}\right)+\text{P}\left(\text{C}\right)$ ### Intersection The intersection of two or more sets is the set of elements that are common to each of the sets. An element is in the intersection if it belongs to all of the sets. The symbol for intersection is $\cap$, and is associated with the word “and”, because $\text{A}\cap{\text{B}}$ is the set of elements that are in $\text{A}$ and $\text{B}$ simultaneously. To find the intersection of two (or more) sets, include only those elements that are listed in both (or all) of the sets. In terms of a Venn Diagram, the intersection of two sets $\text{A}$ and $\text{B}$ can be shown at the shaded region in the middle of two interlocking circles . Intersection of Two Sets: Set A is the circle on the left, set B is the circle on the right, and the intersection of A and B, or $\text{A}\cap{\text{B}}$, is the shaded portion in the middle. In mathematical notation, the intersection of $\text{A}$ and $\text{B}$ is written as$\text{A}\cap{\text{B}}=\{\text{x}:\text{x}\in{\text{A}}$ and $\text{x}\in{\text{B}}\}$. For example, if $\text{A}=\{1,3,5,7\}$ and $\text{B}=\{1,2,4,6\}$, then $\text{A}\cap{\text{B}}=\{1\}$ because $1$ is the only element that appears in both sets $\text{A}$ and $\text{B}$. When events are independent, meaning that the outcome of one event doesn’t affect the outcome of another event, we can use the multiplication rule for independent events, which states: $\text{P}\left(\text{A}\cap{\text{B}}\right)=\text{P}\left(\text{A}\right)\text{P}\left(\text{B}\right)$ For example, let’s say we were tossing a coin twice, and we want to know the probability of tossing two heads. Since the first toss doesn’t affect the second toss, the events are independent. Say is the event that the first toss is a heads and $\text{B}$ is the event that the second toss is a heads, then $\text{P}\left(\text{A}\cap{\text{B}}\right)=\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}$. ## Complementary Events The complement of $\text{A}$ is the event in which $\text{A}$ does not occur. ### Learning Objectives Explain an example of a complementary event ### Key Takeaways #### KEY POINTS • The complement of an event $\text{A}$ is usually denoted as $\text{A}′$, $\text{A}^\text{c}$ or $\bar{\text{A}}$. • An event and its complement are mutually exclusive, meaning that if one of the two events occurs, the other event cannot occur. • An event and its complement are exhaustive, meaning that both events cover all possibilities. #### KEY TERMS • exhaustive: including every possible element • mutually exclusive: describing multiple events or states of being such that the occurrence of any one implies the non-occurrence of all the others ### What are Complementary Events? In probability theory, the complement of any event $\text{A}$ is the event $[\text{not A}]$, i.e. the event in which $\text{A}$ does not occur. The event $\text{A}$ and its complement $[\text{not A}]$ are mutually exclusive and exhaustive, meaning that if one occurs, the other does not, and that both groups cover all possibilities. Generally, there is only one event $\text{B}$ such that $\text{A}$ and $\text{B}$ are both mutually exclusive and exhaustive; that event is the complement of $\text{A}$. The complement of an event $\text{A}$ is usually denoted as $\text{A}′$, $\text{A}^c$ or $\bar{\text{A}}$. ### Examples #### Simple Examples A common example used to demonstrate complementary events is the flip of a coin. Let’s say a coin is flipped and one assumes it cannot land on its edge. It can either land on heads or on tails. There are no other possibilities (exhaustive), and both events cannot occur at the same time (mutually exclusive). Because these two events are complementary, we know that $\text{P}\left(\text{heads}\right)+\text{P}\left(\text{tails}\right)=1$. Coin Flip: Often in sports games, such as tennis, a coin flip is used to determine who will serve first because heads and tails are complementary events. Another simple example of complementary events is picking a ball out of a bag. Let’s say there are three plastic balls in a bag. One is blue and two are red. Assuming that each ball has an equal chance of being pulled out of the bag, we know that $\text{P}\left(\text{blue}\right)=\frac{1}{3}$ and $\text{P}\left(\text{red}\right)=\frac{2}{3}$. Since we can only either chose blue or red (exhaustive) and we cannot choose both at the same time (mutually exclusive), choosing blue and choosing red are complementary events, and $\text{P}\left(\text{blue}\right)+\text{P}\left(\text{red}\right)=1$. Finally, let’s examine a non-example of complementary events. If you were asked to choose any number, you might think that that number could either be prime or composite. Clearly, a number cannot be both prime and composite, so that takes care of the mutually exclusive property. However, being prime or being composite are not exhaustive because the number 1 in mathematics is designated as “unique. ”	{"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": 0.9344878792762756, "perplexity": 159.1604843834456}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232254731.5/warc/CC-MAIN-20190519081519-20190519103519-00061.warc.gz"}
http://math.stackexchange.com/questions/76665/distance-between-two-dna-molecules-x-y	# Distance between two DNA molecules $x, y$ A DNA molecule can be represented as a string of symbols $A, C, G$ and $T$, such as $GGATAATTCTAG\ldots GACCGTACCC$. For the purposes of this question, we will assume that all DNA molecules contain the same (large!) number $N$ of symbols. Thus, a DNA molecule is an $N$-tuple $x = (x_1,\ldots , x_N)$ where $x_i\in\{A,C,G,T\}$ for each $i$. Define the distance between two DNA molecules $x, y$ as the number of elements $i$ in ${1, \ldots ,N}$ such that $x_i\neq y_i$. Prove that this defines a metric on the set of DNA molecules. Tutor defined a set $S(x,y)=\{1,3,6\}$. Why is this? The rest of the working after that is fine. - When editing the question the linebreaks implied that this is somehow copy-pasted from somewhere else. This feels more so as $S(x,y)$ is not defined in the question given here. – Asaf Karagila Oct 28 '11 at 13:04 See Hamming distance. – Did Oct 28 '11 at 13:18 Hint: for a problem like this, where you are asked to show that some object (this measure of distance) belongs to some class (metrics) you are asked to verify the definition of the class. I found them very useful to check the understanding of a definition. So just go down the requirements for a metric and see if it works. Clearly the distance between two strings is $\ge 0$, is only $0$ when the strings are identical, and is symmetric. The only one that takes a bit of work is the triangle inequality. Presumably that was the list of positions where they disagree. Then the distance would be the size of $S$, which is $3$. Does that work? – Ross Millikan Nov 4 '11 at 16:46	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9132229685783386, "perplexity": 189.248006545551}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448398458553.38/warc/CC-MAIN-20151124205418-00119-ip-10-71-132-137.ec2.internal.warc.gz"}
https://proofwiki.org/wiki/Category:Definitions/Negative_Binomial_Distribution	# Category:Definitions/Negative Binomial Distribution Jump to navigation Jump to search This category contains definitions related to Negative Binomial Distribution. Related results can be found in Category:Negative Binomial Distribution. Let $X$ be a discrete random variable on a probability space $\struct {\Omega, \Sigma, \Pr}$. There are two forms of the negative binomial distribution, as follows: ### First Form $X$ has the negative binomial distribution (of the first form) with parameters $n$ and $p$ if: $\Img X = \set {0, 1, 2, \ldots}$ $\map \Pr {X = k} = \dbinom {n + k - 1} {n - 1} p^k \paren {1 - p}^n$ where $0 < p < 1$. It is frequently seen as: $\map \Pr {X = k} = \dbinom {n + k - 1} {n - 1} p^k q^n$ where $q = 1 - p$. ### Second Form $X$ has the negative binomial distribution (of the second form) with parameters $n$ and $p$ if: $\Img X = \set {n, n + 1, n + 2, \dotsc}$ $\map \Pr {X = k} = \dbinom {k - 1} {n - 1} p^n \paren {1 - p}^{k - n}$ where $0 < p < 1$. It is frequently seen as: $\map \Pr {X = k} = \dbinom {k - 1} {n - 1} q^{k - n} p^n$ where $q = 1 - p$. ## Pages in category "Definitions/Negative Binomial Distribution" The following 6 pages are in this category, out of 6 total.	{"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": 0.975387454032898, "perplexity": 1094.049626040709}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585186.33/warc/CC-MAIN-20211018000838-20211018030838-00091.warc.gz"}
http://www.scholarpedia.org/article/Power-law	# Scale-Free Networks (Redirected from Power-law) Curator and Contributors 1.00 - Albert-Laszlo Barabasi A network that has a power-law degree distribution, regardless of any other structure, is called a scale-free network. ## Power-Law degree distribution The degree of a node is the number of links adjacent to it. If we call the degree of a node $$k\ ,$$ a scale-free network is defined by a power-law degree distribution, which can be expressed mathematically as $$P(k)\sim k^{-\gamma}$$ From the form of the distribution it is clear that when: • $$\gamma<2$$ the average degree diverges. • $$\gamma<3$$ the standard deviation of the degree diverges. It has been found that most scale-free networks have exponents between 2 and 3. Thus, they lack a characteristic degree or scale, and therefore their name. High degree nodes are called hubs. ## Scale-free network models There are several models that are able to create a scale-free network. But most of them introduce in one way or another two main ingredients, growth and preferential attachment. By growth we mean that the number of nodes in the network increases in time. Preferential attachment refers to the fact that new nodes tend to connect to nodes with large degree. One can naively argue that for large networks this is nonsense because it requires a global knowledge of the network, i.e., knowing which are the high degree nodes, but this is not the case. There are several local mechanisms that introduce preferential attachment (see below). In mathematical terms, preferential attachment means that the probability that a node with degree $$k_i$$ acquires a link goes as $$P(k_i)=\frac{k_i}{\sum_{i}k_i}$$ ## The Barabasi-Albert model The Barabasi-Albert model (a.k.a. BA model) introduced in 1998 explains the power-law degree distribution of networks by considering two main ingredients: growth and preferential attachment (Barabasi and Albert 1999). The algorithm used in the BA model goes as follows. 1. Growth: Starting with a small number ($$m_0$$) of connected nodes, at every time step, we add a new node with $$m(<m_0)$$ edges that link the new node to $$m$$ different nodes already present in the network. 2. Preferential attachment: When choosing the nodes to which the new node connects, we assume that the probability $$P$$ that a new node will be connected to node $$i$$ depends on the degree $$k_i$$ of node $$i\ ,$$ such that $P\sim \frac{k_i}{\sum_{i}k_{i}}$ Numerical simulations and analytic results indicate that this network evolves into a scale-invariant state with the probability that a node has $$k$$ edges following a power law with an exponent $$\gamma=3\ .$$ The scaling exponent is independent of $$m\ ,$$ the only parameter in the model. ## Analytical solution for the BA model This model can be solved analytically by setting up a differential equation in which the rate at which a node acquires links is equal to the number of links added times the probability of acquiring a link $\frac{dk_i}{dt}= m\frac{k_i}{\sum_{j}k_j}$ This equation can be simplified by realizing that at each time step $$m$$ links are added, thus $\sum_j k_j = 2mt$ and $\frac{dk_i}{dt}=\frac{k_i}{2t}$ $\ln{k_i}=\frac{1}{2}\ln{t} + C$ where $$C$$ is an integration constant that can be determined by using the fact that the $$i^{th}$$ node arrived to the network at time $$t^i$$ having degree $$m\ .$$ Thus $k_i = m(\frac{t}{t_i})^{1/2}\ .$ The degree distribution can be calculated by finding the probability that a node has a degree smaller than $$k$$ $P(k_i(t) > k) = P(t_i < \frac{mt}{k^2}) = 1 - P(t_i > \frac{mt}{k^2})$ Without loss of generality we can assume that nodes are added at a constant rate thus $P(t_i) =\frac{1}{m_0 + t}$ where $$m_0$$ is the total degree of the nodes that got the network started. Using this distribution $P(k_{i} - k) = 1 -\frac{mt}{k^2}\frac{1}{m_0 + t}$ Finally we get the degree distribution by differentiating and conclude that $\frac{d}{dk}P(k_i < k) = P(ki = k) = \frac{2m^2 t}{k^3}\frac{1}{m_0 + t}$ This mechanism was first introduced by Yule in the early 20th century to explain the distribution of different Taxa and was later generalized by Price in the 70’s and coined as cumulative advantage. The example shown here is not the most general version of the Price model that can be found in the original paper as well as in Newman (2005). In any case, the lesson that should be learned from this is that whenever we found a system in which the probability of increasing is proportional to the actual value, we should expect its distribution to follow a power-law. ## Local alternatives for preferential attachment We could imagine that when nodes join the network they follow a link. For example, you move to a new town and an old friend tells you that you should visit a friend of him. If we consider that link to be a randomly chosen one, the probability $$\pi$$ that you were referred to a person with degree $$k$$ is $\pi(k)= k P(k)$ which is the precise definition of preferential attachment. This is because $$k$$ links end up in an agent of degree $$k\ .$$ ### Duplication and divergence model In a biological context the "duplication and divergence" model has been proposed as an explanation for the scale-free nature of protein-protein interaction networks. Let us consider a protein that interacts with $$k_p$$ other proteins. Eventually some of the genes that encode a protein get duplicated and now 2 copies become available. This redundancy allows one copy of the gene to mutate without changing the fitness of the organism. This process ends up with different proteins which are likely to share some interactions. If these duplication and divergence process occurs at random, proteins that have a high degree are more likely to have one of its neighbors change, and again, the probability of winning a neighbor through this process is proportional to the actual number of neighbors. This is another example of linear preferential attachment. ### Limited information A different scenario that one can imagine is the one in which a node incorporates to the network with a limited or local information about it. If linear preferential attachment is used as the rule to create links in the local information context, a power-law degree distribution is also recovered, regardless of having limited information. ## Properties of scale-free networks Scale-free networks have qualitatively different properties from strictly random, Erdos and Renyi, networks. These are: • Scale-free networks are more robust against failure. By this we mean that the network is more likely to stay connected than a random network after the removal of randomly chosen nodes. • Scale-free networks are more vulnerable against non-random attacks. This means that the network quickly disintegrates when nodes are removed according to their degree. • Scale-free networks have short average path lengths. In fact the average path length goes as $$L\sim \log{N}/\log{\log{k}}$$ ## Scale-free networks in nature Scale-free networks have been observed in social, technological and biological systems. These include the citation and co-author scientific networks, the internet and world-wide web, and protein-protein interaction and gene regulatory networks (Albert and Barabasi 2002). ## References • R. Albert, A.-L. Barabási (2002) Statistical mechanics of complex networks. Reviews of Modern Physics 74, 47-97. • A.-L. Barabási and R. Albert (1999) Emergence of scaling in random networks. Science 286, 509-512 • M. E. J. Newman (2005) Contemporary Physics 46, 323-351. Internal references	{"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": 0.9466807842254639, "perplexity": 381.3259996142083}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1405997862711.66/warc/CC-MAIN-20140722025742-00073-ip-10-33-131-23.ec2.internal.warc.gz"}
http://mathhelpforum.com/new-users/217017-new-useless-guy-thank-you.html	# Math Help - New Useless Guy: Thank You 1. ## New Useless Guy: Thank You Hi--I'm new to the forum (and suck at maths) and need to construct an equation for the following: [(Sum of(A + B))/C] x [D/E] x[(Sum of (F)) to the power of the mean of G]=? Can someone please present me with the correct notation? I've tried it as follows but have no idea if it is accurate. ((∑_(k=0)^n▒〖(A+B)〗)/T)(D/E) (∑_(k=0)^n▒F)^G=? Thanks! Attached Thumbnails t 3. ## Re: New Useless Guy: Thank You Originally Posted by speedbird168 Hi--I'm new to the forum (and suck at maths) and need to construct an equation for the following: [(Sum of(A + B))/C] x [D/E] x[(Sum of (F)) to the power of the mean of G]=? Can someone please present me with the correct notation? I've tried it as follows but have no idea if it is accurate. ((∑_(k=0)^n▒〖(A+B)〗)/T)(D/E) (∑_(k=0)^n▒F)^G=? Thanks! Two things. 1. There are characters in your question that aren't coming out. 2. You need to post this in an appropriate forum. The New Users forum is to introduce yourself, not post a problem. -Dan	{"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": 0.9332016706466675, "perplexity": 2826.769228249084}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-30/segments/1469257823133.4/warc/CC-MAIN-20160723071023-00073-ip-10-185-27-174.ec2.internal.warc.gz"}
https://byjus.com/commerce/utility/	# Utility - All you need to know ## What is Utility? A customer is the one who normally determines his demand for goods on the basis of satisfaction (utility) that he procures from it. So, what is a utility? The utility of goods is its want-satisfying capability. The more the stronger the aspiration to have it, the more is the utility procured from the goods. The utility is instinctive. Distinct people can get different degrees of utility from the same goods. For instance, someone who likes sweets will get much higher utility from a sweet than someone who doesn’t sweets. The utility that an individual obtains from the goods can differ with the change in location and time. For instance, utility from the use of an Air conditioner certainly relies upon whether the person is in Srinagar or Jaipur(location) or whether it is winter or summer (season). ## Approaches that elucidate consumer behaviour • Cardinal Utility Analysis – Cardinal utility is defined as – the perspective is put forward by the economists, who presume that utility is quantifiable and the consumer can convey his or her contentment in fundamental or measurable numbers, such as 2,3,4 and so on. Measures of Utility : • Total Utility – Total utility of a determined quantity of goods or commodity (TU) is the total contentment procured from utilizing the given amount of some goods p. • Marginal Utility – MU is the difference in total utility due to the utilization of one extra unit of goods or commodity. • Ordinal Utility Analysis – The customer does not quantify utility in numerals. All the theory of customer decision-making under constraints of certitude can be and mostly is, conveyed in the terms of ordinal utility. The above mentioned is detailed information about the concept Utility, To know more about other Commerce concepts, stay tuned to BYJU’S. Important Topics in Commerce:	{"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": 0.9314165711402893, "perplexity": 1834.4909979954277}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400279782.77/warc/CC-MAIN-20200927121105-20200927151105-00380.warc.gz"}
https://itectec.com/englishusage/learn-english-should-i-use-either-or-any-in-this-sentence/	Learn English – Should I use either or any in this sentence any-everygrammarsome-any So, I want to ask the students to contact me if they are interested in topics A and B. Which one is better? or Or, do you think there is a better way to put that?	{"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": 0.8095701336860657, "perplexity": 244.16349443276644}, "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-2022-21/segments/1652662558015.52/warc/CC-MAIN-20220523101705-20220523131705-00223.warc.gz"}
http://math.stackexchange.com/questions/213818/how-to-evaluate-this-integral-int-infty-infty-fracx2ex1ex2d/213938	How to evaluate this integral $\int_{-\infty}^{+\infty}\frac{x^2e^x}{(1+e^x)^2}dx$? I need to evaluate $$\int_{-\infty}^{+\infty}\frac{x^2e^x}{(1+e^x)^2}dx$$ I think the answer is $\frac{\pi^2}{3}$, but I'm not able to calculate it. - Related – Pedro Tamaroff Oct 14 '12 at 19:52 $$f(x) = \dfrac{x^2 \exp(x)}{(1+\exp(x))^2} = \dfrac{x^2 \exp(-x)}{\left(1 + \exp(-x) \right)^2}$$ Recall that $$\dfrac{a}{(1+a)^2} = a -2a^2 + 3a^3 - 4a^4 + 5a^5 \mp \cdots = \sum_{k=1}^{\infty}(-1)^{k+1}k a^k$$ For $x > 0$, $$f(x) = x^2 \sum_{k=1}^{\infty} (-1)^{k+1} k \exp(-kx)$$ Now for $a > 0$, $$\int_0^{\infty} x^2 \exp(-ax) = \dfrac2{a^3}$$ Hence, $$\int_0^{\infty} f(x) dx = \sum_{k=1}^{\infty} (-1)^{k+1} \dfrac{2k}{k^3} = 2 \sum_{k=1}^{\infty} \dfrac{(-1)^{k+1}}{k^2} = \dfrac{\pi^2}6$$ Hence, $$\int_{-\infty}^{\infty} f(x) dx = 2 \int_0^{\infty} f(x) dx = \dfrac{\pi^2}3$$ - Thank you! It is very clever! I am just wondering if it can be solved by residue calculus. – Urukec Oct 14 '12 at 20:28 In a more general setting, let $$F\left( s \right) = \int\limits_0^\infty {\frac{{{x^{s - 1}}}}{{{{\left( {{e^x} + 1} \right)}^2}}}{e^x}dx}$$ We have, whenever $\|e^{-x}\|<1$, that is, for $x>0$, that $$\frac{1}{{1 + {e^{ - x}}}} = \sum\limits_{k = 0}^\infty {{{\left( { - 1} \right)}^k}{e^{ - kx}}}$$ and convergence is uniform. Thus $$\frac{{{e^{ - x}}}}{{{{\left( {1 + {e^{ - x}}} \right)}^2}}} = \frac{d}{{dx}}\frac{1}{{1 + {e^{ - x}}}} = \sum\limits_{k = 0}^\infty {{{\left( { - 1} \right)}^k}\frac{d}{{dx}}{e^{ - kx}}} = \sum\limits_{k = 1}^\infty {{{\left( { - 1} \right)}^{k + 1}}k{e^{ - kx}}}$$ Note that after multipication and division by $e^{2x}$ one gets $$\frac{{{e^x}}}{{{{\left( {1 + {e^x}} \right)}^2}}} = \sum\limits_{k = 1}^\infty {{{\left( { - 1} \right)}^{k + 1}}k{e^{ - kx}}}$$ Thus $$\frac{{{e^x}{x^{s - 1}}}}{{{{\left( {1 + {e^x}} \right)}^2}}} = \sum\limits_{k = 1}^\infty {{{\left( { - 1} \right)}^{k + 1}}k{e^{ - kx}}{x^{s - 1}}}$$ Now, we look at individual terms $$\int\limits_0^\infty {{e^{ - kx}}{x^{s - 1}}dx} = \frac{1}{{{k^s}}}\int\limits_0^\infty {{e^{ - kx}}{{\left( {kx} \right)}^{s - 1}}d\left( {kx} \right)} = \frac{1}{{{k^s}}}\int\limits_0^\infty {{e^{ - u}}{u^{s - 1}}du} = \frac{1}{{{k^s}}}\Gamma \left( s \right)$$ Thus$$\tag 1 \int\limits_0^\infty {\frac{{{e^x}{x^{s - 1}}}}{{{{\left( {1 + {e^x}} \right)}^2}}}} = \Gamma \left( s \right)\sum\limits_{k = 1}^\infty {{{\left( { - 1} \right)}^{k - 1}}\frac{1}{{{k^{s-1}}}}} = \Gamma \left( s \right)\eta \left( s-1 \right)$$ Specializing for $s=3$, we get: $$\int\limits_0^\infty {\frac{{{e^x}{x^2}}}{{{{\left( {1 + {e^x}} \right)}^2}}}} = \Gamma \left( 3 \right)\eta \left( 2 \right) = 2!\frac{{{\pi ^2}}}{{12}} = \frac{{{\pi ^2}}}{6}$$ so that$$\int\limits_{ - \infty }^\infty {\frac{{{e^x}{x^2}}}{{{{\left( {1 + {e^x}} \right)}^2}}}} = 2\frac{{{\pi ^2}}}{6} = \frac{{{\pi ^2}}}{3}$$ You might be interested in this question of mine where I wonder about integrals of the form $$\int\limits_0^\infty {F\left( {\frac{1}{{{e^x} - 1}},{e^x},{x^s}} \right)dx}$$ and provide some similar expressions to $(1)$. - +1. To add to Peter's generalization, $$\int_0^{\infty} \dfrac{x^{s-1}}{(\exp(ax) + 1)^2} \exp(ax) dx = \dfrac{\Gamma(s) \eta(s-1)}{a^s}$$ – user17762 Oct 14 '12 at 20:50 Gee. It's hard to find answers that will endow you with badges :-) – Martin Jun 12 at 3:10 Another approach \begin{align} \int\limits_{-\infty}^{+\infty}\frac{x^2 e^x}{(1+e^x)^2}dx &=\lim\limits_{a\to 0}\int\limits_{-\infty}^{+\infty}\frac{x^2 e^{(a+1)x}}{(1+e^x)^2}dx\\ &=\lim\limits_{a\to 0}\frac{d^2}{da^2}\int\limits_{-\infty}^{+\infty}\frac{e^{(a+1)x}}{(1+e^x)^2}dx\\ &=\lim\limits_{a\to 0}\frac{d^2}{da^2}\int\limits_{0}^{+\infty}\frac{t^a}{(1+t)^2}dt\\ &=\lim\limits_{a\to 0}\frac{d^2}{da^2}B(1+a,1-a)\\ &=\lim\limits_{a\to 0}\frac{d^2}{da^2}\frac{\Gamma(1+a)\Gamma(1-a)}{\Gamma(2)}\\ &=\lim\limits_{a\to 0}\frac{d^2}{da^2}a\Gamma(a)\Gamma(1-a)\\ &=\lim\limits_{a\to 0}\frac{d^2}{da^2}\pi a\csc(\pi a)\\ &=\lim\limits_{a\to 0}\frac{d^2}{da^2}\pi a\left(\frac{1}{\pi a}+\frac{\pi a}{6}+o(a)\right)\\ &=\lim\limits_{a\to 0}\frac{d^2}{da^2}\left(1+\frac{\pi^2 a^2}{6}+o(a^2)\right)\\ &=\frac{\pi^2}{3} \end{align} Slightly more generally, consider $$J(P,R) = \oint_\Gamma \frac{P(z)\; e^z\; dz}{(1+e^z)^2}$$ where $P$ is a polynomial and $\Gamma$ is the positively oriented rectangular contour from $-R$ to $R$ to $R+2\pi i$ to $-R+2\pi i$. Then $J(P,R) = 2 \pi i \text{Res}(P(z)\; e^z/(1+e^z)^2,z=\pi i) = - 2 \pi i P'(\pi i)$. On the other hand, it is easy to see that the contributions to the integral from the vertical sections go to $0$ as $R \to \infty$, and $$\lim_{R \to \infty} J(P,R) = \int_{-\infty}^\infty \frac{(P(x) - P(x+2\pi i)) e^x}{(1+e^x)^2}\ dx$$ Now $P(x) - P(x + 2 \pi i) = x^2$ for $P(z) = -\dfrac{\pi i}{3} x + \dfrac{1}{2} x^2 + \dfrac{i}{6 \pi} x^3$, which makes $- 2 \pi i P'(\pi i) = \dfrac{\pi^2}{3}$.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9942289590835571, "perplexity": 936.5343452707131}, "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-2013-48/segments/1387345772826/warc/CC-MAIN-20131218054932-00071-ip-10-33-133-15.ec2.internal.warc.gz"}
https://socratic.org/questions/how-many-molecules-are-there-in-6c-6h-12o-6-how-do-you-know	Chemistry Topics # How many molecules are there in 6C_6H_12O_6? How do you know? Jun 9, 2017 6 Molecules. #### Explanation: The big number in the front it called the coefficient, and it indicates the number of molecules of a compound there are in a formula. In this case, the coefficient is 6 and so there are 6 molecules of ${C}_{6} {H}_{12} {O}_{6}$. The smaller, lower numbers after each element symbol are called subscripts and they indicate how many of each atom are in a single molecule.For example, there are 12 Hydrogen atoms in one molecule of ${C}_{6} {H}_{12} {O}_{6}$ and so there are 48 in $6 {C}_{6} {H}_{12} {O}_{6}$. ##### Impact of this question 297 views around the world You can reuse this answer	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 3, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8887407779693604, "perplexity": 645.2482812309651}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027316718.64/warc/CC-MAIN-20190822022401-20190822044401-00389.warc.gz"}
http://securimar.com/the-generalized-square-error-regularized-lms-algorithm/	# The Generalized Square-error-regularized Lms Algorithm RECOMMENDED: If you have Windows errors then we strongly recommend that you download and run this (Windows) Repair Tool. A new computing approach for power signal modeling using fractional adaptive algorithms. – We design FrASP algorithms based on recently introduced variants of generalized least mean square (LMS) adaptive strategies for parameter estimation of the model. The performance of the proposed fractional adaptive schemes is. Generalized square-error- regularized LMS algorithm. In this section, we summarize several algorithms including NLMS, GNGD algorithm [6], and Choi's regularized NLMS [3]. We then present the generalized square-error-regularized LMS. International Journal of Engineering Research and General Science Volume 2, Issue. Regularized Constant Modulus Algorithm: An improvement on Convergence. of poor convergence and high value of mean square error, embedded with various. Adaptive algorithms like RLS, LMS, etc then updates the equalizer filter. National Health and Nutrition Examination Survey – Our primary objective was to apply the LMS method to create gender-specific reference growth. with smoothing. Ruby Standard Error Methods Ruby’s Exception vs StandardError: What’s the difference? May. object containing info about the error. end. classes that ship with Ruby’s standard. Ruby Exceptions <Access. These are the "normal" exceptions that typical Ruby programs try to. The Exception class defines two methods that return details about. If you are inside a method, you do not need (2013) Large-scale Tikhonov regularization of total least squares. Journal of. ( 2008) The Kernel Least-Mean-Square Algorithm. IEEE Transactions on Signal. Abstract: We consider adaptive system identification problems with convex constraints and propose a family of regularized Least-Mean-Square (LMS) algorithms. Simulation results demonstrate the advantages of the proposed filters in both convergence rate and steady-state error under sparsity. Abstract—The purpose of a variable step-size normalized LMS filter is to solve the dilemma of fast convergence or low steady-state error associated The idea is to introduce the inverse of weighted square-error as the regularization parameter. Our new regularized NLMS algorithm outperforms. ysis of mean-square convergence, i.e., the power of the error is. bounded, involves a lot of algebra for the general multichannel. result in a bias on , such that under the nominal condition. FRAANJE et al.: Robustness of the filtered-x LMS algorithm, part II. achieve a lower mean square error than the standard LMS and AP algorithms. I. INTRODUCTION. In general, the problem of system identification involves. With this as the baseline, the adaptive LMS filter examples use the adaptive LMS algorithms to identify this filter in a system identification role. Using the sign-data algorithm changes the mean square error calculation by using the sign of the input data to change the filter coefficients. convergence condition and steady-state excess mean square error. (MSE). It shows that. well known least mean square (LMS) algorithm [1], [2] as well as their variants. regularization or the generalized inverse of the diag- onal matrix.	{"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": 0.8291099071502686, "perplexity": 2069.1623968977524}, "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/1512948567785.59/warc/CC-MAIN-20171215075536-20171215095536-00726.warc.gz"}
http://mathhelpforum.com/algebra/137095-example-irrational-irrational-rational.html	Math Help - example of irrational + irrational = rational 1. example of irrational + irrational = rational plzz give me examples when two DIFFERENT irrational numbers add up to form a rational number. a + b = c where a and b are two different irrational numbers and c is a rational numbers. i will be very much helpful if i get more than one example. plzz help thanks in advance. 2. Originally Posted by saha.subham plzz give me examples when two DIFFERENT irrational numbers add up to form a rational number. a + b = c where a and b are two different irrational numbers and c is a rational numbers. i will be very much helpful if i get more than one example. plzz help thanks in advance. Dear saha, Take any irrational number and it's negetive, e.g; $a=\sqrt{2}~and~b=-\sqrt{2}$ Then, $\sqrt{2}-\sqrt{2}=0\in{Q}$ 3. One more example: $a=\sqrt{2}~and~b=1-\sqrt{2} $ $\sqrt{2}+1-\sqrt{2}=1\in{Q} $ 4. Originally Posted by MJ* hi earboth....i dont think so thats true..22/7 is an irrational number ;it is infact pi. (3.1411592...) And if u ay that irrational numbers cant be expressed in form of fractions; then sqrt(2) wud not have been an irrtional number;coz it can be expressed in fractional form sqrt(2)/1 for reference see: Pi - Wikipedia, the free encyclopedia Normally I wouldn't get too rough with people who respond to questions they have no clue about but posting a reference to Wikipedia that you didn't even bother to read carefully yourself enrages me. Wikipedia does NOT say that $\pi$ is equal to "22/7". It lists that as one of many rational approximations to $\pi$. A number is rational if and only if it can be written as a fraction with integer numerator and denominator- that is often used as the definition of "rational number". Saying that $\sqrt{2}= \sqrt{2}{1}$ does NOT make it a rational number because the numerator is not an integer. $\pi$ is irrational, 22/7 is rational. They are NOT equal, they are "close"- and not that very close. 22/7= 3.142857142857... and differs from $\pi$ in just the third decimal place.	{"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": 0.8658092021942139, "perplexity": 517.2035590735366}, "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-2015-22/segments/1432207926828.58/warc/CC-MAIN-20150521113206-00260-ip-10-180-206-219.ec2.internal.warc.gz"}
https://selfhelpqa.com/thought-culture-or-practical-mental-training/29/	# Thought-Culture or Practical Mental Training IV. Of the Subalterns: _If the Universal (A or E) be true the Particular (I or O) must be true_. As for instance, if (A) “All A is B” is true, then (I) “Some A is B” must also be true; also, if (E) “No A is B” is true, then “Some A is not B” must also be true. The Universal carries the particular within its truth and meaning. But; _If the Universal is false, the particular may be true or it may be false_. As for instance (A) “All A is B” may be false, and yet (I) “Some A is B” may be either true or false, without being determined by the (A) proposition. And, likewise, (E) “No A is B” may be false without determining the truth or falsity of (O) “Some A is not B.” But: _If the Particular be false, the Universal also must be false_. As for instance, if (I) “Some A is B” is false, then it must follow that (A) “All A is B” must also be false; or if (O) “Some A is not B” is false, then (E) “No A is B” must also be false. But: _The Particular may be true, without rendering the Universal true_. As for instance: (I) “_Some_ A is B” may be true without making true (A) “_All_ A is B;” or (O) “Some A is not B” may be true without making true (E) “No A is B.” The above rules may be worked out not only with the symbols, as “All A is B,” but also with _any_ Judgments or Propositions, such as “All horses are animals;” “All men are mortal;” “Some men are artists;” etc. The principle involved is identical in each and every case. The “All A is B” symbology is merely adopted for simplicity, and for the purpose of rendering the logical process akin to that of mathematics. The letters play the same part that the numerals or figures do in arithmetic or the _a_, _b_, _c_; _x_, _y_, _z_, in algebra. Thinking in symbols tends toward clearness of thought and reasoning. _Exercise_: Let the student apply the principles of Opposition by using any of the above judgments mentioned in the preceding paragraph, in the direction of erecting a Square of Opposition of them, after having attached the symbolic letters A, E, I and O, to the appropriate forms of the propositions. Then let him work out the following problems from the Tables and Square given in this chapter. 1. If “A” is true; show what follows for E, I and O. Also what follows if “A” be _false_. 2. If “E” is true; show what follows for A, I and O. Also what follows if “E” be _false_. 3. If “I” is true; show what follows for A, E and O. Also what follows if “I” be _false_. 4. If “O” is true; show what follows for A, E and I. Also what happens if “O” be _false_. CONVERSION OF JUDGMENTS Judgments are capable of the process of Conversion, or _the change of place of subject and predicate_. Hyslop says: “Conversion is the transposition of subject and predicate, or the process of immediate inference by which we can infer from a given preposition another having the predicate of the original for its subject, and the subject of the original for its predicate.” The process of converting a proposition seems simple at first thought but a little consideration will show that there are many difficulties in the way. For instance, while it is a true judgment that “All _horses_ are _animals_,” it is not a correct Derived Judgment or Inference that “All _animals_ are _horses_.” The same is true of the possible conversion of the judgment “All biscuit is bread” into that of “All bread is biscuit.” There are certain rules to be observed in Conversion, as we shall see in a moment. The Subject of a judgment is, of course, _the term of which something is affirmed_; and the Predicate is _the term expressing that which is affirmed of the Subject_. The Predicate is really an expression of an _attribute_ of the Subject. Thus when we say “All horses are animals” we express the idea that _all horses_ possess the _attribute_ of “animality;” or when we say that “Some men are artists,” we express the idea that _some men_ possess the _attributes_ or qualities included in the concept “artist.” In Conversion, the original judgment is called the Convertend; and the new form of judgment, resulting from the conversion, is called the Converse. Remember these terms, please. The two Rules of Conversion, stated in simple form, are as follows: I. Do not change the quality of a judgment. The quality of the converse must remain the same as that of the convertend.	{"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": 0.9237127304077148, "perplexity": 1794.164814315314}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250598217.23/warc/CC-MAIN-20200120081337-20200120105337-00439.warc.gz"}
https://math.stackexchange.com/questions/1519825/the-rigorousness-of-this-proof-about-greatest-common-divisors/1519832	# The rigorousness of this proof about greatest common divisors. My task is to prove that gcd(n, n+1)=1 for all n>0. It is obvious that 1 is a common divisor of both n and n+1 since $$1\|n → 1x=n$$ if x=n, and $$1\|n+1 → 1y=n+1$$ if y=n+1. To prove that 1 is the greatest common divisor, I did as follows: From the integers n and n+1 the other must be an even integer, and the other an odd integer. Since one of them must be odd, the gcd of the two can't be an even number, because the odd one doesn't have any even divisors. If the gcd of the two were an odd integer greater than 1, for example 3, then $$3\|n → 3x=n → n=3x$$ and $$3\|n+1 → 3y=n+1 → n=3y-1,$$ and there are no integers x and y such that both of these equations would hold at the same time. This happens with every odd integer larger than 1. Is the part about the odd integers greater than 1 rigorous enough for a proper proof? I don't think it is; how would you prove this more rigorously? If anyone has any other improvement ideas of any kind, they're more than welcome! • Seems unnecessarily complex. If $d\|a$ and $d\|b$ then $d\|(a-b)$. Hence, in your case, $d\|(n+1-n)\implies d\|1$. Thus the only positive common divisor of $n$ and $n+1$ is 1. – lulu Nov 8 '15 at 22:34 • $\gcd(a,b)\le\|a-b\|$ – JMP Nov 8 '15 at 22:36 • It should be noted that the solution OP has given really is just a long-winded version of the version @lulu and I have given – ASKASK Nov 8 '15 at 22:37 • (obv) for $a\ne b$ – JMP Nov 8 '15 at 22:46 If you know that $a \mid b$ and $a \mid c$ implies $a \mid b \pm c$, then this theorem is easier than you have laid it out to be. Let $a$ be any common divisor of $n$ and $n+1$. Then $a \mid (n+1)-n$, so $a \mid 1$. Do you think you can take it from here? In regards to the first part, just note that if $ap=b$ and $aq=c$, then $b \pm c = a(p \pm q)$. "3\|n→3x=n→n=3x and 3\|n+1→3y=n+1→n=3y−1, , and there are no integers x and y such that both of these equations would hold at the same time. This happens with every odd integer larger than 1." I really hate to say this, but if you can state that there are no integers that solve these, you could just have easily have stated "There are no integers other than 1 that divide both n and n+1" and saved yourself a lot of trouble. In fact, that's the gyst of the matter, the only number that divides both n and n+1 is 1. So how do you prove that? If you accept that if a\|b and a\|c then a\|(b - c), it is easy, as a\|n+1 and a\|n implies a\|(n+1 -n) = 1. And if a\|1 = 1. So gcd(n, n+1) = 1. If you don't know that if a\|b and a\|c then a\|(b - c). Well you know that if a\|n then n = a*m for some m. So $\frac {n+ 1}{a} = m + \frac{1}{a}$. If a > 1 then this is not an integer and $a$ does not divide $n + 1$. So no integer other than 1 divides both n and n + 1. So gcd(n, n+1) = 1. • Er... no positve integer other than 1 divides both n and n+1..... – fleablood Nov 8 '15 at 22:58 If $$n$$ is even then $$n+1$$ is odd, therefore $$gcd(n,n+1)=1$$ Analogously, if $$n$$ is odd then $$n+1$$ is even, therefore $$gcd(n,n+1)=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": 6, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8975042700767517, "perplexity": 164.52033976882305}, "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/1581875144167.31/warc/CC-MAIN-20200219184416-20200219214416-00424.warc.gz"}
http://clay6.com/qa/26813/find-the-coordinate-of-double-ordinate-of-parabola-y-2-4ax-	# Find the coordinate of double ordinate of parabola $y^2=4ax$? $\begin{array}{1 1}(a)\;(h,2\sqrt{ah}),(h,-2\sqrt{ah})\\(b)\;(h,\sqrt{ah}),(h,-\sqrt{ah})\\(c)\;(h,3\sqrt{ah}),(h,-3\sqrt{ah})\\(d)\;(h,4\sqrt{ah}),(h,-4\sqrt{ah})\end{array}$ Let the abscissa of A be h hence ordinate of A, $y^2=4ah$ $y=2\sqrt{ah}$(for first quadrant) Hence co-ordinate of A and A' are $(h,2\sqrt{ah})$ and $(h,-2\sqrt{ah})$ respectively. Hence (a) is the correct answer.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9438915252685547, "perplexity": 1237.6885248199808}, "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/1512948515311.25/warc/CC-MAIN-20171212075935-20171212095935-00636.warc.gz"}
http://mathhelpforum.com/advanced-algebra/104784-unit.html	# Math Help - unit 1. ## unit Suppose $f(x)$ is relatively prime to $0_F$. Then $\gcd(f(x), 0_F) = 1_F$. So $1_F = f(x)u(x)+0_{F}v(x)$ for some polynomials $u(x)$ and $v(x)$. Thus $1_F = f(x)u(x)$. So this means that $f(x)$ is a unit? 2. Originally Posted by Sampras Suppose $f(x)$ is relatively prime to $0_F$. this never happens! every polynomial obviously divides the zero polynomial. \ EDIT: see my next post! 3. Originally Posted by NonCommAlg this never happens! every polynomial obviously divides the zero polynomial. So since the hypothesis is false, what can be said about $f(x)$? It is not a polynomial? 4. Originally Posted by Sampras So since the hypothesis is false, what can be said about $f(x)$? It is not a polynomial? ok, i was too quick in responding i guess ... sorry about that! as i said, every polynomial divides the zero polynomial and thus $\gcd(0,f(x))=f(x),$ for any polynomial. so $\gcd(0,f(x))=1$ if and only if $f(x)$ and $g(x)=1$ are associates, i.e. if and only if $f(x)$ is a unit. so your conclusion is right.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 17, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9792507290840149, "perplexity": 270.01839679911296}, "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/1394678676855/warc/CC-MAIN-20140313024436-00019-ip-10-183-142-35.ec2.internal.warc.gz"}
http://arxiv-export-lb.library.cornell.edu/abs/2208.03312	astro-ph.HE (what is this?) # Title: How do the dynamics of the Milky Way - Large Magellanic Cloud system affect gamma-ray constraints on particle dark matter? Abstract: Previous studies on astrophysical dark matter (DM) constraints have all assumed that the Milky Way's (MW) DM halo can be modelled in isolation. However, recent work suggests that the MW's largest dwarf satellite, the Large Magellanic Cloud (LMC), has a mass of 10-20$\%$ that of the MW and is currently merging with our Galaxy. As a result, the DM haloes of the MW and LMC are expected to be strongly deformed. We here address and quantify the impact of the dynamical response caused by the passage of the LMC through the MW on the prospects for indirect DM searches. Utilising a set of state-of-the-art numerical simulations of the evolution of the MW-LMC system, we derive the DM distribution in both galaxies at the present time based on the Basis Function Expansion formalism. Consequently, we build $J$-factor all-sky maps of the MW-LMC system in order to study the impact of the LMC passage on gamma-ray indirect searches for thermally produced DM annihilating in the outer MW halo as well as within the LMC halo standalone. We conduct a detailed analysis of 12 years of Fermi-LAT data that incorporates various large-scale gamma-ray emission components and we quantify the systematic uncertainty associated with the imperfect knowledge of the astrophysical gamma-ray sources. We find that the dynamical response caused by the LMC passage can alter the constraints on the velocity-averaged annihilation cross section for weak scale particle DM at a level comparable to the existing observational uncertainty of the MW halo's density profile and total mass. Comments: 21 pages (15 - main body; 6 - appendices), 10 figures; prepared for submission to MNRAS. Comments welcome Subjects: High Energy Astrophysical Phenomena (astro-ph.HE); Astrophysics of Galaxies (astro-ph.GA); High Energy Physics - Phenomenology (hep-ph) Report number: LAPTH-038/22 Cite as: arXiv:2208.03312 [astro-ph.HE] (or arXiv:2208.03312v1 [astro-ph.HE] for this version) ## Submission history From: Christopher Eckner [view email] [v1] Fri, 5 Aug 2022 17:58:16 GMT (15111kb,D) Link back to: arXiv, form interface, contact.	{"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": 0.8289273977279663, "perplexity": 2808.3767289546263}, "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/1664030337723.23/warc/CC-MAIN-20221006025949-20221006055949-00366.warc.gz"}
http://mathhelpforum.com/algebra/223961-help.html	2. ## Re: help Let x be the middle odd positive number, then we have ( x-2) + x + ( x+2) = 1/7 [ x ( x-2)(x+2)] 3x = 1/7 [ x ( x^2 - 4 )] 21x =x ( x^2-4) That gives x [ x^2 - 25 ] = 0 Now I am sure you can conclude that the numbers are 3,5,7	{"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": 0.9710555672645569, "perplexity": 2723.1462019207484}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-40/segments/1474738659753.31/warc/CC-MAIN-20160924173739-00241-ip-10-143-35-109.ec2.internal.warc.gz"}
https://www.physicsforums.com/threads/continuous-function-example.224235/	# Continuous function example • Start date #### gaborfk 53 0 1. The problem statement, all variables and given/known data For each $$a\in\mathbb{R}$$, find a function $$f$$ that is continuous at $$x=a$$ but discontinuous at all other points. 3. The attempt at a solution I guess I am not getting the question. I need to come up with a function, I was thinking of a piecewise defined one, half rational half irrational, which is continuous on one but not the other? Is this possible? Related Precalculus Mathematics Homework News on Phys.org #### sutupidmath 1,631 4 what about f(x)={0, x-rational, x, where x irrational. take a sequence {a} that converges to 0, from this sequence lets take two subsequences {b} of rationals, and {c} of irrationals, since {a} converges to 0 also {b} and {c} should converge to zero. now lets take the corresponding sequence of the function f({a})-->0 f({b})-->x-->0 So this function i guess is continuous at x=0, since also f(0)=0, but it is discontinuous everywhere else. Let's see what other guys have to say on this, cuz, i am not 100% sure that what i did actually works. #### gaborfk 53 0 Thank you! That sound great. #### sutupidmath 1,631 4 Thank you! That sound great. Can you show why the function that i took as an example, from the top of my head, is everywhere else discontinous, because i left this part for you to show.??? #### gaborfk 53 0 Because there are infinitely many irrational numbers which would make the graph continuous on the irrationals, but on an interval there would be rationals mixed in between the irrationals? #### sutupidmath 1,631 4 Because there are infinitely many irrational numbers which would make the graph continuous on the irrationals, but on an interval there would be rationals mixed in between the irrationals? Well, try to use the same logic i used to show that it is continuous at 0. In other words try to use sequences and see if you can come up with sth. It is quite trivial frome here, i guess. "Continuous function example" ### Physics Forums Values We Value Quality • Topics based on mainstream science • Proper English grammar and spelling We Value Civility • Positive and compassionate attitudes • Patience while debating We Value Productivity • Disciplined to remain on-topic • Recognition of own weaknesses • Solo and co-op problem solving	{"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": 0.8996228575706482, "perplexity": 1082.997062476943}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-18/segments/1555578596571.63/warc/CC-MAIN-20190423094921-20190423120921-00468.warc.gz"}
http://www.reference.com/browse/mass+numbers	Definitions # Mass ratio In aerospace engineering, mass ratio is a measure of the efficiency of a rocket. It describes how much more massive the vehicle is with propellant than without; that is, it is the ratio of the rocket's wet mass (vehicle plus contents plus propellant) to its dry mass (vehicle plus contents). A more efficient rocket design requires less propellant to achieve a given goal, and would therefore have a lower mass ratio. The mass ratio is a useful quantity for back-of-the-envelope rocketry calculations: it is an easy number to derive from either $Delta v$ numbers or from rocket and propellant mass numbers, and therefore serves as a handy bridge between the two. It is also a useful number to give an impression of the size of a rocket: while two rockets with mass fractions of, say, 92% and 95% may appear similar, the corresponding mass ratios of 12.5 and 20 clearly indicate that the latter system requires much more propellant. Typical multistage rockets have mass ratios in the range from 8 to 20. The Space Shuttle, for example, has a mass ratio around 16. ## Derivation The definition arises naturally from the Tsiolkovsky rocket equation: $Delta v = v_e ln frac \left\{m_0\right\} \left\{m_1\right\}$ where Δv is the desired change in the rocket's velocity ve is the effective exhaust velocity (see specific impulse) m0 is the initial mass (rocket plus contents plus propellant) m1 is the final mass (rocket plus contents) This equation can be rewritten in the following equivalent form: $frac \left\{m_0\right\} \left\{m_1\right\} = e ^ \left\{ Delta v / v_e \right\}$ The fraction on the left-hand side of this equation is the rocket's mass ratio by definition. This equation indicates that a Δv of $n$ times the exhaust velocity requires a mass ratio of $e^n$. For instance, for a vehicle to achieve a $Delta v$ of 2.5 times its exhaust velocity would require a mass ratio of $e^\left\{2.5\right\}$ (approximately 12.2). One could say that a "velocity ratio" of $n$ requires a mass ratio of $e^n$. ## References Zubrin, Robert (1999). Entering Space: Creating a Spacefaring Civilization. Tarcher/Putnam. ISBN 0-87477-975-8.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 9, "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": 0.9943829774856567, "perplexity": 950.4441179674347}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-27/segments/1435375094957.74/warc/CC-MAIN-20150627031814-00251-ip-10-179-60-89.ec2.internal.warc.gz"}
http://mathhelpforum.com/advanced-algebra/160566-showing-something-subgroup.html	# Math Help - Showing something is a subgroup 1. ## Showing something is a subgroup Let S be a set and let a be a fixed element of S. Show that s is an element of Sym(S) such that s(a)=a is a subgroup of Sym(S). 2. If $\sigma_1$, $\sigma_2$ are permutations which fix an element $a \in X$, $\sigma_i \in S_X$, then you need to prove two things, i. $\sigma_1^{-1}$ also fixes $a$. ii. $\sigma_1 \sigma_2$ also fixes $a$. Can you work out why these two things are sufficient? Now, to prove i. you should note that $\sigma_1: a \mapsto a$. If $\sigma^{-1}: a \mapsto b$ then where must $b$ be sent to in $(\sigma_1^{-1})^{-1} = \sigma_1$? To prove ii. simply plug in $a$ into $\sigma_1 \sigma_2$.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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": 0.9681218862533569, "perplexity": 338.0391022657884}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-42/segments/1413507443598.37/warc/CC-MAIN-20141017005723-00272-ip-10-16-133-185.ec2.internal.warc.gz"}
https://projecteuler.net/problem=323	## Bitwise-OR operations on random integers ### Problem 323 Published on Sunday, 6th February 2011, 07:00 am; Solved by 2965;Difficulty rating: 20% Let y0, y1, y2,... be a sequence of random unsigned 32 bit integers (i.e. 0 ≤ yi < 232, every value equally likely). For the sequence xi the following recursion is given: • x0 = 0 and • xi = xi-1\| yi-1, for i > 0. ( \| is the bitwise-OR operator) It can be seen that eventually there will be an index N such that xi = 232 -1 (a bit-pattern of all ones) for all i ≥ N. Find the expected value of N.	{"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": 0.8934374451637268, "perplexity": 2306.931334351018}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267864776.82/warc/CC-MAIN-20180622182027-20180622202027-00470.warc.gz"}
https://physics.stackexchange.com/questions/190135/capacitance-of-three-plates	# Capacitance of three plates I'm sorry if this has been asked before, but here it goes. Consider three parallel plane conducting plates (they can be infinite) and suppose the middle plate has some thickness $\delta$ and the other plates are at a distance $d_1$ and $d_2$ from the middle plate. Now, consider two situations: 1. The middle plate has negative charge $-Q$ and the two other plates have positive charge $+Q$. 2. The middle plate has no charge and the two other plates have charge $+Q$ and $-Q$. If the medium between the plates is vacuum, then $E = \frac{\sigma}{\epsilon_0}$ is the electric field between the plates (in the vacuum, that is), where $\sigma$ is the surface charge density of one of the $+Q$ plates. Suppose I want to calculate the capacitance $C$ of the system. In the first case, should I consider the system to be composed of two capacitors in parallel, so that $$C = C_1 + C_2 = \frac{Q}{Ed_1} + \frac{Q}{Ed_2}?$$ And in the second case, should I consider it as two capacitors in series, such that $$C = \left(\frac{1}{C_1} + \frac{1}{C_2} \right)^{-1}$$ where $C_i = \frac{Q}{Ed_i}$, $i=1,2$? In the second case I'm pretty sure it's correct, because it can be seen as a single two-plate capacitor, and it matches the value $$C = \frac{Q}{E\cdot(d_1+d_2)}.$$ In the first I'm not so sure. Thanks. • Do you mean that the middle plate in situation 1 has charge $-2Q$, so that everything is neutrally charged overall? Jun 18, 2015 at 17:02 • Well, I was thinking (this is not homework, I'm just trying to understand) in the case that the middle plate has originally a charge $-Q$. Jun 18, 2015 at 17:04 • Or does the overall charge have to be null? Jun 18, 2015 at 17:04 • It doesn't have to be, but it usually is. Large charges (positive or negative) inside a system tend to rip it apart due to strong electrostatic forces, or else create sparks to ground. In any case you're going to have trouble with what you mean by "capacitance" since you are defining $Q$ well but "voltage" poorly. (Voltage has to be measured between two places and you haven't really defined what those two places are. I imagine in situation 1 the outer two plates are connected by a wire and you want the voltage on that wire, but I'm not sure.) Jun 18, 2015 at 17:06 • Yes, that was what I had in mind, the two outer plates connected by a wire, and the middle plate is free. In that case, should I consider the two capacitors (each one being formed by an outer plate + the middle plate) as two capacitors in parallel or in series? Jun 18, 2015 at 17:11 Consider the three-terminal device that is your stacked capacitor: A ----============================================= (dielectric medium ɛ) =============================================---- B (dielectric medium ɛ) C ----============================================= All three plates have the same area A. It's pretty clear what happens when A is left "floating". (Defining floating: before operation we ground A so that $Q_A = 0$, then we disconnect it so that $A$ is not connected to any particular other component.) When we float A and measure the capacitance between B and C, or when we float C and measure the capacitance between A and B, we get an electric field $E = Q / (\epsilon A)$ and capacitance $C = \epsilon A / d$ where $A$ is the area of these plates, $Q$ is the charge on one terminal, and $d$ is the distance between them. The key observation here is that the capacitor has a very large area compared to its width, so the electric field outside of the capacitor tends to 0, so neither A nor C really "matters" when it's floating in that 0 electric field. It is somewhat harder to think about what happens when we float B and then measure the capacitance between A and C. The electric field needs to be 0 inside B, because it is a perfect conductor and any electric field will cause current to flow. At the same time, the overall charge is 0 and the situation is still "capacitor-like" ($A$ is much much larger than $2d$ if it's much much larger than $d$) so the electric field outside of the parallel plates should tend to 0. Since the charge on the plate directly creates a discontinuity in the electric field, we have to come to this conclusion: the field in the dielectric between B and C is the same when we put $+Q$ on A, $-Q$ on C, as when we floated A, putting $+Q$ on B and $-Q$ on C. It has to be, because it's the same jump discontinuity from the same starting point ($E = 0$ outside the stack). The same must also be true between A and B. The field must be $Q / (\epsilon A)$ in both dielectrics. The condition that $E = 0$ within the middle plate means that we induce a surface charge of $+Q$ on the BC side of B, and a surface charge of $-Q$ on the AB side of B. When you include those surface charges, it "looks exactly like" two capacitors in series, and you expect half of the capacitance. And that's exactly what happens if you ignore B, too! If you ignore B, then you've got a constant field of $E$ over twice the distance, so $V = 2 E d$ for the same $Q$, so it takes twice as much voltage to get the same charge on each plate. So while B is doing "something", it's actually doing the nothingest something it can do. So you're right to intuit that it should just work like a single two-plate capacitor, if you ignore the thickness of the B plate in the calculation of how thick the capacitor is. Now that we understand this, here comes situation 1. For situation 1, the easiest way to get an analogous result is to connect A and C with a wire, so they are at the same voltage, each plate holding charge $Q/2$ while the B-plate holds charge $-Q$. Then you are correct to intuit that this just looks like two capacitors in parallel. What happens? Well, the field is still 0 outside the system. The charge $Q/2$ therefore means that we have half the electric field inside the AB and BC dielectrics, which means that the same charge requires only half the voltage, so the capacitance doubles. Now what if, as you say, we put charge +Q on plate A, +Q on plate C, and -Q on plate B? Well, we have a problem: the overall charge is no longer 0. Under the same "parallel plate" assumption that $A$ is much much larger than $d$ we find the fields by the principle of superposition: E = + Q/(2 A ɛ) A ----============================================= E = - Q/(2 A ɛ) =============================================---- B E = + Q/(2 A ɛ) C ----============================================= E = - Q/(2 A ɛ) Now, we can't even define capacitance unless we choose two points to measure a voltage between. Suppose you want the points A and C: the voltage between these plates is 0, and the capacitance is infinite. Actually, since this voltage is 0, we wouldn't change the system fundamentally by connecting A to C. So then you can consider the voltage between A and B, and you get the same result as before, twice the capacitance as if C weren't there. The surplus charge on A and C, while it superimposes on the electric fields, doesn't affect the capacitances involved. • Thank you, sincerely. I'm happy my intuition isn't as terrible as I thought hehe. 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https://chem.libretexts.org/Courses/BridgeValley_Community_and_Technical_College/Fundamentals_of_Chemistry/01%3A_Measurements/1.07%3A_Converting_Units	# 1.7: Converting Units Learning Objective • Convert from one unit to another unit of the same type. In Section 2.2, we showed some examples of how to replace initial units with other units of the same type to get a numerical value that is easier to comprehend. In this section, we will formalize the process. Consider a simple example: how many feet are there in 4 yards? Most people will almost automatically answer that there are 12 feet in 4 yards. How did you make this determination? Well, if there are 3 feet in 1 yard and there are 4 yards, then there are 4 × 3 = 12 feet in 4 yards. This is correct, of course, but it is informal. Let us formalize it in a way that can be applied more generally. We know that 1 yard (yd) equals 3 feet (ft): $1\, yd = 3\, ft\nonumber$ In math, this expression is called an equality. The rules of algebra say that you can change (i.e., multiply or divide or add or subtract) the equality (as long as you do not divide by zero) and the new expression will still be an equality. For example, if we divide both sides by 2, we get: $\dfrac{1}{2}\,yd= \dfrac{3}{2}\, ft\nonumber$ We see that one-half of a yard equals 3/2, or one and a half, feet—something we also know to be true—so the above equation is still an equality. Going back to the original equality, suppose we divide both sides of the equation by 1 yard (number and unit): $\dfrac{1\,yd}{1\,yd}= \dfrac{3\,ft}{1\,yd}\nonumber$ The expression is still an equality, by the rules of algebra. The left fraction equals 1. It has the same quantity in the numerator and the denominator, so it must equal 1. The quantities in the numerator and denominator cancel, both the number and the unit: $\dfrac{1\,yd}{1\,yd}= \dfrac{3\,ft}{1\,yd}\nonumber$ When everything cancels in a fraction, the fraction reduces to 1: $1= \dfrac{3\,ft}{1\,yd}\nonumber$ ## Conversion Factors We have an expression that equals 1. $\dfrac{3\,ft}{1\,yd}=1\nonumber$ This is a strange way to write 1, but it makes sense: 3 ft equal 1 yd, so the quantities in the numerator and denominator are the same quantity, just expressed with different units. The expression $\dfrac{3\,ft}{1\,yd}\nonumber$ is called a conversion factor and it is used to formally change the unit of a quantity into another unit. (The process of converting units in such a formal fashion is sometimes called dimensional analysis or the factor label method.) To see how this happens, let us start with the original quantity: 4 yd. Now let us multiply this quantity by 1. When you multiply anything by 1, you do not change the value of the quantity. Rather than multiplying by just 1, let us write 1 as: $\dfrac{3\,ft}{1\,yd}\nonumber$ $4\,yd\times \dfrac{3\,ft}{1\,yd}\nonumber$ The 4 yd term can be thought of as 4yd/1; that is, it can be thought of as a fraction with 1 in the denominator. We are essentially multiplying fractions. If the same thing appears in the numerator and denominator of a fraction, they cancel. In this case, what cancels is the unit yard: $4\,yd\times \dfrac{3\,ft}{1\,yd}\nonumber$ That is all that we can cancel. Now, multiply and divide all the numbers to get the final answer: $\dfrac{4\times 3\, ft}{1}= \dfrac{12\,ft}{1}= 12\,ft\nonumber$ Again, we get an answer of 12 ft, just as we did originally. But in this case, we used a more formal procedure that is applicable to a variety of problems. How many millimeters are in 14.66 m? To answer this, we need to construct a conversion factor between millimeters and meters and apply it correctly to the original quantity. We start with the definition of a millimeter, which is: $1\,mm= \dfrac{1}{1000\,m}\nonumber$ The 1/1000 is what the prefix milli- means. Most people are more comfortable working without fractions, so we will rewrite this equation by bringing the 1,000 into the numerator of the other side of the equation: $1000\,mm=1\,m\nonumber$ Now we construct a conversion factor by dividing one quantity into both sides. But now a question arises: which quantity do we divide by? It turns out that we have two choices, and the two choices will give us different conversion factors, both of which equal 1: $\dfrac{1000\,mm}{1000\,mm}= \dfrac{1\,m}{1000\,mm} \nonumber$ or $\dfrac{1000\,mm}{1\,m}= \dfrac{1\,m}{1\,m}\nonumber$ $1=\dfrac{1\,m}{1000\,mm}\nonumber$ or $\dfrac{1000\,mm}{1\,m}=1\nonumber$ Which conversion factor do we use? The answer is based on what unit you want to get rid of in your initial quantity. The original unit of our quantity is meters, which we want to convert to millimeters. Because the original unit is assumed to be in the numerator, to get rid of it, we want the meter unit in the denominator; then they will cancel. Therefore, we will use the second conversion factor. Canceling units and performing the mathematics, we get: $14.66m\times \dfrac{1000\,mm}{1\,m}= 14660\,mm\nonumber$ Note how $$m$$ cancels, leaving $$mm$$, which is the unit of interest. The ability to construct and apply proper conversion factors is a very powerful mathematical technique in chemistry. You need to master this technique if you are going to be successful in this and future courses. Example $$\PageIndex{1}$$ 1. Convert 35.9 kL to liters. 2. Convert 555 nm to meters. Solution 1. We will use the fact that 1 kL = 1,000 L. Of the two conversion factors that can be defined, the one that will work is 1000L/ 1kL. Applying this conversion factor, we get: $35.9\, kL\times \dfrac{1000\,L}{1\,kL}= 35,900\, L \nonumber\nonumber$ 1. We will use the fact that 1 nm = 1/1,000,000,000 m, which we will rewrite as 1,000,000,000 nm = 1 m, or 109 nm = 1 m. Of the two possible conversion factors, the appropriate one has the nm unit in the denominator: $\dfrac{1\,m}{10^{9}\,nm} \nonumber\nonumber$ Applying this conversion factor, we get: $555\,nm\times \dfrac{1m}{10^{9}nm}= 0.000000555\,m= 5.55\times 10^{-7}\,m \nonumber\nonumber$ In the final step, we expressed the answer in scientific notation. Exercise $$\PageIndex{1}$$ 1. Convert 67.08 μL to liters. 2. Convert 56.8 m to kilometers. 6.708 × 10−5 L 5.68 × 10−2 km What if we have a derived unit that is the product of more than one unit, such as m2? Suppose we want to convert square meters to square centimeters? The key is to remember that m2 means m × m, which means we have two meter units in our derived unit. That means we have to include two conversion factors, one for each unit. For example, to convert 17.6 m2 to square centimeters, we perform the conversion as follows: \begin{align} 17.6m^{2} &= 17.6(m\times m)\times \dfrac{100cm}{1m}\times \dfrac{100cm}{1m} \nonumber \\[4pt] &= 176000\,cm \times cm \nonumber \\[4pt] &= 1.76\times 10^{5} \,cm^2\end{align}\nonumber Example $$\PageIndex{2}$$ How many cubic centimeters are in 0.883 m3? Solution With an exponent of 3, we have three length units, so by extension we need to use three conversion factors between meters and centimeters. Thus, we have: $0.883m^{3}\times \dfrac{100\,cm}{1\,m}\times \dfrac{100\,cm}{1\,m} \times \dfrac{100\,cm}{1\,m}= 883000\,cm^{3} = 8.83\times 10^{5}\,cm^{3}\nonumber$ You should demonstrate to yourself that the three meter units do indeed cancel. Exercise $$\PageIndex{2}$$ How many cubic millimeters are present in 0.0923 m3? 9.23 × 107 mm3 Suppose the unit you want to convert is in the denominator of a derived unit—what then? Then, in the conversion factor, the unit you want to remove must be in the numerator. This will cancel with the original unit in the denominator and introduce a new unit in the denominator. The following example illustrates this situation. Example $$\PageIndex{3}$$ Convert 88.4 m/min to meters/second. Solution We want to change the unit in the denominator from minutes to seconds. Because there are 60 seconds in 1 minute (60 s = 1 min), we construct a conversion factor so that the unit we want to remove, minutes, is in the numerator: 1min/60s. Apply and perform the math: $\dfrac{88.4m}{min}\times \dfrac{1\,min}{60\,s}= 1.47\dfrac{m}{s}\nonumber$ Notice how the 88.4 automatically goes in the numerator. That's because any number can be thought of as being in the numerator of a fraction divided by 1. Exercise $$\PageIndex{3}$$ Convert 0.203 m/min to meters/second. 0.00338 m/s or 3.38 × 10−3 m/s Sometimes there will be a need to convert from one unit with one numerical prefix to another unit with a different numerical prefix. How do we handle those conversions? Well, you could memorize the conversion factors that interrelate all numerical prefixes. Or you can go the easier route: first convert the quantity to the base unit—the unit with no numerical prefix—using the definition of the original prefix. Then, convert the quantity in the base unit to the desired unit using the definition of the second prefix. You can do the conversion in two separate steps or as one long algebraic step. For example, to convert 2.77 kg to milligrams: $2.77\,kg\times \dfrac{1000\,g}{1\,kg}= 2770\,g\nonumber$ (convert to the base units of grams) $2770\,g\times \dfrac{1000\,mg}{1\,g}= 2770000\,mg = 2.77\times 10^{6}\,mg\nonumber$ (convert to desired unit) Alternatively, it can be done in a single multi-step process: \begin{align} 2.77\, \cancel{kg}\times \dfrac{1000\,\cancel{g}}{1\,\cancel{kg}}\times \dfrac{1000\,mg}{1\,\cancel{g}} &= 2770000\, mg \nonumber \\[4pt] &= 2.77\times 10^{6}\,mg \end{align}\nonumber You get the same answer either way. Example $$\PageIndex{4}$$ How many nanoseconds are in 368.09 μs? Solution You can either do this as a one-step conversion from microseconds to nanoseconds or convert to the base unit first and then to the final desired unit. We will use the second method here, showing the two steps in a single line. Using the definitions of the prefixes micro- and nano-, $368.0\,\mu s\times \dfrac{1\,s}{1000000\,\mu s}\times \dfrac{1000000000}{1\,s}= 3.6809\times 10^{5}\,ns\nonumber$ Exercise $$\PageIndex{4}$$ How many milliliters are in 607.8 kL? 6.078 × 108 mL When considering the significant figures of a final numerical answer in a conversion, there is one important case where a number does not impact the number of significant figures in a final answer: the so-called exact number. An exact number is a number from a defined relationship, not a measured one. For example, the prefix kilo- means 1,000-exactly 1,000, no more or no less. Thus, in constructing the conversion factor: $\dfrac{1000\,g}{1\,kg}\nonumber$ neither the 1,000 nor the 1 enter into our consideration of significant figures. The numbers in the numerator and denominator are defined exactly by what the prefix kilo- means. Another way of thinking about it is that these numbers can be thought of as having an infinite number of significant figures, such as: $\dfrac{1000.0000000000 \dots \,g}{1.0000000000 \ldots \,kg}\nonumber$ The other numbers in the calculation will determine the number of significant figures in the final answer. Example $$\PageIndex{5}$$ A rectangular plot in a garden has the dimensions 36.7 cm by 128.8 cm. What is the area of the garden plot in square meters? Express your answer in the proper number of significant figures. Solution Area is defined as the product of the two dimensions, which we then have to convert to square meters, and express our final answer to the correct number of significant figures—which in this case will be three. $36.7\,cm\times 128.8\,cm\times \dfrac{1\,m}{100\,cm}\times \dfrac{1\,m}{100\,cm}= 0.472696\,m^{2}= 0.473\,m^{2}\nonumber$ The 1 and 100 in the conversion factors do not affect the determination of significant figures because they are exact numbers, defined by the centi- prefix. Exercise $$\PageIndex{5}$$ What is the volume of a block in cubic meters with the dimensions 2.1 cm × 34.0 cm × 118 cm? 0.0084 m3 Chemistry is Everywhere: The Gimli Glider On July 23, 1983, an Air Canada Boeing 767 jet had to glide to an emergency landing at Gimli Industrial Park Airport in Gimli, Manitoba, because it unexpectedly ran out of fuel during flight. There was no loss of life in the course of the emergency landing, only some minor injuries associated in part with the evacuation of the craft after landing. For the remainder of its operational life (the plane was retired in 2008), the aircraft was nicknamed "the Gimli Glider." The Gimli Glider is the Boeing 767 that ran out of fuel and glided to safety at Gimli Airport. The aircraft ran out of fuel because of confusion over the units used to express the amount of fuel. Source: Photo courtesy of Will F., (CC BY-SA 2.5; Aero Icarus). The 767 took off from Montreal on its way to Ottawa, ultimately heading for Edmonton, Canada. About halfway through the flight, all the engines on the plane began to shut down because of a lack of fuel. When the final engine cut off, all electricity (which was generated by the engines) was lost; the plane became, essentially, a powerless glider. Captain Robert Pearson was an experienced glider pilot, although he had never flown a glider the size of a 767. First Officer Maurice Quintal quickly determined that the aircraft would not be able make it to Winnipeg, the next large airport. He suggested his old Royal Air Force base at Gimli Station, one of whose runways was still being used as a community airport. Between the efforts of the pilots and the flight crew, they managed to get the airplane safely on the ground (although with buckled landing gear) and all passengers off safely. What happened? At the time, Canada was transitioning from the older English system to the metric system. The Boeing 767s were the first aircraft whose gauges were calibrated in the metric system of units (liters and kilograms) rather than the English system of units (gallons and pounds). Thus, when the fuel gauge read 22,300, the gauge meant kilograms, but the ground crew mistakenly fueled the plane with 22,300 pounds of fuel. This ended up being just less than half of the fuel needed to make the trip, causing the engines to quit about halfway to Ottawa. Quick thinking and extraordinary skill saved the lives of 61 passengers and 8 crew members—an incident that would not have occurred if people were watching their units. ## Key Takeaways • Units can be converted to other units using the proper conversion factors. • Conversion factors are constructed from equalities that relate two different units. • Conversions can be a single step or multi-step. • Unit conversion is a powerful mathematical technique in chemistry that must be mastered. • Exact numbers do not affect the determination of significant figures.	{"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": 2, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9007073640823364, "perplexity": 716.4400350375391}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585302.56/warc/CC-MAIN-20211020024111-20211020054111-00282.warc.gz"}
https://indico.cern.ch/event/181055/contributions/308893/	# Quark Matter 2012 12-18 August 2012 US/Eastern timezone ## Gravitational collapse and holographic thermalization 15 Aug 2012, 10:10 20m Empire () ### Empire Oral Presentation ### Speaker Paul Chesler (MIT) ### Description A remarkable result from heavy ion collisions at the Relativistic Heavy Ion Collider and Large Hadron Collider is that, shortly after the collision event, the quark-gluon plasma produced behaves as a nearly ideal liquid. Understanding the dynamics responsible for such rapid "hydroization" is a challenge using traditional perturbative field theory. In recent years holography has emerged as a powerful tool to study non-equilibrium phenomena, mapping the dynamics of certain quantum field theories onto the dynamics of semi-classical gravity. Via holography, the production of quark-gluon plasma maps onto the process of gravitational collapse and black hole formation, with the relaxation of the black hole's gravitational field encoding hydroization of the dual quark gluon plasma. Thermalization of the quark-gluon plasma is encoded in the thermalization of the black hole's Hawking radiation. I will describe several processes which mimic heavy ion collisions and present results for both hydroization and thermalization times and mechanisms. Slides	{"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": 0.8930402398109436, "perplexity": 1540.229794421694}, "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/1581875144167.31/warc/CC-MAIN-20200219184416-20200219214416-00122.warc.gz"}
https://www.semanticscholar.org/paper/Iterated-differential-forms%3A-Tensors-Vinogradov-Vitagliano/ae8041e24318949020dcead305e9a1b653b002d0	# Iterated differential forms: Tensors @article{Vinogradov2006IteratedDF, title={Iterated differential forms: Tensors}, author={Alexandre M. Vinogradov and Luca Vitagliano}, year={2006}, volume={73}, pages={169-171} } • Published 4 May 2006 • Mathematics We interpret tensors on a smooth manifold M as differential forms over a graded commutative algebra called the algebra of iterated differential forms over M. This allows us to put standard tensor calculus in a new differentially closed context and, in particular, enriches it with new natural operations. Applications will be considered in subsequent notes. 6 Citations • Mathematics • 2007 Basic elements of integral calculus over algebras of iterated differential forms �k, k < ∞, are presented. In particular, defining complexes for modules of integral forms are described and the • Mathematics • 2007 For the multiple differential algebra of iterated differential forms [1, 2, 3, 4] on a diffiety ( • Mathematics • 2007 In the preceding note math.DG/0610917 the $\Lambda_{k-1}\mathcal{C}$--spectral sequence, whose first term is composed of \emph{secondary iterated differential forms}, was constructed for a generic Since the discovery of differential calculus by Newton and Leibniz and the subsequent continuous growth of its applications to physics, mechanics, geometry, etc, it was observed that partial • Mathematics L’Enseignement Mathématique • 2015 The Van Est homomorphism for a Lie groupoid $G \rightrightarrows M$, as introduced by Weinstein-Xu, is a cochain map from the complex $C^\infty(BG)$ of groupoid cochains to the Chevalley-Eilenberg ## References SHOWING 1-10 OF 13 REFERENCES From symmetries of partial differential equations to Secondary Calculus Elements of differential calculus in commutative algebras Geometry of finite-order contact structures and the classical theory ### Russian Math . Surveys 44 , n ◦ 3 ( 1989 ) 220 . [ 6 ] A . Verbovetsky • Secondary Calculus and Cohomological Physics , Contemporary Mathematics ### Dokl • Math. 73, n 2 • 2006 • 18 • 1996 • 2003 • Dokl. 13 • 1972 ### Russian Math • Surveys 44, n 3 • 1989 ### J. Soviet Math • J. Soviet Math • 1624 • Dokl. Math • 2006	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.9421207904815674, "perplexity": 4324.575850159449}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500094.26/warc/CC-MAIN-20230204044030-20230204074030-00628.warc.gz"}
https://kerodon.net/tag/017E	# Kerodon $\Newextarrow{\xRightarrow}{5,5}{0x21D2}$ ### 4.3.4 Joins of Topological Spaces The join operation on simplicial sets admits a topological interpretation. Construction 4.3.4.1. Let $X$ and $Y$ be topological spaces, and let $[0,1] = \| \Delta ^1 \|$ denote the unit interval. We let $X \star Y$ denote the topological space given by the iterated pushout $X \coprod _{ (X \times \{ 0\} \times Y)} (X \times [0,1] \times Y) \coprod _{ (X \times \{ 1\} \times Y) } Y.$ We will refer to $X \star Y$ as the join of $X$ and $Y$. Remark 4.3.4.2. Let $X$ and $Y$ be topological spaces. Then the join $X \star Y$ of Construction 4.3.4.1 is equipped with a pair of maps $\iota _{X}: X \hookrightarrow X \star Y$ and $\iota _{Y}: Y \hookrightarrow X \star Y$. It is not difficult to see that these maps are closed embeddings: that is, they induce homeomorphisms from $X$ and $Y$ onto closed subsets of $X \star Y$. We will generally abuse notation by identifying $X$ and $Y$ with their images under $\iota _{X}$ and $\iota _{Y}$, respectively. Remark 4.3.4.3. Let $X$, $Y$, and $Z$ be topological spaces. Then the datum of a continuous function $X \star Y \rightarrow Z$ is equivalent to the datum of a triple $(f_ X, f_ Y, h)$, where $f_ X: X \rightarrow Z$ and $f_ Y: Y \rightarrow Z$ are continuous functions and $h: X \times [0,1] \times Y \rightarrow Z$ is a homotopy from $f_{X} \circ \pi _{X}$ to $f_{Y} \circ \pi _{Y}$; here $\pi _{X}: X \times Y \rightarrow X$ and $\pi _{Y}: X \times Y \rightarrow Y$ denote the projection maps. Remark 4.3.4.4 (Symmetry). Let $X$ and $Y$ be topological spaces. Then there is a canonical homeomorphism $X \star Y \simeq Y \star X$, which is induced by the homeomorphism $X \times [0,1] \times Y \rightarrow Y \times [0,1] \times X \quad \quad (x,t,y) \mapsto (y, 1-t, x).$ Example 4.3.4.5 (Cones). Let $\ast$ denote the topological space consisting of a single point. For any topological space $X$, we write $X^{\triangleleft }$ for the join $\ast \star X$, and $X^{\triangleright }$ for the join $X \star \ast$, given more concretely by the formulae $X^{\triangleleft } = \ast \coprod _{ (\{ 0\} \times X) }( [0,1] \times X) \quad \quad X^{\triangleright } = (X \times [0,1]) \coprod _{ (X \times \{ 1\} ) } \ast .$ We will refer to both $X^{\triangleleft }$ and $X^{\triangleright }$ as the cone on $X$ (note that they are canonically homeomorphic, by virtue of Remark 4.3.4.4). Remark 4.3.4.6. Let $X$ be a locally compact Hausdorff space. Then the functor $\operatorname{Top}\rightarrow \operatorname{Top}_{X/} \quad \quad Y \mapsto X \star Y$ preserves colimits. This follows from the fact that the functors $Y \mapsto X \times Y$ and $Y \mapsto X \times [0,1] \times Y$ preserve colimits. Example 4.3.4.7. For each integer $n \geq 0$, let $\| \Delta ^{n} \| = \{ (u_0, \ldots , u_ n) \in \operatorname{\mathbf{R}}_{\geq 0}: u_0 + \cdots + u_ n = 1 \}$ denote the topological $n$-simplex. For $p,q \geq 0$, we have maps $\| \Delta ^{p} \| \xrightarrow {\iota } \| \Delta ^{p+1+q} \| \xleftarrow {\iota '} \| \Delta ^{q} \|$ given by the formulae $\iota ( u_0, \ldots , u_ p) = (u_0, \ldots , u_ p, 0, \ldots , 0) \quad \quad \iota '( v_0, \ldots , v_ q) = (0, \ldots , 0, v_0, \ldots , v_ q).$ There is a “straight-line” homotopy $h: \| \Delta ^{p} \| \times [0,1] \times \| \Delta ^{q} \| \rightarrow \| \Delta ^{p+1+q} \|$ from $\iota \circ \pi _{\| \Delta ^{p} \| }$ to $\iota ' \circ \pi _{ \| \Delta ^{q} \|}$, given concretely by the formula $h( (u_0, \ldots , u_ p), t, (v_0, \ldots , v_ q) ) = ( (1-t) u_0, (1-t) u_1, \ldots , (1-t) u_ p, t v_0, \ldots , t v_ q ).$ By virtue of Remark 4.3.4.3, the triple $(\iota , \iota ', h)$ can be identified with a continuous function $H_{p,q}: \| \Delta ^{p} \| \star \| \Delta ^{q} \| \rightarrow \| \Delta ^{p+1+q} \|$. Proposition 4.3.4.8. Let $p$ and $q$ be nonnegative integers. Then the function $H_{p,q}: \| \Delta ^{p} \| \star \| \Delta ^{q} \| \rightarrow \| \Delta ^{p+1+q} \|$ of Example 4.3.4.7 is a homeomorphism of topological spaces. Proof. Since $\| \Delta ^ p \| \star \| \Delta ^ q \|$ is compact and $\| \Delta ^{p+1+q} \|$ is Hausdorff, the continuous function $H_{p,q}$ is automatically closed. To complete the proof, it will suffice to show that $H_{p,q}$ is bijective. Fix a point $x$ of $\| \Delta ^{p+1+q} \|$, given by a sequence of nonnegative real numbers $( \overline{u}_0, \ldots , \overline{u}_ m, \overline{v}_0, \overline{v}_1, \ldots , \overline{v_ n} )$ satisfying $\overline{u}_0 + \cdots + \overline{u}_ m + \overline{v}_0 + \cdots + \overline{v}_ n = 0.$ Set $t = \overline{v}_0 + \cdots + \overline{v_ n}$. If $t = 0$, the set $H_{p,q}^{-1} \{ x\}$ consists of a single point of $\| \Delta ^{p} \|$ (regarded as a subset of $\| \Delta ^{p} \| \star \| \Delta ^{q} \|$), given by the sequence $( \overline{u}_0, \ldots , \overline{u}_ m)$. If $t = 1$, the set $H_{p,q}^{-1} \{ x \}$ consists of a single point of $\| \Delta ^ q \|$ (regarded as a subset of $\| \Delta ^{p} \| \star \| \Delta ^{q} \|$), given by the sequence $( \overline{v}_0, \ldots , \overline{v}_ m)$. In the case $0 < t < 1$, the set $H_{p,q}^{-1} \{ x\}$ consists of a single point of $\| \Delta ^{p} \| \star \| \Delta ^{q} \|$, given as the image of the triple $( \frac{\overline{u}_0}{1-t}, \ldots , \frac{ \overline{u}_ m}{1-t} ), t, ( \frac{ \overline{v}_0}{t}, \ldots , \frac{ \overline{v}_ n}{t} ) ) \in \| \Delta ^{p} \| \times [0,1] \times \| \Delta ^{n} \|.$ $\square$ We now compare the join operation on topological spaces (given by Construction 4.3.4.1) to the join operation on simplicial sets (given by Construction 4.3.3.13). Construction 4.3.4.9. Let $X$ and $Y$ be simplicial sets, and let $\sigma : \Delta ^ n \rightarrow X \star Y$ be a morphism. We define a continuous function $f(\sigma ): \| \Delta ^ n \| \rightarrow \|X\| \star \|Y\|$ as follows (see Remark 4.3.3.15): • If $\sigma$ factors through $X$, we let $f(\sigma )$ denote the composition $\| \Delta ^ n \| \xrightarrow { \| \sigma \| } \|X\| \xrightarrow {\iota _{\|X\|}} \|X\| \star \|Y\|,$ where the second map is the inclusion of Remark 4.3.4.2. • If $\sigma$ factors through $Y$, we let $f(\sigma )$ denote the composition $\| \Delta ^ n \| \xrightarrow { \| \sigma \| } \|Y\| \xrightarrow { \iota _{\|Y\|}} \|X\| \star \|Y\|,$ where the second map is the inclusion of Remark 4.3.4.2. • If $\sigma$ factors as a composition $\Delta ^ n = \Delta ^{p+1+q} \simeq \Delta ^{p} \star \Delta ^{q} \xrightarrow { \sigma _{-} \star \sigma _{+} } X \star Y,$ then we let $f(\sigma )$ denote the composite map $\| \Delta ^ n \| = \| \Delta ^{p+1+q} \| \xrightarrow { H_{p,q}^{-1} } \| \Delta ^{p} \| \star \| \Delta ^{q} \| \xrightarrow { \| \sigma _{-} \| \star \| \sigma _{+} \| } X \star Y,$ where $H_{p,q}$ denotes the homeomorphism of Proposition 4.3.4.8. The construction $\sigma \mapsto f(\sigma )$ is compatible with face and degeneracy maps, and therefore determines a morphism of simplicial sets $f: X \star Y \rightarrow \operatorname{Sing}_{\bullet }( \|X\| \star \|Y\| )$. We will identify $f$ with a continuous function $T_{X,Y}: \| X \star Y \| \rightarrow \|X\| \star \|Y\|$, which we will refer to as the join comparison map. Example 4.3.4.10. Let $X = \Delta ^{p}$ and $Y = \Delta ^{q}$ be standard simplices. Then the join comparison map $T_{X,Y}: \| \Delta ^{p} \star \Delta ^{q} \| \rightarrow \|\Delta ^{p}\| \star \| \Delta ^{q} \|$ fits into a commutative diagram $\xymatrix@R =50pt@C=50pt{ \| \Delta ^{p} \star \Delta ^{q} \| \ar [dr]^{ \| \rho \| } \ar [rr]^{ T_{X,Y} } & & \| \Delta ^{p} \| \star \| \Delta ^{q} \| \ar [dl]_{ H_{p,q} } \\ & \| \Delta ^{p+1+q} \|, & }$ where $\rho : \Delta ^{p} \star \Delta ^{q} \simeq \Delta ^{p+1+q}$ denotes the isomorphism of simplicial sets appearing in Example 4.3.3.20 and $H_{p,q}$ is the homeomorphism of Proposition 4.3.4.8. In particular, $T_{X,Y}$ is a homeomorphism. Proposition 4.3.4.11. Let $X$ and $Y$ be simplicial sets. If either $X$ or $Y$ is finite, then the join comparison map $T_{X,Y}: \|X \star Y\| \rightarrow \|X\| \star \|Y\|$ of Construction 4.3.4.9 is a homeomorphism. Proof. Without loss of generality, we may assume that $X$ is finite. Then the geometric realization $\|X\|$ is a compact Hausdorff space (Corollary 3.5.1.10). Using Remarks 4.3.4.6 and 4.3.3.26, we see that the functors $\operatorname{Set_{\Delta }}\rightarrow \operatorname{Top}_{\|X\|/} \quad \quad Y \mapsto \|X \star Y\|, Y \mapsto \|X\| \star \|Y\|$ preserve colimits. Consequently, if we regard $X$ as fixed, then the collection of simplicial sets $Y$ for which $T_{X,Y}$ is a homeomorphism is closed under colimits. Since every simplicial set can be realized as a colimit of standard simplices (Corollary 1.1.8.17), it will suffice to prove Proposition 4.3.4.11 in the special case where $Y = \Delta ^{q}$ is a standard simplex. In this case, $Y$ is also finite. Repeating the preceding argument (with the roles of $X$ and $Y$ reversed), we are reduced to proving that $T_{X,Y}$ is a homeomorphism in the case where $X = \Delta ^{p}$ is also a standard simplex. In this case, the desired result follows from Example 4.3.4.10. $\square$ Corollary 4.3.4.12. Let $X$ be a simplicial set. Then the join comparison maps $T_{\Delta ^0,X}$ and $T_{X, \Delta ^0}$ supply homeomorphisms of topological spaces $\| X^{\triangleleft } \| \simeq \|X\|^{\triangleleft } \quad \quad \| X^{\triangleright } \| \simeq \| X \|^{\triangleright }.$ Here $X^{\triangleleft }$ and $X^{\triangleright }$ denote the left and right cones on $X$ in the category of simplicial sets (Construction 4.3.3.22), while $\|X\|^{\triangleleft }$ and $\|X\|^{\triangleright }$ denote the cone $\|X\|$ in the category of topological spaces (Example 4.3.4.5). The join comparison map $T_{X,Y}: \|X \star Y\| \rightarrow \|X\| \star \|Y\|$ need not be a homeomorphism in general. However, we do have the following: Corollary 4.3.4.13. Let $X$ and $Y$ be simplicial sets. Then the join comparison map $T_{X,Y}: \|X \star Y\| \rightarrow \|X\| \star \|Y\|$ is a bijection. Proof. As a map of sets, we can realize $T_{X,Y}$ as a filtered colimit of join comparison maps $T_{X', Y}$, where $X'$ ranges over the finite simplicial subsets of $X$ (Remark 3.5.1.8). Each of these maps is bijective (even a homeomorphism), by virtue of Proposition 4.3.4.11. $\square$ Warning 4.3.4.14. Let $X$ and $Y$ be simplicial sets, and let $X \diamond Y$ denote the simplicial set given by the iterated coproduct $X \coprod _{ (X \times \{ 0\} \times Y)} (X \times \Delta ^1 \times Y) \coprod _{ (X \times \{ 1\} \times Y) } Y.$ Since the formation of geometric realization commutes with the formation of colimits, we have an evident comparison map of topological spaces $\|X \diamond Y \| \rightarrow \|X\| \star \|Y\|.$ This map is always bijective, and is a homeomorphism if either $X$ or $Y$ is finite (see Corollary 3.5.2.2). In this case, Corollary 4.3.4.13 supplies a homeomorphism of geometric realizations $\| X \diamond Y \| \simeq \| X \star Y \|$. Beware that this homeomorphism does not arise from a morphism of simplicial sets. In the case $X = \Delta ^ p$ and $Y = \Delta ^{q}$, it arises from the homotopy $h: \| \Delta ^{p} \| \times \| \Delta ^1 \| \times \| \Delta ^{q} \| \rightarrow \| \Delta ^{p+1+q} \|$ $h( (u_0, \ldots , u_ p), t, (v_0, \ldots , v_ q) ) = ( (1-t) u_0, (1-t) u_1, \ldots , (1-t) u_ p, t v_0, \ldots , t v_ q ).$ appearing in Example 4.3.4.7, which is not piecewise-linear with respect to the natural triangulation of the polysimplex $\| \Delta ^{p} \| \times \| \Delta ^1 \| \times \| \Delta ^ q \|$.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9906498193740845, "perplexity": 106.69217408684864}, "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/1600402124756.81/warc/CC-MAIN-20201001062039-20201001092039-00168.warc.gz"}
https://cerncourier.com/a/compass-points-to-triangle-singularity/	# COMPASS points to triangle singularity 23 August 2021 The COMPASS experiment at CERN has reported the first direct evidence for a long-hypothesised interplay between hadron decays which can masquerade as a resonance. The analysis, which was published last week in Physical Review Letters, suggests that the “a1(1420)” signal observed by the collaboration in 2015 is not a new exotic hadron after all, but the first sighting of a so-called triangle singularity. “Triangle singularities are a mechanism for generating a bump in the decay spectrum that does not correspond to a resonance,” explains analyst Mikhail Mikhasenko of the ORIGINS Cluster in Munich. “One gets a peak that has all features of a new hadron, but whose true nature is a virtual loop with known particles.” “This is a prime example of an aphorism which is commonly attributed to Dick Dalitz,” agrees fellow analyst Bernhard Ketzer, of the University of Bonn: “Not every bump is a resonance, and not every resonance is a bump!” Triangle singularities take their name from the triangle in a Feynman diagram when a secondary decay product fuses with a primary decay product. If the particle masses line up such that the process can proceed as a cascade of on-mass-shell hadron decays, the matrix element is enhanced by a so-called logarithmic singularity which can easily be mistaken for a resonance. But the effect is usually rather small, requiring a record 50 million πp→ππ+πp events, and painstaking work by the COMPASS collaboration to make certain that the a1(1420) signal, which makes up less than 1% of the three-pion sample, wasn’t an artefact of the analysis procedure. Hadron experiments are reaching the precision needed to see one of the most peculiar multi-body features of QCD Mikhail Mikhasenko “The correspondence of this small signal with a triangle singularity is noteworthy because it shows that hadron experiments are finally reaching the precision and statistics needed to see one of the most peculiar features of the multi-body non-perturbative regime of quantum chromodynamics,” says Mikhasenko. Triangle singularities were dreamt up independently by Lev Landau and Richard Cutkosky in 1959. After five decades of calculations and speculations, physicists at the Institute for High-Energy Physics in Beijing in 2012 used a triangle singularity to explain why intermediate f0(980) mesons in J/ψ meson decays at the BESIII experiment at the Beijing Electron–Positron Collider II were unusually long-lived. In 2019, the LHCb collaboration ruled out triangle singularities as the origin of the pentaquark states they discovered that year. The new COMPASS analysis is the first time that a “bump” in a decay spectrum has been convincingly explained as more likely due to a triangle singularity than a resonance. COMPASS collides a secondary beam of charged pions from CERN’s Super Proton Synchrotron with a hydrogen target in the laboratory’s North Area. In this analysis, gluons emitted by protons in the target excite the incident pions, producing the final state of three charged pions which is observed by the COMPASS spectrometer. Intermediate resonances display a variety of angular momentum, spin and parity configurations. In 2015, the collaboration observed a small but unmistakable “P-wave” (L=1) component of the f0(980)π system with a peak at 1420 MeV and JPC=1++. Dubbed a1(1420), the apparent resonance was suspected to be exotic, as it was narrower, and hence more stable, than the ground-state meson with the same quantum numbers, a1(1260). It was also surprisingly light, with a mass just above the KK threshold of 1.39 GeV. A tempting interpretation was that a1(1420) might be a dsūs̄ tetraquark, and thus the first exotic hadronic state with no charm quarks, and a charged cousin of the famous exotic X(3872) at the DD threshold to boot, explains Mikhasenko. According to the new COMPASS analysis, however, the bump at 1420 MeV can more simply be explained by a triangle singularity, whereby an a1(1260) decays to a KK pair, and the kaon from the resulting K→Kπ decay annihilates with the initial anti-kaon to create a light unflavoured f0(980) meson which decays to a pair of charged pions. Crucially, the mass of f0(980) is just above the KK threshold, and the roughly 300 MeV width of the conventional a1(1260) meson is wide enough for the particle to be said to decay to KK on-mass-shell. A new resonance is not required. That is phenomenologically significant. Ian Aitchison “The COMPASS collaboration have obviously done a very thorough job, being in possession of a complete partial-wave analysis,” says Ian Aitchison, emeritus professor at the University of Oxford, who in 1964 was among the first to propose that triangle graphs with an unstable internal line (in this case the K) could lead to observable effects. This enables the whole process to occur nearly on-shell for all particles, which in turn means that the singularities of the amplitude will be near the physical region, and hence observable, explains Aitchison. “This is not unambiguous evidence for the observation of a triangle singularity, but the paper shows pretty convincingly that it is sufficient to explain the data, and that a new resonance is not required. That is phenomenologically significant.” The collaboration now plans further studies of this new phenomenon, including its interference with the direct decay of the a1(1260). Meanwhile, observation by Belle II of the a1(1420) phenomenon in decays of the tau meson to three pions should confirm our understanding and provide an even cleaner signal, says Mikhasenko.	{"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": 0.8824149370193481, "perplexity": 1692.3460759878742}, "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-2022-49/segments/1669446711151.22/warc/CC-MAIN-20221207085208-20221207115208-00063.warc.gz"}
https://dsp.stackexchange.com/questions/80717/calculating-tranfer-function-poles-zeros-and-impulse-response-given-input-and	# Calculating tranfer function, poles, zeros and impulse response given input and outpul signals in matlab I have been given an input and output signal. input: x(n)=(0.3^n)u(n) + (5^n)u(-n-1) output: y(n)=(3^n)u(-n-1) - ((2^(n+2))/(3^n))u(n). or x=(0.3.^n).(n>=0) + (5.^n). (n<=-1) --> in Matlab y=(3.^n). * (n<=-1) - 4. * ((2/3).^n).*(n>=0) --> in Matlab I am supposed to find the tranfer function, it's poles and zeros and impulse response using matlab . I searched a lot but I couldn't find a way to aproach it in matlab. I managed to find a transfer polynomial solving this on paper and came up with a=[-5 39.16 -74,63 19], as numerator coefficients and b=[0 -4.7 17.23 -9.4], as denominator coefficients. It turns out, whatsoever, that matlab can't make a valid computation of the poles and zeros as well as the impulse resonspe of the system since "the first denominator filter coefficient must be non-zero". Any idea on how can I solve this exclusively using matlab or how can I eliminate the first 0 (zero) denominator coefficient? thank you. The Z-transform of the transfer function is $$H(z) = Y(z)/X(z)$$: the ratio of output to input Z-transforms. Matlab provides a ztrans() method for computing Z-transforms symbolically (I didn't know this existed until now).	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 1, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8359276056289673, "perplexity": 1961.3694162205554}, "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/1664030334871.54/warc/CC-MAIN-20220926113251-20220926143251-00614.warc.gz"}
https://www.physicsforums.com/threads/the-unit-of-flux-density.334061/	The unit of Flux density. 1. Sep 1, 2009 yash25 The unit of Electric flux is N.(m^2)/C. But how is it , that the Flux density(which is nothing but flux per unit area) has a unit of C/(m^2).? Also , I have a book in Electromagnetic Engineering which states that "'electric displacement' and 'electric flux' are the same thing and both are equal to "Q" i.e the charge. It also states that the unit of flux is Coulomb.' Is all of this true? Last edited: Sep 1, 2009 2. Sep 1, 2009 Born2bwire The electric displacement field and the electric flux density are both the same thing, the vector D, which has units of C/m^2 in MKS. This differs from the electric field by the permittivity, the ratio of D/E is the permittiviy with units of F/m. The electric flux is related to charge using Gauss' Law, which states that the flux over a closed surface is equal to the total charge divided by the permittivity. So the electric flux is equal to (V/m)(m^2) = Vm which is the same as Nm^2/C. Since the electric field and electric flux density are related by the permittivity, we can rewrite Gauss' Law to show that the integral of the electric flux density over a closed surface is equal to the total charge enclosed. 3. Sep 2, 2009 yash25 Yes. That's true. But as I mentioned before, some books on Electromagnetic Engineering (like Jordan and Balmain), equate Electric flux to the amount of charge(i.e psi=Q). They have used the Faraday thought experiment of two concentric spheres for proving this equality. I have never heard of this experiment before and I always know that the electric flux was nothing but the dot product of field and surface area over the whole gaussian surface ... I could not understand this equality between flux and charge!!! 4. Sep 2, 2009 Born2bwire Yeah I have seen that too, but the only previous time I have seen it was in the CRC Handbook of all places. The validity of the statement would be what they defined the electric flux to be and what units they are using. In the CRC Handbook, they defined the electric flux to be the integral of the electric flux density, not the electric field, and so that relationship would still be correct. Sometimes, people will choose a system of units where the permittivity and/or permeability of free-space are unity, this is common in the CGS units. Under these units, the same relationship would apply when defining electric flux to be the integral involving the electric field. 5. Sep 2, 2009 yash25 Im sorry, but I couldn't understand what you meant by Have they considered permittivity to be unity in the CRC handbook? So what would be the correct equation of the flux? Would it be the integral of the electric field or the flux density? 6. Sep 2, 2009 DrZoidberg Both definitions are possible. There are 2 kinds of electric flux. The flux of the E field with the unit Vm=N*m^2/C and the flux of the D field with the unit C Btw. the same is true for magnetic fields. There is the B field which is the magnetic equivalent of the D field and the H field - the magnetic equivalent of the E field. So there are also 2 possible ways to define magnetic flux. In practice however engineers always use the flux of the B field. That can be confusing. Because for electric fields they usually use the flux of the E field which correlates with the H field and not the B field. Because of this students may think that magnetic flux behaves fundamentally differently from electric flux but that is not true. They are perfectly symmetrical. Last edited: Sep 2, 2009 7. Sep 2, 2009 yash25 Thanks a million! I was really confused ( being an engineer and a physics enthusiast can therefore lead to clashing definitions of flux!) ... Why, may i then ask, are there two types of electrical fluxes? is there any relation between the two( i.e " E flux" and "D flux")? And why do engineers prefer the flux due to the D field..And what is the physical interpretation of this kind of flux?(faraday's experiment of concentric spheres leads to the definition of the D flux , if I am not wrong, but what does it exactly mean in layman terms?) 8. Sep 2, 2009 DrZoidberg I'm not sure why there are 2 types. I guess each type has it's advantages and disadvantages. There are different ways to think about it. You could think in terms of field lines. Imagine there is a certain number of field lines emerging from each proton and the same number of field lines are entering each electron. Then the field lines would represent the D field i.e. "D flux". If you use field lines to represent the "E flux" you can run into trouble because then the number of field lines doesn't stay constant anymore if they pass through different materials. But it really depends on the situation which one is easier to work with. Of course there is a relation between the two. Look here http://en.wikipedia.org/wiki/Permittivity Last edited: Sep 2, 2009 9. Sep 2, 2009 Born2bwire Take a look at CGS Lorentz-Heaviside units. You will notice that the units have been chosen such that the flux, calculated using the electric field, is equal to the enclosed charge. There are some slightly different ways of working with Maxwell's equations. Sometimes people like to formulate them so that the inherent physics stands out without much distraction. Like the Lorentz-Heaviside units, it is easier to work with flux equal to charge than flux being equal to charge times some weird number. I prefer MKS myself because then there is no conversion between SI units, which are usually used for measurement, and the results of the equations. On Dr. Zoidberg's comments, I am an engineer myself and I have seen that many engineering books like to make Maxwell's equations symmetric. So they call the B field the magnetic flux density and the H field the magnetic field. I have seen on numerous occasions that the reverse is true for physics references. They normally call the B field the magnetic field. But the engineer's way makes the formulas more symmetric. Sometimes we even add a ficticious magnetic charge and current and make them fully symmetric. But it should be noted that this magnetic charge and current are purely a computational tool and are not thought to be truely physical. 10. Sep 2, 2009 cmos I'd just like to point out that, while it is true many people call B the magnetic field, I do not think any reputable source will ever call H the magnetic flux density. When physicists call H the magnetic field, then they tend to call B the magnetic induction. I have seen several places in the literature where people are calling B the magnetic field and H the magnetizing field. I, personally, am fond of this (new?) trend. Indeed, it makes perfect sense if you think about the conventional way of magnetizing a ferromagnetic rod and hysteresis plots. 11. Sep 3, 2009 Born2bwire Yeah that's true, I was thinking in the cofines of both H and B could be called the magnetic field but it didn't come out right. Of course I never heard of the flux being defined using the D field in the integral, but happened to come across that definition in the CRC handbook when I looked it up in that for another thread on the same discussion of the naming nomencalture. Similar Discussions: The unit of Flux density.	{"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": 0.9785393476486206, "perplexity": 456.66838833264666}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934806438.11/warc/CC-MAIN-20171121223707-20171122003707-00231.warc.gz"}
https://www.physicsforums.com/threads/multivariable-calculus-limits.89711/	# Multivariable Calculus Limits 1. Sep 18, 2005 ### eutopia lim of cos((x^2 + y^2) - 1)/(x^2 + y^2) as (x,y) approaches (0,0) I have no clue how to tackle this problem. I tried to find the level set so at least I can have a clue of what the graph looks like, but then, I didn't know how to find the level sets either. If I set c = the equation, I have 2 unknowns so I cannot solve, and its not an obvious graph like a circle or something. On the other hand, I tried l'hopitale but that needs the derivative and what in the world am i taking a derivative in terms of since there are 2 variables? 2. Sep 18, 2005 ### mathmike you have an intermediate form of lahopital's theorey of the form 1/0 do you know how to do these? 3. Sep 18, 2005 ### Hurkyl Staff Emeritus I think he's missing a parenthesis -- the numerator is supposed to be cos(x²+y²) - 1. (P.S. 1/0 is not indeterminate, and AFAIK it's not L'Hôpital theory) eutopia: what techniques have you seen used for similar problems? There is one in particular that makes this problem very simple. 4. Sep 19, 2005 ### MalleusScientiarum For the limit to exist it has to exist regardless of the direction from which you are approaching the point. Parameterize lines passing through the origin and see if you can get that. 5. Sep 19, 2005 ### Hurkyl Staff Emeritus As you said, the limit has to exist (and be the same value) for any way you approach the origin -- just looking at the lines isn't good enough. 6. Sep 19, 2005 ### HallsofIvy Staff Emeritus "Lines through the origin", suggested by MalleusScientiarum, will help show that a limit does not exist by getting, hopefully, different limits on different lines. But they can't prove that a limit DOES exit (or find it) since even if the limit is the same along all lines, there might be other curves, not lines, passing through the origin that give a different limit. The best way to handle ANY limit problem in more than one variable (going to (0,0) or (0,0,0), etc.) is to change to polar (spherical, etc.) coordinates since that way one variable, r (ρ, etc.) measures the distance to (0,0) directly! In this case, that's easy since x and y only appear in x2+ y2= r2. The original function, $$\frac{cos((x^2 + y^2) - 1)}{x^2 + y^2}$$ becomes $$\frac{cos(r^2-1)}{r^2}$$ which clearly goes to infinity as r goes to 0. Hurkyls suggested correction, $$\frac{cos(x^2+y^2)-1}{x^2+y^2}$$ becomes $$\frac{cos(r^2)-1}{r^2}$$ which now has only one variable and can be done by L'Hopital's rule. (The limit is 0.)	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9030014276504517, "perplexity": 666.8529794077656}, "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/1480698543035.87/warc/CC-MAIN-20161202170903-00180-ip-10-31-129-80.ec2.internal.warc.gz"}
http://vashikaranmantraforlove.com/blog/	### Get Lost Love Back some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some 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here some text here some text here some text here some text here some text here some text here some text here some text here some text here ### Intercaste Marriage Problem Solution some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text here some text 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https://engineeringnotesonline.com/2021/06/	### Geothermal Power Generation June 23, 2021 Geothermal Power Generation Geothermal energy is the energy which lies embedded within the earth. Geothermal electric power plants are built on the edges of tectonic […] ### Tidal energy power generation June 13, 2021 Tidal energy power generation Tidal energy is the permanent source of energy. The rise and fall of tides nearly twice a day are associated with […] ### Rectifier : Half wave rectifier and Full wave Rectifier June 4, 2021 Rectifier The electric power is usually available in a.c supply. The supply voltage varies sinusoidal and has a frequency of 50Hz and used for different […] ### Transistor as an Amplifier June 1, 2021 Transistor as an Amplifier The transistor amplifier circuit is shown in the figure ,shows the common emitter npn amplifier circuit. VBB is connected in the […]	{"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": 0.9439480304718018, "perplexity": 3525.4704656404315}, "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/1637964363598.57/warc/CC-MAIN-20211208205849-20211208235849-00322.warc.gz"}
https://drchristiansalas.com/2016/03/20/dirichlet-character-tables-up-to-mod-21/	# Dirichlet character tables up to mod 21 I have written about Dirichlet characters in previous notes (see Advanced Number Theory Note #13 and the references therein). For reference purposes, I will set out the first twenty Dirichlet character tables in the present note and demonstrate in detail the calculation of the first ten tables. Recall that there are $\varphi(k)$ distinct Dirichlet characters $\chi$ modulo $k$, each of which is completely multiplicative and periodic with period $k$. Each character value $\chi(n)$ is a (complex) root of unity if $gcd(n, k) = 1$ whereas $\chi(n) = 0$ whenever $gcd(n, k) > 1$. We also have $\chi(1) = 1$ for all Dirichlet characters. These facts uniquely determine each Dirichlet character table. k = 2 We have $\varphi(2) = 1$ so there is only one Dirichlet character in this case (the principal one), with values $\chi_1(1) = 1$ and $\chi_1(2) = 0$. k = 3 We have $\varphi(3) = 2$ so there are two Dirichlet characters in this case. One of them will be the principal character which takes the values $\chi_1(1) = 1$, $\chi_1(2) = 1$ and $\chi_1(3) = 0$. To work out the second Dirichlet character we consider the two roots of unity $\omega = e^{2 \pi i/2} = e^{\pi i} = -1$ and $\omega^2 = 1$ Note that the set of least positive residues mod $3$ is generated by $2$: $2 \equiv 2$ mod($3$) $2^2 \equiv 1$ mod($3$) Therefore the non-principal Dirichlet character will be completely determined by the values of $\chi(2)$. If we set $\chi_2(2) = \omega = -1$ then $\chi_2(1) = \chi_2(2^2) = \chi_2^2(2) = \omega^2 = 1$ (though this calculation is superfluous here since $\chi_2(1) = 1$ anyway. This is a fundamental property of Dirichlet characters arising from the fact that they are completely multiplicative). We also have $\chi_2(3) = 0$. This completes the second character. (From now on we will omit the statements of the zero values of the Dirichlet characters, which as stated earlier arise whenever $gcd(n, k) > 1$). k = 4 We have $\varphi(4) = 2$ so there are two Dirichlet characters in this case. One of them will be the principal character. (From now on we will always denote the principal character by $\chi_1$). To work out the second Dirichlet character we again consider the two roots of unity $\omega = e^{2 \pi i/2} = e^{\pi i} = -1$ and $\omega^2 = 1$ Note that the set of least positive residues mod $4$ is generated by $3$: $3 \equiv 3$ mod($4$) $3^2 \equiv 1$ mod($4$) Therefore the non-principal Dirichlet character will be completely determined by the values of $\chi(3)$. If we set $\chi_2(3) = \omega = -1$ then $\chi_2(1) = \chi_2(3^2) = \chi_2^2(3) = \omega^2 = 1$ (though again this second calculation is superfluous since $\chi_2(1) = 1$ anyway). This completes the second character. k = 5 We have $\varphi(5) = 4$ so there are four Dirichlet characters in this case. We consider the four roots of unity $\omega = e^{2 \pi i/4} = e^{\pi i/2} = i$ $\omega^2 = e^{4 \pi i/4} = e^{\pi i} = -1$ $\omega^3 = e^{6 \pi i/4} = e^{3 \pi i/2} = -i$ $\omega^4 = e^{8 \pi i/4} = e^{2 \pi i} = 1$ Note that the set of least positive residues mod $5$ is generated by $2$: $2 \equiv 2$ mod($5$) $2^2 \equiv 4$ mod($5$) $2^3 \equiv 3$ mod($5$) $2^4 \equiv 1$ mod($5$) Therefore the non-principal Dirichlet characters will be completely determined by the values of $\chi(2)$. If we set $\chi_2(2) = \omega = i$ then $\chi_2(3) = \chi_2(2^3) = \chi_2^3(2) = -i$ $\chi_2(4) = \chi_2(2^2) = \chi_2^2(2) = -1$ (and we have $\chi_2(1) = 1$). This completes the second character. To compute the third character we can set $\chi_3(2) = \omega^2 = -1$ then $\chi_3(3) = \chi_3(2^3) = \chi_3^3(2) = -1$ $\chi_3(4) = \chi_3(2^2) = \chi_3^2(2) = 1$ (and we have $\chi_3(1) = 1$). This completes the third character. To compute the fourth character we set $\chi_4(2) = \omega^3 = -i$ then $\chi_4(3) = \chi_4(2^3) = \chi_4^3(2) = i$ $\chi_4(4) = \chi_4(2^2) = \chi_4^2(2) = -1$ (and we have $\chi_4(1) = 1$). This completes the fourth character. k = 6 We have $\varphi(6) = 2$ so there are two Dirichlet characters in this case. We consider the two roots of unity $\omega = e^{2 \pi i/2} = e^{\pi i} = -1$ and $\omega^2 = 1$ Note that the set of least positive residues mod $6$ is generated by $5$: $5 \equiv 5$ mod($6$) $5^2 \equiv 1$ mod($6$) Therefore the non-principal Dirichlet character will be completely determined by the values of $\chi(5)$. If we set $\chi_2(5) = \omega = -1$ then $\chi_2(1) = \chi_2(5^2) = \chi_2^2(5) = \omega^2 = 1$ (though again this second calculation is superfluous since $\chi_2(1) = 1$ anyway). This completes the second character. k = 7 We have $\varphi(7) = 6$ so there are six Dirichlet characters in this case. We consider the six roots of unity $\omega = e^{2 \pi i/6} = e^{\pi i/3}$ $\omega^2 = e^{2 \pi i/3}$ $\omega^3 = e^{3 \pi i/3} = e^{\pi i} = -1$ $\omega^4 = \omega \cdot \omega^3 = -\omega$ $\omega^5 = \omega \cdot \omega^4 = -\omega^2$ $\omega^6 = e^{6 \pi i/3} = e^{2\pi i} = 1$ Note that the set of least positive residues mod $7$ is generated by $3$: $3 \equiv 3$ mod($7$) $3^2 \equiv 2$ mod($7$) $3^3 \equiv 6$ mod($7$) $3^4 \equiv 4$ mod($7$) $3^5 \equiv 5$ mod($7$) $3^6 \equiv 1$ mod($7$) Therefore the non-principal Dirichlet characters will be completely determined by the values of $\chi(3)$. If we set $\chi_2(3) = \omega$ then $\chi_2(2) = \chi_2(3^2) = \chi_2^2(3) = \omega^2$ $\chi_2(4) = \chi_2(3^4) = \chi_2^4(3) = \omega^4 = -\omega$ $\chi_2(5) = \chi_2(3^5) = \chi_2^5(3) = \omega^5 = -\omega^2$ $\chi_2(6) = \chi_2(3^3) = \chi_2^3(3) = \omega^3 = -1$ (and we have $\chi_2(1) = 1$). This completes the second character. To compute the third character we can set $\chi_3(3) = \omega^2$ then $\chi_3(2) = \chi_3(3^2) = \chi_3^2(3) = \omega^4 = -\omega$ $\chi_3(4) = \chi_3(3^4) = \chi_3^4(3) = \omega^8 = \omega^2$ $\chi_3(5) = \chi_3(3^5) = \chi_3^5(3) = \omega^{10} = -\omega$ $\chi_3(6) = \chi_3(3^3) = \chi_3^3(3) = \omega^6 = 1$ (and we have $\chi_3(1) = 1$). This completes the third character. To compute the fourth character we can set $\chi_4(3) = \omega^3 = -1$ then $\chi_4(2) = \chi_4(3^2) = \chi_4^2(3) = 1$ $\chi_4(4) = \chi_4(3^4) = \chi_4^4(3) = 1$ $\chi_4(5) = \chi_4(3^5) = \chi_4^5(3) = -1$ $\chi_4(6) = \chi_4(3^3) = \chi_4^3(3) = -1$ (and we have $\chi_4(1) = 1$). This completes the fourth character. To compute the fifth character we can set $\chi_5(3) = \omega^4 = -\omega$ then $\chi_5(2) = \chi_5(3^2) = \chi_5^2(3) = \omega^2$ $\chi_5(4) = \chi_5(3^4) = \chi_5^4(3) = \omega^4 = -\omega$ $\chi_5(5) = \chi_5(3^5) = \chi_5^5(3) = \chi_5^4(3) \cdot \chi_5(3) = \omega^2$ $\chi_5(6) = \chi_5(3^3) = \chi_5^3(3) = -\omega^3 = 1$ (and we have $\chi_5(1) = 1$). This completes the fifth character. Finally, to compute the sixth character we set $\chi_6(3) = \omega^5 = -\omega^2$ then $\chi_6(2) = \chi_6(3^2) = \chi_6^2(3) = \omega^4 = -\omega$ $\chi_6(4) = \chi_6(3^4) = \chi_6^4(3) = \omega^8 = \omega^2$ $\chi_6(5) = \chi_6(3^5) = \chi_6^5(3) = -\omega^{10} = \omega$ $\chi_6(6) = \chi_6(3^3) = \chi_6^3(3) = -\omega^6 = -1$ (and we have $\chi_6(1) = 1$). This completes the sixth character. k = 8 We have $\varphi(8) = 4$ so there are four Dirichlet characters in this case. We consider the four roots of unity $\omega = e^{2 \pi i/4} = e^{\pi i/2} = i$ $\omega^2 = e^{4 \pi i/4} = e^{\pi i} = -1$ $\omega^3 = e^{6 \pi i/4} = e^{3 \pi i/2} = -i$ $\omega^4 = e^{8 \pi i/4} = e^{2 \pi i} = 1$ In this case, none of the four elements of the set of least positive residues mod $8$ generates the entire set. However, the characters must satisfy the following relations, which restrict the choices: $\chi(3) \cdot \chi(5) = \chi(15) = \chi(7)$ $\chi(3) \cdot \chi(7) = \chi(21) = \chi(5)$ $\chi(5) \cdot \chi(7) = \chi(35) = \chi(3)$ Each character’s values must be chosen in such a way that these three relations hold. To compute the second character, suppose we begin by trying to set $\chi_2(3) = \omega = i$ and $\chi_2(5) = \omega^2 = -1$ Then we must have $\chi_2(7) = \chi_2(3) \cdot \chi_2(5) = -i$ but then $\chi_2(3) \cdot \chi_2(7) = 1 \neq \chi_2(5)$ so this does not work. If instead we try to set $\chi_2(5) = -i$ then we must have $\chi_2(7) = \chi_2(3) \cdot \chi_2(5) = 1$ but then $\chi_2(3) \cdot \chi_2(7) = i \neq \chi_2(5)$ so this does not work either. Computations like these show that $\pm i$ cannot appear in any of the characters mod $8$. All the characters must be formed from $\pm 1$. (Fundamentally, this is due to the fact that the group of least positive residues mod $8$ can be subdivided into four cyclic subgroups of order 2, each of which has characters whose values are the two roots of unity, $1$ and $-1$). To compute the second character we can set $\chi_2(3) = 1$ and $\chi_2(5) = -1$ then we must have $\chi_2(7) = -1$ and this works. To compute the third character we can set $\chi_3(3) = -1$ and $\chi_3(5) = -1$ then we must have $\chi_3(7) = 1$ and this works too. Finally, to compute the fourth character we can set $\chi_4(3) = -1$ and $\chi_4(5) = 1$ then we must have $\chi_4(7) = -1$ and this works too. k = 9 We have $\varphi(9) = 6$ so there are six Dirichlet characters in this case. We consider the six roots of unity $\omega = e^{2 \pi i/6} = e^{\pi i/3}$ $\omega^2 = e^{2 \pi i/3}$ $\omega^3 = e^{3 \pi i/3} = e^{\pi i} = -1$ $\omega^4 = \omega \cdot \omega^3 = -\omega$ $\omega^5 = \omega \cdot \omega^4 = -\omega^2$ $\omega^6 = e^{6 \pi i/3} = e^{2\pi i} = 1$ Note that the set of least positive residues mod $9$ is generated by $2$: $2 \equiv 2$ mod($9$) $2^2 \equiv 4$ mod($9$) $2^3 \equiv 8$ mod($9$) $2^4 \equiv 7$ mod($9$) $2^5 \equiv 5$ mod($9$) $2^6 \equiv 1$ mod($9$) Therefore the non-principal Dirichlet characters will be completely determined by the values of $\chi(2)$. If we set $\chi_2(2) = \omega$ then $\chi_2(4) = \chi_2(2^2) = \chi_2^2(2) = \omega^2$ $\chi_2(5) = \chi_2(2^5) = \chi_2^5(2) = \omega^5 = -\omega^2$ $\chi_2(7) = \chi_2(2^4) = \chi_2^4(2) = \omega^4 = -\omega$ $\chi_2(8) = \chi_2(2^3) = \chi_2^3(2) = \omega^3 = -1$ (and we have $\chi_2(1) = 1$). This completes the second character. To compute the third character we can set $\chi_3(2) = \omega^2$ then $\chi_3(4) = \chi_3(2^2) = \chi_3^2(2) = \omega^4 = -\omega$ $\chi_3(5) = \chi_3(2^5) = \chi_3^5(2) = \omega^{10} = \omega^6 \cdot \omega^4 = -\omega$ $\chi_3(7) = \chi_3(2^4) = \chi_3^4(2) = \omega^8 = \omega^6 \cdot \omega^2 = \omega^2$ $\chi_3(8) = \chi_3(2^3) = \chi_3^3(2) = \omega^6 = 1$ (and we have $\chi_3(1) = 1$). This completes the third character. To compute the fourth character we can set $\chi_4(2) = \omega^3 = -1$ then $\chi_4(4) = \chi_4(2^2) = \chi_4^2(2) = 1$ $\chi_4(5) = \chi_4(2^5) = \chi_4^5(2) = -1$ $\chi_4(7) = \chi_4(2^4) = \chi_4^4(2) = 1$ $\chi_4(8) = \chi_4(2^3) = \chi_4^3(2) = -1$ (and we have $\chi_4(1) = 1$). This completes the fourth character. To compute the fifth character we can set $\chi_5(2) = \omega^4 = -\omega$ then $\chi_5(4) = \chi_5(2^2) = \chi_5^2(2) = \omega^2$ $\chi_5(5) = \chi_5(2^5) = \chi_5^5(2) = -\omega^5 = -\omega^3 \cdot \omega^2 = \omega^2$ $\chi_5(7) = \chi_5(2^4) = \chi_5^4(2) = \omega^4 = \omega^3 \cdot \omega = -\omega$ $\chi_5(8) = \chi_5(2^3) = \chi_5^3(2) = -\omega^3 = 1$ (and we have $\chi_5(1) = 1$). This completes the fifth character. Finally, to compute the sixth character we can set $\chi_6(2) = \omega^5 = -\omega^2$ then $\chi_6(4) = \chi_6(2^2) = \chi_6^2(2) = \omega^4 = -\omega$ $\chi_6(5) = \chi_6(2^5) = \chi_6^5(2) = -\omega^{10} = -\omega^6 \cdot \omega^4 = \omega$ $\chi_6(7) = \chi_6(2^4) = \chi_6^4(2) = \omega^8 = \omega^6 \cdot \omega^2 = \omega^2$ $\chi_6(8) = \chi_6(2^3) = \chi_6^3(2) = -\omega^6 = -1$ (and we have $\chi_6(1) = 1$). This completes the sixth character. k = 10 We have $\varphi(10) = 4$ so there are four Dirichlet characters in this case. We consider the four roots of unity $\omega = e^{2 \pi i/4} = e^{\pi i/2} = i$ $\omega^2 = e^{4 \pi i/4} = e^{\pi i} = -1$ $\omega^3 = e^{6 \pi i/4} = e^{3 \pi i/2} = -i$ $\omega^4 = e^{8 \pi i/4} = e^{2 \pi i} = 1$ Note that the set of least positive residues mod $10$ is generated by $3$: $3 \equiv 3$ mod($10$) $3^2 \equiv 9$ mod($10$) $3^3 \equiv 7$ mod($10$) $3^4 \equiv 1$ mod($10$) Therefore the non-principal Dirichlet characters will be completely determined by the values of $\chi(3)$. If we set $\chi_2(3) = \omega = i$ then $\chi_2(7) = \chi_2(3^3) = \chi_2^3(3) = -i$ $\chi_2(9) = \chi_2(3^2) = \chi_2^2(3) = -1$ (and we have $\chi_2(1) = 1$). This completes the second character. To compute the third character we can set $\chi_3(3) = \omega^2 = -1$ then $\chi_3(7) = \chi_3(3^3) = \chi_3^3(3) = -1$ $\chi_3(9) = \chi_3(3^2) = \chi_3^2(3) = 1$ (and we have $\chi_3(1) = 1$). This completes the third character. Finally, to compute the fourth character we set $\chi_4(3) = \omega^3 = -i$ then $\chi_4(7) = \chi_4(3^3) = \chi_4^3(3) = i$ $\chi_4(9) = \chi_4(3^2) = \chi_4^2(3) = -1$ (and we have $\chi_4(1) = 1$). This completes the fourth character. k = 11 We have $\varphi(11) = 10$ so there are ten Dirichlet characters in this case. We consider the ten roots of unity $\omega = e^{2 \pi i/10} = e^{\pi i/5}$ $\omega^2 = e^{2 \pi i/5}$ $\omega^3 = e^{3 \pi i/5}$ $\omega^4 = e^{4 \pi i/5}$ $\omega^5 = e^{5 \pi i/5} = e^{\pi i} = -1$ $\omega^6 = -\omega$ $\omega^7 = -\omega^2$ $\omega^8 = -\omega^3$ $\omega^9 = -\omega^4$ $\omega^{10} = -\omega^5 = 1$ Note that the set of least positive residues mod $11$ is generated by $2$: $2 \equiv 2$ mod($11$) $2^2 \equiv 4$ mod($11$) $2^3 \equiv 8$ mod($11$) $2^4 \equiv 5$ mod($11$) $2^5 \equiv 10$ mod($11$) $2^6 \equiv 9$ mod($11$) $2^7 \equiv 7$ mod($11$) $2^8 \equiv 3$ mod($11$) $2^9 \equiv 6$ mod($11$) $2^{10} \equiv 1$ mod($11$) Therefore the non-principal Dirichlet characters will be completely determined by the values of $\chi(2)$. If we set $\chi_2(2) = \omega$ then $\chi_2(3) = \chi_2(2^8) = \chi_2^8(2) = \omega^8 = -\omega^3$ $\chi_2(4) = \chi_2(2^2) = \chi_2^2(2) = \omega^2$ $\chi_2(5) = \chi_2(2^4) = \chi_2^4(2) = \omega^4$ $\chi_2(6) = \chi_2(2^9) = \chi_2^9(2) = \omega^9 = -\omega^4$ $\chi_2(7) = \chi_2(2^7) = \chi_2^7(2) = \omega^7 = -\omega^2$ $\chi_2(8) = \chi_2(2^3) = \chi_2^3(2) = \omega^3$ $\chi_2(9) = \chi_2(2^6) = \chi_2^6(2) = \omega^6 = -\omega$ $\chi_2(10) = \chi_2(2^5) = \chi_2^5(2) = \omega^5 = -1$ (and we have $\chi_2(1) = 1$). This completes the second character. To compute the third character we can set $\chi_3(2) = \omega^2$ then $\chi_3(3) = \chi_3(2^8) = \chi_3^8(2) = \omega^{16} = -\omega$ $\chi_3(4) = \chi_3(2^2) = \chi_3^2(2) = \omega^4$ $\chi_3(5) = \chi_3(2^4) = \chi_3^4(2) = \omega^8 = -\omega^3$ $\chi_3(6) = \chi_3(2^9) = \chi_3^9(2) = \omega^{18} = -\omega^3$ $\chi_3(7) = \chi_3(2^7) = \chi_3^7(2) = \omega^{14} = \omega^4$ $\chi_3(8) = \chi_3(2^3) = \chi_3^3(2) = \omega^6 = -\omega$ $\chi_3(9) = \chi_3(2^6) = \chi_3^6(2) = \omega^{12} = \omega^2$ $\chi_3(10) = \chi_3(2^5) = \chi_3^5(2) = \omega^{10} = 1$ (and we have $\chi_3(1) = 1$). This completes the third character. To compute the fourth character we can set $\chi_4(2) = \omega^3$ then $\chi_4(3) = \chi_4(2^8) = \chi_4^8(2) = \omega^{24} = \omega^4$ $\chi_4(4) = \chi_4(2^2) = \chi_4^2(2) = \omega^6 = -\omega$ $\chi_4(5) = \chi_4(2^4) = \chi_4^4(2) = \omega^{12} = \omega^2$ $\chi_4(6) = \chi_4(2^9) = \chi_4^9(2) = \omega^{27} = -\omega^2$ $\chi_4(7) = \chi_4(2^7) = \chi_4^7(2) = \omega^{21} = \omega$ $\chi_4(8) = \chi_4(2^3) = \chi_4^3(2) = \omega^9 = -\omega^4$ $\chi_4(9) = \chi_4(2^6) = \chi_4^6(2) = \omega^{18} = -\omega^3$ $\chi_4(10) = \chi_4(2^5) = \chi_4^5(2) = \omega^{15} = -1$ (and we have $\chi_4(1) = 1$). This completes the fourth character. To compute the fifth character we can set $\chi_5(2) = \omega^4$ then $\chi_5(3) = \chi_5(2^8) = \chi_5^8(2) = \omega^{32} = \omega^2$ $\chi_5(4) = \chi_5(2^2) = \chi_5^2(2) = \omega^8 = -\omega^3$ $\chi_5(5) = \chi_5(2^4) = \chi_5^4(2) = \omega^{16} = -\omega$ $\chi_5(6) = \chi_5(2^9) = \chi_5^9(2) = \omega^{36} = -\omega$ $\chi_5(7) = \chi_5(2^7) = \chi_5^7(2) = \omega^{28} = -\omega^3$ $\chi_5(8) = \chi_5(2^3) = \chi_5^3(2) = \omega^{12} = \omega^2$ $\chi_5(9) = \chi_5(2^6) = \chi_5^6(2) = \omega^{24} = \omega^4$ $\chi_5(10) = \chi_5(2^5) = \chi_5^5(2) = \omega^{20} = 1$ (and we have $\chi_5(1) = 1$). This completes the fifth character. To compute the sixth character we can set $\chi_6(2) = \omega^5 = -1$ then $\chi_6(3) = \chi_6(2^8) = \chi_6^8(2) = 1$ $\chi_6(4) = \chi_6(2^2) = \chi_6^2(2) = 1$ $\chi_6(5) = \chi_6(2^4) = \chi_6^4(2) = 1$ $\chi_6(6) = \chi_6(2^9) = \chi_6^9(2) = -1$ $\chi_6(7) = \chi_6(2^7) = \chi_6^7(2) = -1$ $\chi_6(8) = \chi_6(2^3) = \chi_6^3(2) = -1$ $\chi_6(9) = \chi_6(2^6) = \chi_6^6(2) = 1$ $\chi_6(10) = \chi_6(2^5) = \chi_6^5(2) = -1$ (and we have $\chi_6(1) = 1$). This completes the sixth character. To compute the seventh character we can set $\chi_7(2) = \omega^6 = -\omega$ then $\chi_7(3) = \chi_7(2^8) = \chi_7^8(2) = \omega^8 = -\omega^3$ $\chi_7(4) = \chi_7(2^2) = \chi_7^2(2) = \omega^2$ $\chi_7(5) = \chi_7(2^4) = \chi_7^4(2) = \omega^4$ $\chi_7(6) = \chi_7(2^9) = \chi_7^9(2) = -\omega^9 = \omega^4$ $\chi_7(7) = \chi_7(2^7) = \chi_7^7(2) = -\omega^7 = \omega^2$ $\chi_7(8) = \chi_7(2^3) = \chi_7^3(2) = -\omega^3$ $\chi_7(9) = \chi_7(2^6) = \chi_7^6(2) = \omega^6 = -\omega$ $\chi_7(10) = \chi_7(2^5) = \chi_7^5(2) = -\omega^5 = 1$ (and we have $\chi_7(1) = 1$). This completes the seventh character. To compute the eighth character we can set $\chi_8(2) = \omega^7 = -\omega^2$ then $\chi_8(3) = \chi_8(2^8) = \chi_8^8(2) = \omega^{16} = -\omega$ $\chi_8(4) = \chi_8(2^2) = \chi_8^2(2) = \omega^4$ $\chi_8(5) = \chi_8(2^4) = \chi_8^4(2) = \omega^8 = -\omega^3$ $\chi_8(6) = \chi_8(2^9) = \chi_8^9(2) = -\omega^{18} = \omega^3$ $\chi_8(7) = \chi_8(2^7) = \chi_8^7(2) = -\omega^{14} = -\omega^4$ $\chi_8(8) = \chi_8(2^3) = \chi_8^3(2) = -\omega^6 = \omega$ $\chi_8(9) = \chi_8(2^6) = \chi_8^6(2) = \omega^{12} = \omega^2$ $\chi_8(10) = \chi_8(2^5) = \chi_8^5(2) = -\omega^{10} = -1$ (and we have $\chi_8(1) = 1$). This completes the eighth character. To compute the ninth character we can set $\chi_9(2) = \omega^8 = -\omega^3$ then $\chi_9(3) = \chi_9(2^8) = \chi_9^8(2) = \omega^{24} = \omega^4$ $\chi_9(4) = \chi_9(2^2) = \chi_9^2(2) = \omega^6 = -\omega$ $\chi_9(5) = \chi_9(2^4) = \chi_9^4(2) = \omega^{12} = \omega^2$ $\chi_9(6) = \chi_9(2^9) = \chi_9^9(2) = -\omega^{27} = \omega^2$ $\chi_9(7) = \chi_9(2^7) = \chi_9^7(2) = -\omega^{21} = -\omega$ $\chi_9(8) = \chi_9(2^3) = \chi_9^3(2) = -\omega^9 = \omega^4$ $\chi_9(9) = \chi_9(2^6) = \chi_9^6(2) = \omega^{18} = -\omega^3$ $\chi_9(10) = \chi_9(2^5) = \chi_9^5(2) = -\omega^{15} = 1$ (and we have $\chi_9(1) = 1$). This completes the ninth character. Finally, to compute the tenth character we set $\chi_{10}(2) = \omega^9 = -\omega^4$ then $\chi_{10}(3) = \chi_{10}(2^8) = \chi_{10}^8(2) = \omega^{32} = \omega^2$ $\chi_{10}(4) = \chi_{10}(2^2) = \chi_{10}^2(2) = \omega^8 = -\omega^3$ $\chi_{10}(5) = \chi_{10}(2^4) = \chi_{10}^4(2) = \omega^{16} = -\omega$ $\chi_{10}(6) = \chi_{10}(2^9) = \chi_{10}^9(2) = -\omega^{36} = \omega$ $\chi_{10}(7) = \chi_{10}(2^7) = \chi_{10}^7(2) = -\omega^{28} = \omega^3$ $\chi_{10}(8) = \chi_{10}(2^3) = \chi_{10}^3(2) = -\omega^{12} = -\omega^2$ $\chi_{10}(9) = \chi_{10}(2^6) = \chi_{10}^6(2) = \omega^{24} = \omega^4$ $\chi_{10}(10) = \chi_{10}(2^5) = \chi_{10}^5(2) = -\omega^{20} = -1$ (and we have $\chi_{10}(1) = 1$). This completes the tenth character. The remaining ten Dirichlet character tables which follow in this note can be calculated analogously (the details are not shown). k = 12 k = 13 k = 14 k = 15 k = 16 k = 17 k = 18 k = 19 k = 20 k = 21	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 372, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9909519553184509, "perplexity": 440.38767184331317}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-18/segments/1555578806528.96/warc/CC-MAIN-20190426133444-20190426155444-00206.warc.gz"}
https://astarmathsandphysics.com/gcse-maths-notes/638-pythagoras-theorem.html	## Pythagoras Theorem With Pythagoras theorem we can find the lengths of a side in a right angled triangle given the other two sides. With the sides of the triangle as labelled above, we use the formula For example It is important to label the sides of the triangle a, b and c in the way shown above. If you remember that c is the longest side then labelling becomes easier. Then you must use these values in their correct position in the equation	{"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": 0.9037883877754211, "perplexity": 194.76765278211053}, "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/1560627998755.95/warc/CC-MAIN-20190618143417-20190618165417-00517.warc.gz"}
http://alpheratz.net/about/	Aperiodic Wanderings are occasional topics that rise to my conscious brain or catch my fancy, often of an astronomical nature. The timing is not quite random, not quite chaotic, not quite periodic. Aperiodic. (Sometimes it is helpful to learn about a set of things by learning about a set of things that it is not.) #### YAB So, the fundamental question: why add yet another blog to the clotted blogosphere? This is my answer. How did I arrive at this point? This is that story. #### Mathematics I use math, where it seems natural to do so. It is a compact, efficient, therefore useful language. You, Dear Reader, have no need to fear it, and for heaven’s sake do not assume you are somehow inferior if you don’t quite catch all of the “words.” Not a math person, you say? I beg to differ. If you are human (and not an ideologue), you are a math person. “But math is hard!” So is learning to ride a bike, or to be in a relationship, or to throw a frisbee, or to cook an egg sunny side up, or to read—all harder than learning basic arithmetic and a little algebra, some trig, maybe a bit of calculus. On the other hand, you should be able to understand these posts even if you skip the math bits. Generally, but not always, math is confined to blue-ish boxes. If the article is still not making sense to you, then it is likely due to unclear writing. Complain. I’ll fix it. #### Pictures and Graphics… …are by the author (me!), unless otherwise specified. #### Typography The font I’ve settled on is Mercury ScreenSmart from typography.com.	{"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": 0.8275959491729736, "perplexity": 2508.8056523558344}, "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-2017-13/segments/1490218189472.3/warc/CC-MAIN-20170322212949-00265-ip-10-233-31-227.ec2.internal.warc.gz"}
http://mathoverflow.net/questions/13784/is-there-a-way-to-analytically-compute-the-recurrence-time-of-a-finite-markov-pro	# Is there a way to analytically compute the recurrence time of a finite Markov process? Let $X_t$ be an ergodic (time-homogeneous) Markov process (in discrete or continuous time) on a finite state space $\{1,\dots,n\}$. Let $T(X_0)$ be the stopping time given by the infimum of times such that $X$ has covered the space (i.e., for all $j$ with $1 \le j \le n$ there exists some $t_j \le T(X_0)$ s.t. $X_{t_j} = j$) and $X_{T(X_0)} = X_0$. Clearly $T(X_0)$ dominates the cover time of $X$. I would expect it to be dominated in turn by the sum of the cover time and the expected hitting time of $X_0$ starting from a state chosen w/r/t the invariant distribution $p$. Define the recurrence time as $\sum_{X_0} p(X_0) \cdot \mathbb{E}_{X_0} T(X_0)$, where again the first term is the invariant distribution of $X$. Now it has been quite a while (early 2000s) since I looked at cover and hitting times, but I recall that while the fundamental matrix (in discrete time) or the "deviation matrix" (in continuous time) give lots of information about hitting times, computing the cover time is hard. I am aware of the Matthews bound, but I do not know of a simpler way to compute the cover time than by simulating the chain. In particular, I don't know of an analytical approach to this quantity. I am in the same situation w/r/t the recurrence time, and it is this quantity that interests me much more than the cover time per se. But both are of some interest/utility to me. So my questions are: 1. Has the recurrence time (or a similar quantity besides the cover time or first return time) been treated anywhere? 2. Are there known analytical results on computing or at least (besides the Matthews bound) bounding cover times or recurrence times? - The coupon-collecting approach is exponential. Does that count as analytic? – Douglas Zare Feb 3 '10 at 16:04 Can you provide a reference? – Steve Huntsman Feb 3 '10 at 17:04 arxiv.org/abs/1001.0609 – Steve Huntsman Mar 2 '10 at 16:13 arxiv.org/abs/1004.4371 – Steve Huntsman Jul 29 '10 at 21:26 This is a response to a comment. The coupon collector's problem is elementary. I don't have a particular scholarly reference in mind, but rather the technique of the proofs. There are a few proofs of the $n H_n$ expected time to collect all coupons. One possibility is that you can compute the expected time to collect the $k$th new coupon, $n/(n-k+1)$. That uses a lot of symmetry you don't have for a general Markov process. Here, you have transition probabilities and times on (current location, subset visited so far). Analogous to what I did here, you can use inclusion-exclusion. The expected time to cover everything (with discrete time) is the sum of the probability that you haven't covered everything by time $t-1$, which you can express as $$\sum_t \sum_{S\subset V} -1^{\|S\|+1}Prob(\{X_i\}_{i\lt t}\cap S = \emptyset)$$ where $V$ = $\{1,...,n\}$. You can switch the order of summation to get about $2^n$ analytically solvable problems about avoiding particular subsets. $$\sum_{S\subset V} -1^{\|S\|+1} A(S)$$ where $A(S)$ = expected time before you first enter $S$. The same holds for continuous time. - Those sums should be over the nonempty subsets. – Douglas Zare Feb 4 '10 at 8:07 Douglas, thanks. Do you know of a reference that deals with this approach for generic finite ergodic Markov processes? – Steve Huntsman Feb 4 '10 at 14:49 Sorry, I don't. The method of estimating the time to visit the kth new vertex of a graph is present in "Short Random Walks On Graphs" by Greg Barnes and Uriel Feige. – Douglas Zare Feb 4 '10 at 15:46 http://arxiv.org/abs/1004.4371 might be useful. A note regarding your remark: "Clearly T(X0) dominates the cover time of X. I would expect it to be dominated in turn by the sum of the cover time and the expected hitting time of X0 starting from a state chosen w/r/t the invariant distribution p." This is false (unless I'm misreading it). If you start from 0 on the discrete interval {0,1,..,n}, then after covering you have to return to 0 from n, which takes longer than returning to 0 from a uniformly random point. - Thanks; I caught this myself, see comments to my question. – Steve Huntsman Sep 1 '10 at 9:22 Also I had in mind a cover time of the form used in this paper, otherwise you're quite correct. – Steve Huntsman Sep 1 '10 at 9:23	{"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": 0.8676962852478027, "perplexity": 412.5468368081442}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448398464536.35/warc/CC-MAIN-20151124205424-00170-ip-10-71-132-137.ec2.internal.warc.gz"}
https://www.physicsforums.com/threads/pratice-exam-confusion-showing-the-sumation-is-equal-to-the-other-wee-fun.117176/	# Pratice exam confusion! Showing the sumation is equal to the other! wee fun! 1. Apr 10, 2006 ### mr_coffee Hello everyone, i'm studying for my exam. He gave us anice little pratice exam. Am i thinking too much about this one? Part (a) seems to easy. http://img235.imageshack.us/img235/1554/lastscan9dx.jpg [Broken] I think i'm trying to make x^(n-c) be x^(n) well to do this, i would add c to (n-c), but if u do this, you will have to subtract c from the indicies (the n under the sumation). So that would change, n = b, to n=b-c. Is that all i had to do here? Also in part b, i'm lost now! The directions say: show that a2 = 0, a3 = 0, and that the recurrance relation for this differential equation is: a_(n+4) = -a_n/((n+3)(n+4)) I've been using this as a referance to help me:http://tutorial.math.lamar.edu/AllBrowsers/3401/SeriesSolutions.asp [Broken] http://img126.imageshack.us/img126/2601/lastscan29ir.jpg [Broken] THanks! Last edited by a moderator: May 2, 2017 2. Apr 11, 2006 ### HallsofIvy Staff Emeritus I think it is simpler to follow if you change indices- don't use "n" for both. Let j= n-c. Then n= j+ c so an= aj+c, f(n)= f(j+ c) and xn-c= xj. Also, when n= b, j= b-c. Of course, when n= $\infty$, j= $\infty$. That is: $$\Sigma_{n=b}^\infty a_n f(n) x^{n-c}= \Sigma{j= b-c}^{\infty}a_{j-c}f(j-c}x^j$$. Now, just change the index in the second sum from j to n- it doesn't change the value. If $y= \Sum_{n=0}^\infty a_n x^n$ then $$y"= \Sum_{n=2}^\infty n(n-1)a_n x^{n-2}$$ (the sum starts at n= 2 because if n= 0 or n= 1 the term n(n-1) is 0). Putting that into the equation y"- x2y= 0, we get: $$\Sigma_{n=2}^\infty n(n-1)a_n x^{n-2}- \Sigma_{n=0}^\infty a_n x^{n+2}$$ In the first sum, let j= n-2 so n= j+ 2. The sum becomes $$\Sigma_{j=0}^\infty (j+2)(j+1)a_{j+2}x^j$$ In the second sum, let j= n+2 so n= j- 2. The sum becomes $$\Sigma_{j= 2}^\infty a_{j-2}x^j$$ The point of that is, of course, to have the same exponent on x in both sums so we can "combine like terms". Since the second sum doesn't start until j= 2, we can write j=0 and j= 1 terms in the first sum separately: $$2a_2+ 6a_3x+ \Sigma_{j=2}^{\infty}((j+2)(j+1)a_{j+2}- a{j-2})x^j= 0$$ From that, it should be obvious that 2a2= 0 and 6a3= 0. 3. Apr 13, 2006 ### mr_coffee Thanks again Ivey!! it makes perfect sense, sorry about the delayed responce! Similar Discussions: Pratice exam confusion! Showing the sumation is equal to the other! wee fun!	{"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": 0.9343361258506775, "perplexity": 2867.213686684593}, "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-20170922022225-20170922042225-00081.warc.gz"}
https://socratic.org/questions/593d66e17c0149793e5061b6	Chemistry Topics A 25*g mass of a mixture of calcium carbonate and sodium hydroxide is subjected to fierce heat. What are the masses of each salt in the mixture? Jun 14, 2017 You have not given us all of the question......... Explanation: We know that we can decarboxlyate calcium carbonate with fierce heat....... $C a C {O}_{3} \left(s\right) + \Delta \rightarrow C a O \left(s\right) + C {O}_{2} \left(g\right) \uparrow$ We have a $25 \cdot g$ initial mass of $\text{calcium carbonate}$ and $\text{sodium hydroxide}$. We need a FINAL mass that represents the mass of $\text{calcium oxide}$ and $\text{sodium hydroxide}$ to address your question. The LOST mass represents what? Impact of this question 323 views around the world You can reuse this answer	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 6, "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": 0.9232546091079712, "perplexity": 1965.5470045434495}, "config": {"markdown_headings": false, "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-2020-40/segments/1600400191780.21/warc/CC-MAIN-20200919110805-20200919140805-00388.warc.gz"}
http://mathhelpforum.com/trigonometry/124677-proving-trig-identity.html	# Math Help - Proving Trig. Identity 1. ## Proving Trig. Identity Hey Guyzzz, I have two Identities given and I have to prove either one. 1) $\frac{1 - tan^2 x}{1 + tan^2 x} = cos 2x$ OR 2) $sin x +sin x cot^2 x = sec x$ Thank you. 2. Originally Posted by MordernWar2 2) $sin x +sin x cot^2 x = sec x$ I get. $\sin (x) +\sin (x) \cot^2 (x)$ $\sin (x) +\sin (x) \frac{\cos^2 (x)}{\sin^2(x)}$ $\sin (x) + \frac{\cos^2 (x)}{\sin(x)}$ $\frac{\sin^2 (x)}{\sin(x)} + \frac{\cos^2 (x)}{\sin(x)}$ $\frac{\sin^2 (x) + \cos^2 (x)}{\sin(x)}$ $\frac{1}{\sin(x)}$ $ \csc(x) $ 3. Hello, MordernWar2! $1)\;\;\frac{1 - \tan^2\!x}{1 + \tan^2\!x} \:=\: \cos 2x$ We have: . $\frac{1-\dfrac{\sin^2\!x}{\cos^2\!x}}{1 + \dfrac{\sin^2\!x}{\cos^2\!x}}$ Multiply by $\frac{\cos^2\!x}{\cos^2\!x}; \quad \frac{\cos^2\!x\left(1 - \dfrac{\sin^2\!x}{\cos^2\!x}\right)}{\cos^2\!x\lef t(1 + \dfrac{\sin^2\!x}{\cos^2\!x}\right)}$ . . . . . . . . $=\;\;\frac{\overbrace{\cos^2\!x - \sin^2\!x}^{\text{This is }\cos2x}}{\underbrace{\cos^2\!x + \sin^2\!x}_{\text{This is 1}}} \;\;=\;\; \cos2x$ $2)\;\;\sin x +\sin x\cot^2\!x \:=\: \sec x$ This is not an identity . . . I suspect a typo. It is probably: . $\sin x + \sin x\cot^2\!x \:=\: {\color{blue}\csc x}$ $\text{Factor the left side: }\:\sin x\underbrace{(1 + \cot^2\!x)}_{\text{This is }\csc^2x} \;\;=\;\;\sin x\csc^2\!x$ . . . . . . . . . . . . $=\;\;\underbrace{(\sin x\csc x)}_{\text{This is 1}}\csc x \;\;=\;\;csc x$	{"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": 18, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9982369542121887, "perplexity": 1696.6304049426244}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-30/segments/1469257824201.28/warc/CC-MAIN-20160723071024-00312-ip-10-185-27-174.ec2.internal.warc.gz"}
http://mathhelpforum.com/calculus/1638-concentration-brine-solution-print.html	# Concentration in brine solution • Jan 15th 2006, 10:13 PM jacs 1 Attachment(s) Concentration in brine solution A tank contains 100 litres of brine whose concentration is 3 grams/litres. Three litres of brine whose concentration is 2 grams/litre flow into the tank each minute and at the same time 3 litres of mixture flow out each minute. Show that the quantity of salt, Q gram, in the tank at any time t is given by: Q = 200 + 100e^(-0.03t) i have managed to get Q = 200 - 100e^(-0.03t) and cannot figure out why the minus is there, i have included the pdf of the working i have done so far, any help appreciated thanks jacs • Jan 16th 2006, 11:25 AM CaptainBlack Quote: Originally Posted by jacs A tank contains 100 litres of brine whose concentration is 3 grams/litres. Three litres of brine whose concentration is 2 grams/litre flow into the tank each minute and at the same time 3 litres of mixture flow out each minute. Show that the quantity of salt, Q gram, in the tank at any time t is given by: Q = 200 + 100e^(-0.03t) i have managed to get Q = 200 - 100e^(-0.03t) and cannot figure out why the minus is there, i have included the pdf of the working i have done so far, any help appreciated thanks jacs In your solution you get to a line: $t=-\frac{100}{3}\ln(600-3Q)+c$. But now at $t=0,\ Q=300$ putting this into the above equation gives a term $\ln(-300)$ not $\ln(300)$ which you have. To avoid this change your integral to: $t=-\frac{100}{3} \int \frac{3}{3Q-600}dQ$, or: $-0.03t=\ln(3Q-600)+C$. Which if I have done this right :( should now allow you to get the	{"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": 0.8514691591262817, "perplexity": 923.7209855184772}, "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-2016-44/segments/1476988721067.83/warc/CC-MAIN-20161020183841-00274-ip-10-171-6-4.ec2.internal.warc.gz"}
https://pastel.archives-ouvertes.fr/pastel-00749853	# Evaluation of the pi0 background for the nu_e appearance search in the T2K experiment. First study of the leptonic CP violation phase. 1 T2K LLR - Laboratoire Leprince-Ringuet Abstract : The T2K experiment is an off-axis long baseline neutrino flavour oscillation experiment. In T2K, the neutrino beam produced by an accelerator at J-PARC, Tokai, is detected using the Super Kamiokande (SK) detector, Kamioka, located 295 km away from Tokai. The accelerator produce a very pure nu_mu beam. During the propagation of neutrinos between their production point and SK they may change flavour. This effect is the neutrino flavour oscillation. The main objective of T2K is to measure some parameters describing the neutrino flavour oscillation, in particular of theta13 and deltaCP. This determination is performed by measuring the nu_mu -> nu_e oscillation probability, via the nu_e appearance search. In 2011, T2K was the first experiment to observe nu_e appearance and published a 2.5 sigma evidence that theta13 is not null. In this thesis we present the work done on the evaluation of the systematic error on the pi0 background reconstruction efficiency. The pi0 background is one of the main background sources affecting the nu_e appearance search through the detection of nu_e charged current quasi-elastic events at SK. Two different and complementary approaches have been developed. First, we have started developing a new device that partially reproduces the topology of physics events. Second, we have created a specific sample to estimate the systematic uncertainty of pi0 events reconstruction efficiency. This result was used in the official 2011 nu_e search. Furthermore, results from this sample will be used in future extended oscillation analysis. We also present the first study of the leptonic CP violation phase (deltaCP) using T2K nu_e data, assuming the recently measured value of theta13 by reactor experiments. Keywords : Document type : Theses High Energy Physics - Experiment [hep-ex]. Ecole Polytechnique X, 2012. English https://pastel.archives-ouvertes.fr/pastel-00749853 Contributor : João Pedro Athayde Marcondes de André <> Submitted on : Thursday, November 8, 2012 - 2:52:54 PM Last modification on : Monday, October 13, 2014 - 3:43:25 PM ### Identifiers • HAL Id : pastel-00749853, version 1 ### Citation João Pedro Athayde Marcondes de André. Evaluation of the pi0 background for the nu_e appearance search in the T2K experiment. First study of the leptonic CP violation phase.. High Energy Physics - Experiment [hep-ex]. Ecole Polytechnique X, 2012. English. <pastel-00749853> Consultation de la notice ## 380 Téléchargement du document	{"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": 0.8545692563056946, "perplexity": 4919.1291245073335}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448398446286.32/warc/CC-MAIN-20151124205406-00323-ip-10-71-132-137.ec2.internal.warc.gz"}
http://mathhelpforum.com/calculus/24072-proof-equal-area-under-curve.html	# Thread: Proof of equal area under curve 1. ## Proof of equal area under curve 1.) Without evaluating the integrals, prove that the following areas under the function's curves are equal: $f(x) = \frac{x}{(1+x^4)}$ on the interval of 0 to 2 and $g(x) = \frac{1}{2(1+x^2)}\$ on the interval of 0 to 4 I've no idea where to start on this one... not without evaluating integrals. I do have their graphs to look at, but I'm not seeing anything obvious. Any help is appreciated... 2. Originally Posted by ebonyscythe $f(x) = \frac{x}{(1+x^4)}$ on the interval of 0 to 2 and $g(x) = \frac{1}{2(1+x^2)}\$ on the interval of 0 to 4 It's just proving that $\int_0^2 {\frac{x} {{\left( {1 + x^4 } \right)}}\,dx} = \frac12\int_0^4 {\frac{{dx}} {{1 + x^2 }}\,dx} .$ And this does not require big effort, so for the left integral substitute $u=x^2$ and you're done. 3. So that's not evaluating the integrals then? That's what I was mostly wondering, but now that I think about it, you're not technically evaluating the integrals, just showing that they are equal to each other... Alright, thank you very much. I wanted to make sure there wasn't some "other" way to do it. 4. Sure, it's not necessary to evaluate them. Just a little substitution shows that they're equal to each other.	{"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": 0.8695221543312073, "perplexity": 381.034513362212}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934806447.28/warc/CC-MAIN-20171122012409-20171122032409-00137.warc.gz"}
http://mathhelpforum.com/calculus/76628-confusing-integral.html	# Math Help - Confusing integral 1. ## Confusing integral How do I start this? 2. Let $t=\sin^{-1}x\Rightarrow dt = \frac{dx}{\sqrt{1-x^2}}$ so $\int \frac{xe^{\sin^{-1}x}}{\sqrt{1-x^2}}dx=\int \sin x e^x dx=I$ and $I=e^x\sin x-\int \cos x e^xdx = e^x \sin x - (\cos xe^x + \int \sin x e^x dx)$ $=e^x(\sin x - \cos x)-I\Rightarrow I =e^x\frac{\sin x -\cos x }{2}+C$. 3. ^ I lost you at I ? 4. $I=e^x\sin x -e^x \cos x -I \Rightarrow 2I=e^x\sin x - e^x \cos x$ 5. Hello, VkL! Abu-Khalil is correct . . . (he could have been clearer) $20)\;\;\int\frac{xe^{\sin^{-1}\!x}}{\sqrt{1-x^2}}\,dx$ We have: . $\int x\cdot e^{\sin^{1}x}\cdot\frac{dx}{\sqrt{1-x^2}}$ Let: . $t \:=\: \sin^{-1}\!x \quad\Rightarrow\quad dt \:=\:\frac{dx}{\sqrt{1-x^2}}$ Note that: . $x \:=\:\sin t$ Substitute: . $\int \sin t\cdot e^t\cdot dt$ We will integrate by parts: . $I \;=\;\int e^t\sin t\,dt$ . . $\begin{array}{ccccccc}u &=& \sin t & & dv &=& e^t\,dt \\ du &=& \cos t\,dt & & v &=& e^t\end{array}$ We have: . $I \;=\;e^t\sin t - \int e^t\cos t\,dt$ . . By parts: . $\begin{array}{ccccccc}u &=& \cos t & & dv&=& e^t\,dt \\ du &=& -\sin t\,dt & & v &=& e^t \end{array}$ And we have: . $I \;=\;e^t\sin t - \bigg[e^t\cos t + \int e^t\sin t\,dt\bigg]$ . . . . . . . . . $I \;=\;e^t\sin t - e^t\cos t - \underbrace{\int e^t\sin t\,dt}_{\text{This is }I} + C$ So we have: . $I \;=\;e^t\sin t - e^t\cos t - I + C$ . . . . . . . . . $2I \;=\;e^t\sin t - e^t\cos t + C$ . . . . . . . . . . $I \;=\;\tfrac{1}{2}\left(e^t\sin t - e^t\cos t\right) + C$ Hence: . $\int e^t\sin t\,dt \;=\;\tfrac{1}{2}\,e^t(\sin t - \cos t) + C$ Back-substitute: . $t \,=\,\sin^{-1}\!x$ We have: . $\frac{1}{2}e^{\sin^{-1}\!x}\bigg[\sin\left(\sin^{-1}\!{x}\right) - \cos\left(\sin^{-1}\!x\right)\bigg] + C$ . . . . . . . $= \;\frac{1}{2}\,e^{\sin^{-1}\!x}\bigg[x - \sqrt{1-x^2}\bigg] + C$	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 23, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9988518357276917, "perplexity": 689.9739967812544}, "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-2014-49/segments/1416931012025.85/warc/CC-MAIN-20141125155652-00052-ip-10-235-23-156.ec2.internal.warc.gz"}
https://zbmath.org/?q=an%3A0654.58031	× ## Elliptic operators, topology and asymptotic methods.(English)Zbl 0654.58031 Pitman Research Notes in Mathematics Series, 179. Harlow (UK): Longman Scientific & Technical; New York: John Wiley & Sons. 184 p. \$ 15.50 (1988). This book grew out of lecture notes, and presents a proof of the Atiyah- Singer index theorem for Dirac operators in the spirit of E. Getzler [Topology 25, 111-117 (1986; Zbl 0607.58040)]. The second approach alluded to in Getzler’s paper which combines his scaling argument and Patodi’s generalization of the well-known Minakshisundaram- Pleijel expansion of the heat kernel is completely worked out. Since there is more in this book let us briefly survey its content. Chapter 1 recalls the basic tools from differential geometry: the definition of connexions (connections) on vector bundles and principle bundles, the existence of “good” geodesic coordinates (which give simple representations for the metric tensor and the Christoffel symbols locally around a point), and the definitions of the Hodge-*-operator, the differential, and the codifferential. Chapter 2 introduces Clifford bundles and Dirac operators as developed by M. Gromov and H. B. Lawson jun. [Publ. Math., Inst. Hautes Etud. Sci. 58, 295-408 (1983; Zbl 0538.53047)]. The two main results are the formal self-adjointness of Dirac operators and the Bochner-Weitzenböck formula. Chapter 3 gives the elliptic theory of Dirac operators using Gårding’s inequality and classical a priori estimates in Sobolev spaces. Here the author follows the common approach via analysis on the tori $$T^ n$$ [cf. R. S. Palais et al., Seminar on the Atiyah- Singer index theorem (Ann. Math. Studies 57) (1965; Zbl 0137.170)]. The chapter ends with the functional calculus for bounded functions on the spectrum. This is applied in the next chapter to prove the Hodge decomposition theorem. Chapter 5 is central with its treatment of the heat and the wave equation. The main results are the existence and uniqueness of the heat kernel and the construction of its asymptotic expansion. The integral formula for the trace of the associated trace class operator $$e^{- t\Delta}$$ is given in chapter 6, and the famous Weyl formula for the asymptotic distribution of eigenvalues of the Laplacian on a compact manifold is deduced in chapter 7. Chapter 8 presents a proof of the Atiyah-Bott-Lefschetz-formula, and chapter 9 gives the supertrace formula for the index of a graded Dirac operator (which goes back to McKean and Singer). Chapter 10 collects the rudiments of characteristic classes which are necessary to state the index theorem properly, introduces the notion of spin manifolds together with Lichnerowicz’ specialization of the Bochner- Weitzenböck formula in this case, and provides the last ingredient for the proof of the index theorem, i.e. Mehler’s formula. The proof of the index theorem for Dirac operators makes up chapter 11. Also included is a proof of the Hirzebruch signature theorem and a sketch of a proof of the Chern-Gauß-Bonnet theorem as an exercise. It should be mentioned that each of these chapters is supplemented with a set of exercises which are all quite manageable for the graduate student. The last two chapters contain Witten’s approach to Morse theory and Atiyah’s $$L^ 2$$-index theorem for Galois coverings. This book can very well serve as a guide to a one-semester course introducing and proving the Atiyah-Singer index theorem for Dirac operators. It requires some background in differential geometry and functional analysis since the corresponding parts in the book are too selective on purpose. Reviewer: H.Schröder ### MSC: 58J20 Index theory and related fixed-point theorems on manifolds 58-02 Research exposition (monographs, survey articles) pertaining to global analysis 58A30 Vector distributions (subbundles of the tangent bundles) 58A10 Differential forms in global analysis 58A14 Hodge theory in global analysis 58J35 Heat and other parabolic equation methods for PDEs on manifolds 58C50 Analysis on supermanifolds or graded manifolds ### Citations: Zbl 0607.58040; Zbl 0538.53047; Zbl 0137.170	{"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": 0.8560826778411865, "perplexity": 569.3441531284676}, "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/1664030334644.42/warc/CC-MAIN-20220926020051-20220926050051-00502.warc.gz"}
https://electronics.stackexchange.com/questions/182694/how-do-we-have-a-fixed-current-in-this-case	# How do we have a fixed current in this case? If we are doing a current-to-voltage converter we can use negative feedback on a common-emitter configuration. Using npn transistors would look like so: We are using a fixed current input and it forces the base-emitter voltage to be the corresponding value according to the Ebers Moll model. So my question is, how can we make the current input exactly what we want so it is mirrored on the other side? I don't really know how it is possible to feed the exact current we want to the dual-transistors without having to calculate the voltage V1 in equilibrium for $I_c$. If we increase $I_c$ then $V_{BE}$ would increase, which would make the voltage drop across $R_1$ to decrease (in a smaller amount). In order for the control of the current to be exact, when would need to calculate the precise increase we need in $V_1$ to reach the desired $I_c$ current so the base-emitter voltage is exactly what is needed to drive the second transistor. How can we keep the current exactly how we need 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": 0.9188013672828674, "perplexity": 258.8993995193237}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585916.29/warc/CC-MAIN-20211024081003-20211024111003-00284.warc.gz"}
https://portonmath.wordpress.com/2017/03/27/conjecture-about-funcoids/	Conjecture Let $S$ be a set of binary relations. If for every $X, Y \in S$ we have $\mathrm{up} (X \sqcap^{\mathsf{FCD}} Y) \subseteq S$ then there exists a funcoid $f$ such that $S = \mathrm{up}\, f$.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 5, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9993201494216919, "perplexity": 56.600203566917}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267867304.92/warc/CC-MAIN-20180624234721-20180625014721-00351.warc.gz"}
https://www.physicsforums.com/threads/electric-current.2369/	Electric current • Start date Dx Hiya! I have a problem thats got me confused to solve. I wanna use ohms law but not right. Here is the problem. A nichrome wire has a radius 0f .5 mm and resistivity of 100 x 10^-8 ohm.m. If the wire carries a current of .5 A what is the voltage across the wire? The length of the wire is .32m E = .5A x 100 x 10^-8 ohm is what i have for setting it up but its not the answer. what do i need to do to solve for this? Dx Related Introductory Physics Homework Help News on Phys.org Tom Mattson Staff Emeritus Gold Member 5,475 20 Originally posted by Dx A nichrome wire has a radius 0f .5 mm and resistivity of 100 x 10^-8 ohm.m. Note: You were given the radius and the resistivity. You were not given the resistance. E = .5A x 100 x 10^-8 ohm is what i have for setting it up but its not the answer. The problem is that you plugged the resistivity in for the resistance. what do i need to do to solve for this? Calculate the resistance, then use E=IR. Dx Re: Re: electric current The problem is that you plugged the resistivity in for the resistance. Calculate the resistance, then use E=IR. [/B][/QUOTE] Ok i have the .5A but no E or R to perform ohms law with how do i solve for this please Tom. Dx, Tom Mattson Staff Emeritus Gold Member 5,475 20 Re: Re: Re: electric current Originally posted by Dx Ok i have the .5A but no E or R to perform ohms law with how do i solve for this please Tom. 1. They ask you for E. 2. They give you enough information to find R. Look up the deinfition of resistivity, and you will see how it is related to the dimensions of the wire. Come on, this is easy. You can do it. Dx So Happy! Thank You, Tom! You know i looked at that section numerous times but after taking a days break and reading what you wrote it was like a lightbulb lit up in my head. I cant believe I didnt see it earlier always the simple ones that seem to give me the most trouble. I found the formula R = p(L/A) then converted it to amps using ohms law. Much appreciated, you helped me alot! Dx • Last Post Replies 6 Views 512 • Last Post Replies 1 Views 11K • Last Post Replies 6 Views 2K • Last Post Replies 2 Views 581 • Last Post Replies 16 Views 2K • Last Post Replies 1 Views 6K • Last Post Replies 1 Views 1K • Last Post Replies 4 Views 1K	{"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": 0.8780256509780884, "perplexity": 1600.5750396737728}, "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-2019-47/segments/1573496669183.52/warc/CC-MAIN-20191117142350-20191117170350-00550.warc.gz"}
https://asmedc.silverchair.com/OMAE/proceedings-abstract/OMAE2017/57762/V008T11A013/282504	Stick-slip is one of the typical phenomenon which is observed in offshore drilling and considered as a critical problem for the drilling operation. The stick-slip makes a large fluctuation of drill bit rotation, even though the top of the drill pipe is rotating at a steady velocity and sometimes causes the damage of the drill bit. Additionally, it leads a crushing of the sediment layer which is a big problem especially for the scientific drilling [1][2][3]. The main purpose of the scientific drilling is to correct high quality core samples of sediment layers under the seabed. However, once the stick-slip occurs, it makes difficult to recover a high-quality sediment layer core sample. Therefore, it is necessary to detect the occurrence of stick-slip and its fundamental characteristics such as oscillation periods and amplitudes by simulation with the aid of surface drilling data, which can be monitored during the drilling operation to mitigate or prevent stick-slip. It would be advantageous to identify the characteristics of the stick-slip from the surface drilling data. The past study [4][5][6] investigated a numerical method to analyze the stick-slip by solving the NDDE (Neutral Delay Differential Equation) which is derived from torsional vibration equation. A small-scale model experiment was conducted in a water tank to observe the stick-slip phenomenon, and the result from the analytical model is evaluated with that obtained from the experiments. In this study, the numerical model is applied for the stick-slip analysis not only of the drill pipe model but also the actual drill pipe in operation. The solutions of the NDDE is depend on not the initial value but the initial history of the solution, because NDDE contains a delayed function term. Especially, the initial history settings have much effect on the numerical solution of NDDE in case of the actual drill pipe. Additionally, to solve the NDDE for stick-slip analysis, we must set some model parameters concerned with the frictional torque on drill bit. The present study investigated the effects of the initial history and the model parameters settings on numerical solutions in detail and presented an procedure to determine the appropriate settings of the initial history and the model parameters by reference to the measured top drive torque. This content is only available via PDF. You do not currently have access to this content.	{"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": 0.8403385281562805, "perplexity": 946.5795072276354}, "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-2023-06/segments/1674764500628.77/warc/CC-MAIN-20230207170138-20230207200138-00325.warc.gz"}
http://tex.stackexchange.com/questions/98554/how-to-increase-space-between-bibliography-without-using-biblatex	# How to increase space between bibliography without using biblatex Is there a way to increase space between bibliography without using biblatex? - I don't understand. Would you mind explaining a bit more? – Marc van Dongen Feb 17 '13 at 6:29 If you mean the space between the various references (\bibitems) this can be done as follows using etoolbox to patch the \thebibliography command. In normal styles the bibliography is just a list, but the parameters of the list are given when the bibliography is called (\begin{bibliography} which calls \thebibliography). \usepackage{etoolbox} \apptocmd{\thebibliography}{\setlength{\itemsep}{18pt}}{}{} The same idea can be used for other parameters of the list implementing the bibliography. An alternative is to use natbib which provides \bibsep (\setlength{\bibsep}{18pt}) to change the vertical space between references. \usepackage{natbib} \setlength{\bibsep}{18pt} PS replace 18pt with the measure you want. -	{"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": 0.990776777267456, "perplexity": 1952.4255607925554}, "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-2014-49/segments/1416400380409.19/warc/CC-MAIN-20141119123300-00241-ip-10-235-23-156.ec2.internal.warc.gz"}
https://www.gradesaver.com/textbooks/science/physics/physics-for-scientists-and-engineers-a-strategic-approach-with-modern-physics-3rd-edition/chapter-11-work-exercises-and-problems-page-305/44	## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition) (a) $v = \sqrt{\frac{2~F~h~cot(\theta) - 2mgh}{m}}$ (b) v = 4.0 m/s (a) We can find an expression for the distance $d$ that the crate moves along the slope. $\frac{h}{d} = sin(\theta)$ $d = \frac{h}{sin(\theta)}$ The work done by the force will be equal to the sum of the potential energy and the kinetic energy. $PE+KE = W$ $mgh + \frac{1}{2}mv^2 = F~d~cos(\theta)$ $\frac{1}{2}mv^2 = F~(\frac{h}{sin(\theta)})~cos(\theta) - mgh$ $v^2 = \frac{2~F~h~cot(\theta) - 2mgh}{m}$ $v = \sqrt{\frac{2~F~h~cot(\theta) - 2mgh}{m}}$ (b) We can use the expression in part (a) to find the speed at the top of the slope. $v = \sqrt{\frac{2~F~h~cot(\theta) - 2mgh}{m}}$ $v = \sqrt{\frac{(2)(25~N)(2.0~m)~cot(20^{\circ}) - (2)(5.0~kg)(9.80~m/s^2)(2.0~m)}{5.0~kg}}$ $v = 4.0~m/s$	{"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": 0.8986244797706604, "perplexity": 96.18895695958125}, "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/1531676591332.73/warc/CC-MAIN-20180719222958-20180720002958-00378.warc.gz"}
http://www.graphoilogy.com/2006/02/how-to-track-oil-production-curve.html	## Thursday, February 02, 2006 ### How to Track an Oil Production Curve Last time we explained how to build a logistic oil production profile using a Stochastic Bass Model which can be seen as a stochastic equivalent of the logistic curve used by peakoilers. This post is an attempt to apply this stochastic model so as to literally "track" the production curve and then project future production for the coming years. An interesting framework within which this goal can be achieved is called the particle filtering technique. Particle filtering can be seen as a generalization of the Kalman filter and is sometimes encountered under various names such as the bootstrap filter, the condensation method, the Bayesian filter or the sequential Monte-Carlo Markov Chain (MCMC). These techniques have been very sucessful in many applications such as the visual tracking of targets. Particle filtering is quite a complex technique and there is a fairly large body of literature on it. I don't wish to flood this post with equations rather try to provide some intuitive insights. In order to explain the particle filtering, I will use the following simple analogy: Imagine that you are in a totally dark room looking for a wall switch for the lights. The best way to find it is to randomly sample the walls with your hands in order to find an object that has a similar feeling under the touch. Of course, you have some prior knowledge on where to find the switch, its shape and your trajectory will probably evolve along the walls. Well, particle filtering is exactly the same. The equivalent of the particles are the random sampling of the wall with your hands around your current position. The information you are getting with your hands is the equivalent of the particles' importance weights. Now, let's look at some equations. Each particle will be an hypothetical oil production profile evolving according to a stochastic Bass model (SBM): Note that only the SBM growth rate (or adoption rate) will be a random variable. We also need to discretize the production growth so that it can only grow by a fix amount of oil (ex 0.1 Gb/year). The particle filtering is an iterative procedure where each iteration is composed of the following four steps: 1. Sampling step: we randomly choose N candidate particles from a proposal distribution (i.e. I choose random positions on the wall where to put my hands) 2. Reweighting step: we weight each particle with an importance weight (i.e. do I feel something under my hands?) 3. Resampling step: this step is optional but gives better results and avoid degeneracy of the particles. 4. Filtering step: we finally compute a mean state based on the new particle set (Monte-Carlo integration) The design of the proposal distribution for the sampling step is a delicate task and can be simplified by equating the proposal distribution to the Markovian transition probability of the SBM. In that case, the particle filter is called Bayesian Bootstrap Filter. Consequently, each particle n is independently evolving according to the following equations: The weighting step consists in assigning an importance weight to each particle according to their fit compared to the observed data: where Prod(t) is the observed production at time t. One hurdle is how to extend the tracking for future production where there is no data available. For now, we simply replace the particle weights by an uniform weight which is equivalent to simulate a stochastic Bass model. The resampling step consists in redistributing in place the initial set of particles according to their weight so that particles with bigger weights will be eventually duplicated. Once the particles have been resampled, we can estimate the mean state using a Monte-Carlo approach: The two figures below give an example of curve tracking on the US production using N=1,000 particles (alpha=5e-5, m= 223 Gb). Because the growth rate is the only stochastic parameter it has to fluctuate widely in order to follow closely the oberved curve variations. Fig. 1 Proposed method applied to the US production (EIA data) where the dark dots are the actual data, the thick orange line is the mean state of the particles and the two blue lines delineate the one sigma of the various particles' positions. Fig. 2 The growth rate beta_t is the only parameter of the particles which is stochastic and reflects how particles have to accelerate production growth in order to match the data. Production seemed to have grown faster in the 50s. After 2005, we have no data so the growth rate stays around its mean value. In summary, we are proposing to use a sequential Monte-Carlo Markov Chain for the modeling of the oil production. • The algorithm allows for a dynamic adaptation of the production parameters (growth rate and URR) instead of a static and global view of the parameters as in traditional curve fitting. • production shocks can be modeled. • the underlying Markov process is still based on a logistic model but a stochastic one. There are many improvements possible in particular around the design of the proposal distribution and the choice of the priors for the URR and the growth rate. For a good overview on particle filtering techniques: M. Sanjeev Arulampalam, Simon Maskell, Neil Gordon, and Tim Clapp, A Tutorial on Particle Filters for Online Nonlinear/Non-Gaussian Bayesian Tracking, IEEE TRANSACTIONS ON SIGNAL PROCESSING, VOL. 50, NO. 2, FEBRUARY 2002.	{"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": 0.851306140422821, "perplexity": 650.5093673248663}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-11/segments/1424936464088.46/warc/CC-MAIN-20150226074104-00316-ip-10-28-5-156.ec2.internal.warc.gz"}
https://infoscience.epfl.ch/record/146328	Infoscience Journal article A Novel Criterion for Classifiers Combination in Multistream Speech Recognition In this paper we propose a novel information theoretic criterion for optimizing the linear combination of classifiers in multi stream automatic speech recognition. We discuss an objective function that achieves a trade-off between the minimization of a bound on the Bayes probability of error and the minimization of the divergence between the individual classifier outputs and their combination. The method is compared with the conventional inverse entropy and minimum entropy combinations on both small and large vocabulary automatic speech recognition tasks. Results reveal that it outperforms other linear combination rules. Furthermore we discuss the advantages of the proposed approach and the extension to other (non-linear) combination rules.	{"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": 0.8195831179618835, "perplexity": 600.4726697177578}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-30/segments/1500549425751.38/warc/CC-MAIN-20170726022311-20170726042311-00719.warc.gz"}
https://stats.stackexchange.com/questions/253066/estimating-the-intercept-and-its-variance-in-a-moving-average-model	# Estimating the intercept and its variance in a moving average model Let $y_1,y_2,\dots,y_{10}$ be a time series generated by $$y_t=\delta+\epsilon_t+\theta\epsilon_{t-2}$$ where $\epsilon_t$ is white noise with $E[\epsilon_t]=0$, $Var(\epsilon_t)=\sigma^2$ and $\delta$, $\theta$ are unknown parameters. a) Propose a estimator for $\delta$ and find the variance of this estimator. I don't know well how to do the estimation in moving average models (MA), but I think that the process converge to the mean. $$y_t\rightarrow \mu$$ $$y_t\rightarrow \delta$$ Then the estimator is $$\hat{\delta}=\frac{1}{n}\sum_{t=1}^n y_t$$ $$Var(\hat{\delta})=Var\Big(\frac{1}{n}\sum_{t=1}^n y_t\Big)=\frac{n \mathrm{\gamma}(0) + \sum_{t=1}^{n-1} 2(n-t) \mathrm{\gamma}(t)} {n^2}$$ where $$\gamma(h)=Cov(y_t,y_{t-h})=Cov(\delta+\epsilon_t+\theta\epsilon_{t-2},\delta+\epsilon_{t-h}+\theta\epsilon_{t-h-2})$$ $$=Cov(\epsilon_t,\epsilon_{t-h})+\theta Cov(\epsilon_t,\epsilon_{t-h-2})+\theta Cov(\epsilon_{t-2},\epsilon_{t-h})+\theta^2Cov(\epsilon_{t-2},\epsilon_{t-h-h})$$ $$\gamma(h)=\begin{cases}\sigma^2(1+\theta^2) \qquad h=0\\ \theta\sigma^2 \qquad h=2 \\ 0 \qquad otherwise\end{cases}$$ Is it right? Is there a way to find explicity the mean and $\theta$ estimator at hand? • I think this is tricky: Note e_(t-1) is missing. So this is like a second order MA process with one parameter to be restricted to 0. There are issues relating to stationarity and finite moving averages have a autoregressive representation with an infinite number of parameters. So you can't rely on the AR(1) model to estimate the variance. Also in general there are restrictions on the parameters for stationarity and invertibility. Maybe you should look at Box and Jenkins classic book. – Michael Chernick Dec 23 '16 at 21:36 • If you search google using box-jenkins method, the wikipedia article cites the first edition (1970) of Box and Jenkins book along with three other useful references. – Michael Chernick Dec 23 '16 at 21:42 • @MichaelChernick Is necessary find the autoregressive representation to estimate the variance? I can't just use the covariance properties to find this variance, since that I have just $y_1,\dots,y_{10}$? – user72621 Dec 23 '16 at 23:04 • You can't estimate infinitely many AR parameters based on an n=10. My point is conceptually that the AR(1) model is not the way to approach estimating the process variance. – Michael Chernick Dec 23 '16 at 23:17 • @MichaelChernick It is not a general approach for any process? I mean, it's not just a variance propertie? $Var(\hat{\delta})=Var\Big(\frac{1}{n}\sum_{t=1}^n y_t\Big)=\frac{n \mathrm{\gamma}(0) + \sum_{t=1}^{n-1} 2(n-t) \mathrm{\gamma}(t)} {n^2}$ – user72621 Dec 23 '16 at 23:29	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8333244323730469, "perplexity": 459.81457851337154}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195526337.45/warc/CC-MAIN-20190719182214-20190719204214-00191.warc.gz"}
https://conan777.wordpress.com/2010/11/22/the-moving-needle-problem/	## The moving needle problem November 22, 2010 First, let’s clarify that this post has nothing to do with the Kakeya conjecture (except for the word ‘needle’ in it). Anyways, I was asked the following question via an e-mail from Charles this summer: (It turns out that the question was invented by Jonathan King and then communicated to Morris Hirsch and that’s where Charles heard about it from) In any case, I find the problem quite cute: Problem: Given a smoothly embedded copy of $\mathbb{R}$ in $\mathbb{R}^3$ containing $\{ (x,0,0) \ \| \ x \in (-\infty,-C] \cup [C, \infty) \}$. Is it always possible to continuously slide a unit length needle lying on the ray $(-\infty, -C]$ to the ray $[C, \infty)$, while keeping the head and tail of the needle on the curve throughout the process? i.e. the curve is straight once it passes the point $(-C, 0)$ and $(C,0)$, but can be bad between the two points: We are interested in sliding the needle from the neigative $x$-axis to the positive $x$-axis: Exercise: Try a few examples! It’s quite amusing to see that sometimes both ends of the needle needs to go back and forth along the curve many times, yet it always seem to get through. One should note that this is not possible if we just require the curve to be eventually straight and goes to infinity at both ends. As we can see on a simple ‘hair clip’ curve: The curve consists of two parallel rays of distance $<1$ apart, connected with a semicircle. A unit needle can never get from one ray to the other since the needle would have to rotate $180$ degrees and hence it has to be vertical at some point in the process, but no two points on the curve has vertical distance $1$. After some thought, I think I can show for each given curve, 'generic' needle length can pass through: Claim: For any $C^2$ embedding as above, there is a full measure and dense $G_\delta$ set $\mathcal{L} \subseteq \mathbb{R}$ of lengths where the needle of any length $L \in \mathcal{L}$ can slide through the curve. Proof: Let $\gamma: \mathbb{R} \rightarrow \mathbb{R}^3$ be a smooth parametrization of the curve s.t. $\gamma(t) = t$ for $t \in (-\infty, -C] \cup [C, \infty)$. Define $\varphi: \mathbb{R}^2 \rightarrow \mathbb{R}$ where $\varphi: (s, t) \mapsto d(\gamma(s), \gamma(t))$. Hence $\varphi$ vanishes on the diagonal and takes positive value everywhere else. Since $\gamma(t) = t$ for $t \notin (-C, C)$, hence $\varphi(s,t)=\|s-t\|$ for $\|s\|, \|t\|>C$. $\varphi^{-1}(L)$ contains four rays $\{ \|s-t\| = L \ \| \ s, t \notin (-C, C) \}$: Observation: a needle of length $L$ can slide through the curve iff there is a continuous path in the level set $\varphi(p) = L$ connecting the two rays above the diagonal. (This is merely projection onto the $x$ and $larex y$-axis.) We also have $\varphi$ is $C^2$ other than on the diagonal. (It behaves like the absolute value function near the diagonal). By Sard’s theorem, since $\varphi$ is $C^2$ on $\{s < t\}$, the set of critical values is both measure $0$ and first category. Let $\mathcal{L}$ be the set of regular values of $\varphi$. For any $L \in \mathcal{L}$, by implicit function theorem, the level set $\varphi^{-1}(L)$ is a $C^2$ sub-manifold. Since the arc $\gamma([-C, C])$ is compact, we can find large $R$ where $\gamma([-C, C]) \subseteq B(\bar{0}, R)$. Hence for $\|t\| > R + L$ and $s \in [-C, C]$, we have $\varphi (s, t) = d(\gamma(s), \gamma(t)) > d(\gamma(t), B(\bar{0}, R)) > L$ The same holds with $\|s\| > R + L$ and $t \in [-C, C]$ i.e. $\varphi$ takes value $>L$ in the shaded region below: Hence for $L \in \mathcal{L}$, $\varphi^{-1}(L) \cap \{x \leq y\}$ is a $1$ dimensional sub-manifold, outside a bounded region it contains only two rays. We also know that the level set is bounded away from the diagonal since $\varphi$ vanishes on the diagonal. By an non-ending arc argument, one connected component of $\varphi^{-1}(L)$ must be a curve connecting the end points of the two rays. Establishes the claim. Remarks: This problem happens to come up at the very end (questio/answer part) of Charles’s talk in the midwest dynamics conference last month (where he talked about our joint work about funnel sections). A couple weeks later Michal Misiurewicz e-mailed us a counter-example when the curve is not smooth (only continuous). Initially I tried to use the above argument to get the length $1$ needle. Everything works fine until a point where one has a continuum in the level set connecting the end-points of the two rays. We want the continuum to be path connected. I got stuck on that. In the continuous curve case, Michal’s counter-example corresponds to the continuum containing a $\sin(1/x)$ curve, hence is not path connected. I believe such thing cannot happen for smooth. The hope would be that the length $1$ needle can slide through any $C^2$ (or $C^1$) curve. (Note that once the length $1$ needle can pass through, then all length can pass through just by rescaling the curve.) In any case, still trying…	{"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": 63, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.944434404373169, "perplexity": 250.37757239743567}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-17/segments/1492917122996.52/warc/CC-MAIN-20170423031202-00102-ip-10-145-167-34.ec2.internal.warc.gz"}
https://www.physicsoverflow.org/26337/is-vacuum-entanglement-properly-so-called?show=26452	# Is "vacuum entanglement" properly so called? + 4 like - 0 dislike 1130 views When I first read about vacuum entanglement, I understood it in exactly the same way as Ron wrote in this post. As can be very clearly seen, in a free scalar theory and in Schroedinger wavefunctional picture, vacuum state is a not a product state in terms of position-space wavefunction, but is a product state in momentum-space, and even more trivial(in terms of product structure) a state in the Fock-space representation. They should be all mathematically equivalent, I suppose. However, the question of whether "entanglement" is properly so called depends on how we identify subsystems that are in a sense "physical". Changing from position-space wavefunctional representation to the momentum-space one, or to the Fock-space one, is clearly not a change of basis of the Hilbert space, but a change of identification of subsystems. To elaborate my point, consider a scenario in ordinary quantum mechanics, two spin-$\frac{1}{2}$ particles positioned at two points, are in a entangled state $\|+-\rangle-\|-+\rangle$. However, we can relabel the basis of the 4-dimensional Hilbert space as $\|A\rangle:= \|+-\rangle-\|-+\rangle\\\|B\rangle:=\|+-\rangle+\|-+\rangle\\\|C\rangle:=\|++\rangle\\\|D\rangle:=\|--\rangle .$ In this case, it is surely absurd to say our state is not entangled because it can be written as a single ket $\|A\rangle$, because it is clear what must be identified as the physical subsystems. The scalar field theory I discussed seems to be an infinite-dimensional analog mathematically, but in the field theory case I'm no longer sure what should be identified as the physical subsystem. In fact, in the post I linked, twistor59 raised the same question under Xiao-Gang Wen's answer, I quote, as you say, the vacuum isn't a product state, but I was curious about what the subsystems were in order even to discuss whether it is a product or not. but I see no satisfactory discussion in that post. edited Jan 24, 2015 To be a product, the states (subsystems) must be written in terms of separated variables. Then the subsystems look as "non-interacting" with each other although they may belong to one compound system. In the latter case they are still entangled due to conservation laws for the whole system, in my humble opinion. + 3 like - 0 dislike You're perhaps hitting problems partly because the operators in QFT are not $\hat\phi(x)$ and $\tilde\phi(k)$, which are operator-valued distributions — which we could say use the improper position and momentum basis sets for whatever test function space is used. For a scalar free field operator $\hat\phi_f$ constructed as a sum of creation and annihilation operators, without introducing distributions, $\hat\phi_f=a_{f^}+a_f^\dagger$, we can define the vacuum as the zero eigenstate of the annihilation operators $a_f$, $a_f\|0\rangle=0$, for all test functions $f\in\mathcal{S}$, a Schwartz space of test functions that decrease faster than exponentially in both position and momentum space, not including Dirac delta-functions. The Hilbert space structure is fixed by a choice of positive semi-definite commutator/inner product $[a_f,a_g^\dagger]=(f,g)$. There are many choices of countable basis for Schwartz space (including, say, sets of Hermite polynomials times a Gaussian centered on a chosen point), for each of which the vacuum state is a product state, but none is local in either position or momentum coordinates. Indeed, there is no such thing as an eigenstate of the momentum operator in the GNS-constructed Fock-Hilbert space that could be called a (Wigner) particle, there are only very close approximations (momentum eigenstates can be introduced as operator-valued distributions but they are not in the Hilbert space). Insofar as entanglement refers to a state not being a product state in momentum coordinates, we might say that the concept of entanglement is not defined for the vacuum state in QFT, but insofar as we work carefully with operator-valued distributions in the momentum basis the vacuum is not entangled in the Wigner particle sense. If we take entanglement to refer to a state that cannot be expressed as a product state constructed using only localized test functions (in other words if we adopt a non-Wigner, non-momentum basis definition of particle in terms of localization, which often creeps silently into discussions), then the vacuum is entangled. If we look at Stephen Summers "Yet More Ado About Nothing: The Remarkable Relativistic Vacuum State", http://arxiv.org/abs/0802.1854, which is linked to in the PDF you link to in your other question, we find a definition of a product state: "Given a pair (M,N) of algebras representing the observable algebras of two subsystems of a given quantum system, a state φ is said to be a product state across (M,N) if φ(MN) =φ(M)φ(N) for all M ∈ M, N ∈ N." An entangled state relative to (M,N) is any that cannot be written as sum of normal product states across (M,N). If M and N are local algebras (local in position space, an essential starting point for AQFT), then the free field QFT vacuum is entangled relative to (M,N). If P and Q are algebras that are local in momentum space, which are not an essential part of AQFT, then the free field QFT vacuum is not entangled relative to (P,Q). answered Jan 24, 2015 by (1,210 points) Thanks, so your stand is basically that for QFT vacuum there's no preferred way of identifying subsystems? @JiaYiyang There's no preferred countable basis for the test function space, but the (positive semi-definite) inner product on the test function space is diagonal in the momentum space coordinates (as it has to be to ensure translation invariance), $$(f,g)=\hbar\int \tilde f^(k)2\pi\delta(k^2-m^2)\tilde g(k)\frac{\mathrm{d}^4k}{(2\pi)^4},$$ so I think I'd say that the Wigner definition of a particle is natural, for the free field, in the sense implied by this equation, but limited by the difficulties that arise if we use the improper momentum basis carelessly. "Wigner definition of a particle is natural, for the free field" I'd agree for free scalar field, but thinking about EM field obscures my certainty. Since here E and B can be understood as local observables, and it looks also "natural" to identify local field values to be subsystems. @JiaTiyang: E and B are strongly interacting subsystems. Their equations are coupled. It is like two constituent particles connected with a spring. But you can introduce separated variables - the center of mass coordinate and the relative distance coordinate. Equations for the latter are not coupled, they describe collective normal modes of the total system whose QM states make a product. @VK, Yes it is a system of infinite coupled harmonic oscillators, but we call it a free theory if there's no terms with order higher than 2. It's just semantics, and it doesn't change my question, the free EM ground state, as well as free scalar ground state, are Gaussian in wavefuncional picture. BTW the downvote is not from me. @JiaYiyang Free field EM is still translation invariant, crucially, so the positive semi-definite inner product is still diagonal in momentum space. The presentation can be just the same as for the free scalar field, four equations, $$\hat F_f=a_{f^}+a_f^\dagger, a_f\|0\rangle=0,\langle 0\|0\rangle=1, [a_f,a_g^\dagger]=(f,g),$$ as it can be for any free real boson field, but with the test function space being antisymmetric tensors and with the positive semi-definite inner product being $$(f,g)=-\hbar\int \tilde f^{\alpha\mu}(k)k_\alpha k_\beta g_{\mu\nu}2\pi\delta(k^2)\tilde g^{\beta\nu}(k)\frac{\mathrm{d}^4k}{(2\pi)^4}$$(with the metric $g_{\mu\nu}$ and the test function $\tilde g^{\beta\nu}(k)$ hopefully being distinguishable). This is positive semi-definite because $k_\alpha\tilde f^{\alpha\mu*}(k)$ and $k_\beta\tilde g^{\beta\nu}(k)$ are both orthogonal to the null 4-vector $k$ and space-like. We can do something similar for the nonobservable EM potential, but there are some long-winded details. @PeterMorgan, I understand that in momentum space the free field vacuum is a product state, both for EM and scalar. My mentioning of EM is to argue there seems to be at least a merit to consider position space wavefunctional to be more "natural" than momentum space functional or Fock space: when we measure the local E-field strength(you may smear it bit if you prefer rigor) out of vacuum, the functional $\Psi[E(x)]$ gives the probability amplitude to obtain a value $E(x)$ at the point $x$. This argument wouldn't work for scalar since it's hard to say if the (smeared) field operator is an observable or not. @JiaYiyang I take position space to be natural insofar as the symmetry group of the equations is the Poincaré group, including translations in position space. Translations in momentum space are not a symmetry. I don't see the EM and scalar fields as so very much different, at least not in the way you suggest. + 3 like - 0 dislike One sense in which the vacuum is entangled is in position space. Take a scalar field in 1+1 dimensions for simplicity, so that the spatial dimension is just a line. We have operators \phi(x) on each spatial lattice site, that commute with the operators on all other sites. So we can divide the Hilbert space up into two Hilbert spaces, one of which consists of the operators \phi(x) for x>0 and the other one with x<0. The two subsystems are the positive and negative half-lines. Now one can trace out the left subsystem, to get a density matrix $\rho$ for the right subsystem. The statement that the vacuum is entangled is equivalent to saying that $\rho$ is not a pure state, and has some entanglement entropy. This entanglement entropy is in fact UV divergent due to contributions from modes very close to x=0 on the left and right. For example for a CFT it's log divergent (and determined in terms of the central charge). All of this is equivalent to the statement that the Rindler vacuum is an excited state in terms of the Minkowski vacuum. answered Jan 29, 2015 by Matthew Hi Matthew, thanks for the new perspective! + 1 like - 0 dislike One sense in which the vacuum is entangled is in position space. Take a scalar field in 1+1 dimensions for simplicity, so that the spatial dimension is just a line. We have operators \phi(x) on each spatial lattice site, that commute with the operators on all other sites. So we can divide the Hilbert space up into two Hilbert spaces, one of which consists of the operators \phi(x) for x>0 and the other one with x answered Jan 29, 2015 by Matthew Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. 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To avoid this verification in future, please log in or register.	{"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": 0.8409135341644287, "perplexity": 493.0779485209065}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439737883.59/warc/CC-MAIN-20200808135620-20200808165620-00527.warc.gz"}
http://solvingtheuniverse.blogspot.com/2010/07/n-field-part-2.html	Wednesday, July 21, 2010 The "N" Field - part 2 To discuss my thoughts of the N in this section, I will not use Lagrangians, rather I will use Hamiltonians. The Lagrangian (L) is related to the Hamiltonian (H) as follows: $L=H-\vec{p}\cdot\vec{v}$ The Hamiltonian represents the total energy of a system. The Hamiltonian of a free particle is merely the kinetic energy of the particle. $H=E=\frac{1}{2}mv^2 = \frac{1}{2}\vec{p}\cdot\vec{v}$ There is a very simple trick to find the Hamiltonian of a charged particle in the influence of an electromagnetic field. The extension can be made by replacing the components of the relativistic four momentum with the components of the canonical four momentum. $p\rightarrow p+eA$ Where A is the electromagnetic four potential. In terms of space and time pieces, this is $E\rightarrow E+eV$ $\vec{p}\rightarrow \vec{p}+e\vec{A}$ Making this transformation on the free particle Hamiltonian yields $E+eV=\frac{1}{2}\left(\vec{p}+e\vec{A}\right)\cdot\vec{v}$ Solving for E gives $E=E_{free}+e\vec{A}\cdot\vec{v}+eV$ So we started with the energy of a free particle, and transformed the four momentum to get the energy of a particle that is subject to the electromagnetic field. Start with a free particle. Extend the momentum. End up with interaction terms. The N field seems to be a similar type of extension - except in this case we aren't extending from four momentum to canonical four momentum, but rather from four potential to "canonical four potential", like this $A\rightarrow A+\frac{m}{e}u$ Where u is the four velocity of the particle. So, whereas F is defined as $F=c\partial\overline{A}$ We can define N as $N=c\partial\left(\overline{A+\frac{m}{e}u}\right)$ So the point I am driving at here is this: on one hand we can derive the Hamiltonian of a particle in an electric field by starting with a free Hamiltonian, and extending the four momentum to the canonical momentum. So, what might happen if we were to start with the "free" Maxwell equations (free here means sourceless) and extend the four potential to the "canonical four potential"? My guess is: Start with a sourceless field. Extend the potential. End up with source terms. The free Maxwell equations are written as $\partial\overline{F}=0$ Extending A to the canonical form converts F into N and the equation becomes $\partial\overline{N}=0$ If the free vs. interaction Hamiltonian analogy holds true here, then this homogeneous equation may represent the inhomogeneous maxwell equations. Splitting this up into field terms and source terms we have $\partial\overline{F}+\frac{mc}{e}\partial\overline{\partial\overline{u}}=0$ $\partial\overline{F}=-\frac{mc}{e}\partial\overline{\partial\overline{u}}$ The empirical form of the inhomogeneous Maxwell equations with source terms is given in terms of currents as $\partial\overline{\left[F\right]_V}\right=\frac{1}{\epsilon}\overline{j}$ We can make the correlation $\overline{j}=-\frac{m}{e}\sqrt{\frac{\epsilon}{\mu}}\partial\left[\,\overline{\partial\overline{u}}\,\right]_V$ Of course, as stated previously, when the empirical current shows up as a source term, this represents approximating local disturbances in the potential wave as point disturbances, i.e. delta function sources. If we do not make this approximation, then j is always zero, and the local disturbances manifest themselves through the derivatives of the scalar EM field. Thus, if we do not make the approximation then we have $\partial\overline{u}\partial=0$ Or, more importantly, taking into account the fact that the four vector potential also obeys this equation, we can make a homogenious wave from the four momentum. $\partial\overline{p}\partial=0$	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 18, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8916207551956177, "perplexity": 401.02158651494597}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886102819.58/warc/CC-MAIN-20170817013033-20170817033033-00596.warc.gz"}
http://thesnaplink.com/a2snr/xzne7.php?3859c3=simplifying-expressions-calculator	Simplifying Radical Expressions Calculator is a free online tool that displays a simplified form of the given radical expression or radical number. Simplifying surds calculator: simplify_surd. From Simplifying Rational Expression Calculator to math review, we have got all of it discussed. This is the currently selected item. a + b has two terms. THE POWER OF A BOX. For each expression within parentheses, follow the rest of the PEMDAS order: First calculate exponents and radicals, then multiplication and division, and finally addition and subtraction. SIMPLIFYING EXPRESSIONS WITH RATIONAL EXPONENTS. Show Instructions. Simplifying rational expressions: grouping. Simplifying rational expressions means to reduce the value of a rational expression to its lowest form or simplified form. result in an error. Trigonometric Calculator: trig_calculator. algebra trigonometry statistics calculus matrices variables list. A Trivia Quiz On Simplifying Algebraic Expressions 7 Questions \| By Dtullo \| Last updated: Jun 21, 2019 \| Total Attempts: 11582 Questions All questions 5 questions 6 questions 7 questions GCD (x,y) is the greatest common denominator between x and y. $simplify\:\frac {5x} {6}+\frac {3x} {2}$. ... a rational expression is defined as an algebraic expression which can be written as a ratio of two polynomial expressions. To simplify an algebraic expression that consists of both like and unlike terms, it might be helpful to first move the like terms together. Create in Standard from the equation of the cubic defined. How to simplify your expression. You cannot use the built-in calculator, because the expression contains variables such as "x". a) 14x + 5x = (14 + 5)x = 19x. Simplifying rational expressions: opposite common binomial factors. Simplifying Rational Expression Calculator. Example 2: to simplify 2â3i2+3i. See … Enter the expression you want to simplify into the editor. Algebrator really helps you out in simplify expression calculator. When an algebraic expression is composed of parts connected by + or - signs, these parts, along with their signs, are called the terms of the expression. To perform math operations on fractions before you simplify them try our Fractions Calculator. Example 1: to simplify (1 +i)8 type (1+i)^8. The simplify calculator will then show you the steps to help you learn how to simplify your algebraic expression on your own. Example: Original expression (LaTeX) \overline{a \land b \land (c \lor \bar{d})} \lor \bar{b} â¦ It is very logical. Come to Pocketmath.net and figure out rationalizing, operations and a large amount of additional math subjects For example, to express x2, enter x^2. Simplification of expressions is a very useful mathematics skill because, it allows us to change complex or awkward expression into more simple and compact form. simplify x2 + 14x + 49 49 − x2. We keep a tremendous amount of really good reference tutorials on matters varying from equations by factoring to algebraic expressions Sometimes I see expressions like tan^2xsec^3x: this will be parsed as tan^(23)(x sec(x)). This calculator will simplify fractions, polynomial, rational, radical, exponential, logarithmic, trigonometric, and hyperbolic expressions. Division, square root, radicals, and fractions are not supported. Simplifying Expressions Calculator is a free online tool that displays the simplification of the given algebraic expression. This simple, one-sided worksheet is designed to help students learn/recall how to simplify expressions. Roots: 3 with multiplicity two and -2 with multiplicity one. A variety of PowerPoints, worksheets and activities I have made over the past year to teach simplifying expressions. The following table contains the supported operations and functions: If you like the website, please share it anonymously with your friend or teacher by entering his/her email: In general, you can skip the multiplication sign, so 5x is equivalent to 5x. Point: (2,4). We offer a whole lot of high-quality reference materials on subjects ranging from power to subtracting polynomials en. BYJUâS online simplifying expressions calculator tool makes the calculation faster and it displays the simplified form of the algebraic expression in a fraction of seconds. Exponents are supported on variables using the ^ (caret) symbol. Students are asked to simplify Radicals expressions with and without variables either individually or as radicals being multiplied and divided by e. Subjects: Algebra, Algebra 2, â¦ simplify 5x 6 + 3x 2. simplify-calculator. Simplify Calculator. Suppose you begin with the expression 5x(2x2 â 3x + 7). $simplify\:\frac {x^2+14x+49} {49-x^2}$. First, determine the GCD (x,y) Determine the greatest common denominator between x and y. This helped me a lot with difference of cubes, quadratic equations and roots. All you have to do is provide the input exponent expression and then click on the calculate button to display the concerned output in no time. Simplify calculator. Out of these cookies, the cookies that are categorized as necessary are stored on your browser as they are essential for â¦ Able to display the work process and the detailed explanation. When you enter an expression into the calculator, the calculator will simplify the expression by expanding multiplication and combining like terms. Calculate the Simplest Radical Form for given whole number which has the N th radical value. The following properties of exponents can be used to simplify expressions with rational exponents. I however would warn you not to just paste the answers from the software. Jul 19, 2017 - Explore Jennifer Love's board "Simplifying Expressions" on Pinterest. If you get an error, double-check your expression, add parentheses and multiplication signs where needed, and consult the table below. Links. All you have to do is provide the input exponent expression and then click on the calculate button to display the concerned output in no time. simplifying algebraic fractions calculator Related topics: two numbers with 871 as greatest common factor \| evaluate algebraic expressions lesson plans \| easy math test \| 9th class maths quiz \| how to solve algebraic solution of linear equations-substitution,elimination,determinants \| "ratio formula" \| math yr 8 \| ti - 85 roots of â¦ Right from algebraic fraction calculator to radical expressions, we have everything covered. To simplify your expression using the Simplify Calculator, type in your expression like 2(5x+4)-3x. Simplifying Radical Expressions Calculation. Free math problem solver answers your algebra homework questions with step-by-step explanations. X/Y is the original fraction. Fall: Simplifying Radical Expressions Maze Activity Sets are the perfect activity for your students to sharpen their understanding of reducing Radicals. Simplifying an expression often means removing a pair of parentheses; factoring an expression often means applying them. Use our handy & instant online simplifying exponents calculator and get the exact answer after simplifying the two exponents' expressions. Simplifying calculator to simplify radical expressions of variables. simplify x2 + 4x − 45 x2 + x − 30. Simplifying rational expressions (advanced) Practice: Simplify rational expressions: common binomial factors. The calculator works for both numbers and expressions containing variables. Type a math problem. Click on "advanced expressions" tab to simplify expressions such as 2x2 â4x+2x2 +1 Algebraic Expressions Calculator. Example 1: to simplify $(\sqrt{2}-1)(\sqrt{2}+1)$ type (r2 - 1)(r2 + 1). Parentheses ( ) and brackets [ ] may be used to group terms as in a standard expression. Note: exponents must be positive integers, no negatives, decimals, or variables. Exponents are supported on variables using the ^ (caret) symbol. Simplifying Expressions with Rational Exponents - Examples. Simplifying calculator to simplify radical expressions of variables. In algebra, simplifying and factoring expressions are opposite processes. When you click text, the code will be changed to text format. Any time you actually will be needing support with math and in particular with expand and simplify calculator or quadratic function come visit us at Mathmastersnyc.com. Use the following rules to enter expressions into the calculator. If the calculator did not compute something or you have identified an error, please write it in By using this website, you agree to our Cookie Policy. Trigonometry. Simplify: Submit BYJUâS online simplifying radical expressions calculator tool makes the calculation faster and it displays simplified form in a fraction of seconds. Evaluate. This calculator will also simplify improper fractions into mixed numbers. Welcome to your Grade 8 Simplifying Expressions Quiz. The simplification process is carried out automatically in just one click. Practice your math skills and learn step by step with our math solver. For addition and subtraction, use the standard + and - symbols respectively. If f is a one-to-one function and f(0) = 3, f(1) = 1, and f(2) = 2, find the following: Evaluate the exponential function for the given x-values. Any lowercase letter may be used as a variable. Mark favorite. It begins by explaining when terms can be put together, then there are 12 pairs of terms for students to consider, combining them where appropriate. write sin x (or even better sin(x)) instead of sinx. From the table below, you can notice that sech is not supported, but you can still enter it using the identity sech(x)=1/cosh(x). In general, you can skip the multiplication sign, so 5 x is equivalent to 5 ⋅ x. This website uses cookies to improve your experience while you navigate through the website. The simplification of Boolean Equations can use different methods: besides the classical development via associativity, commutativity, distributivity, etc., Truth tables or Venn diagrams provide a good overview of the expressions.. When you simplify an expression, you're basically trying to write it in the simplest way possible. That is the reason the x 3 term was missing or not written in the original expression. c) p â 3p = (1 â 3)p = â 2p. To get tan^2(x)sec^3(x), use parentheses: tan^2(x)sec^3(x). I have researched all Algebra software on the net. Similarly, tanxsec^3x will be parsed as tan(xsec^3(x)). The coefficient zero gives 0x 3 = 0. Variables. If you skip parentheses or a multiplication sign, type at least a whitespace, i.e. The calculator follows the standard order of operations taught by most algebra books - Parentheses, Exponents, Multiplication and Division, Addition and Subtraction. Trigonometry (from Greek trigōnon, "triangle" and metron, "measure") is a branch of mathematics that studies relationships between side lengths and angles of triangles. This calculator will also simplify improper fractions â¦ Calculator wich can simplify an algebraic expression online. Simplifying Expressions â Explanation & Examples Learning how to simplify expression is the most important step in understanding and mastering algebra. Simplifying Algebraic Expressions Calculator. Algebra Calculator - get free step-by-step solutions for your algebra math problems. In this radical calculator enter the whole number and get Simplest Radical Form. Mathematical Journeys: An Exercise in Averages, Mathematical Journeys: A Tale of Two Contexts, Simplifying Exponents of Polynomials Worksheet, Simplifying Exponents of Variables Worksheet, Fraction / Mixed Number Comparison Calculator, Simplifying Multiple Positive or Negative Signs, Simplifying Variables With Negative Exponents, Simplifying Fractions With Negative Exponents, Factoring a Difference Between Two Squares. ... Chemistry periodic calculator. (When moving the terms, we must remember to move the + or â attached in front of them). To get tan(x)sec^3(x), use parentheses: tan(x)sec^3(x). Recall the three expressions in division: If we are asked to arrange the expression in descending powers, we would write . If your improper fraction numbers are large you can use the Long Division with Remainders Calculator to find whole number and remainder values when simplifying fractions by hand. Find online algebra tutors or online math tutors in a couple of clicks. 3m + 5.07m 15.21m 8.07m. Simplifying rational expressions multiplying dividing rational expressions adding subtracting rational expressions â¦ The * is also optional when multiplying parentheses, example: (x + 1)(x - 1). Simplify. Come to Mathfraction.com and learn math, quiz and loads of additional math subject areas . type (2+3i)/ (2-3i). Simplify trigonometric expressions Calculator Get detailed solutions to your math problems with our Simplify trigonometric expressions step-by-step calculator. This calculator will simplify fractions, polynomial, rational, radical, exponential, logarithmic, trigonometric, and hyperbolic expressions. This java program code will be opened in a new pop up window once you click pop-up from the right corner. The following calculator can be used to simplify ANY expression with complex numbers. Also, be careful when you write fractions: 1/x^2 ln(x) is 1/x^2 ln(x), and 1/(x^2 ln(x)) is 1/(x^2 ln(x)). Then do operations with exponents, where you can. Simplify Expression Calculator. You can select the whole java code by clicking the select option and can use it. This calculator performs the reducing calculation faster than other calculators you might find. In general, you can skip parentheses, but be very careful: e^3x is e^3x, and e^(3x) is e^(3x). Roots are often written using the radical symbol â. The calculator allows with this computer algebra function of reducing an algebraic expression. Remember PEMDAS. â¦ Simplifying Expressions. Simplifying rational expressions This calculator factor both the numerator and denominator completely then reduce the expression by canceling common factors. I've also included the &'Algebra Cards&'; ppt, which when cut up, were brilliant for helping my bottom set year 8 to â¦ To simplify this expression, you remove the parentheses by â¦ All suggestions and improvements are welcome. ... System of Equations System of Inequalities Basic Operations Algebraic Properties Partial Fractions Polynomials Rational Expressions Sequences Power Sums Induction â¦ Use our handy & instant online simplifying exponents calculator and get the exact answer after simplifying the two exponents' expressions. From simplify exponential expressions calculator to division, we have got every aspect covered. This website uses cookies to ensure you get the best experience. How to simplify a fraction. Ex: 25^4 + 6^3 (or) 45^5 - 36^3 (or) 25^4 * 24^3 This simplifying algebraic expressions calculator will give you the result automatically but for manual calculation, follow the steps given below. For instance: 2 * x can also be entered as 2x. simplify 6 x − 1 − 3 x + 1. Free trigonometric simplification calculator - Simplify trigonometric expressions to their simplest form step-by-step This website uses cookies to ensure you get the best experience. Related Concepts. Simplify the expression 3 × 5.07m. Simplifying complex expressions. The Rational Expression Calculator is a free online tool that displays the rationalized form for the given expression. Click on "advanced expressions" tab … Online surds calculator that allows you to make calculations in exact form with square roots: sum, product, difference, ratio. Any lowercase letter may be used as a variable. The simplify calculator will then show you the steps to help you learn how to simplify your algebraic expression on your own. also be entered as 2(x + 5); 2x * (5) can be entered as 2x(5). In case you actually need assistance with math and in particular with simplifying algebraic expressions calculator or factors come visit us at Solve-variable.com. You can solve multiplication and division during the same step in the math problem: after solving for parentheses, exponents and radicals and before adding and subtracting. A * symbol is optional when multiplying a number by a variable. b) 5y â 13y = (5 â13)y = â8y. This calculator factor both the numerator and denominator completely then reduce the expression by canceling common factors. To simplify your expression using the Simplify Calculator, type in your expression like 2 (5x+4)-3x. Notes. Simplify numerical expression calculator that shows work to solve the numerical expressions having array of operations like addition, subtraction, multiplication and division altogether or any of the combination in an expression. Able to display the work process and the detailed explanation . The only exception is that division is not supported; attempts to use the / symbol will The calculator will simplify the equation step-by-step, and display the result. It gives you step by step answers along with explanations. Use the following rules to enter expressions into the calculator. It will not help you in understanding the subject. The simplification calculator allows you to take a simple or complex expression and simplify and reduce the expression to it's simplest form. Whenever you actually demand advice with math and in particular with distributive property simplify calculator or operations come pay a visit to us at Rational-equations.com. Simplifying rational expressions. 1) Remove parentheses by multiplying factors. This calculator simplifies ANY radical expressions. Simplifying algebraic expressions, addition and subtraction forumulas for inverse sine, ti-84 calculator download, graphing inequalities calculator online, grade 10 algerbra, mcdougal littell find the value of x. write your answer in simplest radical form, dividing rational expressions calculator. Students can use a calculator, but if they understand the foundations of integer rules, things will inherently go smoother. They have been taught to a variety of abilities, so find the PPT that suits your class best. This ensures they understand which symbol âfollowsâ or goes with â¦ Simplify an expression or cancel an expression means reduce it by grouping terms. 3) Combine the constants. 2) Combine like terms by adding coefficients. Looking for someone to help you with algebra? Some test questions ask you to simplify algebraic expressions. Simplifying Expressions Calculator is a free online tool that displays the simplification of the given algebraic expression. Use distributive property to simplify the expression: 5 × (x + 0.1) ... by calculatorâ¦ For multiplication, use the * symbol. This calculator factor both the numerator and denominator completely then reduce the expression by canceling common factors. Simplifying an expression is just another way to say solving a math problem. Simplifying expressions. When you enter an expression into the calculator, the calculator will simplify the expression by expanding multiplication and combining like terms. First simplify within parentheses. This java programming code is used to find the simplifying radical expressions . Come to Polymathlove.com and study multiplying and dividing fractions, inverse and countless other math topics Exponents. Solve. Prefer to meet online? Simplifying rational expressions. Start here or give us a call: (312) 646-6365, © 2005 - 2020 Wyzant, Inc. - All Rights Reserved. Similarly, 2 * (x + 5) can comments below. Then do the arithmetic â¦ Check out all of our online calculators here! Simplifying calculator to simplify radical expressions of variables. ... Matrix Calculator. BYJU’S online simplifying expressions calculator tool makes the calculation faster and it displays the simplified form of the algebraic expression in a fraction of seconds. Graphs. Used with the function expand, the function simplify can expand and collapse a literal expression. Come to Algebra-equation.com and read and learn about operations, mathematics and … Get the free "Simplify an expression" widget for your website, blog, Wordpress, Blogger, or iGoogle. You just give your problem and it will create a complete step-by-step report of the solution. Algebrator is a good program to solve simplifying expressions distributive property calculator questions. The best way of simplifying expressions is to use our online simplify calculator. Find more Mathematics widgets in Wolfram\|Alpha. Any time you require assistance with math and in particular with simplify rational expressions calculator or mathematics content come pay a visit to us at Algebra-equation.com. At the end, there shouldn't be any more adding, subtracting, multiplying, or dividing left to do. (x+1) (x+2) (Simplify Example), 2x^2+2y @ x=5, y=3 (Evaluate Example) y=x^2+1 (Graph Example), 4x+2=2 (x+6) (Solve Example) Algebra Calculator is a calculator that gives step-by-step help on algebra problems. $simplify\:\frac {6} {x-1}-\frac {3} {x+1}$. Grade 8 Simplifying Expressions Quiz. MATH FOR KIDS. Enter your expression, click the "Simplify" button and you will get the answer in a moment.See example-buttons below. Examples: 1+2 , 1/3+1/4 , 2^3 * 2^2. At Wyzant, connect with algebra tutors and math tutors nearby. A/B= X/ (GCD (x,y)) / Y/ (GCD (x,y) Where A/B is the simplified fraction. ... Next (Expression Factoring Calculator) >> Algebra. We offer a huge amount of good reference information on matters starting from math review to solving exponential Able to display the work process and the detailed explanation . Oops I did it again!! See more ideas about simplifying expressions, middle school math, combining like terms. For example, take this expression: Right from calculator to simplifying expressions to quadratic function, we have all kinds of things covered. When I teach simplifying expressions, I always have my students write down the problem, then box each term. The final section contains 16 expressions for students to practise simplifying. If you wish to solve the equation, use the Equation Solving Calculator. We keep a ton of great reference information on subjects starting from practice to intermediate algebra syllabus Variables. Ex: 25^4 + 6^3 (or) 45^5 - 36^3 (or) 25^4 * 24^3 Any lowercase letter may be used as a variable. When the numerator and denominator of a rational number have no common factor other than 1, we consider it in its simplified form. 2x + 5y - 3 has three terms. Click on "advanced expressions" tab to simplify expressions such as 2x2 â4x+2x2 +1. Please leave them in comments. Exponents may not be placed on numbers, brackets, or parentheses. + 1 ) that allows you to simplify expressions such as 2x2 â4x+2x2 +1 algebraic expressions calculator or factors visit! Positive integers, no negatives, decimals, or variables the multiplication sign, type at a. Â 3p = ( 14 + 5 ) x = 19x, rational,,. Identified an error... Next ( expression Factoring calculator ) > > algebra x! Both numbers and expressions containing variables ( 2x2 â 3x + 7 ) Solving calculator the simplify calculator used! Or complex expression and simplify and reduce the expression by expanding multiplication and combining terms... 19, 2017 - Explore Jennifer Love 's board simplifying expressions we! Th radical value 5x = ( 5 â13 ) y = â8y also simplify improper fractions into mixed numbers 5x+4... Fractions into mixed numbers / symbol will result in an error, please write in. Warn you not to just paste the answers from the software Cookie Policy also be entered as 2x visit! Use parentheses: tan ( x ) sec^3 ( x ) sec^3 ( x ) (! + 1 ) ( x + 1 your algebra homework questions with step-by-step explanations ⋅ x. simplify calculator because. 3 term was missing or not written in the simplest radical form both numbers and expressions containing.! Denominator between x and y 4x − 45 x2 + x − 1 3. And math tutors in a fraction of seconds, blog, Wordpress, Blogger, or parentheses, click ! To display the work process and the detailed explanation, example: ( x ) to practise.. Perform math operations on fractions before you simplify them try our fractions calculator algebraic! - Explore Jennifer Love 's board simplifying expressions is to use the built-in,. Learn how to simplify any expression with complex numbers simplify fractions, polynomial, rational, radical exponential..., connect with algebra tutors or online math tutors in a fraction of seconds and collapse a literal expression a! Online tool that displays the simplification calculator - simplify trigonometric expressions to quadratic function we. Simplification calculator allows you to take a simple or complex expression and simplify and reduce the expression variables... Have all kinds of things covered and -2 with multiplicity one in simplify expression calculator simplifying... ) ( x, y ) determine the greatest common denominator between x y! Exponents, where you can skip the multiplication sign, so 5 x is equivalent to 5 ⋅ simplifying expressions calculator calculator! That displays the simplification calculator - simplify trigonometric expressions to quadratic function we! Attached in front of them ) ensure you get an error, please write it in its simplified.! 5X = ( 14 + 5 ) x = 19x ) > > algebra with math in. -2 with multiplicity one i however would warn you not to just the! Often means applying them perform math operations on fractions before you simplify them try our calculator... Out in simplify expression is just another way to say Solving a math problem solver answers your algebra questions! One click once you click simplifying expressions calculator, the code will be changed text... Along with explanations multiplying, or parentheses as x '' ( xsec^3 x... The website of it discussed way to say Solving a math problem 49-x^2 } \$ often means them... Parsed as tan ( x ) sec^3 ( x ) sec^3 x! No negatives, decimals, or variables the parentheses by â¦ simplifying expressions is to use handy... 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https://skm-eleksys.com/search/label/Basics	Showing posts with label Basics. Show all posts Showing posts with label Basics. Show all posts Autotransformer Introduction Previously we already discussed basics of single phase transformer and three phase transformer. We also discussed about three phase transformer connections or Vector Groups. Now, here we will discuss about auto-transformers. Auto transformers have several applications. But first we will develop some basic concepts of autotransformer. Before proceeding further one should have some basic knowledge of transformer. Auto transformers can be made in two ways. In one way it can be realized by additively connecting the primary and secondary windings of the two winding transformer. In other way the autotransformers can be thought of built as a single unit with one continuous winding. We will start from first approach and move to second. Single Phase Autotransformer Let us first consider a normal single phase two winding transformer. The schematic diagram of a transformer is shown in Fig-A(i). We will obtain an autotransformer from this usual transformer. Let the primary side and secondary side voltage ratings are respectively V1, I and V2, I respectively. Number of turns in primary and secondary side are N1 and Nrespectively. Now let us connect the transformers as shown in Fig-A(ii) with additive polarity. For the analysis purpose for better visibility we rearrange the windings in Fig-A(ii) as shown in Fig-B. Of course here we have shown a load connected across secondary For the purpose of simplifying the analysis the transformer is considered as ideal. (In case of power transformer the ideal transformer analysis gives quite accurate result). The autotransformer analysis can be very simple if you recall two important concepts of a transformer. • The voltage developed in the windings are dependent on the flux linkages. The windings are wound on the same magnetic core so they link the same flux. Hence V/ N1= V/ N2 So whenever voltage V1 exist across primary winding, then voltage V2 will be induced across the secondary winding irrespective of changes in connections. • Similarly the magnetic circuit demands that mmf should be balanced. It implies the primary side ampere turn should equal the secondary side ampere turn. Hence I. N1= I. N2 It means current I2 that flows in secondary winding is associated with current Iin primary winding according to above mmf balance formula. In Fig-B the primary of autotransformer is taken across both the windings where as secondary is across N2 winding. The autotransformer is so loaded that the secondary current is I1+I2 . It makes the current flowing in the windings as I1 and I2 which are the rated values. using KCL and KVL if the primary side voltage and current of the auto transformer is Vp and Ip and secondary side voltage and current of the auto transformer is Vs and Is then, V= V1+V Ip = I Vs = V Is = I1+I2. Now the capacity of the autotransformer is (V1+V2).I1 or (I1+I2).V2 Using the voltage ratio and mmf balance formula it is quite easy to show that the autotransformer capacity formula can be simplified as The capacity of autotransformer = Sa = (V+ V2).I= (I+ I2).V2 = V1 I1 (1 + N/ N1) = V2 I2 (1 + N/ N1) Where as the capacity of our original transformer = S =V1 I1 = V2 I2 So, Sa = S(1 + N/ N1) But (1 + N2 /N1) is always greater than 1. Hence by forming an autotransformer the capacity of the resulting autotransformer is always more than the original isolated winding type transformer. In the so formed autotransformer it should be remembered that the voltage and current through the windings remains as before i.e the rated values. From the above formula it is clear that larger value of N2/N1 gives larger capacity of the autotransformer. In this case the voltage ratio of autotransformer = (V1+V2)/ V2= V1 / V +1 = (N1 / N) + 1 Hence V/ Vs = (N1 / N) +1 = (N1 / N) +1 From this formula it is clear that if N2 /N1 is made large to increase the capacity then the voltage ratio between primary and secondary of autotransformer approaches 1. For this reason auto transformers are advantageous for use in power network when the voltage ratio between both sides is near unity. It is used in grid substations as interconnecting transformers (ICT). The autotransformers are used to interconnect two different voltage levels. For example interconnection of 400kV and 220kV, 735kV and 345kV and 765kV and 400kV etc.. The voltage ratio should be less than 3:1 for more advantageous use. Again look at the formula, I. N1= I. N2 As we said N2 /N1 should be large for adavantageous use. To achieve it, N2 should be much greater than N1 It implies that I2 will be much less than I1 The current I2 flows in the common winding of the autotransformer. Hence the auto transformer can be designed with N2 number of turns made of conductor of smaller cross section area, so resulting in a big saving.The autotransformers are cheaper and lighter in comparison to two winding transformers. Just looking at the sketch you may think that instead of connecting two windings in additive ways. Why should not it be made of single winding and one terminal brought out from the middle as per requirement. Yes this is true and the autotransformer can be thought of made of a single winding having a part of winding common to primary and secondary . Sometimes this method is used to obtain a variable secondary voltage. This case it is so designed that the middle contact can smoothly slides over the coil. It is commonly used in the academic electrical laboratories. This is usually called as Variac(Fig-C) or Dimmerstat. There are some other terminologies adopted by different manufacturers. In this design it is not possible to adopt conductors of two different cross sectional area as in case of ICT where turns ratio is fixed(due to fixed voltage ratio) between primary and secondary. Autotransformers are also used for voltage regulation in distribution networks, for starting of induction motors and as lighting dimmers. Autotransformers are also used in electric traction. One main disadvantage about autotransformer is that the primary and secondary are electrically connected. So the electrical disturbance i.e high voltage transients from one side can be easily transmitted to the other side. The other disadvantage is that the impedance of the autotransformer is considerably low, so the short circuit current will be more. More over an open circuit in common winding results in full primary side voltage across the load which is harmful. Three Phase Autotransformer First thing is that the theory of single phase autotransformer is the basis of three phase autotransformer. Three single phase autotransfor bank can be used for forming a three phase transformer or a single unit three phase autotransformer can be built. The three phase autotransformers (see Fig-D) are connected in star-star(Wye-Wye). If the autotransformers are connected as Delta-Delta, then phase difference between primary and secondary exist which is not desired (See Vector Groups). In three phase Y-Y connected power autotransformers an additional delta connected winding is used to take care of zero sequence currents (for unbalanced systems), and third harmonic currents. Although we discussed here is for one particular case still we revealed the general approach. If you wish to connect load across the other winding, then you can proceed in a similar way. Moreover the above analysis is for step down case. You can easily analyze for step up case by interchanging the position of source and load. In this case the direction of I1+I2 is reversed so also the directions of Iand I2. It should be recalled again that the change of direction of current in series winding is associated with change of direction of current in common winding to satisfy mmf balance. Complex Power and Power Triangle Introduction In the last article we discussed the basics of Electrical Power and related terminologies. Here we will develop the concept of Complex Power and Power Triangle. These two are very important concepts used frequently by power engineers. As this article requires the knowledge of previous article so it is advised to at least have a look at the last article. From the previous article the following points are clear. • The instantaneous power p is composed of real power and reactive power. • The time average value of instantaneous power is the real power (true power) is \|V \| \| I \| cos φ • The instantaneous reactive power oscillates about the horizontal axis, so its average value is zero • The maximum value of the reactive power is \|V \| \| I \| sin φ It should be remembered that real power is the average value and the reactive power is maximum value. Complex Power In power system analysis the concept of Complex Power is frequently used to calculate the real and reactive power. This is a very simple and important representation of real and reactive power when voltage and current phasors are known. Complex Power is defined as the product of Voltage phasor and conjugate of current phasor. See Fig-A Let voltage across a load is represented by phasor V and current through the load is I. If S is the complex power then, S = V . I* V is the phasor representation of voltage and I* is the conjugate of current phasor. So if V is the reference phasor then V can be written as \|V\| ∠0. (Usually one phasor is taken reference which makes zero degrees with real axis. It eliminates the necessity of introducing a non zero phase angle for voltage) Let current lags voltage by an angle φ, so I = \| I \| ∠-φ (current phasor makes -φ degrees with real axis) I= \| I \| ∠φ So, S = \|V\| \| I \| ∠(0+φ) = \|V\| \| I \| ∠φ (For multiplication of phasors we have considered polar form to facilitate calculation) Writting the above formula for S in rectangular form we get S = \|V\| \| I \| cos φ + j \|V\| \| I \| sin φ The real part of complex power S is \|V\| \| I \| cos φ which is the real power or average power and the imaginary part \|V\| \| I \| sin φ is the reactive power. So, S = P + j Q Where P = \|V\| \| I \| cos φ and Q = \|V\| \| I \| sin φ It should be noted that S is considered here as a complex number. The real part P is average power which is the average value, where as imaginary part is reactive power which is a maximum value. So I do not want to discuss further and call S as phasor. If you like more trouble I also advise you to read my article about phasor or some other articles on phasor and complex numbers. Returning to the main point, from the above formula it is sure that P is always more than zero. Q is positive when φ is positive or current lags voltage by φ degrees. This is the case of inductive load. We previously said that inductance and capacitance do not consume power. The power system engineers often say about reactive power consumption and generation. It is said that inductive loads consume reactive power and capacitors produce reactive power. This incorrect terminology creates confusion. The fact is that most of the loads are inductive and they unnecessarily draw more current from source. Although in each cycle both inductance and capacitance draw power from the source and return same amount of power to the source but the behavior of inductance and capacitance are opposing to each other. When capacitors are connected in parallel to inductive load the power requirement of inductive load is supplied by capacitor in half cycle and in next half cycle the reverse happens. Depending upon the values of capacitor this power requirement of inductance in the load may be fully or partially satisfied. If partially satisfied the rest will be drawn from the distant source. By properly selecting the capacitance the maximum value of reactive power (Q) drawn from the distant source (or returned to the distant source) is reduced. This reduction in reactive power results in reduction of line current so the reduction of losses in transmission line and improvement in voltage at load end. Power Triangle Returning to the complex power formula, P, Q and S are represented in a power triangle as shown in figure below. S is the hypotenuse of the triangle, known as Apparent Power. The value of apparent power is \|V\|\| I \| or \|S\| = \|V\|\| I \| It is measured in VoltAmp or VA. P is measured in watt and Q is measured in VoltAmp-Reactive or VAR. In power systems instead of these smaller units larger units like Megawatt, MVAR and MVA is used. The ratio of real power and apparent power is the power factor of the load. power factor = Cos φ = \|P\| / \|S\| = \|P\| / √(P 2+Q 2) The reactive power Q and apparent power S are also important in power system analysis. As just shown above the control of reactive power is important to maintain the voltage within the allowed limits. Apparent power is important for rating the electrical equipment or machines. Total Power of Parallel Circuits In real world the loads are usually connected in parallel. Here we will show the total power consumed by parallel branches. See Figure-C. It has two branches. First we have to draw the individual power triangles for each branch. Next the power triangles are arranged back to back keeping real power in positive x direction as shown. The total power consumed is obtained by connecting starting point O to the tip of last triangle. This is actually the result of addition of complex numbers. If S= P+j Q1 S= P+j Q2 Then, S = S1+ S or S = (P1+ P ) + j (Q1+ Q) P = P1+ P Q = Q1+ Q In the above diagram S1 P1 , Q1 and φ 1 correspond to branch1 and S P2 , Q2 and φ correspond to branch2. S, P, Q and φ correspond to total power consumption as seen by the generator Instantaneous, Average, Real and Reactive Power Forward We discussed and developed some important concepts of transmission lines in last few articles. Last time we discussed about long transmission lines. Here we discuss a simple but important basic concept Electric Power. This will refresh our knowledge before we move further. Electric Power has same meaning as mechanical power but here the power or energy that we are concerned is in Electrical form. We often encounter terms like instantaneous, average, total, real, reactive, apparent and complex power or simply power. What they mean? how are they related ? That we will discuss here and in next article. DC Circuit As long as our analysis is restricted to Direct Current(DC) circuit the power consumed by the resistance load is the product of voltage across the resistance and current flowing through the resistance. It is really simple. P = V . I The power consumed by the load is the product of voltage across the load and current drawn by the load (Fig-A). Or the Power supplied by the DC source (battery/cell) is the product of voltage across the cell and current supplied by the cell. Both are equal in our example figure(considering ideal battery of zero internal resistance). The law of energy conservation implies power supplied by the source must be same as power consumed by the circuit. In DC circuit case instantaneous power is same as average power. AC Circuit In AC circuit analysis, what is this power that we talk about. The main problem is that the AC voltage and current varies sinusoidally with time. Moreover the presence of circuit reactive elements like Inductor and capacitor shift the current wave with respect to voltage wave (angle of phase difference). Power is rate at which energy is consumed by load or produced by generator. Whether it is DC circuit or AC circuit, the value of instantaneous power is obtained by multiplying instantaneous voltage with instantaneous current. If at any instant of time t the voltage and current values are represented by sine functions as v = Vm sin ωt i = Im sin (ωt-φ) Vm and Im are the maximum values of the sinusoidal voltage and current. Here ω=2 π f f is the frequency and ω is the angular frequency of rotating voltage or current phasors. It should be clear that for a power system f is usually 50 or 60 Hz φ is the phase difference between the voltage and current. As we said the instantaneous power is the product of instantaneous voltage and current, if we name instantaneous power as p then p = v.i = Vm sin ωt . Im sin (ωt-φ) or p = Vm Im sin ωt sin (ωt-φ) Applying trigonometric formula 2.sin A.sin B = cos(A-B) - cos (A+B) we get ${/color{Cyan}p =/frac{V_{m} I_m}{2}[cos/phi -cos(2/omega t-/phi )]}$ It can be written as ${/color{Cyan}p =/frac{V_{m} I_m}{2}cos/phi -/frac{V_{m} I_m}{2}cos(2/omega t-/phi )}$ This is the equation of instantaneous power In the Fig-C is drawn all the three waves corresponding to v, i and p. Graphically also we can get the value of instantaneous power (p) at any instant of time t by simply multiplying the value of current i and voltage v at that particular instant t. (You can verify that in the diagram p is negative when either v or i is negative otherwise p is positive. See the points where p is zero). In the graph we have shown horizontal axis as angle φ instead of time t for easy visualization. It should be clear that both way it is correct. Clearly the instantaneous power p is composed of two terms. The first term is constant because for a given load the phase angle φ is fixed. It does not change unless the load is changed. The second term is varying with time sinusoidally due to the presence of the term cos (2ωt-φ). Look that the instantaneous power frequency is twice the frequency of voltage or current. So the instantaneous power in a single phase circuit varies sinusoidally. The instantaneous power, p = constant term + sinusoidal oscillating term. In one complete period the average of oscillating term is zero. Then what is the average power within a given time, say one Time Period of the wave? It is the constant term. Here is another way to think about the average power. Just observe that the instantaneous power is negative for a small time. For any time interval you just find the total +ve area A+ (above horizontal-axis (blue line) and below p curve) and total -ve area A- (below horizontal axis and above p curve). The net area is obtained by subtracting A- from A+. By dividing this net area ( by the time interval T we get the average power(P). You can do this using calculus. What you will ultimately get is only the first term in the above formula for instantaneous power p. In still another way it is easier to realize that the formula for instantaneous power p has a constant term (Vm.Im / 2) cos φ and the other sinusoidal term (Vm.Im / 2) cos (2 wt - φ). Actually p is the oscillating power which oscillates about the average constant term (Vm.Im / 2) cos φ . So the average power is ${/color{Cyan}P =/frac{V_{m} I_m}{2}cos/phi}$ The above formula can be written as ${/color{Cyan} P=/frac{V_m}{/sqrt{2}}./frac{I_m}{/sqrt{2}}. cos/phi }$ Or, ${/color{Cyan} P=/left \| V /right \|/left \| I /right \|cos/phi }$ here, ${/color{Cyan} /left \|V /right \|=/frac{V_m}{/sqrt{2}}}$ ${/color{Cyan} /left \|I /right \|=/frac{I_m}{/sqrt{2}}}$ V and I are the phasor representation of RMS values of voltage and current sinusoids. The symbols \|V\| and \|I\| are the magnitudes of phasors V and I. (See at the buttom for definition of RMS value). This above formula is your favorite formula for useful power that we are most concerned about. This average power formula is used to find the power consumed by the load. The monthly electric energy bill at home is based on this power. The engineers and technicians in power or electrical industry simply use the term power instead of average power. So whenever we simply call power it means average power. Of course the instantaneous power is oscillating in nature. As we already said it does not oscillates about the horizontal-axis rather about the average power P (cyan color horizontal line). P will be zero when cos φ =0 or φ = 90 degree, that is when the phase angle between voltage and current waves is 90 degrees. It is only when the load is pure inductive or capacitive. In this case the second term only remains in the instantaneous power formula. From the above figure for some time the power becomes negative that means the load supply energy to source for this period. This is due to the presence of reactive element in load. The above formula for instantaneous power can be written in another form. This form actually is an attempt to distinguish the oscillating reactive power from the instantaneous power formula. Rearranging the terms in equation for instantaneous power above we get p = \|V\| \| I \| cos φ (1-cow2ωt) - \|V\| \| I \| sin φ sin2ωt In this equation the first term \|V\| \| I \| cos φ (1-cow2ωt) is oscillatory whose average value is \|V\| \| I \| cos φ. We already talked about this average power. The second term \|V\| \| I \| sin φ sin2ωt which is also oscillatory but with zero average value. The maximum value of this term is \|V\| \| I \| sin φ. This is the so called Reactive power. So Reactive power is the maximum value of a oscillatory power that is repeatedly drawn from the source and again returned to the source within each cycle. So the average of this reactive power is zero. The average power P is called as Real Power. It is also sometimes called active power. Real power = P = \|V\| \| I \| cos φ It is usually written as P = VI cos φ. But it should be remembered that V and I are the rms values of voltage and current. For example when we say single phase 220 volt AC it means the rms value of voltage is 220 volts ( it is not maximum value of voltage sinusoid) Reactive power = Q = \|V\| \| I \| sin φ Real power is measured in Watt and the reactive power is measured in VAR (VoltAmpereReactive). In power sector these units are too small so real power is measured in Megawatt (MW) and reactive power in Megavar (MVAR). The letter R at the end denotes reactive power. Many times students and practicing engineers are confused about the average power (often simply called power). They think that what they get by multiplying RMS voltage and RMS current is RMS power. No that is wrong. There is no RMS power. RMS power has no meaning or not defined. (Also see definition of RMS value, below at the end). It is average power or real power or true power. Power In Three phase Balanced System Let us consider a three phase balanced system. A three phase balanced system is analysed considering only one phase and neutral return. This is called per phase analysis. So the above analysis for single phase is true for balanced three phase case. Let the total power here is Pt. Then we get total three phase power as thrice of single phase case. P= 3 \|V\| \| I \| cos φ It should be remembered that \|V\| and \| I \| are the per phase values. and φ is the phase angle of load in per phase analysis. The above formula for balanced three phase system can be written as P= √3 \|Vl\| \| Il \| cos φ In the above formula Vl and Il are line voltage and current (Fig-D). This equation is independent of type of three phase load connection i.e delta or star connected load. You have to know the line voltage, line current and phase angle φ as above. This form is very convenient and used often in power calculation. There is one main difference between the single phase and total three phase power. The instantaneous single phase power is pulsating. In the balanced three phase case, each phase instantaneous power is pulsating but the three pulsating power waves are 120 degrees displaced from each other. At any instant of time the total of these three instantaneous power waves is a constant which is 3 \|V\| \| I \| cos φ. So the total power consumed in three phase balanced system is not pulsating. Non-pulsating power also imply the desired non-pulsating torque in case of three phase rotating machines. In large 3-phase motors this is really desired. RMS value of AC Sinusoids The value of AC voltage or current that produces the same heating (or same energy) that is produced if DC voltage or current numerically equal to RMS value of AC is applied instead of AC. This concept helps make the formula for power similar for both DC and AC circuits. You should read the next article about Power Triangle and Complex Power Operators j and a in Electrical Engineering We have already discussed about phasors and its simple properties. Perhaps now it is the time that we want to explore a little more. Every effort is made to keep it as simple as our previous article. Before proceeding further I want to clarify that here we are mainly concerned about phasor multiplication and 'j' and 'a' operators.This article will also help us better appreciate the use of symmetrical components ( for analysis of unbalanced 3-phase systems) and subsequently other phenomena in transformer and AC circuits. We know that phasor in the form x+j y is drawn as an arrow from origin to (x, y) point. Till now I represented the phasor in x+j y form also called rectangular form. A phasor can also be represented in polar form. In the polar form we also need two parameters, these are length of phasor (r) and angle(phi) it makes with the +ve horizontal axis . See the Figure-A. Phasor Multiplication I have already discussed the use of j in phasor representation. We know that j is equal to square root of -1. or j = sqrt(-1) so j.j = -1 Now consider two phasors A = 2 + j 3 and B = -1 + j 2 Let us multiply A and B A.B = (2+j 3) . (-1 + j2) = -2 + j 4 -j 3 + j.j (3.2) = -2 + j 1 - 6 = -8 + j 1 Directly multiply each of real and imaginary parts from A with each of B. It is simple! It is even easier to multiply in polar form. See the example below. As illustrated in figure-A, we represent below the phasors A and B in the polar form. For phasor A, 4 is its length and it makes 20 degrees with x-axis. Similarly B is of length 3 units and it makes 40 degrees with +ve horizontal axis ( 20, 40 and 60 are angles in degree) Representing in polar form, the multiplication has become extremely easy. Just multiply the lengths and add the angles to get the new phasor. You can convert it back to the rectangular form. A.B = 12(cos 60 + sin 60) j and a Operators What we will get, if a phasor is multiplied with j? for example if A = 3 + j 4 Then j A =j(3 + j 4) = j 3 + j.j 4 = -4 + j 3 ( As j.j =-1) Now draw the phasor -4 + j 3. It will be observed that the angle between 3 + j 4 and -4 + j 3 is 90 degrees. Any phasor when multiplied by j will rotate the original phasor by 90 degrees in anticlockwise direction. Now if the resultant phasor is again multiplied by j then the phasor is again rotated by 90 degrees in anticlockwise direction, so on. In our example j(-4 + j 3) = -j4 -3 = -(3 + j 4), which is in opposite direction to (p + j q). So clearly the phasor has again undergone 90 degrees anticlockwise rotation. See the figure. Every time we apply j, we rotate the phasor counterclockwise by 90 degrees. Now let us consider about another operator ' a ' (standard symbol). It has the capacity to rotate a phasor counterclockwise by 120 degrees. applying ' a ' twice the phasor is rotated by 240 degrees, by applying thrice the original phasor is rotated 360 degrees or one complete rotation, so the original phasor. It is clear that as the phasor is rotated 120 degrees (magnitude remains the same) then in polar form a = 1/120deg in rectangular form a = 1.cos 120 + j 1.sin 120 or a = -0.5 + j 0.866 see the Fig-C how a phasor A is rotated by 120 degrees when applied with operator a. I colored them red green and blue to recall our balanced three phase system. Clearly we are able to get the phasors B and C by applying the operator a repeatedly on phasor A. Otherwise we can say that, the balanced system of A-B-C sequence can be equally represented in terms of 'a' and A only. The operator a will be used more in our article symmetrical components. Fourier Series in Electrical Engineering The Fourier Series deals with periodic waves and named after J. Fourier who discovered it. The knowledge of Fourier Series is essential to understand some very useful concepts in Electrical Engineering.Fourier Series is very useful for circuit analysis, electronics, signal processing etc. . The study of Fourier Series is the backbone of Harmonic analysis. We know that harmonic analysis is used for filter design, noise and signal analysis. etc.. Harmonic analysis is also very important in power system studies. In power network, harmonics are mainly generated by non-linear elements and switching equipment. Although it is a applied mathematics topic but like our previous article here also we will try to minimise the maths and depict in a simpler way. The Fourier Series deals with periodic waves and named after J. Fourier who discovered it. Fourier Series The Fourier series is concerned with periodic waves. The periodic wave may be rectangular, triangular, saw tooth or any other periodic form(single valued). Here I will also call this periodic wave as signal wave.Any periodic signal wave can be represented as a sum of a series of sinusoidal waves of different frequencies and amplitudes. Otherwise we can also say that a series of sinusoidal waves of different frequencies and amplitudes add up to give a periodic wave of non-sinusoidal form. Let us consider a periodic signal wave v . According to the definition of Fourier series, this periodic signal wave can be written as sum of sinusoidal waves as below. Now the question is how can we find the coefficients I am not going to show you the details of how I obtained the formulas, but you can remember the three general formulas as shown below for obtaining the coefficients. Many of you may not require to find the harmonics by using the formulas below, but those who are interested can use these formulas. Of course, students are required to remember these formulas. n = 1, 2, 3, 4 ..... Putting the values of n we obtain different coefficients. For example if n=1 we get a1 and b1. Some knowledge about the properties of the Fourier series will immensely help you. In most cases signal waves maintain symmetry. Depending on the symmetry of the wave we may not be always required to find all the sine and cosine terms coefficients. So now I will guide you through some important properties that you should remember so that just looking at the signal wave you can immediately say which coefficients should be present in the series. Let us first of all talk a little about harmonics. What is harmonic ? Every periodic wave has a Time period(T) which is one complete cycle. The whole signal is the repetition of this period. Frequency (N) of the wave is the number of complete cycles in one second. clearly N = 1 / T Observe the general Fourier series, it has a component (a1 sin wt), this sinusoid has the same frequency as the actual signal wave and it is called the fundamental component. The next component is (a2 sin 2wt), its frequency is twice that of fundamental (or original signal wave), this sinusoid is called second harmonic. So also the third, fourth etc. As an example if the fundamental wave has a frequency of 60 Hz, then the frequency of second harmonic is 120 (2 60) Hz, third harmonic 180 Hz, 9th harmonic will be 540 Hz etc. The frequency of nth harmonic is (n.60). Properties of Fourier Series We will now consider some important and very useful properties of Fourier series • In the fourier Series the constant term a0 will not appear if the signal wave average value in one period is zero. (one period is T which is equal to 2PI) Looking at the figure it is clear that area bounded by the Square wave above and below t-axis are A1 and A2 respectively. Here A1=A2, so the average is zero. The wave has a0 equal to zero. You can also confirm by using the formula above. But shifting the same wave in vertical direction as redrawn in Fig-B will automatically introduce a0 in Fourier series. Yes by integrating using the formula above you will find a0. By shifting I have actually disturbed the symmetry around horizontal axis and introduced the average value a0. Here in this figure a0 is exactly how much I shifted the whole wave in vertical direction. This ais also called DC component of the signal wave. • The Fourier series having sine and cosine terms can be combined. We know that if, p = a1 sin x +b1 cos x then it can be written in the form p = r sin (x + phi) phi is the angle displacement from sin x. Using school trigonometry r & phi are found from a1 and b1 (Remember that same harmonics are combined in this form. sin x with cos 2x or cos x with sin 2x etc. are not combined in above form). • Considering the square wave in the above Fig-A we will get only terms with sine function. There will be no cosine terms because the coefficients a1, a2 etc will be zero. This you can verify from above formula also. The Fourier series is: As the above square wave maintains symmetry about origin so it will be composed of sine waves only. (sine waves are symmetric about origin). In Fig-C, I have reproduced the square wave. Here I have calculated only three terms of the Fourier series ' v ', the fundamental, third harmonic and fifth harmonic. The blue curve is the sum of these three terms of the Fourier series of the square wave shown. You can see how just by considering three terms of the Fourier series, the blue curve approximate the square wave. In Fig-D the sum of Fourier series(in blue) is drawn by calculating more coefficients( 9 ). Observe how the blue curve approximates the suare wave so that I had to remove the square wave from the figure for clarity. Taking more terms the curve will be even smoother. • Again consider the square wave. I have now chosen the vertical axis by shifting pi/2 to the right and redrawn in Fig-E. Now applying the above equations for getting the coefficients, we will find that the a0 and the sine term coefficients b1, b2...etc has vanished and only cosine terms are there in the series. The Fourier series is: It is very important to compare Figure-C and Figure-E. In both cases the square waves are identical. Only the vertical axes are chosen differently in each cases. Comparing the individual harmonics and their sum (blue) in both the figures it is clear that actually choosing the position of vertical axis does not change the component harmonics but what previously you were getting as sine waves have now become cosine waves automatically. This is only due to the choice of vertical axis. Both are sinusoidal waves and both are correct. If some intermediate position of axis is chosen, we will get both sine and cosine terms. The sine and cosine terms can be combined as above to get only either phase shifted sine(or cosine) terms for each harmonic. So the choice of axis does not change the shape or number of harmonics, only mathematically the series is different. So it is important to realize that for a particular shape of signal wave the harmonics are fixed, whether you want to show them as series of sine terms or cosine terms or both, it does not matter. Of course in exam you may not be allowed to choose axis. • If the signal wave has quarter wave symmetry, then the Fourier expansion will contain only odd harmonics like 3rd, 5th etc.. To test quarter wave symmetry look at the signal wave at Fig-A. Draw an imaginary vertical line passing through T/4. For the +ve half wave if one side of this vertical line is mirror image of other side then the wave shows quarter wave symmetry. Note: A sine wave is symmetric with respect to origin (odd function) and cosine is symmetric with respect to y-axis (even function). Combination of only sine waves will be symmetric with respect to origin (odd function) and combination of cosine waves will remain symmetric with respect to y-axis (even function) Linear and Nonlinear Systems in Electrical Engineering Like any physical system, the Electrical Systems also work based on some well defined principles. For the study of the characteristics of any system, device or element, the block diagram notation (as shown below ) is used universally. In the figure the element or system is represented by a rectangular block. Naturally the device will receive some input and give the output. The arrow is shown to represent the direction of input and output signal flow. Basically any system can be reduced to this simple form for studying its behavior. When the block diagram represents a single element or simple device then instead of 'input' and 'output' we better use the terms 'cause' and 'effect' respectively (even the terms 'excitation' and 'response' are equally used). For example the cause may be voltage applied across the element and effect is the current that flows in the element. Here we will mainly use the terms 'input', 'output' and 'cause', 'effect'. I feel that the terms excitation and response are more specialized. Here we also use different terms like element, device and system interchangeably to satisfy people of different mindset. It should be remembered that these terms are not exactly of the same meaning. For example different elements or devices may be connected in a particular way to form a system etc. The simplest way of studying the behaviour of a element or system is by applying a series of inputs and observing the respective outputs. That is what is our approach here. Linear System When the input(cause) and corresponding output(effect) are tied by the rules of linearity, then the device or system behavior is said to be linear, otherwise nonlinear. Then what is exactly this linearity ? Suppose we apply a series of inputs say x1, x2, x3 and x4. When input is x1 the output obtained is y1. Similarly when inputs are x2, x3 and x4 the respective outputs are y2, y3 and y4. Now a graph is drawn taking points P1(x1, y1), P2(x2, y2), P3(x3, y3), P4(x4, y4). See the points marked with small circles in X-Y plane. In the X axis we take input values and in Y- axis output values. Now join the points by smooth free hand curve, the result is figure-B(ii). The more points we take, the more accurate will be the curve. Depending on the device behavior the curve of input versus output can be of any shape. some more cases are shown in fig-C(i) to (iv). Each of the curves may fit to some mathematical linear or nonlinear equation. For many devices the relationship between input and output can be found directly analytically. Remember that the graph (or curve) is drawn with input( X axis) versus output (Y axis) ( not input vs. time or output vs. time). If a system or device or element satisfy the following two properties then it is called linear. • Superposition Suppose when x1 input is applied the y1 output is obtained, and when x2 input is applied the y2 output is obtained. Now, if input applied is (x1+x2), then the output obtained will be y1+y2 . (equivalently we say that if x1 and x2 are applied simultaneously then out put will be the sum of the outputs obtained individually) • Homogeneity when x1 is the input applied the y1 output is obtained, Then if (kx1) input is applied, then output obtained will be ky1. Here k is any real number. By examining all the input(X) vs output(Y) curves one can be sure that above two properties will be satisfied by only curve c(ii). The equation of this curve is in the form y = mx. Here the constant 'm' may be any real number. It is easy to remember that the input vs. output characteristic curve of a linear device should be straight line passing through the origin (center). Only in this case the above two laws of linearity are satisfied. All the other curves in fig-C are non-linear and they do not satisfy the above two properties of linearity (check it). It should be realized that even the straight line in fig-C(i) does not represent a linear system. Students are often confused and consider any straight line relationship between the input and output as linear. But it is not. One can also verify that the above two properties are not satisfied by every straight line relationship. In maths the linear equation is represented by the form y = mx + c, which is a straight line whose orientation depends on the values of m and c. The constants 'm' and 'c' can be any real numbers. In case of physical systems, all represented by these linear equations are not strictly linear systems. For the system to be linear, c must be zero. Linear and Nonlinear elements/systems in Electrical engineering Now is the time to consider few examples. The best example of a linear element is an ordinary resistance. If the voltage applied across the resistance is 'cause' and current flowing through the resistance as 'effect'. Then the graph of voltage versus current is a straight line (Ohms Law) passing through the origin. see fig-D. You can verify that it satisfies the linearity laws. The next example is the simple Diode circuit (Fig-E). Look at the voltage versus current curve here. Although it passes through the origin it is not a straight line, hence one can immediately say that it's a nonlinear element. Now we will consider the example of a magnetic circuit. Consider the magnetic core in a transformer. Due to the current 'i' applied in the coil surrounding the core, flux 'phi' is established in the magnetic core. A curve is drawn between field intensity and flux density. The arrow direction in the curve indicates that two different values of flux density obtained for same value of field intensity, one is for when the excitation current is increasing and other when the excitation current is decreasing. The o-a-b path is only at beginning of magnetisation. This type of behavior in magnetic circuit of transformer is called hysteresis. The closed loop formed by the curve is called hysteresis loop. Now just looking at the diagram one can be sure that the field intensity vs. flux density curve is non-linear. This non-linearity in transformer characteristic gives rise to harmonics, which requires some techniques to handle. Nonlinearities are often encountered in electrical devices and systems. Most of the elements show non-linearities to some degree beyond certain input range. There are some methods to deal with the nonlinearities. We can also say that every element or system behaves like a linear element or system within a small range of input variation. For small variations of input around the operating point, the curve can be approximated by a small straight line. The middle of this small straight line is to be treated as origin. See Fig-G. In the figure the portion a-b can be treated as linear and Q as operating point. The truth is that in real world sometimes even the systems with much pronounced nonlinear characteristics is analysed using linear techniques. this is because the variation of system input and output is small enough around the operating point. 'x' and 'y' are small changes around the operating point that is our new origin 'Q'. Again remember that the curve with respect to 'x' & 'y' coordinates is linear (within a-b range) while with respect to 'X' and 'Y' coordinates it is still nonlinear. (While solving problems considering small changes around the new origin 'Q', we only concern about 'x', 'y' and 'Q', forgetting original 'X', 'Y' and 'O') Let us consider one more example. (This portion requires little more knowledge in curve tracing and electronics) Consider the characteristics of a transistor (see Fig-H). The output 'Ic' versus 'Vce' (collector to emitter voltage) is drawn for different values of base current 'Ib'. Ic = collector current Vce = collector to emitter voltage Ib = base current Examining the transfer curve (between Ib(cause) and Ic(effect)) it is clear that the curve of Ib versus Ic is non-linear. Close examination reveals that the curve seems to behave linearly in part of the curve (point a to point b). Then our aim is to fix 'Q' in such a position so that the variations ib and ic around Q lies within this linear region. ( Ib and Ic are total values but ib and ic are variations around Q) Just look at the point Q that we fixed in the curve. If the circuit is allowed to operate around this point Q, so that its input value say Ib varies in small values, then this variation with respect to this operating point Q is linear variation. It should be noted that the overall variation with respect to origin 'O' is not linear. The linearity is with respect to point Q. Then why do we need this linear variation? Yes, in the linear portion the output wave shape is not distorted. That means the shape of the wave is preserved. Operating in somewhat linear range gives birth to very little noise. This is one important reason why we need the biasing in the transistor and set the quiescent point Q. In the linear range the superposition law works. If sinusoidal input signal is applied to the transistor, then actually the transistor operates at base value plus sinusoidal input. So actually whether you apply input (ib) or not, IB,IC and VCE are always present as long as Vcc battery is connected as shown. These are the quiescent point values due to biasing, which helps the transistor operate in linear range. From the total collector current Ic, the output ic is filtered out using a capacitor or signal transformer. So, we applied ib and got ic. However ic is several times ib and so we got ic as amplified version of ib. The linearisation method is used in many situations. This technique is also helpful in power system small signal stability studies and other small signal oscillation studies. Phase Sequence in Electrical Systems In Electrical Systems, sometimes without identification of phase sequence it is impossible to proceed further. Both students and practicing engineers often find it confusing. There are some important situations where identification of phase sequence is a must. These are when: • One synchronous generator is to be synchronized to the grid. • Two systems are to operate in parallel. • Two transformers are to operate in parallel. • Connecting two different lines originating from the same source. In a three phase system the voltage or current sinusoid attain peak values periodically one after another. The sinusoid are displaced 120 degrees from each other. So also phasors representing the three sinusoids for voltage or current waves of three lines are phase displaced by 120 degrees. Now the question arises what is the sequence ? In which order the voltage or current waves attain the peak values cyclically. In the diagram just look at the ABC Anti Clock Wise phase sequence. Here the phasors are rotating in anti clock wise direction. An imaginary viewer(see figure-A, left ) will encounter first phase A, then B, then C again A, then B.....like wise.The sequence is ABCABCABC....... or ABC Anti Clock Wise sequence . You might imagine about the possibility of phase sequence ACB Anti Clock Wise. Yes it can be! In this case the phasors rotating the same anti clock wise direction the imaginary viewer(see figure-A, right ) will encounter first phase A, then C, then B. So here the sequence continues like ACBACBACB...... . You might think that in anti clock wise rotation, do the other sequences possible? you may think why not BCA or BAC or CBA or ......? From the above ABC sequence if you start from B then you can see that BCA is nothing but the same ABC sequence. Similarly BAC and CBA sequence are the same as ACB, only we started from other phase. Note: If you studied permutation and combination maths then it is easier to appreciate the case. Hence it is clear that for anti clock wise rotation there are two possible phase sequences ABC or ACB. Why anti clock wise ? yes it is the convention mostly used. Just recall the school maths when you always measured the trigonometric angle starting from positive x-axis in anticlockwise direction and called it positive angle and in clock wise direction the angle is negative. Accordingly the sine wave is drawn. This is the reason why anti clock wise rotation is so prevalent. However the Clock Wise ABC and ACB phase sequence can also be used. The phase sequence identification is purely a convention. It helps in identifying the sequence in which three phase voltage or current attain the peak values. The Anti Clock Wise ABC is equivalent to Clock Wise ACB. Just analyze by placing the imaginary viewer and rotating the phases in respective directions. When any two of the three phase conductors connecting to the three phase induction motor is interchanged the phase sequence of the supply to motor is changed. This results in the rotation of motor in opposite direction. Actually this is the principle used in mechanical phase sequence detectors. The same direction of rotation means same phase sequence. These days solid state sequence detectors are increasingly used. In some regions of the world other letters may be used for phase sequence, like L1L2L3 or RYB. It is really confusing when synchronization of two different systems are considered. In real world, before synchronization or paralleling, the two sides phase sequence is identified by using the same sequence detector. If the same direction of rotation is observed for both sides by the detector, then they are marked accordingly for same sequence. Of course only same sequence is not sufficient. It is also ascertained that terminals of same phase are connected together.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 7, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8922045826911926, "perplexity": 1035.5059137538035}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964358973.70/warc/CC-MAIN-20211130110936-20211130140936-00195.warc.gz"}
https://hal.archives-ouvertes.fr/hal-00568166	# Free and forced modes responses of fractional operators based on non-identical RLC cells Abstract : In this article, we study the behaviour of the RLC cells for the four configurations that we presented earlier in Abi Zeid Daou et al. (2009a). An electric circuit is used in order to study the fractional behaviour and the robustness of these RLC operators and compare their responses to the behaviour of the fractance which is an ideal fractional operator (Moreau et al., 2003). This analysis is conducted for both natural and forced responses. In more details, the initial conditions of the capacitors and inductances are neglected in the first case and they are taken into consideration in the second one. The number of initial conditions is related to the number of RLC cells used. The robustness of all arrangements is analysed by varying the unsteady parameter value which is represented by an inductance in the electrical circuit. This inductance represents a different variable parameter in each field of application. For example, in the hydropneumatic domain, this inductance refers to the mass of the vehicle as the mass has the main influence on the dynamics and the robustness when designing the active suspension (Moreau et al., 2001). A conclusion will sum up the results for all four arrangements and a confirmation that the phase constancy and the robustness are present in both modes. Keywords : Type de document : Article dans une revue International Journal of Adaptive and Innovative Systems (IJAIS), 2010, Volume 1 (Issue 3/4), pp.318-333. 〈10.1504/IJAIS.2010.034807〉 https://hal.archives-ouvertes.fr/hal-00568166 Contributeur : Valérie Abel <> Soumis le : mardi 22 février 2011 - 17:09:15 Dernière modification le : lundi 19 février 2018 - 15:50:04 ### Citation Roy Daou, Clovis Francis, Xavier Moreau. Free and forced modes responses of fractional operators based on non-identical RLC cells. International Journal of Adaptive and Innovative Systems (IJAIS), 2010, Volume 1 (Issue 3/4), pp.318-333. 〈10.1504/IJAIS.2010.034807〉. 〈hal-00568166〉 ### Métriques Consultations de la notice	{"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": 0.8474429249763489, "perplexity": 1610.2029982795686}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912202188.9/warc/CC-MAIN-20190320004046-20190320030046-00367.warc.gz"}
http://www.physicsforums.com/showthread.php?t=460171	Splitting field of a polynomial over a finite field by resolvent1 Tags: field, finite, polynomial, splitting P: 24 1. The problem statement, all variables and given/known data Assume F is a field of size p^r, with p prime, and assume $$f \in F[x]$$ is an irreducible polynomial with degree n (with both r and n positive). Show that a splitting field for f over F is $$F[x]/(f)$$. 2. Relevant equations Not sure. 3. The attempt at a solution I know from Kronecker's theorem that f has a root in some extension field of F, but I don't know that this root is necessarily in F[x]/(f). If I could obtain this, I could use the fact that finite extensions of finite fields are Galois, therefore normal (and separable), so f splits in F[x]/(f). I also know that finite extensions of finite fields are simple, so $$F[x]/(f) \cong F(\alpha)$$ for some $$\alpha$$. Then the substitution homomorphism ($$g \rightarrow g(\alpha)$$) might help, if I knew that $$\alpha$$ is a root of f. Thanks in advance.	{"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": 0.9484944939613342, "perplexity": 194.06653659885006}, "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-2014-35/segments/1409535924501.17/warc/CC-MAIN-20140909014409-00481-ip-10-180-136-8.ec2.internal.warc.gz"}
http://mathoverflow.net/questions/28127/how-can-i-minimise-n-y-mod-x-for-known-x-and-y-and-for-a-given-range-of-n/29177	# how can I minimise (n * y) (mod x) for known x and y, and for a given range of n? How can I minimise (n.y) (mod x), for known x and y, and for a given range of n? ($x$ and $y$ are actually the components of a 2D vector for a line for which I'm trying to generate a set of bounding integer points) So, for example, if x = 61, y = 17, and n must be in the range 0 < n < 12, then minimum value of the modulo operation is at n = 11, i.e. (11 * 17) (mod 61) = 4. If we changed the range to 0 < n < 9, the minimum value is then at n = 4, i.e. (4 * 17) (mod 61) = 7. I need to be able solve this for arbitrary values, but within a known range (around +/- 3000000). This is a practical question so if there is no direct solution (or if a direct solution is very complicated) then a numerical method may be preferrable. - Let me point you towards the proof of Thue's theorem in elementary number theory, where a variant of this problem comes up. – Franz Lemmermeyer Jun 30 '10 at 6:11 Consider the map $n\longmapsto ny/x$. You want to find a value of $n$ in a given range such that this is almost an integer. Such a point is encoded by an integral point of $\mathbb Z^2$ very close to the linear subspace generated by $(1,y/x)$. Closest points of this form are given by continued fraction approximations: Develop $x/y$ as a continued fraction and choose a convergent $a/b$ with $n=\lambda b$ in your range for small $\lambda$. This $n$ does the job. If you want positive minimal values, then only every other convergent works. In your example, one gets convergents 1/4 and 2/7 and $4\cdot 17=68\equiv 7\pmod 61,\ 7\cdot 17\equiv -3\pmod 61$. Thus $n=7$ is a better solution but the smallest representant modulo $61$ of $7\cdot 17$ is negative. - Roland, isn't it simply the Euclidean algo? – Wadim Zudilin Jun 14 '10 at 15:11 I am not sure. Continued fraction expansions and the Euclidean algorithm are of course very close but I do not quite see how to apply the Euclidean algorithm for effectively solving this problem. – Roland Bacher Jun 14 '10 at 15:16 Hmm, it's indeed not obvious... – Wadim Zudilin Jun 14 '10 at 15:27 Wadim: The Extended Euclidean Algorithm produces a sequence of equations $d_m = a_m x + b_m y$, where the $d_m$ are strictly decreasing until reaching the GCD of $x$ and $y$. If I recall correctly, the quotients $-\frac{b_m}{a_m}$ are the covergents of the continued fraction expansion of $\frac{x}{y}$. Hence, if $-\frac{b_m}{a_m} > 0$, one can choose $n = \lambda \|a_m\|$ for a small $\lambda$, as suggested by Roland. – felix Jun 23 '10 at 0:13 What I'm going to say is somewhat similar to Roland's answer but more precise in the case when the range for $n$ is given in the form of upper bound, i.e., $0 < n < N$. Notice that $ny\bmod x = ny - kx$ for some integer $k$. We want to minimize $ny-kx$ that, if we disregard for a moment the sign, can be formulated as minimizing $$\left\|n\frac{y}{x} - k\right\|$$ over integer $n$ in the given range and arbitrary integer $k$. It is known that if some $n,k$ give better approximation (in the sense of the above absolute value) than any other $n',k'$ with $n' < n$, then $\frac{k}{n}$ with necessity represents a convergent to $\frac{y}{x}$. Therefore, a first good candidate for the anticipated $n$ is the largest denominator of a convergent $\frac{k}{n}$ for $\frac{y}{x}$ that fits the given range (i.e., $n < N$). For such $n$, if we have $n\frac{y}{x} - k > 0$ (equivalently, $\frac{k}{n}<\frac{y}{x}$), then it is indeed a solution. However, if $n\frac{y}{x} - k < 0$ (equivalently, $\frac{k}{n}>\frac{y}{x}$), then the solution is given by largest allowed denominator of a semi-convergent located between the preceding and subsequent convergents of $\frac{k}{n}$. That is, if $\frac{k'}{n'}, \frac{k}{n}, \frac{k''}{n''}$ are consecutive convergents, then $\frac{k'}{n'} < \frac{k''}{n''} < \frac{y}{x}$ and $n' < N \leq n''$. Then one needs to find a semi-convergent between $\frac{k'}{n'}$ and $\frac{k''}{n''}$ with the largest denominator smaller than $N$. - Ok, I thought a bit about the problem, and here is another idea. It does not provide an answer, but might give a new idea. Maybe even the sketched algorithm turns out to work well in practice. Assume we want to find some $n \in \mathbb{Z}$ satisfying $C \le n \le D$ (for some constants $C$ and $D$, which can be assumed to be integers as well) such that $n\cdot y \pmod{x}$ is minimal under this condition. For that, first use the Extended Euclidean Algorithm to compute the GCD $d$ of $x$ and $y$, as well as integers $A, B$ with $d = A x + B y$. Then we can write $d' = A' x + B' y$ with $d', A', B'$ if, and only if, $d' = d t$ for some $t \in \mathbb{Z}$, and $A' = A t + s y/d$, $B' = B t - s x/d$ with $s \in \mathbb{Z}$. Hence, we want to make $t \in \mathbb{N}_{\ge 0}$ as small as possible, while keeping $C \le B t - s x/d \le D$ for some $s \in \mathbb{Z}$. Such an $s$ exists if, and only if, the closed interval $[(B t - D) \frac{d}{x}, (B t - C) \frac{d}{x}]$ contains an integer, or equivalently, if $\lceil(B t - D) \frac{d}{x}\rceil \le (B t - C) \frac{d}{x}$. Now $\lceil\frac{a}{b}\rceil = \frac{a + (-a \pmod{b})}{b}$, whence $\lceil(B t - D) \frac{d}{x}\rceil = \frac{(B t - D) d + (-(B t - D) d \pmod{x})}{x}$. This is $\le (B t - C) \frac{d}{x}$ if, and only if, $(D - B t) d \pmod{x} \le (D - C) d$. Therefore, an equivalent problem is finding the smallest $t \ge 0$ such that $$D - B t \pmod{\tfrac{x}{d}} \le D - C.$$ Note that without loss of generality, we can assume that $0 \le B \le \frac{x}{d}$; in fact, in almost every case, we have $B < \frac{x}{d}$ (the only exception is $d = x$ and $B = 1$, $A = 0$, in which $n \cdot y \pmod{x}$ is zero for all $n$). Hence, we can assume that $B d < x$. Moreover, since $1 = A \frac{x}{d} + B \frac{y}{d}$, we see that $B$ and $\frac{x}{d}$ are coprime. In particular, $-B t \pmod{\frac{x}{d}}$, $t \in \mathbb{N}_{\ge 0}$ iterates over every integer the interval $[0, \frac{x}{d})$, including $0$ itself; therefore, we can always find a solution $t$ satisfying $0 \le t < \frac{x}{d}$, which is not surprising when considering the original problem. One could now proceed as follows, which might lead to an algorithm which is fast in practice (in case $x > (D - C) d$): compute several solutions $t$ by choosing some random $T \le B - C$ and computing $t$ such that $D - B t \equiv T \pmod{\frac{x}{d}}$ (i.e. choose $t \equiv (-T + D) \frac{y}{d} \pmod{\frac{x}{d}}$, since $\frac{y}{d}$ is the modular inverse of $B$ modulo $\frac{x}{d}$) and take the minimum $t'$ over all such $t$. Hoping that at least one of these solutions is small, we are left only with a small interval $[0, t']$ to check for smaller solutions. [Note that this is a similar problem to the one we started with: we want to find $T \in [0, D - C]$ such that $(-T + D) \frac{y}{d} \pmod{\frac{x}{d}}$ is minimal, instead of finding $n \in [C, D]$ such that $n \frac{y}{d} \pmod{\frac{x}{d}}$ is minimal.] When we assume that $T \mapsto (-T + D) B^{-1} \pmod{\frac{x}{d}}$ is "random", we can assume that the $t$'s we obtain are randomly distributed in the interval $[0, \frac{x}{d})$, whence $t'$ can be expected to be small. Hence, this algorithm is only faster than just tying all values for $n$ if $t'$ is less than $D - C$, but this can be determined by a simple comparism. -	{"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": 0.9780178666114807, "perplexity": 119.51096826141875}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-27/segments/1435375098072.11/warc/CC-MAIN-20150627031818-00133-ip-10-179-60-89.ec2.internal.warc.gz"}
https://physics.stackexchange.com/questions/589623/explaining-rest-electron-magnetic-field-through-special-relativity	# Explaining rest electron magnetic field through Special Relativity I was reading this post. It says that the electron produces a magnetic field due to its internal magnetic dipole which is given by $${\boldsymbol \mu}= \frac{eg}{2m} {\bf S}$$ In Purcell's book, he gives an explanation of how the magnetic field arises due to wrong frames. Although this seems to be disputed here. But I don't know how you would explain the magnetic field due to its internal magnetic dipole through Special Relativity because the charge is not moving at all. • From your own link: “A magnetic field is not just an electric field with relativity applied, i.e. an electric field viewed from the wrong reference frame. In reality, a magnetic field is a fundamental field which can exist in a certain reference frame without needing any help from an electric field.” Oct 26, 2020 at 4:06 • Yes, that's correct. but what's the source of this magnetic field? the article says that we can get into the frame where there is no electric field but entirely the magnetic field then what's the source of this field? Like electric field have their souch particle.. Oct 26, 2020 at 4:26 • The cause is the intrinsic angular momentum (known as “spin”, although nothing is spinning) of the electron. Oct 26, 2020 at 4:43 • It says you can go to a frame where there is a magnetic field but not an electric field. What that means because If an electron is the one produce an electric field, you can not go somewhere where you can vanish its field because that's mean there is no electron at all. Oct 26, 2020 at 4:55 • That quote isn’t talking about an electron. The point is simply that not all magnetic fields arise from being in the “wrong” frame. Oct 26, 2020 at 4:57 In Purcell's book, he gives an explanation of how the magnetic field arises due to wrong frames. This approach seems to confuse many students. I do not recommend it for learning. It is nice to come back to once you already know the material. If you look carefully at Purcell's actual derivation, he is not deriving the magnetic field, but rather the magnetic force. This is an important distinction because there is always a frame where the magnetic force goes to zero (the rest frame of the test charge), but there is not always a frame where the magnetic field goes to zero. In fact, one of the invariants of the electromagnetic field is $$E^2-B^2$$ (in natural units). So if that invariant is negative then there is no frame where $$B=0$$. This is what is discussed in your second reference. But the point is that since there is not always a frame where $$B=0$$ you cannot consider the $$B$$ field to be merely the $$E$$ field as seen in a different frame. Instead both $$E$$ and $$B$$ vector fields are equally valid components of the overall electromagnetic field tensor $$F$$. The electromagnetic field is governed by Maxwell's equations, which describes the origin of the $$B$$ field component as being due to current density $$j$$. This leads to your confusion: But I don't know how you would explain the magnetic field due to its internal magnetic dipole through Special Relativity because the charge is not moving at all. Actually, even a stationary electron has a probability current density. The probability current density essentially acts as a probabilistic velocity distribution, and hence the electron has a probabilistic electrical current density. The dipole moment can be thought of as arising from this $$j$$, with suitable QM mathematical machinery. Physics is the discipline of studying nature and modelling it with mathematics, so that one can predict new observations and measurements. In the process there are several frameworks where physics models exist, because of kinematic and other variables, and it can be shown mathematically that in the overlap region they merge or emerge from each other. To have a mathematical model one has to impose extra axioms, called laws, principles,postulates in order to pick up those solutions that describe and predict data. In the classical electrodynamics of Maxwell, it is not necessary for electric and magnetic fields to be related with motion in order to exist. There exist monopoles in the theory . The electric monopoles have been discovered in particle data, and it is the electron to start with (and quarks and muons ..) , but no magnetic monopole has been discovered (as yet?). In classical theory it is not necessary to have a special relativity kinematics source for a magnetic field . The world of particles is modeled with quantum mechanics, and within its axiomatic assumptions there is the table of particles with their charges and spins. Axiomatic spin definiton for particles is necessary in order to have the law of angular momentum conservation at the particle level. But I don't know how you would explain the magnetic field due to its internal magnetic dipole through Special Relativity because the charge is not moving at all. Can you eliminate the electric field of the electron by special relativity motion? The confusion comes by using classical electromagnetism where quantum modeling is needed. In the quantum frame , given a spin , nothing has to move, but it has been found that particle spin can be treated classically ( as there is continuity between classical and quantum) and since there is spin and mass, a magnetic moment a la classical EM can be calculated. Note that the classical calculation needs a correction factor. It is useful to check the classical model behavior at the quantum level, it is not binding, the quantum is a different framework with different mathematics, and one has to be careful to keep to the correct framework when using mathematical tools from a larger one. • The article says you can go to a frame where there is a magnetic field but not an electric field. What that means because If an electron is the one produce an electric field, you can not go somewhere where you can vanish its field and that's mean there is no electron at all. Oct 26, 2020 at 6:04 • If you mean the lat link inyour question, I do not see where it says this. if in classical EM it is ignoring poles. Oct 26, 2020 at 6:26 • As far I know now, there are three ways the magnetic field can exist, first through charge motion, second through intrinsic angular momentum (or internal magnetic momentum of the particle), and third with changing the electric field. Is this correct? Can any of them deducted through changing the frame or all are fundamental ways? Oct 26, 2020 at 18:04 • If you consider solutions of Maxwell's equations as fundamental. They model well the observations. It is the intrinsic angular momentum at a particle level and the intrinsic charge at the particle level that have to be accepted axiomatically. Oct 26, 2020 at 18:11 • You mean that Maxwell had taken the magnetic dipole and electric monopole as fundamental sources of field?? Oct 26, 2020 at 18:35 Electrons are photons are both just quanta of fields. The angular momentum of both can be understood as a bulk circular motion of the field. The electron field also has electric charge, and its magnetic field can be understood as arising from the same circular motion. This is covered by Hans C. Ohanian, "What is spin?", Am. J. Phys. 54 (6), June 1986 (online here). He credits F. J. Belinfante (1939) and W. Gordon (1928) for the original derivations. The photon field was studied (as the electromagnetic field) before quantum mechanics was discovered, and was known to have angular momentum, and the angular momentum of photons is obviously just the quantized version of that classical property. Electrons were never understood as a classical field before quantum mechanics, and they were treated as point particles with "intrinsic spin" in the early days of QM, before quantum field theory came along. I suppose that's why you so commonly hear claims that electron spin is a mysterious property of point particles, while you rarely hear that about photons, even though in the modern picture they're closely analogous.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 12, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9056957364082336, "perplexity": 280.9133211358274}, "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/1659882573172.64/warc/CC-MAIN-20220818063910-20220818093910-00189.warc.gz"}
https://www.jobilize.com/trigonometry/test/algebraic-other-types-of-equations-by-openstax	2.6 Other types of equations (Page 5/10) Page 5 / 10 Solve the following rational equation: $\text{\hspace{0.17em}}\frac{-4x}{x-1}+\frac{4}{x+1}=\frac{-8}{{x}^{2}-1}.$ We want all denominators in factored form to find the LCD. Two of the denominators cannot be factored further. However, $\text{\hspace{0.17em}}{x}^{2}-1=\left(x+1\right)\left(x-1\right).\text{\hspace{0.17em}}$ Then, the LCD is $\text{\hspace{0.17em}}\left(x+1\right)\left(x-1\right).\text{\hspace{0.17em}}$ Next, we multiply the whole equation by the LCD. $\begin{array}{ccc}\hfill \left(x+1\right)\left(x-1\right)\left[\frac{-4x}{x-1}+\frac{4}{x+1}\right]& =& \left[\frac{-8}{\left(x+1\right)\left(x-1\right)}\right]\left(x+1\right)\left(x-1\right)\hfill \\ \hfill -4x\left(x+1\right)+4\left(x-1\right)& =& -8\hfill \\ \hfill -4{x}^{2}-4x+4x-4& =& -8\hfill \\ \hfill -4{x}^{2}+4& =& 0\hfill \\ \hfill -4\left({x}^{2}-1\right)& =& 0\hfill \\ \hfill -4\left(x+1\right)\left(x-1\right)& =& 0\hfill \\ \hfill x& =& -1\hfill \\ \hfill x& =& 1\hfill \end{array}$ In this case, either solution produces a zero in the denominator in the original equation. Thus, there is no solution. Solve $\text{\hspace{0.17em}}\frac{3x+2}{x-2}+\frac{1}{x}=\frac{-2}{{x}^{2}-2x}.$ $x=-1,$ $x=0$ is not a solution. Access these online resources for additional instruction and practice with different types of equations. Key concepts • Rational exponents can be rewritten several ways depending on what is most convenient for the problem. To solve, both sides of the equation are raised to a power that will render the exponent on the variable equal to 1. See [link] , [link] , and [link] . • Factoring extends to higher-order polynomials when it involves factoring out the GCF or factoring by grouping. See [link] and [link] . • We can solve radical equations by isolating the radical and raising both sides of the equation to a power that matches the index. See [link] and [link] . • To solve absolute value equations, we need to write two equations, one for the positive value and one for the negative value. See [link] . • Equations in quadratic form are easy to spot, as the exponent on the first term is double the exponent on the second term and the third term is a constant. We may also see a binomial in place of the single variable. We use substitution to solve. See [link] and [link] . Verbal In a radical equation, what does it mean if a number is an extraneous solution? This is not a solution to the radical equation, it is a value obtained from squaring both sides and thus changing the signs of an equation which has caused it not to be a solution in the original equation. Explain why possible solutions must be checked in radical equations. Your friend tries to calculate the value $\text{\hspace{0.17em}}-{9}^{\frac{3}{2}}$ and keeps getting an ERROR message. What mistake is he or she probably making? He or she is probably trying to enter negative 9, but taking the square root of $\text{\hspace{0.17em}}-9\text{\hspace{0.17em}}$ is not a real number. The negative sign is in front of this, so your friend should be taking the square root of 9, cubing it, and then putting the negative sign in front, resulting in $\text{\hspace{0.17em}}-27.$ Explain why $\text{\hspace{0.17em}}\|2x+5\|=-7\text{\hspace{0.17em}}$ has no solutions. Explain how to change a rational exponent into the correct radical expression. A rational exponent is a fraction: the denominator of the fraction is the root or index number and the numerator is the power to which it is raised. Algebraic For the following exercises, solve the rational exponent equation. Use factoring where necessary. ${x}^{\frac{2}{3}}=16$ ${x}^{\frac{3}{4}}=27$ $x=81$ $2{x}^{\frac{1}{2}}-{x}^{\frac{1}{4}}=0$ ${\left(x-1\right)}^{\frac{3}{4}}=8$ $x=17$ ${\left(x+1\right)}^{\frac{2}{3}}=4$ ${x}^{\frac{2}{3}}-5{x}^{\frac{1}{3}}+6=0$ ${x}^{\frac{7}{3}}-3{x}^{\frac{4}{3}}-4{x}^{\frac{1}{3}}=0$ For the following exercises, solve the following polynomial equations by grouping and factoring. ${x}^{3}+2{x}^{2}-x-2=0$ $x=-2,1,-1$ $3{x}^{3}-6{x}^{2}-27x+54=0$ $4{y}^{3}-9y=0$ ${x}^{3}+3{x}^{2}-25x-75=0$ ${m}^{3}+{m}^{2}-m-1=0$ $m=1,-1$ $2{x}^{5}-14{x}^{3}=0$ $5{x}^{3}+45x=2{x}^{2}+18$ $x=\frac{2}{5}$ For the following exercises, solve the radical equation. Be sure to check all solutions to eliminate extraneous solutions. $\sqrt{3x-1}-2=0$ $\sqrt{x-7}=5$ $x=32$ $\sqrt{x-1}=x-7$ $\sqrt{3t+5}=7$ $t=\frac{44}{3}$ $\sqrt{t+1}+9=7$ $\sqrt{12-x}=x$ $x=3$ $\sqrt{2x+3}-\sqrt{x+2}=2$ $\sqrt{3x+7}+\sqrt{x+2}=1$ $x=-2$ $\sqrt{2x+3}-\sqrt{x+1}=1$ For the following exercises, solve the equation involving absolute value. $\|3x-4\|=8$ $x=4,\frac{-4}{3}$ $\|2x-3\|=-2$ $\|1-4x\|-1=5$ $x=\frac{-5}{4},\frac{7}{4}$ $\|4x+1\|-3=6$ $\|2x-1\|-7=-2$ $x=3,-2$ $\|2x+1\|-2=-3$ $\|x+5\|=0$ $x=-5$ $-\|2x+1\|=-3$ For the following exercises, solve the equation by identifying the quadratic form. Use a substitute variable and find all real solutions by factoring. ${x}^{4}-10{x}^{2}+9=0$ $x=1,-1,3,-3$ $4{\left(t-1\right)}^{2}-9\left(t-1\right)=-2$ ${\left({x}^{2}-1\right)}^{2}+\left({x}^{2}-1\right)-12=0$ $x=2,-2$ ${\left(x+1\right)}^{2}-8\left(x+1\right)-9=0$ ${\left(x-3\right)}^{2}-4=0$ $x=1,5$ Extensions For the following exercises, solve for the unknown variable. ${x}^{-2}-{x}^{-1}-12=0$ $\sqrt{{\|x\|}^{2}}=x$ All real numbers ${t}^{10}-{t}^{5}+1=0$ $\|{x}^{2}+2x-36\|=12$ $x=4,6,-6,-8$ Real-world applications For the following exercises, use the model for the period of a pendulum, $\text{\hspace{0.17em}}T,$ such that $\text{\hspace{0.17em}}T=2\pi \sqrt{\frac{L}{g}},$ where the length of the pendulum is L and the acceleration due to gravity is $\text{\hspace{0.17em}}g.$ If the acceleration due to gravity is 9.8 m/s 2 and the period equals 1 s, find the length to the nearest cm (100 cm = 1 m). If the gravity is 32 ft/s 2 and the period equals 1 s, find the length to the nearest in. (12 in. = 1 ft). Round your answer to the nearest in. 10 in. For the following exercises, use a model for body surface area, BSA, such that $\text{\hspace{0.17em}}BSA=\sqrt{\frac{wh}{3600}},$ where w = weight in kg and h = height in cm. Find the height of a 72-kg female to the nearest cm whose $\text{\hspace{0.17em}}BSA=1.8.$ Find the weight of a 177-cm male to the nearest kg whose $\text{\hspace{0.17em}}BSA=2.1.$ 90 kg A laser rangefinder is locked on a comet approaching Earth. The distance g(x), in kilometers, of the comet after x days, for x in the interval 0 to 30 days, is given by g(x)=250,000csc(π30x). Graph g(x) on the interval [0, 35]. Evaluate g(5) and interpret the information. What is the minimum distance between the comet and Earth? When does this occur? To which constant in the equation does this correspond? Find and discuss the meaning of any vertical asymptotes. The sequence is {1,-1,1-1.....} has how can we solve this problem Sin(A+B) = sinBcosA+cosBsinA Prove it Eseka Eseka hi Joel June needs 45 gallons of punch. 2 different coolers. Bigger cooler is 5 times as large as smaller cooler. How many gallons in each cooler? 7.5 and 37.5 Nando find the sum of 28th term of the AP 3+10+17+--------- I think you should say "28 terms" instead of "28th term" Vedant the 28th term is 175 Nando 192 Kenneth if sequence sn is a such that sn>0 for all n and lim sn=0than prove that lim (s1 s2............ sn) ke hole power n =n write down the polynomial function with root 1/3,2,-3 with solution if A and B are subspaces of V prove that (A+B)/B=A/(A-B) write down the value of each of the following in surd form a)cos(-65°) b)sin(-180°)c)tan(225°)d)tan(135°) Prove that (sinA/1-cosA - 1-cosA/sinA) (cosA/1-sinA - 1-sinA/cosA) = 4 what is the answer to dividing negative index In a triangle ABC prove that. (b+c)cosA+(c+a)cosB+(a+b)cisC=a+b+c. give me the waec 2019 questions	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 74, "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": 0.9421543478965759, "perplexity": 695.7832556112245}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027318243.40/warc/CC-MAIN-20190823083811-20190823105811-00294.warc.gz"}
https://www.physicsforums.com/threads/properties-of-estimators-question.380289/	# Homework Help: Properties Of Estimators Question 1. Feb 21, 2010 ### dipset1011 1. The problem statement, all variables and given/known data suppose that 14, 10, 18, 21 constitute a random sample of size 4 drawn from a uniform pdf defined over the interval [0, theta], where theta is unknown. Find an unbiased estimator for theta and y'3, the third order statistic. What numerical value does the estimator have for these particular observations? Is it possible that we would know that an estimate for theta based on y'3 was incorrect, even if we have no idea what the true value of theta might be? explain. 2. Relevant equations I am not entirely sure how to go about solving this problem and help or a kick in the right direction would be great. 3. The attempt at a solution Is the third order statistic 18? isn't the estimator just the summation of all the observations divided by n, the number of samples or observations. Please, any help would be greatly appreciated.	{"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": 0.865502119064331, "perplexity": 301.07390171827086}, "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-2018-39/segments/1537267158691.56/warc/CC-MAIN-20180922201637-20180922222037-00045.warc.gz"}
https://www.physicsforums.com/threads/rotational-motion-and-equilibrium-problems-work-checking.96731/	# Rotational Motion and Equilibrium Problems, Work Checking 1. Oct 25, 2005 ### kris24tf I have a few Rotational Motion and Equilibrium problems that I hav been working on but I am not sure if I completed them correctly. I will show what I have done so far and if anyone could point me in the right direction I would appreciate it. Please, numerical explanations are easier for me to understand. 1) A variation of Russell traction supports the lower leg in a cast. Suppose the leg and cast have a total mass of 15 kg and m1 is 4.5 kg. a) what is the reaction force of the leg muscles to the traction? For this I took m1=F/2g, 4.5kg=F/19.6, so F=88.2N? b) what must m2 be to keep the leg horizontal? For this I took m1g+m2g-15kg(g)=0, m2=15-4.5kg=10.5kg 2)Two masses are supended by a pulley (Atwood Machine). The pulley itself has a mass of .2kg, radius of .15m and constant torque of .35mN. What is the magnitude of the acceleration of the suspended masses if m1=.4kg and m2=.8kg? I don't kno how to show a picture here, so I'll try to describe it. It is an atwood machine with m2 on the left with forces m2g downward and T2 upward, m1 with m1g downward and T1 upward, and a pulley with R, tf, and T2 and T1 downward on m2 and m1 respectively. I ended up with an equation of a/2=(m2-m1)g/m1+m2+M/2+T and I got 1.2 m/s^s which is the correct answer but I have a feeling my equation might be incorrect. Any advice on what equations to go off of here would help 3)An ice skater spinning with outstretched arms has an angular speed of 4 rad/s. She tucks her arms and decreases moment of inertia by 7.5%. a) what is the resulting angular speed? b) By what factor does the skater's kinetic energy change? This is where I am confused most. I know I should use the rotational Work Energy theorem somehow, but I don't know how... Any advice on these woudl be great. Thank you for looking! Last edited: Oct 25, 2005 2. Oct 26, 2005 ### kris24tf Anyone? Hello? 3. Oct 26, 2005 ### andrevdh kris24tf, concerning problem 2 - some guidelines: The friction in the pulley increases the tension on the downwards moving side and decreases it on the upwards moving side with the magnitude of the friction. That is if the tension would be T without any friction it will be $$T_2=T+T_f$$ and $$T_1=T-T_f$$ The resulting torque of these two tensions on the pulley will cause an angular acceleration according to $$\tau=I\alpha$$ The connection between the acceleration of the masses and that of the pulley is that it's tangential acceleration is given by $$a=a_t=r\alpha$$ Last edited: Oct 26, 2005	{"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": 0.9152642488479614, "perplexity": 927.2330848787396}, "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/1480698542932.99/warc/CC-MAIN-20161202170902-00054-ip-10-31-129-80.ec2.internal.warc.gz"}
http://www.conservapedia.com/First-order_language	# First-order language A First-order language in Zermelo-Fraenkel set theory consists of the following symbols: 1. A set of constants, such as A, B, C, ... 2. A set of n-ary relations such as >(x, y). 3. A set of n-ary functions such as +(x, y). 4. An infinite set of variables such as x, y, z,... 5. The connectives $\neg$, $\wedge$. 6. quantifiers: $\forall$, $\exists$. 7. parenthesis: (, ). 8. The equality symbol = . First-order languages are used to describe mathematics in mathematical notation with mathematical formulae.	{"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": 4, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9912686944007874, "perplexity": 2040.4156744047168}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-27/segments/1435376073161.33/warc/CC-MAIN-20150627033433-00056-ip-10-179-60-89.ec2.internal.warc.gz"}

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