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http://openstudy.com/updates/4e9cddfd0b8b3194bed07496 | • anonymous
Write a function named dayCount() that accepts a month, day, and year as its input arguments; calcuates an integer representing the total number of days from the turn of the century to the date that's passed; and returns the calculated integer to the calling function. For this problem, assume each year has 365 days and each month has 30 days. You need to write a main function to prompt for year, month and day. You need to validate the input and ask user reenter one if not correct. Note you should use exact function name in your program.
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Not the answer you are looking for? Search for more explanations. | 2017-03-30 09:30:58 | {"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": 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": 0.3116894066333771, "perplexity": 1114.1948619265054}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1490218193288.61/warc/CC-MAIN-20170322212953-00322-ip-10-233-31-227.ec2.internal.warc.gz"} |
https://codegolf.meta.stackexchange.com/questions/linked/2140?sort=newest | 241 questions linked to/from Sandbox for Proposed Challenges
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### Is “A solution wins if there's no strictly better solution in multiple measures” a valid winning criterion?
To cite a recently sandboxed challenge by l4m2: Solution win if there's no strictly better solution, aka. no solution that take less or same amount of items in mapping table, less or same amount of ...
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It seems now few questions show enough serious sense, and lots are just random requirements. Where do good ones go, are they dilutioned, deleted or just posted less?
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### Should the Info Box Link to the Active Sorted Sandbox?
The link to the sandbox in the info box is nice, but I was wondering if it might be better to link to change the target of the link from ... | 2020-07-10 20:15:20 | {"extraction_info": {"found_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": 0.46768423914909363, "perplexity": 3670.8303622632006}, "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-2020-29/segments/1593655911896.73/warc/CC-MAIN-20200710175432-20200710205432-00042.warc.gz"} |
https://hal.inria.fr/hal-01657005 | # Square on Deterministic, Alternating, and Boolean Finite Automata
Abstract : We investigate the state complexity of the square operation on languages represented by deterministic, alternating, and Boolean automata. For each k such that $1 \le k \le n-2$, we describe a binary language accepted by an n-state DFA with k final states meeting the upper bound $n2^n - k2^{n-1}$ on the state complexity of its square. We show that in the case of $k=n-1$, the corresponding upper bound cannot be met. Using the DFA witness for square with $2^n$ states where half of them are final, we get the tight upper bounds on the complexity of the square operation on alternating and Boolean automata.
Document type :
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### Citation
Ivana Krajňáková, Galina Jirásková. Square on Deterministic, Alternating, and Boolean Finite Automata. 19th International Conference on Descriptional Complexity of Formal Systems (DCFS), Jul 2017, Milano, Italy. pp.214-225, ⟨10.1007/978-3-319-60252-3_17⟩. ⟨hal-01657005⟩
Record views | 2022-08-18 13:26:19 | {"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": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6376892328262329, "perplexity": 2433.093862734856}, "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/1659882573197.34/warc/CC-MAIN-20220818124424-20220818154424-00583.warc.gz"} |
https://www.math.tugraz.at/mathc/new/index.php?link=talks | ## Seminar Talks
List of seminar talks (pdf) since 1999.
#### Strukturtheorie-Seminar
Title: Central Limit Theorem for the capacity of the range of stable random walks
Speaker: Dr. Stjepan Šebek (Univ. Zagreb)
Date: Thursday, 25. April 2019, 15:15
Room: Seminar room A306, Steyrerg. 30/3
Abstract:
In this talk, we will establish a central limit theorem for the capacity of
the range process for a class of d-dimensional symmetric alpha-stable random
walks with the index satisfying $d \geq 3\alpha$. Our approach is based on
controlling the limit behavior of the variance of the capacity of the range
process which then allows us to apply the Lindeberg-Feller theorem.
#### Strukturtheorie-Seminar
Title: Jacobi Polynomials and the Discrete Laguerre Operator
Speaker: Aleksey Kostenko (Univerza v Ljubljani / Universität Wien)
Date: 28.3.2019, 14:00 c.t.
Room: SR AE02, Steyrergasse 30, ground floor
Abstract:
The talk is focused on Bernstein-type estimates for Jacobi polynomials and their applications to various branches in mathematics. This is an old topic but we want to add a new wrinkle by establishing some intriguing connections with dispersive estimates for a certain class of Schrödinger equations whose Hamiltonian is given by the generalized Laguerre operator, i.e., the Jacobi matrix associated with generalized Laguerre polynomials. These operators feature prominently in the recent study of nonlinear waves in (2+1)-dimensional noncommutative scalar field theory since they appear as the linear part in the nonlinear Klein--Gordon and the nonlinear Schrödinger equations investigated in the recent of Chen, Fröhlich and Walcher (2003) and Krueger and Soffer (2015), respectively.
We show that dispersive estimates for the evolution group are connected with Bernstein-type inequalities for Jacobi polynomials. We use known uniform estimates for Jacobi polynomials to establish some new dispersive estimates. In turn, the optimal dispersive decay estimates lead to new Bernstein-type inequalities.
The talk is based on joint work with T. H. Koornwinder (Amsterdam) and G. Teschl (Vienna).
#### Vortrag Habilitationswerberin
Title: Cluster growth models and fractals
Speaker: Dr. Ecaterina Sava-Huss (TU Graz)
Date: Donnerstag, 14.2.2019, 11:00 s.t.
Room: Seminarraum AE06, Steyrergasse 30, EG
Abstract:
A significant part of my research deals with understanding the behavior of the following cluster growth models: internal diffusion limited aggregation, the rotor model, and the divisible sandpile model.
These models can be run on any infinite state space, and they are based on particles moving around according to some rule (that can be either random or deterministic) and aggregating. Describing the limit shape of the cluster whjich these particles produce is one of the main questions one would like to answer. For some of the models, the limit shape is hard to understand, and according to simulations, the fractal nature of the sets they produce is, from the mathematical point of view, far away from being understood. I will present several results concerning the limit shape of the clusters. In particular, I will present a limit shape universality result on the Sierpinski gasket graph, and conclude with some future research directions one can pursue within this topic.
#### Strukturtheorie-Seminar (Master-Vortrag)
Title: The Wiener index of Schreier graphs of the basilica automaton
Speaker: Stefan Hammer (TU Graz)
Date: 7.2.2019, 11:00 s.t.
Room: Seminar room AE02, Steyrergasse 30, ground floor
Abstract:
Automata and graphs are associated in many ways. For invertible automata one can
define the automaton group and observe its action on the set of finite words over the
input alphabet. This leads to the construction of Schreier graphs.
The sum of all distances in a graph, called Wiener index, is a graph property of wide interest. Harry Wiener showed that the properties of molecules are related to the Wiener index of chemical structural formulas. In my presentation I am going to introduce all necessary tools and prove an upper bound for the Wiener index of Schreier graphs of the Basilica automaton.
#### Workshop
Title: Groups, Automata and Graphs
Speaker: https://www.math.tugraz.at/GAG/ ()
Date: February 11-12, 2019
Room: Seminarroom AE06
Abstract:
More information concerning the talks and the speakers can be found on the webpage: https://www.math.tugraz.at/GAG/
If you want to attend the workshop, please let us know in order to organize the coffee breaks. | 2019-05-20 15:48:51 | {"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": 0.5216811895370483, "perplexity": 1280.8178359412072}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232256040.41/warc/CC-MAIN-20190520142005-20190520164005-00284.warc.gz"} |
http://server3.wikisky.org/starview?object=IC+5217 | WIKISKY.ORG
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# IC 5217
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IC 5217 as a double-shell, point-symmetric planetary nebula with a very narrow waistAims.Identification of the structural components and analysis of theinternal kinematics in the planetary nebula IC 5217. Methods:.Narrow-band images and high resolution long-slit spectra in theHα, [N ii] and [O iii] emission lines, and VLA 6 cm continuumdata. Results: .IC 5217 is composed of a very bright equatorialring, open bipolar lobes, off-axis point-symmetric features, on-axisdistant faint regions, and an off-axis very elongated bipolar structure.The ring, open lobes, point-symmetric features and distant faint regionsappear as elements of a single point-symmetric bipolar shell with a verynarrow waist. This shell presents an axis ratio of ~37 and an aspectratio of ~5. The 6 cm data show that the ring is an extremely flat diskwith a central hole. Expansion velocity in the ring ranges from ≤10km s-1 in He ii up to ≃27 km s-1 in [N ii],whereas a velocity of ~460 km s-1 is estimated for the polarregions of the bipolar shell. Strong acceleration of the outer regionsof the ring is observed. The elongated bipolar structure probablyrepresents a highly collimated (aspect ratio ~12), high velocitycylindrical-like shell. A collimated agent (wind or jet) would accountfor the shaping of the bipolar shell if this agent has operated in thedirection perpendicular to the equatorial disk. The point-symmetricfeatures and cylindrical shell are probably related to collimatedejections but that occurred when the basic nebular shape had alreadybeen established. Galactic Planetary Nebulae with Wolf-Rayet Nuclei III. Kinematical Analysis of a Large Sample of NebulaeExpansion velocities (V_{exp}) of different ions and line widths at thebase of the lines are measured and analyzed for 24 PNe with [WC]-typenuclei (WRPNe), 9 PNe ionized by WELS (WLPNe) and 14 ordinary PNe. Acomparative study of the kinematical behavior of the sample clearlydemonstrates that WRPNe have on average 40-45% larger V_{exp}, andpossibly more turbulence than WLPNe and ordinary PNe. WLPNe havevelocity fields very much like the ones of ordinary PNe, rather than theones of WRPNe. All the samples (WRPNe, WLPNe and ordinary PNe) showexpansion velocities increasing with age indicators, for example is larger for low-density nebulae and also it is largerfor nebulae around high-temperature stars. This age effect is muchstronger for evolved WRPNe, suggesting that the [WC] winds have beenaccelerating the nebulae for a long time, while for non-WRPNe theacceleration seems to stop at some point when the star reaches atemperature of about 90,000 - 100,000. Non-WR nebulae reach a maximumV_{exp} ≤ 30 km s(-1) evolved WRPNe reach maximum V_{exp} about 40km s(-1) . For all kinds of objects (WRPNe and non-WRPNe) it is foundthat on average V_{exp}(N(+) ) is slightly larger than V_{exp}(O(++) ),indicating that the nebulae present acceleration of the external shells. A Spectrophotometric Survey of K-Band Emission Lines in Planetary NebulaeWe present observations of 16 planetary nebulae (PNs) in the 2 μm (Kband) spectral region, obtained with a long-slit near-infraredspectrometer at McDonald Observatory. In general, the strongest featuresin our spectra are recombination lines of H I, He I, and (in some cases)He II. Half the sample shows emission from vibrationally excitedH2. Some of the observed PNs (e.g., M 1-13) displayH2 line ratios characteristic of shocked, thermalized gas,while others (e.g., BD +30 3639) have ratios intermediate between pureradiative (UV) and shock excitation, consistent with either acombination of the mechanisms or UV illumination of dense material. Ourspectra of J900 and M 1-13 confirm that published narrowband imagestrace the H2 emission, and we find that the H2emission in SwSt 1 has a larger spatial extent than previously reported.In IC 5117, SwSt 1, and NGC 40 we detect the [Kr III] 2.199 μm lineidentified by Dinerstein in 2001, with strengths indicating that kryptonis enriched relative to the solar abundance, most markedly so in NGC 40.We also detect several lines from the 3G term of [Fe III] inVy 2-2, SwSt 1, and marginally in Cn 3-1. The [Kr III] and [Fe III]lines fall near in wavelength to H2 transitions, which areoften used as diagnostics for UV excitation because they arise fromhigher vibrationally excited levels (v=2, 3). For moderate spectralresolving power, R<=600, these lines may be blended with, or evenmistaken for, the corresponding H2 lines, leading tomisinterpretation of the H2 emission. The strength of boththe Kr and Fe nebular emission lines can be enhanced by specialcircumstances, Kr because of nucleosynthetic self-enrichment in theprogenitor star and Fe due to inefficient initial dust condensation orpartial destruction of the dust after formation, causing a largerfraction of the elemental iron to reside in the gas phase. The abundance discrepancy - recombination line versus forbidden line abundances for a northern sample of galactic planetary nebulaeWe present deep optical spectra of 23 galactic planetary nebulae, whichare analysed in conjunction with archival infrared and ultravioletspectra. We derive nebular electron temperatures based on standardcollisionally excited line (CEL) diagnostics as well as the hydrogenBalmer jump and find that, as expected, the Balmer jump almost alwaysyields a lower temperature than the [OIII] nebular-to-auroral lineratio. We also make use of the weak temperature dependence of helium andOII recombination line ratios to further investigate the temperaturestructure of the sample nebulae. We find that, in almost every case, thederived temperatures follow the relation , which is the relationpredicted by two-component nebular models in which one component is coldand hydrogen-deficient. Te(OII) may be as low as a fewhundred Kelvin, in line with the low temperatures found for thehydrogen-deficient knots of Abell 30 by Wesson, Liu and Barlow.Elemental abundances are derived for the sample nebulae from both CELsand optical recombination lines (ORLs). ORL abundances are higher thanCEL abundances in every case, by factors ranging from 1.5 to 12. Fiveobjects with O2+ abundance discrepancy factors greater than 5are found. DdDm 1 and Vy 2-2 are both found to have a very largeabundance discrepancy factor of 11.8.We consider the possible explanations for the observed discrepancies.From the observed differences between Te(OIII) andTe(BJ), we find that temperature fluctuations cannot resolvethe abundance discrepancies in 22 of the 23 sample nebulae, implyingsome additional mechanism for enhancing ORL emission. In the oneambiguous case, the good agreement between abundances derived fromtemperature-insensitive infrared lines and temperature-sensitive opticallines also points away from temperature fluctuations being present. Theobserved recombination line temperatures, the large abundancediscrepancies and the generally good agreement between infrared andoptical CEL abundances all suggest instead the existence of a coldhydrogen-deficient component within the normal' nebular gas. The originof this component is as yet unknown. The Chemical Composition of Galactic Planetary Nebulae with Regard to Inhomogeneity in the Gas Density in Their EnvelopesThe results of a study of the chemical compositions of Galacticplanetary nebulae taking into account two types of inhomogeneity in thenebular gas density in their envelopes are reported. New analyticalexpressions for the ionization correction factors have been derived andare used to determine the chemical compositions of the nebular gas inGalactic planetary nebulae. The abundances of He, N, O, Ne, S, and Arhave been found for 193 objects. The Y Z diagrams for various Heabundances are analyzed for type II planetary nebulae separately andjointly with HII regions. The primordial helium abundance Y p andenrichment ratio dY/dZ are determined, and the resulting values arecompared with the data of other authors. Radial abundance gradients inthe Galactic disk are studied using type II planetary nebulae. Helium recombination spectra as temperature diagnostics for planetary nebulaeElectron temperatures derived from the HeI recombination line ratios,designated Te(HeI), are presented for 48 planetary nebulae(PNe). We study the effect that temperature fluctuations inside nebulaehave on the Te(HeI) value. We show that a comparison betweenTe(HeI) and the electron temperature derived from the Balmerjump of the HI recombination spectrum, designated Te(HI),provides an opportunity to discriminate between the paradigms of achemically homogeneous plasma with temperature and density variations,and a two-abundance nebular model with hydrogen-deficient materialembedded in diffuse gas of a normal' chemical composition (i.e.~solar), as the possible causes of the dichotomy between the abundancesthat are deduced from collisionally excited lines and those deduced fromrecombination lines. We find that Te(HeI) values aresignificantly lower than Te(HI) values, with an averagedifference of = 4000 K. Theresult is consistent with the expectation of the two-abundance nebularmodel but is opposite to the prediction of the scenarios of temperaturefluctuations and/or density inhomogeneities. From the observeddifference between Te(HeI) and Te(HI), we estimatethat the filling factor of hydrogen-deficient components has a typicalvalue of 10-4. In spite of its small mass, the existence ofhydrogen-deficient inclusions may potentially have a profound effect inenhancing the intensities of HeI recombination lines and thereby lead toapparently overestimated helium abundances for PNe. A reexamination of electron density diagnostics for ionized gaseous nebulaeWe present a comparison of electron densities derived from opticalforbidden line diagnostic ratios for a sample of over a hundred nebulae.We consider four density indicators, the [O II]λ3729/λ3726, [S II] λ6716/λ6731, [Cl III]λ5517/λ5537 and [Ar IV] λ4711/λ4740 doubletratios. Except for a few H II regions for which data from the literaturewere used, diagnostic line ratios were derived from our own high qualityspectra. For the [O II] λ3729/λ3726 doublet ratio, we findthat our default atomic data set, consisting of transition probabilitiesfrom Zeippen (\cite{zeippen1982}) and collision strengths from Pradhan(\cite{pradhan}), fit the observations well, although at high electrondensities, the [O II] doublet ratio yields densities systematicallylower than those given by the [S II] λ6716/λ6731 doubletratio, suggesting that the ratio of transition probabilities of the [OII] doublet, A(λ3729)/A(λ3726), given by Zeippen(\cite{zeippen1982}) may need to be revised upwards by approximately 6per cent. Our analysis also shows that the more recent calculations of[O II] transition probabilities by Zeippen (\cite{zeippen1987a}) andcollision strengths by McLaughlin & Bell (\cite{mclaughlin}) areinconsistent with the observations at the high and low density limits,respectively, and can therefore be ruled out. We confirm the earlierresult of Copetti & Writzl (\cite{copetti2002}) that the [O II]transition probabilities calculated by Wiese et al. (\cite{wiese}) yieldelectron densities systematically lower than those deduced from the [SII] λ6716/λ6731 doublet ratio and that the discrepancy ismost likely caused by errors in the transition probabilities calculatedby Wiese et al. (\cite{wiese}). Using our default atomic data set for [OII], we find that Ne([O II]) Ne([S II]) ≈Ne([Cl III])< Ne([Ar IV]). Planetary nebula distances re-examined: an improved statistical scaleThe distances of planetary nebulae (PNe) are still quite uncertain.Although observational estimates are available for a small proportion ofPNe, based on statistical parallax and the like, such distances are verypoorly determined for the majority of galactic PNe. In particular,estimates of so-called statistical' distance appear to differ byfactors of ~2.7.We point out that there is a well-defined correlation between the 5-GHzluminosity of the sources, L5, and their brightnesstemperatures, TB. This represents a different trend to thoseinvestigated in previous statistical analyses, and permits us todetermine independent distances to a further 449 outflows. Thesedistances are shown to be closely comparable to those determined using aTB-R correlation, providing that the latter trend is taken tobe non-linear.This non-linearity in the TB-R plane has not been noted inprevious analyses, and is likely responsible for the broad (andconflicting) ranges of distance that have previously been published.Finally, we point out that there is a close accord between observedtrends within the L5-TB and TB-Rplanes, and the variation predicted through nebular evolutionarymodelling. This is used to suggest that observational biases areprobably modest, and that our revised distance scale is reasonablytrustworthy. On the O II Ground Configuration Energy LevelsThe most accurate way to measure the energy levels for the O II2p3 ground configuration has been from the forbidden lines inplanetary nebulae. We present an analysis of modern planetary nebuladata that nicely constrain the splitting within the 2D termand the separation of this term from the ground4S3/2 level. We extend this method to H II regionsusing high-resolution spectroscopy of the Orion Nebula, covering all sixvisible transitions within the ground configuration. These data confirmthe splitting of the 2D term while additionally constrainingthe splitting of the 2P term. The energies of the2P and 2D terms relative to the ground(4S) term are constrained by requiring that all six linesgive the same radial velocity, consistent with independent limits placedon the motion of the O+ gas and the planetary nebula data. Electron temperatures and densities of planetary nebulae determined from the nebular hydrogen recombination spectrum and temperature and density variationsA method is presented to derive electron temperatures and densities ofplanetary nebulae (PNe) simultaneously, using the observed hydrogenrecombination spectrum, which includes continuum and line emission. Bymatching theoretical spectra to observed spectra around the Balmer jumpat about 3646 Å, we determine electron temperatures and densitiesfor 48 Galactic PNe. The electron temperatures based on this method -hereafter Te(Bal) - are found to be systematically lower thanthose derived from [OIII] λ4959/λ4363 and [OIII] (88 μm+ 52 μm)/λ4959 ratios - hereafterTe([OIII]na) andTe([OIII]fn). The electron densities based on thismethod are found to be systematically higher than those derived from[OII] λ3729/λ3726, [SII] λ6731/λ6716,[ClIII] λ5537/λ5517, [ArIV] λ4740/λ4711 and[OIII] 88 μm/52 μm ratios. These results suggest that temperatureand density fluctuations are generally present within nebulae. Thecomparison of Te([OIII]na) and Te(Bal)suggests that the fractional mean-square temperature variation(t2) has a representative value of 0.031. A majority oftemperatures derived from the Te([OIII]fn) ratioare found to be higher than those of Te([OIII]na),which is attributed to the existence of dense clumps in nebulae - those[OIII] infrared fine-structure lines are suppressed by collisionalde-excitation in the clumps. By comparingTe([OIII]fn), Te([OIII]na)and Te(Bal) and assuming a simple two-density-componentmodel, we find that the filling factor of dense clumps has arepresentative value of 7 × 10-5. The discrepanciesbetween Te([OIII]na) and Te(Bal) arefound to be anticorrelated with electron densities derived from variousdensity indicators; high-density nebulae have the smallest temperaturediscrepancies. This suggests that temperature discrepancy is related tonebular evolution. In addition, He/H abundances of PNe are found to bepositively correlated with the difference betweenTe([OIII]na) and Te(Bal), suggestingthat He/H abundances might have been overestimated generally because ofthe possible existence of H-deficient knots. Electron temperatures anddensities deduced from spectra around the Paschen jump regions at 8250Åare also obtained for four PNe: NGC 7027, NGC 6153, M 1-42 andNGC 7009. Electron densities derived from spectra around the Paschenjump regions are in good agreement with the corresponding values derivedfrom spectra around the Balmer jump, whereas temperatures deduced fromthe spectra around the Paschen jump are found to be lower than thecorresponding values derived from spectra around the Balmer jump for allthe four cases. The reason remains unclear. A reanalysis of chemical abundances in galactic PNe and comparison with theoretical predictions New determinations of chemical abundances for He, N, O, Ne, Ar and Sare derived for all galactic planetary nebulae (PNe) so far observedwith a relatively high accuracy, in an effort to overcome differences inthese quantities obtained over the years by different authors usingdifferent procedures. These include: ways to correct for interstellarextinction, the atomic data used to interpret the observed line fluxes,the model nebula adopted to represent real objects and the ionizationcorrections for unseen ions. A unique good quality' classical-typeprocedure, i.e. making use of collisionally excited forbidden lines toderive ionic abundances of heavy ions, has been applied to allindividual sets of observed line fluxes in each specific position withineach PN. Only observational data obtained with linear detectors, andsatisfying some quality' criteria, have been considered. Suchobservations go from the mid-1970s up to the end of 2001. Theobservational errors associated with individual line fluxes have beenpropagated through the whole procedure to obtain an estimate of theaccuracy of final abundances independent of an author's prejudices'.Comparison of the final abundances with those obtained in relevantmulti-object studies on the one hand allowed us to assess the accuracyof the new abundances, and on the other hand proved the usefulness ofthe present work, the basic purpose of which was to take full advantageof the vast amount of observations done so far of galactic PNe, handlingthem in a proper homogeneous way. The number of resulting PNe that havedata of an adequate quality to pass the present selection amounts to131. We believe that the new derived abundances constitute a highlyhomogeneous chemical data set on galactic PNe, with realisticuncertainties, and form a good observational basis for comparison withthe growing number of predictions from stellar evolution theory. Owingto the known discrepancies between the ionic abundances of heavyelements derived from the strong collisonally excited forbidden linesand those derived from the weak, temperature-insensitive recombinationlines, it is recognized that only abundance ratios between heavyelements can be considered as satisfactorily accurate. A comparison withtheoretical predictions allowed us to assess the state of the art inthis topic in any case, providing some findings and suggestions forfurther theoretical and observational work to advance our understandingof the evolution of low- and intermediate-mass stars. 12C/13C Ratio in Planetary Nebulae from the IUE ArchivesWe investigated the abundance ratio of 12C/13C inplanetary nebulae by examining emission lines arising from C III2s2p3Po2,1,0-->2s21S0.Spectra were retrieved from the International Ultraviolet Explorerarchives, and multiple spectra of the same object were co-added toachieve improved signal-to-noise ratio. The 13C hyperfinestructure line at 1909.6 Å was detected in NGC 2440. The12C/13C ratio was found to be ~4.4+/-1.2. In allother objects, we provide an upper limit for the flux of the 1910Å line. For 23 of these sources, a lower limit for the12C/13C ratio was established. The impact on ourcurrent understanding of stellar evolution is discussed. The resultinghigh-signal-to-noise ratio C III spectrum helps constrain the atomicphysics of the line formation process. Some objects have the measured1907/1909 Å flux ratio outside the low-electron densitytheoretical limit for 12C. A mixture of 13C with12C helps to close the gap somewhat. Nevertheless, someobserved 1907/1909 Å flux ratios still appear too high to conformto the currently predicted limits. It is shown that this limit, as wellas the 1910/1909 Å flux ratio, are predominantly influenced byusing the standard partitioning among the collision strengths for themultiplet1S0-3PoJaccording to the statistical weights. A detailed calculation for thefine-structure collision strengths between these individual levels wouldbe valuable. Sulfur, Chlorine, and Argon Abundances in Planetary Nebulae. IV. Synthesis and the Sulfur AnomalyWe have compiled a large sample of O, Ne, S, Cl, and Ar abundances thathave been determined for 85 Galactic planetary nebulae in a consistentand homogeneous manner using spectra extending from 3600 to 9600Å. Sulfur abundances have been computed using the near-IR lines of[S III] λλ9069, 9532 along with [S III] temperatures. Wefind average values, expressed logarithmically with a standarddeviation, of log(S/O)=-1.91+/-0.24, log(Cl/O)=-3.52+/-0.16, andlog(Ar/O)=-2.29+/-0.18, numbers consistent with previous studies of bothplanetary nebulae and H II regions. We also find a strong correlationbetween [O III] and [S III] temperatures among planetary nebulae. Inanalyzing abundances of Ne, S, Cl, and Ar with respect to O, we find atight correlation for Ne-O, and loose correlations for Cl-O and Ar-O.All three trends appear to be colinear with observed correlations for HII regions. S and O also show a correlation, but there is a definiteoffset from the behavior exhibited by H II regions and stars. We suggestthat this S anomaly is most easily explained by the existence ofS+3, whose abundance must be inferred indirectly when onlyoptical spectra are available, in amounts in excess of what is predictedby model-derived ionization correction factors in PNe. Finally for thedisk PNe, abundances of O, Ne, S, Cl, and Ar all show gradients whenplotted against Galactocentric distance. The slopes are statisticallyindistinguishable from one another, a result which is consistent withthe notion that the cosmic abundances of these elements evolve inlockstep. Characteristics of Planetary Nebulae with [WC] Central StarsWe have analyzed the plasma diagnostics (electron densities andtemperatures and abundance ratios), and the kinematics of a large sampleof planetary nebulae around [WC] stars by means of high resolutionspectra. The results have been compared with characteristics ofplanetary nebulae around WELS and non-WR central stars. We find that theproportion of nitrogen rich nebulae is larger in WRPNe than innon-WRPNe. None of the 9 nebulae around WELS in our sample showsN-enrichment. WRPNe have larger expansion velocities and/or largerturbulence than non-WRPNe demonstrating that the mechanical energy ofthe massive [WC] stellar wind largely affects the kinematical behaviorof nebulae. A weak relation between stellar temperature and expansionvelocities has been found for all kind of nebulae, indicating that oldernebulae expand faster. The effect is more important for WRPNe. Thiscould be useful in testing the evolutionary sequence [WC]-late ->[WC]-early, proposed for [WC] stars. The relation between Zanstra temperature and morphology in planetary nebulaeWe have created a master list of Zanstra temperatures for 373 galacticplanetary nebulae based upon a compilation of 1575 values taken from thepublished literature. These are used to evaluate mean trends intemperature for differing nebular morphologies. Among the most prominentresults of this analysis is the tendency forη=TZ(HeII)/TZ(HeI) to increase with nebularradius, a trend which is taken to arise from the evolution of shelloptical depths. We find that as many as 87 per cent of nebulae may beoptically thin to H ionizing radiation where radii exceed ~0.16 pc. Wealso note that the distributions of values η and TZ(HeII)are quite different for circular, elliptical and bipolar nebulae. Acomparison of observed temperatures with theoretical H-burning trackssuggests that elliptical and circular sources arise from progenitorswith mean mass ≅ 1 Msolar(although the elliptical progenitors are probably more massive).Higher-temperature elliptical sources are likely to derive fromprogenitors with mass ≅2 Msolar, however, implying thatthese nebulae (at least) are associated with a broad swathe ofprogenitor masses. Such a conclusion is also supported by trends in meangalactic latitude. It is found that higher-temperature ellipticalsources have much lower mean latitudes than those with smallerTZ(HeII), a trend which is explicable where there is anincrease in with increasing TZ(HeII).This latitude-temperature variation also applies for most other sources.Bipolar nebulae appear to have mean progenitor masses ≅2.5Msolar, whilst jets, Brets and other highly collimatedoutflows are associated with progenitors at the other end of the massrange (~ 1 Msolar). Indeed it ispossible, given their large mean latitudes and low peak temperatures,that the latter nebulae are associated with the lowest-mass progenitorsof all.The present results appear fully consistent with earlier analyses basedupon nebular scale heights, shell abundances and the relativeproportions of differing morphologies, and offer further evidence for alink between progenitor mass and morphology. Galactic Planetary Nebulae and their central stars. I. An accurate and homogeneous set of coordinatesWe have used the 2nd generation of the Guide Star Catalogue (GSC-II) asa reference astrometric catalogue to compile the positions of 1086Galactic Planetary Nebulae (PNe) listed in the Strasbourg ESO Catalogue(SEC), its supplement and the version 2000 of the Catalogue of PlanetaryNebulae. This constitutes about 75% of all known PNe. For these PNe, theones with a known central star (CS) or with a small diameter, we havederived coordinates with an absolute accuracy of ~0\farcs35 in eachcoordinate, which is the intrinsic astrometric precision of the GSC-II.For another 226, mostly extended, objects without a GSC-II counterpartwe give coordinates based on the second epoch Digital Sky Survey(DSS-II). While these coordinates may have systematic offsets relativeto the GSC-II of up to 5 arcsecs, our new coordinates usually representa significant improvement over the previous catalogue values for theselarge objects. This is the first truly homogeneous compilation of PNepositions over the whole sky and the most accurate one available so far.The complete Table \ref{tab2} is only available in electronic form atthe CDS via anonymous ftp to cdsarc.u-strasbg.fr (130.79.128.5) or viahttp://cdsweb.u-strasbg.fr/cgi-bin/qcat?J/A+A/408/1029} Quantitative classification of WR nuclei of planetary nebulaeWe analyse 42 emission-line nuclei of Planetary Nebulae (PNe), in theframework of a large spectrophotometric survey of [WC] nuclei of PNeconducted since 1994, using low/medium resolution spectra obtained atESO and at OHP. We construct a grid of selected line-intensities(normalized to C Iv-5806 Å= 100) ordered by decreasing ionisationpotential going from 871 to 24 eV. In this grid, the stars appear tobelong clearly to prominent O (hot [WO1-4] types) or C (cooler [WC4-11]types) line-sequences, in agreement with the classification of massiveWR stars applied to Central Stars of Planetary Nebulae (CSPNe) byCrowther et al. \cite{crowther98} (CMB98). We propose 20 selected lineratios and the FWHM of C Iv and C Iii lines as classificationdiagnostics, which agree well with the 7 line ratios and the FWHMproposed by CMB98. This classification based on ionisation is related tothe evolution of the temperature and of the stellar wind, reflecting themass-loss history. In particular, inside the hot [WO4]-class, wediscover four stars showing very broad lines over the whole spectralrange. These stars possibly mark the transition from the initialmomentum-driven phase to the later energy-driven phase of the CSPNealong their evolution from the post-Asymptotic Giant Branch (post-AGB)phase through [WC] late, [WC4] and [WO]-types. The HR diagram and thediagram linking the terminal velocity and the temperature indicatehighly dispersed values of the stellar mass for our sample, around amean mass higher than for normal CSPNe. The distribution of the 42 starsalong the ionisation sequence shows 24% of [WO1-3], 21% of [WO4], 17% of[WC4] hot stars, and 26% of [WC9-11] cool stars. The [WC5-8] classesremain poorly represented (12%). This distribution is confirmed on thebasis of a large compilation of the 127 known emission-lines CSPNe,which represent about 5% of the known PNe.Based on observations obtained at the European Southern Observatory(ESO), La Silla (Chile), and at the Observatoire de Haute-Provence (OHP,France).Table \ref{liste} is only available in electronic form athttp://www.edpsciences.org Gas temperature and excitation classes in planetary nebulaeEmpirical methods to estimate the elemental abundances in planetarynebulae usually use the temperatures derived from the [O III] and [N II]emission-line ratios, respectively, for the high- and low-ionizationzones. However, for a large number of objects these values may not beavailable. In order to overcome this difficulty and allow a betterdetermination of abundances, we discuss the relationship between thesetwo temperatures. Although a correlation is not easily seen when asample of different PNe types is used, the situation is improved whenthey are gathered into excitation classes. From [OII]/[OIII] andHeII/HeI line ratios, we define four excitation classes. Then, usingstandard photoionization models which fit most of the data, a linearrelation between the two temperatures is obtained for each of the fourexcitation classes. The method is applied to several objects for whichonly one temperature can be obtained from the observed emission linesand is tested by recalculation of the radial abundance gradient of theGalaxy using a larger number of PNe. We verified that our previousgradient results, obtained with a smaller sample of planetary nebulae,are not changed, indicating that the temperature relation obtained fromthe photoionization models are a good approximation, and thecorresponding statistical error decreases as expected. Tables 3-5, 7 and9 are only available in electronic form at http://www.edpsciences.org Study of electron density in planetary nebulae. A comparison of different density indicatorsWe present a comparison of electron density estimates for planetarynebulae based on different emission-line ratios. We have considered thedensity indicators [O Ii]lambda 3729/lambda 3726, [S Ii]lambda6716/lambda 6731, [Cl Iii]lambda 5517/lambda 5537, [Ar Iv]lambda4711/lambda 4740, C Iii]lambda 1906/lambda 1909 and [N I]lambda5202/lambda 5199. The observational data were extracted from theliterature. We have found systematic deviations from the densityhomogeneous models, in the sense that: Ne(ion {N}i) <~Ne(ion {O}{ii}) < Ne(ion {S}{ii}, ion {C}{iii},ion {Cl}{iii} or ion {Ar}{iv}) and Ne(ion {S}{ii}) ~Ne(ion {C}{iii}) ~ Ne(ion {Cl}{iii}) ~Ne(ion {Ar}{iv}). We argue that the lower [O Ii] densityestimates are likely due to errors in the atomic parameters used. Sulfur, Chlorine, and Argon in Planetary Nebulae. I. Observations and Abundances in a Northern SampleThis paper is the first of a series specifically studying the abundancesof sulfur, chlorine, and argon in type II planetary nebulae (PNe) in theGalactic disk. Ratios of S/O, Cl/O, and Ar/O constitute important testsof differential nucleosynthesis of these elements and serve as strictconstraints on massive star yield predictions. We present newground-based optical spectra extending from 3600-9600 Å for asample of 19 type II northern PNe. This range includes the strongnear-infrared lines of [S III] λλ9069,9532, which allowsus to test extensively their effectiveness as sulfur abundanceindicators. We also introduce a new, model-tested ionization correctionfactor for sulfur. For the present sample, we find average values ofS/O=1.2×10-2+/-0.71×10-2,Cl/O=3.3×10-4+/-1.6×10-4, andAr/O=5.0×10-3+/-1.9×10-3. Helium contamination from the progenitor stars of planetary nebulae: The He/H radial gradient and the ΔY / ΔZ enrichment ratioIn this work, two aspects of the chemical evolution of 4He inthe Galaxy are considered on the basis of a sample of disk planetarynebulae (PN). First, an application of corrections owing to thecontamination of 4He from the evolution of the progenitorstars shows that the He/H abundance by number of atoms is reduced by0.012 to 0.015 in average, leading to an essentially flat He/H radialdistribution. Second, a determination of the helium to heavy elementenrichment ratio using the same corrections leads to values in the range2.8 < ΔY / ΔZ < 3.6 for Y p = 0.23 and 2.0< ΔY / ΔZ < 2.8 for Y p = 0.24, in goodagreement with recent independent determinations and theoretical models. Abundances in the Planetary Nebula IC 5217High-resolution optical wavelength spectroscopic data were secured inthe optical wavelengths, 3700-10050 Å, for the planetary nebula IC5217 with the Hamilton Echelle Spectrograph at Lick Observatory. Theseoptical spectra have been analyzed along with the near-UV and UV archivedata. Diagnostic analyses indicate a nebular physical condition withelectron temperature of about 10,700 K (from the [O III] lines) and thedensity of Nɛ=5000 cm-1. Ionicconcentrations have been derived with the representative diagnostics,and with the aid of a photoionization model construction, we derived theelemental abundances. Contrary to the previous studies found in theliterature, He and C appear to be depleted compared to the averageplanetary nebula and to the Sun (and S marginally so), while theremaining elements appear to be close to the average value. IC 5217 mayhave evolved from an O-rich progenitor, and the central star temperatureof IC 5217 is likely to be 92,000 K. An analysis of the observed radio emission from planetary nebulaeWe have analysed the radio fluxes for 264 planetary nebulae for whichreliable measurements of fluxes at 1.4 and 5 GHz, and of nebulardiameters are available. For many of the investigated nebulae, theoptical thickness is important, especially at 1.4 GHz. Simple modelslike the one specified only by a single optical thickness or spherical,constant density shells do not account satisfactorily for theobservations. Also an r-2 density distribution is ruled out.A reasonable representation of the observations can be obtained by atwo-component model having regions of two different values of opticalthickness. We show that the nebular diameters smaller than 10arcsec areuncertain, particularly if they come from photographic plates orGaussian fitting to the radio profile. While determining theinterstellar extinction from an optical to radio flux ratio, cautionshould be paid regarding optical thickness effects in the radio. We havedeveloped a method for estimating the value of self absorption. At 1.4GHz self absorption of the flux is usually important and can exceed afactor of 10. At 5 GHz self absorption is negligible for most of theobjects, although in some cases it can reach a factor of 2. The Galacticbulge planetary nebulae when used to calibrate the Shklovsky method givea mean nebular mass of 0.14 Msun. The statistical uncertaintyof the Shklovsky distances is smaller than a factor of 1.5. Table 1 isonly available in electronic form at http://www.edpsciences.org. Galactic planetary nebulae with Wolf-Rayet nuclei. II. A consistent observational data setWe present high resolution spectrophotometric data for a sample of 34planetary nebulae with [WC] spectral type central stars (WRPNe) in ourGalaxy. The observed objects cover a wide range in stellarcharacteristics: early and late [WC] type stars, as well asweak-emission line stars (WELS). Physical conditions in the nebulae(electron density and temperatures) have been obtained from variousdiagnostic line ratios, and chemical abundances have been derived withthe usual empirical scheme. Expansion velocities were estimated in aconsistent manner from the line profiles for most objects of the sample.A statistical study was developed for the derived data in order to findfundamental relationships casting some light on the evolutionary statusof WRPNe. We found evidence for a strong electron temperature gradientin WRPNe which is related to nebular excitation. Such a gradient is notpredicted in simple photoionization models. Abundance ratios indicatethat there seems to be no preferential stellar mass for the Wolf-Rayetphenomenon to occur in the nucleus of a planetary nebula. Two objects, M1-25 and M 1-32, were found to have a very small Ne/O ratio, a propertydifficult to understand. We reexamined the relation between the nebularproperties of the WRPNe and the spectral types of the central stars. Ourdata confirm the trend found by other authors of the electron densitydecreasing with decreasing spectral type, which was interpreted asevidence that [WC] stars evolve from late to early [WC] types. On theother hand, our data on the expansion velocities do not show theincrease of expansion velocity with decreasing spectral type, that onemight expect in such a scenario. Two objects with very late [WC] typecentral stars, K 2-16 and PM 1-188, do not follow the general densitysequence, being of very low density for their spectral types. We suggestthat the stars either underwent a late helium flash (the born again''scenario) or that they have had a particularly slow evolution from theAGB. The 6 WELS of our sample follow the same density vs. [WC]-typerelation as the bona fide WRPNe, but they tend to have smaller expansionvelocities. Considerations about the evolutionary status of WELS mustawait the constitution of a larger observational sample. The analysis ofthe differences between the WRPNe in the Magellanic Clouds (distributionof [WC] spectral types, N/O ratios) and in the Galaxy indicates thatmetallicity affects the [WR] phenomenon in central stars of planetarynebulae. Based on data obtained at the Observatorio AstronómicoNacional, SPM, B.C., México Tables 2 and 3 are only available athttp://www.edpsciences.org The Galactic disc distribution of planetary nebulae with warm dust emission features - IWe investigate the Galactic disc distribution of a sample of planetarynebulae characterized in terms of their mid-infrared spectral features.The total number of Galactic disc PNe with 8-13μm spectra is broughtup to 74 with the inclusion of 24 new objects, the spectra of which wepresent for the first time. 54 PNe have clearly identified warm dustemission features, and form a sample that we use to construct thedistribution of the C/O chemical balance in Galactic disc PNe. The dustemission features complement the information on the progenitor massesbrought by the gas-phase N/O ratios: PNe with unidentified infraredemission bands have the highest N/O ratios, while PNe with the silicatesignature have either very high N enrichment or close to none. We find atrend for a decreasing proportion of O-rich PNe towards the third andfourth Galactic quadrants. Two independent distance scales confirm thatthe proportion of O-rich PNe decreases from 30\pm 9 per cent inside the solar circle to 14\pm 7 per cent outside. PNe with warm dustare also the youngest. PNe with no warm dust are uniformly distributedin C/O and N/O ratios, and do not appear to be confined to C/O\sim 1. They also have higher 6-cmfluxes, as expected from more evolved PNe. We show that the IRAS fluxesare a good representation of the bolometric flux for compact andIR-bright PNe, which are probably optically thick. Selection of objectswith \fontshape{it}{F}(12\hphantom{0}\mu m)>0.5\hphantom{0} Jyshould probe a good portion of the Galactic disc for these young, denseand compact nebulae, and the dominant selection effects are rooted inthe PN catalogues. Gravity distances of planetary nebulae II. Aplication to a sample of galactic objects.Not Available On the abundance gradient of the galactic diskEstimates of the gas temperature in planetary nebulae obtained from the[O III] emission line ratio and from the Balmer discontinuity indicatedifferences reaching up to 6000 K (Liu & Danziger 1993). The [O III]temperature is commonly used to obtain the ionic fractions of highlyionized ions, particularly the O++ and Ne++ ions when using theempirical method to calculate the elemental abundances of photoionizedgas from the observed emission line intensities. However, if the gastemperature is overestimated the elemental abundances may beunderestimated. In particular this may lead to an incorrect elementalabundance gradient for the Galaxy, usually used as a constraint for thechemical evolution models. Using Monte Carlo simulations, we calculatethe systematic error introduced in the abundance gradient obtained fromplanetary nebulae by an overestimation of the gas temperature. Theresults indicate that the abundance gradient in the Galaxy should besteeper than previously assumed. The dust content of planetary nebulae: a reappraisalWe have performed a statistical analysis using broad band IRAS data onabout 500 planetary nebulae with the aim of characterizing their dustcontent. Our approach is different from previous studies in that it usesan extensive grid of photoionization models to test the methods forderiving the dust temperature, the dust-to-gas mass ratio and theaverage grain size. In addition, we use only distance independentdiagrams. With our models, we show the effect of contamination by atomiclines in the broad band IRAS fluxes during planetary nebula evolution.We find that planetary nebulae with very different dust-to-gas massratios exist, so that the dust content is a primordial parameter for theinterpretation of far infrared data of planetary nebulae. In contrastwith previous studies, we find no evidence for a decrease in thedust-to-gas mass ratio as the planetary nebulae evolve. We also showthat the decrease in grain size advocated by Natta & Panagia(\cite{NattaPanagia}) and Lenzuni et al. (\cite{Lenzuni}) is an artefactof their method of analysis. Our results suggest that the timescale fordestruction of dust grains in planetary nebulae is larger than theirlifetime. Table~1 is only accessible in electronic form at the CDS viaanonymous ftp to cdsarc.u-strasbg.fr (130.79.128.5) or viahttp://cdsweb.u-strasbg.fr/Abstract.html Infrared Planetary Nebulae in the NRAO VLA Sky SurveyIn order to construct a sample of planetary nebulae (PNe) unbiased bydust extinction, we first selected the 1358 sources in the IRAS PointSource Catalog north of J2000 declination delta=-40^deg having measuredS(25 μm)>=1 Jy and colors characteristic of PNe: detections orupper limits consistent with both S(12 μm)<=0.35S(25 μm) andS(25 μm)>=0.35S(60 μm). The majority are radio-quietcontaminating sources such as asymptotic giant branch stars. Free-freeemission from genuine PNe should make them radio sources. The 1.4 GHzNRAO VLA Sky Survey (NVSS) images and source catalog were used to rejectradio-quiet mid-infrared sources. We identified 454 IRAS sources withradio sources brighter than S~2.5 mJy beam^-1 (equivalent to T~0.8 K inthe 45" FHWM NVSS beam) by positional coincidence. They comprise 332known PNe in the Strasbourg-ESO Catalogue of Galactic Planetary Nebulaeand 122 candidate PNe, most of which lie at very low Galactic latitudes.Exploratory optical spectroscopic observations suggest that most ofthese candidates are indeed PNe optically dimmed by dust extinction,although some contamination remains from H II regions, Seyfert galaxies,etc. Furthermore, the NVSS failed to detect only 4% of the known PNe inour infrared sample. Thus it appears that radio selection can greatlyimprove the reliability of PN candidate samples withoutsacrificingcompleteness. Shaping Bipolar and Elliptical Planetary Nebulae: Effects of Stellar Rotation, Photoionization Heating, and Magnetic FieldsWe present two-dimensional hydrodynamical and magnetohydrodynamicalsimulations of the evolution of planetary nebulae formed through theinteraction of two succeeding, time-independent stellar winds. Bothwinds are modeled according to a consistent physical prescription forthe latitudinal dependence of their properties. We propose that singlestars with initial masses above ~1.3 M_solar can achieve near-criticalrotation rates during their superwind'' phase at the tip of theasymptotic giant branch (AGB). We show that the resulting equatoriallyconfined winds and their subsequent inflation to a double lobe structureby the post-AGB wind leads to the typical hourglass shape found in manyplanetary nebulae, such as MyCn 18. Following Chevalier & Luo andRóżyczka & Franco, we then combine the effect of amagnetic field in the post-AGB wind with rotating AGB winds. We obtainhighly collimated bipolar nebula shapes, reminiscent of M2-9 or He2-437. For sufficiently strong fields, ansae and jets, similar to thoseobserved in IC 4593 are formed in the polar regions of the nebula.Weaker fields are found to be able to account for the shapes ofclassical elliptical nebulae, e.g., NGC 6905, in the case of sphericallysymmetric AGB winds, which we propose for single stars with initialmasses below ~1.3 M_solar. Photoionization, via instabilities in theionization-shock front, can generate irregularities in the shape of thesimulated nebulae. In particular, it leads to the formation of cometaryknots, similar to those seen in the Helix nebula (NGC 7293). This effectmay also be responsible for large-scale irregularities like those foundin Sh 2-71 or WeSb 4. We arrive at a scenario in which the majority ofthe planetary nebula with their diverse morphologies is obtained fromsingle stars. This scenario is consistent with the Galactic distributionof the different nebula types, since spherical and ellipticalnebulae-which have a distribution with a large scale height above theGalactic plane-are ascribed to progenitor masses below ~1.3 M_solar,with magnetic effects introducing ellipticities. Bipolar nebulae, on theother hand-which are on average closer to the Galactic plane-are foundto stem from progenitors with initial masses above ~1.3 M_solar.
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http://mathoverflow.net/feeds/question/4953 | Super-linear time complexity lower bounds for any natural problem in NP? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T08:16:31Z http://mathoverflow.net/feeds/question/4953 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/4953/super-linear-time-complexity-lower-bounds-for-any-natural-problem-in-np Super-linear time complexity lower bounds for any natural problem in NP? Rune 2009-11-11T00:27:06Z 2010-06-10T15:44:25Z <p>Do we know any problem in NP which has a super-linear time complexity lower bound? Ideally, we would like to show that 3SAT has super-polynomial lower bounds, but I guess we're far away from that. I'd just like to know any examples of super-linear lower bounds.</p> <p>I know that the time hierarchy theorem gives us problems which can be solved in O(n^3) but not in O(n^2), etc. Thus I put the word "natural" in the question.</p> <p>I ask for problems in NP, because otherwise someone would give examples of EXP-complete problems. </p> <p>I know there are time-space tradeoffs for some problems in NP. I don't know if any of them imply a super-linear time complexity lower bound though.</p> <p>(To address a question below about machine models, consider either multitape Turing machines or the RAM model.)</p> http://mathoverflow.net/questions/4953/super-linear-time-complexity-lower-bounds-for-any-natural-problem-in-np/4957#4957 Answer by Sam Nead for Super-linear time complexity lower bounds for any natural problem in NP? Sam Nead 2009-11-11T00:47:34Z 2009-11-11T00:47:34Z <p>Running time depends on the model of computation. Eg the problem "Given (m, n, p) does m + n = p?" takes at least quadratic time on a one-tape Turing machine. So you can give a more reasonable model of computation... and the next person to come along will explain how it is much too strong, etc.</p> <p>Hmmm. Perhaps your question might be rephrased in terms of using some other resource (which in turn requires at least a set amount of time). So perhaps invent a decision problem around sorting, and appeal to the nlog(n) lower bound...</p> http://mathoverflow.net/questions/4953/super-linear-time-complexity-lower-bounds-for-any-natural-problem-in-np/4986#4986 Answer by Gil Kalai for Super-linear time complexity lower bounds for any natural problem in NP? Gil Kalai 2009-11-11T06:10:48Z 2009-11-11T06:10:48Z <p>The short answer is NO. The best general lower bounds are linear.</p> http://mathoverflow.net/questions/4953/super-linear-time-complexity-lower-bounds-for-any-natural-problem-in-np/9081#9081 Answer by Ryan Williams for Super-linear time complexity lower bounds for any natural problem in NP? Ryan Williams 2009-12-16T06:07:12Z 2009-12-16T06:50:04Z <p>Sorry I am so late to the discussion, but I just registered...</p> <p>There are non-linear time lower bounds on multitape Turing machines for NP-complete problems. These lower bounds follow from the fact that the class of problems solvable in nondeterministic linear time is not equal to the class of those solvable in (deterministic) linear time, in the multitape Turing machine setting. This is proved in:</p> <blockquote> <p>Wolfgang J. Paul, Nicholas Pippenger, Endre Szemerédi, William T. Trotter: On Determinism versus Non-Determinism and Related Problems (Preliminary Version) FOCS 1983: 429-438</p> </blockquote> <p>In fact, unraveling the proof shows that there must be some problem solvable in nondeterministic linear time that is not solvable in $o(n \cdot (\log^* n)^{1/4})$ time (again, on a multitape Turing machine). Note the * in the logarithm; this is just "barely" above linear. One known application of this result is that a natural NP-complete problem in automata theory cannot be solved in $o(n \cdot (\log^* n)^{1/4})$ time:</p> <blockquote> <p>Etienne Grandjean: A Nontrivial Lower Bound for an NP Problem on Automata. SIAM J. Comput. 19(3): 438-451 (1990)</p> </blockquote> <p>Unfortunately the lower bound of Paul et al. relies crucially on the geometry that arises from accessing one-dimensional tapes. We don't know how to prove a non-linear lower bound even if you allow the Turing machine to have a constant number of <b>two</b>-dimensional tapes. We can prove time lower bounds for NP problems on general computational models <i>if</i> you severely restrict the extra workspace used by the machine. (This is getting into my own work so I won't say more unless you're truly interested.)</p> <p>As for the comment above me: the sorting lower bound holds only in a comparison-based model, which is extremely restricted. The claim that sorting requires Omega(n log n) time on general computational models is false. There are faster algorithms for sorting integers. See for example:</p> <blockquote> <p>Yijie Han: Deterministic sorting in O(n log logn) time and linear space. J. Algorithms 50(1): 96-105 (2004) </p> </blockquote> http://mathoverflow.net/questions/4953/super-linear-time-complexity-lower-bounds-for-any-natural-problem-in-np/27711#27711 Answer by Timothy Chow for Super-linear time complexity lower bounds for any natural problem in NP? Timothy Chow 2010-06-10T15:44:25Z 2010-06-10T15:44:25Z <p>If you restrict <i>space</i> usage to be sublinear then you can prove superlinear <i>time</i> lower bounds on SAT. See <a href="http://rjlipton.wordpress.com/2009/04/13/sat-is-not-too-easy/" rel="nofollow">this article on Lipton's blog</a> for a nice exposition.</p> | 2013-06-20 08:16:34 | {"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": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7256712913513184, "perplexity": 1310.2673977662334}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1368711005723/warc/CC-MAIN-20130516133005-00045-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://academy.jahia.com/documentation/developer/jahia/8/extending-and-customizing-jahia-ui/customizing-and-extending-content-editor/configuring-and-customizing-ckeditor | Written by The Jahia Team
Developers
## CKEditor integration
This sections shows you how to customize CKEditor, the WYSIWYG rich text editor used for authoring content in Jahia. Jahia 8 includes CKEditor 4.13 embedded in React forms and uses the React pickers from Content Editor that come with the latest version of Jahia (for selecting files, folders, and content inside CKEditor).
There are different ways to customize CKEditor to fit your needs:
• Using one of the predefined CKEditor toolbars
## Overriding the CKEditor configuration
The CKEditor configuration is loaded in the following order of priority:
1. Any configuration at the property level in your content type definition.
2. A configuration defined in your template set.
3. The configuration defined at the platform level.
If none of these configurations is defined, the default CKEditor configuration applies.
### Configuration at the property level in content type definitions
You can define a custom configuration at the property level directly in the definition of a content type. With the following definition, the summary field of the myArticle nodetype will display the CKEditor with a minified toolbar:
[jnt:myArticle] > jnt:bigText, jmix:basicContent, jmix:editorialContent
- summary (string, richtext[ckeditor.toolbar='Mini']) i18n
You can also define the custom configuration file per field in a similar way, for example:
[bootstrap4mix:text] mixin
- text (string, richtext[ckeditor.toolbar='Tinny',ckeditor.customConfig='\$context/modules/bootstrap4-components/javascript/ckconfig.js']) i18n
The above definition will force CKEditor to use the defined configuration file when authoring content for that field.
Note: By default, CKEditor displays the toolbar depending on the permission the current user has in Jahia. A user can have Full or Basic permission and will see a Full or Basic toolbar, depending on permission they are granted. If a user has no explicit permission, the Light toolbar displays. Overriding the CKEditor toolbar definition, as in the example above, will also override this default logic.
### Configuration at the template set level
By including a /javascript/ckeditor_config.js file into a template set bundle, you can extend or override the CKEditor configuration for all rich text fields in your web projects that use this template set. The format of that file is the same as CKEditor's default config.js file. For more information, see Setting CKEditor Configuration - Using the config.js File.
### Configuration at the platform level
You can customize or override the CKEditor configuration globally, by providing it in an OSGi fragment bundle whose host is the "ckeditor" bundle.
To automate the bundle packaging and deployment process, you can use the CKEditor configuration utility, which is located in the Tools Area (http://[digital-factory-server]:[port]/tools/ckeditorConfig.jsp).
The tool allows you to build and download the configuration bundle or build and deploy the bundle into that Jahia instance to test the configuration immediately. The format of the provided configuration is the same as the one in the CKEditor's config.js file. For more information, see Setting CKEditor Configuration - Using the config.js File.
For example, by building and deploying the following configuration:
CKEDITOR.editorConfig = function( config ) {
config.extraPlugins='mathjax';
config.toolbar_Full[8]=['Image','Flash','Table','HorizontalRule','Smiley','SpecialChar','PageBreak','Mathjax'];
}
We enable the Mathematical Formulas plugin and add the corresponding button into the Full toolbar:
CKEditor is a modular framework that can be extended with many plugins. Version compatibility between CKEditor and the plugins is something to monitor to ensure a smooth integration. Please get in touch with the plugin editor if necessary. A list of the embedded plugins can be found here
To enable one of the embedded plugins, you only have to activate it in your js config, defined in any module.
• In the templateSet module, add the plugin.js under the folder: /javascript/ckeditor/plugins/<nameOfThePlugin>/plugin.js
• In the CKEditor configuration file, add the following line to load the plugin.js file: CKEDITOR.plugins.addExternal('<nameOfThePlugin>', '/modules/nameOfYourTemplateSet/javascript/ckeditor/plugins/nameOfThePlugin/plugin.js');
• In the config function, add your plugin to the extraPlugins variable: CKEDITOR.editorConfig = function( config ) { ... config.extraPlugins = '<nameOfThePlugin>'; ... }
That's it. The additional plugin should be available to all rich text editors inside Jahia.
### Examples
You can also provide template-set specific CKEditor configurations:
There is an example of the config file in the sample-bootstrap-templates module (ckconfg.js) that provides a custom toolbar, and defines custom styles and templates. The custom file is used in several fields from the sample-bootstrap-templates module:
## Toolbar definitions
The toolbar loaded in CKEditor depends on the permission of the user. There are predefined toolbars which are linked to a built-in permission in Jahia.
The predefined toolbars are: Full, Basic, Light, Mini and User. Users with:
• the View Full WYSIWYG Editor permission see the full toolbar
• the View Basic WYSIWYG Editor permission see the basic toolbar
• With no explicit permission see the Light toolbar in CKEditor.
If you have an override on a property level in place, this will be loaded in priority of the toolbar according to the permission check. | 2021-03-03 17:28:02 | {"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": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.18129311501979828, "perplexity": 6577.034195419746}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1614178367183.21/warc/CC-MAIN-20210303165500-20210303195500-00405.warc.gz"} |
https://www.tutorialspoint.com/the-runs-scored-in-a-cricket-match-by-11-players-is-as-follows-6-15-120-50-100-80-10-15-8-10-15find-the-mean-mode-and-median-of-this-data-are-th | # The runs scored in a cricket match by 11 players is as follows:6, 15, 120, 50, 100, 80, 10, 15, 8, 10, 15Find the mean, mode and median of this data. Are the three same?
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Given:
The runs scored in a cricket match by 11 players are as follows:
6, 15, 120, 50, 100, 80, 10, 15, 8, 10, 15
To do:
We have to find the mean, mode and median of this data and whether the three are same.
Solution:
Total number of players$=11$
Scores of the players $=$6, 15, 120, 50, 100, 80, 10, 15, 8, 10, 15
Arranging the scores into ascending order, we get
6, 8, 10, 15, 15, 15, 50, 80, 100, 120
$\boxed{ \text { Mean }=\frac{ \text { Sum of all scores }}{ \text { Total number of players }}}$
$=\frac{6+8+10+15+15+15+50+80+100+120}{11}$
$=\frac{429}{11}$
$=39$
Thus, mean $=39$
Mode is the observation that occurs the highest number of times.
Here, $15$ occurs $3$ times in the given data.
$\therefore$ Mode $=15$
The arranged data is: 6, 8, 10,10, 15, 15, 15, 50, 80, 100, 120
Median is the middlemost observation of the given data
There are $11$ observations here. Thus the middle value is the 6th observation.
Median$=15$ [6th observation]
Therefore, Mean$=39$, Mode$=15$ and median$=15$
No, the mean, mode, and median are not the same.
Updated on 10-Oct-2022 13:33:20 | 2022-12-09 22:27:34 | {"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": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3239297866821289, "perplexity": 2220.8864544571065}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1669446711552.8/warc/CC-MAIN-20221209213503-20221210003503-00672.warc.gz"} |
https://homework.cpm.org/category/CCI_CT/textbook/pc/chapter/5/lesson/5.1.1/problem/5-4 | ### Home > PC > Chapter 5 > Lesson 5.1.1 > Problem5-4
5-4.
On graph paper, sketch $y = \frac { 1 } { x }$, then sketch $y = \frac { 2 } { x }$
Complete the tables in the eTool below to help you with this problem.
Click the link at right for the full version of the eTool: 5-4 HW eTool | 2021-10-23 12:09:11 | {"extraction_info": {"found_math": true, "script_math_tex": 2, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.7139373421669006, "perplexity": 5530.299916061854}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585671.36/warc/CC-MAIN-20211023095849-20211023125849-00334.warc.gz"} |
http://ameroyer.github.io/python,/topology,/meshing/2018/03/01/alphacomplexes.html | This post is a short introduction and Python implementation of the Delaunay and Alpha Complexes triangulation in 2D. For additional reference, see for instance Computational Topology: An Introduction, H. Edelsbrunner and J. Harer, Chapter 3, Section 4.
### Introduction
Mesh generation is the problem of generating a mesh composed of polygons (or polyhedrons in higher dimensions) from a cloud of points. This has typical applications for instance in Computer Graphics, when one wants to produce a model of a surface matching a given point cloud. In particular, an interesting issue is how well the generated mesh respect the original shape that the point cloud is induced from. This obviously depends on the cloud density, but also on the meshing algorithm used.
In this post, I will focus on the 2D case for simplicity and introduce Delaunay Triangulation and Alpha Complexes for mesh generation. We will see that Alpha Complexes are a subset of the Delaunay Triangulation and that they are able to preserve topological information from the initial shape: Contrary to the Delaunay triangulation, it can generates meshes with holes thanks to a radius constraint.
### Basics: The convex hull
#### Point cloud generation
In the rest of this post, I will use point clouds generated by randomly sampling the space defined by a binary mask as inputs. In particular we will consider shapes displaying various topological content (i.e. in the 2D contents, different numbers of holes and connected components)
def generate_cloud_from_image(N, image):
# Threshold
image[image > 125] = 255
image[image < 125] = 0
# Sample from the white pixels
indices = np.random.choice(np.where(image.flatten() != 0)[0], size=N)
samples = np.zeros((N, 2))
samples[:, 0] = indices % image.shape[1]
samples[:, 1] = indices / image.shape[1]
#### The convex hull
A first possible very simple meshing algorithm is to take the convex hull of the points cloud. The convex hull of a point set is simply the tightest convex set that contains the whole set. While simple to build, this model is heavily constrained by the convex requirement, and is a bad approximation for any shape displaying non-convex features.
### The Delaunay triangulation
#### Definition
A triangulation of a 2D point cloud $S \in \mathbb{R}^2$ is triangulation of its convex hull, i.e. a partition of the hull in triangles whose vertices are points of $S$. Additionally, a Delaunay triangulation $DT(S)$ is such that no points in $S$ is inside any of the circumscribed circles to any of the triangles in $DT(S)$, which guarantees a certain regularity to it; In particular it typically prevents very elongated triangles.
Note: According to the definition, the Delaunay Triangulation also has a limiting convex constraint. In order to avoid this, a classical trick is to had some boundary points to form a bounding box around the point clouds, forming a new convex hull. Then after the triangulation is done, we simply remove the triangles for which any vertex lies on the boundary. That way, we retrieve a potentially non-convex triangluation of the original point cloud $S$.
#### Voronoi Diagram
An easy to compute the Delaunay triangulation is to characterize it as the dual of the Voronoi diagram of $S$. More specifically, for each point $x \in S$, its Voronoi cell is defined as the set of points in the space which are closer to $x$ than any other points in $S$: \begin{align} V_x = \left[ y \in \mathbb{R}^2,\ \mbox{s.t. } \forall x’ \in S,\ || y - x||_2 \leq ||y - x’||_2 \right] \end{align}
def voronoi_diagram(samples, ax=None):
# Extract Voronoi regions (sicpy)
vor = Voronoi(samples, qhull_options="Q0")
# vor_ridges; Maps edge index -> Center of incident Voronoi cells
n = len(vor.vertices)
vor_ridges = {min(edges) * n + max(edges):
((centers[0], vor.points[centers[0]]),
(centers[1], vor.points[centers[1]]))
for edges, centers in zip(vor.ridge_vertices, vor.ridge_points)}
#### Building the triangulation
Finally, the Delaunay triangulation is built as the dual of the Voronoi diagram, i.e. we form an edge between any two points $x, x' \in S$ if and only if their respective cells $V_x$ and $V_{x'}$ touch (have a common edge) in the Voronoi diagram.
\begin{align} DL(S): x \leftrightarrow x’ \iff V_x \mbox{ adjacent to } V_{x’} \end{align}
# Build vertices and triangles list
vertices = {}
for (ip, p), (iq, q) in vor_ridges.values():
vertices[ip] = p
vertices[iq] = q
# Build triangles for adjacent cells
triangles = []
for i, q in enumerate(neighbours):
for r in neighbours[i+1:]:
if max(q,r) in adjacency[min(q, r)] and
triangles.append(mp.Polygon(
[vertices[p], vertices[q], vertices[r]], closed=True))
### Alpha Complexes
As we have seen in the previous example, the Delaunay triangluation yields a much more interesting result shape than the convex hull. However, it produces a dense partition of the space and in particular doesn’t recover topological information from the shape such as holes or connected components.
Alpha complexes are a susbset of the Delaunay Triangulation that tackles this issue. As previously, we will use a dual structure. More specifically, Alpha complexes are defined as the dual construction of the restricted Voronoi diagram, $Vor_r(S)$. Which is simply the intersection of the Voronoi diagram with balls of radius $r$ centered on every point in $S$.
\begin{align} Vor_r(S) = \left[ V_x \cap B_{r}(x),\ \forall x \in S \right] \end{align}
#### Line-circle intersection
In order to build the restricted Voronoi diagram, we need to start from the initial Voronoi diagram and compute its intersections with balls of radius $r$. In 2D, this means we need to compute intersections between circles and the lines composing the diagram.
The first two easy cases are when the line segment of the diagram, [p, q] is fully inside or fully outside the circle.
# Case 1: If p and q are both in the circle -> clip to [p, q]
if is_in_circle(p, center, r) and is_in_circle(q, center, r):
return [(p, q)], True, True
# Intersection with line y = ax + b
# Express the line equation as y = slope * x + intersect
slope = (q[1] - p[1]) / (q[0] - p[0])
intersect = q[1] - slope * q[0]
# Express the intersection problem as a quadratic equation ax2 + bx + c
# we need to solve the system:
# (1) y = slope * x + b
# (2) (x - center[0])**2 + (y - center[1])**2 = r
a = slope**2 + 1
b = 2 * (slope * (intersect - center[1]) - center[0])
c = center[0]**2 + (intersect - center[1])**2 - r**2
# Case 2: No intersection
delta = b**2 - 4*a*c
if delta <= 0:
return [], False, False
When the segment does intersect with the circle, we need to take into consideration whether it intersects from the “left”, from the “right” or from both sides (here we order the segment extremities by increasing order of their x-coordinate).
# Intersection -> clip to [p2, q2] n [p, q]
else:
pt1 = p; pt2 = q
is_in_pq = lambda z: (z >= p[0]) and (z <= q[0])
check = False # check will be True iff [p2, q2] n [p, q] is empty
# Case 3: p is not in the circle
if not is_in_circle(p, center, r):
x = (- b - np.sqrt(delta)) / (2 * a)
pt1 = np.array([x, slope*x + intersect])
check = not is_in_pq(x)
# Case 4: q is not in the circle
if not is_in_circle(q, center, r):
x = (- b + np.sqrt(delta)) / (2 * a)
pt2 = np.array([x, slope*x + intersect])
check = (check or cp) and (not is_in_pq(x))
# Case 5: neither p or q are inside the circle
return ([], False, False) if check else
([(pt1, pt2)], is_in_circle(p, center, r), is_in_circle(q, center, r))
#### Building the restricted Voronoi diagram
Once we have this construction, we can build the restricted Voronoi Diagram. We consider every segment $[p, q]$ of the Voronoi diagram. Let us denote $V_x$ and $v_y$ the two Voronoi cells that lie on both sides of $[p, q]$; we say $[p, q]$ is a ridge between $V_x$ and $V_y$.
We need to compute the intersections between $[p, q]$ and $B(x, r)$ (or equivalently, $B(y ,r)$, since by definition of the Voronoi diagram, any point on the ridge is equidistant from $x$ and $y$). A restricted Voronoi cell is represented as a sequence of edges $[(p_0, q_0), \dots (p_n, q_n)]$, where $[p_i, q_i]$ is a segment. Furthermore, either $q_i = p_{i + 1}$, or $q_i \neq p_{i + 1}$ in which case $q_i$ and $p_{i + 1}$ are joint by a circle segment.
for center, region, region_indices, edge_indices in vor_regions:
restr_region = []
for i, p in enumerate(region):
q = region[(i + 1) % len(region)]
inter, clip_p, clip_q = line_circle_intersection(p, q, center, r)
restr_region.extend(inter)
restricted_voronoi_cells.append((center, restr_region))
#### Building the alpha complex
Building the alpha complex from the restricted Voronoi diagram is straightforward. We will split it in two sets: the triangles, which are triangles of the Delaunay triangulation and occur whenever a point lying at the intersection of three Voronoi cells still exist in the restricted Voronoi diagram (i.e. if it belongs to one the balls of radius $r$). The second set are the edges and are the leftover edges which still exist in the alpha complex but are not part of a full triangle due to the radius constraint.
... # continued
if clip_p:
triangles[region_indices[i]].append(edge_indices[i])
if clip_q:
triangles[region_indices[(i + 1) % len(region)]].append(edge_indices[i])
else:
alpha_complex_cells[0].append(edge_indices[i])
# Form triangles
for vertex, incident_edges in triangles.items():
if len(incident_edges) == 3:
vertices = [(x, y) for e in incident_edges for (x, y) in vor_ridges[e]]
_, aux = np.unique([x[0] for x in vertices], return_index=True)
alpha_complex_cells[1].append([vertices[i][1] for i in aux])
### Demo
The main observation is that by tuning the radius constraint correctly, we can obtain a mesh of the point clouds that respect topological features of the original shape (holes), contrary to the Delaunay Triangulation. However, finding a good value for this parameter can be quite difficult as it heavily depends on the input samples.
To highlight this, I also generate an animation of the restricted Voronoi diagram and the alpha complex for growing value of $r$. For very small $r$, no cells collide and the complex is equal to the point cloud; Inversely for large $r$, the complex collapses to the Delaunay Triangulation. For values in between, we get different density meshes of the cloud. | 2018-03-19 10:19:36 | {"extraction_info": {"found_math": true, "script_math_tex": 39, "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": 3, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8309099674224854, "perplexity": 1170.0071294879378}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257646875.28/warc/CC-MAIN-20180319101207-20180319121207-00643.warc.gz"} |
https://in.mathworks.com/help/sdl/ref/limitedslipdifferential.html | # Limited-Slip Differential
Reduce velocity difference between two connected shafts
• Library:
• Simscape / Driveline / Gears
## Description
The Limited-Slip Differential block represents a limited-slip differential (LSD), which is a gear assembly that limits the velocity difference between two connected shafts. The block models the LSD mechanism as a structural component that combines a differential and a clutch.
The differential component in the LSD block is an open differential. An open differential is a gear mechanism that allows two driven shafts to spin at different speeds. In an automobile, a differential allows the inner wheels to spin more slowly than the outer wheels when the vehicle is cornering. A vehicle that has wheel shafts connected by an open differential can get stuck when one of the wheels slips and spins freely due to traction loss. This vehicle stops moving because the driveshaft supplies less power to the wheel with traction than it supplies to the spinning wheel.
In the same scenario, a vehicle that has an LSD is less likely to get stuck because it contains a clutch assembly that can transmit power to the wheel that retains traction. The clutch component in the LSD block is a friction clutch that has two sets of flat friction plates. The clutch engages when applied pressure exceeds the engagement threshold pressure. In an LSD, a spring preload that separates the sun gears presses the plates in both sets together. When the shafts experience a traction differential, the planet pinion gears exert an additional force in the direction of the high-traction shaft. If the additional pressure exceeds the engagement threshold, the clutch assembly engages. The engagement allows the driveshaft to transmit more power to the slower-spinning high-traction wheel. The additional power reduces the difference in velocity of the two shafts. Because the high-traction wheel continues to rotate, the vehicle continues to move.
The figure shows the orientation of the major components in an LSD mechanism. The driveshaft pinion gear is not visible is this representation.
The Limited-Slip Differential block models the LSD mechanism as a structural component based on the Simscape™ Driveline™ Differential and Disk Friction Clutch blocks. The differential mechanism modeled by the Differential block is a structural component based on two other Simscape Driveline blocks, the Simple Gear and the Sun-Planet Bevel. The block diagram shows the structural components of the LSD.
The ports of the Limited-Slip Differential block are associated with the driveshaft (port D) and the two driven shafts (ports S1 and S2), which connect the sun gears to the wheels.
The Limited-Slip Differential block enables you to specify inertias only for the gear carrier and internal planet gears. By default, the inertias of the outer gears are assumed negligible. To model the inertias of the outer gears, connect Simscape Inertia blocks to the D, S1, and S2 ports.
The table shows the rotation direction of the driven shaft ports for different block parameterizations and input conditions.
Rotation Direction of the Driven Shaft Ports (S1 and S2)Crown Gear Location Relative to the CenterlineRotation Direction of Driveshaft Port DRelative Slippage Across the Differential
PositiveRightPositive0
• Positive for the nonslipping port
• Negative the slipping port
RightPositive> 0
NegativeRightNegative0
• Negative for the nonslipping port
• Positive the slipping port
RightNegative> 0
NegativeLeftPositive0
• Negative for the nonslipping port
• Positive the slipping port
LeftPositive> 0
PositiveLeftNegative0
• Positive for the nonslipping port
• Negative the slipping port
LeftNegative> 0
### Model
To examine the mathematical models for the structural components of the Limited-Slip Differential block, see:
### Thermal Model
You can model the effects of heat flow and temperature change by enabling the optional thermal port. To enable the port, set Friction model to ```Temperature-dependent efficiency```.
## Ports
### Conserving
expand all
Rotational mechanical conserving port associated with the driveshaft.
Rotational mechanical conserving port associated with the sun gear 1 shaft.
Rotational mechanical conserving port associated with the sun gear 2 shaft.
Thermal conserving port associated with heat flow.
#### Dependencies
To enable this port, in the Differential tab, set Friction model to ```Temperature-dependent efficiency```.
## Parameters
expand all
### Differential
Location of the bevel crown gear relative to the centerline of the gear assembly.
Fixed ratio, gD, of the number of crown gear teeth NC to the number of driveshaft pinion gear teeth ND. This gear ratio must be strictly greater than `0`. The crown gear is rigidly mounted to the carrier.
Friction model for the block:
• ```No meshing losses - Suitable for HIL simulation```— Gear meshing is ideal.
• `Constant efficiency`— Transfer of torque between the gear wheel pairs is reduced by a constant efficiency, η, such that $0<\eta \le 1$.
• `Temperature-dependent efficiency`— Transfer of torque between the gear wheel pairs is defined by the temperature lookup table
Vector of torque transfer efficiencies, [ηSS, ηCD]. Here,
• ηSS is the output-to-input power ratio that describes the power flow from the driving sun gear to the driven sun gear.
• ηCD is the output-to-input power ratio that describes the power flow from the crown gear to the driveshaft pinion gear.
The carrier is rigidly mounted to the crown gear. The vector elements must be in the range (0,1].
#### Dependencies
To enable this parameter, set Friction model to `Constant efficiency`.
Vector of temperatures used to construct a 1-D temperature-efficiency lookup table. The vector elements must increase from left to right.
#### Dependencies
To enable this parameter, set Friction model to `Temperature-dependent efficiency`.
Vector of output-to-input power ratios that describe the power flow from the driving sun gear to the driven sun gear, ηSS. The block uses the values to construct a 1-D temperature-efficiency lookup table.
Each element is an efficiency that relates to a temperature in the Temperature vector. The length of the vector must be equal to the length of the Temperature vector. Each element in the vector must be in the range (0,1].
#### Dependencies
To enable this parameter, set Friction model to ```Temperature-dependent efficiency```.
Vector of output-to-input power ratios that describe the power flow from the crown gear to the driveshaft pinion gear, ηCD. The block uses the values to construct a 1-D temperature-efficiency lookup table. The carrier gear is rigidly mounted to the crown gear.
Each element is an efficiency that relates to a temperature in the Temperature vector. The length of the vector must be equal to the length of the Temperature vector. Each element in the vector must be in the range (0,1].
#### Dependencies
To enable this parameter, set Friction model to ```Temperature-dependent efficiency```.
Vector of power thresholds, pth, for sun-carrier and longitudinal driveshaft-casing [pS, pD], respectively. The full efficiency loss applies above these values. Below these values, a hyperbolic tangent function smooths the efficiency factor.
When you set Friction model to `Constant efficiency`, the block lowers the efficiency losses to zero when no power is transmitted. When you set Friction model to `Temperature-dependent efficiency`, the block smooths the efficiency factors between zero when at rest and the values provided by the temperature-efficiency lookup tables at the power thresholds.
#### Dependencies
To enable this parameter, set Friction model to `Constant efficiency` or ```Temperature-dependent efficiency```.
Vector of viscous friction coefficients [μS, μD] for the sun-carrier and longitudinal driveshaft-casing gear motions, respectively.
Inertia model for the block:
• `Off` — Model gear inertia.
• `On` — Neglect gear inertia.
Moment of inertia of the planet gear carrier assembly including the crown gear. This value must be positive.
#### Dependencies
To enable this parameter, set Inertia to `On`.
Moment of inertia of the combined planet gears. This value must be positive.
#### Dependencies
To enable this parameter, set Inertia to `On`.
### Clutch
Number, N, of friction-generating contact surfaces inside the clutch.
Effective moment arm radius, reff, that determines the kinetic friction torque inside the clutch.
Force that the spring preload exerts on the clutch plate assemblies. Must be greater than or equal to zero.
Static or peak value of the friction coefficient. The static friction coefficient must be greater than the kinetic friction coefficient.
#### Dependencies
To enable this parameter, set Friction model to ```No meshing losses - Suitable for HIL simulation``` or ```Fixed kinetic friction coefficient```.
Vector of static or peak values of the friction coefficient for a given temperature. The vector must be the same length as the Temperature. Each element must be greater than the maximum value of the corresponding row in the Kinetic friction coefficient matrix.
#### Dependencies
To enable this parameter, set Friction model to ```Temperature-dependent efficiency```.
Vector of input values for the relative velocity. The values in the vector must increase from left to right. The minimum number of values depends on the interpolation method that you select. For linear interpolation, provide at least two values per dimension. For smooth interpolation, provide at least three values per dimension.
Vector of output values for the kinetic friction coefficient. All values must be greater than zero.
#### Dependencies
To enable this parameter, in the Differential tab, set Friction model to ```No meshing losses - Suitable for HIL simulation``` or `Constant efficiency`.
Matrix of output values for the kinetic friction coefficient. All values must be greater than zero.
#### Dependencies
To enable this parameter, in the Differential tab, set Friction model to ```Temperature-dependent efficiency```.
Interpolation methods for approximating the output value when the input value is between two consecutive grid points. To optimize performance, select `Linear`. To produce a continuous curve with continuous first-order derivatives, select `Smooth`.
Extrapolation methods for approximating the output value when the input value is outside the range specified in the argument list. To produce a curve with continuous first-order derivatives in the extrapolation region and at the boundary with the interpolation region, select `Linear`. To produce an extrapolation that does not go above the highest point in the data or below the lowest point in the data, select `Nearest`.
Maximum slip velocity at which the clutch can lock. The slip velocity is the signed difference between the base and follower shaft angular velocities, $w={w}_{F}-{w}_{B}$. When the kinetic friction torque is nonzero and the transferred torque is within the static friction torque limits, the clutch locks if the actual slip velocity falls below the velocity tolerance.
Clutch state at the start of simulation. The clutch can be in one of two states, locked and unlocked. A locked clutch constrains the base and follower shafts to spin at the same velocity, that is, as a single unit. An unlocked clutch allows the two shafts to spin at different velocities, resulting in slip between the clutch plates.
### Thermal Port
To enable these settings, set Friction model to `Temperature-dependent efficiency`.
Thermal energy required to change the component temperature by a single temperature unit. The greater the thermal mass, the more resistant the component is to temperature change.
#### Dependencies
To enable this parameter, setFriction model to ```Temperature-dependent efficiency```.
Block temperature at the start of simulation. The initial temperature sets the initial component efficiencies according to their respective efficiency vectors.
#### Dependencies
To enable this parameter, set Friction model to `Temperature-dependent efficiency`.
expand all
## Version History
Introduced in R2017a | 2023-02-07 15:34:15 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5480700135231018, "perplexity": 2850.0168474738152}, "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/1674764500619.96/warc/CC-MAIN-20230207134453-20230207164453-00311.warc.gz"} |
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Not Knowing is OK23
Not Knowing is OK
14
(OP)
It is OK to not know something.
This message goes out to all my novice engineers. It might seem like the above statement is anathema to engineers. After all, we are paid to know things. What we do is enormously difficult. No one expects anyone, not even senior engineers, to know everything. What your peers do expect is that you identify gaps in your understanding and then get help or do your homework.
As a novice, if you come to me and say, I don't understand something. I am going to respond with either a short answer (do this), a long answer (this is why you do this), or some references to authoritative resources (you can learn about this in these books). I might also say that, "I do not know. Let's go talk to this other engineer."
Your senior engineers want to teach you. They want the product of your labors to be good quality engineering. At a minimum they want the business to succeed. At a higher level, they believe that the practice of engineering serves people.
If you pretend to understand something, a senior engineer will sniff you out in two seconds flat. Depending on their character, they might be amused or they might be annoyed. It is better to simply state that you don't understand. I can work with that.
Here is the thing that you may not understand about this topic if you are novice. The issue is trust. If a novice engineer engages in a pattern of making assertive statements without a strong rationale, then you are going to start losing trust. The senior engineer will form the opinion that you make claims that you can't substantiate.
RE: Not Knowing is OK
This goes for intermediate and senior etc. level engineers as well.
Geotechs are notorious for making confident assertions about settlement calculations when never, in their entire careers, have they ever actually had measurements taken to compare their calculations to reality. Ditto for many other civil engineering disciplines.
RE: Not Knowing is OK
When I graduated from engineering school, some 50 years ago this past week, we were told that the most important thing to remember was that you don't know everything, and it's important to know what those limitations were, but that you also needed to know how to go about finding what you need to know when you actually needed it. In the real world, you're not going to lose points for having to stop and look-up what you need. This was a very good bit of insight to have, and it served me well the rest of my professional career. I only have one regret, that back in 1971, when I was started that first job as a working engineer, that we didn't have the internet to help us look-up all that stuff we needed to find
John R. Baker, P.E. (ret)
EX-'Product Evangelist'
Irvine, CA
Siemens PLM:
UG/NX Museum:
The secret of life is not finding someone to live with
It's finding someone you can't live without
RE: Not Knowing is OK
I've been working with my youngest son on this. He's inquisitive and imaginative. But he needs to learn to build the puzzle from the pieces he actually has.
RE: Not Knowing is OK
To know that one does not know is the greatest knowledge. Sadly this knowledge is missing quite often. With today's information tsunami, one need not know everything but definitely should know where to get that.
RE: Not Knowing is OK
Here's how I often put it:
"I don't know for sure. My gut feel is this- but here's how we can find out the right answer".
RE: Not Knowing is OK
As I have said before, the 1 hour course no college teaches is "What we did not teach you but one day you may need to know". People graduate every year with no real idea of what they should and should not know at that point in time. If they unfortunately get an employer who has no idea also, life will be difficult. That first mentor may be more important than your college or grades.
RE: Not Knowing is OK
We did get a class on doing engineering presentations; amusingly, we used overhead projectors that didn't even use transparencies, wha??!!
Of course, needed the refresher from Corporate Communications minions much further down the line, which was indeed much needed.
TTFN (ta ta for now)
I can do absolutely anything. I'm an expert! https://www.youtube.com/watch?v=BKorP55Aqvg
FAQ731-376: Eng-Tips.com Forum Policies forum1529: Translation Assistance for Engineers Entire Forum list http://www.eng-tips.com/forumlist.cfm
RE: Not Knowing is OK
As a previous boss was fond of saying,
"How do you know what you don't know, if you don't know that you don't know it?"
My glass has a v/c ratio of 0.5
Maybe the tyranny of Murphy is the penalty for hubris. - http://xkcd.com/319/
RE: Not Knowing is OK
2
I attended a presentation speech by an Engineering VP of some company. He made the funny and profound declaration:
I am the luckiest manager in my company!
I am SURROUNDED by young engineers WHO KNOW EVERYTHING.
As I get older, I understand that there is more and more that I DO NOT KNOW.
I have kept that little jewel with me for decades.
TygerDawg
Blue Technik LLC
Manufacturing Engineering Consulting
www.bluetechnik.com
RE: Not Knowing is OK
Quote (That first mentor may be more important than your college or grades.)
I didn't realise that until a few decades after working... a tip of the hat to Arnold Crosier and Paul Krauss... both passed and long remembered.
Rather than think climate change and the corona virus as science, think of it as the wrath of God. Feel any better?
-Dik
RE: Not Knowing is OK
Quote:
It is OK to not know something....It might seem like the above statement is anathema to engineers.
IME that varies a LOT by office/employer. When high standards are enforced there's little room for bs. Unfortunately they often aren't, and many mistake an enjoyable culture and the ol' profit flowing in & work out for "good engineering" bc they don't recognize what isn't being done correctly. There's a ton of minutia in every detail of what we do from actual design to the processes we follow, so its easy to lose some.
My standard advice to juniors is to remain uncomfortable. Bounce around the industry and especially the mega-corps for the first decade learning the various processes. Don't be afraid to be pigeon-holed bc it means you are learning something at a deep level, you can always job-hop if it gets boring after 2-3 years. Don't be afraid of difficult bosses with high expectations including the ones who seem like asses, they will make you a better engineer even if you're uncomfortable or overwhelmed at times. Take advantage of the opportunities given, travel as much as you can to other locations, suppliers, customers, and even competitors to learn how others' work. Trust your mentors but also challenge them if they take shortcuts. Call out rules of thumb and unvalidated analysis as guesstimation, not engineering. Find and read the niche texts and studies for everything you do - get a library card and leverage worldcat loans or save a search on eBay/amazon for the $10 copy of that$500 text - old books are usually great too. Reading a text on bolted-joint design or the latest quality process isn't particularly enjoyable but may keep you from looking foolish or making a costly mistake. Most importantly - recognize that there are critical details in every task we do. Engineering aside, how you write a report and whether a single part is one CAD .prt or an assembly (ie. weldments, castings, forgings etc) is often critical to folks downstream. Ask yourself "where did I learn this?" and "is there a better way of accomplishing this task?" every single day, and review both your work and potential improvements with those around you. Ethical engineering is a team sport, there's no getting around that.
RE: Not Knowing is OK
At my first job, a director in the R&D division was famous for asking questions he already knew the answer to, just to embarrass the presenter. Later, when he applied for a director position at my new company, I interviewed him, and he lied to my face about leading a tiger team that solved a serious yield problem with our sister division's debit card chip, not knowing that I was actually on the tiger team, and it was my effort that found out the root cause to the yield problem.
I think BS'rs exist continue to exist; what's her face that started Theranos, for example.
TTFN (ta ta for now)
I can do absolutely anything. I'm an expert! https://www.youtube.com/watch?v=BKorP55Aqvg
FAQ731-376: Eng-Tips.com Forum Policies forum1529: Translation Assistance for Engineers Entire Forum list http://www.eng-tips.com/forumlist.cfm
RE: Not Knowing is OK
There's always high-level exceptions. Within otherwise great companies I've seen several instances of a fairly low-level fraud getting an overly ambitious but ignorant executive behind a project on the basis that it was going to earn gazillions, and the exec bullying others into going along with it until things fizzle. That's essentially what Theranos did - convinced quite a few powerful people that they were going to earn billions based on a magic box, and bank on the powerful people not wanting to admit their ignorance and mistake. IME with the lower levels in everyday business tho, frauds and their shoddy work stand out like a sore thumb if everybody else is doing world-class work.
RE: Not Knowing is OK
Quote:
everybody else is doing world-class work
I think that's a utopian dream; we all know the 80/20 rule; I think that applies to people as well. Additionally, everyone has strengths and weaknesses; even the best teams do not have fully interchangeable "world-class"ness in every endeavor. Even then, world-class mail-room clerks are hard to come by; in other cases, too many world class cooks spoil the stew with their egos. A team that works well together can often do better than a team of "world class" stars; "Money Ball" comes to mind.
TTFN (ta ta for now)
I can do absolutely anything. I'm an expert! https://www.youtube.com/watch?v=BKorP55Aqvg
FAQ731-376: Eng-Tips.com Forum Policies forum1529: Translation Assistance for Engineers Entire Forum list http://www.eng-tips.com/forumlist.cfm
RE: Not Knowing is OK
Somewhat agree. A century of quality studies have shown that process drives outcome more than the people do. A company full of mediocre staff that are held to high standards by process may not revolutionize the world with new technology, but what they do will be done very well. I've known a few companies at that level, and usually there's at least a few folks doing revolutionary things among them.
RE: Not Knowing is OK
People "held to high standards by process" are generally disincentivized against innovation.
As Scott Adams of Dilbert fame aptly put it, a successful innovation gets you a certificate of appreciation in a handsome plastic frame. A failed innovation is rewarded with a pink slip. So, what's in it for the innovator?
Chris DeArmitt (demon3 here) wrote an excellent book called the Innovation Abyss. A very good read. He takes this on in detail and I agree with many of his conclusions.
There are some people in process-driven organizations who TRY to do "revolutionary things", because that's in their nature. They usually burn out and become part of the herd, or they move on- often after being given a push- because they don't fit the "workplace culture".
It's been my experience that employees aren't idiots and don't like being treated as if they were idiots. Some structure and process is both good and necessary. Too much process turns people into robots, with only the "programmers" getting to exercise any creativity. And that desire for control is frequently the actual source of the problem.
RE: Not Knowing is OK
2
I will have to check out his book and report back.
IME when folks start griping about structured process hampering innovation its usually because there isn't enough process, not too much. If there was an understood process that fit the situation at hand then folks wouldn't be griping (they'd be working), hence the need for more process. Many companies/managers don't understand that its perfectly ok to have multiple processes accomplishing the same goal. Process steps are rather irrelevant so long as the team understands them and produces common deliverables at the end - high quality documentation, high quality products - high standards. Many also fail to recognize the need to adapt process over time, when process inhibits work they blame the existing process rather than realizing that they are usually missing one - regular (biweekly or monthly) process change control boards.
The major innovators in any industry are typically the large, highly process-driven companies. Having worked for a few of their research depts, a pet peeve of mine relevant to both innovation and process is the term "R&D" simply because those are two distinct processes with different deliverables, and companies with a "R&D" dept usually do neither well. Personally, I have never felt treated like a robot or idiot due to either process, quite the opposite actually. A thoroughly organized company allows employees the freedom to focus on their work which leads to more innovation. In those environments my workday is usually predictable a week or three in advance and though they may involve long hours (ie international projects), the stress is much lower and the work-life balance better. Its like a spreadsheet - it allows me to do more, faster, with less stress and better quality than simply using pencil & paper alone. Less-organized companies are employee-killers because you're constantly in a rush to put out the latest fire. Stress remains comparably high and work-life balance usually sucks so you end up with apathetic employees who have to say "f-them, I'm going" because theres no good way to plan a vacation or take a kid to the dentist. Good luck finding time daily to work on that next innovation in those environments, assuming you're relaxed enough to let the mind wander to creativity.
RE: Not Knowing is OK
On knowing and not knowing, love this zen reference, huge relevance to the ways of engineering:
Quote:
The Way is not a matter of knowing or not knowing. Knowing is delusion; not knowing is confusion. When you have really reached the true Way beyond doubt, you will find it as vast and boundless as space. How can it be talked about on the level of right and wrong?
RE: Not Knowing is OK
Thomas Telford started life as as stone mason; the guy who designed the crystal palace was a gardener, Larry Page and Sergey Brin started google in a garage; Bill Gates and Paul Allen for Microsoft; Jeff Bezos started Amazon in a garage; Elon Musks way of thinking / engineering is basiaclly the antithesis to most 'top of the industry' engineering; the Wright Brother's, high school drop outs, certainly weren't using 'top of the industry' process.
Wonder where the modern world would be if all of these guys had been crushed by MBA-driven big corporate processes?
RE: Not Knowing is OK
Good engineering process is driven locally by engineers not management, finance, or otherwise.
It doesn't take a big company to have a revolutionary idea, but big companies are generally in a much better position to capitalize on them. Stereotypically, they have an expert specialized in every component, they test every detail to the nth degree, and they have comparably massive supply chains that will react quickly for low-cost. The net result is the ol' better, cheaper, & faster to market. Had HP listened to Steve Wozniak for example and gone into PCs early then his side-gig Apple would've been doomed bc they couldn't compete with the mega-corps capabilities. Ferrari famously found out similar when Ford developed the GT40 during the 1960s - process driving assets and capability win. The usual mistake big companies make in this regard is (like HP on PCs) that they often make overly conservative choices to stay out of small niches until those markets have proven themselves significantly profitable and matured to a fairly common direction. That's not to say they dont prepare for market possibilities, that's where research depts come in - evaluating the cost and capabilities of future technologies against their limitations, and drafting a rough plan for product development to leverage those technologies in 5-15 years. Unfortunately bc research is kept secret at big companies the public often develops these strange notions that established companies are on the way out vs upstart rivals. They also often develop an even stranger hero-worship for entrepreneurs selling a (usually severely overvalued) brand.
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Register now while it's still free! | 2021-12-02 18:45:41 | {"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": 0.20692965388298035, "perplexity": 3304.2315264801477}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964362287.26/warc/CC-MAIN-20211202175510-20211202205510-00635.warc.gz"} |
https://dsp.stackexchange.com/questions/31586/testing-filter-code-w-octave | # Testing filter code w/ Octave
I want to test some code generated by this site. I selected Bessel LP 1st sample rate $600\textrm{ Hz}$ corner $8\textrm{ Hz}$ long $10\textrm{ bit}$. If I adjust the code for octave to be:
clear all;
close all;
N = 1024
v0=256;
v1=256;
m = [];
for mi = 1:128
ip=round(rand(N,1)*512);
for i = 1:N
v0 = v1;
tmp = ((((ip(i) * 2699550)/32)
+ ((v0 * 3856860)/2))
+ 1048576) / 2097152;
v1 = tmp;
op(i) = v0 + v1;
endfor
op = fft(op,N);
m = [m; abs(op(2:N/2))];
endfor
x=[2:N/2];
y = mean(m);
semilogx(x,20*log(y)-170);
I get this plot:
which looks vaguely correct but the slope is $\approx 13\textrm{ dB/octave}$ and not the expected $6\textrm{ dB/octave}$.
I have limited experience with DSP so I need a method to verify the actual code. Can someone recommend a procedure for verifying code like this (that works)?
I think the solution is to use $\log_{10}(y)$ instead of $\log(y)$. log(y) in octave is the natural $\log$, but you want a log base $10$, which is log10(y).
$20\log_{10}(0.5) = -6 \textrm{ dB/octave}$
$20\ln(0.5) = \frac{-13.86}{20} \textrm{ nepers/octave}$
I am using $\textrm{ln}$ for natural log to avoid any confusion.
• Without the factor $20$ in front, the natural logarithm of a ratio would have the unit nepers. – Matt L. Jun 17 '16 at 20:24 | 2019-11-20 21:45:32 | {"extraction_info": {"found_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": 0.6871668100357056, "perplexity": 1633.418199453777}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1573496670635.48/warc/CC-MAIN-20191120213017-20191121001017-00472.warc.gz"} |
https://zbmath.org/?q=an%3A0728.11039 | # zbMATH — the first resource for mathematics
On functional properties of incomplete Gaussian sums. (English) Zbl 0728.11039
Incomplete Gaussian sums are exponential sums $$\sum_{n\in \omega}\exp (2\pi ian^ 2/q)$$, where $$(a,q)=1$$ and $$\omega$$ is an interval of length at most q. These can be expressed in terms of the function $H(x_ 1,x_ 2)=\sum_{n\neq 0}\frac{\exp (2\pi i(n^ 2x_ 2+nx_ 1))}{2\pi in},$ where the sum of the series is understood as the limit of the symmetric partial sums. By an analysis of this series, it is proved, for instance, that if $$\omega$$ runs over a system of nonoverlapping intervals in the interval [0,q], then for each positive $$\epsilon >0$$, the number of intervals such that the modulus of the corresponding incomplete Gaussian sums exceeds $$\epsilon\sqrt{q}$$ is at most $$c\epsilon^{-2}$$, where c is an absolute positive constant. Also, the size of the implied constant in the classical estimate $$\ll \sqrt{q}$$ for incomplete Gaussian sums is discussed.
Reviewer: M.Jutila (Turku)
##### MSC:
11L05 Gauss and Kloosterman sums; generalizations
##### Keywords:
Incomplete Gaussian sums; number of intervals; constant
Full Text: | 2021-06-14 03:46:39 | {"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": 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": 0.9533727169036865, "perplexity": 350.81685164819334}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1623487611320.18/warc/CC-MAIN-20210614013350-20210614043350-00073.warc.gz"} |
https://en.wikibooks.org/wiki/LaTeX/Glossary | # LaTeX/Glossary
LaTeX
Getting Started
Common Elements
Mechanics
Technical Texts
Special Pages
Special Documents
Creating Graphics
Programming
Miscellaneous
Help and Recommendations
Appendices
edit this boxedit the TOC
Many technical documents use terms or acronyms unknown to the general population. It is common practice to add a glossary to make such documents more accessible.
The glossaries package can be used to create glossaries. It supports multiple glossaries, acronyms, and symbols. This package replaces the glossary package and can be used instead of the nomencl package.[1] Users requiring a simpler solution should consider hand-coding their entries by using the description environment, or the longtabu environment provided by the tabu package.
## Jump start
Place \usepackage{glossaries} and \makeglossaries in your preamble (after \usepackage{hyperref} if present). Then define any number of \newglossaryentry and \newacronym glossary and acronym entries in your preamble (recommended) or before first use in your document proper. Finally add a \printglossaries call to locate the glossaries list within your document structure. Then pepper your writing with \gls{mylabel} macros (and similar) to simultaneously insert your predefined text and build the associated glossary. File processing must now include a call to makeglossaries followed by at least one further invocation of latex or pdflatex.
## Using glossaries
To use the glossaries package, you have to load it explicitly:
\usepackage{glossaries}
if you wish to use xindy (recommended) for the indexing phase, as opposed to makeindex (the default), you need to specify the xindy option:
\usepackage[xindy]{glossaries}
\usepackage[toc]{glossaries}
Finally, place the following command in your document preamble in order to generate the glossary:
\makeglossaries
Any links in resulting glossary will not be "clickable" unless you load the glossaries package after the hyperref package.
In addition, users who wish to make use of makeglossaries will need to have Perl installed — this is not normally present by default on Microsoft Windows platforms. That said, makeglossaries simply provides a convenient interface to makeindex and xindy and is not essential.
## Defining glossary entries
To use an entry from a glossary you first need to define it. There are few ways to define an entry depending on what you define and how it is going to be used.
Note that a defined entry won't be included in the printed glossary unless it is used in the document. This enables you to create a glossary of general terms and just \include it in all your documents.
## Defining terms
To define a term in glossary you use the \newglossaryentry macro:
\newglossaryentry{
<label> is a unique label used to identify an entry in glossary, <settings> are comma separated key=value pairs used to define an entry.
For example, to define a computer entry:
\newglossaryentry{computer} { name=computer, description={is a programmable machine that receives input, stores and manipulates data, and provides output in a useful format} }
The above example defines an entry that has the same label and entry name. This is not always the case as the next entry will show:
\newglossaryentry{naiive} { name=na\"{\i}ve, description={is a French loanword (adjective, form of naïf) indicating having or showing a lack of experience, understanding or sophistication} }
When you define terms, you need to remember that they will be sorted by makeindex or xindy. While xindy is a bit more LaTeX aware, it does it by omitting latex macros (\"{\i}) thus incorrectly sorting the above example as nave. makeindex won't fare much better, because it doesn't understand TeX macros, it will interpret the word exactly as it was defined, putting it inside symbol class, before words beginning with naa. Therefore it's needed to extend our example and specify how to sort the word:
\newglossaryentry{naiive} { name=na\"{\i}ve, description={is a French loanword (adjective, form of naïf) indicating having or showing a lack of experience, understanding or sophistication}, sort=naiive }
You can also specify plural forms, if they are not formed by adding “s” (we will learn how to use them in next section):
\newglossaryentry{Linux} { name=Linux, description={is a generic term referring to the family of Unix-like computer operating systems that use the Linux kernel}, plural=Linuces }
Or, for acronyms:
\newacronym[longplural={Frames per Second}]{fpsLabel}{FPS}{Frame per Second}
This will avoid the wrong long plural: Frame per Seconds.
So far, the glossary entries have been defined as key-value lists. Sometimes, a description is more complex than just a paragraph. For example, you may want to have multiple paragraphs, itemized lists, figures, tables, etc. For such glossary entries use the command longnewglossaryentry in which the description follows the key-value list. The computer entry then looks like this:
\longnewglossaryentry{computer} { name=computer } {is a programmable machine that receives input, stores and manipulates data, and provides output in a useful format}
### Defining symbols
Defined entries can also be symbols:
\newglossaryentry{pi} { name={\ensuremath{\pi}}, description={ratio of circumference of circle to its diameter}, sort=pi }
You can also define both a name and a symbol:
\newglossaryentry{real number} { name={real number}, description={include both rational numbers, such as $42$ and $\frac{-23}{129}$, and irrational numbers, such as $\pi$ and the square root of two; or, a real number can be given by an infinite decimal representation, such as $2.4871773339\ldots$ where the digits continue in some way; or, the real numbers may be thought of as points on an infinitely long number line}, symbol={\ensuremath{\mathbb{R}}} }
Note that not all glossary styles show defined symbols.
### Defining acronyms
To define a new acronym you use the \newacronym macro:
\newacronym{
where <label> is the unique label identifying the acronym, <abbrv> is the abbreviated form of the acronym and <full> is the expanded text. For example:
\newacronym{lvm}{LVM}{Logical Volume Manager}
Defined acronyms can be put in separate list if you use acronym package option:
\usepackage[acronym]{glossaries}
## Using defined terms
When you have defined a term, you can use it in a document. There are many different commands used to refer to glossary terms.
### General references
A general reference is used with \gls command. If, for example, you have glossary entries defined as those above, you might use it in this way:
\Gls{naiive} people don't know about alternative \gls{computer} operating systems: \glspl{Linux}, BSDs and GNU/Hurd. Naïve people don't know about alternative computer opera- ting systems: Linuces, BSDs and GNU/Hurd.
Description of commands used in above example:
\gls{
This command prints the term associated with <label> passed as its argument. If the hyperref package was loaded before glossaries it will also be hyperlinked to the entry in glossary.
\glspl{
This command prints the plural of the defined term, other than that it behaves in the same way as gls.
\Gls{
This command prints the singular form of the term with the first character converted to upper case.
\Glspl{
This command prints the plural form with first letter of the term converted to upper case.
\glslink{
This command creates the link as usual, but typesets the alternate text instead. It can also take several options which changes its default behavior (see the documentation).
\glssymbol{
This command prints what ever is defined in \newglossaryentry{<label>}{symbol={Output of glssymbol}, ...}
\glsdesc{
This command prints what ever is defined in \newglossaryentry{<label>}{description={Output of glsdesc}, ...}
### Referring acronyms
Acronyms behave a bit differently than normal glossary terms. On first use the \gls command will display "<full> (<abbrv>)". On subsequent uses only the abbreviation will be displayed.
To reset the first use of an acronym, use the command:
\glsreset{
or, if you want to reset the use status of all acronyms:
\glsresetall
If you just want to print the long version of an acronym without the abbreviation "<full>", use :
\acrlong{
If you just want to print the long version of an acronym with the abbreviation "<full> (<abbrv>)", use :
\acrfull{
If you just want to print the abbreviation "<abbrv>", use :
\acrshort{
# Displaying the Glossary
To display the sorted list of terms you need to add:
\printglossaries
at the place you want the glossary and the list of acronyms to appear.
If all entries are to be printed the command
\glsaddall
can be inserted before \printglossaries. You may also want to use \usepackage[nonumberlist]{glossaries} to suppress the location list within the glossary.
### Separate Glossary and List of Acronyms
\printglossaries will display all the glossaries in the order in which they were defined.[2] If no custom glossaries are defined, the default glossary and the list of acronyms will be displayed.
The glossary and the list of acronyms can be displayed separately in different places[3]:
\usepackage[acronyms]{glossaries} \printglossary[type=\acronymtype] % prints just the list of acronyms Some text between the list of acronyms and the glossary. \printglossary % if no option is supplied the default glossary is printed.
#### Dual entries with reference to a glossary entry from an acronym
It may be useful to have both an acronym and a glossary entry for the same term. To link these two, define the acronym with a reference to the glossary entry like this:
\newglossaryentry{gls-OWD} { name={One-Way Delay}, description={The time a packet uses through a network from one host to another}, } \newacronym[see={[Glossary:]{gls-OWD}}]{OWD}{OWD}{One-Way Delay\glsadd{gls-OWD}}
Refer to acronym with \gls{OWD} and the glossary with \gls{gls-OWD}
To make this easier, we can use this command (modified from example in the official docs):
Syntax: \newdualentry[glossary options][acronym options]{label}{abbrv}{long}{description}
\usepackage{xparse} \DeclareDocumentCommand{\newdualentry}{ O{} O{} m m m m } { \newglossaryentry{gls-#3}{name={#5},text={#5\glsadd{#3}}, description={#6},#1 } \makeglossaries \newacronym[see={[Glossary:]{gls-#3}},#2]{#3}{#4}{#5\glsadd{gls-#3}} }
then, define new (dual) entries for glossary and acronym list like this:
\newdualentry{OWD} % label {OWD} % abbreviation {One-Way Delay} % long form {The time a packet uses through a network from one host to another} % description
### Custom Name
The name of the glossary section can be replaced with a custom name or translated to a different language. Add the option title to \printglossary to specify the glossary's title. Add the option toctitle to specify a the title used in the table of content (if not used, title is used as default). [4]
\printglossary[title=List of Terms,toctitle=Terms and abbreviations]
### Remove the point
To omit the dot at the end of each description, use this code:
\usepackage[nopostdot]{glossaries}
### Changing Glossary Entry Presentation Using Glossary Styles
A number of pre-built styles are available, and can be changed easily using
% Must be issued before \printglossaries \glossarystyle{}
Commonly used styles include list
My Term Has some long description 7, 9
altlist (inserts newline after term and indents description)
My Term
Has some long description 7, 9
altlistgroup or listgroup (group adds grouping based on the first letters of the terms)
M
My First Term
Has some long description 7, 9
My Second Term
Has some long description 7, 9
A|B|C|D|F|G|I|M|O|R|S|C|D|G|M|P
A
A First term
Has some long description 7, 9
B
Barely missed first
Has some long description 7, 9
Building your document and its glossary requires three steps:
1. build your LaTeX document — this will also generate the files needed by makeglossaries
2. invoke makeglossaries — a script which selects the correct character encodings and language settings and which will also run xindy or makeindex if these are specified in your document file
3. build your LaTeX document again — to produce a document with glossary entries
Thus:
latex doc
makeglossaries doc
latex doc
where latex is your usual build call (perhaps pdflatex) and doc is the name of your LaTeX master file.
If your entries are interlinked (entries themselves link to other entries with \gls calls), you will need to run steps 1 and 2 twice, that is, in the following order: 1, 2, 1, 2, 3.
If you encounter problems, view the doc.log and doc.glg files in a text editor for clues.
# Example for use in windows with Texmaker
## Compile glossary with xindy - In Windows with Texmaker
In TeX Live and since June 2015 in MikTeX xindy is already included.
There is only one issue with path of the install directory of MikTeX containing spaces. It can be solved via the following edit: http://tex.stackexchange.com/questions/251221/miktex-and-xindy-problems/251801#251801
You need to restart Texmaker after installation of xindy, to update PATH references to xindy and Perl binaries.
Then, in Texmaker, go to User -> User Commands -> Edit User Commands.
Choose command 1
2. Command = makeglossaries %
Now push Alt+Shift+F1and then ->F1
Note, for use with the "use build directory" option of Texmaker: makeglossaries needs to find the aux file. Thankfully, while Texmaker does not help there, the option -d <dir> of makeglossaries provides for the subdirectory case. Hence the Command in this case should be:
Command = makeglossaries -d build % instead.
## Document preamble
In preamble should be included (note, hyperref should be loaded before the glossaries):
\usepackage[nomain,acronym,xindy,toc]{glossaries} % nomain, if you define glossaries in a file, and you use \include{INP-00-glossary}
\makeglossaries
\usepackage[xindy]{imakeidx}
\makeindex
## Glossary definitions
Write all your glossaries/acronyms in a file: Ex: INP-00-glossary.tex
\newacronym{ddye}{D$_{\text{dye}}$}{donor dye, ex. Alexa 488}
\newacronym[description={\glslink{r0}{F\"{o}rster distance}}]{R0}{$R_{0}$}{F\"{o}rster distance}
\newglossaryentry{r0}{name=\glslink{R0}{\ensuremath{R_{0}}},text=F\"{o}rster distance,description={F\"{o}rster distance, where 50\% ...}, sort=R}
\newglossaryentry{kdeac}{name=\glslink{R0}{\ensuremath{k_{DEAC}}},text=$k_{DEAC}$, description={is the rate of deactivation from ... and emission)}, sort=k}
## Include glossary definitions and print glossary
Include glossary definitions in the preamble (Before "\begin{document}")
\loadglsentries[main]{INP-00-glossary}
% or using \input:
%\input{INP-00-glossary}
\begin{document}
Print glossaries, near end
\appendix
\bibliographystyle{plainnat}
\bibliography{bibtex}
\printindex
\printglossaries
\end{document}
`
## References
• Using LaTeX to Write a PhD Thesis, Nicola L.C. Talbot, [1]
• Glossaries, Nomenclature, Lists of Symbols and Acronyms, Nicola L. C. Talbot, link | 2016-08-30 18:58:22 | {"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": 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": 0.8672272562980652, "perplexity": 5229.373328156009}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1471983001995.81/warc/CC-MAIN-20160823201001-00148-ip-10-153-172-175.ec2.internal.warc.gz"} |
https://indico.cern.ch/event/634426/contributions/3090550/ | # Hard Probes 2018: International Conference on Hard & Electromagnetic Probes of High-Energy Nuclear Collisions
30 September 2018 to 5 October 2018
Aix-Les-Bains, Savoie, France
Europe/Zurich timezone
PROCEEDINGS OPEN UNTIL DECEMBER 15th 2018
## Sorting out energy loss for medium-modified jets
2 Oct 2018, 14:20
20m
Aix-Les-Bains, Savoie, France
#### Aix-Les-Bains, Savoie, France
Aix-Les-Bains, Congress Center Student Lectures Day: September 30 at CERN
2a) Jets and high-pT hadrons (TALK)
### Speaker
Jasmine Brewer (Massachusetts Institute of Technology)
### Description
Most studies of medium-induced jet modification rely on the comparison of jet properties measured in heavy ion collisions with a proton-proton baseline of the same reconstructed jet $p_T$. Migration of jets from higher to lower $p_T$ due to energy loss, together with the steepness of the jet spectrum, lead to a heavy ion jet sample with a given $p_T$ range which is dominated by the jets that were least modified. We introduce a new strategy to compare heavy ion jet measurements to proton-proton baselines which views energy loss as being monotonic in $p_T$. In this strategy, the jets in a heavy ion collision ordered by $p_T$ can be viewed as modified versions of the same number of highest energy jets in proton-proton collisions. We validate, at MC level, the correlation between the $p_T$ of the parton that initiates a heavy ion jet with the $p_T$ of the vacuum jet which corresponds to it via our novel binning procedure. We show that this strategy mitigates the effect of bin migration and provides a complementary way to study jet modification in heavy ion collisions.
### Primary authors
Jasmine Brewer (Massachusetts Institute of Technology) Dr Guilherme Milhano (LIP-Lisbon & CERN TH) Jesse Thaler (MIT) | 2021-01-15 23:59:22 | {"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": 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": 0.47717368602752686, "perplexity": 4024.857275534931}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1610703497681.4/warc/CC-MAIN-20210115224908-20210116014908-00502.warc.gz"} |
https://www.i2m.univ-amu.fr/events/uniformly-expanding-markov-maps-of-the-real-line-exactness-and-infinite-mixing/ | # Uniformly expanding Markov maps of the real line: exactness and infinite mixing
Marco Lenci
Università di Bologna
http://www.dm.unibo.it/~lenci/
Date(s) : 02/02/2016 iCal
11 h 00 min - 12 h 00 min
We characterize the exact components of a large class of uniformly expanding Markov maps of R. Using this result, for a class of Z-invariant maps and finite modifications thereof, we prove certain properties of infinite mixing (i.e., mixing for systems possessing an infinite invariant measure).
https://arxiv.org/abs/1404.2212
Catégories | 2022-06-28 06:30:24 | {"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": 0.823948085308075, "perplexity": 2626.793039981333}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1656103355949.26/warc/CC-MAIN-20220628050721-20220628080721-00285.warc.gz"} |
https://stats.stackexchange.com/questions/499007/is-it-better-to-do-per-class-anomaly-detection-on-px-y-or-px-y | # Is it better to do per-class anomaly detection on P(x, y) or P(x | y)?
(Not an expert in anomaly detection.)
I'd like to experiment with per-class anomaly detection.
That is, we have a feature vector $$x$$, and a classifier that predicts its class $$\hat{y}$$. I'd like to see if the combination $$(x, \hat{y})$$ is an anomaly, given some training set of non-anomalous $$(x, y)$$ pairs.
It seems that I can train one joint anomaly detector on $$P(x,y)$$, or multiple independent detectors on $$P(x|y)$$.
I think the latter is easier and sufficient. Are there any downsides? Also, is there a name for this technique?
• It is not clear, do you observe the class $y$, or do you predict it using a classifier? – Konstantin Dec 12 '20 at 17:09
• The true class is observed at training time, and predicted when using the anomaly detector. It doesn't matter for your answer, though. – kennysong Dec 13 '20 at 5:44
• I'd say, that the predicted label $\hat y$ then simply aggregates information from $x$, adding nothing new, and there is no point in using it in normally detection. – Konstantin Dec 13 '20 at 9:35
According to the Bayes law the joint probability of a sample observation $$(x,y)$$ can be decomposed into the familiar product:
$$P(x,y) = P(x|y)P(y).$$
In other words, we can always partition the set of features, $$(x,y)$$, into two subsets, $$x$$ and $$y$$, and then decompose the probability of the sample into the marginal probability of one subset $$P(y)$$ and the conditional probability of the other $$P(x|y)$$.
Note that the joint probability of the sample observation might fall below the anomaly threshold for two reasons:
• the probability of observed values for a subset of features, $$P(y)$$ is low, or
• the probability of observed values for the complementary subset, conditional on the values of the first subset, $$P(x|y)$$ is low.
When you only use the $$P(x|y)$$ model, you forego the information contained in the realization of $$y$$. In other words you implicitly agree with the observation of $$y$$, be it a modal (that's okay) or an extremely rare value (that's not okay, this already may be a sufficient reason for rejection).
• Thanks, I agree with your analysis. There's also an orthogonal consideration of whether P(x,y) or P(x|y) is easier to learn. I think usually, P(x|y) is simpler since it's not a mixture, but P(x,y) may be easier if the classes are imbalanced & there is reusable structure between the classes. – kennysong Dec 13 '20 at 5:54
• Yes, I like the point on the balance between classes. Depending on the size of your data, you can also face a trade-off: smaller models may be easier to train, but they may have worse performance due to insufficiency of training data. – Konstantin Dec 13 '20 at 9:31 | 2021-05-16 09:51:47 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 18, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8216228485107422, "perplexity": 485.1084253374769}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1620243992516.56/warc/CC-MAIN-20210516075201-20210516105201-00072.warc.gz"} |
http://www.gradesaver.com/textbooks/math/algebra/intermediate-algebra-6th-edition/chapter-9-section-9-5-logarithmic-functions-exercise-set-page-572/50 | Intermediate Algebra (6th Edition)
$x=2$
If $b\gt0$ and $b\ne1$, then $y=log_{b}x$ means $x=b^{y}$ for every $x$ and every real number $y$. Therefore, $log_{x}8=3$ is equivalent to $x^{3}=8$. We can now solve for x. Take the cube root of both sides. $(x^{3})^{\frac{1}{3}}=(8)^{\frac{1}{3}}$ $x=2$ | 2017-10-24 11:33:32 | {"extraction_info": {"found_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": 0.8601352572441101, "perplexity": 61.55089213771276}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187828411.81/warc/CC-MAIN-20171024105736-20171024125736-00895.warc.gz"} |
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A CONTINUATION ALGORITHM FOR PIECEWISE-LINEAR ANALYSIS OF NONLINEAR RESISTIVE NETWORKS by SHUEH-MIEN LEE, B.S.E. A THESIS IN ELECTRICAL ENGINEERING Submitted to the Graduate Faculty of Texas Tech University in Partial Fulfillment of the Requirement for the Degree of MASTER OF SCIENCE IN ELECTRICAL ENGINEERING Approved Chairman"of the Committee "^ ^^JWI/IAJ^ May, 1980
Transcript
OF NONLINEAR RESISTIVE NETWORKS
ELECTRICAL ENGINEERING
Submitted to the Graduate Faculty of Texas Tech University in Partial Fulfillment of the Requirement for
the Degree of
MASTER OF SCIENCE
ACKNOWLEDGEMENTS
I am deeply indebted to Professor Kwong Shu Chao for his
direction of this thesis and to the other members of my committee.
Dr. Donald Gustafson and Dr. A. K. Mitra, for their helpful
criticism. I am also happy to thank my wife and my family for
the constant encouragement throughout this study.
UL
BY A RELATED DIFFERENTIAL EQUATION
2.1 Introduction 4
2.3 Existence and Uniqueness of Solution Curve . 7
III. MULTIPLE SOLUTIONS AND SOLUTION CURVE DEFINED
BY A CONTINUOUS PIECEWISE-LINEAR FUNCTION
3.1 Introduction 9
3.4 Starting Point 21
3.5 Numerical Algorithm 23
JACOBIAN MATRICES
4.2 Singular Jacobian Matrices Problem 32
V. APPLICATIONS
Network 36
of Multivalued Resistive Nonlinear Networks . . . 41
VI. CONCLUSION 45
3.2. The Ebers-Moll Model of a PNP transistor 11
3.3. A solution curve determined by Katznelson's algorithm 13
3.4. A solution curve crosses the boundaries 19
3.5a The circuit of example 3.1 24
3.5b The piecewise-linear approximation of function h for
the tunnel diode 25
3.9a The characteristic of a nonlinear element 28
3.9b The characteristic of a nonlinear element 28
3.10. The solution curve of example 3.2 29
4.1. The solution curve hits a corner 31
4.2. The solution curve enters a singular region 33
5.1. The circuit of example 5.1 38
5.2. Output voltages for different starting podjits 39
5.3. Output currents for different starting points 40
5.4. Multivalued input-output plot for the circuit in
example 5.2 43
example 5.2 44
INTRODUCTION
One of the most basic problems in the area of nonlinear circuit
analysis is the determination of the solutions (equilibrium points)
of nonlinear resistive networks. Research in nonlinear resistive
networks serves as a prerequisite to the understanding of general
nonlinear circuits and systems. This problem is important not only
from a circuit theoretic point of view but even more so from a com
putational point of view. Indeed, an essential part of most nonlinear
transient analysis computer programs is a DC analysis subprogram
[l-3]. In addition, similar problems are encountered in many related
fields such as mathematical economics, flow networks, power systems,
and least-square approximations.
Much work has been done in obtaining multiple solutions of
resistive nonlinear networks [4-9]. Some effort has been devoted
to using the technique of piecewise-linear approximation of a non
linear function [10-16]. For example, in the analysis of a elec
tronic circuit, diode and transistor characteristics can be repre
sented by continuous linear segments. This often gives considerable
insight to the problem and yields quick solutions. In general, by
peicewise-linear approximations, a nonlinear resistive network can
be characterized by the equation
f(x) = y (1.1)
where f(.) is a continuous piecewise-linear mapping of the real
2
n-dimentional Euclidean space R into itself, x is a point in R and
represents a set of chosen variables of a given network and y is an
arbitrary point in R and represents the inputs. Usually, f(.) is
expressed in the form
f(x) = J ' x + w^^^ = y i = 1, 2, , N (1.2)
where J is a constant nxn matrix called the Jacobian matrix and
w is a constant n-vector, both defined in region R . The whole
space R is divided into a finite number N of polyhedral regions
by a finite number of hyperplanes.
Although various techniques have been deveolped for solving
piecewise-linear equations, relatively little has been done in
locating multiple solutions. It is the purpose of this thesis to
develop a systematic search technique for obtaining multiple soluti
ons of (1.1). The continuation method devised by Chao, Liu, and
Pan [4J plays an important role in the scheme developed in this
report.
For convenience and to make the present thesis self-contained,
the continuation method [4] is outlined in Chapter II. In Chapter
III, the general theory of the new algorithm is developed with two
assumptions, namely, all Jacobian matrices may have arbitrary signs
but are not singular and the solution curves never hit a comer.
The dynamic behavior of the search curve is discussed. Approaches
showing how a solution curve can be reached are presented. Formu-
lation of equation (1.1) is also briefly introduced.
In Chapter IV, the corner problem and the problem of singular
Jacobian matrices are considered. Applications to other fields are
discussed in Chapter V. These include AC nonlinear resistive network
analysis and driving point or transfer characteristic curves of mul
tivalued nonlinear resistive networks. Finally, some concluding
remarks are given in Chapter VI.
CHAPTER II
BY A RELATED DIFFERENTIAL EQUATION
2.1 Introduction
g(x) = y (2.1)
or f(x) = g(x) - y = 0 (2.2)
where f or g is a continuous differentia le function from R onto
itself, X and y are both n-dimensional vectors to represent a set
of chosen variables and the inputs respectively.
A systematic search method has been developed by Chao, Liu
and Pan for locating multiple solutions of (2.2). For convenience,
this method is outlined here.
2.2 Iterative Solution Method
Consider a system of differential equations of the form
^ f.(x(t)) = -f,(x(t)) , f,(x(0)) = 0 i = 1, 2, , n-1 dt 1 1 1
where the initial conditions in f. are such that the starting point
X must lie on a space curve L of intersection f.(x) = 0 , i = 1, 2,
- - -, n-1. Since f is not a function of t explicitly, -r— can be dt
written as
dt ax dt "
and i n the x-space (2.3) reduces to a d i f f e r e n t i a l equat ion
X = J""^( -f , , - f , . , - f , , ±f ) ^ , iP ^ x(Q)e L (2.5) i ^ n—i n
where T denotes matrix transpose. Solved by the Zuler intsgraticn
method, equation (2.5) reduces to an iterative scheme
\ ^ l ' \ ^ hJ- x )( -f (x ), - - -, -f .,(V> ^n^V )'
x^ L, k - 1, 2, (2.6)
where h is the step size. The transistion in sign should occur at
the points where the Jacobian J changes sign and the solution points.
The solutions of (2.3) in the f-space are given as
f,(x(t)) = f^(x(0))e'^ = 0 i - 1, 2, , n-1
0 A. These solutions show that for any x * x(0) L, the corresponding
trajectory x(t) from (2,5) remains on L. Theoretically, the signs
of f, for i » l , 2,---,n-l are not important, because the
corresponding trajectory will remain on L so long as f.(x(0)) = 0,
for i = l , 2 , - - - , n-1. But in computational practice, t"r.e f. 1
at each successive iteration may not 'oe precisely zero; the signs
of f. for i = 1, 2-, - - -, n-1 are therefore kept negative to ensure
that the computed trajectory x(t) does not stray too far from the
solution curve L.
Depending on whether the minus or plus sign is being used,
the function f is forced to be attracted to or diverge away n
from zero along L. To ensure the continuation of tracing along
the solution curve L, the sign of f has to be changed at the
solution points of (2.2) and the points on L where the Jacobian
changes sign. The key point lies in a relation between the
determinant of Jacobian J and the directional derivative of f n
in the tangential direction of L
f = ^n = s^(Vf ) (2.8) ^ 3L~
where s is the unit vector in the tangential direction of L
and Vf is the gradient of f (x). The following theorems [4]
determine the conditions of the sign change of f .
Theorem 2.1; If f'(x) is the directional derivative of f (x) n n
in the tangential direction of L at x defined by
f,(x) = 0 , i = 1, 2, n-1 (2.9)
and A., is the (ij)th cofactor of the Jacobian matrix J of
f = (f^, f^, , f^)^, then
f' = det J / I I V I I n
where ||.|| denotes the Euclidean norm and
T ' = ( nl- ^ 2 - 'nr?
Theorem 2.2: The directional derivative of f^(x) in the tangential
direction of L changes sign if and only if the corresponding
Jacobian of f on L changes sign.
Other aspects concerning the convergence of the method and
starting point problem are also detailed in [4j.
2.3 Existence and Uniqueness of Solution Curve
Success in finding all multiple solutions depends hea-vily on
whether (2.9) defines a unique simple curve L. If L is a simple
curve, i.e., a continuously differentiable curve which does not
intersect itself, then a complete tranversal of it enables one to
find all the multiple solutions. On the other hand, if L contains
several branches, then application of the method may lead to those
solutions that lie on the branch containing this starting point only.
The following theorems gives some sufficient conditions for the
existence and uniqueness of the solution curve [9].
Theorem 2.3: Let f be a real function of n real variables and of
class C . If
and (ii) Lim | | f (x) | | = «, as x -»- °°
then any (n-1) equations off. = 0 , i = l , 2 , - - - , n defines a
unique continuously differentiable space curve.
T Theorem 2.4: Let h(x) = ( fj^(x), f2(x), , f^-l^^^ ^ ^® °^
class C . If there exists an (n-1) vector
x_ = ( ^v ^2' - -' \-l' \+l' - -' \ ^^ ^" """ k, 1 <. k < n,
such that
-k (ii) Lim | |h(x) [ | = «, as x , ->• °°
then h(x) = 0 defines a unique simple curve L.
8
Definition: Let U C R • An n x n matrix A(x) is said to be almost
positive definite on U if A(x) is positive definite for all x
in U except at most a set of isolated points on which A(x) is
positive semidefinite.
Theorem 2.5: Let f:R ^R be a C function. If J(x) is almost
positive definite on R , then any (n-1) equations of f. = 0,
i = 1, 2, , n define a unique continuous space curve.
Theorem 2.6: Let f:R->R,n?^2, b e a C map. Let
S = {x^R^I det J(x) = 0} and T = {x € R^ j x e S}
If (i) det J(x) > 0 (or det J(x) < 0 ) for all x^T
and S is at most a set of isolated points,
(ii) limi I f (x) II = °° , as | | x | | - «
then any n-1 equations o f f . = 0 , i = l , 2, - - - , n define
a unique continuous space curve.
CHAPTER III
BY A CONTINUOUS PIECEWISE-LINEAR FUNCTION
3.1 Introduction
In chapter II a systematic search method for solving multiple
solutions of (2.2) is outlined. Considerable computation time is
needed to completely traverse the solution curve, and the success in
finding all the multiple solutions depends heavily on whether or not
the set of equations f.(x)=0, i = l , 2 , - - - , n-1 defines a simple
curve L, i.e., a continuously differentiable curve which does not
intersect itself.
Instead of sol-vrLng a system of nonlinear equations, the piecewise-
linear approximation is obtained for all the nonlinear elements in
a network. As a result, a set of piecewise-linear equations (1.1)
is to be solved. A method due to Katznelson ElO] was originally
applied only to the networks with 2-terminal elements which are
strictly monotonic. Fujjsawa and Kuh [ll] showed that if f is homeo-
morphic then the algorithm due to Katznelson always converges. Here
a continuous function f is said to be homeomorphic from R onto
itself if and only if the equation f(x) = y has a unique solution for
all y.
Kuh, Ohtsuki, Fujjsawa Il23 have shown that as long as all
the Jacobian matrix determinants det J , i = l , 2, - - - - , N
10
in (1.2) have the same sign, there exists at least one solution to
the equation f(x) = y and the Katznelson's algorithm also converges.
This property is refered as the sign condition. Kuh and Chien [l3]
removed the restriction of the sign condition and demonstrated that
if all the Jacobian determinants in the unbounded region have the
same sign, the equation (1.1) has at least one solution. Chien has
also developed an algorithm that the domain of the x-space is divided
into simplices and the beha-vior of the system is approximated by a
linearized system in each simplex [l3]. Chua [l5] also introduced an
iterative scheme for finding multiple solutions. However, the method
is not suitable for networks with a large number of nonlinear elements.
In the following sections, a formulation of (1.2) and a systematic
search method for obtaining its multiple solutions are presented,
where the Jacobian matrices can have abitrary sign. A theoretical
basis for the development of the method is also given.
3.2 Piecewise-linear Approximation
obtaining the piecewise-linear approximation. If the characteristic
of a nonlinear multiport is given by
z = h(x) (3.1)
then h(x) = h(xQ) -H | | | ^ ^ ^ ( X - X Q ) = J ( X Q ) X + w^ 3 ^
where J(x^) = T~| ^ cl w is a constant vector. (J dX I XFX_ U
II
In the case of 2-terminal elements without coupling, this
reduces to the familiar tangent approximation. The Jacobian at
X- denoted by JCX-.) becomes the slope of the nonlinear function
at X-. and w. becomes the vertical axis intercept of the approximate
line segment. If the network contains n nonlinear resistive
elements, each characterized by an n -segment v.
there are N distinct segment combinations where n
i. curve, then 3
Consider a typical resitive nonlinear network as shown symbloicailv
in Fig 3.1 where the n nonlinear elements have been extracted and
are connected across the ports of a linear n-port network. As
indicated by So [17] and Chua [l5]. Fig 3.1 can be completely
characterized by an n x n hybrid matrix A and an n x 1 source vector
u, namely
Ax + u C3.4}
where the components of the n x 1 port vectors 2 and x can be any
combination of port voltage and port current pro-vided that the
corresponding components of 2 and x have the same port number.
N
Linear
Resistive
Network
rig. 3.2 The Toers—Moll model of a ?">!? transistor
12
If each nonlinear resitive element is either voltage controlled of
current controlled, then (3.4) can be written in the form
z(x) = Ax + u (3.5)
T n
z(x) = (z.(x), z„(x), , z (x)) is a nonlinear mapping from R
into R and z.(x) represents the nonlinear characteristics of the
ith nonlinear resistor. In either case, the terminal constraint
imposed by n nonlinear resistors can be expressed by its piecewise-
linear approximation, which is substituted for z in (3.4) or (3.5).
But the piecewise-linear approximation of multiterminal
elements with coupling is much more complicated. Fortunately, most
of coupling elements are of the type that can be expressed as sums
of nonlinear functions V7ith simple variables such as n
z.= E h..(x.) j = 1,2, , n -J 1 = 1 -•
= J^h'..(x.„) + «0 (3.6)
wh ere h'. .(x.^) is the slope of the function h.,(.) at x. ji lO Ji lU.
by ^e = ES< ^ T - » -"r^Cs( ^ - '^
For example, the Ebers-Moll model of a PNP transistor is represented
eb/v^ , r r ^ cb/v^ _ ^
The equivalent circuit model is shown in Fig (3.2) where
^22 = ^CS( ^^"^'"^ - ' ^ ' ^22(\b> ^'-'^
13
Therefore, by specifying the piecewise-linear approximation
to all the nonlinear elements in the circuit, it is clear that
within each region the equation is linear and is of the form
f(x) = J^^^x + w^^^ = y i = 1, 2, ,N (3.10)
where J is a constant nxn matrix, y is the input vector.
The domain of f(.) in R is divided into N polyhedral regions
by a finite number of hyperplanes (or boundaries). A typical
boundary in the x-space can be characterized by the equation
n^x = 0 (3.11)
where n is the normal vector of the hyperplane (or boundary)
and c is a real constant. In (3.10), for example, if x is the
node-to-datum voltage vector and y is the input current source
vector, then J is the node admittance matrix.
3,3 Dynamic Behavior of Solution Curves
Let R , R be continuous regions with a common boundary charac-
14
T
terized by the hyperplane n x = c. Since f is continuous piecewise-
linear, there exists an n-vector r such that
j(l) . j(j) = ,J (3 3)
where J and J ^ are Jacobian matrices of adjacent regions R
and R respectively.
f(x) = J ^ x + w^ = y, i = 1,2, , N
where y is a given input. Fig. 3.3, a continuous curve in the
x-space is determined, based on Katznelson's algorithm
x^(X) = x^ + AJ ^ (y - y= )
where
and X is a scalar parameter, If X = 1, then
i/iN 1 ^ r(l) f T(i) 1 1\
x ( l ) = x + J (y-J x - w )
= J ^ (y-w^).
It is clear that x (1) is a desired solution if it happens to be
in R" . Otherwise, the value of X has to be determined such that
x"''(X) lies on the boundary of R and R . Such a value of X is
denoted by X^. Define x^ = x^(X^) and y" = f(x ), the line
segment connecting x and x is then a portion of the desired
solution curve. We may extend the solution curve beyond x into
R""" in the similar manner.
15
ohtsuki, Kuh and Fujjsawa [l2] showed that if det J and
det J -" have the same sign, the solution curve in the x-space will
indeed enter R . And the solution curve will never enter a region
which has been traced if all the Jacobian matrices determinants
have the same sign. If we remove the condition that all Jacobian
determinants have the same sign, the problem of the continuous traverse
of the complete solution curve becomes much more complicated. Kuh
a d Fujisawa [ll] have derived a equation on the boundary
det j(i>n^j("-l= det A^^'K'"/'^'^ "^ (3.13)
This equation gives a theoretical basis of the development of the
new algorithm presented here.
In what follows, a new algorithm for locating multiple solu
tions of (1.1) is developed. The solution curve is defined and
the local behavior of the solution curve across the boundary is
also investigated. It is assumed that det J ^ 0 and the solution
curve does not hit a comer, but det J may have arbitrary sign.
The comer problem and singular Jacobian problem will be discussed
in next chapter.
Let gCx) = fCx) - y = J ^ x + w^ - y, i = 1,2, , N (3.14)
It is of interest to locate multiple solutions of g(x) = 0. Instead
of applyign Katznelson's algorithm directly to (3.14) starting
from a point selected in one way or another, the iterative scheme
i+1 i,,i..(i) . . i. , iv / i\ . / INNT X = X +X J (-g (x ), -g2(x ), , -gj . Cx ), *g^(x ))
16 x^ € L, i - 0 , 1 , 2 , (3.15)
i s used where the s t a r t i n g po in t x must l i e on a space curve L of
i n t e r s e c t i o n g . (x) = 0 , i = 1,2, , n-1 and X^ > 0 i s the l a r g e s t
p o s i t i v e number such t h a t x and x are in the same reg ion .
0 1 2 As long as X i s chosen to l i e on L, the i t e r a t i o n po in t s x , x ,
~ — remain on L.
Lemma 2.1: If x lies on a space curve L defined by g (x) * 0
1 2 i = 1,2, , n-1, then the successive points x , x , resulted
from (3.15) and the.'line segments joining these points also lie
on L.
Proof: Without lo s ing g e n e r a l i t y , assume x € L i s in region R .
Consider the following d i f f e r e n t i a l equat ion in R
4 r g . ( x ( t ) ) = - g , ( x ( t ) ) , g , (x (0 ) ) = 0, i = 1,2, , n - l d t 1 1 1
4 - 3 ( x ( t ) ) » ± g ^ ( x ( t ) ) , g„(x(0)) » g^^ x(o) = x°c . L (3.16) dt n n n no ,
3y the chain-rule and Euler integration formula,
1..0, 0 ,0,(0)-l, , 0 , , 0 . . 0. , 0. • X (X ) - X + X J^ {~%^{-3i )> -g2(^ )» > "^n-l^*"^ - ' -^n^^ -
(3.17)
where 0 < X < X , X i s the l a r g e s t p o s i t i v e cons tant such t h a t — — m m
x^(X°) i s s t i l l in R^, i . e . x^(X^) l i e s on the boundary of R and m 01
R . The s o l u t i o n of (3.16) in g-space i s given by
g ^ ( x ( t ) ) - g^(x^) e ' ^ » 0, i « 1,2, , a-1
g _ ( x ( t ) ) » g^n^-^. (3.18) o. n 'nO"
This shows t ha t i i x £ L, the corresponding s o l u t i o n cur-ze x ( t ) ,
1 0 i . e . X (A ) in ( 3 . 1 7 ) , remains on L. Following tne same procedure ,
17
in region R , x (X ) is used as the starting point for the next
2 1 iteration. Then it is clear that the line segment joining x (X )
m
and X (X ) remains on L. m
In general, if x = x (X ) is on L, then by the iterative
scheme (3.15), x = x (X ) and the line segment joing x"""
and X also lies on L. By deduction. Lemma 1. is proved.
Let the line segment joining x and x be denoted by L..
Then in fact, L = L^ U L^ U . The minus sign of g for i =
1,2, , n-1 is used for reducing the computation error in each
successive iteration.
Let d^ 4 J^^^"^-g^(xS, , -gn_i(xS, ±g^(xS)^, i = 0,1,2,
(3.19)
Then that the solution curve L passes through each region R ,
i = 0,1,2, in the direction of d where the positive or
negative sign before g (x ) in each region can be determined by n
the following two lemmas. It is assumed that the solution curve
L passes through R in the direction of d then enters region R
at X and traverses R in the direction of d
Lemma 2.2: If there exists no solution in region R and (i) the
determinants of Jacobians J and J have different signs,
then in region R the sign of g should be changed: (ii) if on
the other hand, the determinants of J and J have the same
sign, then in region R the sign of g should not be changed.
18
* * I 1
Proof: Since x and x are on the solution curve L and there * ' * I 1
exists no solution in R , then g (x ) and g (x ) must have the
same sign. Without loss of generality, assume -g (x ) is used
for obtaining d"*" in R . As shown in Fig. 3.4, T 1 T 1+1
n d > 0 and n d > 0 (3.20)
are satisfied on the boundary between R and R where n is
the normal vector of the boundary. Substituting d into (3.20)
(3.19a)
Since x is a point on the solution curve L, (3.19a) becomes
^Tj(i)"l[ 0, 0, , -g^(xS]^ (3.19b)
From the condition of (3.13)
det J n J = det J n J
it is clear that
det J ^ n^J^^^'^O, 0, - -, -g^(x^)]^
- , , i, , , i+lxij .T(i+1) T^ (i+l)"" r n n f i+l^^T = {g (x )/g (x )}detJ' ' J ' L 0, 0, - -, -g^(x )J
(3.13a)
Now, if det J and det J have different signs, then
from (3.13a) n^J^^^ [ 0, 0, , -g^(x^)]^
, T,Ci+l)'''"rn n / i+l^lT and n J^ [0, 0, -—, "gj (.x )J
T i+1 have different signs. To assure n d > 0 on the boundary,
the sign of g should therefore be changed.
19
On the other hand, if det J and det J have the same
sign, then n^J^^^ [0,0, , -g^(x^)]^
and T_(i+1) n J
-1 [0,0, — . -g^(x^"'^]^
will have the same sign and
,i+l ,(i+l)~ r^ f. f i+lxlT d = J L0,0, , -g^ (x )J
T i+1 should be used to assure n d >0 on the boundary. Similarly,
if +g (x ) is used for obtaining d in R the same conclusion result
Lemma 2.3: If there exists a solution in region R , then the
transition in sign of g should occur if det J and det J o °n
have the same sign of g should not occur if det J and det J
have different signs.
Proof: It is clear that g will change sign when the solution
cuirve L passes through a solution point. Hence g (x ) and g (x )
should have different signs.
20
From the same argument as that given in Lemma 2., and
conditions
Tji - T^i _ n d >0, n d >0
.r.A A . ,{1) T Al)-1 . , .(i+l) T ^(i+l)-l and det J n J = det J n J ' ,
Lemma 2.3. is proved immediately.
In view of the above, the solution curve can easily be
taken care of at a local point on the boundary when Jacobian
determinants have arbitrary signs. Therefore the iterative
scheme of (3.15) allows continuation of tracing along the solution
curve L and, as a result, all the solutions on L can be obtained.
If g.(x) = 0,1,2, , n-1 defines a unique simple curve L,
that is, one with only a single branch, the L passes through
all the solutions of (3.14). Once a starting point on L is
obtained, all of the multiple solutions of (1.1) can be located
by using the technique described in this section. The following
lemma gives a sufficient condition for the existence and unique
ness of this solution curve.
Lemma 2.4: Let g(x) be a continuous piecewise-linear mapping of
R^ into itself and let J ^ denote the matrix by deleting the
kth row and nth column of Jacobian matrix J of region R , and
let J , denote the matrix consisting of the first m rows and km
first m columns of the constant Jacobian J, . If there exists
a k, l<JiC^n, such that for each m = 1,2, , n-1, J, , J
21
km • "km ' ""' (N)
J j do not vanish and have the same sign then g.(x) = 0, j = 1,
2, , n-1 defines a unique simple curve.
Proof: Let h(x) = (g^(x), g2(x), , g _;L '' ^
and x_^ = (x^, x^, — , x^.i^x^+i. — , x^)^
If the value of x, is fixed, then the mapping h of R ~ into itself
is still a continuous piecewise-linear mapping. For any x, ,
h(x_ .,u) = \ x_^ + v^(u) - y_^, i = 1,2, , N
where y_^ = (y^.yj, — , yi . , 7^+1. — . y„)^
Since for each m, m = 1,2, , n-1, J, , , J, do not Icm Km
vanish and have the same sign, the mapping h with fixed x, is
a homeomophism of R onto itself. This implies that for any
given input y and any number u of x, there exists one and only
one point T
x_j (u) = (x^(u), X2(u), , X^_;L^^'^' ^+1^^-^' n - ^
such that h(x i,u) = 0. This guarantees the existence of a unique
continuous piecewise-linear function F: R ->R for h(x) = 0 such
that X ,= F(x, ). Hence F(x,) defines a unique simple curve L(x, )
in R for all
A starting point lying anywhere on the intersection L of
g.(x) = 0, i = 1,2, , n-1 has to be found before the root-finding
technique described in the previous section is applied.
22 Let M be a positive number such that for any x » (x., x-,
T , X ) with X, >_ M, X is in an unbounded region. The union of
such unbounded regions is denoted by U. An arbitrary point x in
k k
R is chosen such that R U. With a fixed value of x, = u,
u >.M, a point x = Cx^(u), , Xj^_^(u), u, x^+^(u), , x^(u))^
which s a t i s f i e s g^(x) - 0, i » 1,2, , n-1 i s searched i n the (n-1) dimensional space R
Let g(x) be a continous p i e c e w i s e - l i n e a r mapping of R onto
i t s e l f defined i n ( 3 . 1 4 ) , and l e t J ^ ^ denote the matr ix by
d e l e t i n g the kth row and nth column of Jacobian mat r ix J of
the reg ion R .. In the following a lemma which was proved by
Kuh, Ohtsuki , and Fuijsawa Cl2] i s s t a t e d .
Lemma 2 . 5 . Let g be a continous p i e c e w i s e - l i n e a r mapping from
R i n t o i t s e l f defined by ( 3 . 1 4 ) . For an a r b i t r a r y v, i f det J
i = 1,2, ,N have same s i g n , then the re e x i s t s a t l e a s t one
s o l u t i o n X to the equat ion g(x) = 0. An immediate consequence of
Lemma 2.5 • i s as fo l lows.
Coro l l a ry : I f f o r a given k , 1 _<k<_ n , such t h a t in each region
R- , R" c ,^» ^ ]. <io ^ot vanish and have the same s ign then for
any x, , x, >^M, there e x i s t s a t l e a s t one s o l u t i o n for
g^(x) - 0 , i « 1, 2 , , n - 1 .
23
This corollary gives a sufficient condition for the existence
of starting point x . The convergence to x from an arbitrary
initial guess in R-', R-'c U is guaranteed by the Katznelson's
algorithm.
in the following steps.
Algorithm A:
Step 1. Find an initial point x in an unbounded region R according
to section 3.4. Set j = 0.
Step 2. Compute g(x ).
Step 3. If j=f 0, compute d according to Lemma 2. and Lemma 3.
^ 1- . ^0 .(O)"-'-, , 0, , 0.,T Otherwise, compute d = J ( -g,(x ) , - - - , -g (x )) .
i i i * Step 4. Compute x = x* + d . If x is in R then x = x is a
solution.
Step 5. Compute x-' = x" + X~ d , where X-" > 0 is the maximum value • • * * *
such that x(X) = x-"' + Xd" , 0 <_ X <_ X^ is in R^. If X^- » go
to step 7.
Step 6. Otherwise, identify region R-" . Set j = j +1 and go to
step 2.
Step 7. Set j = 0. Repeat step 2. to step 6. except in step 3. that
if j = 0 compute d° = J^°^ ( -g^{^^) . . +g^(x°))'^
gand in step 5. that if X-' ->•=», stop.
Two examples are given here for illustration,
24
Example 3.1. Consider the nonlinear circuit of Fig. 3.5a, consisting
of two tunnel diodes and two constant current sources and a linear
resistor. The piecewise-linear approximation of function h for the
tunnel diode is given in Fig. 3.5b with
' 2x X < 3
1.5x-6.5 x > 5
8^^\x) J ^ ^ ^ + w^ - y 0, i - 1, 2, , 9
are obtained. All regions are shown in Fig. 3.6. The equation
associated with each region is given as:
g^»Cx) 3
25
1= 2] +
-6 .5 '
-6 .5
In this example, y, i.e. the input current source is given by
y =
13.5
13.5
h(x)i
h for the tunnel diode
26
R3
R2
R1
R^
R5
R*
R '
R3
R^
X
Fig. 3.6. The regions in the x-space
If the solution curve L of g,»0 is traced, then the points
on L with g^» 0 are the desired solutions. For x.. >, 5, in each
7 8 9 unfaoxjnded region R , R , R oo"* 2.5 has the same sign and does
not vanish. Hence from Lemma 5, for a given value of x. = 5.5 > 5
0 T a corresponding solution x » Cy'5, 6.25) is found by solving
7 8 9 0
g_. Cxi = 0 in the unbounded regions R , R , R . Using x as a
starting point, the search along the curve L of g =0 leads to
five solutions C5.714, 5.714)^, (6.316, 4.211)^, (7.154, 2.115)*,
(4.211, 6.316)^, C2.115, 7..154) .
In fact, a simple solution curve does exist in this example
even though Lemma 4, is not satified. Similarly, if g O is
traced, the same five solutions are obtained. The solution curves
and all solutions are shown in Fig. 3.7.
27
IG.OO
F i g . 3 . 7 . So lu t ion curves of ejcample 3 .1
28
Example 3.2: In Fig. 3.8, the circuit consists of three nonlinear
resistive elements, two linear resistors and a constant voltage
source. The characteristic of each nonlinear element is given by
Fig. 3.9a and Fig. 3.9b with
h^Cx) =
X - 6 X < -1
5x -10 X > 3
Applying the proposed algorithm, three solutions
(1.200, -0.800, 3.200)^, (1.225,-0.900, 2.725)^, (1.323,-1.290,1.452)-^
are obtained. A solution curve L and all solutions on L are shown
in Fig. 3.10.
hi(x) hoCx)
10
Fig. 3.9a The characteristic Fig. 3.9b The characteristic of a nonlinear element of a nonlinear element
29
CHAPTER IV
SINGULAR JACOBIAN MATRICES
^'1 The Corner Problem
This section is devoted to the study of the corner problem
[13, 16]. When a solution curve reaches a corner, the previous
algorithm cannot determine the next region which is to be entered.
Therefore, the proposed algorithm must be modified accordingly.
A boundary hyperplane in x-space can be represented by
I T ^ x ~ (x|nx = c } where n is the normal vector of the hyperplane
and c is a constant. A corner is a subset of the intersection
of two or more such hyperplanes. Since g = J x + w"*"- y
I T and H = { x | n x = c } , the image of any boundary hyperplane H
of region R is a hyperplane H in the g-space where
H = {g|Jj(i)"'g = c+nTj(^>'' (w^-y )}
Furthermore, the image of a corner in the x-space is a subset
of the intersection of two or more hyperplanes in the g-space.
Thus in the g-space HA.^^g|pg = q}= ^g|^-^ g=c+n J (w -y)}
where p is a nxl vector and q is a scalar, a corner can be
I T represented as a subset o f U = { g | P g = Q } where P is a (nx2)
matrix and Q is a (2x1) vector.
30
31
Let
be a corner on the boundary of region R where P. and Q. are known
Since U is contained in Sp(U . ) , the linear manigold spanned by
U . which is an (n-1) dimensional subspace, there exists a (nxl)
unit vector h^ such that (h^) g =0 for all g U . . Let
i i 1 t>T,b, - - - b , constitute a basis for Sp (U . ) . Then h can 1 2 n-1 gi
be found by
where 0 is an (n-1) dimensional vector.
I T Thus, if a corner represented by U ,= {g| P. g=Q.} which is
on the boundary of region R is hit, then the solution line
segment L. in R must be perturbed such that the point x
on the boundary will just miss the comer (Fig. 4.1). A small
vector V = mh is computed, where m is a small number, such that
g(x ) is perturbed by v. It is always possible to obtain a new
i i point X in R such that
g(x ) = g ( x ) + v and x = x + J v
i' i The point x is then used, instead of x , for the iteration in
R ^
32
After the solution curve moves away from the boundary of the comer
i.e., the region extended has been identified, x is used again
to regain the original solution curve L.
From the above arguments, it is clear that the solution curve
L can be extended into a new region when a corner is hit. Further
more, the piecewise linearity of g implies that no two line
segments in a region possessing a nonsingular Jacobian matrix can
be portions of a solution curve. In the following section, the
problem caused by singular Jacobian matrix and the means for
overcoming this problem will be discussed.
4.2 Singular Jacobian Matrices Problem
An important property which plays a key role in the discussion
of the problem of singular Jacobian matrix is introduced. [l3, 16]
Lemma 4.1: If two regions R and R^ with Jacobian matrix J
and J • respectively are adjacent, i.e., have a common boundary
hyperplane, then the rank of J and J -' differ at most by
one. [l6]
through a region R with nonsingular Jacobian matrix. It is
assumed that L enters a singular region R , as shown in Fig. 4.2
It follows from Lemma 4.1 that J of region R is of rank n-1.
Therefore a nonzero vector d' such that J d'= 0 is uniquely
33
segment L. in R is obtained by
x(X) = x^ + Xd' where 0 <_ X ; X^
and X is determined such that x(X ) = x lies on another
hyperplane H. as indicated in Fig. 4.2 where H„ is between R
and R , By examining the mapping of L. into g-space, it is
found
g(x(X)) = J^^^x(X) + w" = J ' ( x^ + d') + w^
_(i) i , i , i. = J x + w =g(x)
i.e., the whole line segment L. is mapped into the single point
gCx ). Hence all the points on L. satify the Cn-1) equations
r \ n ' ^ o i T T j T i+l) 0 in the g.(x) = 0 , 1 = 1 , 2 , - - - , n-1. If det J
* I 1 * I 1
region of R in which x is lying, then according to Lemma 3.4
and Lemma 3.5, the solution curve can be extended into next region
R^+\ .i+1
In the special case when the next region R has also a
lingular Jacobian matrix, i.e., det J = 0 , then it follows
(i+l) from Lemma 4.1 that the rank of J is either n-1 or n-2
det J^^-^^^ 0
34
since J is of rank n-1. Suppose that J is of rank n-2,
then there exists a 2-dimensional plane S , containing x , such
2 i+1 2 that g(H2ns ) = { g (x )}. Since the hyperplane S H^ containing
X is at least one-dimensional, hence, if x' ? x is a point
9 ' .1.1
on S H^ then g(x') = g(x ). For two linearly independent
vectors (x'- x^) and (x " -x^), J^^\x'-x^) = J^^'^x^'^^-x^)
is satisfied in region R . Thus the rank of J is at most n-2,
must be n-1. The solution line segment L. .. can be extended
into region R from x by the same technique stated above,
and the singular Jacobian problem can thus be overcome.
Algorithm A can now be modified so that both the corner
problem and the problem of singular Jacobian problem are taken
care of. The new algorithm is summarized as follows.
Algorithm B:
Step 1. Use Algorithm A to trace the solution curve L.
Step 2. If a comer is hit by the solution curve L, then a
small vector v = mh is computed according to the
perturbation method stated in Section 4.1. A new point
i' i X is then computed and used, instead of x , to iterate
again in region R^. Once the solution curve move away
i+1 from the boundary of the corner, the point x on
the corner is used again to regain the solution curve.
Go to Step 1.
35
Step 3. If a singular region R is identified by the present
solution curve L, then a vector d' is computed such that
J d' = 0 where J is the Jacobian matrix of region
R . Next iterative point x^ is determined by the
direction d'. Go to Step 1.
CHAPTER V
sources are known as as-resitive networks. In this case, the
network equation can be written in the form of
g(x(t), u(t)) = 0 (5.1)
where u(t) is the time varying input vector and x represents
the network variables. Thus, the operating point of the network
is a function of t. One way of solving (5.1) is to solve the
problem at different instants of time, which would be a time-
consuming practice. A continuation algorithm for finding the
solutions (5.1) for all t automatically has been proposed in [22],
Consider the differential equation
^ u(t) = w(t) , u(tQ) = u° (5.2)
where u is assumed to be continously differentiable and its
time derivative, w, is a known function of t. The initial
condition on g is such that s(t-.) = x is a solution of (5.1)
subject to u(tQ) = u . In the x-space, (5.2) is equvalent to
x = -J^^g(x, u) + | ^ w ) , x(tQ) = x° ^^'^^
36
37
where J = r- . Thus, equation (5.3) can be integrated to fine x(t) d X
automatically.
the case where g is a piecewise-linear function. Equation (5.3)
with a piecewise-linear g(x) can now be written as
-1
dt " ^^^^' "' ' au "'' ^^^0 47=-J^^^ (g(x, u) + | 4 w ) , x(tj = x° (5.4)
where J is the Jacobian matrix in the region R in which x
lies. Here it is assumed that J is nonsingular and u(t) is
a small signal vector such that x(t) will remain in R . As it
is noted that since J is a constant matrix, the solution x(t)
f ? can be represented as the superposition of two vectors x' and x
where x' is the solution of g(x, 0) = 0 and x'' is the response
only by the time-varying sources u(t). It is clear that x''(t)
possesses similar waveforms as u(t) since g(x, u) = 0 is a linear
function in region R .
If g(x, u ) = 0 has multiple solutions, then all the solution
points must be used as starting points of (5.4) in order to obtain
a complete family of solution curves. In the following example
a simple circuit is given to illustrate the approach just described,
Example 5.1: Consider the nonlinear resistive circuit in Fig. 5.1
which consists of an ac source in series with a linear resistor
R = 600 ohm and a tunnel diode. The characteristic of the tunnel
38
function.
' 0.006 V - 0.021 V < 4
At t = 0, three operating points are first located by using the
method described in chapter III. The trajectories of v(t) and
i(t) for different Q points are plotted in Fig. 5.2 and Fig. 5.3
respectively.
If u(t) is large enough so that x(t) will not remain in the
region in which x(t-) lies, then the new region that x(t) will
enter must be identified and the Jacobian matrix in that region
should be substituted in (5.4). Proper J • ' s are switched back
and forth in (5.4) for different time intervals and the trajectory
for all t can be obtained accordingly.
600r i(t)
o t
0 .00 4 .00 8 . 0 0 12-00 TIME (SECOND)
16.00 2 0 . 0 0
Fig. 5.2. Output voltages for different starting points"
40
CTiD
:zui"
o o o. 0 .00 ^.OC 2 . 0 0 1 2 . 0 0 16 .00 2 0 . 0 0
T I M E ( S E C O N D ]
Fig. 5.3. Output currents for different startL- g points
41
Multivalued Resistive Nonlinear Networks
Driving point plots and t ransfer cha rac te r i s t i c plots are
important in the analysis of nonlinear r e s i t ive networks because
they can pro-vide a deep insight to the nature of the c i r c u i t .
Bas ica l ly , the two concepts are the same; they are input-output
c h a r a c t e r i s t i c plots showing the relat ionship between a driving
source u. and a cer ta in output variable x . . Several methods 3 1
£18-20,23J are available for finding the input-output plots of
nonlinear r e s i t i v e networks.
Here the technique of continuation scheme for piecewise-linear
analysis of r e s i t i v e nonlinear networks developed in chapter I I I
i s applied to obtain the input-output p lo t s . The network equation,
in general , can be represented by
g(x, u) = 0 (5.6)
where g i s a continous piecewise-linear function, x i s an n-vector
of network va r i ab le s , and u denotes the m-vector of input source.
Since in finding input-output cha rac te r i s t i c plots a l l input
sources except one, say u. , are considered to be fixed, thus
(5.6) i s reduced to
g(x, u.) = 0. (5.7)
Suppose i t i s now required to find the input-output cha rac t e r i s t i c
. - + p lo t X versus u for u 1 u < u , Following the development
42
given in [23], a set of (n+1) equations consisting of (5.7) and
an auxiliary equation g = 0 of the form
g(x, u.) = 0
III, a continuous piecewise-linear curve L can be traced automatically
by the iterative scheme
i"» '' i+1^ X
u
g(x , u.) = 0
where J is the Jacobian matrix in the region of R and x is
the solution point of g(x, u,) = 0. The sign change should be
made when L traverses across a boundary and enters a new region
in which the determinant of the Jacobian matrix has a different
sign,
The advantage of this method is that the input-output plot is
not required to be either input or output controlled, i.e., the
plot may be obtained for multivalued networks. An example is
given in the following to illustrate the approach.
Example 5.2: Consider the tunnel diode problem studied in the
previous example. The input-output characteristic plots, v versus
E and i versus E, is graphed as the input E varies from 0 to 10
volts. The result is shown in Fig. 5.4 and Fig. 5,5.
43
c?
[ N P U T E ( V O L T S ]
Fig . 5 .4 . Multivalued input -output p lo t for the c i r c u i t
.n exaiTDle 5 .2 .
S--00 3 . 0 0 (VOLTS)
10=00
Fig . 5 . 5 . J^ultlvalued input-output p lo t for the. c i r cu i t Ln.esjample. 5.2^
CHAPTER VI
presented for the piecewise-linear analysis of nonlinear resistive
networks. A set of continuous piecewise-linear equations g(x) = 0
is solved by tracing the solution curve L of any n-1 equations of
g(x) = 0 . As a result multiple solutions on L can be obtained.
Furthermore, a sufficient condition for the existence of a unique
simple curve L is given. It is clear that if L is simple, then all
the umltiple solutions of g(x) = 0 can be foimd by the proposed
search scheme.
In Chapter III, the local behavior of L on the boundary is
studied. A theory that guarantees the continuation of the search
algorithm when acrossing the boundary has developed. A method for
finding the starting point also been suggested. Methods for
overcoming the comer problem and singular Jacobian matrix problem
have been presented in Chapter IV. Applications to AC nonlinear
resitive network analysis and driving or transfer characteristic
curve of multivalued nonlinear network are given in Chapter V.
It is found that the algorithm developed here provides an efficient
and systematic way for solving the piecewise-linear nonlinear resis
tive network problems. However, if the solution curve consists of
multiple branches, only the solutions on the branch that contains
45
46
the initial point can be found. A systematic method for locating
a starting point on every branch is worth of further research.
LIST OF REFERENCES
[l] A. F. Malmberg, F. L. Comwell, and F. N. Hofer, "NET-1 network
analysis program", Los Alamos Sciences Lab., Rep. LA-3119, 7090/7094
version, August 1964.
[2] L. D. Milliman, W. A. Massena, and R. H. Dickhaut, "CTRCUS, A
digital computer program for transient analysis of electronic circuit
- user's g\iide", Harry Diamond Labs., The Boeing Co., Seattle, Wash.,
[3] H. W. Mathers, S. R. Sedore, and J. R. Sents, "Automated digital
computer program for determining responses of electronic circuits
(SCEPTRE)", vol. 1, IBM Electronic System Center, Owego, N.Y., File
66-928-611, February, 1969.
[4] K. S. Chao, D. K. Liu, and C. T. Pan, "A systematic search method
for obtaining multiple solutions of simultaneous nonlinear equations",
IEEE Trans. Circuits Syst,, vol. CAS-22, pp. 748-753, Sept. 1975.
[5] F. H. Branin, Jr., "Widely convergent method for finding multiple
solutions of simultaneous nonlinear equations", IBM J. Res. Develop.,
vol. 16, no. 5, pp. 506-522, Sept. 1972.
[6] L. 0. Chua and A. Ushida, "A switching-parameter algorithm for
finding multiple solutions of nonlinear resitive circuits". Int. J.
Circuit Theory and Applications, vol. 4, no. 3, pp. 215-239, July 1976.
[7] R. P. Brent, "On the Davidento-Branin method for solving simu
ltaneous nonlinear equations", IBM J. Res. Develop., vol. 16, no. 4,
pp. 434-436, July 1972.
47
ar
48
Lsj E. S. Kuh and I. N. Hajj, "Nonlinear circuit theory: Resistive
networks", Proc. IEEE, vol. 59, pp. 340-355, March 1971.
19J C. T. Pan, "A discrete approach for solving nonlinear equations"
Master's thesis, Texas Tech University, 1974.
ilQj J. Katznelson, "An algorithm for sol-ving nonlinear resistive
networks" Bell System Tech. J. vol. 44, pp. 1605-1620, 1965.
[ll] T. Fujjsawa and E. S. Kuh, "Piecewise-linear theory of nonline
networks", SIAM J. Appl. Math. vol. 22, pp. 307-328, 1972.
[12] T. Ohtsuki, T. Fujjsawa, and E. S. Kuh, "A sparse matrix method
for analysis of piecewise-linear resistive networks", IEEE Trans.
Circuit Theory, CT-19, pp. 571-584, 1972.
[13] M. J. Chien and E. S. Kuh, "Solving piecewise-linear equations
for resistive networks". Int. J. Circuit Theory and Appl., vol. 4,
no. 1, pp. 3-24, January 1976.
[14J M. J. Chien and E. S. Kuh, "Solving nonlinear resistive networks
using piecewi e-linear analysis and simplicial subdivision", IEEE
Trans. Circuits Syst., vol. CAS-24, no. 6, pp. 305-317, June 1977.
[15] L. 0. Chua, "Efficient computer algorithm for piecewise-linear
analysis of resistive nonlinear networks", IEEE Trans. Circuits
Theory, vol. CT-18, pp. 73-85, January 1971.
[I6] T. Ohtsuki, T. Fujjsawa and S. Kumagai, "Existence theorems
and a solution algorithm for piecewise-linear resistive networks",
SIAM J. Appl. Math. 1977.
I17] H. C. So, "On the hybrid description of a linear n-port result
ing from the extraction of arbitrarily specified elements", IEEE Trans
49
[18] L. 0. Chua, Introduction to Nonlinear Network Theory, McGraw-
Hill, New York, N.Y., 1969.
[19J G. H. Meyer, "On Solving nonlinear equations with a one-parameter
operator imbedding", SIAM J. Numer. Analy., vol. 5, pp. 739-752, 1968.
[20] V. C. Prasad and A. Prabhakar, "Input-output plots of piecewise-
linear resistive networks", Proc. 1974 IEEE Int. Symp. Circuits and
Systems, April 1974, pp. 55-59.
12l] R. Saeks and R. A. Decarlo, Interconnected Dynamical Systems,
New York, Marcel Dekker (to appear).
[22] C. T. Pan, "Integration Methods in System Analysis", PH.D.
dissertation, Texas Tech University, 1976.
[23] K. S. Chao and R. Saeks, "Continuation Methods in Circuit
Analysis", Proc. IEEE, vol. 65, no. 8, August 1977.
Recommended | 2023-03-21 04:25:40 | {"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": 0.8352312445640564, "perplexity": 2868.713160257033}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1679296943625.81/warc/CC-MAIN-20230321033306-20230321063306-00732.warc.gz"} |
https://doc.qt.io/squish/automated-batch-testing.html | # Automated Batch Testing
This chapter discusses all aspects of automating testing, also known as batch testing. The coverage includes automatically executing tests, distributing tests to different machines, and processing the results produced by the test runs. (See also, How to Do Automated Batch Testing.)
## Automated Test Runs
Squish provides command line tools that make it possible to completely automate the running of tests. The tool for executing tests is squishrunner, but for it to work properly a squishserver must also be running—the squishrunner makes use of the squishserver to start AUTs and communicate with them.
Automated batch tests can be created on any of the platforms that Squish supports, including Windows and Unix-like platforms.
For example, here is a simple Unix shell script to execute the complete test suite ~/suite_myapp and save the results to ~/squish-results-date:
#!/bin/bash
export SQUISH_PREFIX=/path/to/squish
export PATH=$SQUISH_PREFIX/bin:$PATH
# Dated directory to store report results:
LOGDIRECTORY=~/squish-results-date -I
# Execute the test suite with a local squishserver:
squishrunner --testsuite ~/suite_myapp --local --reportgen xml3.5,\$LOGDIRECTORY
Since we are using --local, it is not necessary to start a separate squishserver process in the background from our script.
We are using --reportgen to specify the kind of output to generate. For a list of all report-generator formats, see squishrunner --reportgen: Generating Reports.
Here is a similar example, written for Windows using the standard cmd.exe shell. Because the Windows variable, %DATE% sometimes uses a period and other times a dash or some other character to separate the numbers, it can't be used in some locales to form a directory name.
REM add Squish to PATH
set SQUISH_PREFIX=C:\path\to\squish
set PATH=%SQUISH_PREFIX%\bin;%PATH%
REM The Log Directory does not (but perhaps should) have today's date in it:
set LOGDIRECTORY=%USERPROFILE%\squish-results
REM Execute the test suite with a --local squishserver:
squishrunner --testsuite %USERPROFILE%\suite_myapp --local --reportgen "xml3.5,%LOGDIRECTORY%"
One disadvantage of using shell scripts and batch files like this is that for cross-platform testing we must maintain separate scripts for each platform. We can avoid this problem by using a cross-platform scripting language which would allow us to write one script and run it on all platforms we were interested in. Here is an example of such a script written in Python, which you can find in SQUISHDIR/examples/regressiontesting/squishruntests-simple.py.
#!/usr/bin/env python
# -*- encoding=utf8 -*-
# Copyright (C) 2018 - 2021 froglogic GmbH.
# Copyright (C) 2022 The Qt Company Ltd.
#
# This file is part of an example program for Squish---it may be used,
# distributed, and modified, without limitation.
#
# For a script with more options and configurability, see 'squish6runtests.py'
# This script assumes squishrunner and squishserver are both in your PATH
import os
import sys
import subprocess
import time
def is_windows():
platform = sys.platform.lower()
return platform.startswith("win") or platform.startswith("microsoft")
# Create a dated directory for the test results:
LOGDIRECTORY = os.path.normpath(
os.path.expanduser("~/results-%s" % time.strftime("%Y-%m-%d"))
)
if not os.path.exists(LOGDIRECTORY):
os.makedirs(LOGDIRECTORY)
# execute the test suite (and wait for it to finish):
exitcode = subprocess.call(
[
"squishrunner",
"--testsuite",
os.path.normpath(os.path.expanduser("~/suite_myapp")),
"--local",
"--reportgen",
"xml3.5,%s" % LOGDIRECTORY,
],
shell=is_windows(),
)
if exitcode != 0:
print("ERROR: abnormal squishrunner exit. code: %d" % exitcode)
This script does the same job as the Unix shell script and Windows batch file shown earlier, and makes the assumption that squishrunner and squishserver are in the PATH. It should run on Windows, macOS and other Unix-like systems without needing any changes. It also checks the return code of squishrunner for abnormal termination.
squishrunner will run the specified test with all the required initializations and cleanups. The resulting report can then be post-processed as necessary—see Processing Test Results for details.
Once we have the script running on all desired platforms, we must ensure that it can be run automatically, say once a day. How to do this is beyond the scope of this manual, but if you require help you can always contact froglogic's commercial support to assist you. If you would rather try to set it up on your own first, there is a lot of information on the Internet—for Unix it is a matter of setting up a cron job. (Since Windows Services don't support a display for running GUI applications, it is not possible to execute the squishserver as a Windows Service.)
## Distributed Tests
Throughout the manual it is generally assumed that all testing takes place locally. This means that the squishserver, squishrunner, and the AUT, are all running on the same machine. This scenario is not the only one that is possible, and in this section we will see how to remotely run tests on a different machine. For example, let's assume that we work and test on computer A, and that we want to test an AUT located on computer B.
The first step is to install Squish and the AUT on the target computer (computer B). Note though, that we do not need to do this step for Squish for Web. Now—except if we are using Squish for Web on computer B—we must tell the squishserver the name of the AUT's executable and where the executable is located. This is achieved by running the following command:
squishserver --config addAUT <name_of_aut> <path_to_aut>
Later we will connect from computer A to the squishserver on computer B. By default the squishserver only accepts connections from the local machine, since accepting arbitrary connections from elsewhere might compromise security. So if we want to connect to the squishserver from another machine we must first register the machine which will try to establish a connection for executing the tests (computer A in this example), with the machine running the AUT and squishserver (computer B). Doing this ensures that only trusted machines can communicate with each other using the squishserver.
To perform the registration, on the AUT's machine (computer B) we create a plain text file in ASCII encoding called /etc/squishserverrc (on Unix or Mac) or c:\squishserverrc (on Windows). If you don't have write permissions to /etc or c:\, you can also put this file into SQUISH_ROOT/etc/squishserverrc on either platform. (And on Windows the file can be called squishserverrc or squishserverrc.txt at your option.) The file should have the following contents:
ALLOWED_HOSTS = <ip_addr_of_computer_A>
<ip_addr_of_computer_A> must be the IP address of computer A. (An actual IP address is required; using a hostname won't work.) For example, on our network the line is:
ALLOWED_HOSTS = 192.168.0.3
This will almost certainly be different on your network.
If you want to specify the IP addresses of several machines which should be allowed to connect to the squishserver, you can put as many IP addresses on the ALLOWED_HOSTS line as you like, separated by spaces. And if you want to allow a whole group of machines which have similar IP addresses, you can use wildcards. For example, to allow all those machines which have IP addresses that start with 192.168.0, to connect to this squishserver, you can specify an IP address of 192.168.0.*.
Once we have registered computer A, we can run the squishserver on computer B, ready to listen to connections, which can now come from computer B itself or from any of the allowed hosts, for example, from computer A.
We are now ready to create test cases on computer A and have them executed on computer B. First, we must start squishserver on computer B (calling it with the default options starts it on port 4322—see squishserver for a list of available options):
squishserver
For convenience, by default, the Squish IDE starts squishserver locally on startup and connects to this local squishserver to execute the tests. But it is also possible to connect to a squishserver on a remote machine, such as computer B, from within the Squish IDE. We can control this behavior through the preferences dialog. Click Edit > Preferences to invoke the Preferences dialog, then click Squish in the tree of preferences and choose the Remote Testing item to show the Remote Testing preferences page. Uncheck the Start local Squish server automatically checkbox, and enter the IP address of the machine running the remote squishserver (computer B) in the squishserver host line edit. The port number should only be changed if the squishserver is started with a non-standard port number, in which case the port number should be set to match whichever one is used on the remote machine (computer B).
The Squish IDE's Preferences Dialog
Now we can execute the test suite as usual. One immediately noticeable difference is that the AUT is not started locally, but on computer B instead. After the test has finished, the results become visible in the Squish IDE on computer A as usual.
It is also possible to do remote testing using the command line. The command is the same as described earlier, only this time we must also specify a host name using the --host option:
squishrunner --host computerB.froglogic.com --testsuite suite_addressbook
The host can be specified as an IP address or as a name.
This makes it possible to create, edit, and run tests on a remote machine via the Squish IDE. And by adding the --host option to the shell script, batch file, or other script file used to automatically run tests, it is possible to automate the testing of applications located on different machines and platforms as we saw earlier—Automated Test Runs.
Note: When Squish tools are executed, they always check their license key. This shouldn't matter when using a single machine, but might cause problems when using multiple machines. If the default license key directory is not convenient for using with automated tests, it can be changed by setting the SQUISH_LICENSEKEY_DIR environment variable to a directory of your choice. This can be done in a shell script or batch file. See Environment Variables.
## Processing Test Results
In the previous section we saw how to execute an AUT and run its tests on a target machine under the control of a separate machine, and we also saw how to automatically execute test runs using scripts and batch files. In this section we will look at processing the test results from automatic test runs by uploading XML reports to Squish Test Center.
By default, squishrunner prints test results to stdout as plain text. To make squishrunner use the XML report generator, specify --reportgen xml3.5 on the command line. If you want to get the XML output written into a file instead of stdout, specify --reportgen xml3.5,directorypath, e.g.:
squishrunner --host computerB.froglogic.com --testsuite suite_addressbook_py --reportgen xml3.5,/tmp/results
The XML report will be written into the directory /tmp/results.
Next, we can upload the report to Squish Test Center for further analysis, using the testcentercmd command-line tool.
testcentercmd --url=http://localhost:8800 --token=MyToken upload AddressBook /tmp/results --label=MyLabelKey1=MyLabelValue1 --label=OS=Linux --batch=MyBatch
The command above has the following options.
• The URL of a running instance of Squish Test Center http://localhost:8800.
• An upload token, MyToken. See Creating and Managing Upload Tokens for more details.
• A command, upload.
• A project name, in this case AddressBook.
• A directory (or a zip file) of XML test results to upload
• 0 or more labels, or key=value pairs for tagging the result in the database
• An optional batch name, MyBatch, also for the purposes of organizing in the Squish Test Center database.
Project Names, Batches and Labels are concepts used to organize and select test reports in the web interface, and are explained in more detail here.
### The xml Report Format
The document starts with the <?xml?> tag which identifies the file as an XML file and specifies the encoding as UTF-8. Next comes the Squish-specific content, starting with the SquishReport tag which in Squish 6.6, has a version attribute set to 3.4. This tag may contain one or more test tags. The test tags themselves may be nested—i.e., there can be tests within tests—but in practice Squish uses top-level test tags for test suites, and nested test tags for test cases within test suites. (If we export the results from the Test Results view there will be no outer test tag for the test suite, but instead a sequence of test tags, one per test case that was executed.)
The test tag has a type attribute used to store the type of the test. Every test tag must contain a prolog tag as its first child with a time attribute set to the time the test execution started in ISO 8601 format, and must contain an epilog tag as its last child with a time attribute set to the time the test execution finished, again in ISO 8601 format. In between the prolog and epilog there must be at least one verification tag, and there may be any number of message tags (including none).
Every verification tag may contains several sub elements.
The uri tag contains the relative path and filename of the test script or verification point that was executed, and the lineNo tab contains the number of the line in the file where the verification was executed. If uri starts with x-testsuite: or x-testcase: or x-results: the path is relative to respectively the test suite, test case or results directories. If file path is outside of the mentioned directories, the uri tag will contain the abosulute file path.
The scriptedVerificationResult tag is used to specify the verification point type. There are other possible types too: screenshotVerificationResult, propertyVerificationResult or tableVerificationResult.
The screenshotVerificationResult is for screenshot verifications or "propertyVerificationResult" for property verifications (e.g. calls to the Boolean test.vp(name) function) or an empty string for any other kind of verification (such as calls to the Boolean test.verify(condition) function), and tableVerificationResult is for table verification points results.
Every verificaton point's result tag has two attributes: a time attribute set to the time the result was generated in ISO 8601 format, and a type attribute whose value is one of PASS, FAIL, XPASS, XFAIL, FATAL, or ERROR. In addition the scriptedVerificationResult tag should contain at least one detail tag whose text describes the result. Normally, two tags text and detail are present, one that describes the result and the other whose text gives a more detailed description of the result. For screenshot verifications there will be additional tags, one objectName whose content is the symbolic name of the relevant GUI object, and one failedImage whose content is either the text "Screenshots are considered identical" (for passes), or the URI of the actual image (for fails, i.e., where the actual image is different from the expected image).
In addition to verification tags, and at the same level (i.e., as children of a test tag), there can be zero or more message tags. These tags have two attributes, a time attribute set to the time the message was generated in ISO 8601 format, and a type attribute whose value is one of LOG, WARNING, or FATAL. The message tag's text contains the message itself.
Here is an example report of a test suite run. This test suite had just one test case, and one of the screenshot verifications failed. We have changed the line-wrapping and indentation for better reproduction in the manual.
<?xml version="1.0" encoding="UTF-8"?>
<SquishReport version="3.4" xmlns="http://www.froglogic.com/resources/schemas/xml3">
<test type="testsuite">
<prolog time="2015-06-19T11:22:27+02:00">
<name><![CDATA[suite_test]]></name>
<location>
</location>
</prolog>
<test type="testcase">
<prolog time="2015-06-19T11:22:27+02:00">
<name><![CDATA[tst_case1]]></name>
<location>
<uri><![CDATA[x-testsuite:/tst_case1]]></uri>
</location>
</prolog>
<verification>
<location>
<uri><![CDATA[x-testcase:/test.py]]></uri>
<lineNo><![CDATA[2]]></lineNo>
</location>
<scriptedVerificationResult time="2015-06-19T11:22:27+02:00" type="PASS">
<scriptedLocation>
<uri><![CDATA[x-testcase:/test.py]]></uri>
<lineNo><![CDATA[2]]></lineNo>
</scriptedLocation>
<text><![CDATA[Verified]]></text>
<detail><![CDATA[True expression]]></detail>
</scriptedVerificationResult>
</verification>
<verification>
<location>
<uri><![CDATA[x-testcase:/test.py]]></uri>
<lineNo><![CDATA[3]]></lineNo>
</location>
<scriptedVerificationResult time="2015-06-19T11:22:27+02:00" type="FAIL">
<scriptedLocation>
<uri><![CDATA[x-testcase:/test.py]]></uri>
<lineNo><![CDATA[3]]></lineNo>
</scriptedLocation>
<text><![CDATA[Comparison]]></text>
<detail><![CDATA['foo' and 'goo' are not equal]]></detail>
</scriptedVerificationResult>
</verification>
<epilog time="2015-06-19T11:22:27+02:00"/>
</test>
<epilog time="2015-06-19T11:22:27+02:00"/>
</test>
</SquishReport>
In examples/regressiontesting you can find some example scripts which execute the addressbook test suite on different machines and present the daily output on a Web page by post processing the XML and generating HTML. The How to Do Automated Batch Testing section explains how to automate test runs and process the test results to produce HTML that can be viewed in any web browser. | 2022-11-28 22:40:40 | {"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": 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": 0.3543935716152191, "perplexity": 2859.172281225976}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1669446710662.60/warc/CC-MAIN-20221128203656-20221128233656-00621.warc.gz"} |
https://matplotlib.org/3.0.2/api/_as_gen/matplotlib.pyplot.semilogx.html | # matplotlib.pyplot.semilogx¶
matplotlib.pyplot.semilogx(*args, **kwargs)[source]
Make a plot with log scaling on the x axis.
Call signatures:
semilogx([x], y, [fmt], data=None, **kwargs)
semilogx([x], y, [fmt], [x2], y2, [fmt2], ..., **kwargs)
This is just a thin wrapper around plot which additionally changes the x-axis to log scaling. All of the concepts and parameters of plot can be used here as well.
The additional parameters basex, subsx and nonposx control the x-axis properties. They are just forwarded to Axes.set_xscale.
Parameters: basex : scalar, optional, default 10 Base of the x logarithm. subsx : array_like, optional The location of the minor xticks. If None, reasonable locations are automatically chosen depending on the number of decades in the plot. See Axes.set_xscale for details. nonposx : {'mask', 'clip'}, optional, default 'mask' Non-positive values in x can be masked as invalid, or clipped to a very small positive number. lines A list of Line2D objects representing the plotted data. **kwargs All parameters supported by plot. | 2021-09-25 04:17:34 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4274834394454956, "perplexity": 8270.584710055658}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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-20210925021713-20210925051713-00464.warc.gz"} |
https://mattermodeling.stackexchange.com/questions/4436/unable-to-recreate-data-for-benzene-from-paper-in-gromacs-2020-5 | # Unable to recreate data for benzene from paper in GROMACS 2020.5
I have been trying to simulate benzene in OPLSS-AA forcefield as per this paper: https://pubs.acs.org/doi/10.1021/ct2002122. I have been at a loss, since I believe I have everything running as per the paper, except for the vdw and coulombic cutoffs, but I am not generating the results the paper is showing.
My simulation is running bug-free on GROMACS, however, I am not getting the results they are showing. I energy minimized my system, performed an NVT equilibration for 50 ps, then I performed an NPT equilibration for a 100 ps. Then I did a production run for 5 ns.
I have rcoulomb = rvdw = 1.3 in my simulation. That is because GROMACS does not allow twin cutoff ranges anymore.
My density is coming out in the neighborhood of 823 kg/m3 when it should be around 867 kg/m3.
I obtained my GRO and ITP file from LigParGen server, however, the charges on the LigParGen Server were different from the charges in the paper, so I changed them.
The only thing the paper hasn't provided are the harmonic bond and angle potential coefficients, and torsional coefficients.
I am really stuck as to how to go about this problem. Where could my bug be lying?
For the sake of completeness, I have attached my primary input files. This is my .gro file (coordinate):
LIGPARGEN GENERATED GRO FILE
12
1UNK C00 1 0.100 0.100 0.000
1UNK C01 2 -0.040 0.100 0.000
1UNK C02 3 -0.109 0.100 0.121
1UNK C03 4 -0.039 0.100 0.242
1UNK C04 5 0.100 0.100 0.241
1UNK C05 6 0.170 0.100 0.121
1UNK H06 7 0.154 0.100 -0.094
1UNK H07 8 -0.094 0.100 -0.094
1UNK H08 9 -0.218 0.100 0.121
1UNK H09 10 -0.093 0.100 0.336
1UNK H0A 11 0.155 0.100 0.335
1UNK H0B 12 0.278 0.100 0.121
1.00000 1.00000 1.00000
This is my .itp file:
;
; GENERATED BY LigParGen Server
; Jorgensen Lab @ Yale University
;
[ atomtypes ]
opls_804 C804 12.0110 0.000 A 3.55000E-01 2.92880E-01
opls_803 C803 12.0110 0.000 A 3.55000E-01 2.92880E-01
opls_811 H811 1.0080 0.000 A 2.42000E-01 1.25520E-01
opls_809 H809 1.0080 0.000 A 2.42000E-01 1.25520E-01
opls_810 H810 1.0080 0.000 A 2.42000E-01 1.25520E-01
opls_805 C805 12.0110 0.000 A 3.55000E-01 2.92880E-01
opls_802 C802 12.0110 0.000 A 3.55000E-01 2.92880E-01
opls_806 H806 1.0080 0.000 A 2.42000E-01 1.25520E-01
opls_808 H808 1.0080 0.000 A 2.42000E-01 1.25520E-01
opls_800 C800 12.0110 0.000 A 3.55000E-01 2.92880E-01
opls_807 H807 1.0080 0.000 A 2.42000E-01 1.25520E-01
opls_801 C801 12.0110 0.000 A 3.55000E-01 2.92880E-01
[ moleculetype ]
; Name nrexcl
UNK 3
[ atoms ]
; nr type resnr residue atom cgnr charge mass
1 opls_800 1 UNK C00 1 -0.115 12.0110
2 opls_801 1 UNK C01 1 -0.115 12.0110
3 opls_802 1 UNK C02 1 -0.115 12.0110
4 opls_803 1 UNK C03 1 -0.115 12.0110
5 opls_804 1 UNK C04 1 -0.115 12.0110
6 opls_805 1 UNK C05 1 -0.115 12.0110
7 opls_806 1 UNK H06 1 0.115 1.0080
8 opls_807 1 UNK H07 1 0.115 1.0080
9 opls_808 1 UNK H08 1 0.115 1.0080
10 opls_809 1 UNK H09 1 0.115 1.0080
11 opls_810 1 UNK H0A 1 0.115 1.0080
12 opls_811 1 UNK H0B 1 0.115 1.0080
[ bonds ]
2 1 1 0.1400 392459.200
3 2 1 0.1400 392459.200
4 3 1 0.1400 392459.200
5 4 1 0.1400 392459.200
6 1 1 0.1400 392459.200
7 1 1 0.1080 307105.600
8 2 1 0.1080 307105.600
9 3 1 0.1080 307105.600
10 4 1 0.1080 307105.600
11 5 1 0.1080 307105.600
12 6 1 0.1080 307105.600
6 5 1 0.1400 392459.200
[ angles ]
; ai aj ak funct c0 c1 c2 c3
1 2 3 1 120.000 527.184
2 3 4 1 120.000 527.184
3 4 5 1 120.000 527.184
2 1 6 1 120.000 527.184
2 1 7 1 120.000 292.880
1 2 8 1 120.000 292.880
2 3 9 1 120.000 292.880
3 4 10 1 120.000 292.880
4 5 11 1 120.000 292.880
1 6 12 1 120.000 292.880
6 5 11 1 120.000 292.880
6 1 7 1 120.000 292.880
5 6 12 1 120.000 292.880
5 4 10 1 120.000 292.880
1 6 5 1 120.000 527.184
4 3 9 1 120.000 292.880
4 5 6 1 120.000 527.184
3 2 8 1 120.000 292.880
[ dihedrals ]
; IMPROPER DIHEDRAL ANGLES
; ai aj ak al funct c0 c1 c2 c3 c4 c5
12 6 1 5 4 180.000 10.460 2
11 5 4 6 4 180.000 10.460 2
10 4 3 5 4 180.000 10.460 2
9 3 2 4 4 180.000 10.460 2
8 2 1 3 4 180.000 10.460 2
7 1 2 6 4 180.000 10.460 2
[ dihedrals ]
; PROPER DIHEDRAL ANGLES
; ai aj ak al funct c0 c1 c2 c3 c4 c5
4 3 2 1 3 30.334 0.000 -30.334 -0.000 -0.000 0.000
5 4 3 2 3 30.334 0.000 -30.334 -0.000 -0.000 0.000
6 1 2 3 3 30.334 0.000 -30.334 -0.000 -0.000 0.000
4 5 6 1 3 30.334 0.000 -30.334 -0.000 -0.000 0.000
6 5 4 3 3 30.334 0.000 -30.334 -0.000 -0.000 0.000
5 6 1 2 3 30.334 0.000 -30.334 -0.000 -0.000 0.000
12 6 1 2 3 30.334 0.000 -30.334 -0.000 -0.000 0.000
7 1 6 5 3 30.334 0.000 -30.334 -0.000 -0.000 0.000
11 5 6 1 3 30.334 0.000 -30.334 -0.000 -0.000 0.000
10 4 5 6 3 30.334 0.000 -30.334 -0.000 -0.000 0.000
11 5 4 3 3 30.334 0.000 -30.334 -0.000 -0.000 0.000
12 6 5 4 3 30.334 0.000 -30.334 -0.000 -0.000 0.000
9 3 2 1 3 30.334 0.000 -30.334 -0.000 -0.000 0.000
7 1 2 3 3 30.334 0.000 -30.334 -0.000 -0.000 0.000
8 2 1 6 3 30.334 0.000 -30.334 -0.000 -0.000 0.000
10 4 3 2 3 30.334 0.000 -30.334 -0.000 -0.000 0.000
8 2 3 4 3 30.334 0.000 -30.334 -0.000 -0.000 0.000
9 3 4 5 3 30.334 0.000 -30.334 -0.000 -0.000 0.000
10 4 3 9 3 30.334 0.000 -30.334 -0.000 -0.000 0.000
12 6 1 7 3 30.334 0.000 -30.334 -0.000 -0.000 0.000
12 6 5 11 3 30.334 0.000 -30.334 -0.000 -0.000 0.000
9 3 2 8 3 30.334 0.000 -30.334 -0.000 -0.000 0.000
8 2 1 7 3 30.334 0.000 -30.334 -0.000 -0.000 0.000
11 5 4 10 3 30.334 0.000 -30.334 -0.000 -0.000 0.000
[ pairs ]
1 4 1
2 5 1
3 6 1
3 7 1
1 9 1
5 7 1
4 8 1
2 10 1
1 11 1
6 8 1
5 9 1
3 11 1
2 12 1
7 8 1
6 10 1
4 12 1
8 9 1
9 10 1
7 12 1
10 11 1
11 12 1
This is my topol.top file:
; include forcefield parameters
#include "oplsaa.ff/forcefield.itp"
; Include mol topology
#include "UNK_5D626B_rev.itp"
[ system ]
; Name
liquid-benzene-simulation
[ molecules ]
;compound number of molecules
UNK 600
And this is my final production run md.mdp file:
title = liquid benzene oplss simulation
; Run parameters
integrator = md ; leap-frog integrator
nsteps = 5000000 ; 1 * 1000000 = 1000 ps (1 ns)
dt = 0.001 ; 1 fs
; Output control
nstxout = 0 ; suppress bulky .trr file by specifying
nstvout = 0 ; 0 for output frequency of nstxout,
nstfout = 0 ; nstvout, and nstfout
nstenergy = 5000 ; save energies every 5.0 ps
nstlog = 5000 ; update log file every 5.0 ps
nstxout-compressed = 5000 ; save compressed coordinates every 5.0 ps
compressed-x-grps = System ; save the whole system
; Bond parameters
continuation = yes ; Restarting after NPT
constraint_algorithm = lincs ; holonomic constraints
constraints = h-bonds ; bonds involving H are constrained
lincs_iter = 1 ; accuracy of LINCS
lincs_order = 4 ; also related to accuracy
; Neighborsearching
cutoff-scheme = Verlet ; Buffered neighbor searching
ns_type = grid ; search neighboring grid cells
nstlist = 10 ; 20 fs, largely irrelevant with Verlet scheme
rcoulomb = 1.3 ; short-range electrostatic cutoff (in nm)
rvdw = 1.3 ; short-range van der Waals cutoff (in nm)
; Electrostatics
coulombtype = PME ; Particle Mesh Ewald for long-range electrostatics
pme_order = 8 ; order-8 interpolation
fourierspacing = 0.12 ; grid spacing for FFT
; Temperature coupling is on
tcoupl = V-rescale ; modified Berendsen thermostat
tc-grps = System ; two coupling groups - more accurate
tau_t = 0.5 ; time constant, in ps
ref_t = 298 ; reference temperature, one for each group, in K
; Pressure coupling is on
pcoupl = Parrinello-Rahman ; Pressure coupling on in NPT
pcoupltype = isotropic ; uniform scaling of box vectors
tau_p = 2.0 ; time constant, in ps
ref_p = 1.0 ; reference pressure, in bar
compressibility = 4.5e-5 ; isothermal compressibility of water, bar^-1
; Periodic boundary conditions
pbc = xyz ; 3-D PBC
; Dispersion correction
DispCorr = Ener ; account for cut-off vdW scheme
; Velocity generation
gen_vel = no ; Velocity generation is off
My nvt equilibration mdp:
title = liquid benzene nvt equilibration
; define = -DPOSRES ; position restrain the protein
; Run parameters
integrator = md ; leap-frog integrator
nsteps = 50000 ; 2 * 50000 = 100 ps
dt = 0.002 ; 2 fs
; Output control
nstxout = 500 ; save coordinates every 1.0 ps
nstvout = 500 ; save velocities every 1.0 ps
nstenergy = 500 ; save energies every 1.0 ps
nstlog = 500 ; update log file every 1.0 ps
; Bond parameters
continuation = no ; first dynamics run
constraint_algorithm = lincs ; holonomic constraints
constraints = h-bonds ; bonds involving H are constrained
lincs_iter = 1 ; accuracy of LINCS
lincs_order = 4 ; also related to accuracy
; Nonbonded settings
cutoff-scheme = Verlet ; Buffered neighbor searching
ns_type = grid ; search neighboring grid cells
nstlist = 10 ; 20 fs, largely irrelevant with Verlet
rcoulomb = 1.0 ; short-range electrostatic cutoff (in nm)
rvdw = 1.0 ; short-range van der Waals cutoff (in nm)
DispCorr = EnerPres ; account for cut-off vdW scheme
; Electrostatics
coulombtype = PME ; Particle Mesh Ewald for long-range electrostatics
pme_order = 4 ; cubic interpolation
fourierspacing = 0.16 ; grid spacing for FFT
; Temperature coupling is on
tcoupl = V-rescale ; modified Berendsen thermostat
tc-grps = System ; one coupling groups - more accurate
tau_t = 0.1 ; time constant, in ps
ref_t = 298 ; reference temperature, one for each group, in K
; Pressure coupling is off
pcoupl = no ; no pressure coupling in NVT
; Periodic boundary conditions
pbc = xyz ; 3-D PBC
; Velocity generation
gen_vel = yes ; assign velocities from Maxwell distribution
gen_temp = 300 ; temperature for Maxwell distribution
gen_seed = -1 ; generate a random seed
My npt equilibration file:
title = liquid-benzene-simulation
; define = -DPOSRES ; position restrain the protein
; Run parameters
integrator = md ; leap-frog integrator
nsteps = 100000 ; 1 * 100000 = 100 ps
dt = 0.001 ; 1 fs
; Output control
nstxout = 500 ; save coordinates every 1.0 ps
nstvout = 500 ; save velocities every 1.0 ps
nstenergy = 500 ; save energies every 1.0 ps
nstlog = 500 ; update log file every 1.0 ps
; Bond parameters
continuation = yes ; Restarting after NVT
constraint_algorithm = lincs ; holonomic constraints
constraints = h-bonds ; bonds involving H are constrained
lincs_iter = 1 ; accuracy of LINCS
lincs_order = 4 ; also related to accuracy
; Nonbonded settings
cutoff-scheme = Verlet ; Buffered neighbor searching
ns_type = grid ; search neighboring grid cells
nstlist = 10 ; 20 fs, largely irrelevant with Verlet scheme
rcoulomb = 1.0 ; short-range electrostatic cutoff (in nm)
rvdw = 1.0 ; short-range van der Waals cutoff (in nm)
DispCorr = EnerPres ; account for cut-off vdW scheme
; Electrostatics
coulombtype = PME ; Particle Mesh Ewald for long-range electrostatics
pme_order = 4 ; cubic interpolation
fourierspacing = 0.16 ; grid spacing for FFT
; Temperature coupling is on
tcoupl = V-rescale ; modified Berendsen thermostat
tc-grps = System ; two coupling groups - more accurate
tau_t = 0.1 ; time constant, in ps
ref_t = 298 ; reference temperature, one for each group, in K
; Pressure coupling is on
pcoupl = Parrinello-Rahman ; Pressure coupling on in NPT
pcoupltype = isotropic ; uniform scaling of box vectors
tau_p = 2.0 ; time constant, in ps
ref_p = 1.0 ; reference pressure, in bar
compressibility = 4.5e-5 ; isothermal compressibility of water, bar^-1
refcoord_scaling = com
; Periodic boundary conditions
pbc = xyz ; 3-D PBC
; Velocity generation
gen_vel = no ; Velocity generation is off
$$$$
`
• One thing I noticed is that you have 600 benzene molecules in the topology file, but the .gro file only has one benzene molecule defined. Did you create a new .gro file with more than one benzene? Mar 3 at 14:53
• I can't edit my old comment now, but after digging up the paper, I can see that there are some differences in the cutoffs and other options in your md.mdp file. The paper used a cutoff of 1.5, you used 1.3 nm, the paper used Nose-Hoover thermostat, you used modified Berendsen, and there are some other minute differences. Would it be possible for you to post a link to the actual .gro file that you used for the NVT or NPT steps, so that everything is clear? Mar 3 at 15:15
• right, yes, I can do that. The thing I posted was my initial gro file, which I editconf-ed, then I applied insert-molecule to get the 600 molecules. Mar 3 at 15:40
• initial 600 molecule gro file: ctxt.io/2/AACgtax-FA, 600 molecule gro file after NVT: ctxt.io/2/AACgtax-Eg, 600 molecule gro file after NPT: ctxt.io/2/AACgpV-XEg Mar 3 at 15:47
• @ShoubhikRMaiti, I agree about the 5nm. I simply did that to ensure that all the molecules fit in the box, since i was going to equilibrate anyway with NPT later. But I will change the bond-constraints Mar 3 at 17:46 | 2021-09-23 17:43:51 | {"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": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 1, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6010779738426208, "perplexity": 6515.110216136328}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1631780057427.71/warc/CC-MAIN-20210923165408-20210923195408-00579.warc.gz"} |
https://socratic.org/questions/what-is-the-difference-between-photosynthesis-and-cellular-respiration | What is the difference between photosynthesis and cellular respiration?
Mar 16, 2016
Photosynthesis primarily occurs in plants, photosynthetic bacteria and algae. Respiration occurs in animals. Remember that these are both cellular processes and as such happen in individual cells.
Explanation:
Photosynthesis
The balanced equations for photosynthesis is
$6 C {O}_{2} + 12 {H}_{2} O \rightarrow {C}_{6} {H}_{12} {O}_{6} + 6 {O}_{2} + 6 {H}_{2} O$
This biochemical process is vital for pretty much all flora and fauna. Photosynthesis changes elements and compounds into nutrients.Without this process most food chains and ecosystems would collapse. Sometimes photosynthesis is thought of as an anabolic process.
Respiration
The balanced equation for cellular respiration is
C_6H_12O_6 + 6O_2 → 6CO_2 + 6 H_2O + ATP(energy)
This is the fundamental process by which organisms obtain energy from foodstuffs. It is of two types: anaerobic and aerobic. Aeroboic respiration produces more energy and is more efficient. | 2022-11-28 00:49:13 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 2, "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": 0.47144487500190735, "perplexity": 7852.4449961733035}, "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-2022-49/segments/1669446710462.59/warc/CC-MAIN-20221128002256-20221128032256-00142.warc.gz"} |
https://learn.careers360.com/ncert/question-find-the-equations-of-the-hyperbola-satisfying-the-given-conditions-foci-plus-minus-3-root-of-5-0-the-latus-rectum-is-of-length-8/ | # 12. Find the equations of the hyperbola satisfying the given conditions. Foci $(\pm 3\sqrt5, 0)$, the latus rectum is of length 8.
P Pankaj Sanodiya
Given, in a hyperbola
Foci $(\pm 3\sqrt5, 0)$, the latus rectum is of length 8.
Here, focii are on the X-axis so, the standard equation of the Hyperbola will be ;
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
By comparing standard parameter (length of latus rectum and foci) with the given one, we get
$c=3\sqrt{5}$ and
$\frac{2b^2}{a}=8\Rightarrow 2b^2=8a\Rightarrow b^2=4a$
Now, As we know the relation in a hyperbola
$c^2=a^2+b^2$
$c^2=a^2+4a$
$a^2+4a=(3\sqrt{5})^2$
$a^2+4a=45$
$a^2+9a-5a-45=0$
$(a+9)(a-5)=0$
$a=-9\:or\:5$
Since $a$ can never be negative,
$a=5$
$a^2=25$
$b^2=4a=4(5)=20$
Hence, The Equation of the hyperbola is ;
$\frac{x^2}{25}-\frac{y^2}{20}=1$
Exams
Articles
Questions | 2020-03-30 11:08:09 | {"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": 17, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9323520064353943, "perplexity": 2134.918237187922}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1585370496901.28/warc/CC-MAIN-20200330085157-20200330115157-00488.warc.gz"} |
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### How can I write a conditional coloring command that distinguishes variables within sectioning commands vs those within the body text?
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### How to compare the output of \ref to a number?
I want to compare the output of \ref to a number in LaTeX. I thought it would be easy but after spending lots of hours surfing the net I still don't know the answer. I found similar questions on this ... | 2015-01-26 20:49:25 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9707447290420532, "perplexity": 3210.2984763070326}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-06/segments/1422122191023.37/warc/CC-MAIN-20150124175631-00244-ip-10-180-212-252.ec2.internal.warc.gz"} |
http://www.gradesaver.com/textbooks/math/calculus/calculus-early-transcendentals-8th-edition/chapter-3-section-3-4-the-chain-rule-3-4-exercises-page-204/29 | ## Calculus: Early Transcendentals 8th Edition
$H'(r)=\dfrac{2(r^{2}-1)^{2}(r^{2}+3r+5)}{(2r+1)^{6}}$
$H(r)=\dfrac{(r^{2}-1)^{3}}{(2r+1)^{5}}$ Differentiate using the quotient rule: $H'(r)=\dfrac{[(2r+1)^{5}][(r^{2}-1)^{3}]'-[(r^{2}-1)^{3}][(2r+1)^{5}]'}{[(2r+1)^{5}]^{2}}=...$ Use the chain rule to find $[(r^{2}-1)^{3}]'$ and $[(2r+1)^{5}]'$: $...=\dfrac{[(2r+1)^{5}][3(r^{2}-1)^{2}(r^{2}-1)']-[(r^{2}-1)^{3}][5(2r+1)^{4}(2r+1)']}{[(2r+1)^{5}]^{2}}=...$ $...=\dfrac{[(2r+1)^{5}][3(r^{2}-1)^{2}(2r)]-[(r^{2}-1)^{3}][5(2r+1)^{4}(2)]}{[(2r+1)^{5}]^{2}}=...$ Simplify: $...=\dfrac{6r(2r+1)^{5}(r^{2}-1)^{2}-10(2r+1)^{4}(r^{2}-1)^{3}}{(2r+1)^{10}}=...$ Take out common factors $2$, $(2r+1)^{4}$ and $(r^{2}-1)^{2}$, then continue simplifying: $...=\dfrac{2(2r+1)^{4}(r^{2}-1)^{2}[3r(2r+1)-5(r^{2}-1)]}{(2r+1)^{10}}=...$ $...=\dfrac{2(2r+1)^{4}(r^{2}-1)^{2}(6r^{2}+3r-5r^{2}+5)}{(2r+1)^{10}}=...$ $...=\dfrac{2(2r+1)^{4}(r^{2}-1)^{2}(r^{2}+3r+5)}{(2r+1)^{10}}=...$ $...=\dfrac{2(r^{2}-1)^{2}(r^{2}+3r+5)}{(2r+1)^{6}}$ | 2017-03-27 23:02:53 | {"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": 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": 0.9327801465988159, "perplexity": 208.21250685628422}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1490218189534.68/warc/CC-MAIN-20170322212949-00136-ip-10-233-31-227.ec2.internal.warc.gz"} |
https://forums.rpgmakerweb.com/index.php?threads/caethyrils-mz-plugins.125657/#post-1106562 | # Caethyril's MZ Plugins
### Caethyril's RMMZ Plugins
#### Features
I've made some plugins for RPG Maker MZ! Here's a list of them with brief feature descriptions.
• The dates are when each plugin was last updated.
• For plugins marked script, JavaScript knowledge is strongly recommended.
• "See also" lists similar and/or alternative plugins; these are not guaranteed to be compatible.
Plugins:
Suggestions for improvements or additional features are welcome!
#### How to Use
2. Place it in your project's folder, under js/plugins.
3. Open your project in RPG Maker MZ.
4. Go to Tools > Plugin Manager.
5. Double-click an empty line to add a new plugin.
6. Select the plugin from the Name list.
To edit the values of any of the parameters displayed to the right of the drop-down, double-click them in the list. The more complicated plugins have details for use in their help description; you can view this help via the Plugin Manager. Additional notes:
• Avoid renaming my plugins.
(Renamed plugins won't see their plugin parameters.)
• Save your project after making changes in the Plugin Manager.
(Plugin changes are only applied to the game after saving.)
#### Troubleshooting
If you experience problems with any of these plugins that you want to report, you can get support:
• By posting in this thread;
• By posting a new thread on the JavaScript/Plugin Support board (link); or
• By starting a private conversation with me (link).
This information can help to fix the problem faster:
• Instructions stating how to reproduce the problem in a new project.
• A screenshot of the console when the unexpected behaviour occurs.
(You can open the console by pressing F8 during test-play.)
These plugins are all free to use, modify, and redistribute!
For the standalone plugins ("General" category), these conditions apply:
1. None of the original plugin header may be removed.
2. Credit should be given to Caethyril for the original work.
Happy RPG Making~!
Last edited:
#### Milennin
##### "With a bang and a boom!"
Thanks, might try out some of these. By the way, the WindowMotion link plugin has no url attached to it.
#### caethyril
##### ^_^
@Milennin thanks for pointing that out! Fixed.
#### wrigty12
##### Just a QAer playing with Javascript
Can I make a request? This might tie your WindowMotion and FaceOnRight plugins together, but can either:
1) The FaceOnRight plugin also allow you to move the NameBox to the same side as the Face?
2) The WindowMotion have an option to specifically move a window to alight to a specific edge of the screen, like the right edge or top edge?
Also I checked and your FaceOnRight plugin looks to be compatible with at least the VisuStella Message Core, so it looks promising it'll be compatible with others!
#### caethyril
##### ^_^
WindowMotion lets you specify a JS formula for the target position, so it's possible to specify window positions relative to the screen edges using some scripting:
• Left: 0
• Right: Graphics.boxWidth - this.width
• Top: 0
• Bottom: Graphics.boxHeight - this.height
You can tell I developed it for myself rather than for a request, haha. It's very versatile if you are familiar with scripting, though!
The FaceOnRight name box idea is great! I've included a new plugin parameter in v1.1 to sync the name box position with the face position. Also fixed a small bug with mirroring not happening sometimes~
Also good to hear my stuff is playing nicely with other plugins! I downloaded the VisuStella demo today but haven't tried it out yet.
#### wrigty12
##### Just a QAer playing with Javascript
Thanks for updating!
The FaceOnRight now throws an error. I have the parameter set to Never, but it looks like setting it to any value doesn't change it. I turned off all other plugins and I still see it happening.:
Code:
js/plugins/Cae_FaceOnRight.js:207 Cae_FaceOnRight.js unrecognised face mirror rule "".
#### caethyril
##### ^_^
Whoops, yea, I accidentally put two lines the wrong way round. The error was harmless, I think, but it's gone now (still v1.1, same link)~
Also pushed an update for PictureTouch: scripted binds now save/load correctly and picture clicks should no longer count as "move here" map clicks.
Last edited:
#### RK DracoRoy
##### Fire Emblem RPG Gamer
@caethyril Are the rest of your plugins from MV coming over someday? I had found great use for your Cae_ConsumeEval plugin, which made things really convenient with item conditions.
#### caethyril
##### ^_^
@RK DracoRoy yep, sorry, got distracted with some other things. Just released an MZ port of ConsumeEval; I did a couple of basic tests, seems it didn't need much tweaking. Let me know if it bugs out, though!
If you (or anyone else) have any more specific requests I'd be happy to focus on those first, otherwise I'll look into trying to merge/extend stuff like I've done with AutoSwitchVars~
#### chocopanda
##### Veteran
A Quest Plugin is what I look for everywhere T.T
PS: I can't afford Visustella's plugins
#### caethyril
##### ^_^
@chocopanda yea, quest logs are generally a lot of work, since it's an entirely new scene and (usually) a bunch of new windows, too. I'll consider it, but don't expect a quest log from me any time soon.
You can achieve a basic quest log using the Select Item command, e.g.
• Make one "Hidden Item A" type item (or B if you're already using A) per logged quest.
• Make an item called "Journal" or something, for the player to access the quest info.
• Make the journal call a common event like this:
Code:
◆Comment：Player selects quest from inventory.
： ：(It's set to zero if they cancel.)
◆Select Item：quest, Hidden Item A
◆Comment：What did they pick?
◆If：quest = 4
◆Comment：Selected item ID 4! Quest = rat slayer!
： ：Now check quest progress and show appropriate info~
◆Text：None, None, Window, Bottom
： ：Bob wants me to kill 5 rats.
： ：I've killed \v[11] so far.
◆
：Else
◆Text：None, None, Window, Bottom
： ：I've killed 5 rats!
： ：I should go back and let Bob know.
◆
：End
◆
：End
◆If：quest = 15
◆Comment：Selected item ID 15!
： ：etc...
◆
：End
• Then add/remove the hidden items from the player when they start/finish each quest~
Not as fancy as a special quest log plugin, but it gets the message across!
#### Tentacurls
##### Villager
For the keyboard inputs plugin, I wanted to map w a s d to up left right down. I changed the 'name' of the respective codes to the respective direction, and all the keys work except for W. Do you happen to know why? Or do you need more info?
#### Jhin
##### Veteran
Hi,
First off, thank you for your awesome plugins.
I'm amazed with this plugin. Very useful and I thind it should deserve a lot of love.
• v1.1 Cae_CustomTextCodes 2020-08-30 script
Script your own text codes like \V[x] or \$. Includes premade examples. Could you please provide an example to make an /item[x] code work to show item icon and item name? I'm trying to use but this code does not work. Code: const itemId = parseInt(args[1], 10); return this.drawItemName($dataItems[itemId].iconIndex, 20, 20)
Thanks.
#### RK DracoRoy
##### Fire Emblem RPG Gamer
@caethyril I do have an additional request for MZ.
This idea is ExtendedSkills where if I select a skill in battle, a window will come up around the center (location and size (for the sake of text) can be adjusted) showing this set of skills based on the skill notetag.
Code:
<Extended Skill: x, x, x>
MZ allows replacing the Attack command with a skill through a trait "Attack Skill". The extended skill window can come from both the Actor command window and the Skill window.
#### caethyril
##### ^_^
@Tentacurls: by default W is mapped to "pagedown", that's the reason for the difference. KeyboardInputs has a safety parameter called Preserve Original Keys: if you set that to false in the Plugin Manager, it should allow you to remap W to "up" like you've described.
In that case I'd recommend removing any keys you're not using from the parameter list, if you haven't already...otherwise other things (like Z for "ok", or X for "escape") may also be overwritten.
JavaScript:
const itemId = parseInt(args[1], 10);
const item = \$dataItems[itemId];
if (!item) return '';
return '\x1bI[' + item.iconIndex + '] ' + item.name;
I.e. get the item, then return "\I[x] item name" as the replacement text. (It has to be \x1b here rather than a simple \ because backslashes get converted just before processing the text.)
[Edit: also, it should be used like \item[12], i.e. a \backslash, not a /forward slash. ]
@RK DracoRoy: nice idea, I'll stick it on my list~
Last edited:
#### Jhin
##### Veteran
Thank you for the help @caethyril
It is very nice the way you provide support and you help other people.
Thank you again.
#### Raith
##### Squire
Hey, thank you for the great plugin!
For the Cae_PictureTouch I'd like to report this "bug" (well, i'd like to call it expected behavior rather than a bug, anyway):
The picture respond nicely to the supposed input and execute the following bound script/common event. However, when you click-spam or press-spam on the picture while the bound common even is still running, the bound common event/script will be executed again and again after the the first execution is done. The common event/script will loop and loop and loop until you stop clicking on the picture. This has more devastating effect on the "press" option, rendering the game unplayable.
My current workaround is by reducing picture opacity to zero before the main sequence of script/common event and finally increase it again in the end of the common event. However, this method also make the click-based movement being triggered, resulting player running into the clicked tile coordinate.
EDIT: edited for grammatical error
Last edited:
#### SimProse
##### Veteran
I'm trying your custom title screen plugin, adding a Quit Game command using another plugin. I put in 4 commands without scrolling, and it seems to ignore it, and still only displays 3 commands in the box at once and needing to scroll. Using LunaTechs Quit2Desktop plugin if that matters.
#### caethyril
##### ^_^
@Raith: oh, yep. Not sure what the best way around that would be...maybe unbind the event at the start of its common event, then rebind it at the end?
I'll look into a more elegant solution! Maybe a "disable bind" plugin command and/or an option to suppress a picture bind if it's still processing that picture's bind~
@SimProse: have you tried loading my plugin after LunaTechs Quit2Desktop? (Click and drag to rearrange plugins in the Plugin Manager.) Seems to work OK for me!
#### SimProse
##### Veteran
Here's my plugin list. Does not seem to work still.
Hmm...once I turn off the VisuStella Core plugin, it works fine. That's something many people will use, so you might want to check on that.
### Latest Profile Posts
I think Kipling said it best.
If you can make one heap of all your winnings
And risk it on one turn of pitch-and-toss,
And lose, and start again at your beginnings
(...)
Yours is the Earth and everything that's in it,
And—which is more—you'll be a Man, my son!
Do you feel like your modern cities are too clean? I made some dumpsters and trash variations. Visit my MZ resource thread (link in signature).
Doing clean-up on tall-ified sprites, filling in pixel by pixel where the stretching warped the image, feels incredibly tedious but also highly rewarding when one zooms out and the sprite looks even better than before. | 2022-01-22 03:38:46 | {"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": 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": 0.25984710454940796, "perplexity": 6415.797845805467}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320303729.69/warc/CC-MAIN-20220122012907-20220122042907-00091.warc.gz"} |
https://tex.stackexchange.com/questions/624542/verbatim-lstlisting-environment-with-background-color-highlighting | # Verbatim/Lstlisting environment with background color highlighting
Use case: Currently we are reporting security findings using MS Word. These findings contain large parts of code snippets, command outputs and HTML code. We highlight relevant parts of this code so the client knows where to look. I'm trying to convert this to latex.
Problem: I cannot find a way to perform highlighting in a verbatim environment, each of the solutions I have found on this site have a drawback. The requirements are:
• Highlighting using a background color (not changing the text color).
• The highlighting should work across line breaks.
• The highligted text should have the same style as the rest of the verbatim environment.
• The highlighted text should be able to contain special characters.
• The highlighting should work correctly, even if the verbatim environment has a background.
• Highlighting should work correctly across pages
The following is what I would like the end result to look like: (screenshot from MS Word).
I have the following latex document:
\documentclass{article}
\usepackage{color}
\definecolor{codebg}{RGB}{217,217,217}
\definecolor{highlight}{RGB}{255,255,0}
%Listings
\usepackage{listings}
\usepackage{mdframed} %For background and frame
\newcommand{\listingstyle}{\ttfamily\fontsize{8pt}{9pt}\selectfont}
\lstset{
basicstyle={\listingstyle},
breakatwhitespace=false,
escapeinside={*@}{@*},
breaklines=true,
keepspaces=true,
xleftmargin=-5pt,
xrightmargin=-5pt,
aboveskip=0pt,
belowskip=0pt,
}
\lstnewenvironment{code}{
\mdframed[backgroundcolor=codebg,linecolor=black,linewidth=1pt]
}{
\endmdframed
}
\begin{document}
\begin{code}
!@#$%^&*(){}[]\| test$${{highlight!@#%^&*(){}[]\|}}$$abcd \end{code} \end{document} The result of the latex document is as follows: I would like the text between $${{ and }}$$ highlighted. It is okay if different separators have to be used, but these separators is what we currently use in our tooling. I have tried many solutions on this site, notably: • Welcome to TeX.SE! Nov 30, 2021 at 21:33 ## 1 Answer If you are willing to use LuaLaTeX, you can do this with my lua-ul package. Basically you can define a command which enables highlighting for the rest of the current group by copying the definition of \highLight from lua-ul while omitting outer braces, then that command can be used with moredelims=** to enable delimiters which can be combined with other style elements: \documentclass{article} \usepackage{color} \usepackage{plex-mono}% To make bold more visible in the exmple \definecolor{codebg}{RGB}{217,217,217} \definecolor{highlight}{RGB}{255,255,0} %Listings \usepackage{listings} \usepackage{mdframed} %For background and frame \usepackage{luacolor,lua-ul}% LuaLaTeX based improved color and underlining/highlighting support \newcommand{\listingstyle}{\ttfamily\fontsize{8pt}{9pt}\selectfont} % By default, lua-ul's \highLight has to be used as \highLight[color]{text}, % but moredelim=** needs a command which gets used as {\highLight[color] text} % Therefore we define \highLightSwitch as such a variant \makeatletter \DeclareDocumentCommand\highLightSwitch{m}{% \luaul@setcolor{#1}% \@highLight } \makeatother \lstset{ basicstyle={\listingstyle}, breakatwhitespace=false, escapeinside={*@}{@*}, breaklines=true, keepspaces=true, xleftmargin=-5pt, xrightmargin=-5pt, aboveskip=0pt, belowskip=0pt, language=C, % To demonstrate keyword highlighting moredelim=**[is][\highLightSwitch{yellow}]{$$\{\{}{\}\}$$}, % Tell listings about our markers % ** indicate that it should be combined with other styles. } \lstnewenvironment{code}{ \mdframed[backgroundcolor=codebg,linecolor=black,linewidth=1pt] }{ \endmdframed } \begin{document} \begin{code} !@#$%^&*(){}[]\|
test$${{highlight!@#% ^&*(){}[]\|}}$$abcd
\end{code}
\begin{code}
int main(int argc, char **argv) {
$${{int ret = 42}}$$;
return 42;
}
\end{code}
\end{document}
• Thank you for your answer! This works great for me, I have not been able to find problems with it yet. Dec 1, 2021 at 3:41 | 2022-05-20 16:13:48 | {"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": 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": 0.8086532950401306, "perplexity": 1701.589891971077}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1652662533972.17/warc/CC-MAIN-20220520160139-20220520190139-00787.warc.gz"} |
https://socratic.org/questions/5841ee94b72cff2bd006d78f | # Question #6d78f
Dec 3, 2016
$2 \cdot 4 \equiv 1 \text{ (mod "7")}$, thus $4 = {2}^{- 1}$ in ${\mathbb{Z}}_{7}$.
#### Explanation:
(This answer assumes a basic understanding of modular arithmetic)
When working in ${\mathbb{Z}}_{n}$, if $k \in {\mathbb{Z}}_{n}$ is coprime with $n$, that is, if $\text{GCD} \left(k , n\right) = 1$ then there exists a unique ${k}^{- 1} \in {\mathbb{Z}}_{n}$ such that $k \cdot {k}^{- 1} \equiv 1 \text{ (mod "n")}$. We call ${k}^{- 1}$ the multiplicative inverse of $k$ in ${\mathbb{Z}}_{n}$.
To see that ${2}^{- 1} \equiv 4$ in ${\mathbb{Z}}_{7}$, then, we need only show that $2 \cdot 4 \equiv 1 \text{ (mod "7")}$. Indeed,
$2 \cdot 4 \equiv 8 \equiv 1 + 7 \equiv 1 \text{ (mod "7")}$
Thus ${2}^{- 1} = 4$ in ${\mathbb{Z}}_{7}$ | 2022-08-11 04:55:43 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 18, "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": 0.9955132603645325, "perplexity": 155.17559185286325}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1659882571234.82/warc/CC-MAIN-20220811042804-20220811072804-00215.warc.gz"} |
https://www.physicsforums.com/threads/needle-probability-distribution-problem.493469/ | # Needle probability distribution problem
## Homework Statement
part a(complete)
The needle on a broken car speedometer is free to swing, and bounces
perfectly off the pins at either end, so that if you give it a flick it is equally likely to
come to rest at any angle between 0 and $$\pi$$.
part b(attempting)
We consider the same device as the previous problem, but this time
we are interested in the x-coordinate of the needle point--that is, the "shadow", or
"projection", of the needle on the horizontal line.
1. what is $$\rho$$(x)?
## Homework Equations
x=rcos($$\theta$$)
## The Attempt at a Solution
assuming that my solution for the probability function for $$\theta$$ is correct, (1/$$\theta$$)
(i'm going to cheat and use a unit needle for a second)
I took x=cos($$\theta$$)
and found arccos(x)=$$\theta$$
and d$$\theta$$=-dx/($$\sqrt{1-x^{2}}$$
but I am unsure of the usefulness of this relationship
so I took dx=-sin($$\theta$$)d$$\theta$$
and got p(x)dx=(-dx/$$\pi$$)(sin(arccos(x)))^-1 | 2020-02-19 10:15:52 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.7535170912742615, "perplexity": 2512.9184161633425}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1581875144111.17/warc/CC-MAIN-20200219092153-20200219122153-00532.warc.gz"} |
https://www.gradesaver.com/textbooks/math/precalculus/precalculus-mathematics-for-calculus-7th-edition/chapter-10-section-10-3-matrices-and-systems-of-linear-equations-10-3-exercises-page-710/42 | ## Precalculus: Mathematics for Calculus, 7th Edition
Published by Brooks Cole
# Chapter 10 - Section 10.3 - Matrices and Systems of Linear Equations - 10.3 Exercises - Page 710: 42
#### Answer
( $10-4t$ , $\displaystyle \frac{7}{2}+\frac{1}{2}t$ , $t$ ), $\ \ t\in \mathbb{R}$
#### Work Step by Step
Write the augmented matrix and, using row transformations, arrive at the row-reduced echelon form. $\left[\begin{array}{llll} 1 & -2 & 5 & 3\\ -2 & 6 & -11 & 1\\ 3 & -16 & 20 & -26 \end{array}\right] \ \ \begin{array}{l} .\\ +2R_{2}.\\ -3R_{1}. \end{array}$ $\left[\begin{array}{llll} 1 & -2 & 5 & 3\\ 0 & 2 & -1 & 7\\ 0 & -10 & 5 & -35 \end{array}\right] \ \ \begin{array}{l} +R_{2}.\\ \div 2.\\ -5R_{2}. \end{array}$ $\left[\begin{array}{llll} 1 & 0 & 4 & 10\\ 0 & 1 & -1/2 & 7/2\\ 0 & 0 & 0 & 0 \end{array}\right] \ \ \begin{array}{l} .\\ .\\ . \end{array}$ Last row has all zeros, so the system is dependent (infinite number of solutions). Taking $w=t\in \mathbb{R}$ (arbitrary), back-substituting: $x=10-4t$ $y=\displaystyle \frac{7}{2}+\frac{1}{2}t$ Solution set: $\{$( $10-4t$ , $\displaystyle \frac{7}{2}+\frac{1}{2}t$ , $t$ ), $t\in \mathbb{R} \}$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 2019-01-22 06:06:01 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9548972249031067, "perplexity": 1647.235333854838}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-04/segments/1547583829665.84/warc/CC-MAIN-20190122054634-20190122080634-00024.warc.gz"} |
https://testbook.com/question-answer/what-isrm-int_0a-fracfa-xfxfa--607d791942596f6dbdd774b1 | # What is $$\rm \int_0^a \frac{f(a-x)}{f(x)+f(a-x)}\ dx$$ equal to?
This question was previously asked in
NDA (Held On: 18 Apr 2021) Maths Previous Year paper
View all NDA Papers >
1. a
2. 2a
3. 0
4. $$\rm \frac{a}{2}$$
Option 4 : $$\rm \frac{a}{2}$$
Free
Electric charges and coulomb's law (Basic)
41133
10 Questions 10 Marks 10 Mins
## Detailed Solution
Concept:
Definite Integrals:
$$\rm \int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx$$.
If f(x) = f(2a - x), then $$\rm \int_0^{2a}f(x)\ dx=2\int_0^af(x)\ dx$$.
A function f(x) is:
• Even, if f(-x) = f(x). And $$\rm \int_{-a}^ {\ \ a}f(x)\ dx=2\int_{0}^af(x)\ dx$$.
• Odd, if f(-x) = -f(x). And $$\rm \int_{-a}^ {\ \ a}f(x)\ dx=0$$.
• Periodic, if f(np ± x) = f(x), for some number p and n ∈ Z.
Calculation:
We know that $$\rm \int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx$$.
∴ I = $$\rm \int_0^a \frac{f(a-x)}{f(x)+f(a-x)}\ dx$$
$$\rm \int_0^a \frac{f[(a+0)-(a-x)]}{f[(a+0)-x]+f[(a+0)-(a-x)]}\ dx$$
$$\rm \int_0^a \frac{f(x)}{f(a-x)+f(x)}\ dx$$
And, 2I = $$\rm \int_0^a \frac{f(a-x)}{f(x)+f(a-x)}\ dx$$ + $$\rm \int_0^a \frac{f(x)}{f(a-x)+f(x)}\ dx$$
$$\rm \int_0^a 1\ dx$$
= a.
⇒ I = $$\rm\frac a2$$. | 2021-10-24 10:38:34 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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": 0.2517957389354706, "perplexity": 11289.13254998338}, "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-2021-43/segments/1634323585916.29/warc/CC-MAIN-20211024081003-20211024111003-00005.warc.gz"} |
https://www.gradesaver.com/textbooks/math/algebra/college-algebra-6th-edition/chapter-p-prerequisites-fundamental-concepts-of-algebra-exercise-set-p-5-page-74/63 | ## College Algebra (6th Edition)
$(4x+3)(16x^{2}-12x+9)$
If $a$ and $b$ are real numbers, we know that $a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2})$. Therefore, we know that $64x^{3}+27=(4x+3)((4x)^{2}-4x\times3+(3)^{2})=(4x+3)(16x^{2}-12x+9)$. In this case, $a=4x$ and $b=3$. | 2020-03-29 21:17:38 | {"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": 0.8173036575317383, "perplexity": 75.67155516052988}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1585370496227.25/warc/CC-MAIN-20200329201741-20200329231741-00180.warc.gz"} |
https://benwhale.com/blog/stable-causality-implied-by-causality-/ | # Stable causality implied by causality + ?
I’ve been reading some interesting papers showing that stable causality holds on a spacetime assuming that causality + some other condition holds. This is some really interesting stuff.
Causality conditions on spacetimes describe how nicely behaved physics is. Causality is the condition that there are no closed causal curves. That is, no travelling back in time to the same location even if you travel at the speed of light. Stable causality says that even if the speed of light increased you couldn’t form closed causal curves, that is go back in time.
Anyway Stable Causality is also interesting as it implies the existence of a, non-unique, global time. Which is quite a powerful implication.
Go to my CiteULike page and search for Stable Causality for the papers, check the notes before you read. | 2020-09-29 04:16:11 | {"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": 0.8304163813591003, "perplexity": 611.0687061451654}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1600401624636.80/warc/CC-MAIN-20200929025239-20200929055239-00657.warc.gz"} |
https://ita.skanev.com/13/01/02.html | # Exercise 13.1.2
Draw the red-black tree that results after TREE-INSERT is called on the tree in Figure 13.1 with key $36$. If the inserted node is colored red, is the resulting tree a red-black tree? What if it is colored black?
The new element is going to be the gray one:
If we color it red, it will violate property 4, that is, red nodes need to have black children. In this case, it's parent, 35 is red, so it must be black.
If we color it black, it will violate property 5. The path to the descendants of 36 from the root will have 4 black nodes, but the one to the descendants of 39 will have 3.
That is, TREE-INSERT does not produce a valid red-black tree in this case, regardless of how we color the node. | 2022-01-21 05:20:32 | {"extraction_info": {"found_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": 0.731105625629425, "perplexity": 676.4330759981488}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320302723.60/warc/CC-MAIN-20220121040956-20220121070956-00040.warc.gz"} |
https://math.stackexchange.com/questions/2452106/in-the-quaternion-group-why-does-1k-k-or-any-other-element-for-that-mat | # In the Quaternion group, why does $-1*k = -k$, or any other element for that matter?
The Quaternion group is given as
$$Q_8 = \{ \pm 1, \pm i, \pm j, \pm k \}$$
with $(-1)^2 =1$,$i^2 = j^2 = k^2 = ijk = -1$. Is it by definition, then, that we have $-1*k = -k$, or the same multiplication for any other element for that matter?
• Have you tried making a Cayley table? – Cameron Buie Sep 30 '17 at 22:34
Yes, that is what $-k$ means, by definition. We have the element $-1$ and the element $k$, they commute and we call their product $-k$. It is called that partly because of what it represents in the ring of quaternions (where you can prove that $(-1)\cdot k$ fulfills the defining property of $-k$, the additive inverse of $k$), and partly for brevity. | 2019-05-23 07:21:42 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9214668869972229, "perplexity": 233.6859557318578}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232257156.50/warc/CC-MAIN-20190523063645-20190523085645-00280.warc.gz"} |
https://www.johnmackintosh.net/blog/2022-09-22-picture/ | # Cover Up
Adding full sized cover images to Word documents programatically with officer, officedown, and purrr
I needed to add full size (A4, portrait) images to the first page of my Word document ( there are 13, all 30+ pages, which I’m generating in RMarkdown, as per previous post).
I didn’t want to do this by hand, especially when I discovered that due to margins and page settings, each image would have to be resized to fit on the page. (I tried adding the full sized image to a blank Word document but it was too large, and resulting attempts to resize it did not work well. I did not want to do this multiple times).
Here was the plan:
• get all the full size images my colleagues created and save them in a source folder
• automate the process of creating resized copies, and save those in another folder
• create a blank, one page, Word document for each image
• loop through and add the resized images to the corresponding blank document
• use {officer} to add these blank documents as my ‘cover’ for each report.
I saved all the original , full size images into a ‘resources’ folder inside my ‘img’ folder, then looped through them and resized them as below.
library(here)
library(purrr)
library(magick)
setwd(here("img", "resources"))
files <- dir()
# taking dimensions at 90%
geometry = geometry_size_percent(width = 90)))
outdir <- paste0(here("img","resized", files))
walk2(files_resized, outdir, ~ image_write(.x, path = .y))
The next bit was to create blank documents, and add the resized images.
I created 2 functions; create_covers and add_images
library(officer)
library(officedown)
library(here)
library(magrittr)
create_covers <- function(areas) {
# get list of resized images to insert into blank doc
areas <- dir(here("img","resized"), full.names = TRUE)
outnames <- gsub(".png", ".docx", areas)
# save these to a 'covers' folder outside the main 'img' folder
outnames <- gsub("img/resized","covers",outnames)
# read the blank source doc
# write out the blank files
walk(outnames,~ print(temp, .x))
}
add_images <- function(outnames, images, width = 9.3,
height = 13.15625,
position_at = "after") {
for (i in seq_along(outnames)) {
cursor_begin() %>%
body_remove() %>%
width = 9.3, # was 9.5 @92%
height = 13.15625, # was 13.4375 @ 92%
pos = position_at) %>%
print(.,outnames[i])
}
}
With those functions in place, here’s how I created the covers and added the images:
areas <- dir(here("img","resized"), full.names = TRUE)
# create the blank covers - takes some time
create_covers(areas)
# replace the file extension
outnames <- gsub(".png", ".docx", areas)
outnames <- gsub("img/resized","covers", outnames)
images <- dir(here("img","resized"), full.names = TRUE)
# for loops FTW
The final step was to add this to my Rmd file
cover_page_margin <- officer::page_mar(bottom = 0.00787402, # .02 cm
top = 0.00787402,
right = 0.015748, #.04 cm
left = 0.015748,
footer = 0,
gutter = 0.00393701) # .01 cm
cover_section <- officer::prop_section(page_margins = cover_page_margin,
page_size = (page_size(orient = "portrait")),
type = "continuous")
And also :
officer::block_pour_docx(here("covers",eval(paste0(params\$HSCP,".docx"))))
officer::block_section(cover_section)
This set the margins for the cover ( I had to have some space assigned), then created the section properties for the cover.
The block_pour_docx inserted the relevant cover page, and the block_section applied the cover section layout. This has to be placed after the content, as opposed to before it, as you may have expected.
With all this in place, the front covers look good, with a fairly even border all round.
Doing this work involves constantly converting between inches, centimetres, and pixels - here are some links that will help:
https://www.a4-size.com/a4-size-in-pixels/
https://convertermaniacs.com/px-to-inch/convert-1263-pixels-to-inches.html | 2023-01-28 03:33:30 | {"extraction_info": {"found_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": 0.3813447952270508, "perplexity": 10888.488783537754}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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-20230128023233-20230128053233-00425.warc.gz"} |
https://micro.giov.dev/categories/technology/ | • For emacs lovers: I extended @dgold’s code to post to micro.blog from emacs. Now you can include images in the markdown buffer to be posted to micro.blog, and they will be uploaded and linked properly in your published post!
And here’s the code (or, alternatively, see gist):
(require 'request)
(setq mb-emacs-app-token "123456789") ;; create this in account -> app tokens -> edit apps
(setq mb-micropub-endpoint "https://micro.blog/micropub")
(defun mb-get-media-endpoint ()
(cdr (assoc "media-endpoint"
(let (result)
(request
mb-micropub-endpoint
:params '(("q" . "config"))
:type "GET"
:sync t
:timeout: 10
:headers (("Content-Type" . "application/json")
("Authorization".,(format "Bearer %s" mb-emacs-app-token)))
:complete (cl-function
(lambda (&key data &allow-other-keys)
(setq result data))))
(if result
result
(error "Can't get media endpoint")))
#'string=)))
"Post inline images."
(cdr (assoc "url"
(let ((result))
(request
(concat media-endpoint
:type "POST"
:files (("file" . ,img-path))
:headers (("Content-Type" . "multipart/form-data")
("Authorization".,(format "Bearer %s" mb-emacs-app-token)))
:sync t
:success (cl-function
(lambda (&key data &allow-other-keys)
(setq result data))))
(if result
result
#'string=)))
(interactive)
(save-excursion
(save-restriction
(let ((media-endpoint (mb-get-media-endpoint)))
(widen)
(goto-char (point-min))
(let ((start (match-beginning 0))
(imagep (match-beginning 1))
(end (match-end 0))
(file (match-string-no-properties 6)))
(when (and imagep
(not (zerop (length file))))
(when (file-exists-p file)
(let* ((abspath (if (file-name-absolute-p file)
file
(concat default-directory file)))
(replace-match img-upload-url t t nil 6))))))))))
(defun mb-post-buffer ()
"Post current buffer to micro.blog (possibly as draft)."
(interactive)
(if (yes-or-no-p "Are you sure you want to post this?")
(save-restriction
(widen)
(let ((buffer-contents (buffer-substring-no-properties (point-min) (point-max)))
(mb-post-name (read-string "Enter post name (leave empty if none):"))
(mb-post-status (post-status . [,(if (yes-or-no-p "Post as draft?") "draft" "published")])))
;; copy content of current buffer to new buffer,
(with-current-buffer (generate-new-buffer "post2mb")
(goto-char (point-min))
(insert buffer-contents)
(goto-char (point-min))
(request
(concat mb-micropub-endpoint
:type "POST"
:data (json-encode ((type . ["h-entry"])
(properties
(content . [,(buffer-substring-no-properties (point-min) (point-max))])
(name . [,mb-post-name]) ,mb-post-status)))
("Authorization".,(format "Bearer %s" mb-emacs-app-token)))
:success (cl-function
(lambda (&key data &allow-other-keys)
(message "Success.")))))))))
• ## Stay away from AutoML (...if you can't do ML)
Image taken from this post on reddit.
These are thoughts in response to an article on dataiku about empowering more people in an organization to use ML.
Often there’s just not enough data scientists to support the needs of an entire organization. The typical recommendations - and trend - to alleviate this problem is to put effort in setting up self-service analytics, and automatic model creation with AutoML.
While I do think self-service analytics have a big role to play, I am not at all convinced about AutoML. What’s that?
To quote from my third Google result for “AutoML”:
AutoML provides methods and processes to make Machine Learning available for non-Machine Learning experts, to improve efficiency of Machine Learning and to accelerate research on Machine Learning. Machine learning (ML) has achieved considerable successes in recent years and an ever-growing number of disciplines rely on it. However, this success crucially relies on human machine learning experts to perform manual tasks. As the complexity of these tasks is often beyond non-ML-experts, the rapid growth of machine learning applications has created a demand for off-the-shelf machine learning methods that can be used easily and without expert knowledge. We call the resulting research area that targets progressive automation of machine learning AutoML.
That is, people with no data science expertise should be able to roll out their own automatically generated Machine Learning model, to support decisions in what the dataiku article would categorize as “simpler projects”.
## Why AutoML is not the solution
I am extremely skeptical of using AutoML when you don’t have expertise in Machine Learning. With the right data, and the right problem? Sure, you might get a decent model out. The issue is that you need training to be able to tell what the right data is, what a decent model is, and even how to define the problem correctly in the first place!
Otherwise you better expect garbage.
It is so very easy to fool yourself into a data-backed story that crumbles down as soon as you start poking it, and you need a lot of training to avoid fooling yourself (as an aside, this seems to be a general rule for life). With this in mind, when you are not trained to reason on data or ML, AutoML is just a recipe for disaster.
It doesn’t really matter whether the idea is to only apply it to “simpler projects”. Decision based on garbage are still based on garbage, independent from project complexity.
## Train your people, make them effective
Don’t get me wrong, empowering more people to play with data in an organization is a brilliant idea. But you have to train your people first!
You can take away the software engineering pains of data science, and even a lot of Machine Learning boilerplate processes; In that, tools are extremely useful. But you can’t take away the data science. You need people to be able to reason about problems, data and models. No amount of AutoML is going to help if you don’t have that.
If you have to invest in something to make your organization data-driven, invest in your people. Invest in training. Don’t try to substitute tools for expertise.
• ## Linux embedded development stays a second-class citizen
Arduino debugging can be a pain, as you need to get a (quite expensive) specialized piece of hardware, that you then need to hook in the right way to the board in order to get going.
Now, Microsoft has introduced a very nifty feature in the Arduino extension for VSCode, which allows you to debug some Arduino boards without any additional hardware. That is, just plug the board to the USB port, put some breaking points, and step through your code as if it’s nobody’s business. That’s great news if you happen to be using one of the supported boards, which I am. No more print statements, sounds like heaven!
So I set the environment up in VSCode, followed the steps, started the debugging session and… Error. Crap. I must have misconfigured something. Tha’s fine, let’s google it.
And, lo and behold, nope. No misconfiguration. The feature is broken in Linux; actually, it seems that any debugging on Arduino boards (with or without segger) just won’t work because of a small VSCode bug, which has been reported back in 2017.
I wish this was a fluke. Unfortunately embedded development on Linux keeps lagging behind, and I keep having to switch back and forth between OS’s depending on which specific part of development I am at. | 2021-12-03 00:30:05 | {"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": 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": 0.25854727625846863, "perplexity": 5165.561377262179}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1637964362571.17/warc/CC-MAIN-20211203000401-20211203030401-00437.warc.gz"} |
https://codegolf.stackexchange.com/questions/126589/dungeon-of-botdom | Dungeon of botdom
– of crisis and martyrdom
(that's the subtitle because subtitles are cool)
In this challenge in (yup; you need go no farther to know you may not submit in java), you need to create a bot that plays a game very similar to welcome to the dungeon
Game Rules
(note that this is not the original game)
There is a deck, an item set, and some reward cards and death cards. Base HP is 3. The deck consists of 13 monster cards numbered 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 7, 9 to denote their strengths.
Item List
1. Demonic pact: Defeat the demon (strength 7 monster), and the monster below it on the dungeon pile. - (just defeats the demon if the demon was the last in the dungeon)
2. Health potion: When you fall to 0 HP, defeat the monster and return to 3 HP.
3. Holy grail: Defeat monsters of even-numbered strength (in the game, these are undead). If an even-numbered monster occurs after the demonic pact has been used, that acts first, and you will not get an extra-pact kill after this monster.
4. Vorpal dagger: Choose one monster before entering the dungeon; this type of monster is defeated. If the targeted monster occurs after the demonic pact has been used, that acts first, and you will not get an extra-pact kill after this monster.
5. Shield: Add 3 to total HP before spelunking. This does not affect usage of health potion, which will always return health to 3.
6. Armour: Add 5 to total HP before spelunking. This does not affect usage of health potion, which will always return health to 3.
Reward cards are used to keep track of who has succeeded in the dungeon. Death cards track who has failed in the dungeon.
Drawing Phase
Before the drawing phase begins all monster cards are returned to the deck, both players are restored to 3 HP, and all discarded items are restored such that there is one of each.
The first player decides whether to draw a card from the deck, hiding it from the other player. If so, they must choose to either place it on top of the dungeon pile or discard it along with an item of their choice. Discarded items and cards will be unavailable to either player until the next round.
After player one takes their turn, player two does the same. The players alternately decide whether to draw and what to do with the drawn card, until someone decides not to draw or a player takes the last card from the deck. If a player decides not to draw, or draws the last card, the drawing phase ends and the other player now has to enter the dungeon and begin spelunking.
Spelunking Phase
If the Vorpal dagger has not been discarded, the spelunking player must now decide which card to apply it to. There are no active decisions to be made for the rest of this phase.
The first player takes the top card; that is, the last card placed in the dungeon, and see its strength number. If demonic pact is active from the previous turn, the drawn card is discarded. Otherwise, the player's items will be checked in the order 'demonic pact', 'holy grail', 'Vorpal dagger'. The first un-discarded item capable of defeating the drawn card will then be used, and the card discarded. If demonic pact is used, it will now be active for the next card. The used item is not discarded.
If there is no applicable item available, the strength of the card is subtracted from the player's health. If their health is no longer positive they will be restored to 3 HP and the potion discarded if available, otherwise the dungeon crawl ends and they take a death card.
While the player is not defeated and there are cards remaining in the dungeon, this process of drawing the top card is repeated. Upon successfully defeating all cards in the dungeon the dungeon crawl ends and the spelunking player collects a reward card.
Full Game Description
A game consists of a series of rounds, each having a drawing phase and then a spelunking phase. At the end of each round one player will have collected either a death card or a reward card; once a player accumulates 5 of either type the game ends. If they have 5 death cards they lose the game. If they have 5 reward cards, they win. Either way, the other player receives the opposite result. If neither player has 5 cards of one type, play progresses to the next round and the player who went second in the previous round now goes first and vice versa.
KOTH details
Each bot will play 400 games against every other bot according to the rules described above. Which bot is player one (and so goes first in the first round) alternates each game, and all state is reset between games.
Here are the items again:
1. Demonic pact: Defeat the demon (strength 7 monster), and the monster below it on the dungeon pile. - (just defeats the demon if the demon was the last in the dungeon)
2. Health potion: When you fall to 0 HP, defeat the monster and return to 3 HP.
3. Holy grail: Defeat monsters of even-numbered strength (in the game, these are undead). If an even-numbered monster occurs after the demonic pact has been used, that acts first, and you will not get an extra-pact kill after this monster.
4. Vorpal dagger: Choose one monster before entering the dungeon; this type of monster is defeated. If the targeted monster occurs after the demonic pact has been used, that acts first, and you will not get an extra-pact kill after this monster.
5. Shield: Add 3 to total HP before spelunking. This does not affect usage of health potion, which will always return health to 3.
6. Armour: Add 5 to total HP before spelunking. This does not affect usage of health potion, which will always return health to 3.
and the deck: 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 7, 9.
You must implement a bot class that does not use class variables, deriving from the following base class:
class BasePlayer:
def start_turn(self, last_turn):
raise NotImplementedError
def play(self, card):
raise NotImplementedError
def vorpal_choice(self, last_turn):
raise NotImplementedError
def result(self, bot, result, dungeon, vorped):
raise NotImplementedError
This base class shows methods your class needs to implement, and the number of arguments taken by each.
Method Argument Descriptions
• last_turn in vorpal_choice and start_turn is an integer or a None value. A value of 0 to 5 indicates that the enemy discarded the drawn card along with the item indicated by that value (see the list of items above). A value of 6 indicates that the enemy placed the card in the dungeon. A None value indicates that the bot is playing first in this round (not possible for vorpal_choice). In vorpal_choice last_turn is likely to be 7, indicating that they passed that turn. The only circumstance in which it is not 7 is when the enemy drew the last card.
• card is a number representing the strength of one of the cards from the deck as enumerated above.
Now, the arguments for result are a little bit more complex:
• bot indicates the bot which entered the dungeon. 0 indicates entering the dungeon, and 1 indicates that the enemy entered the dungeon.
• result indicates the success of the trip. False indicates the spelunking bot succeeded, while True indicates they failed.
• dungeon is a list of cards/ints representing the cards that were in the dungeon. The dungeon is ordered by order placed; the first card placed in the dungeon is first in the list, and the last card placed in is at the end. You will not receive any info about discarded cards; they are secret from the other bot.
• vorped is an integer representing the vorpal_choice made by the spelunking bot. If bot==0, you already know this, but if bot==1, this can be useful information.
I'll be honest, I don't entirely recall why I made winning result False, but I think it was a good idea at the time.
Return Values
• start_turn: Return 1 to draw a card, or 0 to pass.
• play: Return 0 to 5 to discard the corresponding item and the drawn card, or 6 to place the card in the dungeon (consistent with last_turn input, except for passing, which is done during start_turn).
• vorpal_choice: Return the number of the card you wish to eliminate with the Vorpal dagger (1 to eliminate 1s, 5 to eliminate 5s). Picking a non-existent card kills you (8 is illegal, 10 is illegal, 0 is illegal).
• result: You may return anything, because this is an informing function to update the bot's data.
Additional clarifications, or just repeating some small details you might have skipped past and might want to know quickly:
Bots play 400 games with each other bot.
There is no specific end time for this KOTH, except the end of the bounty for what that is worth. Just try to be winning at each time.
Results so far (sorry for being rather lazy with these guys :P)
1 GrailThief 2732 0.98
2 Steve 2399 0.86
3 DevilWorshipper 1854 0.66
4 RunAway 1336 0.48
5 BoringPlayer 1190 0.42
6 SlapAndFlap 783 0.28
7 DareDevilDumDum 750 0.27
8 RandomMandom 156 0.06
Grailthief "steals" the bounty. not really, because it earned it. Good work, Sleafar!
• I've actually played this game once or twice. Pretty challenging sometimes. – Draco18s Jun 15 '17 at 13:55
• ehhh, tbh I doubt it will be that popular :P – Destructible Lemon Jun 16 '17 at 0:17
• 2. place the item in the dungeon. The item goes in the dungeon (duh) appears to be a typo; there isn't an item mentioned at that point (you just drew a card from the monster deck). The rules should probably be clarified a bit. – user62131 Jun 16 '17 at 0:40
• @ais523 or any information known only to one player. am i being unclear again? the dungeon is only revealed at the end, so cards drawn by a bot is known only to one bot. as a bonus, cards discarded are not revealed ever. if you think "oh well then there is a probabilistic best strategy", opponent prediction is still greatly important so this is also invalid – Destructible Lemon Jun 16 '17 at 0:46
• Ah, it's currently unclear from the rules that you don't know the current contents of the dungeon (only the influence that you've personally had on it). That should almost certainly be clarified. – user62131 Jun 16 '17 at 0:59
GrailThief
An old and experienced dungeon crawler. He knows that most of the others hope, that the holy grail saves them, therefore he makes sure it disappears.
from base import BasePlayer
import copy
class GrailThief(BasePlayer):
class Stats:
def __init__(self):
self.deck = [1, 2, 3, 4, 5] * 2 + [6, 7, 9]
self.items = {0, 1, 2, 3, 4, 5}
self.dungeon_known = []
self.dungeon_unknown = 0
self.monsters_safe = {2, 4, 6, 7}
self.update()
def update(self):
self.dungeon_total = len(self.dungeon_known) + self.dungeon_unknown
deck_factor = float(self.dungeon_unknown) / len(self.deck) if len(self.deck) > 0 else 1.0
self.dungeon_weighted = [(i, 0.0 if i in self.monsters_safe else 1.0) for i in self.dungeon_known] + [(i, 0.0 if i in self.monsters_safe else deck_factor) for i in self.deck]
dungeon_weighted_sums = dict.fromkeys(self.dungeon_known + self.deck, 0.0)
for i in self.dungeon_weighted:
dungeon_weighted_sums[i[0]] += i[0] * i[1]
self.vorpal = max(dungeon_weighted_sums, key = dungeon_weighted_sums.get)
if 3 in self.items:
self.dungeon_weighted = [(i[0], 0.0 if i[0] == self.vorpal else i[1]) for i in self.dungeon_weighted]
new = copy.copy(self)
new.items = {i for i in new.items if i != item}
if item == 0:
new.monsters_safe = {i for i in new.monsters_safe if i != 7}
elif item == 2:
new.monsters_safe = {i for i in new.monsters_safe if i not in {2, 4, 6}}
if card is not None:
new.deck = list(new.deck)
new.deck.remove(card)
new.update()
return new
def to_dungeon(self, card):
new = copy.copy(self)
if card is None:
new.dungeon_unknown += 1
else:
new.deck = list(new.deck)
new.deck.remove(card)
new.dungeon_known = list(new.dungeon_known)
new.dungeon_known.append(card)
new.update()
return new
def effective_hp(self):
hp = 3.0
if 1 in self.items:
hp += 3.0
if self.dungeon_total > 0:
hp += sum([(i[0] - 1) * i[1] for i in self.dungeon_weighted]) / self.dungeon_total
if 4 in self.items:
hp += 3.0
if 5 in self.items:
hp += 5.0
return hp
def effective_damage(self):
damage = sum([i[0] * i[1] for i in self.dungeon_weighted])
if 0 in self.items:
if self.dungeon_total > 1:
damage -= damage / (self.dungeon_total - 1)
return damage
def __init__(self):
self.stats = self.Stats()
def process_last_turn(self, last_turn):
if last_turn in [0, 1, 2, 3, 4, 5]:
elif last_turn == 6:
self.stats = self.stats.to_dungeon(None)
def start_turn(self, last_turn):
self.process_last_turn(last_turn)
if self.stats.effective_hp() > self.stats.effective_damage() + 1.5:
return 1
else:
return 0
def play(self, card):
if 2 in self.stats.items:
return 2
else:
self.stats = self.stats.to_dungeon(card)
return 6
def vorpal_choice(self, last_turn):
self.process_last_turn(last_turn)
return self.stats.vorpal
def result(self, bot, result, dungeon, vorped):
self.stats = self.Stats()
DevilWorshipper
My first attempt at a KOTH challenge:
from base import BasePlayer
#from random import randint
class DevilWorshipper(BasePlayer):
def reset(self):
self.items = [0, 1, 2, 3, 4, 5]
self.turns = 0
self.demon = False
self.dragon = False
def __init__(self):
self.reset()
def start_turn(self, last_turn):
if last_turn in self.items:
self.items.remove(last_turn)
if last_turn is not None:
#self.demon = True if randint(1, 13 - self.turns) <= 2 else False
self.turns += 1
if (((self.demon == True and not (0 in self.items)) or (self.dragon == True)) and not (3 in self.items)):
return 0
if (len(self.items) <= 1):
return 0
return 1
def play(self, card):
self.turns += 1
if (card == 9):
self.dragon = True
return 6
if (card == 7):
self.demon = True
return 6
for i in [3, 0, 2, 1, 5, 4]:
if (i in self.items):
self.items.remove(i)
return i
return 6
def vorpal_choice(self, last_turn):
return 5 #If it works for others maybe it will work for us
def result(self, bot, result, dungeon, vorped):
self.reset()
Essentially, we get rid of the pact and the vorpal dagger, wait for the demon to get into the deck, and pass. Every round the opponent may have drawn the demon, it has the % chance that the last card the opponent drew was a demon to just assume they played the demon already.
Let me know if there's anything I did wrong; I haven't messed with python in a while, this is my first KOTH, and it's 2 am, so there's bound to be something.
EDITS:
Taking out the randomness turns out to help it a lot. With the randomness in it is very dumb. Also, as said in the comments below, it tries to summon the demon or dragon.
Steve
from base import BasePlayer
from random import choice
class Steve(BasePlayer):
def reset(self):
self.items = [0, 1, 2, 3, 4, 5]
self.turns = 0
self.dungeon = []
self.possibledungeon = [1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 7, 9]
self.lastDied = 0
def __init__(self):
self.TRIALS = 10 #How many dungeon runs to do each turn
self.PASS = 8 #How many dungeon runs have to have died to pass
self.SMASHITEMS = 4 #If less dungeon runs died, smash items.
self.reset()
def start_turn(self, last_turn):
if (last_turn is not None):
self.turns += 1
if (last_turn in self.items):
self.items.remove(last_turn)
else:
self.dungeon.append(-1)
#Check if the dungeon is lethal
died = 0
total_hp = 3
if (5 in self.items):
total_hp += 5
if (3 in self.items):
total_hp += 3
vorpal = self.vorpal_choice(None)
for i in range(self.TRIALS):
hp = total_hp
temppossible = self.possibledungeon.copy()
usedpotion = False
killedDemon = False
#Going for a crawl
for j in self.dungeon[::-1]:
if (killedDemon == True): #If last round we killed the Demon
killedDemon = False
j = 0
if (j == -1): #If we don't know what this card is
j = choice(temppossible)
temppossible.remove(j)
if (j == 7 and 0 in self.items): #If we kill demon with the pact
j = 0
killedDemon = True
if (j % 2 == 0 and 2 in self.items) or (j == vorpal): #If we vorpal or grail
j = 0
hp -= j
if (hp <= 0):
if (not usedpotion and 1 in self.items):
hp = 3
usedpotion = True
else:
died += 1
break
if (died >= self.PASS):
return 0
died = self.lastDied
return 1
def play(self, card):
self.possibledungeon.remove(card)
if (self.lastDied < self.SMASHITEMS):
if (7 in self.dungeon) and (0 in self.items):
self.items.remove(0)
return 0
if ( (9 in self.dungeon) or (5 in self.dungeon) ) and (3 in self.items):
self.items.remove(3)
return 3
for i in [2, 1, 5, 4, 3, 0]:
if (i in self.items):
self.items.remove(i)
return i
self.dungeon.append(card)
return 6
def vorpal_choice(self, last_turn):
if (last_turn is not None):
self.turns += 1
if (last_turn in self.items):
self.items.remove(last_turn)
else:
self.dungeon.append(-1)
if (self.dungeon.count(5) == 2):
return 5
if (9 in self.dungeon):
return 9
if (self.dungeon.count(4) == 2 and not 2 in self.items):
return 4
if (7 in self.dungeon and not 0 in self.items):
return 7
for i in range(6)[::-1]:
if (i+1 in self.dungeon):
return i+1
return 5
def result(self, bot, result, dungeon, vorped):
self.reset()
Steve tries to guess whether or not the dungeon is lethal. If he thinks it is, he passes. Other than that, I tried to make him get rid of items smartly. He used to adjust his PASS threshold depending on if he died in the dungeon or the opponent lived, but it ended up making him a lot dumber so I got rid of it.
He still doesn't beat GrailThief on my machine, but at least comes closer.
• you know that 9 is the dragon, not the demon? – Destructible Lemon Jun 24 '17 at 5:57
• ...oh. I hadn't realised that. Well, this code seems to be doing fine as is at least – J. Dingo Jun 24 '17 at 5:58
• I've changed the code to attempt to summon the demon or dragon. – J. Dingo Jun 24 '17 at 6:05
• I would like to say: nice job and also thanks for joining the koth – Destructible Lemon Jun 24 '17 at 6:09
• I will attempt to execute this some time in the future, but I currently don't have my preferred computer so it may be hard... maybe I'll ask someone else nicely to execute it for me? – Destructible Lemon Jun 24 '17 at 6:29
SlapAndFlap
First time in KOTH, so slap me hard for mistakes.
This one simpleton is always trying to remove all good items with low-strength monsters, while keep powerful ones and then just forces opponent to play.
It beats RunAway and DumDum at least :D
My other bot in deleted answer for some time, I need to fix him by tommorow
from base import BasePlayer
class SlapAndFlap(BasePlayer):
def reset(self):
# Monsters that you pushed in
self.know_monster = list(self.deck)
# Items still in game
self.items_in_game = [True, True, True, True, True, True]
# List of items, sorted by value
self.valuables = [3,1,5,0,2,4]
# Counter
self.cards = 13
def __init__(self):
# Deck
self.deck = (1,1,2,2,3,3,4,4,5,5,6,7,9)
# Indexes of item cards
self.items = (0, 1, 2, 3, 4, 5)
self.reset()
def start_turn(self, last_turn):
if last_turn is not None:
self.cards -= 1
# Sneak peak at items removed by opponent
if last_turn is not None and last_turn < 6:
self.items_in_game[last_turn] = False
self.valuables.remove(last_turn)
# Flap!
if self.cards < 6:
return 0
return 1
def play(self, card):
if card < 6 and any(self.items_in_game):
self.know_monster.remove(card)
to_return = self.valuables[0] # remove the best of the rest
self.valuables = self.valuables[1:]
self.items_in_game[to_return] = False
else:
return 6
def vorpal_choice(self, last_turn):
# We can just guess what monster will be there
# But we know ones, we removed
# If we have pact, no need to remove demon
if self.items_in_game[0]:
self.know_monster.remove(7)
# If we have grail, no need to remove even monsters (kinda)
if self.items_in_game[2]:
self.know_monster = [i for i in self.know_monster if i%2]
# Find what threatens us the most, counting its strength multiplied by number
weight = [i * self.know_monster.count(i) for i in self.know_monster]
return weight.index(max(weight)) + 1
def result(self, bot, result, dungeon, vorped):
self.reset() # we live for the thrill, not the result!
• Well, the first thing I'm going to slap you hard for is that it is KOTH, not KOHL. – Gryphon Jun 15 '17 at 16:33
• @Gryphon That one hurts :( – Dead Possum Jun 15 '17 at 16:35
RandomMandom
The obligatory random bot. Appropriately, he loses hard to the default bots, which is good, because it means that the game has at least some strategy.
from base import BasePlayer
from random import randint
from random import choice
class RandomMandom(BasePlayer):
def __init__(self):
self.items = [0,1,2,3,4,5]
def start_turn(self, last_turn):
if last_turn in self.items:
self.items.remove(last_turn)
if len(self.items) > 0:
return randint(0,1)
return 1
def play(self, card):
if len(self.items) > 0:
if randint(0,1) == 1:
selection = choice(self.items)
self.items.remove(selection)
return selection
return 6
def vorpal_choice(self, last_turn):
return choice([1,2,3,4,5,6,7,9])
def result(self, bot, result, dungeon, vorped):
# Never learns
pass
• I'm confused... why does the bot always keep drawing when all items are gone? also you might want to weight it, otherwise it has a 50/50 chance of not placing anything in the dungeon and then you just look silly. PS nice name of bot – Destructible Lemon Jun 18 '17 at 3:12
• (when I said not placing anything I meant not playing any turns) – Destructible Lemon Jun 19 '17 at 4:21
• I didn't analyze the controller very closely, because I assumed that it wouldn't present me with a choice if none existed. Also, it's a super-naive bot, because I wanted to get at least one entry in before my real entry. I will use your feedback in the next bot. – Andrew U Baker Jun 19 '17 at 4:52
• I think you confused the deck with items. you can have no items but still have a deck and have another turn. it is unlikely given how bots will probably play to not be stupid. – Destructible Lemon Jun 19 '17 at 4:54
• also you have 75% chance of winning the bounty at this point if you do make that new bot so that'll probably be fun for you (mostly because of people not answering) – Destructible Lemon Jun 19 '17 at 5:02
DareDevilDumDum
kinda clear. never backs down. the only way you can (consistently; RunAway loses sometimes but it still beats this most of the time) lose to this bot is if you don't remove any items or are a super coward. think of this bot as a reminder to remove items, otherwise even this can win.
from base import BasePlayer
class DareDevilDumDum(BasePlayer):
def start_turn(self, last_turn):
return 1 # damn squiggles. Draw a card
def play(self, card):
return 6+card*0 # put the card in the dungeon, and use card to avoid squiggles :P
def vorpal_choice(self, last_turn):
return 9+last_turn*0 # dragon
def result(self, bot, result, dungeon, vorped):
pass # we live for the thrill, not the result!
And
RunAway
pretty much they remove armour and then run away some time before the end. as daredevildumdum, it doesn't remember anything, except for the number of cards in the deck (which tbh wouldn't be remembered in the actual game (you'd just check)) and whether someone removed the armour (mostly same as before).
from base import BasePlayer
class RunAway(BasePlayer):
def __init__(self):
self.cards = 13
self.armoured = True
def start_turn(self, last_turn):
if last_turn is not None:
self.cards -= 1 # opponents play
if last_turn is 5:
self.armoured = False
if self.cards < 4:
return 0 * last_turn # avoid the ---noid--- squiggles
else:
return 1
def play(self, card):
if self.cards > 11 and self.armoured: # if it is the first turn and armour has not been removed
choice = 5 # remove armour
else:
choice = 6 # put the card in the dungeon
self.cards -= 1 # this play
return choice
def vorpal_choice(self, last_turn):
return 5 # without using more memory, this is the best choice statistically
def result(self, bot, result, dungeon, vorped):
self.cards = 13
self.armoured = True
Also because I am special challenge poster, these bots do not count toward my bot count, because they are example bots that are dumb
• Fixed the second code snippet (the beginning of it) – Mr. Xcoder Jun 15 '17 at 14:11
BoringPlayer
The opposite of RandomMandom, BoringPlayer always makes the same choices. The problem is that it appears that it is too successful for such a simple bot. It scores 3800+ in my local test.
from base import BasePlayer
class BoringPlayer(BasePlayer):
def start_turn(self, last_turn):
return 1
def play(self, card):
return 6
def vorpal_choice(self, last_turn):
return 5
def result(self, bot, result, dungeon, vorped):
# Never learns
pass
• This is almost identical to Daredevil except it chooses the golem for its vorpal sword. weird that you get 3800+. this seems like an obscure bug – Destructible Lemon Jun 20 '17 at 3:05
• ok, it seems like picking 5 over nine is actually a really big advantage... O_o. I guess that this is not a bug – Destructible Lemon Jun 20 '17 at 4:35
• I think that 5 over 9 is significant against runaway (because 9 was overkill anyway), and the other bots kind of need more fine tuning with their numbers anyway – Destructible Lemon Jun 20 '17 at 4:45
• I didn't intend for it to be that close to DareDevil. I was actually working on a more sophisticated bot that tracked the cards more closely, and this was just the stub values. However, since it outperformed everything else, I decided I should submit it anyway. I will try to get my more sophisticated bot working to see if I can come up with a better bot. – Andrew U Baker Jun 20 '17 at 5:02
• Note to people seeing this: This bot does not mean that this simplistic strategy will dominate; the other bots are also very simple and this is why this bot wins – Destructible Lemon Jun 20 '17 at 5:04 | 2019-10-16 07:36:58 | {"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": 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": 0.29343849420547485, "perplexity": 4404.518673588212}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986666467.20/warc/CC-MAIN-20191016063833-20191016091333-00178.warc.gz"} |
https://www.federalreserve.gov/econres/notes/feds-notes/why-have-initial-unemployment-claims-stayed-so-high-for-so-long-20120702.html | July 02, 2021
### Why Have Initial Unemployment Claims Stayed So High for So Long?
Brendan M. Price
#### Introduction
As the labor market recovers from the COVID-19 pandemic, claims for unemployment insurance (UI) have been surprisingly slow to return to conventional levels. As recently as 2021Q1, initial claims for regular UI benefits averaged nearly 800,000 per week (see Figure 1)—more than twice as many as were observed at a comparable point during the recovery from the Great Recession.1 Although initial claims fell substantially this spring, the latest readings are still almost double those observed in the lead-up to the pandemic.2 Throughout the recovery, the UI numbers have been consistently hard to square with the degree of improvement evident in other labor market indicators, such as the unemployment rate, job postings, and layoffs.
##### Figure 1. Weekly initial claims for regular state UI and PUA benefits (in thousands)
Why have initial UI claims remained so high for so long? In this note, I argue that much of the answer lies with the expansions to UI eligibility and generosity implemented in response to the pandemic. I focus primarily on the role of Pandemic Unemployment Assistance (PUA), a federal program that extends income support to self-employed workers and others who are ineligible for traditional UI benefits. Although such individuals would normally have little reason to apply for regular UI, many states have required them to do so as a first step towards obtaining PUA. Using data on UI eligibility currently available through the first quarter of 2021, I estimate that initial claims for regular UI benefits would have been 20 percent lower from 2020Q3 through 2021Q1 if not for increased filing prompted by PUA.
Other changes in UI policy have also kept claims abnormally elevated deep into the recovery. Supplemental benefits of $300 per week (formerly$600) have given unemployed and underemployed workers unusually strong incentives to apply for UI. In addition, almost every state waived its usual job search requirements early in the pandemic. These waivers, most of which were still in effect earlier this year, have made UI benefits available to many individuals for whom caregiving responsibilities or fears of infection have complicated returning to work.
Accounting for last year's UI expansions can thus go a long way towards explaining why initial claims have taken so long to return to familiar levels. By the same token, declining claim volumes in spring 2021 may owe partly to a new round of policy changes, as most states have by now reinstated pre-pandemic search requirements and many are withdrawing from the federal UI programs prior to their nationwide expiration. Initial claims are likely to fall further as the pandemic UI programs phase out and as the labor market continues to heal.
#### The role of the PUA program
Eligibility for regular UI benefits is limited to workers who have accrued sufficient earnings in UI-covered jobs in a base period leading up to their claim.3 Three groups of workers commonly fall short of this requirement: self-employed workers and independent contractors, who do not pay into the UI system; recent labor market entrants, who have not yet established an earnings history; and wage workers whose earnings fall below the required threshold, either because they work part-time at low wages or because they were jobless for much of the base period.
These groups are strikingly overrepresented among claims filed during the pandemic. Figure 2 plots the share of new claims passing the earnings test, as reported in the Department of Labor's quarterly data on UI monetary determinations.4 Prior to the pandemic, this share ranged between 75 and 90 percent, with a modest decline after each recession.5 During the pandemic, however, the passage rate fell much more precipitously than in the past: just over 50 percent of new claims satisfied the earnings test in the latter half of 2020, with only a slight rebound in 2021Q1. These unusually low eligibility rates signify an unprecedented influx of applicants with limited earnings in UI-covered jobs.6
##### Figure 2. Percentage of new claims passing the earnings test for UI eligibility
This dramatic shift in the UI claimant pool owes largely to the federal PUA program, which was created in March 2020 to provide benefits to classes of workers not covered by traditional UI.7 Because federal guidelines restrict PUA to those who are ineligible for regular UI, many states instruct benefit-seekers that they must apply (and be denied) for regular UI before they can apply for PUA. As a result, many self-employed workers and other atypical claimants have been counted towards both regular UI and PUA in successive weeks, even though initial claims for these two programs are tallied separately in government statistics.
As evidence that PUA has fueled the surge in ineligible claimants, Figure 3 plots the share of new claimants passing the earnings test in three groups of states and territories: those that require claimants to apply for regular UI as a precondition for seeking PUA (27 programs); those that allow individuals to apply directly for PUA (16 programs); and those whose websites provide nuanced or ambiguous instructions (10 programs).8 Although passage rates have fallen nationwide, they have fallen much more steeply in states that instruct prospective PUA claimants to first apply for regular UI. These state-level patterns suggest that we would have seen a much more muted decline in eligibility rates nationwide if not for PUA.9
##### Figure 3. Percentage of new claims passing the earnings test, split by state application protocols
A simple back-of-the-envelope calculation can help us gauge what initial claim volumes might have looked like had the PUA program not been created. The first step is to note that initial claims fall into two separate categories, which differ in their relationship to PUA. New initial claims are filed by workers who have not received UI in the recent past. Reopened initial claims are filed by workers who were previously approved for benefits and who are now resuming benefit receipt after an intervening period of employment.10 I estimate counterfactual initial claims in the absence of PUA as
$$\text{reopened initial claims} + \left(\frac{\text{observed passage rate}}{\text{passage rate absent PUA}}\right) \times \text{new inital claims},$$
where I assume that 75 percent of new initial claims would have passed the earnings test in the absence of PUA, in line with the passage rate observed after the Great Recession.
The logic behind this formula is as follows. By definition, reopened claims originate from workers who have already passed the earnings test—in other words, standard claimants who would most likely be seeking benefits even in the absence of PUA. I therefore make no adjustment to the observed number of reopened claims. By contrast, new initial claims reflect a combination of standard and non-standard claimants. To weed out claims filed by non-standard claimants as a stepping-stone to PUA, I scale down the observed number of new claims based on the unusually low share of claimants passing the earnings test.11
Applying this formula quarter by quarter, I estimate that—if not for PUA—initial claims for regular UI benefits would have been about 7 percent lower in 2020Q2, when PUA was first being rolled out, and about 20 percent lower in each of 2020Q3, 2020Q4, and 2021Q1.
Even with this adjustment, initial claims would still have averaged over 600,000 per week in the first quarter of 2021. Such a tally would be comparable to the weekly counts reported in early 2009, even though monthly worker flows from employment into unemployment were about 50 percent higher in 2009 than they were in 2021Q1. Although PUA is a major reason why claims have been so slow to return to typical levels, it is clearly far from the only reason.
A second factor that has boosted initial claims is Federal Pandemic Unemployment Compensation (FPUC), which has provided supplemental benefits of $300 or$600 per week at various points during the pandemic.12 Take-up of UI benefits among eligible individuals is usually far from universal: qualifying workers may decline to file for UI because of a lack of information, stigma associated with benefit receipt, or an expectation that they will quickly be reemployed. FPUC has given unemployed and underemployed workers unusually strong incentives to claim UI, and there is good reason to believe that take-up rates have risen as a result.13 Although take-up rates are difficult to measure in available data, some indirect evidence comes from the unusually high share of 2020–21 UI claimants receiving partial UI benefits while working part-time. Partial UI recipients receive the $300/600 FPUC supplement in full, so FPUC has an especially large proportional impact on UI replacement rates for this group of workers.14 Alongside PUA, FPUC, and other federal programs created during the pandemic,15 many states broadened access to their UI programs at the start of the crisis by suspending waiting periods and by waiving their normal job search requirements. These state-level waivers may have drawn more applicants into the UI system, since individuals unable or unwilling to work—for example, because of childcare constraints associated with virtual schooling—might choose not to apply if they would be required to look for jobs. In addition, UI statistics were heavily impacted in the first months of the crisis by backlogged claims, fraudulent filings, and other irregularities (Cajner et al., 2020; Government Accountability Office, 2020). Although these issues are likely much less consequential than they were early on, they may still be having an appreciable impact on claim volumes in some states. These institutional factors should largely abate over the course of 2021. PUA, FPUC, and other federal programs are scheduled to expire in September, and many states have chosen to withdraw from them prior to their nationwide expiration.16 Most states have also reinstated their pre-pandemic job search requirements, and state and federal authorities have worked hard to resolve backlogs and to deter fraud. As the institutional landscape returns to pre-pandemic norms, policy measures unique to the pandemic should have a diminishing impact on reported claims. #### The state of the labor market A final consideration is that labor market conditions over the past year have likely been somewhat weaker than measured unemployment suggests. The official unemployment rate—which is calculated from household responses to the Current Population Survey (CPS)—excludes or undercounts at least four groups of individuals who may be filing UI claims. First, the Bureau of Labor Statistics has cautioned that some unemployed CPS respondents have been misclassified as employed but absent from work. Second, survey response rates have fallen during the pandemic, especially for groups that tend to have higher unemployment rates. The sampling weights used to estimate aggregate unemployment may not fully account for these missing respondents. Third, millions of workers have exited the labor force because of caregiving needs, fears of the virus, or a lack of suitable job opportunities. And lastly, the number of individuals working part-time for economic reasons is still well above pre-pandemic levels.17 In sum, while much of the recent disconnect between initial UI claims and other labor market indicators can be attributed to changes in UI policy in response to the pandemic, some of it may reflect forms of labor market weakness that other data sources do not adequately capture. As long as initial claims remain far above pre-pandemic levels, they will serve as a weekly reminder that the labor market recovery remains uneven and that much ground is still left to be regained. I close with a cautionary note about how to interpret these findings. Although last year's UI expansions amplified claim volumes given contemporaneous labor market conditions, they have also influenced those conditions in turn. The pandemic UI programs have provided critical income support to millions of American households, but their net effect on the labor market has been hotly debated: supporters of these programs argue that they have spurred the recovery by buttressing aggregate demand, whereas critics maintain that they have impeded the recovery by disincentivizing job search. My analysis is silent on the question of how the labor market would have fared had UI policy not responded to the pandemic. #### References • Cajner, Tomaz, Andrew Figura, Brendan M. Price, David Ratner, and Alison Weingarden. 2020. "Reconciling Unemployment Claims with Job Losses in the First Months of the COVID-19 Crisis." Finance and Economics Discussion Series Working Paper No. 2020-055. • Dube, Arindrajit. 2021. "Aggregate Employment Effects of Unemployment Benefits during Deep Downturns: Evidence from the Expiration of the Federal Pandemic Unemployment Compensation." National Bureau of Economic Research Working Paper No. 28470. • Finamor, Lucas, and Dana Scott. 2021. "Labor Market Trends and Unemployment Insurance Generosity during the Pandemic." Economics Letters, 199(2021): 109722. • Ganong, Peter, Pascal Noel, and Joseph Vavra. 2020. "US Unemployment Insurance Replacement Rates during the Pandemic." Journal of Public Economics, 191(2020), 104273. • Government Accountability Office. 2020. "Urgent Actions Needed to Better Ensure an Effective Federal Response." Report GAO-21-191. • Marinescu, Ioana, Daphne Skandalis, and Daniel Zhao. 2021. "The Impact of the Federal Pandemic Unemployment Compensation on Job Search and Vacancy Creation." National Bureau of Economic Research Working Paper No. 28567. #### Acknowledgements The views expressed here are strictly those of the author and do not necessarily represent the views of the Federal Reserve Board or its staff, nor those of the Department of Labor. I am grateful to Isabel Leigh for excellent research assistance and to Andrew Figura, Charles Fleischman, Ryan Michaels, Seth Murray, and Ivan Vidangos for helpful comments. 1. I compare 2021Q1 with 2014Q2. Using seasonally adjusted data from the Bureau of Labor Statistics, the unemployment rate averaged 6.2 percent in both quarters, and the share of workers transitioning from employment to unemployment was similar as well. Initial claims averaged about 315,000 per week in 2014Q2. Adjusting for subsequent growth in the labor force, this translates into about 330,000 claims per week in today's terms. Return to text 2. At the time of writing, the most recent Department of Labor press release reported that an average of 392,750 initial claims for regular UI benefits were filed weekly in the four weeks ending June 26, 2021. (This number is subject to revision in future releases.) By comparison, initial claims averaged about 220,000 per week in 2019. Return to text 3. Most states define the base period as the first four of the five most recent completed quarters. In addition to the earnings test, claimants must also satisfy a range of other eligibility criteria, such as having a valid reason for being unemployed, being able and available to work, and (absent a statewide waiver) engaging actively in job search. Return to text 4. At the time of analysis, state-level records were unavailable for Alabama throughout 2020 and for Alabama, Colorado, Montana, and New Jersey in 2021Q1. The passage rates plotted in Figures 2 and 3 were computed using data from all available states in each quarter. The patterns look virtually identical if I exclude these four states throughout the analysis period, so that the sample is defined consistently over time. Return to text 5. A plausible explanation for the post-recession decline in UI eligibility is that the longer and more frequent unemployment spells experienced during recessions gradually erode workers' base-period earnings. Some job losers may file claims without realizing they are ineligible, especially if they have been entitled to UI benefits in the past. Return to text 6. A similarly steep decline is evident in the ratio of the number of claims receiving first payments in each month to the number of new claims filed in that month—a proxy for the share of claims that are ultimately awarded benefits. However, because benefit payments are dated to the time of payment rather than the time of filing, this measure is potentially distorted by time lags in the adjudication of claims and the disbursal of benefits. The share of claims passing the earnings test is less susceptible to such distortions because monetary determinations are issued early in the process and because the numerator and denominator pertain to the same point in time. Return to text 7. The unusual sectoral and occupational profile of pandemic job losses may also have contributed to the record share of ineligible claimants. Since low-wage segments of the labor market have been hit hardest by the pandemic, recent job losses may be unusually concentrated among workers with insufficient base-period earnings to receive UI. Return to text 8. These 53 programs represent the 50 states, the District of Columbia, Puerto Rico, and the US Virgin Islands. I classify states based on a review of each state's UI website. In some cases, program websites explicitly indicate whether a two-stage application is required. For example, the Illinois UI website says: "You can file a claim for PUA only after you applied for regular unemployment insurance benefits and have been denied." By contrast, the Massachusetts website says that self-employed workers and others typically ineligible for UI benefits should apply directly to PUA. Other cases are less clear-cut; for example, some states recommend that PUA applicants first apply for regular UI but indicate that this is not a strict requirement. Return to text 9. Even in states where people can apply directly for PUA, some claimants who expect to ultimately receive PUA are likely to start by applying for regular UI. Advocate organizations encourage applicants to try regular UI first, even if they think they may be ineligible. Return to text 10. The date when a claim is first approved marks the beginning of a 12-month "benefit year." Reopened claims (known as "additional claims" in Department of Labor parlance) are initial claims filed within an existing benefit year, whereas new claims are those filed when no benefit year is in progress. The Department of Labor reports monthly breakdowns of initial claims into new versus additional claims in Employment and Training Administration form 5159. New claims accounted for 85 percent of initial claims in 2020Q2, but only about two-thirds of initial claims in subsequent quarters, as many workers have experienced recurrent spells of unemployment. Return to text 11. I derive the scaling factor as follows. We can express the observed number of new initial claims as the sum of three components: claimants who pass the earnings test (denoted P, for "pass"); claimants who fail the earnings test, but would have applied even in the absence of PUA (denoted SD, for "standard denied applicants"); and claimants who fail the earnings test and who would not have applied in the absence of PUA (denoted ND, for "non-standard denied applicants"). The observed passage rate is $\frac{P}{P+SD+ND}$, and the passage rate absent PUA is $\frac{P}{P+SD}$. The ratio of the observed passage rate to the counterfactual passage rate absent PUA is therefore $\frac{P+SD}{P+SD+ND}$, which equals the share of observed claimants who would have applied even in the absence of PUA. Return to text 12. From its creation in March 2020 through July 2020, FPUC provided supplemental benefits of$600 per week. From August to December 2020, FPUC was unavailable, though a short-lived program called Lost Wages Assistance provided $300 or$400 per week for part of this period. From January 2021 until its scheduled expiration in September 2021, FPUC provides \$300 per week. See Ganong, Noel, and Vavra (2020) for an analysis of UI replacement rates during the first phase of FPUC. Return to text
13. Note that the question of whether FPUC has boosted UI take-up rates is distinct from the question of whether FPUC has impeded the labor market recovery by disincentivizing claimants from returning to work. A spate of academic papers have found scant evidence that FPUC constrained employment last year (e.g., Dube [2021], Finamor and Scott [2021], and Marinescu, Skandaris, and Zhao [2021]), though its effects may differ in the tighter labor market of 2021. Regardless of how it may have affected employment, the program may have encouraged filing among unemployed and underemployed individuals who would otherwise not have applied for benefits. Return to text
14. Partial UI offers prorated benefits to workers who remain employed but experience significant declines in weekly earnings—for example, because they have lost one of two jobs, or because their weekly hours have been cut. (These partial benefits are distinct from the Short-Time Compensation or "work-sharing" program, which requires employer participation and is designed for workers experiencing more modest reductions in earnings.) The share of claimants receiving partial benefits surged above 16 percent in 2020Q4, compared with a pre-pandemic record high of about 11 percent. Furthermore, the average benefit amount among partial UI recipients (not counting the FPUC supplement) has fallen sharply during the pandemic. These patterns are consistent with increased take-up among underemployed claimants whose weekly benefits would normally be too meager to induce them to apply. Return to text
15. Other federal programs include Pandemic Emergency Unemployment Compensation (PEUC), which provides extra weeks of benefits to claimants who exhaust regular UI, and Mixed Earner Unemployment Compensation (MEUC), which covers claimants who have a combination of earned and self-employment income. The federal government has also provided full funding of Extended Benefits (EB), which are normally financed jointly with the states. PEUC and EB have accounted for large shares of UI beneficiaries in 2020–21, but neither is likely to be a major driver of initial claims, since they come into play at the end of a worker's UI spell rather than at the beginning. Return to text
16. As of mid-June, 26 states had announced plans to withdraw from FPUC in advance of its nationwide expiration (or had already done so). All but five of these states are withdrawing from PUA and PEUC, as well. See Coral Murphy Marcos (2021), "Here Are the States Eliminating Pandemic Unemployment Benefits, and When," The New York Times, June 15. Return to text
17. In addition to these factors, brief unemployment spells lasting less than four weeks can go unobserved if they fall in the space between the monthly CPS surveys. Such transient spells may be especially common in the pandemic labor market, as businesses alternately close and reopen in response to public-health conditions. Changes in the prevalence of short unemployment spells should not bias estimates of the stock of unemployed workers (the numerator in the unemployment rate), but they can distort measures of layoff activity and other labor market flows. Return to text
Price, Brendan M. (2021). "Why Have Initial Unemployment Claims Stayed So High for So Long?," FEDS Notes. Washington: Board of Governors of the Federal Reserve System July 02, 2021, https://doi.org/10.17016/2380-7172.2932.
Disclaimer: FEDS Notes are articles in which Board staff offer their own views and present analysis on a range of topics in economics and finance. These articles are shorter and less technically oriented than FEDS Working Papers and IFDP papers. | 2022-09-28 13:06:23 | {"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": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.19238919019699097, "perplexity": 4537.063013130371}, "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-2022-40/segments/1664030335254.72/warc/CC-MAIN-20220928113848-20220928143848-00331.warc.gz"} |
http://math.stackexchange.com/questions/80644/calculating-the-limit-of-a-sequence | # Calculating the limit of a sequence
I'm currently studying limits because of my calculus class and i've wondered how for example wolfram alpha computes the limit of a sequence. Is it more a brute force way, or is there an efficient method to calculate/find them?
-
I would guess, since most of the sequences we give are monotonic it would search for a greatest lower bound or least upper bound(if at all the sequence is convergent). – Dinesh Nov 9 '11 at 22:23 | 2015-05-25 10:10:31 | {"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": 0.8179546594619751, "perplexity": 344.4805919405393}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-22/segments/1432207928479.19/warc/CC-MAIN-20150521113208-00239-ip-10-180-206-219.ec2.internal.warc.gz"} |
https://codereview.stackexchange.com/questions/158142/practical-number-algorithm | # Practical number algorithm
I am trying to write a program to find the practical numbers, from an input from $1$ to $n$.
Practical numbers
My code is running correctly but it is extremely slow when calculating numbers around 50 - it gets stuck at 44.
import Foundation
func getInteger() -> Int {
var firstNum:Int = 0
while true {
// get value from user. Using optional input since readLine returns an optional string.
// ensure string is not nil
if let unwrappedInput = input {
if let unwrappedInt = Int(unwrappedInput) {
firstNum = unwrappedInt
break
}
else { // the input doesn't convert into an int
print("\(unwrappedInput) is not an integer. Please enter an integer")
}
}
else { // did not enter anything
}
}
return firstNum
}
func addOne(signArray: [Int]) -> [Int] { // finds the combinations
var signArray2 = [Int]()
for i in 0...signArray.count-1 {
signArray2.append (signArray[i])
}
for i in 0...signArray2.count-1 {
if signArray2[i] == 1 {
signArray2[i] = 0
}
else {
signArray2[i] = 1
break
}
}
return signArray2
}
func signEval (signArray: [Int], divArray: [Int], inNum: Int) -> Bool {// changes 2nd
var counts = 0
for i in 0...divArray.count-1 {
if signArray[i] == 0 {
counts = divArray[i] + counts }
if counts == inNum {
return true
}
}
return false
}
print("Please enter a number to find the summable numbers up to that number:")
var input2 = getInteger()// if num = 1 print 1 if num = 2 print 1 and 2 else print >2 1, 2
var inNum = 0
var numHalf = 0.0
var numRound = 0.0
var numCheck = false
var numCheck2 = false
var numQuarter = 0.0
var numSixth = 0.0
var divArray:[Int] = []
var theirArray = [Int]()
var signArray = [Int]()// array of 0s and 1s
var summableArray:[Int] = [1,2] // need to check if num is bigger than 2!
for input in 1...input2 {
numHalf = Double (input) / 2.0
numRound = round(numHalf)
if numRound == numHalf {
numCheck = true }
if input > 2 && numCheck == false { // odd numbers greater than one are not summable
}
else { // these are possible summable nums
numQuarter = Double (input) / 4.0
numRound = round(numQuarter)
if numRound == numQuarter {
numCheck = true
}
else {
numCheck = false
}
numSixth = Double(input) / 6.0
numRound = round(numSixth)
if numRound == numSixth {
numCheck2 = true }
else { numCheck2 = false}
if numCheck == true || numCheck2 == true {
theirArray = []
divArray = []
signArray = []
summableArray = []
for i in 1...input {
theirArray.append (i)
}
for i in 1...input { // creates an array of all the diviors of inputted number
if input%i == 0 {
divArray.append (i)
}
}
for j in 1...divArray.count {//
signArray.append(0)
}
for i in 1...input{
let x: Int = Int(pow(Double(2),Double(input-1)))// int 2 to the power of input -1
var Boolcheck = false
for q in 1...x-1 { // i to 2^n -1 (sequence to check)
Boolcheck = (signEval(signArray: signArray, divArray: divArray, inNum: i))// checks
if Boolcheck == true {
summableArray.append(i)// creates array of mini summable numbers
break
}
}
if summableArray.count == input {
print ("\(input)")
}
}
}
}
}
You are correct that the readLine() returns an optional and you need to ensure that it is not nil. However, nil is only returned on an end-of-file condition, which means that it makes no sense to call readLine() again. You can only terminate the program in that situation. That is a typical use-case for guard:
func readInteger() -> Int {
while true {
guard let line = readLine() else {
fatalError("Unexpected end-of-file")
}
if let n = Int(line) {
return n
}
}
}
Now let's have a look at
var signArray2 = [Int]()
for i in 0...signArray.count-1 {
signArray2.append (signArray[i])
}
First, this will crash if signArray is empty. Better use a half-open range instead:
for i in 0..<signArray.count { ... }
or iterate over the indices
for i in signArray.indices {
signArray2.append (signArray[i])
}
or just iterate over the elements:
for e in signArray {
signArray2.append(e)
}
But actually, you are just copying the array:
var signArray2 = signArray
Determination of the divisors should be done in a separate function. Your code
divArray = []
for i in 1...input {
if input % i == 0 {
divArray.append (i)
}
}
can be simplified to
divArray = (1...input).filter { input % \$0 == 0 }
There are various ways to make this faster. For example the observation that with each divisor $i$ of a number $n$, $n/i$ is another divisor (see for example Find all divisors of a natural number). That allows to reduce the number of loop iterations to the square-root of the given number, and could look like this:
func divisors(n: Int) -> [Int] {
var divs = [Int]()
for i in 1...Int(sqrt(Double(n))) {
if n % i == 0 {
divs.append(i)
if n/i != i {
divs.append(n/i)
}
}
}
return divs // or divs.sorted(), if necessary
}
signArray = []
for j in 1...divArray.count {//
signArray.append(0)
}
creates an array with the same size as divArray, but filled with zeros. That can be simplified to
signArray = Array(repeating: 0, count: divArray.count)
There is no need to use string interpolation when printing an integer (or any single value),
print ("\(input)")
can be simplified to
print(input)
## Why is your code slow?
There is a logical error in
let x: Int = Int(pow(Double(2),Double(input-1)))// int 2 to the power of input -1
because the number of possible combinations of divisors is just $2 ^ {\text{number of divisors}}$, which can be computed as
let x = 1 << divArray.count
With that change, your program computes all practical numbers up to 200 in 0.2 seconds, and up to 1,000 in 25 seconds (on a MacBook, compiled in Release mode, with optimization).
## A faster algorithm
In order to determine if $n$ is a practical number, you test all numbers $1 \le i \le n$ if they are a sum of distinct divisors of $n$, and for each $i$ that is done by building all possible sums of divisors, until $i$ is found.
It is more efficient to work the other way around. From the list of divisors of $n$, build a list of all sums of distinct divisors. This can be done iteratively: Starting with $0$, the first, second, ... divisor is added to the numbers obtained previously. Then check if all numbers from $1 ... n-1$ are in that list.
The following implementation uses a boolean array to mark all numbers which are confirmed as sums of divisors. In addition, it uses that powers of two are always practical numbers (using the method from Determining if an integer is a power of 2).
func divisors(_ n: Int) -> [Int] {
var divs = [Int]()
for i in 1...Int(sqrt(Double(n))) {
if n % i == 0 {
divs.append(i)
if n/i != i {
divs.append(n/i)
}
}
}
return divs.sorted()
}
func isPractical(n: Int) -> Bool {
// 1 and 2 are practical numbers:
if n == 1 || n == 2 {
return true
}
// Every other practical number must be divisible by 4 or 6:
if n % 4 != 0 && n % 6 != 0 {
return false
}
// Every power of 2 is a practical number:
if n & (n-1) == 0 {
return true
}
var isSumOfDivisors = Array(repeating: false, count: n)
isSumOfDivisors[0] = true
var last = 0 // Index of last true element.
// For all divisors d of n (except n):
for d in divisors(n).dropLast() {
// For all i which are a sum of smaller divisors (in reverse order, so that
// we can simply update the array):
for i in stride(from: min(last, n-1-d), through: 0, by: -1) where isSumOfDivisors[i] {
// Mark i + d:
isSumOfDivisors[i + d] = true
if i + d > last {
last = i + d
}
}
}
// n is "practical" if isSumOfDivisors[i] == true for all i = 0...n-1:
return !isSumOfDivisors.contains(false)
}
Test code:
let startDate = Date()
let practicalNumbers = (1...10_000).filter(isPractical)
let endDate = Date()
print(endDate.timeIntervalSince(startDate))
print(practicalNumbers)
On my MacBook, this computes the practical numbers up to 1,000 in 0.003 seconds, and up to 10,000 in 0.1 seconds.
The indentation of your code looks messy in some places. Let your IDE format the code for you. At least the counts = divArray line needs to be further to the right, and the closing brace at the end of that line needs to be on a line of its own.
The code should have a function called isPractical(x). That function should look approximately like the following, for a very naive approach, but maybe that works already.
func isPractical(num:Int) -> Bool {
var factors = getFactors(num)
for i in 1...num {
if (!isSumPossible(i, factors)) {
return false
}
}
return true
}
I don't understand your variable names. Imagine you would explain this program to a human. Which words would you use to describe the purpose of each of these variables? These words are the basis for good names.
• getInteger should be called readInt
• firstNum is unnecessary, since you can just return unwrappedInt
• input should be called line
• addOne should be called nextCombination
• signArray should not have the word sign in its name, since a 0 in that array means take this number
Instead of if condition { numCheck = true } else { numCheck = false }, you can simply write numCheck = condition.
Swift has the remainder operator %, which can be used like this:
if number % 4 == 0 {
print("Divisible by 4")
}
Instead of using pow(2, n), you can write 1 << n, which has the same result for Int, but is simpler and faster.
All the above will not help you with the time-limit-exceeded, but it will make your program readable enough so that others may want to have a look at it. So when you have fixed all the above issues, feel free to post another new question. | 2019-11-19 14:12:14 | {"extraction_info": {"found_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": 0.1809670478105545, "perplexity": 4175.261844892334}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1573496670151.97/warc/CC-MAIN-20191119121339-20191119145339-00427.warc.gz"} |
http://rspa.royalsocietypublishing.org/content/400/1818/97 | Quantum Theory, the Church-Turing Principle and the Universal Quantum Computer
D. Deutsch
Abstract
It is argued that underlying the Church-Turing hypothesis there is an implicit physical assertion. Here, this assertion is presented explicitly as a physical principle: every finitely realizible physical system can be perfectly simulated by a universal model computing machine operating by finite means'. Classical physics and the universal Turing machine, because the former is continuous and the latter discrete, do not obey the principle, at least in the strong form above. A class of model computing machines that is the quantum generalization of the class of Turing machines is described, and it is shown that quantum theory and the universal quantum computer' are compatible with the principle. Computing machines resembling the universal quantum computer could, in principle, be built and would have many remarkable properties not reproducible by any Turing machine. These do not include the computation of non-recursive functions, but they do include quantum parallelism', a method by which certain probabilistic tasks can be performed faster by a universal quantum computer than by any classical restriction of it. The intuitive explanation of these properties places an intolerable strain on all interpretations of quantum theory other than Everett's. Some of the numerous connections between the quantum theory of computation and the rest of physics are explored. Quantum complexity theory allows a physically more reasonable definition of the complexity' or `knowledge' in a physical system than does classical complexity theory. | 2015-05-28 15:04:54 | {"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": 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": 0.6447126269340515, "perplexity": 503.8329850463924}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-22/segments/1432207929418.92/warc/CC-MAIN-20150521113209-00132-ip-10-180-206-219.ec2.internal.warc.gz"} |
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## anonymous 5 years ago In triangle ABC, BC = a, AC = b, AB = c, and a < (b+c)/2. Prove that angle BAC < (angle ABC + angle ACB)/2.
• This Question is Closed
1. anonymous
|dw:1323301015086:dw|
2. anonymous
Given that $a < \frac{1}{2}(b+c)$ Prove that $\angle BAC < \frac{1}{2}(\angle ABC + \angle ACB)$
3. anonymous
WOW
4. asnaseer
my 1st thought is that you should be able to use the sine rule for this?$\frac{a}{\sin(BAC)}=\frac{b}{\sin(ABC)}=\frac{c}{\sin(ACB)}$
5. anonymous
ok
6. anonymous
i thought it's related to cosine rule? $a^2 = b^2 +c^2 - 2bc (\cos \angle BAC)$
7. asnaseer
I suppose you could use that too.
8. anonymous
$a^2 < \frac{1}{4} (b^2 + 2bc + c^2)$$a < \frac{1}{2}(b+c)$ $a^2 < \frac{1}{4} (b+c)^2$
9. asnaseer
hmmm - needs some thought - let me think about it for a bit...
10. asnaseer
thoughts so far: BAC=180-(ABC+ACB) so what we need to prove can be adjusted to prove:$\angle ABC+\angle ACB>120^0$ also, we can show that:$a=b\cos(ACB)+c\cos(ABC)$
11. asnaseer
its past 1am here in the UK and I need to catch some sleep. I'll take a look at this again sometime tomorrow. It's definitely an interesting problem.
12. anonymous
good night
13. anonymous
what country are you in? Im in Canada and its 8:21 pm
14. asnaseer
building on my original hunch to use the sine rule, I have been able to get far:$\frac{a}{\sin(A)}=\frac{b}{\sin(B)}=\frac{c}{\sin(C)}$which leads to:$b=\frac{a\sin(B)}{\sin(A)}\quad\text{and}\quad c=\frac{a\sin(C)}{\sin(A)}$$\therefore b+c=\frac{a(\sin(B)+\sin(C))}{\sin(A)}$$\therefore \frac{b+c}{2}=\frac{a(\sin(B)+\sin(C))}{2\sin(A)}$and we are given that:$a<\frac{b+c}{2}$$\therefore a<\frac{a(\sin(B)+\sin(C))}{2\sin(A)}$$\therefore 1<\frac{\sin(B)+\sin(C)}{2\sin(A)}$$\therefore\sin(A)<\frac{\sin(B)+\sin(C)}{2}$ need to work out how to get from this result to $$A<\dfrac{B+C}{2}$$
15. myininaya
since each of the angles A,B,C are between 0 and 180 degrees 0<sin(A)<=1 0<sin(B)<=1 0<sin(C)<=1 $\sin(A)<A ; \frac{\sin(B)+\sin(C)}{2}<\frac{B+C}{2}$
16. myininaya
I can't think if this helps or not
17. asnaseer
not sure it does, but I have been able to eliminate A from my final equation as follows. the last result I got was:$\sin(A)<\frac{\sin(B)+\sin(C)}{2}$but we know:$A=180-(B+C)$$\therefore \sin(A)=\sin(180-(B+C))=\sin(B+C)$so we finally get:$\sin(B+C)<\frac{\sin(B)+\sin(C)}{2}$so we now need to prove that this implies B+C>120
18. myininaya
you are so smart
19. asnaseer
We are ALL smart in our own ways. I strongly believe that there no such thing as a dumb person :-)
20. anonymous
|dw:1323473308275:dw| My teacher gave me some hints.
21. asnaseer
I have thought of another way to reason about the problem in order to come up with the desired solution. Lets start off by looking at an equilateral triangle where $$a=b=c$$ and $$\angle A=\angle B=\angle C=60^0$$: |dw:1323541342145:dw| \begin{align} \text{projection of AB onto BC will have length=}\frac{c}{2}\\ \text{projection of AC onto BC will have length=}\frac{b}{2}\\ \text{and we can see that }a=\frac{b+c}{2}\\ \end{align} Now, one way for $$a<\dfrac{b+c}{2}$$ is to move point B towards point C along line BC by a small amount. This would reduce the projected length of AB onto BC. Moving point B towards point C would also increase $$\angle B$$ by a small amount - lets call this increase $$\delta$$. It would also decrease $$\angle A$$ by this same amount. Thus we would end up with:\begin{align} a&<\frac{b+c}{2}\\ \angle A&=60^0-\delta\\ \angle B&=60^0+\delta\\ \angle C&=60^0\\ \therefore \angle B + \angle C&=120^0+\delta\\ \therefore \frac{\angle B+\angle C}{2}&=60^0+\frac{\delta}{2}\\ \therefore \angle A=60^0-\delta&<\frac{\angle B+\angle C}{2}\\ \end{align}
22. anonymous
this proof is awesome. It does not require any trignometric identities
23. asnaseer
I am not sure if it counts as a /true/ mathematical proof. It is just a description of the reasoning I went through to try and prove it to myself.
24. anonymous
Let's ask myininaya when he gets on
25. asnaseer
It might be worth asking JamesJ or Zarkon as well.
26. anonymous
i will look at the hint and see if i can understand it. back later
27. anonymous
ok thank you for trying
28. jhonyy9
moneybird how do you think if you begin from : the sum of angles in one triangle is 180 degree so than angle A + angle B + angle C = 180 so than angle A = 180 -(angle B + angle C)
29. anonymous
Yea so you know that if you can prove either angle A is less than 60 or angle B + angle C is 120, then you are done.
30. jhonyy9
try angle A=180-(angle B + angle C) so divide both sides by 2 and will get angle A/2 = 90 -(angle B + angle C)/2
31. jhonyy9
so (angle B + angle C)/2 = 90 - angle A
32. anonymous
angle A < 90 - angle A That's what we need to prove
33. jhonyy9
what you have wrote there now is so from this resulted that angle A < 45
34. anonymous
A/2 = 90 -(angle B + angle C)/2 angle A = 180 - (angle B + angle C) (angle B + angle C) = 180 - angle A
35. jhonyy9
so than angle B + angle C >= 135
36. jhonyy9
and 135/2 > 45 allways
37. anonymous
myininaya says $\angle BAC < \frac{1}{2} (\angle ABC + \angle ACB)$ $\frac{3}{2} \angle BAC < \frac{1}{2}(180)$ $\angle BAC < 60$
38. jhonyy9
good luck bye
39. anonymous
thanks
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Privacy Policy | 2017-01-22 22:22:48 | {"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": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 2, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8729414343833923, "perplexity": 6786.173750576495}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560281649.59/warc/CC-MAIN-20170116095121-00470-ip-10-171-10-70.ec2.internal.warc.gz"} |
https://plainmath.net/34408/given-rectangles-model-the-area-each-using-quadratic-function-then-eval | # Given 2 rectangles. Model the area of each using a quadratic function, then eval
Given 2 rectangles. Model the area of each using a quadratic function, then evaluate each, considering x=8.
(1st) l=2x+4; h=x+3
(2nd) l=3x-9; h=x+2
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Formula for a rectangle's area:
A=lh
Given:
x=8
Substitute for:
(1st) $$\displaystyle{\left({2}{x}+{4}\right)}{\left({x}+{3}\right)}={20}\times{11}={220}$$
(2nd) $$\displaystyle{\left({3}{x}-{9}\right)}{\left({x}+{2}\right)}={15}\times{10}={150}$$ | 2022-01-21 03:09:13 | {"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": 0.899849534034729, "perplexity": 4342.669371815165}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320302715.38/warc/CC-MAIN-20220121010736-20220121040736-00648.warc.gz"} |
http://www.alexday.me/posts/github-actions/ | Easy code is easy to compile and run. That has and always will be true. However, once the code you write spans across multiple classes, files, or even packages it can be hard to properly test, compile, and release this software. Continuous integration (CI) tries to solve this problem. By defining a pipeline of actions to take your code from source to product that run the same way every time. Docker (and containerization in general) augments this process by providing easy to use clean-slate images that these pipelines can built up on. Always starting at the same point and then running the same actions provides a stable pipeline that can reliably test, compile, and release a piece of software. This, in turn, provides a reliable platform that allows developers to more easily solve more problems.
## GitHub Actions
There are umpteen solutions for continuous integration. This includes Jenkins, Tekton, and GitHub Actions. Over the past couple of weeks I have been using GitHub Actions (first for work and most recently for this website) and I have found that it drastically increases my productivity by reducing the time I spend on either producing a wheel, running pytest, or generating and uploading this Hugo-driven site.
### Pipeline
GitHub uses YAML files to define a CI pipeline. A pipeline is a set of actions in an explicit order. These actions can either be a command to the operating system (e.g. echo "Hello World" or curl www.google.com) or a group of commands developed by someone (e.g. checking out a repo or generating a hugo site). All these actions run on a fresh instance of a virtual machine. There are two choices when it comes to these virtual machines. You can either host your own virtual machine, or use one that GitHub hosts (GitHub hosted platforms are Windows Server 2019, Ubuntu 18.04 and 16.04, and MacOS Catalina).
Below is the pipeline for this site. The pipeline's environment is Ubuntu 18.04 and there are four actions this pipeline takes. The first action (actions/checkout@v2) checks the site repo out. Because the definition for this workflow is in the devel branch that is the branch that this action checks out. The next action (peaceiris/actions-hugo@v2) installs Hugo v0.68.3. The site is then built using a simple action that executes hugo --minify. This builds the site into the public directory. The last action (peaceiris/actions-gh-pages) publishes the public directory into the master branch and adds the CNAME file that allows me to use my own domain name.
# .github/actions/main.yml
name: Build Hugo
on:
push:
branches:
- devel
jobs:
deploy:
runs-on: ubuntu-18.04
steps:
- uses: actions/checkout@v2
- name: Setup Hugo
uses: peaceiris/actions-hugo@v2
with:
hugo-version: '0.68.3'
- name: Build
run: hugo --minify
- name: Deploy
uses: peaceiris/actions-gh-pages@v3
with:
## Conclusions
GitHub Actions are an amazing addition to the platform. They allow small projects to enable a free and easy continuous integration pipeline while also being able to scale to larger projects. While it is not the end-all-be-all for CI, it is certainly a good jumping off point if you want to start investigating this world. There are some features that were not discussed in this post (namely artifacts as I haven't played with them much), but more information is available in the documentation. | 2020-05-29 11:06:38 | {"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": 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": 0.17195729911327362, "perplexity": 2120.2472536959312}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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-20200529085930-20200529115930-00017.warc.gz"} |
https://en.academic.ru/dic.nsf/enwiki/1465407 | # RE (complexity)
RE (complexity)
In computability theory and computational complexity theory, RE (recursively enumerable) is the class of decision problems for which a 'yes' answer can be verified by a Turing machine in a finite amount of time.[1] Informally, it means that if the answer is 'yes', then there is some procedure which takes finite time to determine this. On the other hand, if the answer is 'no', the machine might never halt. Equivalently, RE is the class of decision problems for which a Turing machine can list all the 'yes' instances, one by one (this is what 'enumerable' means).
Similarly, co-RE is the set of all languages that are complements of a language in RE. In a sense, co-RE contains languages of which membership can be disproved in a finite amount of time, but proving membership might take forever.
Each member of RE is a recursively enumerable set and therefore a Diophantine set.
## Relations to other classes
The set of recursive languages (R) is a subset of both RE and co-RE.[2] In fact, it is the intersection of those two classes:
$\mbox{R} = \mbox{RE}\cap\mbox{co-RE}.$
## RE-complete
RE-complete is the set of decision problems that are complete for RE. In a sense, these are the "hardest" recursively enumerable problems. All such problems are nonrecursive. Generally, no constraint is placed on the reductions used except that they must be many-one reductions.
Examples of RE-complete problems:
1. Halting problem: Whether a program given a finite input finishes running or will run forever.
2. By Rice's Theorem, deciding membership in any nontrivial subset of the set of recursive functions is RE-hard. It will be complete whenever the set is recursively enumerable.
3. [Myhill 1955][3] has proven that all creative sets are RE-complete.
4. The uniform word problem for groups or semigroups. [Indeed, the word problem for some individual groups is RE-complete.]
5. Deciding membership in a general unrestricted formal grammar. [Again, certain individual grammars have RE-complete membership problem.]
6. The validity problem for first-order logic.
7. Post correspondence problem: Given a finite set of strings, determine if there is a string that can be factored into a composition of the strings (allowing repeats) in two different ways.
8. Determining if a Diophantine equation has any integer solutions.
## co-RE-complete
co-RE-complete is the set of decision problems that are complete for co-RE. In a sense, these are the complements of the hardest recursively enumerable problems.
Examples of co-RE-complete problems:
1. The Domino Problem for Wang tiles.
2. The satisfiability problem for first-order logic
List of undecidable problems
## References
1. ^ Complexity Zoo: Class RE
2. ^ Complexity Zoo: Class co-RE
3. ^ Myhill, J. "Creative sets," Z. Math. Logik Grundlag. Math., v. 1, pp. 97–108.
Wikimedia Foundation. 2010.
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• complexity — complexity. См. сложность генома. (Источник: «Англо русский толковый словарь генетических терминов». Арефьев В.А., Лисовенко Л.А., Москва: Изд во ВНИРО, 1995 г.) … Молекулярная биология и генетика. Толковый словарь.
• complexity — index complication, confusion (turmoil), enigma, entanglement (confusion), imbroglio, impasse … Law dictionary | 2019-12-12 01:09:50 | {"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": 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": 0.6544138789176941, "perplexity": 4413.878834223875}, "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-2019-51/segments/1575540534443.68/warc/CC-MAIN-20191212000437-20191212024437-00507.warc.gz"} |
https://www.picostat.com/dataset/r-dataset-package-daag-appletaste | # R Dataset / Package DAAG / appletaste
Webform
Category
Webform
Category
Webform
Category
Webform
Category
Webform
Category
Webform
Category
## Visual Summaries
Embed
<iframe src="https://embed.picostat.com/r-dataset-package-daag-appletaste.html" frameBorder="0" width="100%" height="307px" />
Attachment Size
820 bytes
Dataset Help
Documentation
## Tasting experiment that compared four apple varieties
### Description
Each of 20 tasters each assessed three out of the four varieties. The experiment was conducted according to a balanced incomplete block design.
### Usage
data(appletaste)
### Format
A data frame with 60 observations on the following 3 variables.
aftertaste
a numeric vector
Apple samples were rated for aftertaste, by making a mark on a continuous scale that ranged from 0 (extreme dislike) to 150 (like very much).
panelist
a factor with levels a b c d e f g h i j k l m n o p q r s t
product
a factor with levels 298 493 649 937
### Examples
data(appletaste)
appletaste.aov <- aov(aftertaste ~ panelist + product, data=appletaste)
termplot(appletaste.aov)
--
Dataset imported from https://www.r-project.org.
Recent Queries For This Dataset
No queries made on this dataset yet. | 2020-10-29 21:53:44 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.345537394285202, "perplexity": 7267.97153639673}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1603107905965.68/warc/CC-MAIN-20201029214439-20201030004439-00381.warc.gz"} |
https://www.semanticscholar.org/paper/New-physics-effects-in-tree-level-decays-and-the-in-Brod-Lenz/38a878003f354785c394a1389e4434bb11effac6 | # New physics effects in tree-level decays and the precision in the determination of the quark mixing angle γ
@article{Brod2015NewPE,
title={New physics effects in tree-level decays and the precision in the determination of the quark mixing angle $\gamma$},
author={Joachim Brod and Alexander Lenz and Gilberto Tetlalmatzi-Xolocotzi and Martin Wiebusch},
journal={Physical Review D},
year={2015},
volume={92},
pages={033002}
}
• Published 3 December 2014
• Physics
• Physical Review D
We critically review the assumption that no new physics is acting in tree-level B-meson decays and study the consequences for the ultimate precision in the direct determination of the Cabibbo-Kobayashi-Maskawa (CKM) angle γ. In our exploratory study we find that sizeable universal new physics contributions, ΔC1,2, to the tree-level Wilson coefficients C1,2 of the effective Hamiltonian describing weak decays of the b quark are currently not excluded by experimental data. In particular, we find…
84 Citations
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## References
SHOWING 1-10 OF 58 REFERENCES
• Physics
• 2011
We update the Standard-Model predictions for several quantities related to Bs−B¯s and Bd−B¯d mixing. The mass and width differences in the Bs system read ΔMSMs=(17.3±2.6)ps−1 and
$B -> X_d \gamma$ and constraints on new physics
• Physics
• 2011
We combine recent progress in measuring the branching ratio of the decay B->X_d gamma$with the discovery that hadronic uncertainties in the CP-averaged branching ratio drop out to a large extent. New physics in $$b\rightarrow s$$b→s transitions after LHC run 1 • Physics • 2014 We present results of global fits of all relevant experimental data on rare $$b \rightarrow s$$b→s decays. We observe significant tensions between the Standard Model predictions and the data. After Measurement of the semileptonic charge asymmetry in$B^0$meson mixing with the D0 detector • Physics • 2012 We present a measurement of the semileptonic mixing asymmetry for B0 mesons, a^d_{sl}, using two independent decay channels: B0 -> mu+D-X, with D- -> K+pi-pi-; and B0 -> mu+D*-X, with D*- -> antiD0 Precision measurement of CP violation in B(S)(0)→J/ΨK+K- decays. • Physics Physical review letters • 2015 For the first time, the phase ϕs is measured independently for each polarization state of the K+K- system and shows no evidence for polarization dependence. Weak annihilation and new physics in charmless$B \to M M$decays • Physics • 2014 We use currently available data of nonleptonic charmless 2-body$B\to MM$decays ($MM = PP, PV, VV$) that are mediated by$b\to (d, s)\$ QCD- and QED-penguin operators to study weak annihilation and
On new physics in ΔΓd
• Physics
• 2014
A bstractMotivated by the recent measurement of the dimuon asymmetry by the DØ collaboration, which could be interpreted as an enhanced decay rate difference in the neutral Bd-meson system, we
Search for CP violation in B0-B0 mixing using partial reconstruction of B0→D*- Xℓ+ νℓ and a kaon tag.
• Physics
Physical review letters
• 2013
A search for CP violation in B0- B0 mixing with the BABAR detector finds the CP violating asymmetry ACP, corresponding to ΔCP=1-|q/p|=(0.29±0.84(-1.61)(+1.88))×10(-3). | 2022-05-19 08:30:35 | {"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": 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": 0.8232336044311523, "perplexity": 5065.102208878939}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1652662526009.35/warc/CC-MAIN-20220519074217-20220519104217-00515.warc.gz"} |
https://www.reddit.com/r/math/comments/bos31k/whats_the_most_abstract_mathematics_youve/ | Press J to jump to the feed. Press question mark to learn the rest of the keyboard shortcuts
2
Posted by6 days ago
## What’s the most abstract mathematics you’ve encountered?
I’d probably have to say higher category theory. If you’re a working category theorist, what’s a part of category theory that’s really out there?
60% Upvoted
Sort by
level 1
9 points · 6 days ago
I'd happily argue that the concept of an abstract derivation in operator algebras is just as abstract as anything anyone in category theory runs across.
Ofc, there is a reasonably straightforward dictionary to translate back and forth between OA and CT so, to a category theorist, they ought to be somehow one and the same anyway.
level 2
Algebraic Geometry3 points · 5 days ago
I legit don't know what you mean in your last paragraph. Can you explain?
level 2
2 points · 5 days ago
Can you explain what a deviation is or point towards a resource on it?
level 3
Algebraic Topology3 points · 5 days ago
A derivation is just something that satisfies the same product law derivatives do. So f is a derivation if f(ab)=f(a)b+af(b).
One important application of this formula is that they basically characterize partial derivatives. If you ask for all the derivations of smooth functions at a point, it is a vector space spanned by all the partial derivatives.
level 4
6 points · 5 days ago
So it's a tangent vector in the language of differential geometry, only possibly generalized to all sorts of other weird structures?
level 4
1 point · 5 days ago
A theorem of Tarasov proves that any fractional differintegral operator consistent with integer order differentiation cannot simultaneously satisfy R-Linearity, the product rule and the derivative of a constant is zero.
A professor of mine once told me that he viewed studying fractional calculus as kind of a waste of time because it was better to just study operator algebras instead, but I never got around to it as my interests moved elsewhere.
I'm curious how interesting or uninteresting the above result is from the perspective of someone in operator algebras.
level 1
Algebraic Topology6 points · 5 days ago
Well -- a lot of questions in higher category theory, especially recently, are motivated by extended topological field theory. This is ... also pretty abstract, but various aspects of it, including the higher-categorical ones, keep showing up in the physicists' papers!
Some of the others are still just mathematics, but have been useful for formalizing things in physics. For example, the classification of invertible extended topological field theories goes through a paper of Schommer-Pries really using (∞, n)-categories, and the result (less categorical) is used in theoretical physics papers.
level 1
5 points · 5 days ago
As someone who’s into analysis, stuff is usually quite close to the ground and non abstract. But ergodic theory.. that stuff is messed up man.
level 1
Mathematical Physics3 points · 5 days ago
Model theory maybe.
But I'm sure just pure logic is almost by definition the most abstract mathematics.
level 1
3 points · 5 days ago
C-systems, which Voevodsky was using to provide a formal model for the structure underlying type theories.
And these type theories, Martin-Löf dependent type theory in particular, was to provide a foundation for mathematics, assuming additionally the univalence axiom.
So it's some category theory to model some type theory, which was to provide a foundation for mathematics that could be done in proof assistant software.
level 1
1 point · 5 days ago
I'm studying Group Theory.. Is it gonna help with Category Theory later?
level 2
1 point · 5 days ago
It is gonna help with (almost) everything later.
level 2
1 point · 3 days ago
It is going to help motivate category theory. All math needs group theory at some point so it is definitely worth learning.
level 1
1 point · 5 days ago
Representation Theory
level 1
1 point · 5 days ago
ricci flow with surgery is crazy as well
level 1
Category Theory1 point · 5 days ago
The Ext-functor. It is that weird child of diagram chase and group theory...
level 1
Control Theory/Optimization0 points · 5 days ago
category theory i guess
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Dynamical Systems | 2019-05-21 02:41:59 | {"extraction_info": {"found_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": 0.4048108756542206, "perplexity": 7977.137792390602}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232256215.47/warc/CC-MAIN-20190521022141-20190521044141-00473.warc.gz"} |
https://www.gradesaver.com/textbooks/math/precalculus/precalculus-6th-edition-blitzer/chapter-10-section-10-5-the-binomial-theorum-exercise-set-page-1092/55 | ## Precalculus (6th Edition) Blitzer
The middle term of the expansion of ${{\left( \frac{3}{x}+\frac{x}{3} \right)}^{10}}$ is $252$. | 2021-05-17 12:39:56 | {"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": 0.8175835609436035, "perplexity": 781.4756551202157}, "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-2021-21/segments/1620243991772.66/warc/CC-MAIN-20210517115207-20210517145207-00149.warc.gz"} |
https://www.questionsolutions.com/effects-atmospheric-resistance-accounted/ | # If the effects of atmospheric resistance are accounted for
If the effects of atmospheric resistance are accounted for, a falling body has an acceleration defined by the equation $a = 9.81[1-v^2(10^{-4})]$ m/$s^2$, where is v in m/s and the positive direction is downward. If the body is released from rest at a very high altitude, determine (a) the velocity when t = 5 s, and (b) the body’s terminal or maximum attainable velocity (as $t\rightarrow\infty$).
#### Solution:
Remember that we can write acceleration as:
$a(v)=\dfrac{dv}{dt}$
$dt=\dfrac{dv}{a(v)}$
Let us take the integral of both sides of the equation:
$\,\displaystyle \int^{t_2}_{t_1}dt=\int^{v}_{v_0}\dfrac{1}{a(v)}dv$
Substitute our acceleration equation:
$\,\displaystyle \int^{t}_{0}dt=\int^{v}_{0}\dfrac{1}{9.81[1-v^2(10^{-4})]}dv$
$t=\dfrac{50}{9.81}\text{ln}\left(\dfrac{1+0.01v}{1-0.01v}\right)$
Isolate for v:
$v=\dfrac{100(e^{0.1962t}-1)}{e^{0.1962t}+1}$
At t=5 seconds, we have:
$v=\dfrac{100(e^{(0.1962)(5)}-1)}{e^{(0.1962)(5)}+1}$
$v=45.5$ m/s
As $t\rightarrow\infty$, we have:
$v=100$ m/s
$v=45.5$ m/s
Maximum velocity, $v=100$ m/s | 2022-12-09 04:42:31 | {"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": 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": 15, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7937896847724915, "perplexity": 1048.3468311831848}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1669446711390.55/warc/CC-MAIN-20221209043931-20221209073931-00733.warc.gz"} |
https://themodularperspective.com/2019/02/25/the-arithmetic-of-polynomial-rings-over-finite-fields-theyre-like-the-integers/ | # The Arithmetic of Polynomial Rings over Finite Fields
Today we’re going to talk about something quite exciting and unexpected: the arithmetic of polynomial rings over finite fields and its similarity to the arithmetic of the integers. We’ll first run though some preliminary observations about the arithmetic of the polynomial rings which will suggest that it behaves similar to the arithmetic of the integers. Then we’ll look at analogues of the $\phi$-function as well as the classic theorems of Euler and Fermat, and conclude with some discussion of a zeta-function over the polynomial rings.
As usual, let $\mathbb{F}_{q}[T]$ be the polynomial ring in $T$ over a finite field $\mathbb{F}_{q}$. We note here that the role of positive integers in $\mathbb{F}_{q}[T]$ are played by monic polynomials, and prime integers are played by irreducible monic polynomials. In fact, we can interchange “irreducible” with “prime” because like $\mathbb{Z}$, $\mathbb{F}_{q}[T]$ is also a PID. The proof for both cases are standard facts in an introductory abstract algebra course, and are even analogous; they both deal with looking at a minimal element of an ideal, with respect to a norm, and then using the division algorithm. Both rings also happen to have infinitely many primes. The proofs here are analogous as well, if you use Euclid’s argument for infinitely many primes. Moreover, the residue class rings modulo a non-zero ideal of both rings are finite. This follows quickly by using the division algorithm to find a complete set of representatives for the residue class ring. If we restrict the ideal to be generated by a prime element, then the residue class rings become residue class fields in both cases. There is also the discussion of units. Both $\mathbb{Z}$ and $\mathbb{F}_{q}[T]$ have a finite system of units, the former being $\{\pm1\}$ and the latter being $\mathbb{F}_{q}^{*}$ by degree considerations. It is here that we see the role of $2$ being played by $q-1$ in $\mathbb{F}_{q}[T]$. These basic properties suggest that the arithmetic of $\mathbb{F}_{q}^{*}$ should be similar to that of $\mathbb{Z}$. As said above, this is the case. However, it is surprisingly easier to understand and requires the use of much less machinery than $\mathbb{Z}$. For example, there is an analogous zeta-function for $\mathbb{F}_{q}[T]$, and therefore an analogous Riemann Hypothesis. Here is where $\mathbb{F}_{q}[T]$ differs from $\mathbb{Z}$, because we have proven that the hypothesis is true for $\mathbb{F}_{q}[T]$. Before digging myself into that hole too quickly, let’s talk about some simpler analogous number theoretic statements in $\mathbb{F}_{q}[T]$.
#### The $\phi$-function, Euler, and Fermat
Before looking at analogues to the $\phi$-function and theorems of Euler and Fermat there is a small technical note to be made aware of. If $f \in \mathbb{F}_{q}[T]$, then we define $|f| = q^{deg(f)}$. By finding a full set of representatives for $A/fA$ using the division algorithm it’s easy to check that the order of $A/fA$ is $|f|$.
Let’s start with Euler’s $\phi$-function. Euler proved a product formula for $\phi(n)$, rightly named Euler’s product formula, which says
$\displaystyle{\phi(n) = n\prod_{p|n}(1-\frac{1}{p})}$,
where $p$ is a prime. The proof rests on the fact that the $\phi$-function is multiplicative and that for a prime $p$ and positive integer $k$ we have $\phi(p^{k}) = p^{k}(1-\frac{1}{p})$. The analogous function for $\mathbb{F}_{q}[T]$ is $\Phi$ where $\Phi(f)$, for $f$ non-zero, returns the order of the group $(A/fA)^{*}$ or equivalently the number of polynomials of degree less than $f$ and relatively prime to $f$. The analogous product formula is
$\displaystyle{\Phi(f) = |f|\prod_{P|f}(1-\frac{1}{|P|})}$,
where $P$ is an irreducible monic polynomial. This looks strikingly similar to the Euler product formula above, and keeping in a similar fashion as before, the proof is somewhat analogous. The Chinese remainder theorem in $\mathbb{F}_{q}[T]$ is used to show $\Phi$ is multiplicative. Yet, we need a different argument to compute $\Phi(P^{k})$. It is essentially a counting argument; the ring $A/P^{k}A$ is shown to be local, and so by counting the complement of a maximal ideal in $A/P^{k}A$ we have also counted $(A/P^{k}A)^{*}$. Like in the case of the Euler product formula, the proof follows easily from these two facts.
We also have an analogues to the famous number theory statements of Euler and Fermat. In $\mathbb{F}_{q}[T]$ Euler’s theorem takes the form
$g^{\Phi(f)} \equiv 1 \pmod{fA}$,
where $f \in \mathbb{F}_{q}[T]$, $f \ne 0$, and $g$ is relatively prime to $f$. As usual, the proof is completely analogous since the order of $(A/fA)^{*}$ is by definition $latex \Phi(f)$, and since $g$ is relatively prime to $f$ it’s an element of $(A/fA)^{*}$ (this last statement follows from the Chinese remainder theorem in $\mathbb{F}_{q}[T]$). If we take $f$ to be some irreducible monic polynomial $P$ then we obtain the analogue to Fermat’s little theorem:
$g^{|P|-1} \equiv 1 \pmod{PA}$.
Indeed, $\Phi(P)$ is the order of $(A/PA)^{*}$ which is \$latex $|P|-1$ because $A/PA$ is a field as we have discussed above.
#### The Zeta-Function $\zeta_{\mathbb{F}_{q}[T]}$
As mentioned, there is an analogue to the zeta-function in $\mathbb{Z}$. In $\mathbb{F}_{q}[T]$, the zeta-function $\zeta_{\mathbb{F}_{q}[T]}$ on complex numbers $s$ with $\mathcal{R}(s) > 1$ is defined as
$\displaystyle{\zeta_{\mathbb{F}_{q}[T]}(s) := \sum_{\substack{f \in \mathbb{F}_{q}[T] \\ f monic}}\frac{1}{|f|^{s}} = \prod_{P}(1-\frac{1}{|P|^{s}})^{-1}} = \frac{1}{1-q^{1-s}}$.
The product form of $\zeta_{\mathbb{F}_{q}[T]}(s)$ can be obtained in the exact same style of how Riemann did for the usual zeta-function by sieving out multiples of $\frac{1}{|P|^{s}}$ for all monic irreducible polynomials $P$. The closed form expression can be deduced by noticing that the summation form is an infinite geometric series by using the definition of $|f|$ and collecting terms by degree. Similar to the usual zeta-function $\zeta$, $\zeta_{\mathbb{F}_{q}[T]}$ can be extended to a meromorphic function on $\mathbb{C}$ with a simple pole at $1$. However, $\zeta_{\mathbb{F}_{q}[T]}$ is significantly more simple to work with.
That’s all for today. I hope to continue this topic in the future with a heavier discussion about $\zeta_{\mathbb{F}_{q}[T]}$, so keep an eye out. Thanks for reading!
References
Number Theory in Function Fields – Michael Rosen | 2019-05-21 06:50:32 | {"extraction_info": {"found_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": 87, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.903661847114563, "perplexity": 117.79389570627076}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232256281.35/warc/CC-MAIN-20190521062300-20190521084300-00176.warc.gz"} |
https://cracku.in/5-in-triangle-abc-altitudes-ad-and-be-are-drawn-to-t-x-cat-2022-slot-2-quant | Question 5
# In triangle ABC, altitudes AD and BE are drawn to the corresponding bases. If $$\angle BAC = 45^{\circ}$$ and $$\angle ABC=\theta\$$, then $$\frac{AD}{BE}$$ equals
Solution
It is given, Angle BAE = 45 degrees
This implies AE = BE
Let AE = BE = x
In right-angled triangle ABD, it is given $$\angle ABC=\theta\$$
$$\sin\theta=\frac{AD}{AB}\$$
$$\sin\theta=\frac{AD}{x\sqrt{\ 2}}\$$
$$\sqrt{\ 2}\sin\theta=\frac{AD}{BE}\$$ | 2023-04-01 10:18:00 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.8546053767204285, "perplexity": 2442.062005386504}, "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-2023-14/segments/1679296949958.54/warc/CC-MAIN-20230401094611-20230401124611-00474.warc.gz"} |
https://biblioteca.portalbolsasdeestudo.com.br/?q=ARBITRAGE&pagination=2 | Página 2 dos resultados de 1137 itens digitais encontrados em 0.004 segundos
## Intertemporal arbitrage, speculative balances, and the liquidity effect
Fortowsky, Elaine Barbara
Português
Relevância na Pesquisa
27.32%
This thesis explores money manager intertemporal arbitrage as an explanation of the liquidity effect. We develop a theoretical model of optimal portfolio adjustment by professional money managers, and show that they engage in intertemporal asset price arbitrage; they reduce their holdings of financial assets when they expect asset prices to fall, and increase their holdings when they expect prices to rise. Since a reduction in financial assets can only be accomplished through an increase in money holdings, a connection exists between intertemporal price arbitrage and speculative balances. We show that in equilibrium, money manager behavior causes market liquidity shocks to be accompanied by a form of asset price overshooting in which asset prices first rise above their long-run value and then slowly return as speculative balances are lent out to borrowers and absorbed into transactions balances. Such asset price overshooting is precisely the liquidity effect, stated in terms of asset prices rather than interest rates. This shadows the result established by Hartley (1990), who showed that the combination of sector-specific liquidity shocks and trading rigidities across sectors will cause general price overshooting in those sectors closest to the money supply injection. The second part of this thesis attempts to identify an empirical relationship between speculative balances and asset prices as a means of verifying the hypothesis that money manager intertemporal price arbitrage generates the liquidity effect. It is not possible to estimate this relationship on an aggregate level because no means exist to identify speculative balances relative to the total money supply. However...
## Nonparametric and arbitrage-free construction of call surfaces using l1-recovery
Tipo: Artigo de Revista Científica
Relevância na Pesquisa
27.32%
This paper is devoted to the application of an l1-minimisation technique to construct an arbitrage-free call-option surface. We propose a novel approach to obtaining model-free call option surfaces that are perfectly consistent with market quotes and free of static arbitrage. The approach is inspired from the compressed-sensing framework that is used in signal processing to deal with under-sampled signals. We address the problem of fitting the call-option surface to sparse option data. To illustrate the methodology, we first apply the methodology to a slice of the call-option surface of the S&P500 to recover a finer slice free of butterfly arbitrage and matching the market quotes. We then proceed to the construction of the whole call-price surface of the S&P500 options, taking into account the arbitrage possibilities in the time direction. The resulting object is a surface free of both butterfly and calendar-spread arbitrage that matches the original market points.; Comment: 16 pages, 9 figures
## Characterization of arbitrage-free markets
Strasser, Eva
Tipo: Artigo de Revista Científica
Relevância na Pesquisa
27.32%
The present paper deals with the characterization of no-arbitrage properties of a continuous semimartingale. The first main result, Theorem \refMainTheoremCharNA, extends the no-arbitrage criterion by Levental and Skorohod [Ann. Appl. Probab. 5 (1995) 906-925] from diffusion processes to arbitrary continuous semimartingales. The second main result, Theorem 2.4, is a characterization of a weaker notion of no-arbitrage in terms of the existence of supermartingale densities. The pertaining weaker notion of no-arbitrage is equivalent to the absence of immediate arbitrage opportunities, a concept introduced by Delbaen and Schachermayer [Ann. Appl. Probab. 5 (1995) 926-945]. Both results are stated in terms of conditions for any semimartingales starting at arbitrary stopping times \sigma. The necessity parts of both results are known for the stopping time \sigma=0 from Delbaen and Schachermayer [Ann. Appl. Probab. 5 (1995) 926-945]. The contribution of the present paper is the proofs of the corresponding sufficiency parts.; Comment: Published at http://dx.doi.org/10.1214/105051604000000558 in the Annals of Applied Probability (http://www.imstat.org/aap/) by the Institute of Mathematical Statistics (http://www.imstat.org)
## Double Exponential Instability of Triangular Arbitrage Systems
Cross, Rod; Kozyakin, Victor
Tipo: Artigo de Revista Científica
Português
Relevância na Pesquisa
27.32%
If financial markets displayed the informational efficiency postulated in the efficient markets hypothesis (EMH), arbitrage operations would be self-extinguishing. The present paper considers arbitrage sequences in foreign exchange (FX) markets, in which trading platforms and information are fragmented. In Kozyakin et al. (2010) and Cross et al. (2012) it was shown that sequences of triangular arbitrage operations in FX markets containing 4 currencies and trader-arbitrageurs tend to display periodicity or grow exponentially rather than being self-extinguishing. This paper extends the analysis to 5 or higher-order currency worlds. The key findings are that in a 5-currency world arbitrage sequences may also follow an exponential law as well as display periodicity, but that in higher-order currency worlds a double exponential law may additionally apply. There is an "inheritance of instability" in the higher-order currency worlds. Profitable arbitrage operations are thus endemic rather that displaying the self-extinguishing properties implied by the EMH.; Comment: 22 pages, 22 bibliography references, expanded Introduction and Conclusion, added bibliohraphy references
## Universal Arbitrage Aggregator in Discrete Time Markets under Uncertainty
Burzoni, Matteo; Frittelli, Marco; Maggis, Marco
Tipo: Artigo de Revista Científica
Português
Relevância na Pesquisa
27.42%
In a model independent discrete time financial market, we discuss the richness of the family of martingale measures in relation to different notions of Arbitrage, generated by a class $\mathcal{S}$ of significant sets, which we call Arbitrage de la classe $\mathcal{S}$. The choice of $\mathcal{S}$ reflects into the intrinsic properties of the class of polar sets of martingale measures. In particular: for S=${\Omega}$ absence of Model Independent Arbitrage is equivalent to the existence of a martingale measure; for $\mathcal{S}$ being the open sets, absence of Open Arbitrage is equivalent to the existence of full support martingale measures. These results are obtained by adopting a technical filtration enlargement and by constructing a universal aggregator of all arbitrage opportunities. We further introduce the notion of market feasibility and provide its characterization via arbitrage conditions. We conclude providing a dual representation of Open Arbitrage in terms of weakly open sets of probability measures, which highlights the robust nature of this concept.
## Volatility smile and stochastic arbitrage returns
Fedotov, Sergei; Panayides, Stephanos
Tipo: Artigo de Revista Científica
Relevância na Pesquisa
27.32%
The purpose of this work is to explore the role that random arbitrage opportunities play in pricing financial derivatives. We use a non-equilibrium model to set up a stochastic portfolio, and for the random arbitrage return, we choose a stationary ergodic random process rapidly varying in time. We exploit the fact that option price and random arbitrage returns change on different time scales which allows us to develop an asymptotic pricing theory involving the central limit theorem for random processes. We restrict ourselves to finding pricing bands for options rather than exact prices. The resulting pricing bands are shown to be independent of the detailed statistical characteristics of the arbitrage return. We find that the volatility smile'' can also be explained in terms of random arbitrage opportunities.; Comment: 15 pages, 3 figures. The paper was accepted for publication in Physica A
## Arbitrage theory without a num\'eraire
Tehranchi, Michael R.
Tipo: Artigo de Revista Científica
Português
Relevância na Pesquisa
27.32%
This note develops an arbitrage theory for a discrete-time market model without the assumption of the existence of a num\'eraire asset. Fundamental theorems of asset pricing are stated and proven in this context. The distinction between the notions of investment-consumption arbitrage and pure-investment arbitrage provide a discrete-time analogue of the distinction between the notions of absolute arbitrage and relative arbitrage in the continuous-time theory. Applications to the modelling of bubbles is discussed.; Comment: 27 pages
## Diversity and no arbitrage
Herczegh, Attila; Prokaj, Vilmos; Rásonyi, Miklós
Tipo: Artigo de Revista Científica
Português
Relevância na Pesquisa
27.32%
A stock market is called diverse if no stock can dominate the market in terms of relative capitalization. On one hand, this natural property leads to arbitrage in diffusion models under mild assumptions. On the other hand, it is also easy to construct diffusion models which are both diverse and free of arbitrage. Can one tell whether an observed diverse market admits arbitrage? In the present paper we argue that this may well be impossible by proving that the known examples of diverse markets in the literature (which do admit arbitrage) can be approximated uniformly (on the logarithmic scale) by models which are both diverse and arbitrage-free.; Comment: 14 pages, final version
## Statistical Arbitrage Mining for Display Advertising
Zhang, Weinan; Wang, Jun
Tipo: Artigo de Revista Científica
Relevância na Pesquisa
27.32%
We study and formulate arbitrage in display advertising. Real-Time Bidding (RTB) mimics stock spot exchanges and utilises computers to algorithmically buy display ads per impression via a real-time auction. Despite the new automation, the ad markets are still informationally inefficient due to the heavily fragmented marketplaces. Two display impressions with similar or identical effectiveness (e.g., measured by conversion or click-through rates for a targeted audience) may sell for quite different prices at different market segments or pricing schemes. In this paper, we propose a novel data mining paradigm called Statistical Arbitrage Mining (SAM) focusing on mining and exploiting price discrepancies between two pricing schemes. In essence, our SAMer is a meta-bidder that hedges advertisers' risk between CPA (cost per action)-based campaigns and CPM (cost per mille impressions)-based ad inventories; it statistically assesses the potential profit and cost for an incoming CPM bid request against a portfolio of CPA campaigns based on the estimated conversion rate, bid landscape and other statistics learned from historical data. In SAM, (i) functional optimisation is utilised to seek for optimal bidding to maximise the expected arbitrage net profit...
## The Mirage of Triangular Arbitrage in the Spot Foreign Exchange Market
Fenn, Daniel J.; Howison, Sam D.; McDonald, Mark; Williams, Stacy; Johnson, Neil F.
Tipo: Artigo de Revista Científica
Relevância na Pesquisa
27.38%
We investigate triangular arbitrage within the spot foreign exchange market using high-frequency executable prices. We show that triangular arbitrage opportunities do exist, but that most have short durations and small magnitudes. We find intra-day variations in the number and length of arbitrage opportunities, with larger numbers of opportunities with shorter mean durations occurring during more liquid hours. We demonstrate further that the number of arbitrage opportunities has decreased in recent years, implying a corresponding increase in pricing efficiency. Using trading simulations, we show that a trader would need to beat other market participants to an unfeasibly large proportion of arbitrage prices to profit from triangular arbitrage over a prolonged period of time. Our results suggest that the foreign exchange market is internally self-consistent and provide a limited verification of market efficiency.
## Transit Fare Arbitrage: Case Study of San Francisco Bay Area Rapid Transit (BART) System
Haque, Asif
Tipo: Artigo de Revista Científica
Relevância na Pesquisa
27.32%
Transit fare arbitrage is the scenario when two or more commuters agree to swap tickets during travel in such a way that total cost is lower than otherwise. Such arbitrage allows pricing inefficiencies to be explored and exploited, leading to improved pricing models. In this paper we discuss the basics of fare arbitrage through an intuitive pricing framework involving population density. We then analyze the San Francisco Bay Area Rapid Transit (BART) system to understand underlying inefficiencies. We also provide source code and comprehensive list of pairs of trips with significant arbitrage gain at github.com/asifhaque/transit-arbitrage. Finally, we point towards a uniform payment interface for different kinds of transit systems.
## Virtual Arbitrage Pricing Theory
Ilinski, Kirill
Tipo: Artigo de Revista Científica
Relevância na Pesquisa
27.32%
We generalize the Arbitrage Pricing Theory (APT) to include the contribution of virtual arbitrage opportunities. We model the arbitrage return by a stochastic process. The latter is incorporated in the APT framework to calculate the correction to the APT due to the virtual arbitrage opportunities. The resulting relations reduce to the APT for an infinitely fast market reaction or in the case where the virtual arbitrage is absent. Corrections to the Capital Asset Pricing Model (CAPM) are also derived.; Comment: Latex, 12 pages
## Simple arbitrage
Bender, Christian
Tipo: Artigo de Revista Científica
Relevância na Pesquisa
27.38%
We characterize absence of arbitrage with simple trading strategies in a discounted market with a constant bond and several risky assets. We show that if there is a simple arbitrage, then there is a 0-admissible one or an obvious one, that is, a simple arbitrage which promises a minimal riskless gain of \epsilon, if the investor trades at all. For continuous stock models, we provide an equivalent condition for absence of 0-admissible simple arbitrage in terms of a property of the fine structure of the paths, which we call "two-way crossing." This property can be verified for many models by the law of the iterated logarithm. As an application we show that the mixed fractional Black-Scholes model, with Hurst parameter bigger than a half, is free of simple arbitrage on a compact time horizon. More generally, we discuss the absence of simple arbitrage for stochastic volatility models and local volatility models which are perturbed by an independent 1/2-H\"{o}lder continuous process.; Comment: Published in at http://dx.doi.org/10.1214/11-AAP830 the Annals of Applied Probability (http://www.imstat.org/aap/) by the Institute of Mathematical Statistics (http://www.imstat.org)
## Stochastic arbitrage return and its implications for option pricing
Fedotov, Sergei; Panayides, Stephanos
Tipo: Artigo de Revista Científica
Relevância na Pesquisa
27.32%
The purpose of this work is to explore the role that arbitrage opportunities play in pricing financial derivatives. We use a non-equilibrium model to set up a stochastic portfolio, and for the random arbitrage return, we choose a stationary ergodic random process rapidly varying in time. We exploit the fact that option price and random arbitrage returns change on different time scales which allows us to develop an asymptotic pricing theory involving the central limit theorem for random processes. We restrict ourselves to finding pricing bands for options rather than exact prices. The resulting pricing bands are shown to be independent of the detailed statistical characteristics of the arbitrage return. We find that the volatility "smile" can also be explained in terms of random arbitrage opportunities.; Comment: 14 pages, 3 fiqures
## Critical transaction costs and 1-step asymptotic arbitrage in fractional binary markets
Cordero, Fernando; Perez-Ostafe, Lavinia
Tipo: Artigo de Revista Científica
Relevância na Pesquisa
27.46%
We study the arbitrage opportunities in the presence of transaction costs in a sequence of binary markets approximating the fractional Black-Scholes model. This approximating sequence was constructed by Sottinen and named fractional binary markets. Since, in the frictionless case, these markets admit arbitrage, we aim to determine the size of the transaction costs needed to eliminate the arbitrage from these models. To gain more insight, we first consider only 1-step trading strategies and we prove that arbitrage opportunities appear when the transaction costs are of order $o(1/\sqrt{N})$. Next, we characterize the asymptotic behavior of the smallest transaction costs $\lambda_c^{(N)}$, called "critical" transaction costs, starting from which the arbitrage disappears. Since the fractional Black-Scholes model is arbitrage-free under arbitrarily small transaction costs, one could expect that $\lambda_c^{(N)}$ converges to zero. However, the true behavior of $\lambda_c^{(N)}$ is opposed to this intuition. More precisely, we show, with the help of a new family of trading strategies, that $\lambda_c^{(N)}$ converges to one. We explain this apparent contradiction and conclude that it is appropriate to see the fractional binary markets as a large financial market and to study its asymptotic arbitrage opportunities. Finally...
## Gauge Invariance, Geometry and Arbitrage
Vazquez, Samuel E.; Farinelli, Simone
Tipo: Artigo de Revista Científica
Relevância na Pesquisa
27.38%
In this work, we identify the most general measure of arbitrage for any market model governed by It\^o processes. We show that our arbitrage measure is invariant under changes of num\'{e}raire and equivalent probability. Moreover, such measure has a geometrical interpretation as a gauge connection. The connection has zero curvature if and only if there is no arbitrage. We prove an extension of the Martingale pricing theorem in the case of arbitrage. In our case, the present value of any traded asset is given by the expectation of future cash-flows discounted by a line integral of the gauge connection. We develop simple strategies to measure arbitrage using both simulated and real market data. We find that, within our limited data sample, the market is efficient at time horizons of one day or longer. However, we provide strong evidence for non-zero arbitrage in high frequency intraday data. Such events seem to have a decay time of the order of one minute.; Comment: 45 pages, 15 figures
## Arbitrage Opportunities and their Implications to Derivative Hedging
Panayides, Stephanos
Tipo: Artigo de Revista Científica
Português
Relevância na Pesquisa
27.32%
We explore the role that random arbitrage opportunities play in hedging financial derivatives. We extend the asymptotic pricing theory presented by Fedotov and Panayides [Stochastic arbitrage return and its implication for option pricing, Physica A 345 (2005), 207-217] for the case of hedging a derivative when arbitrage opportunities are present in the market. We restrict ourselves to finding hedging confidence intervals that can be adapted to the amount of arbitrage risk an investor will permit to be exposed to. The resulting hedging bands are independent of the detailed statistical characteristics of the arbitrage opportunities.; Comment: 10 pages, 2 figures added references, corrected typos
## Market models with optimal arbitrage
Chau, Huy N.; Tankov, Peter
Tipo: Artigo de Revista Científica
Relevância na Pesquisa
27.49%
We construct and study market models admitting optimal arbitrage. We say that a model admits optimal arbitrage if it is possible, in a zero-interest rate setting, starting with an initial wealth of 1 and using only positive portfolios, to superreplicate a constant c>1. The optimal arbitrage strategy is the strategy for which this constant has the highest possible value. Our definition of optimal arbitrage is similar to the one in Fernholz and Karatzas (2010), where optimal relative arbitrage with respect to the market portfolio is studied. In this work we present a systematic method to construct market models where the optimal arbitrage strategy exists and is known explicitly. We then develop several new examples of market models with arbitrage, which are based on economic agents' views concerning the impossibility of certain events rather than ad hoc constructions. We also explore the concept of fragility of arbitrage introduced in Guasoni and Rasonyi (2012), and provide new examples of arbitrage models which are not fragile in this sense.
## Geometric Arbitrage Theory and Market Dynamics
Farinelli, Simone
Tipo: Artigo de Revista Científica
Português
Relevância na Pesquisa
27.38%
We have embedded the classical theory of stochastic finance into a differential geometric framework called Geometric Arbitrage Theory and show that it is possible to: --Write arbitrage as curvature of a principal fibre bundle. --Parameterize arbitrage strategies by its holonomy. --Give the Fundamental Theorem of Asset Pricing a differential homotopic characterization. --Characterize Geometric Arbitrage Theory by five principles and show they they are consistent with the classical theory of stochastic finance. --Derive for a closed market the equilibrium solution for market portfolio and dynamics in the cases where: -->Arbitrage is allowed but minimized. -->Arbitrage is not allowed. --Prove that the no-free-lunch-with-vanishing-risk condition implies the zero curvature condition. The converse is in general not true and additionally requires the Novikov condition for the instantaneous Sharpe Ratio Dynamics to be satisfied.
## The dynamics of financially constrained arbitrage
Gromb, Denis; Vayanos, Dimitri
Fonte: The London School of Economics and Political Science Systematic Risk Centre Publicador: The London School of Economics and Political Science Systematic Risk Centre
Tipo: Monograph; NonPeerReviewed Formato: application/pdf | 2020-09-20 08:25:56 | {"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": 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": 0.46679913997650146, "perplexity": 1843.951698592924}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1600400196999.30/warc/CC-MAIN-20200920062737-20200920092737-00782.warc.gz"} |
https://www.physicsforums.com/threads/neumann-functions.213797/ | # Neumann Functions
1. Feb 7, 2008
### castusalbuscor
So for an assignment I have to write a program to find the roots of the Neumann function $$N_{n}(x)$$. However the only Neumann function I have in my class notes is:
Which is not overly helpful, and its the only one that was "boxed" in class.
Any hints on how I can incorporate that into a computer program to find the roots would be great!
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
2. Feb 7, 2008
### Dick
What system are you using to write the program? If it's something like Mathematica you should find that the Neumann function is already defined (as a form of Bessel function).
3. Feb 7, 2008
### castusalbuscor
I've never used Mathematica, though there seems to be a lot of people mentioning it. The prof wants us to use either C/C++ or FORTRAN. I have some experience with C++ so thats what I would be using.
Also I would be compiling it in Unix/Linux, if that helps...
Found an equation on Wikipedia:
$$Y_{\alpha} = \frac{J_{\alpha}cos(\alpha \pi) - J_{- \alpha}}{sin(\alpha \pi)}$$
This one seems more promising, but not sure how to use it to find the first five roots for $$N_{1}$$, $$N_{2}$$, and $$N_{3}$$
4. Feb 7, 2008
### Dick
You should find built in functions in the C math libraries, things like jn and yn. Setting up decent approximations for transcendental functions like this is a job for a numerical analysis type person. Just finding roots once the functions are defined isn't so hard.
5. Feb 7, 2008
### castusalbuscor
Sounds doable.. will. report back with success or failure. | 2017-01-18 18:32:18 | {"extraction_info": {"found_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": 0.4138432741165161, "perplexity": 788.2014022944351}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1484560280310.48/warc/CC-MAIN-20170116095120-00550-ip-10-171-10-70.ec2.internal.warc.gz"} |
https://physics.stackexchange.com/questions/336449/if-two-commuting-observables-do-not-share-eigenspaces-cant-one-measurement-aff | # If two commuting observables do not share eigenspaces, can't one measurement affect the other due to an extra wave function collapse?
When we learn quantum mechanics, we are told that the only way to extract information from a system is to conduct measurements. We are told that if two observables commute then performing one measurement does not affect the outcome of the other.
This is how we are told measurements work: if $a$ is an eigenvalue of $\hat A$, then if we perform the measurement corresponding to $\hat A$, the probability of measuring $a$ is $\langle \psi| \hat P_a | \psi \rangle$ where $\hat P_a$ is the projection operator onto the $a$-eigenspace. $| \psi \rangle$ then collapses into the state $\frac{\hat P_a | \psi \rangle}{\sqrt{\langle \psi | \hat P_a |\psi \rangle}}$.
What if, however, we have two operators that commute but do not share eigenspaces? Take, for example, the two operators
$\hat A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{pmatrix}, \hat B = \begin{pmatrix} 3 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{pmatrix}.$
Certainly, $[\hat A, \hat B] = 0$. However, these operators do not share eigenspaces for any eigenvalues. Let's say we conduct the measurement $\hat B$ right before we conduct the measurement $\hat A$. When we measure $\hat B$, we will measure an eigenvalue $b$ with probability $\langle \psi| \hat P_b | \psi \rangle$ and the state will collapse into $\frac{\hat P_b | \psi \rangle}{\sqrt{\langle \psi | \hat P_b |\psi \rangle}}$. The probability to measure $a$ after this is just $\frac{\langle \psi | \hat P_a \hat P_b | \psi \rangle }{\langle \psi | \hat P_b | \psi \rangle }$. Afterwards, the state will collapse into $\frac{ \hat P_a \hat P_b | \psi \rangle}{\sqrt{\langle \psi | \hat P_a \hat P_b |\psi \rangle}}$.
Now, the probability of measuring some eigenvalue $a$ does not change depending on whether or not we measured $\hat B$ beforehand. This can be easily checked, as the probability of measuring an eigenvalue $a$ is just, summing over all possible $b$ measurements,
$\sum_b \langle \psi | \hat P_b| \psi\rangle \frac{\langle \psi | \hat P_a \hat P_b | \psi \rangle }{\langle \psi | \hat P_b | \psi \rangle } = \sum_b \langle \psi | \hat P_a \hat P_b | \psi \rangle = \langle \psi | \hat P_a | \psi \rangle$
which is the same as before. Indeed, measuring $\hat B$ before $\hat A$ does not affect the probability of measuring any particular $a$, which is what we were told in class.
HOWEVER, the collapsed state after the $\hat A$ measurement is proportional to $\hat P_a \hat P_b |\psi \rangle$, NOT proportional to $\hat P_a | \psi \rangle$. If $\hat A$ and $\hat B$ share no eigenspaces, then this will always be a different state then if we had conducted no $\hat B$ measurement at all. In other words, the fact that we conducted the $\hat B$ measurement lives on in the fact that our state has collapsed further than it would have otherwise. How do we know that we won't be able to detect this? For example, maybe we could wait some time, let the system evolve in some way, and then pick up some interference effect. How does this extra collapse of the wave function affect causality in QFT with regards to commuting space-like separated local operators? And why is this never talked about in introductory QM books?
Of course they share an eigenspace. Indeed the vectors $$\vert 1\rangle=\left(\begin{array}{c} 1 \\ 0 \\0 \end{array}\right) \qquad \vert 2\rangle=\left(\begin{array}{c} 0 \\ 1 \\0 \end{array}\right) \qquad \vert 3\rangle=\left(\begin{array}{c} 0\\ 0 \\1 \end{array}\right)$$ are each common eigenvectors of both operators. The fact they do not have the same eigenvalues does not affect the fact they are common eigenstates. | 2023-03-28 17:38:11 | {"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": 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": 0.9115312695503235, "perplexity": 156.68370757274167}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1679296948868.90/warc/CC-MAIN-20230328170730-20230328200730-00540.warc.gz"} |
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# On Security Assumptions Underpinning Recent Schnorr Threshold Schemes
## Chelsea Komlo | August 5, 2022
In this post, we discuss differences in the security assumptions underpinning four Schnorr threshold signature schemes. In particular, we will review the two-round FROST signing protocol by Komlo and Goldberg1 that we refer to as FROST1, as well as an optimized variant FROST2 by Crites, Komlo, and Maller2. We refer to these schemes in conjunction as FROST 1/2. We contrast these schemes with two three-round signing protocols: SimpleTSig, also by Crites, Komlo, and Maller2, as well as the three-round scheme by Lindell3, which we call Lindell22.
TLDR.
• FROST1/2 requires One-More Discrete Logarithm (OMDL) and Programmable Random Oracle Model (PROM) assumptions.
• SimpleTSig can be proven using only discrete logarithm (DLP) and PROM assumptions.
• Lindell22 can be proven using only DLP+PROM. The protocol employs the Fischlin Transform4 in lieu of Schnorr signatures for proofs of knowledge.
These assumptions refer only to the security of threshold signing and not to the distributed key generation process. Thanks to Elizabeth Crites and Mary Maller for feedback on this post.
Let's dig more into the details now.
## Part One: What are security models, and why do they matter?
Security models are a useful tool to allow for proving cryptography schemes while also indicating potential assumptions which may or may not hold in practice. Each security model encodes certain assumptions such as an adversary's capabilities, the ability to perfectly simulate certain functionality, or the properties of the underlying mathematical assumptions. For example:
• The Standard Model. The adversary is limited only by time and computational power.
• The Random Oracle Model (ROM). Assumes outputs from a hash function are indistinguishable from random values.
• The Programmable Random Oracle Mode (PROM). Allows the random oracle to be programmed by the execution environment (which runs the adversary and simulates responses to the adversary's oracle queries), with the restriction that the programming must be indistinguishable from all other truly random responses.
## Part Two: What are security assumptions, and why do they matter?
A security assumption simply states the assumed hardness to an adversary of some particular computational problem. For example:
• Discrete Logarithm Problem (DLP). Considered to be a "standard assumption" in cryptography. The problem is simple: given some challenge $Y$ that is in a group $G$ where $g$ is a generator of $G$, output the discrete logarithm relation $x$ between $Y$ and $g$, where $Y = g^x$.
• One More Discrete Logarithm Assumption (OMDL). OMDL was first introduced by Bellare et al.5 and proven secure6, and can be as follows: given $\ell +1$ discrete logarithm challenges $X_0 = g^{\alpha_0}, X_1 = g^{\alpha_1}, \dots, X_\ell = g^{\alpha_\ell}$ and access to a discrete logarithm solution oracle $\mathcal{O}_\text{dlsol} (X_i) \rightarrow \alpha_i$ which can be queried up to $\ell$ times, the challenge is to output $\ell+1$ discrete logarithm solutions $\alpha_i$ for all $i \in \{ 0, \ldots, \ell\}$.
While perhaps not considered a "standard" assumption in the same way that plain Computational Diffie-Hellman (CDH) or other problems that reduce to simply a single discrete logarithm assumption, OMDL underpins the security of many cryptographic schemes in theory and in practice, such as blind signatures.
• Knowledge of Exponent Assumption (KEA). KEA is a white-box assumption, and is not falsifiable (thus a stronger assumption than what we have reviewed thus far). KEA says that for an adversary given a generator $g$ of a group $G$ and random element $X \in G$ such that $X = g^x$ for a random x, then if the adversary outputs a tuple $(A, B)$ such that $(A, B) = (g^a, X^a)$, then there exists an extractor that will output $a$. Informally, this means that the only way for the adversary to produce $(A, B)$ is by exponentiating each element in the tuple $(g, X)$ with the value $a$, thereby demonstrating the adversary's knowledge of $a$ (as opposed to choosing random elements in $G$).
If you would like more context on how these assumptions are used to prove the security a cryptographic scheme, we give more context later in this post.
## Part Three: I thought we were supposed to be talking about Schnorr threshold schemes...
Yes, we are! Finally getting to that.
The reason we wanted to write this post is because there has been some debate about the security of two-round Schnorr threshold signature schemes (FROST1/2) and how they compare to less efficient three-round Schnorr threshold signature schemes. We'll review these schemes now, and clarify their resulting security next.
FROST1 was introduced by Komlo and Goldberg in 20201. In that work, they did two things. They introduced 1) a Distributed Key Generation (DKG) protocol that is a minor improvement upon the Pedersen DKG7 that we will call PedPop, as well as 2) a novel two-round threshold signing protocol that is secure against ROS attacks8 that we refer to as FROST1.
FROST2 is an optimized variant of FROST1 by introduced by Crites, Komlo, and Maller in 20212, and reduces the number of exponentiations required for signing operations and verification from linear in the number of signers to constant.
SimpleTSig is a three-round threshold signature scheme also introduced by Crites, Komlo, and Maller2, and is the threshold analogue of a three-round multisignature scheme called SimpleMuSig, presented in the same work2.
Lindell22 is a three-round threshold signing protocol introduced by Lindell in 20223.
We next show that for threshold signing, SimpleTSig and Lindell22 require the weakest assumptions of all of these schemes. FROST1/2 requires sightly stronger assumptions due to OMDL. However, as mentioned before, OMDL underpins many existing cryptosystems such as blind signatures.
We split this analysis into two parts, that of 1) key generation, and 2) signing. The reason for this split is because the key generation mechanism can be viewed as independent to signing, so long as it produces the expected secret and public key material required for signing operations. Hence, the security assumptions required by a certain key generation are imposed on a scheme only if that particular key generation mechanism is used.
## Part Four: What assumptions underpin various key generation protocols that could be used by FROST1/2, SimpleTSig, or Lindell22?
We now describe three different key generation mechanisms, all of which can be used in conjunction with any of the threshold signature schemes described in part four. Note that this list is not exhaustive.
[Standard Model] Trusted key generation. In this setting, a trusted dealer can simply generate all key material and distribute it to each player via Shamir's secret sharing. Shamir's secret sharing is information-theoretically secure, but if Verifiable Secret Sharing (VSS) is used then the discrete logarithm assumption is required. VSS is generally helpful as it allows each participant to ensure that its share is consistent with other players. In each setting however, the dealer is trusted to perform key generation honestly and delete key material after. This variant is described more in the FROST CFRG draft in Appendix B.
[Standard Model] Pedersen. The security of the Pedersen DKG when used as key generation for FROST1, FROST2, SimpleTSig, or Lindell22 relies on at least half of the participants being honest and the underlying signature scheme being secure.
[KEA+PROM] PedPop. An efficient two-round DKG introduced by Komlo and Goldberg along with FROST1. PedPop is simply Pedersen DKG, with the additional step where each participant additionally publishes a Schnorr signature during the first round to prove knowledge of their secret key material. This extra step ensures that security holds given any threshold of honest parties. The security of PedPop when used as key generation for FROST2 and SimpleTSig was demonstrated2. Note that KEA is required for the environment to extract the adversary's secret keys in the proof of security; alternatively, the Fischlin transform could be used in lieu of Schnorr as the proof of possession (and so would be only in the PROM). See further discussion in Part 9.
[Standard Model] Gennaro et al. A three-round DKG that is secure in the standard model.
## Part Five: Which assumptions does two-round threshold signing protocol FROST1/2 rely on?
FROST1/2 signing can be proven using:
1. One-More Discrete Logarithm Assumption (OMDL)
2. Programmable Random Oracle Model (PROM)
By reducing to OMDL, the environment does not need to rely on extracting secret information from the adversary during its simulation of signing; the adversary is simply required to output a valid forgery. The use of two nonces and the randomizing factor in FROST allows for a true reduction to OMDL, unlike prior related multisignature schemes that had subtle flaws in their attempt to an OMDL reduction9,
The proof for FROST11 required a heuristic assumption and so could not prove these properties directly. The proof for FROST22 provides a direct proof for FROST2 with PedPop as the key generation protocol. Proofs for FROST1 and FROST2 in a recent paper by Bellare, Tessaro, and Zhu10 employ an abstraction of key generation, and so demonstrate a direct reduction to PROM+OMDL.
## Part Six: Which assumptions does three-round SimpleTSig signing rely on?
SimpleTSig signing can be proven using:
1. Discrete Logarithm Problem (DLP)
2. PROM
The reason why SimpleTSig can be proven in ROM+DLP is because it relies upon a commit-open-sign protocol flow. Similarly to FROST 1/2, the environment does not need to extract secret values from the adversary during its simulation of signing; it simply requires that the adversary output a valid forgery at the end of the protocol.
## Part Seven: Which assumptions does three-round Lindell22 signing rely on?
Lindell22 signing can be proven using:
1. DLP
2. PROM
Unlike FROST1/2 and SimpleTSig, Lindell22 employs Schnorr signatures at intermediate steps throughout the signing protocol so that participants can prove possession of their nonces. The proof of security requires the environment to extract the adversary's nonces in order to demonstrate the reduction to DLP. While employing Schnorr signatures is sufficient to perfectly simulate an idealized zero-knowledge and commitment functionality, Schnorr signatures are not sufficient for the environment to perform this extraction step.
Hence, Lindell22 must instead employ the Fischlin Transform in lieu of employing Schnorr signatures for the proof to go through in the PROM. Doing so has a non-zero impact on the performance and complexity of the protocol. See further discussion below on this topic.
## Part Eight. I'm confused about why Fischlin/KEA are even required.
This is going to be dense, so hang on :)
In summary, the Fischlin Transform requires a change to the actual protocol so that the prover brute-forces to find a weak hash function output where the least significant $b$ bits are zero. Why is this transform necessary? In summary, it ensures that in the proof of security, the environment is able to extract the necessary secret information from the adversary for the proof to go through.
KEA simply defines an extractor that is assumed to be able to extract the correct values, given the constraints described above. Hence, this assumption is non-falsifiable and therefore considered a strong assumption.
Notably, KEA and Fischlin are often interchangable for protocols that require online extraction for proofs of possession. Lindell22 employs the Fischlin Transform (and hence is in the PROM), but could easily instead employ KEA. The proof for PedPop2 assumes KEA, but alternatively, could use Fischlin.
Forking+rewinding is how the unforgeability of Schnorr signatures is proven to reduce to the hardness of discrete log in the programmable ROM, when Fiat-Shamir is employed. We describe in more detail this reduction at the end of this post. However, while the proof of unforgeability for Schnorr signatures incurs acceptable tightness loss when forking+rewinding is used, the same is not true when Schnorr signatures are employed as proofs of possession (PoP) and the environment must extract secret information from the adversary, as is the case in PedPop and Lindell22. In the extractability case, the tightness loss incurred is instead exponential. Hence why in the PoP setting where extractability is required, either KEA or Fischlin must instead be employed.
The Fischlin Transform provides an alternative to forking+rewinding, so that the environment can similarly extract secret information in an online manner, hence allowing for a tight(er) proof. Sounds too good to be true? It is, a bit. The Fischlin Transform requires the prover to brute-force finding a challenge where the least significant $b$ bits of the challenge must be all zeros. Hence, since the prover is unlikely to find this challenge immediately, it must make many challenge queries, therefore allowing the environment to extract secret values, similarly to the forking+rewinding case. However, unsurprisingly, doing so is expensive for reasonable security parameters.
## Part Nine: This post is really long. What should I take away from all of this?
Let's summarize the key takeaways.
1. Two-round threshold signing protocols FROST 1 and FROST2 rely on the programmable Random Oracle Model (PROM) and One-More Discrete Logarithm (OMDL) assumptions.
2. Three-round threshold signing protocol SimpleTSig relies on PROM + DL.
3. Three-round threshold signing protocol Lindell22 relies on PROM + DL. The Fischlin Transform imposes some performance costs.
Thanks, and happy threshold signing!
## More on proving the security of cryptographic schemes
Similar to demonstrating that a problem is in NP for complexity theory by reducing the problem to another known NP problem, in cryptography, we use reductions to hard mathematical problems to demonstrate that breaking a cryptography scheme is as hard as breaking some known-to-be-hard mathematical problem. For example, we might say that "an adversary wishing to compromise the security of a key-exchange protocol must solve for the discrete logarithm of a value, where the most efficient way to do it is by brute force, which takes X computational power over Y number of years." We model the adversary as a black-box randomized algorithm, which is run by the execution environment, outputting some value at the end, resulting in either a win or fail for the adversary. We can then provide a lower bound on how long it would take an adversary to eventually win (e.g., solve for an unknown discrete log), and determine parameters for the security.
Showing this reduction can be done a number of ways, but there are two proof techniques that are considered to be best practices in cryptography.
1. Game-based proofs
2. Simulation-based proofs.
Game-based proofs demonstrate that an adversary that wins in some game A can be used as a subroutine by another adversary to win in a different game B. Using our key-exchange example, we could show that an adversary that wins in a game against the key-exchange scheme could be used as a subroutine by another adversary to win in a game against the discrete logarithm problem. By a game, we simply mean some program that initializes an adversary, simulates oracle queries to it, and at the end determines if the adversary has successfully completed its attack.
Simulation-based proofs instead rely on proving that some function in the "real world" is indistinguishable to that function in the "ideal world." For example, we could call an encryption secure if an adversary that learns the output of some ciphertext of a real message (real world) obtains no more information than an adversary that learns a ciphertext of garbage (ideal world). The adversary is allowed to interact with the environment, receiving outputs representing the real world and the ideal world. We say the scheme is secure if the adversary successfully distinguishes between the two with negligible probability.
## More on the reduction of Schnorr signatures to DLP
Above, we talked about proving the security of Schnorr signatures as the result of Fiat-Shamir by reducing to the hardness of DLP. We give this reduction step-by-step now:
1. The environment is given $PK$ as the challenge without knowing the secret key, and must simulate signing to an adversary, whose goal is to compute a forgery.
2. When the adversary successfully produces a forgery (with some probability), the environment then forks its state, and then re-runs the adversary. The adversary again will produce a second forgery, outputting the following two forgeries to the environment.
$(R, c, z), \text{ and } (R, c', z')$
1. In the above, $R$ is the commitment (and importantly, is the same in the two tuples).
2. $c=H(R, m)$ is the challenge the adversary obtains from the challenge oracle (the random oracle H that the environment simulates) before the adversary is forked, and $c' = H(R, m)$ is the challenge after the adverary is forked, where importantly $c\neq c'$.
3. $z$ is the adversary's forgery with respect to $(R, c)$ before the fork, and $z'$ is the adversary's forgery with respect to $(R, c')$ after the fork.
4. The environment can then extract $sk$ simply by deriving $(z-z')/(c-c')$.
## Footnotes
cryptography@ethereum.org | 2023-02-03 10:30:25 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 41, "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": 0.764457643032074, "perplexity": 1754.1668393952114}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1674764500044.66/warc/CC-MAIN-20230203091020-20230203121020-00624.warc.gz"} |
https://stur.wedl.ws/6ubjfvo/viewtopic.php?e8f831=factorization-questions-and-answers-pdf | The correct answer is: A . The prime factors of both 495 and 220 can be displayed using prime factor trees. Finding Prime Factors Establish whether a number up to 100 is prime and recall prime numbers up to 19. The examples are quick and concise with exam style questions, go to GCSE Maths if you need more in-depth explanations. The correct answer is: B . and the factorisation of 495 is,. Prime factors of 25 = 5 x 5. Alternative notation Full worked solutions. Register before starting the test to explore the benefits of Math Quiz profile Type your answers into the boxes provided leaving no spaces. For example, 80 Degree lagging is a power factor that is subject to … In the study of mathematics, the prime factorization of a number is achieved when we figure out what prime numbers multiply for us to find it. How to find the product of prime factors? It is like "splitting" an expression into a multiplication of simpler expressions. 182 DOING RESEARCH Learning how to design and use structured interviews, questionnaires and observation instruments is an important skill for research-ers. T-tests 5.1 Using the data file survey.sav follow the instructions in Chapter 16 of the SPSS Survival Manual to find out if there is a statistically significant … 5. All the numbers have a factor of one. Answer the questions in the spaces provided – there may be more space than you need. Here, some important questions related to factorization topic from NCERT class 8 book are also included. 220=2\times2\times5\times11 . True. Try your best to answer the questions above. Multiples and factors quiz E3 Answers Level A 1 . Answers to worksheets (where available) are included at the bottom of each topic. If the equation is first order then the highest derivative involved is a first derivative. Instructions Use black ink or ball-point pen. 3. Along with Detailed Answers, Timing, pdf download. 8, 16, 36. The numbers that are multiplied together are called factors of the final number. The PERFORMANCE APPRAISAL QUESTION and ANSWER BOOK: A SURVIVAL GUIDE FOR MANAGERS So, the factorisation of 220 is,. A square number 69. You could not on your own going later book deposit or library or borrowing from your associates to admittance them. Which list is made up of multiples of 4? Information The marks for each question are shown in brackets – use this as a guide as to how much time to spend on each question. 2 . If it is too low then cable over heating &equipment overloading will occur. Free PDF Download - Best collection of CBSE topper Notes, Important Questions, Sample papers and NCERT Solutions for CBSE Class 8 Math Factorization. That can be as many as two numbers in an equation or as many as it possibly takes to reach it. Deciding the order of questions 193 Include all potential answer choices 193 11-Seale-4312-CH-11-Part 2.indd 181 22/11/2011 4:03:25 PM. Pick a real example (make sure it’s not about someone working in the company you want to … Which of the following is not a prime factorization? These past paper questions help you to master the 11+ Exam Maths Questions… Click here for Answers. 130 TOP EPIDEMIOLOGY Multiple Choice Questions and Answers pdf. 5 is a factor of 20, not a multiple of 20. D The smallest prime number that can divide 25 is 5. at the same time they constitute a key free factor of pro-duction in the new knowledge economy. a) 20 = 2×10 , b) 14 = 2×7 , c) 64 = 4 3, d) 120 = 2 3 × 15 Solution Prime factorization involves only prime numbers. FACTORS. Example 3: Express 66 as a product of its prime factors. Answer all questions. Common Factors Know and use the vocabulary of prime numbers, prime factors and composite (non-prime) numbers. The prime factors of 100 = 2 x 2 x 5 x 5. Title: Factorization Questions And Answers Class 8 Author: media.ctsnet.org-Franziska Frankfurter-2020-09-28-04-38-33 Subject: Factorization Questions And Answers Class 8 The worksheets (WS) may be of use to teachers looking for a quick source of some extra questions, or to students looking for extra practice. 3. The correct answer is: B . Section 1: Theory 3 1. Create factor trees to find the prime factors of the given numbers. The median of these numbers 70. Low power factor means losses will be more.it is the ratio of true power to apparent power. Lagging power factor: The lagging power factor is called the lagging power factor when the current is behind the voltage or if the inductive load is higher than the capacitive load in the AC circuit. This is a tricky question. Fill in the boxes at the top of this page with your name, centre number and candidate number. Well done! 25/5 = 5. Title: Factorization Questions And Answers Class 8 Author: wiki.ctsnet.org-Doreen Pfeifer-2020-09-29-15-00-01 Subject: Factorization Questions And Answers Class 8 Read Online Factorization Questions And Answers Class 8 Factorization Questions And Answers Class 8 Getting the books factorization questions and answers class 8 now is not type of challenging means. Attempt every question. Common factor: A common factor of two or more given numbers is a number which divides each given number completely. C To make this easier we can break 81 to be 9 x 9 and then find the prime factors of each of these prime numbers. Common factor of 12 and 18 are 1,2,3,6. Practice Questions 4. Factorising expressions, HCF, common factor. 11+ For You – Maths Paper Sample Questions Visit www.11plustestpapers.co.uk for more great papers covering maths, english, verbal reasoning and non … Donʼt spend too long on one question. 5 is a multiple of 20. Factorization. 3. During working hours when load is above 80%, power factor is 0.9 and when load goes down to 30% during off working hours , power factor goes to 0.5-0.6. Fa lse. 50 Common Interview Questions and Answers Page 8 of 25 9. You have reached the end of the test. Two factors of 24 68. Hence The following are not prime factorization. If you have spare time make sure you go back and check over the answers. Tell me about your worst boss. Identify multiples and factors, including finding all factor pairs of a number, and common factors of two numbers. Title: Factorization Questions And Answers Class 8 Author: learncabg.ctsnet.org-Juliane Hahn-2020-09-24-18-02-15 Subject: Factorization Questions And Answers Class 8 Tips on using solutions 6. that we wish to solve to find out how the variable y depends on the variable x. The entire NCERT textbook questions have been solved by best teachers for you. If it is greater than 1 then load will act as capacitor and starts feeding the source and will cause tripping. Read each question carefully before you begin answering it. Access the answers to hundreds of Factorization questions that are explained in a way that's easy for you to understand. FACTORS, MULTIPLES PRIMES Materials required for examination Items included with question papers Ruler graduated in centimetres and Nil millimetres, protractor, compasses, pen, HB pencil, eraser. So far there are answers to all the KS3 topics, and more answers will be added over time. This makes negotiations in the cultural field extremely controversial and difficult. So answer it with a mix of honesty, diplomacy and positivity. a) 20 = … As you work through the exercise regularly click the "check" button. In the NEBOSH IGC Questions and Answers, there are a lot of Command words used like- define, give, explain, suggest, complete, describe etc. Check your answers seem right. As several experts point out, no other industry has generated so much debate on the political, economic and institutional limits of the regional and global integration processes or their legitimacy. Get help with your Factorization homework. Calculators may be used. 3 × 6 = 18, 3 × 8 = 24 and 3 × 10 = 30. 2. Example: factor 2y+6. Take the prime factorization math test! ANSWERS TO EXERCISES AND REVIEW QUESTIONS PART FIVE: STATISTICAL TECHNIQUES TO COMPARE GROUPS Before attempting these questions read through the introduction to Part Five and Chapters 16- 21 of the SPSS Survival Manual. QUESTIONS AND ANSWERS 415 Answer:Thethreemaintypesofcheesethatrelyonmoldsarebluecheese,soft ripened cheese (such as camembert and brie) and rind-washed cheese (such as Standard integrals 5. Factors of 12 = 1,2,3,4,6, and 12. 495=3\times3\times5\times11 . Tracing paper may be used. These command words are used in the NEBOSH IGC questions … These factorization questions will help CBSE students to be well-prepared for class 8 exam. Answer: Power factor should be high in order to get smooth operation of the system. 4. Factoring (called "Factorising" in the UK) is the process of finding the factors: Factoring: Finding what to multiply together to get an expression. Answers 4. This is an no question simple means to specifically get guide by on-line. A prime number END OF TEST Congratulations! There are 4 methods: common factor, difference of two squares, trinomial/quadratic expression and completing the square. Example 1: Express 120 as a product of its prime factors. Theory Consider an ordinary differential equation (o.d.e.) EPIDEMIOLOGY Multiple Choice Questions and Answers pdf :-1) All of the following are true of odds ratio except: A) It is an estimate of relative risk B) It is the only measure of risk that can be obtained directly form a case-control study C) It tends to be biased towards 1 (neither risk or protection at high rates of disease A number may be made by multiplying two or more other numbers together. Must Practice 11 Plus (11+) Prime Factorization Past Paper Questions. 18, 24 and 30 are all multiples of three . Class 8 important questions for chapter 14 factorization provided here cover various short answer type questions, long answer type questions and HOTS questions. Multiples, Factors and Primes Practice Questions Click here for Questions . Exam Style Questions Ensure you have: Pencil, pen, ruler, protractor, pair of compasses and eraser You may use tracing paper if needed Guidance 1. See how well you understand this notion by taking the quiz below! 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https://mymathforum.com/threads/how-to-find-the-maximum-angular-speed-from-a-bar-which-rotates-about-an-axis.347797/ | # How to find the maximum angular speed from a bar which rotates about an axis?
#### Chemist116
The problem is as follows:
A rigid bar of negligible mass has three particles whose masses are the same and are joined to the bar as indicated in the figure. The bar is free to rotate in a vertical plane about an axis with no friction perpendicular to the bar through point $P$ and is released from rest on the horizontal position at $t=0$. Find the maximum angular speed in radians per second attained by the bar. You may consider $g=9.8\frac{m}{s^2}$ and $d=4.2\pi$
The alternatives given are:
$\begin{array}{ll} 1.&1\,\frac{rad}{s}\\ 2.&2\,\frac{rad}{s}\\ 3.&3\,\frac{rad}{s}\\ 4.&4\,\frac{rad}{s}\\ \end{array}$
What I've attempted to solve this problem was to equate the potential energy and the rotational energy of the masses assuming there's a superposition of the moment of inertia of the masses.
This is translated as:
$mgh=\frac{1}{2}I\omega^2$
$mgh=\left(\frac{1}{2}m\left(\frac{4d}{3}\right)^{2}+\frac{1}{2}m\left(\frac{d}{3}\right)^{2}+\frac{1}{2}m\left(\frac{2d}{3}\right)^{2}\right) \times \omega^2$
But after following the procedure I can't find a way to cancel the 4.2$\pi$. Supposedly, the answer is the first choice. How can I get to that result?
#### skeeter
Math Team
$\omega_{max}$ will occur when the bar is at the vertical ...
$U_{g0}-U_{gf} = K_{Rf}-K_{R0}$
$mg \cdot \dfrac{4d}{3} + mg \cdot \dfrac{d}{3} - mg \cdot \dfrac{2d}{3} = mgd = \dfrac{1}{2} \cdot I \cdot \omega^2$
$I = m \cdot \dfrac{16d^2}{9} + m \cdot \dfrac{d^2}{9} + m \cdot \dfrac{4d^2}{9} = m \cdot \dfrac{7d^2}{3}$
$mgd = m \cdot \dfrac{7d^2}{6} \cdot \omega^2 \implies \omega = \sqrt{\dfrac{6g}{7d}}$
using the given value for $d$, which I suspect is mistaken, the result is close to 1 rad/sec
#### Chemist116
$\omega_{max}$ will occur when the bar is at the vertical ...
$U_{g0}-U_{gf} = K_{Rf}-K_{R0}$
$mg \cdot \dfrac{4d}{3} + mg \cdot \dfrac{d}{3} - mg \cdot \dfrac{2d}{3} = mgd = \dfrac{1}{2} \cdot I \cdot \omega^2$
$I = m \cdot \dfrac{16d^2}{9} + m \cdot \dfrac{d^2}{9} + m \cdot \dfrac{4d^2}{9} = m \cdot \dfrac{7d^2}{3}$
$mgd = m \cdot \dfrac{7d^2}{6} \cdot \omega^2 \implies \omega = \sqrt{\dfrac{6g}{7d}}$
using the given value for $d$, which I suspect is mistaken, the result is close to 1 rad/sec
Can you include some drawing to accompany your solution? I'm having a problem at identifying where's the height to establish the potential energy. When you rotate the bar about the axis which is represented by the orange dot, the bar is vertical right? But from where to where you put the height?
Why Initially do we have
$mg \cdot \dfrac{4d}{3} + mg \cdot \dfrac{d}{3}$
and why finally do we have:
$mg \cdot \dfrac{2d}{3}$
@skeeter Can you please represent the justification of this in a drawing?
I got that part of adding the moment of inertia due the principle of superposition of them and the rest is just logical, but...
$\omega = \sqrt{\dfrac{6g}{7d}}$
in this part is where I'm stuck. How on earth did you get an answer close to $1\frac{rad}{s}$?
If I do insert the given values, this would become into:
$\omega = \sqrt{\dfrac{6\times 9.8}{7\times 4.2 \times 3.14}} \approx 0.7980 \frac{rad}{s}$
Can you please explain this part? Or did I misunderstand something from your explanation? Any help please?
#### skeeter
Math Team
0.798 is closest to 1rad/sec, given the available choices ... note that I stated disagreement with your given value for $d = 4.2\pi$
For $\omega = 1 \text{\rad/sec}$, $d$ would need to equal to 8.4
Full explanation of the method used to find the maximum angular velocity is given in the following diagram
topsquark | 2020-02-17 17:08:26 | {"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": 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": 0.7833556532859802, "perplexity": 266.36257006040944}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1581875142603.80/warc/CC-MAIN-20200217145609-20200217175609-00259.warc.gz"} |
https://www.physicsforums.com/threads/tension-of-strings-in-an-elevator.230784/ | # Tension of strings in an elevator
1. Apr 23, 2008
### sodaMay
Two masses are connected by a string and are hanging from the ceiling of an elevator like this:
[string 1]
|
[4kg mass]
|
[string 2]
|
[6kg mass]
Show the tension of each string in the following situations:
i) Stationary
ii) Moving up at 3m/s
iii) Moving down at 3m/s
iv) Moving up at 3m/s/s
v) Moving down at 3m/s/s
Ok, I am kinda stumped. Where do I start? Is there a formula for tension? Is it F=ma? Or F=m(9.8-a) or something?
I'm just so horrible at physics, I'll be so grateful if someone could explain how this works..
Thanks! :)
2. Apr 23, 2008
### Dr. Jekyll
You just need to look at the net force on each string. For example, you have two forces on the first string (weights of the hanging bodies). Other one is even easier. Remember the 1st Newton's law. That's when the elevator is stationary.
The formula $F=ma$ can be used in part (iv) and (v). It's the inertial force. Just be aware of its direction. | 2018-02-22 21:45:42 | {"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": 0.5979471802711487, "perplexity": 1478.4012317389836}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1518891814290.11/warc/CC-MAIN-20180222200259-20180222220259-00697.warc.gz"} |
https://electronics.stackexchange.com/questions/598853/compensating-for-voltage-divider-computing-sensor-board-analog-reading | # compensating for voltage divider computing sensor board analog reading
I'm looking at air quality sensors with the intention of monitoring indoor air quality.
One sensor I'm looking at is the Waveshare Dust Sensor Detector Module with Sharp GP2Y1010AU0F Onboard. See https://www.waveshare.com/wiki/Dust_Sensor for a technical description of the sensor board.
I found this sensor on Amazon and reading over the reviews found one with a fair amount of technical details about how the person is using it.
This review mentions the code used with this sensor and one thing I do not understand is the multiple sampling of the analog signal for 11 times creating a sum of those 11 samples. The reason given for this code is because the line output is routed through a resistor voltage divider that is some 11K ohms (a 10k ohms plus a 1K ohms resistors) on the sensor board before presented to the output connector on the board.
I've also looked at the demo source code from the Waveshare website, a copy of which is posted further below. That code generates an average of 10 samples which is then multiplied by 11.
So it seems that both the Amazon reviewer's code and the demo program are taking the 11 to 1 resister voltage divider into account using two different procedures.
Why are they doing an adjustment rather than just using a reading of the line and taking that value?
Post on Amazon and the code there
The reviewer wrote:
2./ While the Sharp sensor has output Aout on pin 5 in range 0.4-4V the Waweshare board lowers it with the resistor divider R10 10k + R6 1k (11:1);
3./ I simply compensated the voltage divider by adding 11 counts in the measuring loop. In addition the command 'analogReference(INTERNAL);' changes default reference voltage from 5V to 1.1V further increasing counts from the analog to digital converter.
The source code posted is as follows:
long int systemTime; // replaces delay()
int AoutMin = 360; // equal to Aout min in mV for clean air
int AoutMax = 3600; // equal to Aout max in mV (toothpick in measuring window)
float dustConst; // it will be computed from Aout range and Sharp sensor range 0-500 ug/m3
int dustConc; // dust concentration in ug/m3/mV
unsigned int sum11; // sum of 11 counts to compensate 11:1 Aot voltage divider
int AoutAvg; // direct Sharp Vout in mV computed as average from 11 measurements
int ref1024 = 1100; // reference voltage for 10-bit ADC; 1023 counts = 1100 mV
int dustQualityIndex; // it is quite arbitrary
char *dustQualityChar[] = {"1.Excellent", "2.Very good", "3.Good", "4.Fair", "5.Poor"};
void setup() {
Serial.begin(9600);
pinMode(A0, INPUT); // pin selected for 3.3V, GND, AOUT, and ILED
pinMode(A2, OUTPUT); // in order RED, BLACK, BLUE, and YELLOW as on WaveShare board
analogReference(INTERNAL); // for better resolution use 1100mV instead of default 5000mV
analogRead(A0); // activate IOref pin, 1.1V will be present
dustConst = 500 / float(AoutMax - AoutMin); // in ug/m3 per millivolt
Serial.print("\n ------------------------------------------------------\n");
Serial.print(" The WaveShare board divides the Sharp Sensor Vout 11:1.\n");
Serial.print(" The sum of 11 counts will compensate for lowered output.\n");
Serial.print(" Then the sum is converted to mV, and from it\n");
Serial.print(" is computed dust concentration and air quality.");
Serial.print("\n-------------------------------------------------------\n");
Serial.print("500/(");
Serial.print(AoutMax);
Serial.print("-");
Serial.print(AoutMin);
Serial.print(")=");
Serial.print(dustConst);
Serial.print(" Dust density constant in ug/m3/mV");
Serial.print("\n-------------------------------------------------------\n");
}
void loop() {
sum11 = 0;
for (int i = 0; i < 11; i++) {
//Sharp datasheet: pulse cycle 10ms, pulse width 0.32ms, sampling at 0,28ms
digitalWrite(A2, HIGH);
delayMicroseconds(280);
delayMicroseconds(40); // 280+40=320
digitalWrite(A2, LOW);
delayMicroseconds(9680); // 320 + 9680 = 10000
}
if ((millis() - systemTime) > 2000) {
computeAirQuality();
systemTime = millis();
}
}
void computeAirQuality()
{
AoutAvg = float(sum11) * float(ref1024) / 1024; // Sharp sensor direct Aout in mV
int q, i;
q = float(AoutAvg - AoutMin) * dustConst;
if (q < 0) q = 0; // handle non positive values
if (q < 40) i = 0;
else if (q < 80) i = 1;
else if (q < 160) i = 2;
else if (q < 320) i = 3;
else i = 4;
dustConc = q; dustQualityIndex = i;
printAirQuality();
}
void printAirQuality()
{
Serial.print(" Sum_11 ");
Serial.print(sum11);
Serial.print(" \t ");
Serial.print(AoutAvg);
Serial.print(" mV \tdust_C ");
Serial.print(dustConc);
Serial.print(" ug/m3\t\t");
Serial.print(dustQualityChar[dustQualityIndex]);
Serial.print(" air quality\n");
}
Demo code posted on Waveshare site
The demo Arduino program from the Waveshare site is doing a kind of moving average of the last ten values read from the sensor and then multiplying that average by 11.
/*********************************************************************************************************
*
* File : DustSensor
* Hardware Environment:
* Build Environment : Arduino
* Version : V1.0.5-r2
* By : WaveShare
*
* http://www.waveshare.net
* http://www.waveshare.com
*
*********************************************************************************************************/
#define COV_RATIO 0.2 //ug/mmm / mv
#define NO_DUST_VOLTAGE 400 //mv
#define SYS_VOLTAGE 5000
/*
I/O define
*/
const int iled = 7; //drive the led of sensor
/*
variable
*/
float density, voltage;
/*
private function
*/
int Filter(int m)
{
static int flag_first = 0, _buff[10], sum;
const int _buff_max = 10;
int i;
if(flag_first == 0)
{
flag_first = 1;
for(i = 0, sum = 0; i < _buff_max; i++)
{
_buff[i] = m;
sum += _buff[i];
}
return m;
}
else
{
sum -= _buff[0];
for(i = 0; i < (_buff_max - 1); i++)
{
_buff[i] = _buff[i + 1];
}
_buff[9] = m;
sum += _buff[9];
i = sum / 10.0;
return i;
}
}
void setup(void)
{
pinMode(iled, OUTPUT);
digitalWrite(iled, LOW); //iled default closed
Serial.begin(9600); //send and receive at 9600 baud
Serial.print("*********************************** WaveShare ***********************************\n");
}
void loop(void)
{
/*
*/
digitalWrite(iled, HIGH);
delayMicroseconds(280);
digitalWrite(iled, LOW);
/*
covert voltage (mv)
*/
voltage = (SYS_VOLTAGE / 1024.0) * adcvalue * 11;
/*
voltage to density
*/
if(voltage >= NO_DUST_VOLTAGE)
{
voltage -= NO_DUST_VOLTAGE;
density = voltage * COV_RATIO;
}
else
density = 0;
/*
display the result
*/
Serial.print("The current dust concentration is: ");
Serial.print(density);
Serial.print(" ug/m3\n");
delay(1000);
}
• I can see possible over-voltage issues of using a 5V boost supplied signal if Arduino shares the same V+ input. A series current limiting R should have been adequate for the signal. No good reason for 1/11. Averaging random noise improves st .dev by sqrt(n) so not much improvement. Dust tends to have a log scale, so a linear scale with a high threshold is pretty coarse. Dec 10, 2021 at 20:57
• @TonyStewartEE75 what do you mean by "Dust tends to have a log scale"? Do you have an article that would explain dust measurement? I think the underlying reason for the voltage divider is to drop the AOUT voltage to within the range of an A/D converter on an MCU. The Arduino code sets the range for the A/D on the board to be 0.0 v to 1.1 v. I'm not sure about other MCU devices such as STM32 however I suspect the value was chosen to have a range compatible with most A/D converters on common MCUs. Dec 11, 2021 at 5:25
• I have experience using a 16 channel Laser particle counter and the particle count increases inversely to the particle size over a several decades in random dust. However smoke particles cover a small range of large particles. I used this to measure office and manufacturing environments for HDD ranging up to 1M particles per cu.ft down to 100kp/cf. HEPA flow booths are 10kp/cf max, clean room rooms & HDD enclosures with recirculating HEPA filters are 1 to 10 p/cf all for >0.1um yet far more particles exist under this size . This range indicates 6 orders of magnitude. Dec 11, 2021 at 14:57
• I measured mass and counts by using air deflection for a calibrated nozzle air flow by deflecting particles with a transverse pitot tube with IR reflection and counts of particles per second then extrapolated to particles per cubic foot. Dec 11, 2021 at 15:01
Why are they doing an adjustment rather than just using a reading of the line and taking that value?
• The output from the device is 0.4V to 4.0V.
• The waveshare board is meant to be powered from a supply of 2.5V to 5.5V. This probably restricted them to using an ADC reference of 2.5V minus some headroom. For whatever reason, they picked 1.1V.
• With a 1.1V reference the A/D converter can only measure signals up to 1.1V max.
• Therefore, they needed to divide the signal by at least a factor or 3.64:1.
For some reason they have chosen to divide by 11.
• From a signal to noise ratio standpoint, 11 is not optimal because you get 1~2 less bits in your reading.
• The resistor values are nice round numbers and are probably cheap and easy to get.
• An 11:1 divider has high input impedance (so it won't load down the sensor, and protects the board from faults), but lower output impedance going to the ADC. See Why is high input impedance good?
• Adding 11 measurements together reduces overall noise compared to taking a single measurement. Also using addition doesn't require that the MCU can multiply numbers, so they can use a really cheap MCU.
• 11 is an integer. If they divided by a fraction, then reconstructing the original reading may have required floating point math, and therefore a more expensive microcontroller.
• Is "output from the device" referring to the VO line from the Sharp GP2Y1010AU0F dust sensor? Most of Waveshare board seems to be power supply/conversion for the GP2Y1010AU0F and a voltage divider for the VO of the GP2Y1010AU0F to produce AOUT with smaller range for the board? The A/D converter you mention is on the Arduino? I don't recognize an A/D in the Waveshare board schematic. GP2Y1010AU0F max AOUT is 4.0 v so divided by 11 is 0.36 v so using Arduino A/D set analogReference(INTERNAL) sets A/D range to 0.0 v to 1.1 v.. Waveshare chose common, cheap resistors to convert AOUT below 1.1v? Dec 11, 2021 at 1:49 | 2022-08-10 11:56:45 | {"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": 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": 0.22906167805194855, "perplexity": 6565.681683329039}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1659882571153.86/warc/CC-MAIN-20220810100712-20220810130712-00739.warc.gz"} |
http://physics.stackexchange.com/questions/30587/what-makes-the-earth-keep-spinning | # What makes the Earth keep spinning?
I am thinking of what makes the Earth keep spinning? Is there anybody here know the answer?
-
Possible duplicates: physics.stackexchange.com/q/12140/2451 and links therein. – Qmechanic Sep 25 at 5:04
I agree with the previous answer. Angular momentum, something the earth has because of its rotation about its axis, can only be changed when an external torque (twisting motion) is applied to the earth. As far as I know, there are two ways in which this can happen. If there was friction between the earth's surface and space, then that would slow down the earth. However, this is negligible because space is essentially a vacuum. The second way is through the gravitational/tidal force applied by the moon. This force creates the tidal bulge of the earth's oceans, exerting a torque. So, while the earth seems to be spinning at a constant rate, it is actually slowing down slightly over time because of these tidal forces.
-
And someday the earth will stop spinning? – Stiff Jokes Jun 22 '12 at 4:15
I believe so, though I read somewhere that before that happens, the sun will have expanded so much that it will engulf the earth. This just shows how slowly this decrease rotational speed is! – Steven Harris Jun 22 '12 at 4:21
@Forgiver It won't ever stop spinning. If the Sun didn't turn into a red giant the Earth would eventually end up tidally locked to the moon. But as Steven pointed out the sun will eat the Earth/Moon system before that happens. – Dan Neely Jun 22 '12 at 19:32
"And someday the earth will stop spinning?" Yes, but only relative to the Moon (in case of lunar tides). Moon has already stopped spinning relative to Earth. – Leos Ondra Jun 30 '12 at 8:56
Spinning at a constant rate does not require any outside force or torque, due to conservation of angular momentum.
-
But is there a friction force to decelerate the earth rotation? – Stiff Jokes Jun 22 '12 at 3:51
@Forgiver: no, there isn't. – David Z Jun 22 '12 at 3:59 | 2013-12-20 11:56:33 | {"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": 0.839365541934967, "perplexity": 786.0099404569369}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-48/segments/1387345771702/warc/CC-MAIN-20131218054931-00057-ip-10-33-133-15.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/304851/evaluate-sum-limits-k-1-infty-frack2-1k4k21 | # Evaluate $\sum\limits_{k=1}^{\infty} \frac{k^2-1}{k^4+k^2+1}$
How to find $$\sum_{k=1}^{\infty} \frac{k^2-1}{k^4+k^2+1}$$
I try something like this:
\begin{align*}\sum_{k=1}^{\infty} \frac{k^2-1}{k^4+k^2+1}=\sum_{k=1}^{\infty}\frac{k^2}{k^4+k^2+1}-\sum_{k=1}^{\infty}\frac{1}{k^4+k^2+1}.\end{align*}
Using fact that $$\sum_{k = 1}^{n}{\frac{1}{k^4+k^2+1}}=\frac{1}{2}\cdot\frac{n+1}{n^2+n+1}+\frac{1}{2}\cdot\sum_{k = 1}^{n-1}{\frac{1}{k^2+k+1}}$$ we find that \begin{align*}\sum_{k=1}^{\infty}\frac{1}{k^4+k^2+1} &=\frac{1}{2}\cdot\sum_{k=1}^{\infty}{\frac{1}{k^2+k+1}}\\ &=\frac{1}{6}\left(\sqrt{3}\pi \tanh{\left(\frac{\sqrt{3}\pi}{2}\right)}-1\right).\end{align*}
But I don't know how to find $\displaystyle\sum_{k=1}^{\infty}\frac{k^2}{k^4+k^2+1}.$
If someone want to know how to evaluate $\displaystyle\sum_{k=0}^{\infty}\frac{1}{k^2+k+1}$:
First, $$\displaystyle\sum_{k=0}^{\infty} \frac{1}{k^2+k+1}=\sum_{k=0}^{\infty}{\frac{1}{\left(k+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}}.$$ Now, using "well-know" formula $$\displaystyle\cos(\phi)=\prod_{k=0}^{\infty}{\left( 1-\frac{4\phi^2}{(2k+1)^2\pi^2}\right)}$$ we find that $$\displaystyle\log (\cos(\phi))=\sum_{k=0}^{\infty}{\log\left( 1-\frac{4\phi^2}{(2k+1)^2\pi^2}\right)}$$ and then we attack with $\dfrac{d}{d\phi}$ and find $$\displaystyle\tan(\phi)=\sum_{k=0}^{\infty}{\frac{8\phi}{(2k+1)^2\pi^2-4\phi^2}}.$$ Let $\phi=\pi\alpha\cdot i$, then we get $$\displaystyle\tan(\pi\alpha\cdot i)=i\cdot\tanh(\pi\alpha)=i\cdot\sum_{k=0}^{\infty}{\frac{8\pi\alpha}{(2k+1)^2\pi^2+4\pi^2\alpha^2}}=\frac{2\alpha i}{\pi}\cdot\sum_{k=0}^{\infty}{\frac{1}{\left(k+\frac{1}{2}\right)^2+\alpha^2}}.$$ So, we find that $$\displaystyle\sum_{k=0}^{\infty}{\frac{1}{\left(k+\frac{1}{2}\right)^2+\alpha^2}}=\frac{\pi}{2\alpha}\cdot\tanh(\pi\alpha).$$ Let $\alpha=\dfrac{\sqrt{3}}{2}.$ We get $$\displaystyle\sum_{k=0}^{\infty}{\frac{1}{\left(k+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}}=\frac{\sqrt{3}\pi}{3}\cdot\tanh\left(\frac{\sqrt{3}\pi}{2}\right)$$ or $$\displaystyle\sum_{k=0}^{\infty} \frac{1}{k^2+k+1}=\frac{\sqrt{3}\pi}{3}\cdot\tanh\left(\frac{\sqrt{3}\pi}{2}\right).$$
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+1 Nice question: a question and (a partial) answer all wrapped up in one package! (re edit: I simply "highlighted" your question)... – amWhy Feb 15 '13 at 14:42
@amWhy Thank you. Nice edit, I like it :) – Cortizol Feb 15 '13 at 14:55
Hmm... $\frac{1}{k^2+k+1} + \frac{1}{k^2-k+1} = \frac{2(k^2+1)}{k^4+k^2+1}$ and $\frac{1}{k^2-k+1} = \frac{1}{(k-1)^2 + (k-1)+1}$. – achille hui Feb 15 '13 at 15:04
## 1 Answer
Using Partial Fraction Decomposition $$\frac{k^2-1}{k^4+k^2+1}=\frac{Ak+B}{k^2-k+1}+\frac{Ck+D}{k^2+k+1}$$
So, $k^2-1=k^3(A+C)+k^2(A+B-C+D)+k(A+B+C-D)+B+D$
Comparing the coefficients of different powers of $k$ in the above identity,
$A+C=0$
From $A+B+C-D=0,B+D=0$ and $B+D=-1\implies B=-D=-\frac12$
From $A+B-C+D=1\implies A-C=2$ and $A+C=0\implies A=-C=1$
$$\implies\frac{k^2-1}{k^4+k^2+1}=\frac{k-\frac12}{k^2-k+1}-\frac{k+\frac12}{k^2+k+1}$$
$$\frac{2(k^2-1)}{k^4+k^2+1}=\frac{2k-1}{k^2-k+1}-\frac{2k+1}{k^2+k+1}=T(k)\text{ say}$$
$$\implies T(n)=\frac{2n-1}{n^2-n+1}-\frac{2n+1}{n^2+n+1}$$
If we set $U(m)=\dfrac{2m-1}{m^2-m+1},U(m+1)=\dfrac{2(m+1)-1}{(m+1)^2-(m+1)+1}=\dfrac{2m+1}{m^2+m+1}$
$$\implies T(n)=U(n)-U(n+1)$$
Clearly, the first part of any term except the first term is cancelled by the last part of the previous term.
$$\implies2\sum_{k=1}^{\infty} \frac{k^2-1}{k^4+k^2+1}=U(1)=\cdots$$
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Could anybody please verify this? – lab bhattacharjee Feb 15 '13 at 15:01
Verified by hand and by mathematica. – muzzlator Feb 15 '13 at 15:07
@muzzlator, thanks for your feedback. – lab bhattacharjee Feb 15 '13 at 15:08
Note that $\displaystyle \sum_{k=1}^{\infty} \frac{2k - 1}{k^2 - k + 1} = \sum_{k=0}^{\infty} \frac{2(k+1) - 1}{(k+1)^2 - (k+1) + 1}$ – J.H. Feb 15 '13 at 15:10
@muzzlator, I just tried to see what happens after Partial Fraction Decomposition and then identified the telescoping nature. – lab bhattacharjee Feb 15 '13 at 15:17 | 2015-08-01 14:16:21 | {"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": 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": 0.9385306239128113, "perplexity": 1192.8006114818586}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-32/segments/1438042988718.8/warc/CC-MAIN-20150728002308-00199-ip-10-236-191-2.ec2.internal.warc.gz"} |
https://chat.stackexchange.com/transcript/41?m=55652232 | 1:45 AM
At a rough look through I found out that about 90% of all pgfplotstable-questions are due to me. :()
I suspect that the author will invite me to coffee at some point.
5 hours later…
6:47 AM
@cis 28 = 90% of 979 :-)
@DavidCarlisle :)
7:16 AM
@yo' I tried some overleaf promotion:-) tex.stackexchange.com/questions/564112/…
@DavidCarlisle Thanks :) I'll note this down for our Product team, actually...
7:54 AM
@JosephWright Can you sanity check this, always dangerous to tell people they won't lose work... github.com/learnlatex/learnlatex.github.io/issues/…
@DavidCarlisle Yes looks ok: no rebase so no risks
@DavidCarlisle There's always the reflog anyway, although that's a fallback really
@JosephWright thanks
@JosephWright incidentally I went for putting a comment about the update in all the issues rather than send an email as I didn't have email contacts for everyone.
@DavidCarlisle I'd seen :)
@JosephWright it's OK but doesn't really scale and we didn't have open issues for all translations (I used a closed issue and a PR comment for two of them) we need a better plan going forward but that worked OK for now, and hopefully a public prompt may move some of them on.
@DavidCarlisle Yes: I wonder how other projects handle this
@DavidCarlisle In other news, I've restored the docs for l3backend, so people can see all of the comments about how we worked out the undocumented features of each one
8:04 AM
@JosephWright sounds good
@DavidCarlisle Which reminds me I need to hassle Frank about how to safely activate link breaking in dvips: I've got lots of PostScript but nowhere to use it :(
@DavidCarlisle Yes: the issue turns out to be l3docstrip! (Or rather how we have to set it up for l3backend) - I'll be pleased when everything is merged back together in DocStrip
@JosephWright I thought afterwards a better thing would have been to add that comment to the lesson-06 PR but @ ping all the translators in the comment to notify them rather than duplicating the comment to each translators issue.
@DavidCarlisle Ah yes
8:21 AM
Developer: We have a problem.
Manager: Remember, there are no such things as problems, only opportunities.
Developer: Well then, we have a DDoS opportunity.
2
9:00 AM
\newcommand\myMacro{%
\begin{aligned}%
&def%
\end{aligned}%
}
\begin{frame}{Church and Curry Products}
\begin{align*}
\myMacro
\end{align*}
\end{frame}
en/more-01.md:well known for a series of books called the “Art of Computer Programming”
fr/more-01.md:_The Art of Computer Programming_ (« L'Art de la programmation », surnommés
pt/more-01.md:Knuth é conhecido pela série de livros chamada "The Art of Computer Programming"
vi/more-01.md:Donald Knuth. Knuth được biết đến là tác giả của bộ sách "The Art of Computer
@DavidCarlisle :)
@DavidCarlisle it just looks a bit strange with the abbreviation behind it.
@samcarter_is_at_topanswers.xyz nice that ducks can now hug ...
5:27 PM
$git commit -m "The Art" [master d5555c1] The Art 1 file changed, 1 insertion(+), 1 deletion(-) @Skillmonlikestopanswers.xyz ^ @DavidCarlisle :) @DavidCarlisle "It's the Arts" ;-) @DavidCarlisle hmmm 6:25 PM I'm fascinated with the attention ducks get here. 6:37 PM @LaTeXereXeTaL quack @UlrikeFischer As we humans are struggling with the pandemic, the ducks have to do all the hugging :) 4 @samcarter_is_at_topanswers.xyz ooooh ;-) 6:53 PM The new syntax highlighting on TeX.SE looks really meh, or is this just me who doesn't like it? 3 @UlrikeFischer Here is also one for you: \documentclass{standalone} \usepackage{tikzducks,tikzlings} \begin{document} \begin{tikzpicture} \duck[] % \bear[scale=0.7,xshift=1.3cm,back] \node at (0.8,0.6) {\includegraphics[width=1cm]{photo-ulrike}}; \duck[invisible,wing,yscale=-1,scale=0.8,yshift=-1.7cm,xshift=-0.2cm] \duck[invisible,wing,yscale=-1,xscale=-1,scale=0.8,yshift=-1.3cm,xshift=-2.5cm] \end{tikzpicture} \end{document} @Skillmonlikestopanswers.xyz ... also has a lot of glitches Is there some easy way to see, in case of an error/traceback during compilation from a Lua file, where the Lua file is being referenced from? At least the current version of LuaTeX that I'm using doesn't obviously give this information. @samcarter_is_at_topanswers.xyz yes, some macros aren't highlighted for no apparent reason. @Skillmonlikestopanswers.xyz the zebra striped \\\\\ are also funny Okay this is going to sound rather silly, but here goes. I have been under the (terribly wrong) impression that LuaLaTeX required learning an entirely new language (Lua). It doesn't. It's a LaTeX engine that processes LaTeX. I'm sitting here composing mathematical expressions with unicode characters. I get it now. I'm just sorry it took me so long. 7:00 PM @samcarter_is_at_topanswers.xyz awww @LaTeXereXeTaL Rather the other way round: you can more or less use all normal latex things, but if you happen to know lua, you can do a lot more @PauloCereda :D @samcarter_is_at_topanswers.xyz Yes! I completely see that now. @Skillmonlikestopanswers.xyz on the other hand, there are languages which were hit much worse than latex. I just read an rmarkdown Q and at first thought the highlighting did not work at all. At closer inspection it is dark blue mixed with black :) 7:34 PM @LaTeXereXeTaL don't encourage them. @DavidCarlisle quack @Skillmonlikestopanswers.xyz it doesn't know @ is a letter so any command with @ get half coloured, I assume the protocol is to raise a bug at highlight.js @PauloCereda dinner @DavidCarlisle oh no @DavidCarlisle I'm just watching the ducks. quack 4 @samcarter_is_at_topanswers.xyz ^^^ 7:37 PM @LaTeXereXeTaL ooh @LaTeXereXeTaL s/watching/eating/g @UlrikeFischer ooh @DavidCarlisle you are mean @DavidCarlisle Sounds like \makeatletter is missing :) @UlrikeFischer yeah! @LaTeXereXeTaL Will you throw some bread crumbs? @PauloCereda you got the wrong person Sep 7 '17 at 7:36, by Christian Hupfer @Skillmon: You are mean ;-) @DavidCarlisle ooh we need a list of these mentions 7:38 PM Always. I grew up near a duck pond where my father would take me to feed the ducks and geese. Barbara once told me bread is not good for us ducks :) @PauloCereda Auntie BBC said bread is OK as long as you also eat other things @samcarter_is_at_topanswers.xyz ooh bread and Nutella @PauloCereda That's OK, just don't put it on pizza @PauloCereda no break and duck is OK 7:41 PM @samcarter_is_at_topanswers.xyz ooh Nutella pizza is good @DavidCarlisle oh no May 30 at 17:19, by samcarter_is_at_topanswers.xyz @PauloCereda Auntie says bread is OK as long as you mix it with other things https://www.bbc.com/news/newsbeat-50087990 @samcarter_is_at_topanswers.xyz see footnote #2 of the TeXplate documentation (texdoc texplate). :) @samcarter_is_at_topanswers.xyz highlight.js seems to have a lot of issues open since they started using it here two latex ones github.com/highlightjs/highlight.js/… @PauloCereda :) one has to use this just for the entertainment value @samcarter_is_at_topanswers.xyz sounds like the last paragraph of my thesis acknowledgments. :) <markm> c++: the power, elegance and simplicity of a hand grenade [paulo@cambridge ~]$
7:50 PM
@DavidCarlisle there might be a correlation with a sudden increase in user numbers :)
@DavidCarlisle there is also a meta question.
8:06 PM
@UlrikeFischer I braved a miktex comment tex.stackexchange.com/questions/564218/…
@DavidCarlisle we just said the same ;-)
@UlrikeFischer yes but I spoke from ignorance
1 hour later…
9:41 PM
@JosephWright I looked at the iccbased stuff in colorspace. Basically it creates a stream object from an icc, it maps DefaultRGB in the ColorSpace resource to this profile. It doesn't handle outputintents. It is not difficult, but I'm not quite sure where it belongs. From the topic it is color related, but most of the code is l3pdf/pdfresource stuff.
2 hours later…
11:45 PM
@JosephWright Can I run BibTeX with l3build check? Or should I have a pre-built .bbl` file? | 2020-11-27 04:16:26 | {"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": 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": 0.6952950358390808, "perplexity": 5913.513923846547}, "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-2020-50/segments/1606141189038.24/warc/CC-MAIN-20201127015426-20201127045426-00702.warc.gz"} |
https://www.ringomok.com/tag/integration/ | # Tag: integration
## L’Hopital’s Limit From A Phone Game
One of my recently graduated students was playing a phone riddle game that required an answer to this level: Turns out the solution to the game was just simply the word “answer” as the instruction was to simply enter that in. However, the integral and the limit shown there...
## Things to Reconsider: The Integral of $$\frac{1}{x}$$
Absolute Absolutes? Consider the following integral: $\int \frac{1}{x}\,dx = \log_e|x| + C$ When learning this integral, teachers often stress the need to include the absolute value signs. The reasoning behind this is often to do with the fact that the logarithm of a negative number is undefined. However, is... | 2019-05-20 14:37:30 | {"extraction_info": {"found_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": 0.8713565468788147, "perplexity": 790.9387933956987}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232256040.41/warc/CC-MAIN-20190520142005-20190520164005-00347.warc.gz"} |
https://serverug.ru/%D1%86%D0%B5%D1%80%D0%BA%D0%BE%D0%B2%D1%8C/world-in-conflict-mega-trainer-1-011-__top__/ | ### World In Conflict MEGA TRAINER 1.011 __TOP__Номер шаблона: 3268
Демонстрация шаблона: ПОСМОТРЕТЬ
ЦЕНА:
World In Conflict MEGA TRAINER 1.011
A large number of transgenic and knockout mice have been generated to define the pathophysiology of diseases that include retinal diseases. However, there are few reports of mice that express mutant human genes. Introduction of such mouse models is expected to facilitate the understanding of human diseases, since the similarity of the proteins may cause phenotypic changes that are not similar to those of the human disease.
An intravenous (iv) injection of adeno-associated virus-2 (AAV-2) transduces long-term transgene expression in many different tissues in transgenic mice. A similar transduction efficiency is observed in the retina after a local subretinal injection. However, a local subretinal injection of AAV-2 via the vitreous body has several drawbacks, including the small size of the injected tissue, and the need for a surgical procedure. Therefore, an alternative approach is required to achieve efficient and non-surgical transgene expression in the mouse retina.
In this issue, Terasawa et al.^[@i1552-5783-59-3-1396-b01]^ report a new animal model of human retinal diseases. They injected AAV-2 carrying a near-infrared (NIR) fluorescent reporter gene, *eGFP*, subretinally in a mouse model of a dominant *retinitis pigmentosa* (*RP*) family with autosomal dominant RP10 (ADRP10). The authors showed that the construct was effective for the long-term *eGFP* expression in the outer retinal cells, even though AAV-2 did not efficiently transduce the photoreceptors.^[@i1552-5783-59-3-1396-b01]^ In addition, the authors also showed that the subretinal injections induced the local expression of the transgenes in photoreceptor cells in the outer retinal layer.^[@i1552-5783-59-3-1396-b01]^
The ADRP10 is a severe form of autosomal dominant RP (ADRP) that is caused by the p.Gly1896Glu (p.Gly1896Glu) mutation of the X-linked retinitis pigmentosa 1 (*RPGR*) gene,^[@i1552-5783-59-3-1396
It is located on the Appdata folder so it seems that it is the new version 1.011 of the game. Some people are saying that World in Conflict returns to the World War II school of gameplay which is a false claim. It uses a brand new real time strategy engine. It is a sequel to World In Conflict.
List of games with built-in trainers
Category:Windows games
Category:Windows-only games
Category:Windows multimedia software
Category:Video games developed in Russia
Category:1998 video games
Category:World War III speculative fiction
Category:Ubi Soft games
Category:UPlay games
Category:Video games set in Russia
Category:Real-time strategy video gamesQ:
Proving that $f$ is constant on $\{(x,y) : \exists z\in \mathbb{R} ;y=\sin(z)\}$.
Given $f : [0,2\pi] \times \mathbb{R} \rightarrow \mathbb{R}$.
Prove that $f$ is constant on $\{(x,y) : \exists z\in \mathbb{R} ;y=\sin(z)\}$.
My approach: Let $z \in \mathbb{R}$ be arbitrary. Then, $f$ is constant on the set $$\{(x,y) : \exists z\in \mathbb{R} ;y=\sin(z)\} = \{(x,y) : y=\sin(z)\}$$
Now, on the other hand, $f$ is continuous and we can express $f$ as $f(x,y)=g(x,y)+h(x,y)$, where $g(x,y)=f(x,\sin(y))$ and $h(x,y)=f(x,y)-f(x,\sin(y))$.
Since $g$ and $h$ are continuous, we know that they’re bounded by some real numbers $M_1,M_2$ on $\{(x,y) : y=\sin(z)\}$: M_1\leq g(x,y)\leq M_2\quad\forall (x,y)\in\{(x,
55cdc1ed1c | 2022-07-06 15:26:56 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4265872836112976, "perplexity": 3548.6871720031554}, "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-2022-27/segments/1656104675818.94/warc/CC-MAIN-20220706151618-20220706181618-00699.warc.gz"} |
http://nrich.maths.org/public/leg.php?code=5039&cl=2&cldcmpid=5468 | Search by Topic
Resources tagged with Interactivities similar to Factors and Multiples Game:
Filter by: Content type:
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There are 216 results
Broad Topics > Information and Communications Technology > Interactivities
Colour Wheels
Stage: 2 Challenge Level:
Imagine a wheel with different markings painted on it at regular intervals. Can you predict the colour of the 18th mark? The 100th mark?
Stars
Stage: 3 Challenge Level:
Can you find a relationship between the number of dots on the circle and the number of steps that will ensure that all points are hit?
Stage: 2 Challenge Level:
If you have only four weights, where could you place them in order to balance this equaliser?
Factor Lines
Stage: 2 Challenge Level:
Arrange the four number cards on the grid, according to the rules, to make a diagonal, vertical or horizontal line.
Beat the Drum Beat!
Stage: 2 Challenge Level:
Use the interactivity to create some steady rhythms. How could you create a rhythm which sounds the same forwards as it does backwards?
Times Tables Shifts
Stage: 2 Challenge Level:
In this activity, the computer chooses a times table and shifts it. Can you work out the table and the shift each time?
A Dotty Problem
Stage: 2 Challenge Level:
Starting with the number 180, take away 9 again and again, joining up the dots as you go. Watch out - don't join all the dots!
Number Differences
Stage: 2 Challenge Level:
Place the numbers from 1 to 9 in the squares below so that the difference between joined squares is odd. How many different ways can you do this?
Magic Potting Sheds
Stage: 3 Challenge Level:
Mr McGregor has a magic potting shed. Overnight, the number of plants in it doubles. He'd like to put the same number of plants in each of three gardens, planting one garden each day. Can he do it?
Rabbit Run
Stage: 2 Challenge Level:
Ahmed has some wooden planks to use for three sides of a rabbit run against the shed. What quadrilaterals would he be able to make with the planks of different lengths?
Counters
Stage: 2 Challenge Level:
Hover your mouse over the counters to see which ones will be removed. Click to remover them. The winner is the last one to remove a counter. How you can make sure you win?
Coordinate Cunning
Stage: 2 Challenge Level:
A game for 2 people that can be played on line or with pens and paper. Combine your knowledege of coordinates with your skills of strategic thinking.
Multiples Grid
Stage: 2 Challenge Level:
What do the numbers shaded in blue on this hundred square have in common? What do you notice about the pink numbers? How about the shaded numbers in the other squares?
Part the Piles
Stage: 2 Challenge Level:
Try to stop your opponent from being able to split the piles of counters into unequal numbers. Can you find a strategy?
Noughts and Crosses
Stage: 2 Challenge Level:
A game for 2 people that everybody knows. You can play with a friend or online. If you play correctly you never lose!
Multiplication Square Jigsaw
Stage: 2 Challenge Level:
Can you complete this jigsaw of the multiplication square?
Ratio Pairs 2
Stage: 2 Challenge Level:
A card pairing game involving knowledge of simple ratio.
Got it Article
Stage: 2 and 3
This article gives you a few ideas for understanding the Got It! game and how you might find a winning strategy.
Round Peg Board
Stage: 1 and 2 Challenge Level:
A generic circular pegboard resource.
Colour in the Square
Stage: 2 Challenge Level:
Can you put the 25 coloured tiles into the 5 x 5 square so that no column, no row and no diagonal line have tiles of the same colour in them?
Coordinate Tan
Stage: 2 Challenge Level:
What are the coordinates of the coloured dots that mark out the tangram? Try changing the position of the origin. What happens to the coordinates now?
Building Stars
Stage: 2 Challenge Level:
An interactive activity for one to experiment with a tricky tessellation
Domino Numbers
Stage: 2 Challenge Level:
Can you see why 2 by 2 could be 5? Can you predict what 2 by 10 will be?
Code Breaker
Stage: 2 Challenge Level:
This problem is based on a code using two different prime numbers less than 10. You'll need to multiply them together and shift the alphabet forwards by the result. Can you decipher the code?
Seven Flipped
Stage: 2 Challenge Level:
Investigate the smallest number of moves it takes to turn these mats upside-down if you can only turn exactly three at a time.
100 Percent
Stage: 2 Challenge Level:
An interactive game for 1 person. You are given a rectangle with 50 squares on it. Roll the dice to get a percentage between 2 and 100. How many squares is this? Keep going until you get 100. . . .
Square Tangram
Stage: 2 Challenge Level:
This was a problem for our birthday website. Can you use four of these pieces to form a square? How about making a square with all five pieces?
Train
Stage: 2 Challenge Level:
A train building game for 2 players.
Spot Thirteen
Stage: 2 Challenge Level:
Choose 13 spots on the grid. Can you work out the scoring system? What is the maximum possible score?
Which Symbol?
Stage: 2 Challenge Level:
Choose a symbol to put into the number sentence.
Venn Diagrams
Stage: 1 and 2 Challenge Level:
Use the interactivities to complete these Venn diagrams.
Light the Lights Again
Stage: 2 Challenge Level:
Each light in this interactivity turns on according to a rule. What happens when you enter different numbers? Can you find the smallest number that lights up all four lights?
Countdown
Stage: 2 and 3 Challenge Level:
Here is a chance to play a version of the classic Countdown Game.
Cycling Squares
Stage: 2 Challenge Level:
Can you make a cycle of pairs that add to make a square number using all the numbers in the box below, once and once only?
Stage: 2 Challenge Level:
NRICH December 2006 advent calendar - a new tangram for each day in the run-up to Christmas.
Stage: 1 and 2 Challenge Level:
Our 2008 Advent Calendar has a 'Making Maths' activity for every day in the run-up to Christmas.
Memory Game Template
Stage: 2 Challenge Level:
Using angular.js to bind inputs to outputs
See the Light
Stage: 2 and 3 Challenge Level:
Work out how to light up the single light. What's the rule?
Light the Lights
Stage: 2 Challenge Level:
Investigate which numbers make these lights come on. What is the smallest number you can find that lights up all the lights?
Dominoes Environment
Stage: 1 and 2 Challenge Level:
These interactive dominoes can be dragged around the screen.
Power Crazy
Stage: 3 Challenge Level:
What can you say about the values of n that make $7^n + 3^n$ a multiple of 10? Are there other pairs of integers between 1 and 10 which have similar properties?
Makeover
Stage: 1 and 2 Challenge Level:
Exchange the positions of the two sets of counters in the least possible number of moves
Take the Right Angle
Stage: 2 Challenge Level:
How many times in twelve hours do the hands of a clock form a right angle? Use the interactivity to check your answers.
The Path of the Dice
Stage: 2 Challenge Level:
A game for 1 person. Can you work out how the dice must be rolled from the start position to the finish? Play on line.
Junior Frogs
Stage: 2 Challenge Level:
Have a go at this well-known challenge. Can you swap the frogs and toads in as few slides and jumps as possible?
Rod Ratios
Stage: 2 Challenge Level:
Use the Cuisenaire rods environment to investigate ratio. Can you find pairs of rods in the ratio 3:2? How about 9:6?
Nine-pin Triangles
Stage: 2 Challenge Level:
How many different triangles can you make on a circular pegboard that has nine pegs?
Sorting Symmetries
Stage: 2 Challenge Level:
Find out how we can describe the "symmetries" of this triangle and investigate some combinations of rotating and flipping it.
Board Block Challenge
Stage: 2 Challenge Level:
Choose the size of your pegboard and the shapes you can make. Can you work out the strategies needed to block your opponent? | 2013-12-08 20:57:09 | {"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": 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": 0.17722256481647491, "perplexity": 2370.4851531393933}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-48/segments/1386163811461/warc/CC-MAIN-20131204133011-00099-ip-10-33-133-15.ec2.internal.warc.gz"} |
https://stats.stackexchange.com/questions/121854/elastic-net-regularization-mean-square-error-monotonically-increases-with-lambd | Elastic net regularization: mean square error monotonically increases with lambda
This is quite coincidental as my question is nearly identical to this one asked shortly before, but I am also using elastic net regularization with R's glmnet library as a method of variable selection (but in my case it is for a Gaussian and not binomial family).
I have an instance where there does not appear to be any value of lambda which reduces the deviance of the fit. In the left figure below, the $\lambda$ values are selected automatically by cv.glmnet(); in the right figure below I have specified the values to be $\exp(\{-11\ldots-1\})$ (using 250 points evenly spaced between -11 and -1), but in either case it appears there the deviance or mean square error (MSE) is monotonically increasing with $\lambda$.
Does this say something useful about my system, and is there any value of $\lambda$ for which a meaningful interpretation of the selected variables can be extracted?
I would additionally appreciate reference to peer-reviewed literature or conference proceedings which discuss similar cases - I have come up empty-handed in my search.
Many thanks, community.
(For non-R users, the left dotted vertical line in each figure indicates the minimum MSE value - which in my problem corresponds to the minimum value of $\lambda$ selected in both cases - and the right dotted vertical line corresponds to the $\lambda$ of the minimum MSE + 1 standard error solution.)
• The cross-validated MSE seems exactly zero, suggesting zero irreducible error, which is unlikely in real-world data problems. Also, are the MSE values for log(Lambda) from -11 to -5 precisely zero or just very close? Can you perhaps share code and/or data? Feb 26, 2021 at 1:47
The interpretation is exactly the same as I discussed in the Q & A you link to. All your right-hand plot has done is extend the penalty ($\log(\lambda)$) into a region of exponentially decreasing values, i.e. exponentially decreasing amounts of shrinkage.
The CV deviance of the model is flat over a range of values for $\log(\lambda)$ yet there is shrinkage being applied; sufficient shrinkage to remove some variables from the model completely. I'm not sure you can trust the values on the upper axis for your automatic plot - do you really have 1636 covariates in the model? - but for the left-hand plot the simplest/smaller model within 1 SE of the best model has somewhere between 100 and 87 covariates in it.
• @crippledlambda then, in that case, you are seeing huge shrinkage for effectively no impact on the model's ability to explain variance in your sample. That is good! I think at low values of $\lambda$ you are overfitting so go with the 1SE mode fit. Oct 29, 2014 at 3:48
• @crippledlambda your setting is sparse; you have remove ~ 1800 of the predictors from the model by shrinking their coefficients to zero. The CV is a stochastic procedure; samples are assigned to folds at random hence one would expect some variation between runs unless you set the random seed the same each time. If you want to hone in the best 1SE value of $\lambda$ you'll need to use a lot more points over range say -6 to -3. but really it isn't going to matter much in terms of prediction. Oct 29, 2014 at 4:11
• Alternatively, run the CV quite a few times and record the $\lambda$ for the 1SE solution from each of these. Use a representative value for your chosen $\lambda$. Oct 29, 2014 at 4:13 | 2022-05-18 15:34:03 | {"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": 0.7102240324020386, "perplexity": 667.4500209281224}, "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/1652662522284.20/warc/CC-MAIN-20220518151003-20220518181003-00046.warc.gz"} |
http://www.gradesaver.com/textbooks/math/other-math/basic-college-mathematics-9th-edition/chapter-3-adding-and-subtracting-fractions-review-exercises-page-254/25 | ## Basic College Mathematics (9th Edition)
Published by Pearson
# Chapter 3 - Adding and Subtracting Fractions - Review Exercises: 25
#### Answer
$\frac{5}{8}$
#### Work Step by Step
$\frac{5}{12}$ + $\frac{5}{24}$ _____ The least common multiple of 12 and 24 is 24. Rewrite both fractions as fractions with a least common denominator of 24. Rewrite as like fraction $\frac{5}{12}$ + $\frac{5}{24}$= $\frac{10}{24}$ + $\frac{5}{24}$ Add the numerator = $\frac{10 + 5}{24}$ = $\frac{15}{24}$ $\frac{15}{24}$ is not in lowest term, Divide the numerator and denominator by common multiple 3 $\frac{15}{24}$ = $\frac{5}{8}$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 2017-10-21 05:12:24 | {"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": 0.8229838013648987, "perplexity": 744.6484098804561}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187824570.79/warc/CC-MAIN-20171021043111-20171021063111-00855.warc.gz"} |
https://socratic.org/questions/how-do-you-write-f-x-x-2-3x-1-in-vertex-form | # How do you write f(x) = x^2 - 3x + 1 in vertex form?
In vertex form $f \left(x\right) = {\left(x - \frac{3}{2}\right)}^{2} - \frac{5}{4}$ Vertex is $\left(\frac{3}{2} , - \frac{5}{4}\right)$
$f \left(x\right) = {x}^{2} - 3 x + 1 = \left({x}^{2} - 3 x + \frac{9}{4}\right) - \frac{9}{4} + 1 = {\left(x - \frac{3}{2}\right)}^{2} - \frac{5}{4}$ Vertex =$\left(\frac{3}{2} , - \frac{5}{4}\right)$ graph{x^2-3x+1 [-10, 10, -5, 5]}[Ans] | 2021-11-27 06:08:15 | {"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": 0.5284816026687622, "perplexity": 3123.7601470809946}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1637964358118.13/warc/CC-MAIN-20211127043716-20211127073716-00463.warc.gz"} |
https://zbmath.org/?q=an%3A1433.35447 | ## Global Hölder regularity for the fractional $$p$$-Laplacian.(English)Zbl 1433.35447
Summary: By virtue of barrier arguments we prove $$C^\alpha$$-regularity up to the boundary for the weak solutions of a non-local, non-linear problem driven by the fractional $$p$$-Laplacian operator. The equation is boundedly inhomogeneous and the boundary conditions are of Dirichlet type. We employ different methods according to the singular $$(p<2)$$ of degenerate $$(p>2)$$ case.
### MSC:
35R11 Fractional partial differential equations 35B65 Smoothness and regularity of solutions to PDEs 35J92 Quasilinear elliptic equations with $$p$$-Laplacian 47G20 Integro-differential operators
Full Text:
### References:
[1] Aikawa, H., Kilpel”ainen, T., Shanmugalingam, N. and Zhong, X.:Boundary Harnack principle forp-harmonic functions in smooth Euclidean domains. Potential Anal.26 (2007), no. 3, 281–301. · Zbl 1121.35060 [2] Baernstein, A. II: A unified approach to symmetrization. In Partial differential equations of elliptic type (Cortona, 1992), 47–91. Symposia Mathematica 35, Cambridge University Press, Cambridge, 1994. [3] Bjorland, C., Caffarelli, L. and Figalli, A.: Non-local gradient dependent operators. Adv. Math.230 (2012), no. 4-6, 1859–1894. · Zbl 1252.35099 [4] Cabr’e, X. and Sire, Y.:Nonlinear equations for fractional Laplacians I: Regularity, maximum principles, and Hamiltonian estimates. Ann. Inst. H. Poincar’e Anal. Non Lin’eaire31 (2014), no. 1, 23–53. · Zbl 1286.35248 [5] Caffarelli, L. and Silvestre, L.: Regularity theory for fully nonlinear integrodifferential equations. Comm. Pure Appl. Math.62 (2009), no. 5, 597–638. · Zbl 1170.45006 [6] Caffarelli, L. and Silvestre, L.: Regularity results for nonlocal equations by approximation. Arch. Ration. Mech. Anal.200 (2011), no. 1, 59–88. · Zbl 1231.35284 [7] Di Castro, A., Kuusi, T. and Palatucci, G.: Local behavior of fractionalpminimizers. Ann. Inst. H. Poincar’e Anal. Non Lin’eaire33 (2016), no. 5, 1279–1299. · Zbl 1355.35192 [8] Di Castro, A., Kuusi, T. and Palatucci, G.: Nonlocal Harnack inequalities. J. Funct. Anal.267 (2014), no. 6, 1807–1836. · Zbl 1302.35082 [9] Di Nezza, E., Palatucci, G. and Valdinoci, E.: Hitchhiker’s guide to the fractional Sobolev spaces. Bull. Sci. Math.136 (2012), no. 5, 521–573. · Zbl 1252.46023 [10] Iannizzotto, A., Liu, S., Perera, K. and Squassina, M.: Existence results for fractionalp-Laplacian problems via Morse theory. Adv. Calc. Var. 9 (2016), no. 2, 101–125. · Zbl 06567151 [11] Iannizzotto, A., Mosconi, S. and Squassina, M.:HsversusC0-weighted minimizers. NoDEA Nonlinear Differential Equations Appl.22 (2015), no. 3, 477–497. · Zbl 1339.35201 [12] Iannizzotto, A., Mosconi, S. and Squassina, M.:, A note on global regularity for the weak solutions of fractionalp-Laplacian equations. Atti Accad. Naz. Lincei 1392A. Iannizzotto, S. Mosconi and M. Squassina · Zbl 1336.35360
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching. | 2022-08-08 04:03:13 | {"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": 0.47680479288101196, "perplexity": 5026.6496812537835}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1659882570765.6/warc/CC-MAIN-20220808031623-20220808061623-00638.warc.gz"} |
https://wiki.geogebra.org/en/LaTeX | # LaTeX
#### GeoGebra Objects
In GeoGebra you can write formulas as well. To do so, check the box LaTeX formula in the dialog window of the Text Tool and enter your formula in LaTeX syntax.
Note: In order to create text that contains a LaTeX formula as well as text you may enter the text inside \text{}, while LaTex Formula is activated.
Example: \text{The length of the diagonal is } \sqrt{ 2 }
Note: You can also use the FormulaText Command to enter your formula within quotes.
Example: FormulaText["\text{The length of the diagonal is } \sqrt{ 2 }"]
Note: You can simply obtain a LaTeX text containing the value of an object listed in the Algebra View by dragging that object in the Algebra View and dropping it in a selected location of the Graphics View .
You can find the syntax for common formula symbols from the drop-down menu next to the LaTeX checkbox (GeoGebra Desktop Version) or in the Advanced menu below the input field (GeoGebra Web and Tablet Apps Version). This inserts the corresponding LaTeX code into the text field and places the cursor in between a set of curly brackets. The Symbols drop-down menu contains a list of common math symbols, Greek letters and operators. If you would like to create dynamic text within the formula, you need to select the relating objects from the Objects drop-down list, causing GeoGebra to insert their names as well as the syntax for mixed text.
Some important LaTeX commands are explained in following table. Please have a look at any LaTeX documentation for further information.
LaTeX input Result
a \cdot b \mathrm{\mathsf{ a \cdot b }}
\frac{a}{b} \mathrm{\mathsf{ \frac{a}{b} }}
\sqrt{x} \mathrm{\mathsf{ \sqrt{x} }}
\sqrt[n]{x} \mathrm{\mathsf{ \sqrt[n]{x} }}
\vec{v} \mathrm{\mathsf{ \vec{v} }}
\overline{AB} \mathrm{\mathsf{ \overline{AB} }}
x^{2} \mathrm{\mathsf{ x^{2} }}
a_{1} \mathrm{\mathsf{ a_{1} }}
\sin\alpha + \cos\beta \mathrm{\mathsf{ \sin\alpha + \cos\beta }}
\int_{a}^{b} x dx \mathrm{\mathsf{ \int_{a}^{b} x dx }}
\sum_{i=1}^{n} i^2 \mathrm{\mathsf{ \sum_{i=1}^{n} i^2 }} | 2017-05-28 06:54:38 | {"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": 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": 0.9371199607849121, "perplexity": 3813.3544810584863}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-22/segments/1495463609605.31/warc/CC-MAIN-20170528062748-20170528082748-00213.warc.gz"} |
https://intl.siyavula.com/read/science/grade-12/work-energy-and-power/05-work-energy-and-power-05 | We think you are located in South Africa. Is this correct?
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Energy conservation
Exercise 5.3
A $$\text{60,0}$$ $$\text{kg}$$ skier with an initial speed of $$\text{12,0}$$ $$\text{m·s^{-1}}$$ coasts up a $$\text{2,50}$$ $$\text{m}$$-high rise as shown in the figure. Find her final speed at the top, given that the coefficient of friction between her skis and the snow is $$\text{0,0800}$$. (Hint: Find the distance traveled up the incline assuming a straight-line path as shown in the figure.)
We need to determine the length of the slope as this is the distance over which friction acts as well as the normal force of the skier on the slope to determine the magnitude of the force due to friction. The normal force balances the component of gravity perpendicular to the slope, therefore: \begin{align*} F_{\text{friction}}&= - \mu N \\ &= - \mu F_g\cos\theta \\ &= - \mu mg \cos\theta \end{align*} The length of the slope will be $$\Delta x = \dfrac{h}{\sin\theta}$$.
\begin{align*} W_{\text{non-conservative}} + E_{k,i} + E_{p,i} & = E_{k,f} + E_{p,f} \\ -\mu mg\cos\theta \dfrac{h}{\sin\theta} + \frac{1}{2}mv_i^2 +mgh_i &= \frac{1}{2}mv_f^2 + mgh_f \\ -\mu g\cos\theta \dfrac{h}{\sin\theta} + \frac{1}{2}v_i^2 +gh_i &= \frac{1}{2}v_f^2 + gh_f \\ \frac{1}{2}v_f^2 + gh_f &= -\mu g\cos\theta \dfrac{h}{\sin\theta} + \frac{1}{2}v_i^2 +gh_i \\ \frac{1}{2}v_f^2 + &= -\mu g\cos\theta \dfrac{h}{\sin\theta} + \frac{1}{2}v_i^2 +g(h_i -h_f) \\ v_f^2 &= -2\mu g\cos\theta \dfrac{h}{\sin\theta} + v_i^2 +2g(h_i -h_f) \end{align*} \begin{align*} v_f^2 &= -2(0,08)(9,8) \cos(35) \dfrac{2.5}{\sin(35)} + (12)^2 +2(9,8)(-2,5) \\ v_f & = \sqrt{89,4016598} \\ v_f & = \text{9,46}\text{ m·s$^{-1}$} \end{align*} $$\text{9,46}$$ $$\text{m·s^{-1}}$$
How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and its initial speed is $$\text{110}$$ $$\text{km·h^{-1}}$$?
\begin{align*} 110 km/hr&= 110 \times \frac{1000}{3600} \\ &= \text{30,56}\text{ m·s$^{-1}$} \end{align*} \begin{align*} mgh&= \frac{1}{2}mv^2 \\ h&= \frac{v^2}{2g} \\ &= \frac{(\text{30,56})^2}{2(\text{9,8})} \\ &= \text{47,65}\text{ m} \end{align*}
$$\text{47,65}$$ $$\text{m}$$
If, in actuality, a $$\text{750}$$ $$\text{kg}$$ car with an initial speed of $$\text{110}$$ $$\text{km·h^{-1}}$$ is observed to coast up a hill to a height $$\text{22,0}$$ $$\text{m}$$ above its starting point, how much thermal energy was generated by friction?
Kinetic energy was converted into potential energy. The addition of friction as a dissipative force ensures that some of the kinetic energy is lost as thermal energy. The difference in the potential energy gained with and without friction is the energy lost to friction. The gravitational potential energy gained is only related to height so the difference in height allows us to determine the energy lost to friction quickly.
Without friction the car rose to a height of $$\text{47,65}$$$$\text{m}$$, with friction the height was only $$\text{22}$$$$\text{m}$$. The energy lost to friction is equivalent to the energy required to increase the gravitational potential energy by raising the vehicle a height of $$\text{25,65}$$$$\text{m}$$. Therefore the energy lost to friction is:
\begin{align*} W&= mgh \\ &= (750)(9,8)(25.65) \\ & = \text{188 527,5}\text{ J} \end{align*}
We calculated the gravitational potential energy that was lost as friction but the correct answer will be negative this quantity as the work done by friction is negative: -$$\text{188 527,5}$$ $$\text{J}$$
What is the average force of friction if the hill has a slope $$\text{2,5}$$ $$\text{º}$$ above the horizontal?
We will assume a constant force of friction. The disance over which the friction acted is the length of the slope, we know the angle of the slope and the vertical height so we can calculate the distance (hypotenuse). We can use the definition of work to calculate the force, remember that friction acts opposite to the displacement:
\begin{align*} W_{friction}&= F_{friction}\Delta x \cos\theta \\ &= -F \Delta x \\ &= -F \frac{h}{\sin(2.5)} \\ -\text{188 527,5}&= -F \frac{h}{\sin(2.5)} \\ F&= \text{188 527,5} \frac{\sin(2.5)}{22} \\ F&= \text{373,79}\text{ N} \end{align*} $$\text{373,79}$$ $$\text{N}$$
A bullet traveling at 100 m/s just pierces a wooden plank of 5 m. What should be the speed (in m/s) of the bullet to pierce a wooden plank of same material, but having a thickness of 10m?
Final speed and hence final kinetic energy are zero in both cases. From "work- kinetic energy" theorem, initial kinetic energy is equal to work done by the force resisting the motion of bullet. As the material is same, the resisting force is same in either case. If subscript "1" and "2" denote the two cases respectively, then:
5m:
\begin{align*} 0−\frac{1}{2}mv_1^2 & = −Fx_1 \\ \frac{1}{2}m(100)^2=F\times 5 \end{align*}
10m:
\begin{align*} 0−\frac{1}{2}mv_2^2 & = −Fx_2 \\ \frac{1}{2}mv_2^2=F\times 10 \end{align*}
Take the ratio of these two equations:
\begin{align*} \frac{v_2^2}{(100)^2} &=\frac{F\times 10}{F\times 5}=2 \\ v_2^2 & = 2 \times 10000 = 20000 \\ v_2 & = \text{141,4}\text{ m·s$^{-1}$} \end{align*}
$$\text{141,4}$$ $$\text{m·s^{-1}}$$ | 2019-08-18 19:02:49 | {"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": 1.0000098943710327, "perplexity": 1363.6432426939507}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027313996.39/warc/CC-MAIN-20190818185421-20190818211421-00462.warc.gz"} |
https://sirhunte.teachable.com/courses/pure-mathematics-ii/lectures/31432618 | ## - Mathematical Modelling
LESSON 1
(a) A pond is initially empty and is then filled gradually with water. After $$t$$ minutes, the depth of the water, $$x$$ metres, satisfies the differential equation
$$\displaystyle {dx\over dt}={\sqrt{4+5x}\over 5(1+t)^2}$$
Solve this differential equation to find $$x$$ in terms of $$t$$.
(b) Another pond is gradually filling with water, after $$t$$ minutes, the surface of the water forms a circle of radius $$r$$ metres. The rate of change of the radius is inversely proportional to the area of the surface of the water.
(i) Write down a differential equation, in the variables $$r$$ and $$t$$ and a constant of proportionality, which represents how the radius of the surface of the water is changing with time.
(You are not required to solve your differential equation.)
(ii) When the radius of the pond is 1 metre, the radius is increasing at a rate of 4.5 metres per second. Find the radius of the pond when the radius is increasing at a rate of 0.5 metres per second.
SOLUTION
(a) Solve as a separable differential equation
$$\displaystyle {dx\over dt}={\sqrt{4+5x}\over 5(1+t)^2}$$
$$\displaystyle {1\over \sqrt{4+5x}}dx={1\over 5(1+t)^2} dt$$
$$\displaystyle ∫(4+5x)^{-{1\over2}} dx={1\over5} ∫(1+t)^{-2} dt$$
$$\displaystyle {(4+5x)^{{1\over2}}\over 5\left({1\over2}\right)} ={1\over5} [-(1+t)^{-1}]+c$$
$$\displaystyle {2\over5} \sqrt{4+5x}=-{1\over5} \left({1\over 1+t}\right)+c$$
Since the pond is initially empty, this means that at $$t=0, x=0$$. We can use these values to determine $$c$$
When $$x=0, t=0$$
$$\displaystyle {2\over5} \sqrt{4+5(0)}=-{1\over5} \left(1\over 1+0\right)+c$$
$$1=c$$
• Substitute and solve for $$x$$
$$\displaystyle {2\over5} \sqrt{4+5x}=1-{1\over5} \left({1\over 1+t}\right)$$
$$\displaystyle \sqrt{4+5x}={5\over2}-{1\over2} \left(1\over 1+t\right)$$
$$\displaystyle 4+5x=\left[{5\over2}-{1\over 2(1+t)} \right]^2$$
$$\displaystyle 5x=\left[{5\over2}-{1\over 2(1+t)}\right]^2-4$$
$$\displaystyle x={1\over5} \left[{5\over2}-{1\over 2(1+t)} \right]^2-{4\over5}$$
(b) (i) Since $$\displaystyle {dr\over dt}$$ represents the rate of change of the radius and there is inverse
proportionality
$$\displaystyle {dr\over dt}={k\over πr^2}$$
(ii) Use the fact that
$$\displaystyle r=1, {dr\over dt}=4.5$$
$$\displaystyle ⟹{9\over2}={k\over π(1)^2}$$
$$\displaystyle ⟹{9\over2}={k\over π}$$
$$\displaystyle ⟹k={9π\over2}$$
• Rewrite the differential equation with the determined value of $$k$$.
$$\displaystyle {dr\over dt}={9\over 2r^2}$$
• Use equation to determine required value
When $$\displaystyle {dr\over dt}={1\over 2}$$
$$\displaystyle {1\over2}={9\over 2r^2}$$
$$r^2=9$$
$$r=3$$
LESSON 2
The number of bacteria in a liquid culture is observed to grow at a rate proportional to the number of cells present. At the beginning of the experiment there are 10,000 cells and after three hours there are 500,000. How many will there be after one day of growth if this unlimited growth continues? What is the doubling time of the bacteria?
SOLUTION
• Let $$y(t)$$ represent the number of bacteria present at time $$t$$. Then the rate of change is
$$\displaystyle {dy\over dt}=ky$$
where $$k$$ is the constant of proportionality.
• Solve as a separable differential equation
$$\displaystyle {1\over y}\:dy=k\:dt$$
$$\displaystyle ∫{1\over y}\:dy=∫k\: dt$$
$$\ln y=kt+c$$
$$e^{\ln y} =e^{kt}+c$$
$$y=e^{kt+c}$$
$$y=e^c e^{kt}$$
$$y=Ae^{kt}$$ where $$A=e^c$$
$$y(t)=Ae^{kt}$$
• Use the fact that the initial population occurs when $$t=0$$ (initial population) to determine the value of $$A$$
$$y(0)=Ae^{k(0)} =A$$
Therefore $$A$$ is the initial population.
$$⟹A=10000$$
• Use the information from the problem to determine $$k$$
$$500 000=10 000e^{3k}$$
$$50=e^{3k}$$
$$\ln 50=3k$$
$$\displaystyle k={1\over 3} \ln 50\approx 1.304$$
$$y=10 000e^{1.304t}$$
Doubling time refers to the amount of time for the bacteria to double, in number, from its original number.
The original number is 10,000 so we need to determine how long it takes to get to 20,000.
$$20000=10000e^{1.304t}$$
$$2=e^{1.304t}$$
$$\ln 2=1.304t$$
$$\displaystyle t={\ln 2\over 1.304}\approx 0.532$$ hours | 2021-04-20 10:47:54 | {"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": 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": 0.9148581027984619, "perplexity": 248.49435480965826}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1618039388763.75/warc/CC-MAIN-20210420091336-20210420121336-00563.warc.gz"} |
http://mathhelpforum.com/calculus/221350-has-limit-everywhere-but-continuous-nowhere.html | # Math Help - Has limit everywhere but continuous nowhere?
1. ## Has limit everywhere but continuous nowhere?
Hi again MHF members,
originating from the affirmative answer of the problem "Continuous Everywhere but Differentiable Nowhere",
I would like to ask whether or not it is possible to find a function,
which has a limit at each point in its domain but it is not continuous at all?
Thanks.
bkarpuz
2. ## Re: Has limit everywhere but continuous nowhere?
Yes it is possible.Let's say there is a hole at (3,3) on the function f(x)=x. There is a limit, but the function ends up not being continuous. Know the three conditions of continuity, 1. f(x) must exist 2. lim x->a f(x) must exist 3. f(x) must equal lim x->a f(x)
3. ## Re: Has limit everywhere but continuous nowhere?
Yes it is possible.Let's say there is a hole at (3,3) on the function f(x)=x. There is a limit, but the function ends up not being continuous. Know the three conditions of continuity, 1. f(x) must exist 2. lim x->a f(x) must exist 3. f(x) must equal lim x->a f(x)
Your example is continuous almost everywhere on its domain.
Originally Posted by bkarpuz
it is possible to find a function,
which has a limit at each point in its domain but it is not continuous at all?
4. ## Re: Has limit everywhere but continuous nowhere?
Okay, I see my error.
5. ## Re: Has limit everywhere but continuous nowhere?
Originally Posted by bkarpuz
originating from the affirmative answer of the problem "Continuous Everywhere but Differentiable Nowhere", I would like to ask whether or not it is possible to find a function,
which has a limit at each point in its domain but it is not continuous at all?
You seem to have a perverse idea as to what continuity actually means.
Continuity is intrinsically connected to limits. Look at this webpage.
If you will just stop and think about it, then you can see what a nonsense question this really is.
6. ## Re: Has limit everywhere but continuous nowhere?
But what about the Weierstrass function Weierstrass function - Wikipedia, the free encyclopedia?
I've always used this function as a mnemonic to remember that Differentiability implies continuity but not necessarily the other way around. Continuity everywhere, by definition, implies existence of a limit everywhere and furthermore that the limit at each point c, for all c in the domain, equals f(c).
7. ## Re: Has limit everywhere but continuous nowhere?
Originally Posted by Plato
You seem to have a perverse idea as to what continuity actually means.
Continuity is intrinsically connected to limits. Look at this webpage.
If you will just stop and think about it, then you can see what a nonsense question this really is.
I don't follow you, Plato. If a function f is continuous then the limit of that function at c exists and is equal to f(c) for all c in the domain. However, bkarpuz is asking about a case where the limit exists everywhere but the function is not continuous. This can occur for a function whose limit everywhere in the domain exists but is not equal to what the function evaluate to at each respective point.
Existence of a limit alone doesn't guarantee continuity.
8. ## Re: Has limit everywhere but continuous nowhere?
Originally Posted by bkarpuz
Hi again MHF members,
originating from the affirmative answer of the problem "Continuous Everywhere but Differentiable Nowhere",
I would like to ask whether or not it is possible to find a function,
which has a limit at each point in its domain but it is not continuous at all?
Thanks.
bkarpuz
Your question is inconsistent. The function you're interested in is 1. Continuous everywhere and 2. Differentiable nowhere. And then, referring to that function, you ask about a function that has the properties: 1. The limit exists everywhere. 2. Not continuous everywhere.
9. ## Re: Has limit everywhere but continuous nowhere?
Plato, I could do see that it does not sound good for reals. I also know that for discrete topological spaces it is not also possible. This is why I didn't say anything about the domain. It can be some kind of a dense but not complete topological space... | 2014-07-22 08:01:55 | {"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": 0.9182547330856323, "perplexity": 591.8036397876422}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1405997857710.17/warc/CC-MAIN-20140722025737-00105-ip-10-33-131-23.ec2.internal.warc.gz"} |
https://ask.libreoffice.org/en/users/9605/basicmacro/?sort=recent | 2016-01-19 18:12:38 +0100 answered a question Copy by preview SHEET This works even if the previous day was the last day of a month (e.g 30 or 31). Dim iValue As Integer ' Get value from previous sheet' iValue = ThisComponent.Sheets.getByName(Day(Now()-1)).getCellByPosition(0, 0).getValue() ThisComponent.CurrentController.ActiveSheet = ThisComponent.Sheets.getByName(Day(Now())) ' Set value in todays sheet' ThisComponent.Sheets.getByName(Day(Now())).getCellByPosition(0, 0).setValue(iValue+1) .getCellByPosition(nColumn, nRow) nColumn is the column index of the cell. nRow is the row index of the cell. Cell positions start at zero. Examples A1 = (0,0) B1 = (1,0) A2 = (0,1) A5 = (0,4) E8 = (4,7) Other methods .setValue(Number) .setString(Text) .setFormula(Formula) 2016-01-19 18:11:35 +0100 answered a question How may I turn my ODS sheet on its side? With the Sidebar On the right of the screen, click Properties (spanner icon). Then under Page, click Orientation and change to Landscape. If you have an old version of LibreOffice or the sidebar on the right is disabled. With the Menu On the top of the screen, click the menu Format, then Page... In the window that pops-up, click the Page tab along the top. Change Orientation to Landscape. 2016-01-18 14:18:20 +0100 answered a question Is it possible to reduce the height of the UI? Click the menu Tools, then Options. On the left expand LibreOffice then click View. On the right, under User Interface. Icon size and style only changes the Icons. Scaling changes the menu. Update: 22/Jan/2016 The height should automatically change to fit the text. Since it does not then it is likely your LibreOffice user profile is corrupted. Before resetting the whole LO user folder, try resetting the file registrymodifications.xcu For Windows 7 the location is %appdata%\libreoffice\4\user (LibreOffice 4 & 5) %appdata%\libreoffice\3\user (LibreOffice 3) From the Wiki - Resolving corruption in the user profile In many cases it would be sufficient to only rename/(re)move the file registrymodifications.xcu within the user profile to reset all options to their default values, instead of moving the entire user profile directory. Doing so preserves all user installed extensions, gallery, auto-correction entries and auto-text and so on. But of course this would not help if, for example, one of the extensions or some other corrupted file was the culprit. You may want to experiment with that. But mind that this has effect on registration of macro's too, and makes that custom colors are removed (standard.soc overwritten). 2014-03-22 00:01:39 +0100 answered a question **HELP** With Auto Install .BAT file, 99% there..... params I used "LibreOffice_4.1.5_Win_x86.msi" on 32bit Vista. I removed the double quotes after the switch /i and also gm_o_jf_Palm, gm_o_jf_Pocketpc, gm_r_ex_Dictionary_Rs. :15 msiexec /qn /i "F:\sws\Programs.New Setups\LibreOffice 4.2.0.msi" /l* "F:\sws\Programs.New Setups\LibO_install_log.txt" SELECT_WORD=1 SELECT_EXCEL=1 SELECT_POWERPOINT=1 REGISTER_ALL_MSO_TYPES=1 UI_LANGS=en_US RebootYesNo=No ISCHECKFORPRODUCTUPDATES=1 QUICKSTART=1 VC_REDIST=1 ADDLOCAL=ALL REMOVE=gm_r_ex_Dictionary_Af,gm_r_ex_Dictionary_An,gm_r_ex_Dictionary_Ar,gm_r_ex_Dictionary_Be,gm_r_ex_Dictionary_Bg,gm_r_ex_Dictionary_Bn,gm_r_ex_Dictionary_Br,gm_r_ex_Dictionary_Pt_Br,gm_r_ex_Dictionary_Pt_Pt,gm_r_ex_Dictionary_Ca,gm_r_ex_Dictionary_Cs,gm_r_ex_Dictionary_Da,gm_r_ex_Dictionary_Nl,gm_r_ex_Dictionary_Et,gm_r_ex_Dictionary_Gd,gm_r_ex_Dictionary_Gl,gm_r_ex_Dictionary_Gu,gm_r_ex_Dictionary_De,gm_r_ex_Dictionary_He,gm_r_ex_Dictionary_Hi,gm_r_ex_Dictionary_Hu,gm_r_ex_Dictionary_It,gm_r_ex_Dictionary_Ku_Tr,gm_r_ex_Dictionary_Lt,gm_r_ex_Dictionary_Lv,gm_r_ex_Dictionary_Ne,gm_r_ex_Dictionary_No,gm_r_ex_Dictionary_Oc,gm_r_ex_Dictionary_Pl,gm_r_ex_Dictionary_Ro,gm_r_ex_Dictionary_Ru,gm_r_ex_Dictionary_Si,gm_r_ex_Dictionary_Sk,gm_r_ex_Dictionary_Sl,gm_r_ex_Dictionary_El,gm_r_ex_Dictionary_Es,gm_r_ex_Dictionary_Sv,gm_r_ex_Dictionary_Te,gm_r_ex_Dictionary_Th,gm_r_ex_Dictionary_Uk,gm_r_ex_Dictionary_Vi,gm_r_ex_Dictionary_Zu,gm_r_ex_Dictionary_Fr,gm_r_ex_Dictionary_Hr Tips Use the switch /qf to see what options are removed with REMOVE= Info about command line install - Deployment_and_Migration DO NOT warn if not in ODF format may need to be done post install. See the section Some tricks for post installation configuration which is at the bottom of the page in the link above. 2014-03-19 10:29:09 +0100 received badge ● Teacher 2014-03-18 09:10:37 +0100 answered a question How to disable auto save method in libreoffice Click on the menu "Tools", then "Options" On the left click the "+" beside "Load/Save", then click "General". On the right Untick "Save AutoRecovery information every ... Minutes" 2014-03-10 18:42:22 +0100 answered a question Counting/Restart Count based on another column This only works if the numbers in column A are sorted. Put 1 in cell B1 as this cell will never change. In B2 put this formula. =IF(A1=A2,B1+1,1) How it works, IF A1 = A2 THEN add 1 to the previous value in column B, ELSE reset the counter to 1. 2014-03-07 20:42:36 +0100 received badge ● Editor (source) 2014-03-07 20:41:55 +0100 answered a question Copying part of a string after instance of a character A less elegant way using a formula =RIGHT(A1, LEN(A1) - (SEARCH("\",A1,SEARCH("\",A1,SEARCH("\",A1)+1)+1)) ) To find the location of the third backslash - Basically it searches for the location of the first backslash - Then it starts another search from that location to find the second backslash - Then another from that location to find the location of the third backslash Using the RIGHT function - RIGHT( Text , Number of chars at the end of the string to keep ) 2014-03-06 20:42:46 +0100 commented question again about loop in macro Change from a single cell to a range of cells. (Change "$A$2" to "$A$2:$A$10"). It should work for you, but if you still want to loop through each cell individually, then I'll add a loop. | 2020-02-24 09:36:21 | {"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": 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": 0.23169754445552826, "perplexity": 3934.743385342646}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875145910.53/warc/CC-MAIN-20200224071540-20200224101540-00386.warc.gz"} |
https://me.gateoverflow.in/591/gate2016-2-5 | # GATE2016-2-5
Numerical integration using trapezoidal rule gives the best result for a single variable function, which is
1. linear
2. parabolic
3. logarithmic
4. hyperbolic
recategorized
## Related questions
The error in numerically computing the integral $\int_{0}^{\pi }(\sin x+\cos x)dx$ using the trapezoidal rule with three intervals of equal length between $0$ and $\pi$ is ___________
The root of the function $f(x) = x^3+x-1$ obtained after first iteration on application of Newton-Raphson scheme using an initial guess of $x_0=1$ is $0.682$ $0.686$ $0.750$ $1.000$
Solve the equation $x=10\cos(x)$ using the Newton-Raphson method. The initial guess is $x=\pi /4$. The value of the predicted root after the first iteration, up to second decimal, is ________
The best approximation of the minimum value attained by $e^{-x}\sin(100x)$ for $x\geq 0$ is _______
The definite integral $\int_{1}^{3}\dfrac{1}{x}$ is evaluated using Trapezoidal rule with a step size of $1$. The correct answer is _______ | 2021-09-18 00:49:48 | {"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": 0.8163779973983765, "perplexity": 318.63882648489715}, "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/1631780056120.36/warc/CC-MAIN-20210918002951-20210918032951-00012.warc.gz"} |
http://dataspace.princeton.edu/jspui/handle/88435/dsp017s75df78b | Please use this identifier to cite or link to this item: http://arks.princeton.edu/ark:/88435/dsp017s75df78b
Title: Effects of Bounded Rationality in Financial Markets Authors: Kim, Jeong-Ho (John) Advisors: Sims, Christopher A. Contributors: Economics Department Keywords: active managementlearningmutual fundsrational inattentionreaching for yieldunconventional monetary policy Subjects: EconomicsFinanceEconomic theory Issue Date: 2016 Publisher: Princeton, NJ : Princeton University Abstract: The first chapter studies the historical growth of active asset management despite the industry's poor track record and its coincidence with net fund entry. Our quantitative model of demand for active management features learning about heterogeneity in skills and about fund-level decreasing returns to scale. Investors are uncertain about parameters governing fund returns and returns to scale, and they learn about them from realized returns. After observing a fund's negative performance, investors infer that the manager's skill is lower than expected rather than that the value of active management more generally is lower than expected. Optimism about the industry as a whole persists at the expense of disappointment about existing individual funds. On the other hand, fund-level decreasing returns will imply that the average unit cost associated with investing in active management is lower as the number of funds increases and, ceteris paribus, make the industry grow bigger over time. The second chapter, coauthored with Delwin Olivan, studies the effect of monetary policy and ow-performance relationship on risk taking by active equity mutual funds. First, we document that the past decade provided conditions that encouraged these funds to reach for yield’’. Low interest rate periods are associated with both higher measures of performance and risk taking. We then utilize discrete Fed announcements providing forward guidance to inform event studies analyzing these factors. Our results are broadly consistent with these funds reaching for yield, and provide evidence of a strong interaction between unconventional low-rate policy and mutual fund behavior. The third chapter studies the effects of rational inattention to public information in the beauty contest’’ model à la Morris and Shin (2002). If the preference for action complementarity is strong, for intermediate levels of the public signal precision, multiple equilibria, indexed by the level of attention allocation, emerge. I also show that equilibrium attention allocation is inefficient for a wide range of model parameters. Key to these results is the fact that strategic complementarity translates into attention complementarity, when the information cost is positive, but sufficiently small (compared to zero). Quintessentially, almost perfect attention’’ is very different from perfect attention assumption. URI: http://arks.princeton.edu/ark:/88435/dsp017s75df78b Alternate format: The Mudd Manuscript Library retains one bound copy of each dissertation. Search for these copies in the library's main catalog: http://catalog.princeton.edu/ Type of Material: Academic dissertations (Ph.D.) Language: en Appears in Collections: Economics | 2016-12-07 20:25:47 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.318116158246994, "perplexity": 4515.788089751941}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-50/segments/1480698542246.21/warc/CC-MAIN-20161202170902-00335-ip-10-31-129-80.ec2.internal.warc.gz"} |
https://mathematics.huji.ac.il/eventss/events-seminars?page=17 | 2019 May 14
# Dynamics Lunch: Hongming Nie "Preperiodic points of unicritial polynomials”.
12:00pm to 1:00pm
Following the paper “ Preperiodic points and unlikely intersection” by Baker and DeMarco.
2019 May 13
# NT & AG Lunch: Michael Temkin "Elliptic curves with complex multiplication, II"
1:00pm to 2:00pm
## Location:
Faculty lounge, Math building
2019 May 13
# NT & AG Seminar: "A dream desingularization algorithm", Michael Temkin (HU)
2:30pm to 4:00pm
## Location:
Ross 70
Abstract: Any birational geometer would agree that the best algorithm
for resolution of singularities should run by defining a simple invariant of
the singularity and iteratively blowing up its maximality locus.
The only problem is that already the famous example of Whitney umbrella
shows that this is impossible, and all methods following Hironaka had
to use some history and resulted in more complicated algorithms.
Nevertheless, in a recent work with Abramovich and Wlodarczyk we did
2019 May 28
# Dynamics seminar: Toru Sera (Kyoto) "Multidimensional arcsine law for intermittent interval maps"
2:00pm to 3:00pm
Abstract:
Paul L\'evy's classical arcsine law states that the occupation time ratio of one-dimensional Brownian motion for the positive side is arcsine-distributed. The arcsine law has been generalized to a variety of classes of stochastic processes and dynamical systems.
2019 Jun 10
# NT & AG Seminar - Dan Edidin - An intrinsic characterization of cofree representations of reductive groups''
1:00pm to 2:00pm
## Location:
Ross 70
Title: An intrinsic characterization of cofree representations of reductive groups
2019 May 07
# Anatoly Vershik (St. Petersburg) Соmbinatorial (locally finite) еncoding of the Bernoulli processes with infinite entropy.
2:00pm to 3:00pm
Abstract.
The realization of m.p automorphisms as transfer on the space of the
paths on the graded graphs allows to use new kind of encoding
of one-sided Bernoulli shift.
I will start with simple example how to realize Bernoulli shift in
the locally finite space (graph) $\prod_n {1,2,\dots n}$ (triangle
compact.)
Much more complicated example connected to old papers by
S.Kerov-Vershik and recent by Romik-Sniady in which one-sided
Bernoulli shift is
realized as Schutzenberger transfer on the space of infinite Young
2019 Jun 27
# Group and dynamics seminar: Michael Chapman (HUJI): Cutoff on Ramanujan complexes
10:00am to 11:15am
## Location:
Ross 70
Abstract: A Markov chain over a finite state space is said to exhibit the total variation cutoff phenomenon if, starting from some Dirac measure, the total variation distance to the stationary distribution drops abruptly from near maximal to near zero. It is conjectured that simple random walks on the family of $k$-regular, transitive graphs with a two sided $\epsilon$ spectral gap exhibit total variation cutoff (for any fixed $k$ and \$\epsilon). This is known to be true only in a small number of cases.
2019 Jun 03
# Combinatorics - back to back:
11:00am to 1:00pm
## Location:
CS Rothberg bldg, room B-500, Safra campus
First talk:
Title: Voronoi Cells of Varieties
Abstract:
2019 May 23
# Basic Notions: Ehud de Shalit (HUJI) " Hilbert's 12th problem - recent developments"
4:00pm to 5:15pm
## Location:
Ross 70
Hilbert's 12th problem (Kronecker's Jugendtraum) is one of the major open problems
2019 Jun 24
# Combinatorics: Doron Puder (TAU) "Aldous' spectral gap conjecture for normal sets"
11:00am to 1:00pm
## Location:
CS bldg, room B-500, Safra campus
Speaker: Doron Puder, TAU
Title: Aldous' spectral gap conjecture for normal sets
Abstract:
Aldous' spectral gap conjecture, proved in 2009 by Caputo, Liggett and Richthammer, states the following a priori very surprising fact: the spectral gap of a random walk on a finite graph is equal to the spectral gap of the interchange process on the same graph.
2019 May 13
# Combinatorics: Shira Zerbib (U. Michigan, Iowa State University) "Envy-free division of a cake without the “hungry players" assumption"
11:00am to 1:00pm
## Location:
CS bldg, room B-500, Safra campus
Speaker: Shira Zerbib (U. Michigan, Iowa State University)
Title: Envy-free division of a cake without the “hungry players" assumption
Abstract:
The fair division theorem due to Stromquist (1980) ensures that under some conditions it is possible to divide a rectangular cake into n pieces and assign one piece to each of n players such that no player strictly prefers a piece that has not been assigned to him.
2019 May 06
# Combinatorics: Omri Ben Eliezer (TAU) "Finding patterns in permutations"
11:00am to 1:00pm
## Location:
CS building, room B-500, Safra campus
Speaker: Omri Ben Eliezer, TAU
Title: Finding patterns in permutations
Abstract:
For two permutations sigma and pi, we say that sigma contains a copy of
pi, if there is a subset (not necessarily consecutive) of elements in sigma,
whose relative order is the same as in pi. For example, if pi = (1,2,3),
then a copy of pi in sigma amounts to an increasing subsequence in sigma
of length 3.
As shown by Guillemot and Marx, a copy of a constant length pi can be
found in sigma in linear time. However, how quickly can one find such a
2019 May 16
# Basic Notions: Ehud de Shalit (HUJI) " Hilbert's 12th problem - recent developments"
4:00pm to 5:15pm
## Location:
Ross 70
Hilbert's 12th problem (Kronecker's Jugendtraum) is one of the major open problems
2019 Jun 10
# Combinatorics: Eyal Karni (BIU) "Combinatorial high dimensional expanders"
11:00am to 1:00pm
## Location:
CS bldg, room B-500, Safra campus
Speaker: Eyal Karni (BIU)
Title: Combinatorial high dimensional expanders
Abstract:
An eps-expander is a graph G=(V,E) in which every set of vertices X where |X|<=|V|/2 satisfies |E(X,X^c)|>=eps*|X| . There are many edges that "go out" from any relevant set.
2019 May 27
# Combinatorics: Uri Rabinovich (U. Haifa) "SOME EXTREMAL PROBLEMS ABOUT SIMPLICIAL COMPLEXES"
11:00am to 1:00pm
## Location:
CS Rothberg bldg, room B-500, Safra campus
Speaker: Uri Rabinovich (U. Haifa)
Title: SOME EXTREMAL PROBLEMS ABOUT SIMPLICIAL COMPLEXES
Abstract:
We shall discuss the following three issues:
* The existence of Hamiltonian d-cycles, i.e., simple d-cycles containing a spanning d-hypertree of a complete d-complex K_n^d;
* The existence of a distribution D over spanning d-hypertrees T of K_n^d, so that for ANY
(d-1)-cycle C there, the expected size of the d-filling of C with respect to a random T from D is Omega(n^d); | 2020-07-10 09:12:28 | {"extraction_info": {"found_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": 0.6099166870117188, "perplexity": 7649.957394075143}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655906934.51/warc/CC-MAIN-20200710082212-20200710112212-00460.warc.gz"} |
https://matheducators.stackexchange.com/questions/10912/breaking-students-from-the-habit-of-relying-on-examples?answertab=oldest | # Breaking students from the habit of relying on examples
One of the most frustrating things about my experiences teaching math (at the university level, if that matters) is that students seem very reluctant to actually learn the material. Instead, they seem to want to be presented with a series of examples, then generalize from those examples, rather than the other way around. This works well up till calculus and then fails completely afterward, when math becomes more than simply a set of computational algorithms. I've never had any success getting students to think mathematically; any suggestions?
Also, like anyone who's taught mathematics, or even answered questions on stackexchange, I've been frustrated (and irrationally annoyed) by students who come to me with no ideas whatsoever about a problem, saying that they "don't know where to start." A bit of prodding often reveals that they don't understand one of the terms in the problem. I'm not sure how they were planning on solving it without that, but my pet theory is that they tried to compare that problem to their list of examples, found nothing, and gave up. (If they had found such an example, that might have allowed them to solve the problem without the missing definition, effectively treating the example as a black box.) Is there any credence to that idea, or is it something else entirely?
• Welcome to ME.SE. I feel like your question could do with some tidying up, but I can't pin down what specifically would be best, so I'll leave that to someone else. What I will say is that your experience sounds like mine. – Jessica B Apr 25 '16 at 16:54
• Since this is not literally an answer... : I myself strongly prefer examples at every stage. A sufficient stock of examples should indeed suffice to compare to anything within some specified range... almost by definition, I think. And interpolation of a sufficient stock of examples comes very close to completely determining what theorem could be true of them. – paul garrett Apr 25 '16 at 18:04
• I think there are really two questions here; one about examples, the other (more disguised) about how to learn students how to solve exercises. I think that both should be asked, but separately. The first one should be concluded by a more focused question than "any suggestion?": here it feels like you have not yet fully found your question. Concerning the second question, let me insist that the first step to formulate it is to realize that student should be taught how to solve exercise, rather than expected to come up with a good methodology by themselves. – Benoît Kloeckner Apr 25 '16 at 18:16
• I may make a proper answer eventually. For now don't forget that in most other human endeavours you do learn through a series of examples and non-examples (eg language aquisition, cooking). Students are simply doing what comes naturally. – DavidButlerUofA Apr 25 '16 at 23:59
• I like this question. It is a continual frustration how little attention is paid to the theory which is presented. – James S. Cook Apr 26 '16 at 0:36
## 3 Answers
This question is very usefully provocative, as evidenced by the comments, and the pro-example versus [sic] pro-abstraction notions... and the apt comment(s) suggesting that, in particular, the genuine issue is not that students have not "learned the definition", but that their intellectual methodologies are inadequate to the task... or maybe some really are a bit lazy or work-avoidant (sounds better!) while not being actively opposed to the mathematics.
E.g., I was a bit baffled by the comment that implied that it is perhaps dangerously deficient to characterize things as: continuity is that the graph has no jumps, and a differentiable functions' graph has no corners... considering that this is exactly my own intuition, although/and I can conform to the ambient standards-of-proof as need be.
I would claim that the underlying problem is that students are typically rewarded more for compliance and conformity than for other forms of astuteness. Thus, they have often learned to ignore their own intuition, since the instructor/exam/homework is actively trying to prank them in that regard. "Definitions" are therefore not at all codifications of extensive prior experience of serious people, but just set-ups for pratfalls of the victims, ... EDIT: for example, prank questions might have very little mathematical content, but, rather, play upon delicacies of wording, or the tension/conflict/ambiguity between mathematical use and colloquial use. No way to reason these out. Rather, one must know what the examiner is thinking. Similarly, textbooks can create conventions/rules/definitions with little genuine mathematical content, but, nevertheless, with very precise boundaries, the latter lending themselves to questioning.
In particular, in all my experience with standard courses/textbooks/exams, it is not the case that definitions [sic] are helpful, much less clarify pre-existing ambiguities or troubles or confusions. Instead, for exam and such purposes, they are far too often (and, then, this "taints the well") used to prank students in a quasi-legalistic sense. An example of the worst sort of crap is any question about "how many axioms does a group satisfy?". EDIT: again, such a question has essentially no mathematical content, because any finite list can be put together into a single axiom by conjunction, of course. But students may be simply required to remember what a particular book says on a particular page (as opposed to intrinsic mathematical assertions). Similarly, notational conventions can be "tested".
The other issue, which is maybe not literally "laziness", but "passivity", is (I claim) partly a result of years of being beaten down by petty-despot "teachers" of mathematics, who portray the subject as consisting of ineffable, uncontestable rule-sets... It is surely fairly futile, but I think any approach that does not tell kids to "trust, but maybe refine/improve, their intuition", is doomed to disengagement.
The same appears to be true of grad students in mathematics at good universities, etc., btw.
So, yes, I'm resisting answering the question as asked... but eminently willing to edit, considering that I do think this is an important sort of question.
• +1 Because I've been so eager for anyone to try answering this. The "petty-despot... uncontestable rule-sets" part is good. If anything were to be edited, I might suggest the "trying to prank them" paragraph, because I really didn't follow what you were saying in that portion. – Daniel R. Collins Apr 26 '16 at 23:11
• One of the 'pranks' that I like in the exercises, but would never put in a test, is asking for the derivative of pi squared. Or e^7. That may not be the sort of pranking Paul was referring to, though. I think it does test something conceptually important. (Does the student recognize when something is a constant?) But it's too much a 'gotcha!' for me to be willing to put it on a test. – Sue VanHattum May 1 '16 at 23:02
$c^2=a^2+b^2-2ab\cos C\\$
The law of cosines. When I lay out the problem so that a, b, and c don't line up with the equation, half the students are suddenly lost.
Similarly, I offer a right triangle, but play an awful trick, I label the hypotenuse with the letter A, and the 2 legs are B and C. I watch as $a^2+b^2=c^2\\$ is followed regardless, with nonsensical results. It gets marginally better if I start with say, x, y, z, but not much.
Examples are great, until the student is stumped with the similar triangle question, confused, as they solved 4 problems each with a man and flagpole with shadows for each, but now you've presented a building with its shadow.
My best approach is to vary the examples so much that they don't fall into the traps I tried to illustrate. Find ways for the example to be just that, but help them open their mind to apply the example to other concrete examples.
• I think the particular issue of relying on the letters used deserves its own question! – DavidButlerUofA Apr 28 '16 at 17:22
• Thx David. I understand. I thought this was a subset of the 'example' issue, where the example is followed so closely, the very letters confuse students when moving to the next problem. – JoeTaxpayer Apr 28 '16 at 23:38
• A related question would be why, after a discussion of similar triangles, they can't immediately solve the man-and-flagpole-style problem. – anomaly Apr 28 '16 at 23:49
• @JoeTaxpayer It is definitely related, but I think it has a much more specific mechanism that deserves discussion in its own right, that's all. – DavidButlerUofA Apr 29 '16 at 0:17
• @DavidButlerUofA - thanks for the clarification. I'll look at how I can re-word to pose as a question. – JoeTaxpayer Apr 30 '16 at 13:33
Instead, they seem to want to be presented with a series of examples, then generalize from those examples, rather than the other way around.
To be honest: This does not seem that wrong in general. Most of the generalisations offered by mathematics are motivated by examples, e.g., the concept of continuity is motivated by the fact that the vast majority of real-life relations is continuous (at least in very good approximation). Moreover, examples are a good way to discover proofs. That being said, the power and importance of mathematics comes from abstraction and generalisation, but if you cannot illustrate this power by connecting the example level with the abstract one, your students will not appreciate the latter. This holds in particular, if your students are not actually aspiring to become mathematicians, but study physics, computer science, or similar.
Hence:
• Explain to them the way mathematics works:
• examples motivate definitions;
• statements based on suitable definitions have wide, easy, and robust applicability (in a sense, this is why we bother with abstract mathematics in the first place), while appealing to example-based intuition is prone to errors;
• examples are a good way to discover proofs, but they never are proofs themselves.
For a practical example: Once we show that some structure complies with the vector-space axioms, the whole apparatus of linear algebra becomes available to us and due to the rigidity of mathematics, we do not have to spend any thought on whether we can actually apply it – example-based intuition cannot do this.¹
• Guide them by example (yes, yes, I know): Do not only present them polished proofs, but show them how to come up with a proof. Write down all the relevant definitions and known properties, and continue from there. Also try to give an example where you can translate some example-based intuition to a proper proof.
• For students of other disciplines like physics or computer science, debunk the myth that they only need mathematics as a computational tool. For example, explain to physics students that vector spaces are the basic language of quantum mechanics and they cannot properly understand it without understanding vector spaces.
¹ Blatant self-advertising: For an even more practical example for this example, see this didactics paper of mine (preprint). | 2019-06-27 05:07:43 | {"extraction_info": {"found_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": 0.5809179544448853, "perplexity": 851.7299003800056}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560628000613.45/warc/CC-MAIN-20190627035307-20190627061307-00446.warc.gz"} |
https://julialang.github.io/PackageCompiler.jl/stable/refs.html | References
PackageCompiler.create_sysimageFunction
create_sysimage(packages::Vector{String}; kwargs...)
Create a system image that includes the package(s) in packages (given as a string or vector). If the packages argument is not passed, all packages in the project will be put into the sysimage.
An attempt to automatically find a compiler will be done but can also be given explicitly by setting the environment variable JULIA_CC to a path to a compiler (can also include extra arguments to the compiler, like -g).
Keyword arguments:
• sysimage_path::String: The path to where the resulting sysimage should be saved.
• project::String: The project that should be active when the sysimage is created, defaults to the currently active project.
• precompile_execution_file::Union{String, Vector{String}}: A file or list of files that contain code from which precompilation statements should be recorded.
• precompile_statements_file::Union{String, Vector{String}}: A file or list of files that contain precompilation statements that should be included in the sysimage.
• incremental::Bool: If true, build the new sysimage on top of the sysimage of the current process otherwise build a new sysimage from scratch. Defaults to true.
• filter_stdlibs::Bool: If true, only include stdlibs that are in the project file. Defaults to false, only set to true if you know the potential pitfalls.
• include_transitive_dependencies::Bool: If true, explicitly put all transitive dependencies into the sysimage. This only makes a difference if some packages do not load all their dependencies when themselves are loaded. Defaults to true.
• base_sysimage::Union{Nothing, String}: If a String, names an existing sysimage upon which to build the new sysimage incrementally, instead of the sysimage of the current process. Defaults to nothing. Keyword argument incremental must be true if base_sysimage is not nothing.
• cpu_target::String: The value to use for JULIA_CPU_TARGET when building the system image. Defaults to native.
• script::String: Path to a file that gets executed in the --output-o process.
• sysimage_build_args::Cmd: A set of command line options that is used in the Julia process building the sysimage, for example -O1 --check-bounds=yes.
PackageCompiler.create_appFunction
create_app(package_dir::String, compiled_app::String; kwargs...)
Compile an app with the source in package_dir to the folder compiled_app. The folder package_dir needs to contain a package where the package includes a function with the signature
julia_main()::Cint
# Perhaps do something based on ARGS
...
end
The executable will be placed in a folder called bin in compiled_app and when the executable run the julia_main function is called.
Standard Julia arguments are set by passing them after a --julia-args argument, for example:
\$ ./MyApp input.csv --julia-args -O3 -t8
An attempt to automatically find a compiler will be done but can also be given explicitly by setting the environment variable JULIA_CC to a path to a compiler (can also include extra arguments to the compiler, like -g).
Keyword arguments:
• executables::Vector{Pair{String, String}}: A list of executables to produce, given as pairs of executable_name => julia_main where executable_name is the name of the produced executable with the julia function julia_main. If not provided, the name of the package (as specified in Project.toml) is used and the main function in julia is taken as julia_main.
• precompile_execution_file::Union{String, Vector{String}}: A file or list of files that contain code from which precompilation statements should be recorded.
• precompile_statements_file::Union{String, Vector{String}}: A file or list of files that contain precompilation statements that should be included in the sysimage for the app.
• incremental::Bool: If true, build the new sysimage on top of the sysimage of the current process otherwise build a new sysimage from scratch. Defaults to false.
• filter_stdlibs::Bool: If true, only include stdlibs that are in the project file. Defaults to false, only set to true if you know the potential pitfalls.
• force::Bool: Remove the folder compiled_app if it exists before creating the app.
• include_lazy_artifacts::Bool: if lazy artifacts should be included in the bundled artifacts, defaults to false.
• include_transitive_dependencies::Bool: If true, explicitly put all transitive dependencies into the sysimage. This only makes a difference if some packages do not load all their dependencies when themselves are loaded. Defaults to true.
• cpu_target::String: The value to use for JULIA_CPU_TARGET when building the system image.
• sysimage_build_args::Cmd: A set of command line options that is used in the Julia process building the sysimage, for example -O1 --check-bounds=yes.
PackageCompiler.create_libraryFunction
create_library(package_dir::String, dest_dir::String; kwargs...)
Compile a library with the source in package_dir to the folder dest_dir. The folder package_dir should to contain a package with C-callable functions, e.g.
Base.@ccallable function julia_cg(fptr::Ptr{Cvoid}, cx::Ptr{Cdouble}, cb::Ptr{Cdouble}, len::Csize_t)::Cint
try
x = unsafe_wrap(Array, cx, (len,))
b = unsafe_wrap(Array, cb, (len,))
A = COp(fptr,len)
cg!(x, A, b)
catch
Base.invokelatest(Base.display_error, Base.catch_stack())
return 1
end
return 0
end
The library will be placed in the lib folder in dest_dir (or bin on Windows), and can be linked to and called into from C/C++ or other languages that can use C libraries.
Note that any applications/programs linking to this library may need help finding it at run time. Options include
• Installing all libraries somewhere in the library search path.
• Adding /path/to/libname to an appropriate library search path environment variable (DYLD_LIBRARY_PATH on OSX, PATH on Windows, or LD_LIBRARY_PATH on Linux/BSD/Unix).
• Running install_name_tool -change libname /path/to/libname (OSX)
To use any Julia exported functions, you must first call init_julia(argc, argv), where argc and argv are parameters that would normally be passed to julia on the command line (e.g., to set up the number of threads or processes).
When your program is exiting, it is also suggested to call shutdown_julia(retcode), to allow Julia to cleanly clean up resources and call any finalizers. (This function simply calls jl_atexit_hook(retcode).)
An attempt to automatically find a compiler will be done but can also be given explicitly by setting the environment variable JULIA_CC to a path to a compiler (can also include extra arguments to the compiler, like -g).
Keyword arguments:
• lib_name::String: an alternative name for the compiled library. If not provided, the name of the package (as specified in Project.toml) is used. lib will be prepended to the name if it is not already present.
• precompile_execution_file::Union{String, Vector{String}}: A file or list of files that contain code from which precompilation statements should be recorded.
• precompile_statements_file::Union{String, Vector{String}}: A file or list of files that contain precompilation statements that should be included in the sysimage for the library.
• incremental::Bool: If true, build the new sysimage on top of the sysimage of the current process otherwise build a new sysimage from scratch. Defaults to false.
• filter_stdlibs::Bool: If true, only include stdlibs that are in the project file. Defaults to false, only set to true if you know the potential pitfalls.
• force::Bool: Remove the folder compiled_lib if it exists before creating the library.
• header_files::Vector{String}: A list of header files to include in the library bundle.
• julia_init_c_file::String: File to include in the system image with functions for initializing julia from external code.
• version::VersionNumber: Library version number. Added to the sysimg .so name on Linux, and the .dylib name on Apple platforms, and with compat_level, used to determine and set the current_version, compatibility_version (on Apple) and soname (on Linux/UNIX)
• compat_level::String: compatibility level for library. One of "major", "minor". Used to determine and set the compatibility_version (on Apple) and soname (on Linux/UNIX).
• include_lazy_artifacts::Bool: if lazy artifacts should be included in the bundled artifacts, defaults to false.
• include_transitive_dependencies::Bool: If true, explicitly put all transitive dependencies into the sysimage. This only makes a difference if some packages do not load all their dependencies when themselves are loaded. Defaults to true.
• script::String: Path to a file that gets executed in the --output-o process.
• cpu_target::String: The value to use for JULIA_CPU_TARGET when building the system image.
• sysimage_build_args::Cmd: A set of command line options that is used in the Julia process building the sysimage, for example -O1 --check-bounds=yes. | 2023-01-31 04:42:36 | {"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": 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": 0.2465263456106186, "perplexity": 5366.61032156176}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499842.81/warc/CC-MAIN-20230131023947-20230131053947-00324.warc.gz"} |
https://notes.reasoning.page/html/spangles | # A calculus of the absurd
#### 8.3 Spangles
Spangles are "special" angles. They’re special because they show up a lot. Their values are given in this table 5959 There are numerous problems with the formatting of this table which I will one day get around to fixing..
0 $$\frac {\pi }{6}$$ $$\frac {\pi }{4}$$ $$\frac {\pi }{3}$$ $$\frac {\pi }{2}$$ $$\sin (x)$$ 0 $$\frac {1}{2}$$ $$\frac {\sqrt {2}}{2}$$ $$\frac {\sqrt {3}}{2}$$ $$1$$ $$\cos (x)$$ 1 $$\frac {\sqrt {3}}{2}$$ $$\frac {\sqrt {2}}{2}$$ $$\frac {1}{2}$$ 0 $$\tan (x)$$ 0 $$\frac {1}{\sqrt {3}}$$ 1 $$\sqrt {3}$$ undefined
Don’t memorise the table! All you need to remember is that
$$\sin \left ( \frac {\pi }{6} \right ) = \frac {1}{2}$$
From there, you can work out the rest of the values for $$\sin (x)$$, as the number being rooted just goes up by one (from $$\frac {\sqrt {1}}{2}$$ to $$\frac {\sqrt {2}}{2}$$ to $$\frac {\sqrt {3}}{2}$$). The values of $$\cos (x)$$ do the same thing, but the other way round. For $$\tan (x)$$, as
$$\tan (x) = \frac {\sin (x)}{\cos (x)}$$
the values of $$\tan (x)$$ can be computed from the values of $$\sin (x)$$ $$\cos (x)$$. | 2022-11-27 05:42:19 | {"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": 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": 2, "x-ck12": 0, "texerror": 0, "math_score": 0.8760896325111389, "perplexity": 154.6863510015731}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710192.90/warc/CC-MAIN-20221127041342-20221127071342-00686.warc.gz"} |
https://brilliant.org/problems/special-lattice-paths/ | # Special Lattice Paths
Let $$S$$ be the set of $$\{(1,0), (0,1), (1,1), (1,-1), (-1,1)\}$$-lattice path which begin at $$(1,1)$$, do not use the same vertex twice, and never touch either the $$x$$-axis or the $$y$$-axis.
Let $$P_{x,y}$$ be the number of paths in $$S$$ which end at the point $$(x,y)$$. Determine $$P_{2,4}$$.
Details and assumptions
A lattice path is a path in the Cartesian plane between points with integer coordinates.
A step in a lattice path is a single move from one point with integer coordinates to another.
The size of the step from $$(x_1,y_1)$$ to $$(x_2,y_2)$$ is $$(x_2-x_1,y_2-y_1)$$.
The length of a lattice path is the number of steps in the path.
For a set $$S = \{(x_i,y_i)\}_{i=1}^{k}$$, an $$S$$-lattice path is a lattice path where every step has size which is a member of $$S$$.
× | 2017-10-22 01:04:02 | {"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": 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": 0.654231071472168, "perplexity": 168.97160165496808}, "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-2017-43/segments/1508187824931.84/warc/CC-MAIN-20171022003552-20171022023552-00281.warc.gz"} |
http://mathoverflow.net/revisions/3952/list | 4 added 18 characters in body
For any cycle decomposition, we can uniquely order the cycles from smallest length to largest length, breaking ties between cycles of the same length in some fixed arbitrary way (say by maximal elements). Let us do this for concreteness.
Suppose there was an (injective) way to "join" and A-cycle and a B-cycle together to form an A+B cycle whenever |A| and |B| are > 1. Then, given any cycle decomposition as above, one can start bubbling the two largest cycles together to eventually form a single cycle of some length m.
If m = n - 1, we are done.
If m = n, write S = (1............) and then omit the last term.
If m = 1 the problem is very easy.
If n - 1 > m > 2, there is an injective map from m-cycles to elements whose cycle decomposition is a product of an m-cycle with a 2-cycle. For concreteness, one can add the 2-cycle with the two lowest missing entries. Now bubble again to form an m+2 cycle.
If m = 2, and n is at least 5, bubble with a 3-cycle.
If m = 2 and n = 4 (the last case), form whatever bijection you like between the two sets of 6 elements.
The key point is therefore to find a way to bubble an A-cycle and a B-cycle when |A|,|B| > 1. We do this as follows.
Amongst the entries of A and B, there is a unique smallest integer, call it X.
Case 1. If X lies in A, Let Z denote the largest element of B. Then one can (uniquely) write A = (X.....) and B = (.....Y,Z), where Y < Z. Then consider the A+B cycle obtained by concatenating A and B in this form, i.e. (X,......,Y, Z).
Case 2. If X lies in B, let Z denote the smallest element of A. Then one can (uniquely) write B = (X....) and A = (.....,Y,Z), where now Y > Z. Then consider the A+B cycle (X,.....,Y,Z).
Given an A+B cycle, one can uniquely write it in the form (X,....,Y,Z), where X is the smallest entry. Then one can break it up into an A-cycle and B-cycle depending on whether Y < Z or Y > Z.
Since this was apparently a little confusing, suppose that the cycle lengths of S are a_1 <= a_2 <= a_3 <= ..... <= a_r. Here I omit the 1-cycle lengths, so a_1 > 1, and sum a_r = m for some m possibly less than n. Then the cycle lengths of the steps in the algorithm will have lengths:
(a_1, ...., a_(r-1),a_r),
(a_1, ...., a_(r-2),a_(r-1) + a_r),
(a_1, ...., a_(r-3),a_(r-2) + a_(r-1) + a_r),
....
(a_1 + a__2 + ... + a_r) = (m)
(2,m)
(m+2)
(2,m+2)
(m+4)
...
(n-1 or n, depending on m mod 2),
then n-1.
(if m = 2 and n is at least 5, then instead it should go
(2) --> (2,3) --> (5) --> (2,5) --> (7) --> (2,7) --> (9) ...,etc.
3 added 276 characters in body
For any cycle decomposition, we can uniquely order the cycles from smallest length to largest length, breaking ties between cycles of the same length in some fixed arbitrary way (say by maximal elements). Let us do this for concreteness.
Suppose there was an (injective) way to "join" and A-cycle and a B-cycle together to form an A+B cycle whenever |A| and |B| are > 1. Then, given any cycle decomposition as above, one can start bubbling the two largest cycles together to eventually form a single cycle of some length m.
If m = n - 1, we are done.
If m = n, write S = (1............) and then omit the last term.
If m = 1 or 2, the problem is very easy.
If n - 1 > m > 2, there is an injective map from m-cycles to elements whose cycle decomposition is a product of an m-cycle with a 2-cycle. For concreteness, one can add the 2-cycle with the two lowest missing entries. Now bubble again to form an m+2 cycle.
If m = 2, and n is at least 5, bubble with a 3-cycle.
If m = 2 and n = 4 (the last case), form whatever bijection you like between the two sets of 6 elements.
The key point is therefore to find a way to bubble an A-cycle and a B-cycle when |A|,|B| > 1. We do this as follows.
Amongst the entries of A and B, there is a unique smallest integer, call it X.
Case 1. If X lies in A, Let Z denote the largest element of B. Then one can (uniquely) write A = (X.....) and B = (.....Y,Z), where Y < Z. Then consider the A+B cycle obtained by concatenating A and B in this form, i.e. (X,......,Y, Z).
Case 2. If X lies in B, let Z denote the smallest element of A. Then one can (uniquely) write B = (X....) and A = (.....,Y,Z), where now Y > Z. Then consider the A+B cycle (X,.....,Y,Z).
Given an A+B cycle, one can uniquely write it in the form (X,....,Y,Z), where X is the smallest entry. Then one can break it up into an A-cycle and B-cycle depending on whether Y < Z or Y > Z.
Since this was apparently a little confusing, suppose that the cycle lengths of S are a_1 <= a_2 <= a_3 <= ..... <= a_r. Here I omit the 1-cycle lengths, so a_1 > 1, and sum a_r = m for some m possibly less than n. Then the cycle lengths of the steps in the algorithm will have lengths:
(a_1, ...., a_(r-1),a_r),
(a_1, ...., a_(r-2),a_(r-1) + a_r),
(a_1, ...., a_(r-3),a_(r-2) + a_(r-1) + a_r),
....
(a_1 + a__2 + ... + a_r) = (m)
(2,m)
(m+2)
(2,m+2)
(m+4)
...
(n-1 or n, depending on m mod 2),
then n-1.
(if m = 2 and n is at least 5, then instead it should go
(2) --> (2,3) --> (5) --> (2,5) --> (7) --> (2,7) --> (9) ...,etc.
2 added 594 characters in body
For any cycle decomposition, we can uniquely order the cycles from smallest length to largest length, breaking ties between cycles of the same length in some fixed arbitrary way (say by maximal elements). Let us do this for concreteness.
Suppose there was an (injective) way to "join" and A-cycle and a B-cycle together to form an A+B cycle whenever |A| and |B| are > 1. Then, given any cycle decomposition as above, one can start bubbling the two largest cycles together to eventually form a single cycle of some length m.
If m = n - 1, we are done.
If m = n, write S = (1............) and then omit the last term.
If m = 1 or 2, the problem is easy.
If n - 1 > m > 2, there is an injective map from m-cycles to elements whose cycle decomposition is a product of an m-cycle with a 2-cycle. For concreteness, one can add the 2-cycle with the two lowest missing entries. Now bubble again to form an m+2 cycle.
The key point is therefore to find a way to bubble an A-cycle and a B-cycle when |A|,|B| > 1. We do this as follows.
Amongst the entries of A and B, there is a unique smallest integer, call it X.
Case 1. If X lies in A, Let Z denote the largest element of B. Then one can (uniquely) write A = (X.....) and B = (.....Y,Z), where Y < Z. Then consider the A+B cycle obtained by concatenating A and B in this form, i.e. (X,......,Y, Z).
Case 2. If X lies in B, let Z denote the smallest element of A. Then one can (uniquely) write B = (X....) and A = (.....,Y,Z), where now Y > Z. Then consider the A+B cycle (X,.....,Y,Z).
Given an A+B cycle, one can uniquely write it in the form (X,....,Y,Z), where X is the smallest entry. Then one can break it up into an A-cycle and B-cycle depending on whether Y < Z or Y > Z.
Since this was apparently a little confusing, suppose that the cycle lengths of S are a_1 <= a_2 <= a_3 <= ..... <= a_r. Here I omit the 1-cycle lengths, so a_1 > 1, and sum a_r = m for some m possibly less than n. Then the cycle lengths of the steps in the algorithm will have lengths:
(a_1, ...., a_(r-1),a_r),
(a_1, ...., a_(r-2),a_(r-1) + a_r),
(a_1, ...., a_(r-3),a_(r-2) + a_(r-1) + a_r),
....
(a_1 + a__2 + ... + a_r) = (m)
(2,m)
(m+2)
(2,m+2)
(m+4)
...
(n-1 or n, depending on m mod 2),
then n-1.
1 | 2013-05-26 06:51:02 | {"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": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8712072968482971, "perplexity": 936.7470759359934}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1368706635944/warc/CC-MAIN-20130516121715-00022-ip-10-60-113-184.ec2.internal.warc.gz"} |
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# Each edge of the cube shown above has length s. What is the perimeter
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Each edge of the cube shown above has length s. What is the perimeter [#permalink]
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Each edge of the cube shown above has length s. What is the perimeter of ∆ BDE?
(A) 3s
(B) 6s
(C) s√3/2
(D) 3s√2
(E) 2s + s√2
[Reveal] Spoiler:
Attachment:
2017-12-01_0947.png [ 9.8 KiB | Viewed 378 times ]
[Reveal] Spoiler: OA
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Re: Each edge of the cube shown above has length s. What is the perimeter [#permalink]
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30 Nov 2017, 21:56
D
By Pythagoras theorem we have BD=BE=DE=s*(2)^0.5. Hence the perimeter of triangle BDE is 3*s*(2)^0.5
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Re: Each edge of the cube shown above has length s. What is the perimeter [#permalink]
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30 Nov 2017, 22:09
Bunuel wrote:
Each edge of the cube shown above has length s. What is the perimeter of ∆ BDE?
(A) 3s
(B) 6s
(C) s√3/2
(D) 3s√2
(E) 2s + s√2
[Reveal] Spoiler:
Attachment:
2017-12-01_0947.png
In a cube all the sides are equal. Consider ABCD as square
Apply Pythagoras theorem BE=ED= BD = $$\sqrt{2}$$s
Hence Perimeter =3s$$\sqrt{s}$$
Hence D
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Re: Each edge of the cube shown above has length s. What is the perimeter [#permalink]
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01 Dec 2017, 12:39
Bunuel wrote:
Each edge of the cube shown above has length s. What is the perimeter of ∆ BDE?
(A) 3s
(B) 6s
(C) s√3/2
(D) 3s√2
(E) 2s + s√2
[Reveal] Spoiler:
Attachment:
2017-12-01_0947.png
Each side of ∆ BDE is the hypotenuse of a two-dimensional isosceles right triangle.
They are congruent, and there are three: ∆ BCD, ∆ CDE, and ∆ BCE
Each has side length $$s$$ (= BC, CD, and CE)
Because they are one-half of a square:
All have angle measures of 45-45-90 and
corresponding side lengths*
$$x : x : x\sqrt{2}$$
Side/leg length $$x = s$$
Hypotenuse, per ratio, hence is $$s\sqrt{2}$$ , which =
Length of all three sides of ∆ BDE
Perimeter of ∆ BDE
Three sides of length $$s\sqrt{2}$$ =
$$3s\sqrt{2}$$
*Knowing those ratios is key, but if not:
A square cut by a diagonal produces two right isosceles triangles
The relationship between one leg of such a triangle its hypotenuse (also the square's diagonal)
$$h = s\sqrt{2}$$ , derived from Pythagorean theorem:
$$s^2 + s^2 = h^2$$
$$2s^2 = h^2$$
$$\sqrt{2}\sqrt{s^2} = \sqrt{h^2}$$
$$\sqrt{2}s = h$$
$$h = s\sqrt{2}$$
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Re: Each edge of the cube shown above has length s. What is the perimeter [#permalink] 01 Dec 2017, 12:39
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# Each edge of the cube shown above has length s. What is the perimeter
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Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 2018-01-22 10:30:49 | {"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": 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": 0.5709483623504639, "perplexity": 9539.489269817861}, "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-2018-05/segments/1516084891277.94/warc/CC-MAIN-20180122093724-20180122113724-00351.warc.gz"} |
http://berjon.net/blog/2013/11/07/sphinx-installation/ | # Sphinx installation
Posted on
A friend wanted to use CMU Sphinx with some students but had some issues to make it work. We used the French database1 as our English accents weren’t good enough for the recognition. In order to use it, you you would need to respect several steps.
• install CMU Sphinx,
• install Alsa (and possibly Jack)
• a dictionnary (a French one in my case)
• a microphone (with the appropriate configuration)
## Package installation or compilation
There are two possible installation processes:
• Sphinx: the standalone application written in java
• PocketSphinx: the C library used mainly in embedded systems
The starting idea was to use it on a Raspberry PI, so I decided to install PocketSphinx2. If you use any mainstream GNU/Linux distribution, you should find the required packages and all dependencies should manage accordingly. If like me, you’re using Slackware, you will probably have some difficulties to find appropriate Slackbuilds (you can find old ones on my repository).
### Dependencies
If in any case, you need to compile everything by yourself, you’ll need to satisfy the following dependencies:
• Bison
• Alsa
• sphinxbase
• sphinxtrain
If you need to perform the compilation by yourself, I invite you to read carefully the INSTALL file in each library tarball.
## Audio path configuration
Once the installation is done, you need now to configure correctly the audio path. In my case, I used the Jack Audio server in order to have real time privileges3. If you have some trouble to give these rights to Jack, you can refer to this (very light) tutorial. Once Jack is configured, you can run jackd:
jackd -R -d alsa -d hw:0,0 -r 44100
If you don’t know the ID of your sound card (corresponding to the hw:<ID> in the command given above, you can use the aplay -l command). The option -r corresponds to the sample rate.
From this point, in theory you are ready to use Pockersphinx library. Nevertheless, the recognition algorithm is really sensible to non linearities in the signal. It means, that you should properly configure your microphone input in order to avoid any saturation. You can either record a portion of signal and analyse it using Matlab/Octave or Python, or a bit easier (less accurate for sure but enough in this case), you can loop the microphone signal to your headphone and assess “by ear” any saturation that could occur. To perform this operation, you can use jack. First, list the I/O’s of your system using the command jack_lsp. It should give something close to that:
# jack_lsp output
system:capture_1
system:capture_2
system:playback_1
system:playback_2
If we take the example that you microphone is plugged in of the input 1 and headphones on the output 1. We ask jack to connect the I/Os as following:
jack_connect system:capture_1 system:playback_1
If you prefer to have the signal on both ears, do the same operation as described above and connect to playback_2. Now you can adapt the level using the command alsamixer. Once you are inside the mixer, everythin related to the inputs can be accessed by pressing the key “F4”.
You can now really use Pocketsphinx. By default, the system is using an English dictionnary. In our case, we are interested in French, so we are going to download the right dictionnaries:
You need to extract the different archives. The DMP file is a bz2 archive and you should use the following command to extract its content:
bz2 -d file.lm.dmp.bz2
The other two archives are classical tarballs:
tar xvf file.tar.gz
Now you are completely ready to play with Pocketsphinx by giving him as arguments the dictionnaries and acoustic model:
pocketsphinx_continuous -hmm acoutic_model_folder/ -dict frenchWords.dic -lm french.lm.dmp
You should have many information displayed in the console but the essential keyword being “READY…”. If you see it, you are ready to speak and enjoy the speech recognition. On average, we can expect a correlation of 70%. The next step would be to connect the speech recognition engine to something else to control anything you want.
1. The database can be generic or specific. It will be dependent of the use made during production (for example, a hotline). [return]
2. Warning, the installation has been performed on the x86 platform and not on a Raspberry PI, few changes may be required. [return]
3. Usually, this type of application use need low latency buffer and have specific privileges are in order. Nevertheless, the use of the RT kernel isn’t mandatory. [return] | 2018-08-15 19:10:32 | {"extraction_info": {"found_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": 0.41585206985473633, "perplexity": 2181.524720002705}, "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-2018-34/segments/1534221210249.41/warc/CC-MAIN-20180815181003-20180815201003-00224.warc.gz"} |
http://tex.stackexchange.com/questions/41261/mathop-shifts-the-baseline-declaremathoperator-doesnt | # mathop shifts the baseline, DeclareMathOperator doesn't
Here is a little example that illustrates this strange behaviour:
\documentclass{article}
\usepackage{amsmath}
\DeclareMathOperator\foo{\mathbf{l}}
\begin{document}
baseline$\mathbf{l}$baseline$\mathop{\mathbf{l}}$baseline$\mathop{l}$baseline$\foo$baseline
\end{document}
Note that the explict call to \mathop shifts the baseline of the l. But if you use \DeclareMathOperator this doesn't happen. Is this a bug or a feature?
I personally think that the shifted baseline looks really weird if you're discussing an operator in text, so I'd like to shift it up again. Do I have an alternative to endless use of \DeclareMathOperator?
-
Is \DeclareMathOperator not about functions like sin, cos, etc., and \mathop about operators like +, -, etc.? – Martin Scharrer Jan 16 '12 at 15:23
@MartinScharrer if that's the case, the terminology is confusing! there's \mathbin and \mathrel, I thought they were the ones for the infix notation symbols… – Seamus Jan 16 '12 at 15:27
Ok, yes \mathbin is for binary operators! \mathrel is for relations, i.e. = etc. I don't have The TeXBook handy right now, otherwise I would look it up. – Martin Scharrer Jan 16 '12 at 15:29
\mathop is also about \sum and friends, which is why it centers single glyphs vertically on the math axis. – barbara beeton Jan 16 '12 at 15:58
it's a feature. here's an excerpt from amsopn.dtx:
In the interior of the \mathop we need a null object (we choose a zero kern for minimum waste of main mem) in order to guard against the case where #3 is a single letter; TeX will seize it and center it on the math axis if there is nothing else inside the \mathop atom.
and here's the definition of \qopname which underlies \DeclareMathOperator:
\DeclareRobustCommand{\qopname}[3]{%
\mathop{#1\kern\z@\operator@font#3}%
\csname n#2limits@\endcsname}
-
\operatorname? – egreg Jan 16 '12 at 15:33
@egreg -- \operatorname is also defined in amsopn, but it's defined in terms of \qopname. i'm pretty sure that michael preferred \DeclareMathOperator as being more descriptive; he also limited its use to the preamble, and i know that he felt strongly that all such commands should be defined only in the preamble. his documentation (texdoc amsopn) is pretty clear, even to a non-guru. – barbara beeton Jan 16 '12 at 15:46
I was just pointing to \operatorname that's the "user level" version of \qopname (which requires three quite mysterious arguments) and is available also in document. – egreg Jan 16 '12 at 15:50
Why is it considered a feature to have single letter operator names centre on the math axis? – Seamus Jan 20 '12 at 12:18
@Seamus -- it's not single letter but single glyph operators, in particular \sum, \prod, \int, etc., that are responsible for this. in the cm extension font, most, if not all, glyphs are positioned entirely below the baseline, and the \mathop mechanism used to position them correctly. i don't know for sure, but it's likely that knuth didn't need single-letter operator names, and didn't think it worth the "cost" of checking for them. (remember that when tex was written, computers weren't nearly as fast as they are now, and tremendous effort went into making tex efficient.) – barbara beeton Jan 20 '12 at 14:57
The "shifting" obeys TeX rules: when the argument of \mathop consists of one character only it is centered with respect to the math axis (where the fraction line sits); this is laid down in the TeXbook, Appendix G, Rule 13.
For "one shot" operators you can use \operatorname, that cures the problem (and chooses automatically \mathrm.
$\operatorname{l}$ is different from $\mathop{\mathrm{l}}$
The command is available as soon as amsopn is loaded (the same package that supplies \DeclareMathOperator); it is loaded automatically by amsmath.
The *-variant has a similar effect to \DeclareMathOperator*, so defines a math operator with limits.
If the input \operatorname{\mathnormal{l}} for getting a "normal italic l" is too much, then
\mathop{\kern0pt l}
is shorter.
-
If you use \newcommand\foo{\kern0pt l} then \foo still has the baseline shifted. I assume this is some expansion magic, but how might one fix that? – Seamus Jan 28 '14 at 11:16
@Seamus Sorry, but I don't get it: \mathop{\foo} will not shift l from the baseline, while \foo simply prints l. – egreg Jan 28 '14 at 11:51
\documentclass{article} \newcommand\foo{\kern0pt \mathop{l}} \begin{document} $l\foo l$ \end{document} Note baseline is shifted. That is, the solution you suggest doesn't work when mathop is used in defining a new command with only one glyph. – Seamus Jan 29 '14 at 12:08
@Seamus Exactly what's expected: if the list given as argument to \mathop consists of a single character, the character is shifted, as I wrote in the first paragraph of my answer. – egreg Jan 29 '14 at 12:13
@Seamus Sorry, but I don't get it: \kern0pt must go inside the argument to \mathop. – egreg Jan 31 '14 at 12:09
Fixes for \mathop are inserting empty space within the argument, such as by \hspace, \kern, or \mbox.
This has the correct baseline:
\mathop{\mathbf{l}\mbox{}}
Ant this too:
\mathop{\mathbf{l}\hspace{0pt}}
amsmath does it similar, here you can see it in within \operatorname, which is called by \DeclareMathOperator via \@declmathop:
\DeclareRobustCommand{\qopname}[3]{%
\mathop{#1\kern\z@\operator@font#3}%
\csname n#2limits@\endcsname}
`
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https://en.wikipedia.org/wiki/User_talk:JRSpriggs | # User talk:JRSpriggs
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## Conversation on inflation?
Hi, I wonder if you remember that I told you a few years ago that I would seriously reconsider my belief in mainstream New Keynesian macroeconomic theory, if inflation exceeded 10% on an annualized basis in any two consecutive months. I believe the last 5 years has been a unique natural experiment and that it confirms the mainstream macro models that:
1. Price inflation is mainly determined by the relative 'slack' in the economy (measured by unemployment above normal, or by production below capacity) - it decelerates when there is positive slack, and accelerates when there is negative slack.
2. The quantity of money may be good as a long run guide, but it little to do with inflation in the short run, especially when the economy is near the zero nominal lower bound.
3. In a modern economy, it is very difficult for wages to fall (and hence deflation to occur), even in the face of depression levels of unemployment.
Given the inflation record of the last few years, I wonder if you would be willing to discuss your beliefs about monetary theory and 'mark to market' your beliefs? LK (talk) 03:05, 28 May 2014 (UTC)
Let me begin by pointing out that, if I recall correctly, you were the one who cut off most of our previous attempts to discuss economics. Although you were often the one to make the last argument, I only refrained from responding because your language made it clear that you did not want to hear my arguments and I did not feel comfortable continuing to argue (on your talk page) with someone under those conditions. Also, I felt that you often read things into what I said that were not there — especially assuming my total agreement with anyone who I mentioned favorably.
A "natural experiment" is not a scientific experiment and cannot serve as a reliable basis for testing hypotheses. In a scientific experiment, the experimenter controls the inputs and deliberately equalizes (or randomizes) the conditions (other than the variable being analyzed) between the control group and the experimental group. In this "natural experiment", there is no control group (indeed no experimental group other than the one case) and no equalizing of the other inputs. Thus any result could be (and probably is) caused by factors other than the variable being analyzed.
I agree that a large quantity of money does not necessarily result in price inflation; there are other considerations.
The ability of the economy to provide jobs has been impaired by taxes, regulations (especially labor law and obamacare), and government spending (raising the cost of raw materials and capital).
Bearing these in mind, if you still want to talk, go ahead. JRSpriggs (talk) 03:10, 29 May 2014 (UTC)
My memory is very bad nowadays, so I don't remember how our last convo ended. If I was brusque with you, I do apologize. I'm not sure if I agree with your assessment about natural experiments. Granted, it's not as good as a double blind random test, but it can be pretty good, and is often the best we have in the social sciences. Given that some events appear to consistently follow certain natural experiments, we can build up certain 'stylized facts' about the behavior of the economy. From this, we reject theories that contradict the stylized facts and try to construct theories that explain them. That's what economics is (at it's best). Science is not forming a set of beliefs, and then looking at the world through colored lenses, interpreting everything in a way to back up one's preconceptions (at least it shouldn't be like that). The last few years appear to have added (or solidified) a few stylized facts to our arsenal, namely 1, 2, and 3 above. LK (talk) 12:17, 30 May 2014 (UTC)
I was not asking for an apology, especially since some of the things I said may have been offensive to you. Rather, I was trying to remind you that in our conversations I have questioned some your ingrained beliefs in ways which may have been disturbing to you. You might want to look back at our previous conversations and consider that possibility before you decide to proceed.
The near impossibility of doing scientific experiments in social science does not mean that we should give more credence to unreliable methods.
I agree that it would be wrong to re-interpret every piece of evidence as necessary to construe it as confirmation of one's favorite hypotheses. However, no field of knowledge (such as economics) can be understood in isolation from the rest of our knowledge of reality. Economic theory must be reconciled with what we know about: physics, chemistry, biology, evolution, psychology, mathematics, business administration, law, politics, ethics, epistemology, metaphysics, etc..
What do you mean by a "stylized fact"? How is it different from an actual fact?
Regarding (1): Is this not just the theory of the Phillips curve which was discredited by the stagflation of the 1970s?
Regarding (2): This appears to be true.
Regarding (3): What do you mean by "modern economy" that would have any bearing on the flexibility of wage rates? If deflation is unlikely to occur, then why has the Federal Reserve been striving so desperately to avoid it?
JRSpriggs (talk) 03:40, 31 May 2014 (UTC)
Economists use the term "stylized fact" to refer to something that (almost) everyone agrees happens most of the time in a certain part of the economy, and that needs to be part of a good theory about that particular part/aspect of the economy.
• Regarding (1): Not at all! What was discredited was the naive assumption (which few actually believed) that we could 'choose' to stably remain a particular point on a stable Philips curve. What came out of it is the theory of short and long run Philips curves. There obviously is a Philips curve in the short run, but persistently being on one side of the Philips curve (e.g. unemployment persistently lower than the natural rate) will cause the short run Philips curve to shift over time.
• Regarding (3): Non-modern (especially agrarian) economies apparently could experience relatively high rates of deflation. Post WWII developed economies (with complex supply chains, long term contracts, unions, etc) apparently cannot. Instead, they fall into economic depression – associated with very low inflation or slight deflation. It is this situation that the Fed is trying to avoid.
Best, LK (talk) 03:02, 10 June 2014 (UTC)
Regarding (1): If you want the Phillips curve to be a scientific theory, you need to specify how it "shifts". That is, you need a formula for the change in the curve's parameters. If you have such, please provide a link to it.
What really bothers me about this is that you are treating the economy as if it were some kind of recalcitrant engine which needs to be tweaked to make it work correctly. You are ignoring the fact that it consists of living people. Those people have their own goals which they are trying to pursue to the extent possible given the contradictory demands of reality and the government bureaucracy. Essentially, the policies you are advocating are a kind of manipulation or deception by which you try to get people to do something which they would not knowingly and freely do.
I do not believe that there is ever any "slack" in the economy. If the owner of a factor of production chooses not to use it, he does so because it is not profitable to do so. Probably because some co-factors of production are missing or prices/regulations have shifted in a way that ruined his plans for it. Or perhaps he is holding onto it against a possible future contingency (a scarcity of that factor). If you trick him into using it (by causing inflation, lending money at unsustainably low interest rates), then it will not end well. You may get a temporary burst of economic activity, but the result will be waste.
Regarding (3): So government and the unions it protects have created a situation where instead of correcting itself quickly via rapid deflation, economic activity is choked-off when deflation would be appropriate. JRSpriggs (talk) 05:24, 12 June 2014 (UTC)
Hi, sorry for the long delay, I've been distracted by other things. Regarding the question of how the Philips curve shifts, you should know that nowadays, almost everything in mainstream economics is rigorously mathematically modeled. For an intro to New Keynesian inflation theory (with graphs but little maths), I suggest Macroeconomics[1], by Hubbard and O'Brien, Chapter 16. For the state of the art, see Interest and Prices: Foundations of a Theory of Monetary Policy[2], by Michael Woodford (warning, probably only readable to people with PhDs in economics). For something for free, here's "Keynesian Macroeconomics without the LM Curve"[3], by David Romer, about how to teach New Keynesian economics to undergraduates. Here's a more mathematical version of the same[4]. Here's an even more mathematical version about teaching the basic model[5].
About slack in the economy, it has often been observed that crowds and mobs can be irrational. E.g. Stock market bubbles; people trampling each other to death during panics; people at a stadium concert all standing, when they could all have a equally good view sitting down. Why not the same for the macroeconomy? The failure to coordinate actions can cause individually rational agents to be collectively irrational. Milton Friedman had an interesting thought experiment on this. Rationality implies that money and prices should be transparent to the real economy — it should be equivalent if bread costs $1 and hourly wages are$2, or if bread costs $5 and hourly wages are$10. An hour of work buys 2 bread. Hence monetary policy should have no effect. Equivalently, it shouldn't matter if we label the hour when the sun is highest as 11:00, 12:00, or 13:00, people rationally should behave the same, changing activities depending on how bright it is. Hence, daylight savings time should have no effect. However, experience proves otherwise — monetary policy and daylight savings time change behavior. Why? Coordination failure. Given long enough, people will adapt no matter what the nominal numbers are, but in the short run, changing nominal numbers have a real effect.
Lastly, I hope you recognize that your (3) is akin to a statement of religious belief. LK (talk) 11:39, 27 June 2014 (UTC)
The paper by Carlin and Soskice is more understandable than most papers on economics, but it contains some simple mathematical errors. They lost a factor of two when they differentiated the cost function of the central bank (although this does not matter since the derivative is then set to zero). More importantly, they ignored the continuing effects (future costs) of uncorrected deviations of the rate of inflation from its target. They arbitrarily assume at one point that some of their constant coefficients are equal to 1. And they never specify the units of the variables and constants in their equations. Also, like most papers, it is much too wordy and yet manages to leave important matters to the guess-work of the reader.
Any attempt by business firms to coordinate their activities to be more rational risks running afoul of government regulations and especially the anti-trust laws. Also it would make their tax returns into an even bigger nightmare (how do they account for the tax consequences of the contingent costs and benefits in their contracts?).
My recognition of the effects of minimum wage laws and the contracts which businesses are compelled to negotiate with various unions is not religion, it is a fact. JRSpriggs (talk) 04:26, 28 June 2014 (UTC)
I probably shouldn't have pointed you at that paper, it's a suggested reformulation, not the standard model. The two previous are better representations of the standard New Keynesian model. It's also an internal working document and not peer reviewed. If you are sure that you have correctly identified errors, you should write the authors, I'm sure they'll appreciate the chance to fix it. BTW, Krugman just posted an interesting blog entry on the history of macro, it's a good read[6].
Your statement that "government and the unions it protects have created a situation where instead of correcting itself quickly via rapid deflation, economic activity is choked-off when deflation would be appropriate" is not a "fact'" - at least not one that 99% of macroeconomists would agree with. In fact more economists would probably agree with these "facts": "Because it is impossible to get rid of market imperfections, an enlightened bureaucrat can always find a way to improve on a market outcome." (See Theory of the second best). "Modest increases in the minimum wage have negligible or slightly positive effect on employment." (See Card & Kruger 1995). LK (talk) 11:17, 2 July 2014 (UTC)
Here we get to what I see as the basic philosophical difference between us. You have this belief that an "enlightened bureaucrat" will be making the government's decisions. That is magical-thinking. Rationality is not something that comes automatically or free of charge. In fact, it is very rare outside of matters within a person's painful and hard-earned experience. That is why the people making decisions should always be those who are directly affected by the results. Hence my belief in freedom and capitalism. JRSpriggs (talk) 10:06, 3 July 2014 (UTC)
I hope you do recognize that both your views and the ones you attribute to me are essentially beliefs, and little different from catechisms like "Jesus Christ is my personal savior". BTW, the way I see the difference in our views is like this. "Motor vehicles exist, they cause pollution and hurt/kill a lot of people every year, however, they are fundamental to our modern lifestyles. I'm in favour of restricting the amount of motor vehicles, using them only where necessary, and improving the way they work and how they fit into our societies. You believe that they are fundamentally evil and favour getting rid of them altogether." Substitute government for motor vehicles in the above, and that's how I see our disagreement. LK (talk) 04:26, 7 July 2014 (UTC)
1. I have good reasons for what I believe, it is not like a religion.
2. Argument by analogy (to automobiles or religion) is not a valid form of argument.
3. Automobiles provide a very large benefit — carrying people and goods from one place to another where they may be more valuable. On the other hand, government has only one "benefit" that is not available in the private sector, it can provide overwhelming force to ensure that it gets its way. But force is inherently destructive, so it is not appropriate to use it except in situations where the target is a net evil which cannot be eradicated by any lesser method.
4. History has shown that governments are almost invariably captured (or originally established) by corrupt people who use them for evil purposes. JRSpriggs (talk) 10:49, 10 July 2014 (UTC)
## BLP
I suggest that this edit is in contravention of Wikipedia:Biographies of living persons policy, which "applies to all material about living persons anywhere on Wikipedia". You might like to revert it. Deltahedron (talk) 17:36, 10 July 2014 (UTC)
Thanks for that. Deltahedron (talk) 06:24, 11 July 2014 (UTC)
## Thanks for tensor edits!
Thanks (as usual) for the edits, in this case to help correct my additions to tensor. :)
So I don't clog the talk page with dumb questions: does the distinction you are making mean that the co/contra-variant aspects of the tensor cannot be separately assessed (in a meaningful way)? I was thinking dimensions b/c of the contribution to the dimensionality of the final matrix (and in a sense the dimensionality of the tensor as a whole), but I wish to understand where the flaw is in my perception. TricksterWolf (talk) 19:17, 12 August 2014 (UTC)
One does not normally talk about the dimensions of the array, perhaps because of the possibility of confusion with the dimensionality of the underlying space.
Each dimension of the array is associated with one index. The index ranges over a finite number of values — each value corresponding to one of the dimensions of the underlying space. So a tensor with type (1, 1) over a ten-dimensional underlying vector space would consist of a 10×10 matrix having one hundred components. Each row is associated with a distinct element in the basis of the underlying vector space. Each column is associated with an element of the basis of the covector space dual to the vector space.
If the underlying vector space has d dimensions, then a tensor of type (n, m) will have dn+m components each of which is a real number (or a real valued function of location on the manifold, if we are talking about a tensor field). Is that clear? JRSpriggs (talk) 23:55, 12 August 2014 (UTC)
## ISIS & the ISI
Hello, there. Although I wasn't before, I am now concerned about the length of the ISIS page and Gazkthul as you will have seen has suggested doing with the ISI group what he did with ISIS's earlier predecessors, giving it its own article and leaving a resume in "History" in the ISIS page, which would reduce the size of the "History" section considerably. He will need to know whether he has consensus to do this and I wonder if I could ask you to give your response at the end of the thread here, pro or con. I cannot see why the 2014 timeline is duplicated on this page when it already has its own article! Reducing the ISI section and removing the 2014 timeline would drastically reduce the article's size! Thanks for telling the IP how to go about becoming an editor; the page could do with some more input. BTW, there has still been no response to my request at the WP:RSN about the Israel citations! It was even archived at one point, but I retrieved it and put it back in the list. --P123ct1 (talk) 12:08, 23 September 2014 (UTC)
## deleted talk stuff on natural numbers
Hello JR it is nice to make your acquaintance.
Yeah of course I'm not out to delete other contributions. What you don't see is that there was a second half of that conversation already deleted by MjolnirPants. Rather than deleting it directly he said he was moving it to another section, put the pointer for the move, but the text never appeared in the other section. It could have been an accident. I don't mind. He now seems to be excited about my observation that there is a conflict here between formal math definition and what school book writers would like to see. He has opened a talk section on that. I think it will be productive.
Wikipedia does allow for talk page "refactoring", and I do think that section should go according to those guidelines. https://en.wikipedia.org/wiki/Wikipedia:Refactoring_talk_pages
There was an effort to prevent zero from being mentioned as a natural when I arrived at the site. The first sentence was a circular definition and called naturals {1,2, 3... } etc. The point about not linking to other pages is correct. Every time I mentioned zero, my edit was deleted. When I added pointers to the set-theoretic natural number, my edit ws deleted. When I gave the first of Peano's axioms, zero is a natural, it was deleted - despite being on the page about the Peano axioms. When I pointed out that zero is an additive identity needed in arithmetic, my edit was deleted. When I mention Von Neuman's definition my edit was deleted. When I copied another editors comment that "The convention used by set theorists .." it was deleted as was the other editors remark. It is back now. When I fixed the circular definition on whole numbers, that was deleted... These deletes were often when ignoring my talk page entries. Majorpants ignored six talk page entries and the confirmation of two editors when he deleted stuff.
So yeah, I wish people would have more respect about deletes. You didn't ask me about my edit before you reverted it either. I have a talk page.
Now I'm on the fence about "refactoring" and talking out that section about the war on zero because people don't understand it without more context, and it is too much work to put the links to the deletions mentioned above in order to support the point, but most of all it is not on the topic of natural numbers but on the topic of editor etiquette. Besides what good does it do? Furthermore it isn't fair that my text stays but the other text goes.
So I hope you understand why I might refactor that out. — Preceding unsigned comment added by Thomas Walker Lynch (talkcontribs) 18:40, 8 October 2014 (UTC)
## graph cuts
hi you seem to be a math/stats guy. I was trying to make sense of Cut (graph theory) and was struggling, and tagged it for being too technical, and was drilled on per Talk:Cut_(graph_theory)#Technical. Maybe you might be willing to add some discussion to that article to make it more understandable to your average college educated person at least? thanks for considering... Jytdog (talk) 01:22, 24 October 2014 (UTC)
Sorry. Graph theory is not an aspect of mathematics to which I have paid much attention. So anything I said on that subject would be no better than what a layman might say. JRSpriggs (talk) 12:26, 24 October 2014 (UTC)
thanks for having a look and replying! Jytdog (talk) 12:34, 24 October 2014 (UTC)
## Loving relation in FOL-article
Hi! Thank you for looking after my tinkering there. I only by chance stepped into this article and you were so kind to repair my rubbish. Just now I noticed this and tried to improve by marginals again. I am fully OK with substituting formulas for formulae, but like to remark that I deliberately chose someone loves someone over everyone loves someone, because I feel less a priori quantification in the former formulation. This is not to say that I want to revert it, but just give a reason for me doing so. Sorry, if I bothered you. Purgy (talk) 08:00, 11 December 2014 (UTC)
I just felt that the examples in first sentence of First-order logic#Loving relation should agree with one another to avoid confusing the readers. If you want to change both of them, while retaining their agreement, then go ahead. JRSpriggs (talk) 15:44, 11 December 2014 (UTC)
If I ever get the feeling I could do some substantial improvement to this article, I'll reconsider this idea. :) Thanks! Purgy (talk) 17:59, 11 December 2014 (UTC)
## Lagrangian of a free particle in GR
OK, here I will give a detailed discussion. First, some notes on Lagrangian and actions, as well as the geodesic equation:
• The Lagrangian is unphysical.
• The action is unphysical.
• Lagrangians and actions are equivalent if they generate the same equations of motion.
• δS=0 implies the Euler-Lagrange equations, not the other way around.
• Overall factors in the Lagrangian are irrelevant. This includes signs and constants.
• The geodesic equation does not rest on an Euler-Lagrange style argument. It can be derived by other methods.
• The mass shell condition is a consequence of the geodesic equation.
The first three are basic tenets of the Lagrangian prescription. See any textbook on QFT and you will see Lagrangians and actions mangled and beaten into shape to extract physical information. Now, because of the linearity of integration and functional differentiation, if the Lagrangian is multiple of something else, then that multiple cancels in δS=0, leaving the equations of motion unaffected.
• How does one derive the geodesic equation from your Lagrangian? That Lagrangian IS correct, in a sense, as I will show later. However, it is not in a form that I can see leading to the properly parametrized geodesic equation. (A multiple of τ is the correct parameterization.)
• The units really don't matter. I could be using units in which m=c=1. This does not affect anything.
• The Newtonian argument breaks down when we consider massless particles. For such particles your action is zero! By your reasoning, this leads to trivial equations of motion.
First I will show that the geodesic equation is not strictly a consequence of δS=0. For this, I will adapt the discussion found in Weinberg's Gravitation and Cosmology. Let ξ be a locally inertial coordinate system (free-falling) and τ be the proper time along the worldline (the time on a clock as measured by the particle in its rest frame). According to the Principle of Equivalence, there is such a coordinate system in which the equation of motion for SR holds:
${\displaystyle {\frac {d^{2}\xi ^{\alpha }}{d\tau ^{2}}}=0}$
In natural units, the proper time in this frame is
${\displaystyle d\tau ^{2}=-\eta _{\alpha \beta }d\xi ^{\alpha }d\xi ^{\beta }}$
Now suppose we use any other coordinate system x, which may be a cartesian system in a laboratory, curvilinear, accelerated, or whatever. The freely falling coordinates ξ are functions of the x, thus the first equation becomes
${\displaystyle 0={\frac {d}{d\tau }}\left({\frac {\partial \xi ^{\alpha }}{\partial x^{\mu }}}{\frac {dx^{\mu }}{d\tau }}\right)={\frac {\partial \xi ^{\alpha }}{\partial x^{\mu }}}{\frac {d^{2}x^{\mu }}{d\tau ^{2}}}+{\frac {\partial ^{2}\xi ^{\alpha }}{\partial x^{\mu }\partial x^{\nu }}}{\frac {dx^{\mu }}{d\tau }}{\frac {dx^{\nu }}{d\tau }}}$
A line of algebra and the chain rule leads to the geodesic equaton
${\displaystyle {\frac {d^{2}x^{\lambda }}{d\tau ^{2}}}+\Gamma _{\;\;\mu \nu }^{\lambda }{\frac {dx^{\mu }}{d\tau }}{\frac {dx^{\nu }}{d\tau }}=0}$
where
${\displaystyle \Gamma _{\;\;\mu \nu }^{\lambda }={\frac {\partial x^{\lambda }}{\partial \xi ^{\alpha }}}{\frac {\partial ^{2}\xi ^{\alpha }}{\partial x^{\mu }\partial x^{\nu }}}}$
is the affine connection. The proper time may also be written in an arbitrary coordinate system:
${\displaystyle d\tau ^{2}=-\eta _{\alpha \beta }{\frac {\partial \xi ^{\alpha }}{\partial x^{\mu }}}dx^{\mu }{\frac {\partial \xi ^{\beta }}{\partial x^{\nu }}}dx^{\nu }=-g_{\mu \nu }dx^{\mu }dx^{\nu }}$
where g is the metric tensor
${\displaystyle g_{\mu \nu }=\eta _{\alpha \beta }{\frac {\partial \xi ^{\alpha }}{\partial x^{\mu }}}{\frac {\partial \xi ^{\beta }}{\partial x^{\nu }}}}$
Denote by ${\displaystyle \partial _{\mu }}$ the operator ${\displaystyle \partial /\partial x^{\mu }}$. You can have the pleasure of verifying that, given the previous equations
${\displaystyle \partial _{\lambda }g_{\mu \nu }=\Gamma _{\;\;\lambda \mu }^{\rho }g_{\rho \nu }+\Gamma _{\;\;\lambda \nu }^{\rho }g_{\rho \mu }}$
Add to this equation the same equation with μ and λ interchanged and subtract the same equation with ν and λ interchanged. Remembering that the affine connection as defined above is guaranteed to be symmetric, this leads to the usual Christoffel symbols
${\displaystyle \Gamma _{\;\;\mu \nu }^{\lambda }={\frac {1}{2}}g^{\lambda \rho }(\partial _{\mu }g_{\nu \rho }+\partial _{\nu }g_{\mu \rho }-\partial _{\rho }g_{\mu \nu })}$
So it is easy to see that the geodesic equation does not rest upon any particular variational principle of any particular Lagrangian, because you don't need the action principle to derive it.
From here on I will use some coordinate free notation mixed in with the usual Ricci notation. Let γ(τ) be the worline of the particle, parameterized by the proper time τ. Let (M, g) be a 4-dimensional pseudo-Riemannian manifold with a Lorentz metric g and inner product ${\displaystyle \langle .,.\rangle =g(.,.)}$. Let ${\displaystyle \nabla }$ be the Levi-Civita connection on M. Let
${\displaystyle {\dot {\gamma }}={\frac {d}{d\tau }}}$
be the tangent vector to the curve. In this notation, the geodesic equation is
${\displaystyle \nabla _{\dot {\gamma }}{\dot {\gamma }}=0}$
i.e. the tangent vector is autoparallel along the geodesic. Next we use the Ricci identity. See any text on differential geometry for a proof. For any three vectors X, Y and Z
${\displaystyle X\langle Y,Z\rangle =\langle \nabla _{X}Y,Z\rangle +\langle Y,\nabla _{X}Z\rangle }$
So let ${\displaystyle X=Y=Z={\dot {\gamma }}}$. Then
${\displaystyle {\frac {d}{d\tau }}\langle {\dot {\gamma }},{\dot {\gamma }}\rangle ={\dot {\gamma }}\langle {\dot {\gamma }},{\dot {\gamma }}\rangle =2\langle \nabla _{\dot {\gamma }}{\dot {\gamma }},{\dot {\gamma }}\rangle =0}$
where the last equality follows from the geodesic equation. We integrate the very first term along the worldline P. Thus
${\displaystyle \langle {\dot {\gamma }},{\dot {\gamma }}\rangle ={\text{const}}}$
The convention is to take this constant to be -1. This comes about as follows. Suppose a particle in SR travels along a worldline P. Then the change in proper time is given by the line integral
${\displaystyle \Delta \tau =\int _{P}d\tau =\int _{P}{\sqrt {-\eta _{\alpha \beta }d\xi ^{\alpha }d\xi ^{\beta }}}}$
Now suppose we turn on the gravitational field. Then by the Equivalence Principle (I will drop the P from now on)
${\displaystyle \Delta \tau =\int {\sqrt {-g_{\mu \nu }dx^{\mu }dx^{\nu }}}}$
You object that the integral is not in the form ∫f(x)dx. So we choose to parameterize using the proper time and write
${\displaystyle \Delta \tau =\int {\sqrt {-\langle {\dot {\gamma }},{\dot {\gamma }}\rangle }}\,d\tau }$
Comparing with above, we conclude that the choice
${\displaystyle \langle {\dot {\gamma }},{\dot {\gamma }}\rangle =-1}$
is consistent. But what if we choose some other parameter λ? As of this moment we have no restrictions on what can be a parameter. Thus
${\displaystyle \Delta \tau =\int {\sqrt {-\langle {\bar {\gamma }},{\bar {\gamma }}\rangle }}\,d\lambda }$
where
${\displaystyle {\bar {\gamma }}={\frac {d}{d\lambda }}}$
Now I want to bring the action principle into the fold. I think that we can both agree that in SR, if we care about overall signs and factors (c=1 still remains though), the free action for a point particle of mass m is
${\displaystyle S=-m\int {\sqrt {1-{\vec {v}}^{2}}}\,dt=-m\int d\tau }$
The GR generalization is
${\displaystyle S=-m\int d\tau =-m\int {\sqrt {-\langle {\bar {\gamma }},{\bar {\gamma }}\rangle }}\,d\lambda }$
The principle of least action is of course δS=0. Now I hope you see that the factor -m does not do anything at all! Thus the variational principle is simply
${\displaystyle \delta \int {\sqrt {-\langle {\bar {\gamma }},{\bar {\gamma }}\rangle }}\,d\lambda =0}$
I also hope you see that ${\displaystyle \langle {\bar {\gamma }},{\bar {\gamma }}\rangle }$ is in no way restricted or fixed and can thus S can be varied. In this article this variation is performed. I will not do it here. The result is
${\displaystyle {\frac {d^{2}x^{\lambda }}{d\lambda ^{2}}}+\Gamma _{\;\;\mu \nu }^{\lambda }{\frac {dx^{\mu }}{d\lambda }}{\frac {dx^{\nu }}{d\lambda }}=0\quad {\text{or}}\quad \nabla _{\bar {\gamma }}{\bar {\gamma }}=0}$
Once again we can easily show that
${\displaystyle \langle {\bar {\gamma }},{\bar {\gamma }}\rangle ={\text{const}}}$
from which it follows that ${\displaystyle \lambda \propto \tau }$. This follows only from the equations of motion and is not obvious before. As a constraint derived from the equations of motion, it cannot be applied to the Lagrangian! This would be like doing QFT on the mass shell only. (In fact, as I will show, the mass shell condition is a consequence of the geodesic equation.)
I said that the Lagrangian you presented was not incorrect. This is because you have defined the action as
${\displaystyle S=\int L\,dt}$
whereas I have defined it as
${\displaystyle S=\int L\,d\tau }$
This is a simple change of parametrization. However, your Lagrangian is not invariant under diffeomorphisms. The coordinate t has a nontrivial transformation rule as opposed to λ, which allows for invariant parameterization. Some food for thought: try deriving the geodesic equation from your Lagrangian. Maybe you can surprise me and get it.
The current discussion is adapted from Becker, Becker & Schwarz String Theory and M-Theory. The action -mdτ has some problems with it:
• In the path integral sense, it is a disaster. The square root makes this very hard to quantize.
• It is completely unsuited to handling massless particles for obvious reasons.
These problems can be circumvented by introducing an auxiliary (bosonic) field α(τ) in the modified action
${\displaystyle S={\frac {1}{2}}\int d\lambda \,\left({\frac {\langle {\bar {\gamma }},{\bar {\gamma }}\rangle }{\alpha }}-m^{2}\alpha \right)}$
First we determine the equations of motion for α(τ):
${\displaystyle {\frac {\delta S}{\delta \alpha }}=-{\frac {1}{2}}\left({\frac {\langle {\bar {\gamma }},{\bar {\gamma }}\rangle }{\alpha ^{2}}}+m^{2}\right)=0\longrightarrow \langle {\bar {\gamma }},{\bar {\gamma }}\rangle =-m^{2}\alpha ^{2}}$
Plugging this back into S, we see that we recover the original action. Now, it may be shown that this modified action is parameterization invariant if α has certain transformation properties. (BBS exercise 2.3) This leads to a sort of gauge invariance. The gauge α=1 is nice. In this gauge
${\displaystyle \langle {\bar {\gamma }},{\bar {\gamma }}\rangle =-m^{2}}$
is the familiar mass shell condition. This means ${\displaystyle {\bar {\gamma }}}$ is the momentum vector. The additive constant in the Lagrangian can be just thrown away because it contributes nothing to the equations for γ. Thus the action is effectively
${\displaystyle S={\frac {1}{2}}\int \langle {\bar {\gamma }},{\bar {\gamma }}\rangle \,d\lambda }$
Now this makes perfect sense for massless particles. For massive particles, we have τ=mλ. So, up to an overall factor (which we just throw away, the 1/2 is kept for historical reasons)
${\displaystyle S={\frac {1}{2}}\int \langle {\dot {\gamma }},{\dot {\gamma }}\rangle \,d\tau }$
As I showed in the post you deleted, this leads to the geodesic equation
${\displaystyle \nabla _{\dot {\gamma }}{\dot {\gamma }}=0}$
Thus, the Lagrangian of general relativity can be taken to be
${\displaystyle L={\frac {1}{2}}g_{\mu \nu }{\dot {x}}^{\mu }{\dot {x}}^{\nu }}$
Because this leads to the correct equations of motion, it is an acceptable Lagrangian.
I hope this clears up any confusion.
Differential 0celo7 (talk) 22:43, 18 December 2014 (UTC)
It would have been better to hold this discussion at Talk:Lagrangian#What is the correct Lagrangian for an isolated test particle in general relativity? so that other interested people could find it and participate.
Much of what you say is true and already well known to me. However, I disagree on some important points.
I disagree about ignoring constant factors because we may wish to add Lagrangians for various parts of a system together to get the Lagrangian for the total system and this will fail unless their proportions are correct. Also you need the constant factors to get the correct stress-energy tensor from the Lagrangian.
Your Lagrangian was either pulled out of the air and justified by its results (which I think should be avoided when possible) or it is an attempt to generalize ${\displaystyle {\frac {mv^{2}}{2}}}$ in an inappropriate way.
I do not disagree with the geodesic equation, but only with your method of deriving it.
I have shown how to derive a variation of the geodesic equation (using momentum) from my Lagrangian at User:JRSpriggs/Force in general relativity#Derivation from Lagrangian.
I have not yet seen a fully satisfactory derivation of the equation of motion for massless particles except by taking a limit of the equation for massive particles as the mass approaches zero. However, the only particles which might be massless are the photon and the graviton (if it exists).
Using the proper-time τ as a parameter for the particle's path is inappropriate because the value of τ is path-dependent. In particular, the limits of integration might have to be changed as you do the variation which is highly undesirable. JRSpriggs (talk) 12:27, 19 December 2014 (UTC)
I'll move the discussion to the Lagrangian talk page.
Differential 0celo7 (talk) 19:55, 19 December 2014 (UTC)
## Invitation
You've been invited to be part of WikiProject Cosmology Hello. Your contributions to Wikipedia have been analyzed and it seems that this new Wikiproject would be interesting to you. I hope you can contribute to it by expanding the main page and later start editing the articles in its scope. Make sure to check out the Talk page for more information! Cheers Tetra quark (talk) 20:23, 30 December 2014 (UTC)
## Invitation to Participate in a WikiProject Study
Hello JRSpriggs,
We’d like to invite you to participate in a study that aims to explore how WikiProject members coordinate activities of distributed group members to complete project goals. We are specifically seeking to talk to people who have been active in at least one WikiProject in their time in Wikipedia. Compensation will be provided to each participant in the form of a \$10 Amazon gift card.
The purpose of this study is to better understanding the coordination practices of Wikipedians active within WikiProjects, and to explore the potential for tool-mediated coordination to improve those practices. Interviews will be semi-structured, and should last between 45-60 minutes. If you decide to participate, we will schedule an appointment for the online chat session. During the appointment you will be asked some basic questions about your experience interacting in WikiProjects, how that process has worked for you in the past and what ideas you might have to improve the future.
You must be over 18 years old, speak English, and you must currently be or have been at one time an active member of a WikiProject. The interview can be conducted over an audio chatting channel such as Skype or Google Hangouts, or via an instant messaging client. If you have questions about the research or are interested in participating, please contact Michael Gilbert at (206) 354-3741 or by email at mdg@uw.edu.
We cannot guarantee the confidentiality of information sent by email.
The link to the relevant research page is m:Research:Means_and_methods_of_coordination_in_WikiProjects
Ryzhou (talk) 03:55, 28 January 2015 (UTC)
## Remark on politics
On your user page, you write the following:
"A free economy based on private property tends to correct itself by the process of natural selection. Any attempt by a government planner to manage the economy will necessarily be misguided due to his limited knowledge and the destructive force necessary to attempt such management. Invariably the attempt will damage the economy and work against the interests of everyone including the intended beneficiaries.
All entitlement programs (obamacare, social security, medicare, medicaid, welfare, etc.) should be repealed immediately because they are: outside the proper functions of government, promoters of dependency, too costly, and unsustainable."
I agree that natural selection might be one of the forces ruling the market. It might work as follows: Those companies which make the most profits survive, while those making less profits go bankrupt. I imagine that if the government intervenes for instance by taxing CO2 emissions, the companies will produce less CO2, since they make more money then. This, I think, would be an example for a non-misguided government intervention. --Mathmensch (talk) 09:53, 11 February 2015 (UTC)
Businesses making a profit may survive. Businesses which suffer a loss and continue to do so will be forced to liquidate.
You do not know how harmful carbon-dioxide emissions are or even whether they are harmful rather than beneficial. Neither do politicians and bureaucrats. But what is clear is that taxation and the violence which is required to collect taxes is very harmful.
Imagine that someone was going around your neighborhood (at night) siphoning gasoline out of your vehicles. How would you feel? Perhaps you cannot make it to work on time. Perhaps you cannot use your car in an emergency. What if the perpetrator was caught and his defense was that he was protecting the planet from your carbon-dioxide emissions? JRSpriggs (talk) 08:49, 13 February 2015 (UTC)
I only use public transport in order to avoid extensive CO2 emissions. I think I would probably be O.K. with somebody stealing my stuff if they have better use for it.
On the basis of what I currently know, I personally would guess that carbon-dioxide emissions are probably heating up earth because of the greenhouse effect. This is for example depicted in this picture:
There are certain issues connected to this effect, which go against my (utilitarian) ethics. These include increased extreme weather (which can damage people). Extreme weather comes about more often with increased CO2 emissions, because almost all weather phenomenons are triggered by energy in the atmosphere, and this energy in turn is increased by the greenhouse effect. I list tornadoes, for example, as an extreme weather phenomenon triggered by high energy in the atmosphere.
Another issue are the effects on crops. In the equatorial region, for the next 100 or something years we expect a population growth by three billion. However, these regions are also those where crop growth is predicted to go back. I don't know of firm evidence confirming that there will be compensation in higher latitude countries, but even if there is, it is by no means certain that the crops are transported from there to the regions where they are needed.
And further, if one would give weight to the biodiversity of this planet: If the environmental conditions change, many species might not have the chance to migrate to an area where the ecological circumstances which they need prevail. This could in some cases even trigger species extinction (see Climate change and ecosystems). --Mathmensch (talk) 22:04, 13 February 2015 (UTC)
You should be more concerned for your own needs rather than tolerating someone stealing your energy (or regulating or taxing it).
Carbon-dioxide emissions do raise the temperature of the Earth a little, but the effect is very small.
The increased heat in the atmosphere due to the greenhouse effect is merely increasing the temperature of the heat sink, not the heat source. Thus it will reduce the output of the weather 'engine' rather than increasing it. Also it is recentism to think that recent extremes of weather are unusual; extremes of weather (and climate change) have occurred in all epochs of time.
Even before people appeared, species went extinct (and were replaced by new species) with great frequency. Most of those species were no great loss to the world — they were too fragile (inflexible) and too specialized to provide any substantial benefit to others. JRSpriggs (talk) 19:25, 15 February 2015 (UTC)
You might also be interested in "The Climate Change Solution No One Will Talk About" by Stefan Molyneux. JRSpriggs (talk) 04:55, 17 February 2015 (UTC)
Actually, I am more worried about a new Ice age in which the United States and Canada are crushed under a huge ice sheet (glacier) rather than global warming. Northern Eurasia would also be covered in ice. JRSpriggs (talk) 17:29, 23 February 2015 (UTC)
See "What it would take to prove global warming". JRSpriggs (talk) 04:20, 23 April 2015 (UTC)
Hi; sorry if I'm being blind, but by this revert you are implying that you seen something on that page.. currently to me the page User:Mathbot/List of mathematical redlinks lists a single list, which is blue: List of Italian mathematicians.. I agree that list contains some redlinks.. is that what you are talking about? Maybe I'm getting confused, but that page does not seem to do what it claims to do. I would quite like to see a list of redlink, like User:Mathbot/Most wanted redlinks, but alas it has not been updated since 2007. So I'm not sure it's worth linking to.. Mark M (talk) 10:22, 5 March 2015 (UTC)
Yes, you are being blind. Besides the link you mentioned (to Italian mathematicians), that page contains links labelled "0-9", "A", "B", ..., "Z" and below those a link to the other page you mentioned "User:Mathbot/Most wanted redlinks". Try reading the text. JRSpriggs (talk) 00:51, 6 March 2015 (UTC)
Haha, I don't know how I missed that. I see it now. It is unfortunate this list is so out of date.. after a quick (and admittedly small) sampling, it seems that a large proportion (possibly more than half) of the listed articles are either no longer red, or no longer have any incoming links. But I suppose there isn't anything better at the moment. Cheers, Mark M (talk) 09:10, 6 March 2015 (UTC)
I asked Oleg Alexandrov (talk · contribs) to have Mathbot (talk · contribs) update the lists. It last did so in 2007. So there might be a question about whether it still has the capability to do that. JRSpriggs (talk) 15:55, 6 March 2015 (UTC)
## Recursively enumerable set
Hey JR, what's wrong with the change you reverted?
https://en.wikipedia.org/w/index.php?title=Recursively_enumerable_set&oldid=651019471&diff=prev
• If it's correct, it's 3 times shorter and far more easy to understand.
• If it's incorrect, please tell me why. It's not an ego thing; in fact it's the opposite. It would mean that my understanding of recursively enumerable sets is incomplete. To me, that would be appalling (really, and for several reasons).
Thank you for educating me! I love being educated, fast and deep.☺
--VerdanaBøld 10:34, 14 March 2015 (UTC)
As I said in my edit summary, your "definition" of recursively enumerable set is unclear and thus not a definition. It could be interpreted in a way which would make it correct, but it could also be interpreted in other ways which are incorrect.
You said "A set is recursively enumerable if there exists an algorithm that maps each element of the set, in finite time, to a unique natural number.". You failed to restrict the set to be a set of natural numbers. This opens a can of worms about how the correspondence with natural numbers should be done. "Algorithm" is vague, whereas there is a precise mathematical definition of "partial recursive function". "In finite time" could mean finitely many steps of a calculation, but it could mean something else. "To a unique natural number" could mean that only one natural number is in the range (which while equivalent is a different definition from the one we were using) rather than that the function is at most single valued.
In any case, there is already an informal definition like yours in the lead. The section which you changed was supposed to be for the technically correct and precise definition. Also the (old) sources I am familiar with use a definition like mine rather than yours, and we are supposed to follow the sources (in case of doubt). JRSpriggs (talk) 03:02, 16 March 2015 (UTC)
## Conjugate variables in QM
I suppose you have completed this section [7]. For research purposes, if you have the sources I might need them, however if you proved them yourself, I might be very interested in the proofs. Especially the electric potential-charge density uncertainty seems to have some implications in nano-communications. I would be extremely grateful if you could help me out with this. I have a BS in both electrical engineering and physics and an MS in electrical engineering so I might follow your work.
As a side note, I am also a go player. I rank 5k in KGS. I might be a little obsolete, but if you are close to my rank, we may play :) Thank you! Caglarkoca (talk) 00:52, 13 May 2015 (UTC)
The article on the Hamilton–Jacobi equation may be helpful in understanding this. I no longer remember where I got the information on the conjugate variables. But you might see Covariant formulation of classical electromagnetism#Lagrangian for classical electrodynamics, the subsection on "Matter" where it gives the Lagrangian in non-relativistic notation.
I am not an active GO player at this time and I never got above 10 kyu at best. I am probably closer to 18 kyu now. JRSpriggs (talk) 02:05, 14 May 2015 (UTC)
The electromagnetic items on the list are wrong because they are derivatives of the Lagrangian rather than the action. However, if everything that is questionable were removed from the article, then there would not be much, if anything, left. So I gave up on it and took it off my watch list some time back. JRSpriggs (talk) 03:10, 14 May 2015 (UTC)
Thank you for your time. I'll check them anyway. But you are wrong about Go. You being around 18k and me being 5k does not prevent us from playing it. Contact me if you are interested in a teaching or handicap game :) 212.175.32.131 (talk) 08:01, 14 May 2015 (UTC)
## Einstein Field Equations
Hello Mr JRSpriggs, Could you please explain deleting the application section you judged to be no value? The 2 references and application were valuable to me, and I believed to others, especially to new learners. Thank you, --Jcardazzi (talk) 11:49, 21 August 2015 (UTC)jcardazzi
I still believe that the section you added, Einstein field equations#Simple Application and Meaning, has no value. Presently, it says "Solving the equation by using the values for the physical constants on the right side of the field equation, where G=Newton's gravitational constant, c=the speed of light, the equation reduces to: G=0.00000000000000000000000000000000000000000000002*T meaning to get a warp of space-time takes a large amount of mass. For example the Earth's mass at 1 M⊕ = 5.97219 × 1024 kg warps space-time to the strength of gravity which we experience on Earth causing an acceleration of mass of approximately 9.8 meters/second2 at the Earth's surface.". A long string of zeros (instead of an expression in scientific notation) is incomprehensible as most people would have difficulty even counting the number of zeros, let alone figuring out what they mean. Presumably you are assuming some system of units, but you have not identified which system. No scientist talks about "a warp of space-time"; that is straight out of fiction. The "G" in your formula is potentially confusing as some people might read it as referring to the gravitational constant rather than the Einstein tensor. Finally, your section amounts to saying that the mass of the Earth is responsible for the Earth's gravitational field. Well, duh!!! Thus I will remove it again. JRSpriggs (talk) 23:36, 23 August 2015 (UTC)
Hello Mr JRSpriggs, thank you for the comments, they are in informative to me. I will try to rework the wording from the references and your comments for better clarity, and propose them in the talk page in the future.
Regarding the term: "a warp of spacetime";
I read the term and similar uses of it, in many articles, for example: https://en.wikipedia.org/wiki/Introduction_to_general_relativity "...According to general relativity, the observed gravitational effect between masses results from their warping of spacetime."
"High-precision test of general relativity by the Cassini space probe....radio signals sent between the Earth and the probe... are delayed by the warping of spacetime...due to the Sun's mass.
Also, fyi, here is an article on the physics of space warp experiments: https://en.wikipedia.org/wiki/Warp-field_experiments
a book using the term: http://www.its.caltech.edu/~kip/scripts/PubScans/VI-47.pdf General relativity is Einstein’s law of gravity, his explanation of that fundamental force which holds us to the surface of the Earth. Gravity, Einstein asserted, is caused by a warping of space and time—or, in a language we physicists prefer, by a warping of spacetime.
Thank you,--Jcardazzi (talk) 00:33, 24 August 2015 (UTC)jcardazzi
## Lagrangian
Hi James, good to see you here. I tidied up after my bad CSD suggestion. Ooo, while I'm here, now that Mathjax support has been removed, how are you guys getting proper rendering? I came across an abandoned script that enables it again, was thinking of hosting it. Does that sound useful? Regards 15:35, 11 September 2015 (UTC)
## Countable sets
Thanks for the edits, I sometimes get a little too sloppy when composing and it is nice to know that other eyes will catch my slip-ups. In the proposition you modified, I had left the disjoint condition in, just to be faithful to the reference, which leads me to the question I wanted to ask you. I am contemplating putting in a remark at the beginning of the section saying that the functions being used may not be precisely as presented in the references, but that the differences are minor. It should have been, but wasn't, easy to find references that presented things using the conventions of the article. Most of my goto sources do not consider 0 to be a natural number and many of them only define the functions in the denumerable case with N as the domain. Of course it is trivial to fix these problems, but the reader who goes off for more information in the references is not going to see the exact statements that appear in the article. Hence, I see a need for the disclaimer. I would be interested to know what you think about putting that in. Thanks. Bill Cherowitzo (talk) 03:52, 19 September 2015 (UTC)
## ArbCom elections are now open!
Hi,
You appear to be eligible to vote in the current Arbitration Committee election. The Arbitration Committee is the panel of editors responsible for conducting the Wikipedia arbitration process. It has the authority to enact binding solutions for disputes between editors, primarily related to serious behavioural issues that the community has been unable to resolve. This includes the ability to impose site bans, topic bans, editing restrictions, and other measures needed to maintain our editing environment. The arbitration policy describes the Committee's roles and responsibilities in greater detail. If you wish to participate, you are welcome to review the candidates' statements and submit your choices on the voting page. For the Election committee, MediaWiki message delivery (talk) 13:47, 23 November 2015 (UTC)
## Reverted edit to "Lawrence Kohlberg's stages of moral development"
Could you please provide me a more detailed rationale for your reversion of my edit? Is the "girl/boy" idea a direct quote from the source? If so, "kid" probably wasn't a great idea on my part, but with some rephrasing "girl/boy" could be removed entirely without the use of "kid". I haven't been able to get a copy of the source paper, but I have looked into other papers written about the original paper and they all manage to avoid "girl/boy", so I think it is fair to expect Wikipedia to do the same. I am interested in improving the quality of the article and not partaking in an edit war or argument, so I want to know why my edit was reverted in order to potentially find a better way of including it in the article. Thanks for your time and your contributions to the article! FalkirksTalk 15:51, 11 December 2015 (UTC)
I meant to explain in my edit summary, but something went wrong and my edit was saved before I could type in the edit summary.
The word "kid" is too informal for an encyclopedia. Strictly speaking, it refers to baby goats not people. Thus it could be seen as insulting to children generally.
You explained your change to Lawrence Kohlberg's stages of moral development with the summary "Changed 'boy/girl' to 'kid' to be more clear and remove potential transphobic connotations". I think that you are worried unnecessarily. Everyone knows that "boy/girl" in this context is short for "boy or girl". Virtually no one (other than yourself apparently) is even going to think about transgender issues in relation to this article. And if we made a change that caused them to think about it, that would be distracting from the topic of the article and thus wrong. JRSpriggs (talk) 11:36, 13 December 2015 (UTC)
Thanks for your explanation. I am glad that the lack of summary was accidental. My concern is that "boy or girl" enforces the idea of a Gender binary which in my opinion is outside of the scope of this article. If one reads the article on Gender they will notice that Wikipedia discusses identities that can't be defined as "boy or girl", this article excludes those identities. I am not seeking revolution for my way of thinking, I am simply looking for consistency throughout the encyclopedia with regards to inclusion of non-binary gender identities. The issue is much more blatant than you portray it as and I firmly believe that it deserves some remedy. The idea that "kid" refers more to goats than people went completely over my head and I hope that a better word exists. Do you have any suggestions for a gender neutral term that could be used in place of "kid"? FalkirksTalk 18:13, 15 December 2015 (UTC)
You could say "child". However, in the real world people say "boy" or "girl" as appropriate. Recognizing a person's sex is considered polite. The fact that our language leaves out other (extremely rare) alternatives is unfortunate, but it is not the function of Wikipedia to try to revise the world. JRSpriggs (talk) 18:27, 15 December 2015 (UTC)
## Re: Sanders
The sort of people who support Bernie Sanders may (mostly) believe that the United States is a plutocracy. But that is not true for the majority of Americans.
I think you're wrong on this one. Sanders has worked closely with conservatives on many issues, and while they may disagree on the exact path to take to get to the solution, for the most part, they all agree on the nature of the problem. This is also true for the majority of Americans.[8] So I think the facts just don't support you on this. The use and misuse of the word "socialism" seems to be your sore point, but pork spending, corporate welfare, health care and job security are all things Americans are on the same page about. The thing is, when you are too extreme, and too far to the right, mainstream ideas like those of Sanders are perceived as radical, when in fact they are considered center to center left in relation to the rest of the world. Viriditas (talk) 20:16, 27 December 2015 (UTC)
## Category:Relativists
Category:Relativists, which you created, has been nominated for possible deletion, merging, or renaming. If you would like to participate in the discussion, you are invited to add your comments at the category's entry on the Categories for discussion page. Thank you. —Justin (koavf)TCM 06:15, 30 December 2015 (UTC)
## Johann Kudar's elevation to notability
Just a brief note that I do feel strongly Johann Kudar does not rise to the notability level of an article implicitly proposed by the redlink in the Klein-Gordon equation I deleted, evidently without success. Come on: Johann Kudar? What were these guys thinking?? Cuzkatzimhut (talk) 03:28, 26 January 2016 (UTC)
## Reverting for no reason is highly disruptive
You undid an edit of mine without bothering to explain why.[9] This is rude and highly disruptive behaviour. If you had a reason to revert, you should have specified it in the edit summary, and if you didn't, you shouldn't have reverted. 109.155.216.108 (talk) 11:46, 12 February 2016 (UTC)
If you knew anything about the book, you would realize that the information you removed was all true, relevant, and important. You are vandalizing the article and using "policy" as an excuse. JRSpriggs (talk) 05:04, 13 February 2016 (UTC)
Don't make personal attacks. I'll report future occurrences. Later on today I'll go through the changes and explain to you one by one why they are necessary, though it really is incredibly obvious.82.132.213.42 (talk) 15:05, 13 February 2016 (UTC)
You (the IP) are the one with the "rude and highly disruptive behaviour", and should cease and desist. Your only reason so far has been "rm pov, unencyclopaedic content". What is so "unencyclopaedic" about the content? You may find the bits you deleted trivial/obvious, but not everyone will, someone stumbling on the page may know nothing about the book or subject. No one has made an explicit personal attack against you, just a revert. Everyone gets frustrated when they're reverted but unless the edit summary indicates, a revert is not a "personal attack". MŜc2ħεИτlk 08:46, 15 February 2016 (UTC)
It's also possible to track where you are with a visible IP: AS29180 O2-ONLINE-AS Telefonica UK Limited (registered Jun 23, 2003. In principle, IP addresses at the place can be blocked just from the disruptive behavior of one account. MŜc2ħεИτlk 08:50, 15 February 2016 (UTC)
## tensor
${\displaystyle {\hat {T}}_{j'}^{i'}=(R^{-1})_{i}^{i'}R_{j'}^{j}T_{j}^{i}=R_{j'}^{j}T_{j}^{i}(R^{-1})_{i}^{i'}}$
the first equality is directly computing by the above tensor formula, the second just change the order of sum, which is invariant for finite sum, and the righ hand side is equal to ${\displaystyle (RTR^{-1})_{j'}^{i'}}$, That's why we have
${\displaystyle {\hat {T}}=RTR^{-1}.}$
Please check, thanks Wttwcl (talk) 03:40, 29 February 2016 (UTC)
I think you have the order reversed. If a contravariant vector is represented by a column matrix, then you are wrong. JRSpriggs (talk) 02:20, 1 March 2016 (UTC)
## Userpage
Hey JRSpriggs. You have been around here a long time. I am surprised to see that you Userpage appears to violate the WP:Userpage policy, with all this stuff on your personal views on various political and economic issues. Do you really think that is appropriate? I don't want to get all admin/dramaboard=y about this... Jytdog (talk) 17:15, 11 March 2016 (UTC)
I think this question was considered before at Wikipedia:Miscellany for deletion/User:JRSpriggs. Do you have a new argument? JRSpriggs (talk) 09:12, 12 March 2016 (UTC)
I believe it would help if sources and/or wikilinks were provided that support the statements on the user page. For example, the non-contradiction section is a rewording or restating of a something found in Atlas Shrugged, Chapter Seven I doubt Ayn Rand was the first to say that but what you have seems close enough that you could credit her. --Marc Kupper|talk 04:47, 19 March 2016 (UTC)
The ideas on my user page are mine. While I do not claim originality for most of them, neither am I consciously quoting anyone else. There is no one (not even Ayn Rand) with whom I totally agree. So I do not generally give links because that might be construed as an unqualified endorsement.
Regarding Ayn Rand in particular: Her notion that there is a fallacy of the stolen concept is itself mistaken. What she calls a fallacy (in this case) is merely a specific variety of contradiction. Since it is a contradiction, it must be the product of a false premise or some (other) fallacy. But it is not necessarily the first falsity in the chain of reasoning itself.
Her notion tends to undermine the process of reasoning by preventing people from searching for the true cause of the contradiction. As such it is a very dangerous idea.
Another place where the Objectivists go wrong is conflating the objectivity of legal judgements with the public nature of them. As a result, they insist that government is necessary even though this does not follow from their theory of rights. JRSpriggs (talk) 16:52, 19 March 2016 (UTC)
## I responded.
Please look at User_talk:98.195.88.33.98.195.88.33 (talk) 01:00, 30 May 2016 (UTC)
Regarding your edit to Methods of computing square roots. I am aware that your method works and I am not surprised that you can prove that it does. But that does not address the reasons which I and others gave for excluding it from the article. JRSpriggs (talk) 21:02, 30 May 2016 (UTC) | 2016-08-25 20:18:24 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 46, "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": 0.6472599506378174, "perplexity": 1019.7106981029575}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1471982294097.9/warc/CC-MAIN-20160823195814-00249-ip-10-153-172-175.ec2.internal.warc.gz"} |
https://math.stackexchange.com/questions/4455071/valuation-ring-of-completion-of-a-field | # Valuation ring of completion of a field
I am actually quite confused. I have done an exercise that $$\mathbb{Z}_p$$ is a completion of $$\mathbb{Z}$$ w.r.t. the $$p$$-adic norm. Then again I got to know after reading somewhere that $$\mathbb{Z}_p$$ is also a completion of $$\mathbb{Z}_{(p)}$$. So is the completion of a valuation ring of field $$K$$ related to the valuation ring of the completion of $$K$$?
Another question I wanted to ask. Whether any complete field (where Cauchy sequences converge) is always a completion over another non complete field?
• The completion $\widehat{R}$ of a valuation ring $R$ at its valuation is a valuation ring. The valuation extends to the fraction field $Frac(R)$ and the valuation ring of $\widehat{Frac(R)}$ is the same as $\widehat{R}$. Note that it is not the same as the $\mathfrak{m}$-adic completion $\varprojlim R/\mathfrak{m}^n$. May 20 at 19:34
• @reuns So they are same.Now if I define the valuation ring of $(K,|\quad|)$ as the set $R=\{x\in K| |x|\leq1\}$ and that of completion $(\widehat{K},||\quad||)$ as the set $R^{\prime}=\{x\in K| ||x||\leq1\}$Then how can I show that $R^{\prime} = \widehat{R}$? Any hints or anything.I dont know actually what is m-adic completion. May 21 at 2:22 | 2022-08-17 13:54:16 | {"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": 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": 7, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7904530167579651, "perplexity": 140.94312371882344}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1659882572908.71/warc/CC-MAIN-20220817122626-20220817152626-00758.warc.gz"} |
https://chat.stackexchange.com/transcript/message/10151896 | 12:14 AM
@PauloCereda Have you ever played with MacPorts?
@PauloCereda Why am I even asking. You're too busy celebrating. ;-)
@AlanMunn Nope, only brew. My experience with MacPorts was so traumatic. :)
@AlanMunn Now I'm back, sorry. :)
@egreg: I can't believe Italy played at 1PM in Salvador. That's just cruel, very cruel.
12:30 AM
@PauloCereda No apologies needed. I'm trying to help this guy. tex.stackexchange.com/q/121257/2693 and it seems he has MacPorts which is screwing everything up. I'd just like to know how to get rid of it.
@AlanMunn Maybe which -a texhash might help us:
paulo@alexandria ~$which -a texhash /opt/texbin/texhash /usr/bin/texhash /bin/texhash We can see all occurrences of a command in the path and the order. 12:45 AM @PauloCereda Oh, that's useful. I didn't know the -a option of which. @AlanMunn Been there done that. It's a blessing when you are lost in path priority hell. :) @PauloCereda Once I got rid of Fink many years ago I have continued to be in /usr/local heaven. :) @AlanMunn Ah! :) @AlanMunn: success, apparently. :) Good job. :) @PauloCereda Yes, so it seems. The OP isn't very good at explaining things it seems or responding to questions, which is why the question ended up closed. 1:03 AM @AlanMunn Indeed. Sadly, some problems end users face are not TeX related, making them quite difficult to track. We need a lot of communication in this case. 2 hours later… 3:21 AM @AlanMunn So should it be left closed or reopened to answer properly? 3:51 AM @Qrrbrbirlbel I guess it's possible that someone else might have the same problem. There is a sense in which any non-MacTeX distribution always causes problems, so having an answer somewhere that says that might be helpful. 4:01 AM I've cast the last reopen vote and added an answer. 2 hours later… 6:04 AM @egreg what shall I do when my long exact sequence is to long for the page ? 1 hour later… 7:23 AM Hello, I'm trying to define syntax highlighting for XML. I have following snippet: \lstset{ basicstyle=\ttfamily, columns=fullflexible, showstringspaces=false, commentstyle=\color{gray}\upshape } \lstdefinelanguage{XML} { morecomment=[s]{<?}{?>}, morecomment=[s]{<!--}{-->}, morestring=[b]", morestring=[s]{>}{<}, commentstyle=\color{gray}, stringstyle=\color{black}, identifierstyle=\color{darkblue}, keywordstyle=\color{cyan}, morekeywords={xmlns,version,type},% list your attributes here } And it almost works. Just comments (<!-- something -->) are not highlighted in grey colour. Interestingly a string "<? something ?>" has grey colour. Do you have any idea what might be wrong? 7:35 AM I got no idea Does anybody experienced problem with tikzcd and adobe reader ? @MartinScharrer, @StefanKottwitz (@Stefan) Could one of you take a look at tex.stackexchange.com/admin/flags/close-as-off-topic and if you are happy approve the new off-topic text? Bullet points don't seem to work, so I've run the list from Andrew Swann's meta answer inline 8:15 AM @JosephWright Will the "too localized" reason be removed? 8:32 AM @DominicMichaelis You split it. ;-) @egreg Do you have an advice how? Not that I suggest something wrong by the splitting @DominicMichaelis multline, definitely; split at an arrow. 9:00 AM @Stefan Already gone 9:33 AM @egreg do you like writelatex.com/252294wsgmdm @DominicMichaelis No. ;-) You aren't really writing Kef and Cof, are you? Mac Lane does it :/ I am using$\ker$and$\coker \$ for the morphisms
@egreg writelatex.com/252309czwtdj this is the current version of my seminar work
Do you like it else? I just mean the layout
9:59 AM
@DominicMichaelis Not bad. However Kef and so on are simply wrong. Mac Lane or not. ;-) If you want italics, it's \operatorname{\mathit{Ke}}f; but I'd not use italics: "Ke" is an operator, after all.
ok that was what i felt right anyway
1 hour later…
11:30 AM
Has anyone applied the latest patches from TeXLive 2013? seems the inconsolata package has been broken
@ArTourter It's a pity that the new package doesn't have inconsolata.sty for backward compatibility; now the package to be loaded is called zi4 :(
also the otf font has disapeared
I have the vague impression that the package was meant to enhance the original one, not replace it
@egreg What the zi4 package seems to do in fact is insert Type3 font characters for the straight quotes. it is not a modification of the the original font.
it looks more like a hack than anything else
11:50 AM
@ArTourter Complain with the author! He's usually active in answering to users' needs.
New 'off topic- text is installed :-)
@JosephWright Yay!
@egreg I tried to subscribe to the tex-live mailing list and send a message there but although I am subscribed and receiving the messages, my messages don't seem to get anywhere.
@ArTourter The tex-live list is not really the right place for this. Michael Sharpe's email can be found in the README file for newtx: msharpe at ucsd dot edu
@ArTourter When you subscribe to a list, a couple of days can be needed for the first message to show up.
@egreg thanks for the info re the delay in the list.
@egreg well I originally thought the package was replaced by mistake, and that the zi4 should have been a separate one, hence my emailing the list.
I will email the maintainers as well
12:10 PM
12:47 PM
I was looking at my profile:
1 post marked offensive
I marked something offensive or someone marked a post of mine offensive?
12:59 PM
@PauloCereda Another party day for @AlanMunn
@egreg Indeed. :)
@PauloCereda Uh, oh, it happened again. @DavidCarlisle won't be happy.
@egreg quite.
@DavidCarlisle Don't you like the \switch macro?
1:16 PM
@egreg well it is lovely of course but I think actually I prefer my answer (what a surprise:-) semantically += is being used as on operator of that name so \mathrel{{+}{=}} seems closer to that (should be hidden behind a top level macro of course). \mathrel{+}= comes out the same thing but isn't as clear I think.
@DavidCarlisle One knows that = is a relation. ;-)
equivalence is an equivalence relation :D
@DominicMichaelis Sometimes. ;-)
1:40 PM
@egreg David has a bouncy castle in his backyard, no reason to be sad. :)
@PauloCereda I don't actually. The company had one for an afternoon but they sent it back:(
@DavidCarlisle Oh no!
@egreg woohoo!
@PauloCereda We only have (small) greens and a fountain.
1:47 PM
@egreg :)
1:59 PM
@PauloCereda Thanks for the Canada day wishes. :)
@AlanMunn My pleasure. :)
2:16 PM
@AlanMunn Uh! You're still able to put words into the right order. No party yet?
2:30 PM
@egreg I just got up. :) It's a little early for drinking. Although in today's e-mail comes a reason to drink independent of Canada day: an invited speaker to our workshop this weekend has just pulled out.
2:47 PM
@egreg : Hi, Enrico. Why do you consider æ to be wrong?
@egreg !!! Bunga-bunga !!!
@Brent: Hi Brent. :) I have some news for you later on. :)
Jan 15 at 21:25, by Paulo Cereda
@egreg: out of curiosity, I was wondering if all ae occurrences should be typeset as \ae, e.g, Ordinarium Missae -> Ordinarium Missæ
Jan 15 at 21:28, by egreg
@PauloCereda The æ ligature is not even medieval, but later. Never use it in Latin and don't trust anybody who tells you to use it.
:)
@PauloCereda :( I like being different!
@Brent.Longborough Me too. :)
@PauloCereda Are you one of those NSA people?
@PauloCereda Anyway, it's not Latin, it's Swedish
@AlanMunn NSA people don't qualify trust; they mistrust everyone
2:52 PM
@AlanMunn No, I'm clean. :) /I've been spotted! Where's that cyanide pill?
@Brent.Longborough It's a renaissance artifact.
They say that more people have died of vascular incidents caused by annoyance with 'Homeland Security' than by terrorists.
@Brent.Longborough I'll leave bunga-bunga to another one. ;-)
@Brent.Longborough Sorry for only being able to provide a horrible workaround for the DejaVu question.
@egreg LOL... Will he be seeing 'square sunrises' soon?
@Brent.Longborough I'm afraid he won't. He's aged, you know. But the chances he will be kicked out of Parliament are good.
2:57 PM
@Brent.Longborough Titãs has a song called "Vossa Excelência". It sums the general feeling very well. :)
@egreg No, that's fine. It was just for one of my boys' CV. Personally, I'd never think of using DejaVu, any more than you'd think of listening to, say, Stockhausen.
@Brent.Longborough The probability that I consider DejaVu are infinitely greater than me thinking to listening who? At least DejaVu has a rich character set.
3:14 PM
@egreg: Does the command xpatch have a verbose mode where I can see why patchings is failing?
I mean \xpatchcmd
3:31 PM
@egreg: There is a small typo in regexpatch. In the documentation the command checkifpatchable is explained, however the package defines only \checkpatchable (without if).
1
I'm looking for a reliable reference manager and PDF organizer in order to make my research work more efficient and productive. In particular, I would like the software to help me to export references into the Bibtex format, suggest me new papers on a specific topic (or according to the papers) a...
Would appreciate feeling on my comment
@JosephWright seems OK to me (the comment, not the question)
3:51 PM
@egreg Hmm, I hope that's Aleph-0 for DejaVu and Aleph-1 for Stockhausen...
4:17 PM
@MarcoDaniel Thanks
@Brent.Longborough No, it's 0.001 for DejaVu and 0 for who?
@MarcoDaniel You can use \tracingpatches (it uses etoolbox). With regexpatch there's \tracingxpatches that can be turned off with \tracingxpatches[0].
@JosephWright I agree, and just voted to close. But the only reason that showed up was the new one about code/typos. I'm confused.
@AlanMunn I think to get migration paths sorted out we have to ask again: I can migrate anywhere, but 'normal' users cannot. The close-changes seem to have killed the routes we used to have.
@JosephWright But also I thought we were supposed to have a "It's just off topic" and "It's about TeX, but its a typo/package bug/etc." reason to choose from.
@AlanMunn Yes, that's quite true. It seems that the default text only shows up when close reasons don't match, but you can't actively choose it. I suspect that's not the intention: will ask in the mod room.
@AlanMunn Perhaps I should ask on Meta.SO?
@AlanMunn Seems this is deliberate
@AlanMunn So I guess I'll add another custom reason that is just the default one!
4:54 PM
@JosephWright So we get three reasons, each of which we need to specify explicitly and there is no default?
@AlanMunn Erm, I'm not quite sure :-)
Was pointed to
25
Introduction Per-site off-topic reasons were created to reduce the amount of confusion and debate surrounding what specifically is on- or off-topic for a given site. Folks asking questions for the first time are often unclear on what is allowed or expected; by explaining common misconceptions in...
@AlanMunn You only get the default if the close reasons don't match between voters
Idea I think is that each close vote should have a reason, either on the pre-defined ones or a free text version
Thus the idea is that the reasons are 'tight'
We also probably need some new migration paths
I note in particular
> At a bare minimum, off-topic reasons should identify a specific topic considered inappropriate.
I'm beginning to think these custom reasons are bringing more pain than gain. :)
@PauloCereda Oh, you are telling me
@JosephWright But then we definitely need to make the default one of the 3 active ones, I think. It might be useful to have something about MathJax?
@JosephWright :)
4:59 PM
@PauloCereda As I've tried to indicate, getting in with the 'vision' is not so easy
@AlanMunn I did think of this, but am not sure how many MathJax questions we actually get
The other thing raised was that most sites have 'do not ask about' in their FAQ, which we don't
I think I'll need to write all of this up properly and ask yet-another meta question (or perhaps a couple: see point about migration)
As @AlanMunn is on the 'next mod' list, perhaps he can help out ;-)
@PauloCereda Making my life 'interesting', at least
@JosephWright Yay! :)
@JosephWright I see you're NSA plus CIA. You know everything about us and you provoke too.
@JosephWright You need to get an Xbox too. :)
@AlanMunn LOL
I suspect the problem is getting the idea: what the Powers want is good explanations to people who have questions closed. We've been good on that, in the main, in comments, so picking up the idea is trickier
@AlanMunn: pardon, you mean MI7. :)
5:04 PM
@PauloCereda People I know mainly have PS3s
@PauloCereda 7 is there a 7? it must be even more secret than the not very secret 5 and 6?
So do I, as I can get 4OD for free on the PS3, but not on the XBox
@JosephWright I do have one as well. :) Either one, buy it. :)
@PauloCereda Traded in Xbox for PS3 last week :-)
My first experience of consoles: is costing more than expected
@JosephWright Oh my! :)
5:06 PM
@PauloCereda MS wanted money to use TV-on-demand apps which are free on PS3. No contest.
@JosephWright Ah yes. :) That's why the end of the line for me is with my faithful Xbox 360. I won't get a One.
@PauloCereda Also my friends at work have PS3s, so the game recommendations are all inclined that way
@DavidCarlisle Oh, I meant MI6, but apparently there is a 7 as well! They say it's defunct, but maybe <ackbar>IT'S A TRAP!</ackbar>. :)
@JosephWright Yay! :)
Which games will you get? :)
@PauloCereda vim, is a good joke app I've seen
@DavidCarlisle Look ma, no arrow keys! :)
5:16 PM
"Look Ma, this man can twist his fingers as if they were made of rubber,
isn't that amazing? -- Not really, he's been using emacs for years...!"
@topskip ooh! :)
6:01 PM
@PauloCereda Then Heiko pops up with \pdfescape ;-)
@egreg That's totally unfair. :)
6:54 PM
0
The new approach to closing questions means that things are changing in the way we handle 'off-topic' questions. I've already asked about the generic text and for some predefined off-topic close reasons. However, I think I at least didn't quite see how this was going to work. So I'd like to ask a...
Sorry if I'm repeating myself, but the new close system is taking a little getting used to!
@AlanMunn (in particular) might want to restate what he's already contributed!
@JosephWright What, again? I'm getting too old for this. :)
@AlanMunn Yeah, I know!
@AlanMunn Imagine my view: as a mod, it's important I take an interest here, but I didn't decide to alter how closure works!
@AlanMunn I hope the reason I've asked again makes sense
@JosephWright Yes, I think so. But since we only have 3 active reasons and one should be 'This is not about TeX et al." then we're really down to 2 other reasons, right? (Unless we have good migration paths, in which case "Belongs on another site" might be useful.
7:11 PM
@AlanMunn The point about the 'this is not about TeX et al.' closure reason is that the Powers are suggesting that such a case should be 'custom reason' and more detail. For example, taking something current, 'Implementing the Knuth-Plass algorithm in other software does not require technical knowledge about TeX itself'
@AlanMunn Should I try to explain that in the meta question?
should we make a tree of possible paths for closing reasons? I think that's pretty confusing as it is.
@percusse :-)
@percusse I'm taking an interest precisely because I don't yet get the new system
I'm also not sure it will work!
@JosephWright Should we start a rebellion? @percusse and I can take care of that. :)
@JosephWright I think we are missing the frustration of the SO guys. None of them makes sense to us
@percusse Yes, true. They have 7k questions per day to keep on top of. It's quite clear that most other sites don't quite see the need to change closure in the same way.
7:18 PM
You would be burning with rage to really want to use those options.
@percusse: Joseph is a proud owner of a PS3. :)
@PauloCereda First of all you suck. I've flagged you and also your duck is wrong. There are many things wrong with arara too you should be ashamed of yourself. Read some programming books before you touch the computer !!
2
hehehehe this is what SO to me.
@percusse Wow. Wow. I'm gonna sit in the dark corner and cry. I'll see you guys in 10 years of heavy therapy. <3
2
@PauloCereda We really need to get this online gaming in action one day.
@percusse: And look, there's also a French punctuation! You bastard. :)
@percusse I know, it would be awesome. :)
7:20 PM
hahaha
Let's see the rage of a chemist killing people in CoD. :)
Déformation professionnelle is a French phrase, meaning a tendency to look at things from the point of view of one's own profession rather than from a broader perspective. It is often translated as "professional deformation" or "job conditioning". The implication is that professional training, and its related socialization, often result in a distortion of the way one views the world. As a term in psychology, it was likely coined by the Belgian sociologist Daniel Warnotte or Russian-American sociologist Pitirim Sorokin. Alexis Carrel: "Chaque savant, grâce à une déformation professionnel...
@percusse OMG
<3
@percusse You are asking to get that starred :-)
@PauloCereda Would you like my datatooltk.yaml file for arara?
7:22 PM
@PauloCereda It's not very exciting :-) but here it is:
!config
identifier: datatooltk
name: DatatoolTk
command: <arara> datatooltk --out "@{out}" @{options}
arguments:
- identifier: out
flag: <arara> @{parameters.out}
- identifier: options
flag: <arara> @{parameters.options}
@JosephWright :-)
@NicolaTalbot Thank you very much. :)
@AlanMunn I've edited into my meta question an example of how I think the 'free form' off-topic option is meant to work
Oh man come one who starred that? ...
4
7:24 PM
@percusse Guess who. :)
@PauloCereda :-)
@PauloCereda I've added a search function now :-)
@NicolaTalbot ooh! :)
Whoever starred it doesn't know the difference between a password and a padlock...
hehehe SO is really fun.
There is an answer starts with All the other answers are wrong :0)
@percusse Like Camelot, it is a silly place
7:26 PM
@percusse Of course I know! Padlock is that princess girl in the Star War movies, Lea and Luke's mother. :P
@PauloCereda That's not the right Pad. Stop regexping movies
@percusse LOL epic
@JosephWright I can imagine that IT business is equally challenging.
emphasis on the challenging. They have assgined me a career coach and I'm banned from using the word problem. Everything is a challenge now.
@JosephWright Fetchez la vache! :)
@percusse LOL
I said But it's a damn hard problem that we need to solve and got the reply I'm sure you can handle it
And I'm known for my surfer attitude
7:37 PM
I'd be happy to include an emacs section on how to integrate arara with TeX editors. :)
@PauloCereda ???
@MarcoDaniel AucTeX has a lot of commands, doesn't it? We could certainly add an arara wrapper.
@PauloCereda Or avoid emacs.
:-)
@MarcoDaniel My thoughts exactly, but we can't be that evil. :)
7:53 PM
0
I did an update to my LaTeX packages two days ago (I needed XeLaTeX for beamer, and this is why I did the update), then tried to continue writing my notes using LaTeX (MiKTeX2.9; Texniccenter2.0 Beta 1; SumatraPDF), and here came the surprise latex> (Fatal format file error; I'm stymied) I've...
Looks like the first case for 'really TL:=> delete'!
8:14 PM
@JosephWright Voted for "irreproducible problem"
@egreg Ah, good idea
8:29 PM
@JosephWright: I tried to alternatethe XeTeX driver to draw a frame with rounded corner in the foreground of the content but I without luck. I get a funny offset for the frame. However, maybe I addressed it the wrong way. Placing the content using \rlap{} first outside the driver code might be the correct way.
@MartinScharrer Yes: the L3 drivers use \rlap
@MartinScharrer Sounds fun!
9:14 PM
@MartinScharrer if in doubt when hacking back end code add rlap, never fails:-)
@DavidCarlisle Thanks ;-) So far it works well!
@JosephWright, @DavidCarlisle: My mistake was also that I thought to bundle the round clipping and frame together. Actually people might want to draw round frames around unclipped content, too.
9:59 PM
That's it. I can no longer earn more reputation. Perfection has been reached (at least until next year). i.imgur.com/UnMq5UF.png
@SeanAllred Are you really planning to earn 1 rep in one year?
@egreg That's the plan. I hope I can stick to it.
XD
@SeanAllred My plan is to get to 100000 per year. I'm still behind schedule at a mere 86647. :P
@egreg To each his own according to his talents. Although you're going to break 200k sometime soon --- is there a party planned? XD
@PauloCereda Another songs question for you.
10:09 PM
@AlanMunn I'll take a look. :)
@SeanAllred Be around here on July 14, more or less.
@egreg That is frightening growth. You grow a new me every two weeks?
@SeanAllred 8.48 days, on the average. ;-)
@AlanMunn: the answer is quite easy to answer, but I'm in a hurry now. I'll try to answer it later on. :(
@SeanAllred Perhaps we can party on July 13, according to the latest estimation: 11.5 days needed at the current average.
10:18 PM
@egreg What should be done? We should make a <obnoxiously ill-suited program> for it. It's Turing-complete, after all. Make a couple GUI libraries and we're all set.
Should take you what - about an hour? ;-)
I swear, some of the things you pull out of your hat around here!
@SeanAllred Unfortunately my only programming language is TeX.
@egreg You mean you are fortunate enough to know a Turing-complete language capable of everything any other language is. You could probably even rig up an internet browser of sorts (in theory). (Wouldn't that be interesting? Render HTML as TeX, with beamer-like links? That would be crazy.)
@SeanAllred Have you seen this question:
75
I think the questions says it all! :) By TeX, I mean any derivative of TeX as long as the code is clearly a derivative of TeX, i.e. that it could belong on TeX.SX (and not pure Lua for example). I'll give two answers myself so you see what I mean.
@AlanMunn Oh yes. Ooooh yes. I've also stared blankly at the Mars rover implementation in TeX done by some unbeknownst-to-me madman.
One of my favorite questions.
10:53 PM
Meh, I am lost (or blind): looks like "Too localized" is gone as a reason to close...
What's the best way to put text into a fixed size space? (I'm creating name badges).
@AlanMunn adjustbox (and max width option)
@Xavier Neither, but you clearly have been living on another planet. :) We've been having endless discussions on meta about the new close reasons.
@egreg Perfect. Thanks.
@AlanMunn Have been living on "relocation" planet for the last week or two... :)
Though I thought so, as I first search for "Too localized" on meta, but didn't immediately saw the discussion. Is there an explanation somewhere?
11:44 PM
@egreg Again:(
@SeanAllred well there exist html and xml parsers in TeX, and rendering and making links isn't so hard.
@DavidCarlisle This is looking pretty feasible, actually - a fun fall project ;)
I would definitely learn myself some real TeX-as-a-programming-language
@SeanAllred doing basic html isn't so hard (I had a go at html 3, which never was standardised actually, back then with typehtml package) and xmltex is a pretty complete xml parser , hard bit comes when people expect you to do something with a typical web page which is 90% javascript and css and only 10% html markup.
@DavidCarlisle Writing a JS emulator shouldn't be too bad; one just needs to develop on object paradigm for TeX (which I'm sure has been thought through thoroughly). (Edit: geez - say that five times fast!) | 2022-08-15 04:23:28 | {"extraction_info": {"found_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": 0.6095656156539917, "perplexity": 4517.715084705152}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1659882572127.33/warc/CC-MAIN-20220815024523-20220815054523-00337.warc.gz"} |
https://www.hepdata.net/record/ins1353541 | • Browse all
Study of W boson production in pPb collisions at $\sqrt{s_{\mathrm{NN}}} =$ 5.02 TeV
The collaboration
Phys.Lett. B750 (2015) 565-586, 2015
Abstract (data abstract)
CERN-LHC. The first study of W boson production in pPb collisions is presented, for bosons decaying to a muon or electron, and a neutrino. The measurements are based on a data sample corresponding to an integrated luminosity of 34.6 inverse nanobarns at a nucleon-nucleon centre-of-mass energy of $\sqrt{s_{\mathrm{NN}}}$ = 5.02 TeV, collected by the CMS experiment. The W boson differential cross sections, lepton charge asymmetry, and forward-backward asymmetries are measured for leptons of transverse momentum exceeding 25 GeV, and as a function of the lepton pseudorapidity in the $|\eta_{\textrm{lab}}| < 2.4$ range. Deviations from the expectations based on currently available parton distribution functions are observed, showing the need for including W boson data in nuclear parton distribution global fits.
• #### Table 1
Data from Figure 4
10.17182/hepdata.69232.v1/t1
Lepton charge asymmetry, $(N_{\ell}^+ - N_{\ell}^-)/(N_{\ell}^+ + N_{\ell}^-)$ as a function of the lepton pseudorapidity.
• #### Table 2
Data from Table 1 a
10.17182/hepdata.69232.v1/t2
Production cross section for $\textrm{pPb} \to W^+ + X \to \ell \nu + X$ for positively (top) and negatively (bottom)...
• #### Table 3
Data from Table 1 b
10.17182/hepdata.69232.v1/t3
Production cross section for $\textrm{pPb} \to W^- + X \to \ell \nu + X$ for positively (top) and negatively (bottom)...
• #### Table 4
Data from Figure 5 a
10.17182/hepdata.69232.v1/t4
Forward-backward asymmetry, $N_{\ell}(+\eta_{\textrm{lab}})/ N_{\ell}(-\eta_{\textrm{lab}})$, for the negative leptons.
• #### Table 5
Data from Figure 5 b
10.17182/hepdata.69232.v1/t5
Forward-backward asymmetry, $N_{\ell}(+\eta_{\textrm{lab}})/ N_{\ell}(-\eta_{\textrm{lab}})$, for the positive leptons.
• #### Table 6
Data from Figure 6
10.17182/hepdata.69232.v1/t6
Forward-backward asymmetry, $N_{\ell}(+\eta_{\textrm{lab}})/ N_{\ell}(-\eta_{\textrm{lab}})$, for the charged-summed W bosons, as a function of the lepton pseudorapidity. | 2018-11-20 19:43:10 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9070753455162048, "perplexity": 4464.5469243389725}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-47/segments/1542039746639.67/warc/CC-MAIN-20181120191321-20181120213321-00456.warc.gz"} |
http://www.vmj.ru/eng/archive/detail.php?ELEMENT_ID=4843&SECTION_ID=387 | ## Contacts
362027, RNO-A, Russia
Phone: (8672)50-18-06
E-mail: rio@smath.ru
# Cyclic subgroups of second degree full linear group over a field of the zero characteristic
Zhemuhova M. Z. , Pachev U. M.
Vladikavkaz Mathematical Journal 2011. Vol. 13. Issue 3.
Abstract:
A description of cyclic subgroups in a full linear group $$GL_2 (F)$$ over any zero characteristic field $$F$$ is given.
Keywords: differential equations, great high-frequency items, averaging method, asimptotic, substantion. | 2022-07-01 19:19:55 | {"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": 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": 0.4383828043937683, "perplexity": 6896.91009928809}, "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-2022-27/segments/1656103945490.54/warc/CC-MAIN-20220701185955-20220701215955-00459.warc.gz"} |
https://www.physicsforums.com/threads/how-do-i-finish-this-off.341134/ | # How do I finish this off?
1. Sep 28, 2009
### neutron star
I think I'm close but I don't know what the next step is.
1. The problem statement, all variables and given/known data
A ball is dropped off a cliff on a planet with an exact gravity of $$10m/s^2$$, 1.5 seconds later a lead ball is thrown straight down with an initial speed of $$20 m/s$$ If the two balls hit the base of the cliff at the same time, find the height of the cliff.
2. Relevant equations
$$X$$f=$$X$$o+$$V$$o$$t$$ -$$1/2gt^2$$
3. The attempt at a solution$$-d=-1/2gt^2$$
$$t^2=2d/g$$
$$t=\sqrt{2d/g}$$
$$-d=0-20(\sqrt{2d/g}-1.5)-1/2g(\sqrt{2d/g}-1.5)^2$$
Last edited: Sep 28, 2009
2. Sep 28, 2009
### Delphi51
Wow, it took me a while to see that your last expression is d = Vi*t + .5at^2 with t replaced by (t-1.5). So that is your distance for the second ball at time t from when the first ball is dropped. To finish the job, you equate that distance to a similar expression for the first ball's distance.
If I may offer a more general tip, it would be to begin with
ball 1 distance = ball 2 distance
.5*a*t^2 = Vi*(t-1.5) + .5a(t-1.5)^2
so you have the big picture clearly stated, and then it is just detail work to finish it.
Also, you might take "down" to be positive in this problem to reduce the number of troublesome minus signs. I see it isn't really a quadratic equation - the t^2 term cancels a step or two further on. | 2018-02-18 05:51:33 | {"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": 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": 0.6713164448738098, "perplexity": 497.9910574026875}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1518891811655.65/warc/CC-MAIN-20180218042652-20180218062652-00640.warc.gz"} |
https://www.gradesaver.com/textbooks/math/geometry/CLONE-68e52840-b25a-488c-a775-8f1d0bdf0669/chapter-3-section-3-1-congruent-triangles-exercises-page-128/3 | # Chapter 3 - Section 3.1 - Congruent Triangles - Exercises - Page 128: 3
m$\angle$A=72$^{\circ}$
#### Work Step by Step
as stated in question 2 that both the triangles are congurrent means two angles of both the triangles are equal. and according to both the triangles $\angle$A=$\angle$F and it is given in question 3 that m$\angle$F=72$^{\circ}$ therefore, m$\angle$A=72$^{\circ}$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 2022-05-27 13:17:06 | {"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": 0.4066472053527832, "perplexity": 1369.3043234716306}, "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-2022-21/segments/1652662647086.91/warc/CC-MAIN-20220527112418-20220527142418-00578.warc.gz"} |
https://electronics.stackexchange.com/questions/340741/calculate-battery-life | # Calculate battery life
I would like to know how to calculate battery life when using an LED strip. The LED strip is powered by 8 AA batteries in series. LED Current Draw = 260 mA. LED power consumption: 3.1 W. LED Operating Voltage range: 9-14.8 VDC. Please see attached images for battery datasheet.
This is my calculation, but I'm not sure if it's correct:
Battery Life = (Battery Capacity)/(Load Current) = 2000 mAh /260 mA = 7.7 Hours
This is for one battery, and the lights are powered by 8 batteries, so is the total battery life = (7.7)x(8) = 61.6 hours ?
• Thanks for doing your homework before asking. It is a well written question. – mkeith Nov 19 '17 at 22:06
• You should be more precise. How are the LED strip and the 8 batteries are connected? All batteries in parallel is not possible, 1.5 V is too low for a LED strip. If all 8 batteries are in series, the capacity is 2000 mAh for them, no multiplication with 8. With 8 cells in series, the voltage at start will be 12 V, but at finish of discharge only 8 * 0.8 V = 6.4 V. You should look in the datasheet of the LED strip for the minimum operating voltage. If this voltage is 10 V for example, one cell may be discharged only to 1.25 V and the capacity is smaller than 2000 mAh. – Uwe Nov 19 '17 at 22:08
• @Uwe Thank you for your answer. 8 AA batteries are in series and the minimum operating voltage is 9 VDC. – A.K Nov 19 '17 at 22:18
• If minimum voltage is 9 V DC, the cells may be discharged to 1.125 V. In the discharge diagramm not 10 hours at 250 mA, only about 6 hours. – Uwe Nov 19 '17 at 22:29
• We really need to have the datasheet for the LED strip, or at least the V-I curve (which you could measure with your bench top power supply.) We also need to know what kind of a converter you are using, and its efficiency (datasheet would be best, or measurements at least). With your partial question, our attempted answers are also partial, sorry to say. – MicroservicesOnDDD Jan 1 at 20:40
I would like to know how to calculate battery life when using LED strip. The LED strip is powered by 8 AA batteries. LED Current Draw = 260 mA.
Right. The batteries will be discharged at 260 mA. This is close enough to the 250 mA of the third bar on the capacity chart so we can assume 2000 mAh capacity.
This is my calculation, but I'm not sure if it's correct. Battery Life = (Battery Capacity)/(Load Current) = 2000 mAh /260 mA = 7.7 hours.
Correct.
This is for one battery, and the lights are powered by 8 batteries, so is the total battery life = (7.7)x(8) = 61.6 hours?
You were doing fine up until this. Here's the way it works:
• One cell gives 2000 mAh at 1.5 V.
• Two cells in parallel gives 4000 mAh at 1.5 V. (But this is not your configuration.)
• Two cells in series gives 2000 mAh at 3 V.
• Eight cells in series gives 2000 mAh at 12 V.
Run time = 2000mAh / 260mA = 7.7 h.
We can cross-check this with the energy capacity which is given by V x mAh.
• Each cell has an energy capacity of 1.5 x 2000m = 3000 mWh.
• Eight cells have a capacity of 8 x 3000 = 24000 mWh = 24 Wh.
• Using your figure of 3.1 W the run time is 24 / 3.1 = 7.7 h.
The result is 7.7 h in both cases.
• For including the cross-check... Bravo! I love seeing this, as it increases my confidence in the answer, as well as reminding me to do so on my own, as taught in school. Excellent! – MicroservicesOnDDD Jan 1 at 19:35
• For a converter with 80% efficiency, that 7.7 hours gets reduced to 7.7 * 0.8 = 6.16 hours, and a linear solution at 40% efficiency would be much worse at only 3.08 hours. Unfortunately, we are not told the efficiency of the converter, but I leave this for those who aren't thinking about the "efficiency factor". For battery powered applications, efficiency can matter a lot! – MicroservicesOnDDD Jan 1 at 20:05
Calculating battery life is pretty simple, just look at the units. Battery capacity is rated in mAh so 2000mAh can provide 2000mA for 1 hour, or 1mA for 2000 hours, etc.
Battery Life [hours] = Capacity [mAh] / Average Current [mA]
For example
Capacity [mAh] / Average Current [mA] = Battery Life [hours]
2000 mAh / 260 mA = 7.7 hours
You can pretty much ignore the Wh rating on the battery. It's like the mAh rating but it accounts for the fact that the battery voltage isn't constant over the battery's life.
As for series and parallel:
| Capacity | Energy | Voltage | Max Discharge Current
| mAh | Wh | V | mA
----------------------------------------------------------------
For example, using 2000mAh AA batteries:
| Capacity | Energy | Voltage | Max Discharge Current
| mAh | Wh | V | mA
----------------------------------------------------------------
Single | 2000 | 3000 | 1.5V | 500
2xSeries | 2000 | 6000 | 3V | 500
2xParallel | 4000 | 6000 | 1.5V | 1000
Note from the datasheet though that the expected capacity changes based on the discharge rate. You get 3000mAh at 25mA but only 1500mAh at 500mA.
For your situation, you would need 6-8 batteries in series to reach 9-12V which would give you 2000mAh or 7.7 hours. | 2021-04-17 05:59:07 | {"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": 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": 0.3506825864315033, "perplexity": 1732.5806909131104}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1618038101485.44/warc/CC-MAIN-20210417041730-20210417071730-00467.warc.gz"} |
http://www.mathfireworks.com/2015/07/michigans-legislature-and-math-illiteracy/ | # Michigan's Legislature and Math Illiteracy
Maybe you have seen this video. I stumbled upon it while looking at Dylan Kane's blog and found it interesting for a few reasons.
In the video, Deborah Ball is presenting to Michigan's House and Senate Education Committees -- a group of about 15 legislators. She is making the point that teachers need to know more than just the content they teach. She uses the example of whole-number multiplication to show that there is more to teaching than just knowing the content.
She explains that content knowledge just lets you identify if a student is right or wrong. It does not help you diagnose their error or inform how to help them correct their work.[3]One of my graduate school professors shared research on what math teachers need to know. Here, Ball is pointing out that teachers need more than just content knowledge. Another thing teachers need is pedagogical knowledge -- knowing what general practices are best regardless of content.
The research introduced a third type know as pedagogical content knowledge -- knowing the best ways in which to present specific content or support students in learning specific content. As Ball illustrates in this video, this is an area of knowledge often overlooked by non-teachers.
She analogizes this to a doctor who can only identify if a patient is sick but cannot treat them.
As a teacher, it is easy to watch this video and feel good about teaching. Ball does a wonderful job of illustrating the specific skills that teaching simple elementary math requires. However, I am much more interested in two moments that occur in the video. The first occurs after Ball asks the legislators to calculate $49\times25$. One of the legislators asks, "Can I pull out my phone?" Then a few others laugh.
Little moments and comments like these are so incredibly common (I have started posting the ones I find here). On its own, this joke is mostly harmless. I could even see myself making the same joke in a similar situation. But, once you become attuned to hearing and noticing these little statements of math illiteracy, you start to see how they could collectively shape culture.
Here is another one less than two minutes later:
"Ummm, Mr. Chair, I wasn't aware that math was going to be involved." This statement is followed by a few chuckles. The legislators are being asked to think about multiplying whole numbers; and, as Ball says, they all reduce to "jelly." The message from these powerful adults is clear "Math is intimidating and something you want to avoid doing."
I am not overly concerned about the Michigan legislature's math background. I am confident that all involved could answer these questions with more time and less pressure. What concerns me is how normalized being intimidated by math has become everywhere. Everyday, I hear someone say, in some form or another, that they are bad at math or intimidated by math. The implied conclusion being that there is no way they will or could improve.
When students hear these statements coming from elected officials, celebrities, parents, and teachers, it negatively impacts their opinions about a subject that so many already struggle with. Giving up on understanding math really has become the norm in this country and it is painful to encounter reminders of that every day. | 2019-09-22 04:16:17 | {"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": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.28123441338539124, "perplexity": 1448.3975477488343}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514575076.30/warc/CC-MAIN-20190922032904-20190922054904-00103.warc.gz"} |
http://codeforces.com/blog/entry/106837 | ### Aris's blog
By Aris, history, 3 weeks ago, translation,
Hello! Codeforces Round #820 (Div. 3) will start at Sep/12/2022 17:35 (Moscow time). You will be offered 6-8 problems with expected difficulties to compose an interesting competition for participants with ratings up to 1600. However, all of you who wish to take part and have a rating of 1600 or higher, can register for the round unofficially.
The round will be hosted by rules of educational rounds (extended ICPC). Thus, solutions will be judged on preliminary tests during the round, and after the round, it will be a 12-hour phase of open hacks.
You will be given 6-8 problems and 2 hours and 15 minutes to solve them. One of the problems in this round is interactive. Don't forget to read the guide on interactive problems.
Note that the penalty for the wrong submission in this round is 10 minutes.
Remember that only the trusted participants of the third division will be included in the official standings table. As it is written by link, this is a compulsory measure for combating unsporting behavior. To qualify as a trusted participant of the third division, you must:
• take part in at least five rated rounds (and solve at least one problem in each of them)
• do not have a point of 1900 or higher in the rating.
Regardless of whether you are a trusted participant of the third division or not, if your rating is less than 1600, then the round will be rated for you.
Thanks to MikeMirzayanov for the platform, help with ideas for problems and for coordination of our work. Problems have been created and written by ITMO University team: MikeMirzayanov, myav, Gol_D, Aris, Gornak40, SixtyWithoutExam and Vladosiya.
We would like to thank: Kniaz, BledDest, Svyat, Be_dos, Timur2006, FelixDzerzhinsky, alphabet321, FedShat, vsinitsynav, Jostic11, HamletPetrosyan, _Roma_, KKT_89, ABalobanov, lightseba, powergee101, DafuQ_o and AshrafEzz for testing the contest and valuable feedback. List of testers will be updated.
Good luck!
UPD 1:
We are in the middle of the round (and missing Gornak40 and Gol_D).
UPD 2: Editorial
• +301
» 3 weeks ago, # | 0 Auto comment: topic has been updated by Aris (previous revision, new revision, compare).
» 3 weeks ago, # | +5 Good luck for everyone . Do your best to be the best !!
» 3 weeks ago, # | 0 Good Luck everyone! I hope everyone has fun!❤️ Divs 3 are sometimes moody.
» 3 weeks ago, # | +10 Its the 1729th codeforces round! The Hardy-Ramanujan number which is very well known in Mathematics
• » » 3 weeks ago, # ^ | +18 Nice!
• » » 3 weeks ago, # ^ | 0 Great!
» 3 weeks ago, # | -8 For those who are not familiar with interactive problems. This might help you! Link
» 3 weeks ago, # | +14 Finally, my first out of comp-wait this is div.3, not div.4
» 3 weeks ago, # | +28 As a tester, give me a contribution, please
• » » 3 weeks ago, # ^ | -14 As a participant, I will downvote. Spoiler
• » » » 3 weeks ago, # ^ | 0 Repeating a funny thing, makes it unfunny
• » » » » 3 weeks ago, # ^ | ← Rev. 2 → 0 then can you help me in being funny? seriously I think I need help
• » » » » » 3 weeks ago, # ^ | ← Rev. 3 → +50 Ya, I would like to, do not comment!Things will automatically be funny :)
• » » » 3 weeks ago, # ^ | +19 As a tester, I will upvote. Spoiler
• » » » » 3 weeks ago, # ^ | +11 Finally found a kind-hearted man whose name is FelixDzerzhinsky! Thanks for such kindness (❁´◡`❁)
• » » » 3 weeks ago, # ^ | 0 :D HAHA
» 3 weeks ago, # | +3 Good luck to everyone and I hope the problems will be enjoyable!
» 3 weeks ago, # | +7 First unrated DIV3 ❤️, hope to solve 6 or more problems this time!
• » » 3 weeks ago, # ^ | 0 i wish i could be expert!
• » » » 3 weeks ago, # ^ | +1 Surely you will reach Expert one day!Don't lose the hope, grind yourself and love Competitive Programming ❤️
• » » » » 3 weeks ago, # ^ | 0 thank you brothers...take love
• » » 3 weeks ago, # ^ | 0 why unrated ?
• » » » 3 weeks ago, # ^ | 0 It is unrated for those with rating >= 1600
• » » » » 3 weeks ago, # ^ | 0 ohh
• » » » » 3 weeks ago, # ^ | 0 it means it is unreated for me ?
• » » » » » 3 weeks ago, # ^ | 0 No it's not. The ratings will be updated soon.
• » » » » » » 3 weeks ago, # ^ | 0 its currently in the unrated contest list in my profile so will it be updated soon and moved to rated contest ? Ps. This is my first contest so technically 0 rated as of now
• » » » » » » » 3 weeks ago, # ^ | 0 Yes unless there is a separate post about the contest being unrated. Check this post where the contest was unrated.Normally, a contest takes around 24 hours to update the rating. So hopefully, the ratings will be distributed within a few hours.
» 3 weeks ago, # | +4 Hope to be Expert this round!
• » » 3 weeks ago, # ^ | ← Rev. 2 → 0 Hope to be pupil again
» 3 weeks ago, # | +1 Codeforces Language Picker -- chrome extension to fix codeforces language picker.
» 3 weeks ago, # | 0 Hopefully I'm gonna be green this contest.
• » » 3 weeks ago, # ^ | ← Rev. 2 → -10 .
• » » 3 weeks ago, # ^ | 0 Guy's name tho
» 3 weeks ago, # | 0 Div3 Ftw
» 3 weeks ago, # | ← Rev. 3 → 0 I hope that I will be able to solve A!! spoilerIt is really a very very big challenge for me!!!
» 3 weeks ago, # | 0 Auto comment: topic has been updated by Aris (previous revision, new revision, compare).
» 3 weeks ago, # | +2 As a tester, I hope you enjoy the problem set! Good luck and have fun!
» 3 weeks ago, # | -14 Hoping to get a positive delta!! Give me some upvotes. I'm sad! :(
• » » 3 weeks ago, # ^ | 0 okay, i give you up vote..Good Luck for upcoming contest!!!
» 3 weeks ago, # | 0 now the time of div 3 bruuh
• » » 3 weeks ago, # ^ | 0 Hello cheater,let's see if your solutions again gets skipped or not.
» 3 weeks ago, # | -11 Div. 2.5
• » » 3 weeks ago, # ^ | ← Rev. 2 → -29 The comment removed because of Codeforces rules violation
• » » » 3 weeks ago, # ^ | +40 Please don't talk about contests before they are finished.
• » » » » 3 weeks ago, # ^ | +9 oh sorry but i didn't tell anything consider a spoiler + easiness is a mysterious ( sorry my english is not good) term maybe its just easy for me right? + i see people say mathforces speedforcess etc.. in other rounds why this comment consider deferent from them
• » » » » » 3 weeks ago, # ^ | +17 In my opinion ANY comment referencing the contest during it shouldn't be allowed, I'm not sure to what extent the codeforces staff want/can block this type of unfair comments, but in any case saying a problem is easy is a bit worse than saying all problems are easy (although I agree they should be removed as well)
• » » » » » » 3 weeks ago, # ^ | +19 yeah you are right i won't write anything during the contest again
» 3 weeks ago, # | -47 i am not simp but girl in middle look very cute. if anyone know her username
• » » 3 weeks ago, # ^ | ← Rev. 2 → +59 Whenever someone starts a line with, "i am not simp", there is a high chance that they are not simp. Seems legit.
• » » 3 weeks ago, # ^ | +32 I know...
• » » 3 weeks ago, # ^ | +18 no but her insta is @zachscud
• » » » 3 weeks ago, # ^ | 0 what a hypocrisy people got her id because of me and still down vote me and one one who release her id got upvote too many simps here
• » » » » 3 weeks ago, # ^ | +5 that's not her id lmao
• » » » 3 weeks ago, # ^ | 0 This is codeforce not site where you get customer so don't promote your business here
» 3 weeks ago, # | ← Rev. 2 → +1 D is much easier than C.
• » » 3 weeks ago, # ^ | 0 No it is not. Like C took me 7 minutes at most, D took 15-20.
• » » » 3 weeks ago, # ^ | +2 C took 20 minutes to understand the statement, and another 7 to implement it.
• » » » » 3 weeks ago, # ^ | 0 it took me 20 min to solve D and 20 min to solve C but another 30 min to find the mistake in my C
• » » » 3 weeks ago, # ^ | 0 A took 10 min and B took 40 min
• » » 3 weeks ago, # ^ | 0 How did you come up with intuition that only 2 person will go together to have maximal groups and not even a group of 3 will give us the maximal answer?Is it some common idea or what?
• » » » 3 weeks ago, # ^ | ← Rev. 3 → 0 If 3 people can form a group, then there must be 2 people in that group that can form a group. As we only want to maximize the number of groups, if 2 people can already form a group, why add more members to the group? I think basic observation is enough the come up with this idea.
» 3 weeks ago, # | 0 any idea for problem no E?? binary search is one solution but it requires 60 quarries.
• » » 3 weeks ago, # ^ | 0 you can just ask 25 pairs (a, b) and (b, a), and you are able to print the answer once you've got different paths for an arbitrary pair. The probability of getting same path for each pair is 2^(-25).
• » » 3 weeks ago, # ^ | +1 u can just try every pair {1, i} and {i, 1} for i from 2 to 27. In each two query ({1, i} and {i, 1}) u have a 50% chance that u get distance in different directions. If u do, then the answer is ans1 + ans2, where ans1 and ans2 is results for queries. Also, if u get -1 at some point, it means, that the previous i was the greatest vertex in graph, so it should be the answer. So print i — 1 and return
• » » » 3 weeks ago, # ^ | 0 Instead we can just take the pair (1,2) and query it 50 times. There is a probability of (1-2^-50) that we will get two different numbers. Just summing them ( with a +1 ofcourse) will give the result.
• » » » » 3 weeks ago, # ^ | +11 No, I too missed that part in the statement at first: after an answer is determined for a specific query, it wont change, so it will keep outputting the already computed value
• » » » 3 weeks ago, # ^ | ← Rev. 2 → 0 Well, there is a possibility that all pairs are equal, right?! So why is the solution correct!! Could he give a wrong answer in my opinion.
• » » » » 3 weeks ago, # ^ | +3 yes, but that possibility is really really low, so it will almost certainly pass (and if it miraculously doesn't, submit a second time)
• » » » » » 3 weeks ago, # ^ | 0 Thx
• » » » » 3 weeks ago, # ^ | +1 Yes, its possible, but the probability of it happening is equal to 1/2^25, which is veeeery small, so u can hope that it wont happen)
• » » » » » 3 weeks ago, # ^ | 0 Please, can you explain where did the number come from(1/2^25)?
• » » » » » » 3 weeks ago, # ^ | +2 there are 25 pairs, the probability of each of them to equalize is 1/2 and also the probabilities are unrelated, so the probability for all of them to fail at the same time is (1/2)^25
• » » » » » » » 3 weeks ago, # ^ | 0 Moreover, if you want your solution to fail then the average number of times you need to submit to get first WA is 2^25 ~ 10^7.5
• » » » » » » » » 3 weeks ago, # ^ | 0 Actually a bit less, remember there are multiple test cases. But in general yes, that's what I said, the solution won't fail
• » » » 3 weeks ago, # ^ | +1 Gosh, I thought about that, but I was like "hey, this won't pass for whatever test case, there must be another way"...
» 3 weeks ago, # | 0 Was E ternary search?
• » » 3 weeks ago, # ^ | 0 I think no. Try basic ternary give me WA TC 1x
• » » 3 weeks ago, # ^ | 0 I try every pair {1, i} and {i, 1} for i from 2 to 27 ^^
• » » 3 weeks ago, # ^ | +1 I thought so for a while but no, it's much simpler.
» 3 weeks ago, # | +2 About problem E: what is the probability that out of 25 pairs of queries at least one pair of query will give distinct answers, e.g. query(a, b) != query(b, a)?
• » » 3 weeks ago, # ^ | ← Rev. 2 → 0 Roughly 1-2^-25 when n is big enough.
• » » » 3 weeks ago, # ^ | ← Rev. 2 → 0 It is said that it may be equal but how you are so sure that it will be equal with 25 queries
• » » » » 3 weeks ago, # ^ | 0 The interactor chooses one of the two paths with equal probability.
• » » » 3 weeks ago, # ^ | +2 But the probability of WA is not completely 0. Is there any deterministic solution?
• » » » » 3 weeks ago, # ^ | 0 Most likely not. It's the same as the Polynomial Hashes, there is a very small chance of failure, so nothing wrong probably
• » » » » » 3 weeks ago, # ^ | +6 I thought that codeforces round always has some deterministic solution.
• » » » » » » 3 weeks ago, # ^ | +1 Well, again, Polynomial Hashes are not deterministic, yet still it's a popular way to solve problems on strings. There is nothing wrong about it. You just need to make sure the chance of failing is very small
• » » » » » » » 3 weeks ago, # ^ | 0 I get it but usually there is always a deterministic solution. Polynomial hashes has deterministic alternatives.
• » » » » » » » » 3 weeks ago, # ^ | +4 Not always
• » » » » » » » » 3 weeks ago, # ^ | +4 "usually there is always a deterministic solution" well, there IS, it just needs more (than 50) queries. People often confuse a problem with a problem statement. You should not consider a problem statement in isolation, rather all of statement, input format, input constraints as a package. For example, for any NP Hard problem, we all know that there are no polynomial solutions. That does not mean they are not solvable. You can set small constraints and expect an exponential solution with some optimization here and there. In other words, this is the difference between math olympiad and programming contest.I hope that satisfies your itchy feelings about not having a 100% guaranteed solution. This problem was designed and the constraint was set specifically for a probabilistic solution. If more than 60 queries allowed, one can find an exact solution using binary search.
• » » » 3 weeks ago, # ^ | 0 I see, thank you.
• » » 3 weeks ago, # ^ | 0 same bro how you can even guess that number .
» 3 weeks ago, # | +5 I think that problem E is more suitable for div 2,it's really hard !!
• » » 3 weeks ago, # ^ | +4 No, the idea, especially with No hacks, 50 test cases, really says it uses random. Also, as there is no binary search, you can only rely on distances, but they could all be simply distances on a part of a graph, so you cannot guarantee the answer is correct. The only way is to minimise the chance of fail
• » » 3 weeks ago, # ^ | +36 We rarely have "easy" problems with randomized solutions. This is a good way to introduce randomized solutions to beginners
• » » » 3 weeks ago, # ^ | 0 Yup. First time I have ever solved a question with randomized solutions
» 3 weeks ago, # | ← Rev. 2 → -29 Most unbalanced round ever, downvote.
• » » 3 weeks ago, # ^ | 0 I guess you've never heard of mr Green Grape
» 3 weeks ago, # | 0 Can anyone help me with my solution for Problem E? why is it not working and taking more iterations than 50? although my solution should run within log(1e18) < 50 (correct me if i m wrong here).submission
• » » 3 weeks ago, # ^ | +4 log(1e18) > 50
• » » » 3 weeks ago, # ^ | +6 Oops! I calculated log with base e (that is ~ 42). that is why i got confused.thanks mate
• » » » » 3 weeks ago, # ^ | 0 I calculated the log with base 2 LOL and used binary search
» 3 weeks ago, # | +5 cool E
• » » 3 weeks ago, # ^ | 0 Greattt
» 3 weeks ago, # | ← Rev. 2 → 0 In question E how you can be so sure that ans will be find by just finding it upto 25
• » » 3 weeks ago, # ^ | 0 The chance of failure is 2^(-25), which is very small
» 3 weeks ago, # | +1 According to some cursory analysis, my solution for E seems to have a 99.391488785% chance of passing a set of 100 test cases. Don't think this is the intended solution.
• » » 3 weeks ago, # ^ | 0 i guess it is (i did the same btw), because hacks are disabled and there are 50 tests only. Moreover, the problemsetters did even tell us that amount of tests is 50, which is kinda strange.
• » » » 3 weeks ago, # ^ | +15 The problem emphasized "with equal probability" in bold letters, and also made sure to state that "? a b" will not necessarily yield the same result as "? b a". They also mentioned the number of tests in the jury (which I've never seen mentioned before in the problem). So yes, I am pretty sure the intended solution is to try "? a b" and "? b a" until you get something different. You do need to make sure you try distinct pairs each time, but there are easy ways of doing that (including ways that guarantee the correct answer if the result is ever -1). My submission: 171950029
» 3 weeks ago, # | +13 Great round <3 <3 <3
» 3 weeks ago, # | 0 How to solve C?
• » » 3 weeks ago, # ^ | 0 Find all the characters in string that lie between letters s[0] and s[n — 1] and use all of them for jumps.
• » » » 3 weeks ago, # ^ | ← Rev. 2 → 0 sorry can you elaborate further how this achieve the minimum cost ps : sorry i meant the maximum jump in this case for example: abchhhhhhhhhhhhhhd shouldn't your solution fail?
• » » » » 3 weeks ago, # ^ | 0 abchhhhhhhhd it will be as a b c d (b-a)+(c-b)+(d-c)=4 and also as d-a=4
• » » » » » 3 weeks ago, # ^ | 0 oh you are right i was thinking in a scenario where taking a character which is not between s[0] and s[n — 1] can lead to minimum cost too but this example doesn't satisfy it thanks alot
» 3 weeks ago, # | 0 How solve E
• » » 3 weeks ago, # ^ | 0 I try every pair {1, i} and {i, 1} for i from 2 to 27 ^^
• » » » 3 weeks ago, # ^ | 0 why till 27?? can you please explain?
• » » » » 3 weeks ago, # ^ | 0 Because there is a total of 50 queries, and for each vertex you need to use 2 queries
• » » » » » 3 weeks ago, # ^ | 0 Thank you.
• » » » » » 3 weeks ago, # ^ | 0 Hi Noobish_Monk. I have a doubt. If I print all my queries as 1 and 2 for 50 times, then if I get 2 different sets in my answer then it's just the sum . What's wrong with this approach. And the probability that I always get the same answer is 1/power(2,50) which should be approximately 0. Why do we need to print for 27 queries ??
• » » » » » » 3 weeks ago, # ^ | 0 The answer for the same query is the same However queries ? a b and ? b a are different So you can't just ask 1 2 and 2 1
• » » » » » » » 3 weeks ago, # ^ | +3 I should have read that statement :( . Thanks mate :)
• » » » » 3 weeks ago, # ^ | 0 You can also do it from 4 to 29, because the problem states that n ≥ 3.
• » » » » » 3 weeks ago, # ^ | 0 Thank you.
• » » » » 3 weeks ago, # ^ | 0 Cause we have 50 queries bro
» 3 weeks ago, # | +1 Solution for E was probably meant to be different, But here is how I AC it: IdeaI basically just checked 1-[2, 26] vertices in a format:? 1 v? v 1And if the results differ, then the answer is their sum. If you got -1, it means the last valid run was the previous one, so you can output v-1. It worked for some reasons, probably not intended.
• » » 3 weeks ago, # ^ | 0 same thing for me too!I never thought it would work but some how it worked
• » » 3 weeks ago, # ^ | 0 I share the same idea with you ^^
• » » 3 weeks ago, # ^ | +3 IT'S IS THE INTENDED SOLUTION
• » » » 3 weeks ago, # ^ | 0 Can you explain how it is the intended solution like very first thing that comes to mind is Binary search but it didn't work. I also tried the concept of binary search on infinite array but it failed at test 18.
• » » » » 3 weeks ago, # ^ | 0 Because of how many times the statement highlighted the idea of equal probability i think this is the intended solution.
• » » » » 3 weeks ago, # ^ | 0 There is no deterministic solution that guarantees 100% correctness. Suppose I was an adversary that doesn't have a specific graph in mind, or even a value of $n$, but I make answers to screw you over while always ensuring that there exists some graph that is consistent with all my answers. Then I can always ensure that each answer I give will be either -1 or some distance that is below the established lower bound from the program. With 50 queries, you can only gain enough information to distinguish between $2^{50}$ possible answers, However, the range for $n$ goes up to $10^{18}$ (which is much larger than $2^{50}$), so you cannot guarantee a correct solution.The problem dropped a lot of hints suggesting the randomized approach ("equal probability" in bold, clarifying that "? a b" and "? b a" are processed independently, specifying the number of jury tests, etc), so I'm quite sure this is the intended solution.
• » » » » » 3 weeks ago, # ^ | 0 If you want your solution to fail then the average number of times you need to submit to get first WA is 2^25 ~ 10^7.5
• » » » » » » 3 weeks ago, # ^ | 0 Why can't we just use pair 1,2 and 2,1 25 times ? What is the probability that use these 2 pairs wont give me the answer in 25 queries ?
• » » » » » » » 3 weeks ago, # ^ | 0 It is stated that the same pair will give the same result for all requests.
• » » » » » » » 3 weeks ago, # ^ | +1 The problem states that repeating the same query will yield the exact same result every time. There is only a 1/2 chance that "? 1 2" and "? 2 1" will yield different results. If they both produce the same distance, then you will always get the same result for "? 1 2" or "? 2 1", so your probability of success will remain as 1/2. This is only for one jury test, so the probability of getting all 50 jury tests would be $\left(\frac{1}{2}\right)^{50} \approx 0.00000000000008817842\%$, which is not worth trying. But by trying different pairs each time, there is a 1/2 probability of success for each distinct pair. The probability that all 25 pairs fail is $\left(\frac{1}{2}\right)^{25}$, so the probability of being successful at least once is $1 - \left(\frac{1}{2}\right)^{25}$. A single success is enough to guarantee the correct answer. We need to pass 50 jury tests, so the probability of ACCEPTED verdict becomes $\left(1 - \left(\frac{1}{2}\right)^{25}\right)^{50} \approx 99.999850988\%$. It would be extremely unlikely for such a submission to fail, and I would be very interested to see if there is anybody who met this unfortunate fate.
• » » » » » » » » 3 weeks ago, # ^ | 0 Thanks for your explanation. This is helpful.I read this statement of same result produced for same query but didn't realise how important this was.Even after reading this, somehow I thought that using 1,2 & 2,1 still is random. But it seems like that the interactor is caching the result of the query.Similar to these lines, IMAGINE if 1,2 & 2,1 produced different results with 1/2 prob but they can produce different result every time, then we can just query 1,2 & 2,1 50 times right ?
• » » » » » » » » » 3 weeks ago, # ^ | 0 If duplicate queries were rolled independently, then yes, you can just query "? 1 2" 50 times. You can mix in "? 2 1" as well, but it is completely identical to "? 1 2" in this case, so it doesn't matter. In this scenario, the probability of failure becomes $\left(\frac{1}{2}\right)^{49}$ (all 49 queries after the first one match with the first query). Probability of AC becomes much higher then (though it was already high to begin with), and the implementation becomes much easier too (though it was already easy to begin with).
• » » » » » » » 3 weeks ago, # ^ | 0 Even without luck being involved, it can very well be the case that 1 and 2 are on opposite side of the cycle so longer and shorter both lengths are INDEED equal. So you must take different pairs to keep a backdoor in case your queried vertices fall on opposite sides.
• » » » » » » » » 3 weeks ago, # ^ | 0 Ah, you're right; I forgot about that case. We can deal with this by simply performing 25 queries of "? 1 2" and 25 queries of "? 1 3". This slightly hurts the probability of success, but it's still very high.
• » » » » » » » » » 3 weeks ago, # ^ | 0 25 queries of "? 1 2" gives the same answer each time. So you are really just wasting 24 queries in this case. Same goes for "? 1 3".
• » » » » » » » » » 3 weeks ago, # ^ | ← Rev. 2 → 0 OneBit1024 This was part of a discussion on the hypothetical scenario that each query is rolled independently, even if the same query was made in the past. In such a scenario, it would be optimal to repeat the same query (e.g., "? 1 2") 50 times instead of trying to find new pairs. If all 50 queries yield the same result, then the guess should be on twice this value (since it's likely that both paths have the same length).Obviously, for the original problem E, repeating the same query would be a waste, so you have to keep trying distinct pairs (so the possibility that both paths have the same length would not be an issue at all).
• » » » » » » » » » 3 weeks ago, # ^ | +1 Oh ok, sorry, I didn't realize that you were discussing a hypothetical scenario. Well in this scenario, this approach will work. actually, during contest I tried the same thing because I missed the part where (a,b) would give the same result each time.
• » » » » » 3 weeks ago, # ^ | 0 "Suppose I was an adversary" I don't need to read anything else. The solution will fail against an adversary. That's the whole point, that it was not an adversary, and it will honestly return any of the shorter or longer path from a to be with equal probability, without ever trying to make you fail. The moment an adversary comes into play, it will need binary search i.e. 60 queries.
• » » » » » » 3 weeks ago, # ^ | 0 The fact that an adversary will always be able to screw you over is definite proof that there does not exist a deterministic solution that guarantees correctness, which further suggests that the randomized approach is, indeed, the intended solution. Some people were unsure if that was the case, but I don't actually think there should be any doubt.
» 3 weeks ago, # | 0 Can anybody explain how they came up with the number 25 that checking this many pairs will have a different result?? really didn't get this part.
• » » 3 weeks ago, # ^ | 0 It's probability theory. We have a chance of success 1/2, same for failure. As we ask 25 times, we have a chance of failure equal to 2^(-25), which is a VERY small number
• » » 3 weeks ago, # ^ | +3 Since we have at most 50 queries, we can check 25 pairs in both ways. Assume we got a pair (a, b). Checking it both ways means that you chack a pair (a, b) and (b, a) since it can give different results. Since the output path length is randomized between the 2 paths(the short and the long versions, since it's a cycle), it's 50% chance for each to output.Asking 25 queries in this format and checking for -1 or different results for the (a, b) & (b, a) query. If you get -1, it means the last vertice(call it v; 2<=v<=26) is the correct answer, since the current one is out of bounds and the last one wasn't. So we output the last valid run. If the results differ, you can output sum of the results, since you got both paths for (1, v) pair and their sum is a complete cycle.This one failing has quite a low chance, since it's 1/2 for each path to be given for a query, it's 2^-25% chance of this idea failing, if I am not mistaken.I hope this makes sense.
• » » 3 weeks ago, # ^ | 0 You're allowed 50 queries. You make a pair of queries "? a b" and "? b a", and hope that you get different results for these two. If the results are different, then you can add the two results to get the number of vertices. I'm not sure if the guess counts as a query, but if it doesn't, then you have 25 pairs of queries to get this. If it does count, then you only get 24 pairs of queries, plus an extra query, and one for the answer.It's annoying that this is a randomized solution, but the probability of this failing is less than 1%, and there are many hints in the problem statement itself to suggest that this is the intended approach (equal probability in bold letters, mentioning that "? a b" and "? b a" are rolled independently, specifying the number of jury tests, etc).
• » » 3 weeks ago, # ^ | +3 Also if you ask why the numbers of allowed querry is not greater it's because with 60 querry (which is log2(1e18)) you can just get the result with a binary search
» 3 weeks ago, # | 0 Very interesting problems, especially F!I have really enjoyed it!
» 3 weeks ago, # | ← Rev. 2 → 0 I think my A problem is brief enough.Is there any person can give me some advice to simplify my approach further?
» 3 weeks ago, # | 0 Can anyone tell me what i am doing wrong in my solution of problem C ?My submission
• » » 3 weeks ago, # ^ | 0 As the solution uses sorting I might suggest it's that the last index is not the last in your sequence. Or you could possibly go in the wrong order if p >= q or less than q. IDK, but that's most likely the mistake
• » » 3 weeks ago, # ^ | ← Rev. 2 → 0 Try "baba".I fell into the same trap and got WA. Thankfully, I figured it out before the contest was over.I didn't read your code (but I checked "baba" via custom invocation), but it's most likely due to your use of sorting. The assumption that your range covers all instances of first character and all instances of last character only applies if first character < last character. In the scenario where last character < first character, your range likely does not capture all instances of first and/or last character.
• » » 3 weeks ago, # ^ | 0 Here is the counter case for your solution: 1 yydiacorrect ans: 24 5 1 2 4 3 5Your ans: 24 4 1 4 3 5
• » » 3 weeks ago, # ^ | +4 Now it's Accepted!Accepted
• » » 3 weeks ago, # ^ | ← Rev. 3 → 0 In the p > q part, you should include p and q in your answer.
• » » » 3 weeks ago, # ^ | 0 Yeah Now it's accepted.
» 3 weeks ago, # | 0 https://codeforces.com/contest/1729/submission/171949266 Can anybody help to find a short test where my solution for C does not work?
• » » 3 weeks ago, # ^ | 0 one counter test case: 1 puvh correct ans: 8 2 1 4 your output: 8 3 1 2 4
» 3 weeks ago, # | ← Rev. 2 → 0 Worst Div 3
» 3 weeks ago, # | 0 Whats wrong with my solution to C? 171872810
• » » 3 weeks ago, # ^ | 0 Check 1 zzzzza
• » » » 3 weeks ago, # ^ | ← Rev. 2 → 0 Output: 25 6 5 4 3 2 1 6 Edit: Thanks :D
» 3 weeks ago, # | +13 To the author of problem E: how did you generate the tests? I'm curious to know. Btw, awesome contest!
• » » 3 weeks ago, # ^ | +6 Yeh same doubt regarding the E part of the contest, in all the test cases, input is in this form-3 9195979e279504391c49d2f080c1d8c755d044ca what is this — 9195979e279504391c49d2f080c1d8c755d044ca ?Shouldn't this be a long integer? Thanks.
• » » 3 weeks ago, # ^ | 0 I think you store the length of the permutation as well as it's index in sorted order. Not sure though...
» 3 weeks ago, # | 0 i was trying to solve F with condition s(l1,l1+(w-1) ) != s(l2,l2+(w-1)) when i read l1 != l2 then when i realaize my mistake the contest was over :(
» 3 weeks ago, # | +1 Can someone find a mistake in my code for 1729C - Jumping on Tiles?Code: 171911705Thanks!
• » » 3 weeks ago, # ^ | 0 Check 1 zzzzza
• » » 3 weeks ago, # ^ | 0 Try "baba".I fell into the same trap and got WA. Thankfully, I figured it out before the contest was over.I didn't read your code (but I checked "baba" via custom invocation), but it's most likely due to your use of sorting. The assumption that your range covers all instances of first character and all instances of last character only applies if first character < last character. In the scenario where last character < first character, your range likely does not capture all instances of first and/or last character.
» 3 weeks ago, # | ← Rev. 2 → +2 Oh my god I solved E. I have never tried random algorithms in competitive programming ever. Too bad the contest is over by now, but holy fuck I would never believe that it would actually work. Holy Jesus. 171950813
» 3 weeks ago, # | ← Rev. 2 → 0 Nice problem D! It makes me difficult even though it's only div 3. Can anyone give me a solution ?
• » » 3 weeks ago, # ^ | 0 My solutions problem c solution problem D solution
• » » 3 weeks ago, # ^ | 0 For C or D?For C, jumping directly to the end yields the minimum cost, which is |last character — first character|. How to maximize jumps? Observe that if you're in $\alpha$ and you jump to $\beta$ and then to $\gamma$ such that $\alpha <= \beta <= \gamma$, then the cost is the same as jumping directly from $\alpha$ to $\gamma$. So you can jump to any character in the range [current character, final character] and it would not affect the final cost. For example, with "logic", you can go from "l" to "i" to "g" to "c" because "l" > "i" > "g" > "c". To maximize jumps, you need to visit every character in the range [starting character, final character] in order (either non-increasing or non-decreasing order, depending on the relative order between first and last character). For D, if you calculate $y - x$, you basically get the person's balance from paying for their meal. This can be negative if they don't have enough money. You can then sort the balances, and try to pair the one with the lowest (i.e., most negative) balance with the highest balance. If the result is non-negative, it's a valid pair, and you can move on. If not, then the lowest balance is hopeless and you move to the next lowest balance (try with highest balance), and so on.
• » » » 3 weeks ago, # ^ | 0 very ez understand, thanks
» 3 weeks ago, # | 0 Would you please review 171943515? Maybe it is wrong when $n$ is small, but I do not know how to fix it.
• » » 3 weeks ago, # ^ | +1 int isn't big enough to hold the values required.
• » » » 3 weeks ago, # ^ | 0 Damn it you hacked me twice ^_^. Thank you very much!!!!!!!!!! When debugging in a limited time, it is easy to ignore some very basic facts...
» 3 weeks ago, # | ← Rev. 2 → +13 In problem E, all the AC solutions I've checked till now are querying $?$ $1$ $x$ and $?$ $y$ $1$ where ($2 \le y \le 26$). But how is that acceptable when the interactor chooses with a certain probability? Yeah, the probability of getting a correct answer is very close to $1$ ($1 - 2^{-25}$) but certainly not $1$. I thought of such a solution but ignored thinking since it can create a situation where same solutions may get AC and WA which is unexpected. I don't think this was a good/standard problem for a contest.
• » » 3 weeks ago, # ^ | +5 I did feel bothered by the fact that the intended solution is randomized, especially since I was trying much more complicated approaches before the contest ended, and only got AC after it was over.However, I think it's really a matter of opinion on whether this is appropriate. Randomized algorithms have numerous practical applications in the real world, so this kind of recognition that a randomized algorithm would work is a good skill to have, which can be argued as being among the type of skills that competitive programming should cover. Also, the problem did drop a lot of hints that this was the intended solution ("equal probability" in bold, mentioning that "? a b" and "? b a" are processed independently, mentioning the number of jury tests, etc), so I think it's justified in this case.
• » » 3 weeks ago, # ^ | +3 I personally think it was a great problem (maybe I have some bias since I solved it quite quickly in the contest). It makes you think outside of the box and consider non standard methods that you don't regularly see in CP, which ultimately is a great way to test/improve problem solving.
» 3 weeks ago, # | +1 Can someone explain how does the rating work, I'm confused a little bit because I'm new to codeforces.
• » » 3 weeks ago, # ^ | +3 It depends on your rank in the contest and your current rating.
» 3 weeks ago, # | ← Rev. 2 → 0 In problem E ,is it really fair to give such a probability based question in a contest because it's not like the probability of getting them all equal is zero right,just because of that many of us didn't go for that approach and the binary search one was exceeding the queries(which should actually be the intended solution).
• » » 3 weeks ago, # ^ | +25 The probability of success is over 99%, and the problem dropped a lot of hints to go for a randomized approach ("equal probability" in bold letters, clarifying that "? a b" and "? b a" are processed independently, mentioning the number of jury tests), so I think this was justified. I understand how one would assume there is a deterministic solution, but randomized algorithms have numerous practical applications in the real world, and recognizing when a randomized algorithm would be effective is a valuable skill for a programmer to have, so it can be argued that it should fall within the scope of competitive programming. I understand the concerns, but I think it's a matter of subjective opinion and cannot definitely be ruled as "unfair".
» 3 weeks ago, # | 0 Is there a deterministic solution to the problem E, which always works no matter how the generator is configured?
• » » 3 weeks ago, # ^ | +2 No. An adversary can always ensure that the result for each query is either -1 or some distance that is less than the current lower bound that was established. Therefore, there are essentially two possible types of information to be gleaned from a single query. With 50 queries, this only allows identifying up to $2^{50}$ different values, which is less than $10^{18}$. So there is no 100% correct deterministic solution.
» 3 weeks ago, # | 0 Are hacks rated?
• » » 3 weeks ago, # ^ | 0 No
» 3 weeks ago, # | 0 Is square root decomposition the intended solution for problem F?
• » » 3 weeks ago, # ^ | +3 there is no need in sqrt decomposition, the only data structure we actually need is prefix sum array
» 3 weeks ago, # | ← Rev. 2 → +1 In problem E, shouldn't it be "in C++ you should use function fflush(stdout)"?
» 3 weeks ago, # | 0 Any hints for G?
• » » 3 weeks ago, # ^ | ← Rev. 2 → +1 Hint1If you picked a occurrence of $t$ starting at $i$ then the previous occurrence which needs to be picked should have $startingindex <= i - m$ Hint2If the rightmost index that satisfies the above condition starts at $j$ then you can't pick a $startingindex$ which has $endingindex$ before $j$. Hint3The above problem can be solved using dynamic programming Hint4For number of ways you need to consider only $index >= startingindex$ of last occurrence and $index <= endingindex$ of last occurrence which gives the minimum answer for $dp[i]$ where $dp[i]$ is the minimum answer if $ith$ occurrence of $t$ in $s$ is picked. Code string s,t; read(s,t); int n = s.length(); int m = t.length(); if(m > n){ print(0,1); return ; } vec ocr; fr1(i,0,n - m + 1){ if(s.substr(i,m) == t){ ocr.pb(i); } } if(ocr.empty()){ print(0,1); return ; } vec dp(sz(ocr)); vec ways(sz(ocr)); fr1(i,0,sz(ocr)){ bool flag = false; int lim = -1; int mn = n; int cnt = 0; fr2(j,i - 1,0){ if(!flag){ if(ocr[j] + m - 1 < ocr[i]){ flag = true; lim = ocr[j] - m; if(dp[j] < mx){ mn = dp[j]; cnt = ways[j]; } else if(dp[j] == mn){ eadd(cnt,ways[j]); } } } else { if(ocr[j] > lim){ if(dp[j] < mn){ mn = dp[j]; cnt = ways[j]; } else if(dp[j] == mn){ eadd(cnt,ways[j]); } } else break; } } if(flag){ dp[i] = mx + 1; ways[i] = cnt; } else { dp[i] = 1; ways[i] = 1; } } int st = ocr.back() - m; int ans = 0,mn = n; fr2(i,sz(ocr) - 1,0){ if(ocr[i] > st){ if(mn > dp[i]){ mn = dp[i]; ans = ways[i]; } else if(mn == dp[i]){ eadd(ans,ways[i]); } } } print(mn,ans);
» 3 weeks ago, # | +16 An interactive problem without Binary Search ,,,,,, surprising
• » » 3 weeks ago, # ^ | 0 There was one such 1713D - Tournament Countdown
» 3 weeks ago, # | 0 how to solve problem c? I used priority queue but got wa
• » » 3 weeks ago, # ^ | ← Rev. 2 → 0 The maximum number of jumps must be on the characters which in the range between the first and the last characterex) s = "codewars"character 'w' doesn't belong to ['c' -> 's'] so you shouldn't jump on itTake the rest of the characters, sort them, then print their indexes.My submission
• » » 3 weeks ago, # ^ | 0 The minimum distance is simply the distance from jumping directly from the first to the last. How to maximize jumps? If you go from $\alpha$ to $\beta$ to $\gamma$, where $\alpha \leq \beta \leq \gamma$, the total cost is the same as going straight from $\alpha$ to $\gamma$. So you can keep jumping to an intermediate index, provided that the character in it lies between the previous character you were in and the next character that you're going into. So if you need to jump from "b" to "f", then you can jump into all of the characters in the range [b, f], in non-decreasing order. By the same argument, if your final character is actually smaller than the first character, you can still jump into all characters in the range, but in non-increasing order this time.
» 3 weeks ago, # | 0 Anyway，I dont think E is a good problem.
• » » 3 weeks ago, # ^ | 0 Definitely because if you go to exact statement then it will always be a point of discussion whether (2^-25) = 0 or not .
• » » 3 weeks ago, # ^ | +3 Randomized algorithms are not that bad!
• » » » 3 weeks ago, # ^ | +35 Randomized algorithms are the best thing in the whole universe.
• » » » » 3 weeks ago, # ^ | +2 Especially when they're not the intended solution
• » » » » » 3 weeks ago, # ^ | 0 Based.Btw, are u talking about today's E? Wasn't the randomized solution intended?
• » » » » » » 3 weeks ago, # ^ | +9 No, in this E the random solution is 99% intended I was talking about 1310B - Double Elimination, the random solution imo wasn't intended.
• » » 3 weeks ago, # ^ | 0 I agree. Giving such an easy problem at E is confusing. I got the idea in maybe 10-20 seconds but kept wondering whether there are any tricky scenarios or edge cases. It should've been in D.
» 3 weeks ago, # | 0 Isn't intentional code obfuscation forbidden?Please check this: https://codeforces.com/contest/1729/submission/171937226MikeMirzayanov
• » » 3 weeks ago, # ^ | 0 This submission too: https://codeforces.com/contest/1729/submission/171874872MikeMirzayanov
» 3 weeks ago, # | 0 So I was just casually looking at the hacks corner and found a few hacks for problem A which seem way too odd. 171954115, by: vaibhav_1710 to: chaudharyvaibhav184; 171952586, by: andreifilimon to: wavetome;171877008, by: just_to_code to: suiiiiiNow how and why in the world did all these people put an a is equal to some random no. check? I don't know where to report these so I'm just posting it here. There are points for hacks so this should count as a violation right?
• » » 3 weeks ago, # ^ | ← Rev. 2 → 0 those are probably made specialy so people who check a lot of submissions can find those stupid mistakes and take the points. I agree to you thats like a violation but as those are there why not to make a easy hack. (there are a lot of this mistakes in submissions) by the way a good way to hack smth i tried a lot on problem C was big tests...
• » » 3 weeks ago, # ^ | 0 There are some telegram channels which sell these solutions during the contest for as cheap as 20-30 cents per question.
• » » » 3 weeks ago, # ^ | +1 That's not good. That should be reported.
• » » 3 weeks ago, # ^ | 0 Can those be Trojan Horses for cheaters? In edu rounds (Edu, Div 3, Div 4) there are no points given for hacks. Therefore, there's no incentive to plant deliberate mistakes for future hacks.
• » » » 3 weeks ago, # ^ | 0 Thanks for being the only person to clear my doubts. I didn't know hacks didn't account for any points for div. 3 onwards. Now I feel stupid for posting this.
» 3 weeks ago, # | +6 I like problems with randomized solutions. It would be great if there were more such problems as today's E.Do you like randomized algorithms? — Yes — No
• » » 3 weeks ago, # ^ | +4 I don't know what to say about $E$, I had that idea early, but I doubted it would work, then i submitted it in the last minutes (because why not) and somehow it worked, but I got high penality.I'm certain that many people (maybe hundreds) had this solution in their head but they didn't try because it depends on luck :DSo i don't like this random algorithm.
• » » 3 weeks ago, # ^ | 0 me not touching the yes/no button as it will disturb the perfect pattern
» 3 weeks ago, # | 0 I had a doubt regarding the E part of the contest, in all the test cases, input is in this form- 3 9195979e279504391c49d2f080c1d8c755d044ca what is this — 9195979e279504391c49d2f080c1d8c755d044ca ?Shouldn't this be a long integer? Thanks.
» 3 weeks ago, # | 0 Any hints for F? I'm guessing there's some number theory property related to mod9 as $w$ can overflow int64.
• » » 3 weeks ago, # ^ | 0 Hint: any number in decimal mod 9 = the sum of its digits mod 9. It would be a good exercise to prove this statement.
• » » » 3 weeks ago, # ^ | 0 Hint about proving this statement & other similar statementsConsider each digit's contribution to the total modulo. What is $1$, $10$, $100$, $1000$, etc modulo $M$? It is possible to prove statements about any arbitrary modulo using this method.
» 3 weeks ago, # | 0 Why wa on tc 2. problem c. https://codeforces.com/contest/1729/submission/1719150491.
• » » 3 weeks ago, # ^ | 0 Try this input: 2 azz bz The map yields values from 1 to 26. When adding elements to the 2D vector, the first index is given by the map, so the first vector index also ranges from 1 to 26 (1 for 'a', 26 for 'z'). However, when clearing the vector at the end of the test case, only indices 0 to 25 are cleared. So any instances of 'z' saved from earlier test cases (at v[26]) will continue to remain in future test cases. (note that v[26] is technically out of bounds, since v was declared to have size 26; this will not necessarily be detected as an error, and the program may continue to use v[26] as the unallocated memory location that sits 26 steps from the start of the vector, until an issue arises with this location later)
• » » » 3 weeks ago, # ^ | +1 thanks man.
» 3 weeks ago, # | 0 In problem E, hacks are disabled. Is system testing also disabled or can it be a FST?
• » » 3 weeks ago, # ^ | 0 The problem statement says that there will be 50 test cases and there were 50 pretests. So I'm presuming that there will not be any system testing for E
» 3 weeks ago, # | 0 Curious to know if rating changes have been applied or not for this round. It is showing me as a unrated contest atleast for me when I tab the "all contests" dropbox in rating graph. Or this is a normal case when rating is not updated till now.
• » » 3 weeks ago, # ^ | 0 It shows it as an unrated contest if rating changes haven't been applied. As of now, the hacking stage has just gotten over and the system testing still has to commence.
» 3 weeks ago, # | +1 How to do Question EIf n<10^15 or number of queries < 60 my solution of ternary/binary search would have worked. My naïve idea= ask query as "? 1 m" . if this returns -1 decrease value of m or else increasebut this uses 60 queries. How to approach this types of question
• » » 3 weeks ago, # ^ | ← Rev. 2 → 0 The issue with binary search is that you can only eliminate at most 1/2 of the search space consistently; in fact with question E I think there is no way to get a solution that you can be absolutely 100% sure is correct.There is a solution though that has about a 99.99% chance of being correct. It's only possible because of the problem emphasizing that one of two paths will be chosen with equal probability.
» 3 weeks ago, # | 0 MikeMirzayanov is the only one without laptop in the picture.Also I don't know the others. Does anyone else know all of them??
• » » 3 weeks ago, # ^ | 0 It's the ITMO university team. Just read earlier in the post. There are seven in the team, but two are not in the picture (specified in the caption). I'm pretty sure the seven of them all know each other, and there are probably other people (possibly from ITMO University as well) that would also know them, but I'm not sure why you're so interested in them.
• » » » 3 weeks ago, # ^ | 0 I was just curious, since they were all with Mike and probably at the headquarters from where this brilliant site is maintained.
» 3 weeks ago, # | +3 How do you realize when system testing is over in edu rounds and div3 rounds? At the end of hacking stage, it says "Final standings, open hacking phase finished". But don't system tests take place after this, in that case how is it final?
• » » 3 weeks ago, # ^ | 0 System testing in these rounds usually happen 4-5 hours after hacking phase
• » » » 3 weeks ago, # ^ | +2 I am aware that system testing takes place a few hours after hacking. I am just curious as to how you realise whether sytem testing has finished or not. Like for round 820, what indication do I get to understand that system testing is done or not?All it says is "final standings" which isn't true I think because system testing hasn't started.
» 3 weeks ago, # | +1 when will the ratings be assigned?
• » » 3 weeks ago, # ^ | 0 it's already updated
• » » » 3 weeks ago, # ^ | 0 I am not able to see yesterdays contest in the list of contests i have taken part in .
• » » » » 3 weeks ago, # ^ | 0 i was joking, rating will update after 12 more hours
• » » » » » 3 weeks ago, # ^ | 0 ok thanq
» 3 weeks ago, # | 0 Why are ratings still not published, is this round rated or unrated
• » » 3 weeks ago, # ^ | 0 This round had 12 hours of hacking phase, which will be used to re-judge all of the solutions as far as I know. The hacking round finished a few hours ago, now hacks are being analyzed most likely and after re-testing, ratings shall be published, so a few more hours probably.
» 3 weeks ago, # | ← Rev. 2 → 0 I am having trouble with this, please share your insights. What is your comment on the difficulty or standard of problem E? Is it a good problem or an average one? Can hard problems have such simple solutions? If so, can you give some examples?
» 3 weeks ago, # | +1 Cool update. Would love to have these in future rounds as well :)
» 3 weeks ago, # | 0 I think if this problem was placed in the A place,there will be more people solved it.Many people had the corret idea,but they think this solution is not fit to E problem.This kind of problems may be good,but where they should be placed is still a problem.
• » » 3 weeks ago, # ^ | ← Rev. 2 → +1 Well, A through D are easier than E, so everything's fine, imo. And if it had been set as A, more participants would have been determined to dodge this round.I think we are forgetting, that this is a Div3 round. This E is even harder, than usual E Div3 problems, imo again.
» 3 weeks ago, # | 0 When will the rating change? my rating has not changed yet
• » » 3 weeks ago, # ^ | 0 Same stuff, pal. I am so looking forward it :D
» 3 weeks ago, # | 0 Where's the editorial
» 3 weeks ago, # | 0 Problem F video editorial.
» 3 weeks ago, # | 0 when can I see the results? 12 hours-fade is already up...
» 3 weeks ago, # | 0 they look awesome!
» 3 weeks ago, # | 0 Finally ratings are updated :D
• » » 3 weeks ago, # ^ | 0 Hey but it didn't reflected on my profile still now.
• » » 3 weeks ago, # ^ | 0 Waiting for expert :D
» 3 weeks ago, # | 0 Who knows why my raiting is unrated ? why?
» 3 weeks ago, # | 0 why is taking so long to update ratings? :/
• » » 3 weeks ago, # ^ | 0 System testing ended few minutes ago, so you may expect rating change soon!
• » » » 3 weeks ago, # ^ | 0 Ok cool
• » » » 3 weeks ago, # ^ | 0 my rating is yet not changed.neither the contest appears in the list of contests i have taken part in
» 3 weeks ago, # | +8 G can be solved in $\mathcal{O}(m + n\cdot \log(n))$ time and $\mathcal{O}(n + m)$ memory, by using dp + segment tree.First compute starting positions of occurences of $t$ in $s$. This can be done in linear time in many ways (I used z function). It's easy to notice that we can apply operations in order from right to left. Then, we will compute $dp_i$ = answer if the rightmost operation was applied to range $[i,i+m)$.To compute $dp_i$, we could iterate to the left of $i$ on the starting index of the next operation (let's call it $j$). Two things need to happen: If there is an occurence of $t$ in $s$ that is completely between $[j+m, i-m]$, it won't be covered by any operation, so $j$ is not valid. This defines a lowerbound on the value of $j$ (we can easily mantain it when increasing $i$ by using two pointers). if $j+m-1 \geq i$, then $j$ is not a valid candidate, because given that we apply operations from right to left, the operation on index $i$ removed some of the characters of the occurence starting on index $j$. This defines an upperbound on the value of $j$. This two conditions define a valid range of values for $j$. We just need to be able to combine the $dp$ values of this range. We can use segment tree to get this merged value.Overall, we compute occurences of $t$ in $s$ in $\mathcal{O}(m + n)$ and then compute $dp$ values in $\mathcal{O}(n\cdot \log(n))$, so final complexity is $\mathcal{O}(m + n\cdot \log(n))$ with $\mathcal{O}(n + m)$ memory.Here is a link to my submission :)
• » » 3 weeks ago, # ^ | 0 so cool
• » » 3 weeks ago, # ^ | ← Rev. 2 → +3 That's what I thought! This task seems similar to a (slightly harder) problem from POI 2018, stage 2 problem "Konduktor". Link: https://szkopul.edu.pl/p/default/problemset_eng/oi/25 You change the language to English in the top right corner. Also, forgot to say, nice explaination of your solution
• » » 3 weeks ago, # ^ | 0 Since the query range is fixed, you could overkill this further by replacing the Segment Tree with a Minimum Queue for a $O(n+m)$ solution. submission
» 3 weeks ago, # | 0 I solved 2 problems in this round but still, the rating is not updated. what is the estimated time it takes to update the ranking?
• » » 3 weeks ago, # ^ | 0 It will be updated soon.
» 3 weeks ago, # | +1 System Testing has ended way back.... then why so long in rating update?
• » » 3 weeks ago, # ^ | ← Rev. 2 → 0 It takes time (probably around 24 hrs from contest end) for educational round rating to be updated.
• » » » 3 weeks ago, # ^ | 0 But this isn't an Educational round...
• » » » » 3 weeks ago, # ^ | 0 Read the second paragraph of this blog
• » » » » » 3 weeks ago, # ^ | 0 Ohh ok... my bad
• » » 3 weeks ago, # ^ | 0 This Round is showing under Unrated Category now!! Why?
• » » » 3 weeks ago, # ^ | 0 Oh yes :/
• » » » 3 weeks ago, # ^ | 0 how ?
» 3 weeks ago, # | 0 Aris Is it unrated round?
» 3 weeks ago, # | +2 Editorial section still not uploaded
• » » 3 weeks ago, # ^ | 0 Ratings also not updated :(
• » » » 3 weeks ago, # ^ | 0 Let's see I think it may be Unrated thats whay:(
» 3 weeks ago, # | 0 I like E. Open my mind.
• » » 3 weeks ago, # ^ | +3 I didn't. Even though the probablity is 50-50 still there's a chance that the ans for "? a b" is same as that of "? b a" for the first 50 queries. Change my mind :(
• » » » 3 weeks ago, # ^ | ← Rev. 2 → +1 Probability of all 50 queries failing and none giving an useful information is 2^-25, that's basically somewhat like 99.9998% chance of a solution passing.
• » » » » 3 weeks ago, # ^ | 0 I don't get this. It fails for an edge case right. What if I'm the testcase author and I wantedly set such a case?
• » » » » » 3 weeks ago, # ^ | 0 Testcases are random as well, it's not a preset. The answers are generated randomly while testing.
• » » » » » 3 weeks ago, # ^ | ← Rev. 2 → 0 it's not about case work. In statement it is said, that The interactor chooses one of the two paths with equal probability
• » » » » » » 3 weeks ago, # ^ | 0 I get it now. Thanks ;)
• » » » » » » » 3 weeks ago, # ^ | 0 You're welcome!
• » » » 3 weeks ago, # ^ | 0 Probability 0.00000003 is small enough to never happen, that's it
• » » » » 3 weeks ago, # ^ | 0 See I get this but they've mentioned that the answers for the queries are pre-determined right so there is a chance that the solution fails for a case.
• » » » » » 3 weeks ago, # ^ | 0 No, answer is not pre-determined. It becomes determined only after you ask about this pair
• » » » » » 3 weeks ago, # ^ | ← Rev. 2 → 0 Or, if will be better to understand, they are pre-determined, but not with bare hands, but using a randomised algorithm, that randomly picks pre-determined answer for each pair. It is still the same task
» 3 weeks ago, # | 0 I submitted this solution https://codeforces.com/contest/1729/submission/171912216 to the problem D yesterday during the contest, and after the system testing, it gave me a runtime error on test 7. In contrast, after system testing, I submitted the exact same code I submitted yesterday, to the problem and it got accepted. This is the link to that accepted solution https://codeforces.com/contest/1729/submission/172063331. You can check both the solutions are precisely the same line by line. So why does my code fail in system testing? MikeMirzayanov Vladosiya SixtyWithoutExam Gornak40 Aris Gol_D myav
• » » 3 weeks ago, # ^ | ← Rev. 2 → 0 I do not run your code, but I think while(a[max1] >= 0 ) max1--; is dangerous. What if max1==-1? This causes undefined behaviour. I review your code on a mobile phone. Maybe I am wrong.
• » » 3 weeks ago, # ^ | +1 Maybe thats the reason of ratings changing delay
• » » 3 weeks ago, # ^ | ← Rev. 2 → 0 I didn't review it in depth, but considering the above comment, I suppose your code once slipped with out of bounds indexing, while another time it didn't. Might be because of the second time out of bound index still made sense(maybe bcs it was allocated at some other place where the resulting index was valid). You can try to send it again, you might get another runtime or might not, lol. Not sure what are the chances of the same happening again.
• » » 3 weeks ago, # ^ | ← Rev. 2 → 0 This part will just keep doing max1--; while the value in *(a+max1) >= 0 even if max1 less than 0In the first time it worked because maybe the value in the memory just before a[0] is minus.in system testing maybe it was positive, so actually your code is wrong, and you are lucky that you got accepted 2 times. while (a[max1] >= 0){ max1--; } max1++; | 2022-10-04 13:24:13 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.29492634534835815, "perplexity": 1435.53745421672}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337504.21/warc/CC-MAIN-20221004121345-20221004151345-00025.warc.gz"} |
https://www.tcs.tifr.res.in/events/testing-whether-multivariate-polynomial-zero | # Testing whether a Multivariate Polynomial is Zero
## Speaker:
Pranab Sen School of Technology and Computer Science Tata Institute of Fundamental Research Homi Bhaba Road <
## Time:
Tuesday, 16 June 2009 (All day)
## Venue:
• AG-69
Suppose we have a polynomial $P$ in variables $X_1, \\ldots, X_n$ with coefficients from a field $F$, with total degree at most $d$. The polynomial $P$ is given in terms of some algebraic expression involving $X_1, \\ldots, X_n$. We want to know if $P$ is identically zero, without opening up the expression in its full glory. For example, we want to know if $(X_1+X_2)(X_1-X_2) - X_12 + X_22$ is
identically zero, without opening up the expression. This problem is known as {\\em polynomial identity testing}, and is an important problem in algorithmic algebra and theoretical computer science.
In this talk, we shall see some efficient methods for this problem. Starting from Schwartz and Zippel's method that uses a generalisation to multivariate polynomials of the well known fact that a non-zero univariate polyomial of degree $d$ has at most $d$ roots, we shall glimpse recent approaches by Chen and Kao that involve transcendental extensions of $F$, Lewin and Vadhan that use the power series ring over $F$, and finally, Agrawal and Biswas's elementary but powerful approach using Chinese remaindering. The last work was the precursor to Agrawal, Kayal and Saxena's famous efficient algorithm for primality testing. | 2022-12-08 15:42:58 | {"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": 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": 0.6684628129005432, "perplexity": 461.15065485379046}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 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/1669446711344.13/warc/CC-MAIN-20221208150643-20221208180643-00086.warc.gz"} |