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# NGC 2818

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 Optically Thick Radio Cores of Narrow-Waist Bipolar NebulaeWe report our search for optically thick radio cores in 16 narrow-waistbipolar nebulae. Optically thick cores are a characteristic signature ofcollimated ionized winds. Eleven northern nebulae were observed with theVery Large Array (VLA) at 1.3 and 0.7 cm, and five southern nebulae wereobserved with the Australia Telescope Compact Array (ATCA) at 6 and 3.6cm. Two northern objects, 19W32 and M1-91, and three southern objects,He 2-25, He 2-84, and Mz 3, were found to exhibit a compact radio corewith a rising spectrum consistent with an ionized jet. Such jets havebeen seen in M2-9 and may be responsible for shaping bipolar structurein planetary nebulae. High-Resolution Spectra of Bright Central Stars of Bipolar Planetary Nebulae and the Question of Magnetic ShapingWe present ESO New Technology Telescope high-resolution echellespectroscopy of the central stars (CSs) of eight southern bipolarplanetary nebulae (PNe) selected for their asymmetry. Our aim was todetermine or place limits on the magnetic fields of the CSs of thesenebulae, and hence to explore the role played by magnetic fields innebular morphology and PN shaping. If magnetic fields do play a role, weexpect these CSs to have fields in the range102-107 G from magnetic flux conservation on thereasonable assumption that they must evolve into the high-field magneticwhite dwarfs. We were able to place an upper limit of ~20,000 G on themagnetic fields of the central stars of He 2-64 and MyCn 18. Thespectrum of He 2-64 also shows a P Cygni profile in He I λ5876and λ6678, corresponding to an expanding photosphere withvelocity ~100 km s-1. The detection of helium absorptionlines in the spectrum of He 2-36 confirms the existence of a hot stellarcomponent. We did not reach the necessary line detection for magneticfield analysis in the remaining objects. Overall, our results indicatethat if magnetic fields are responsible for shaping bipolar planetarynebulae, these are not required to be greater than a few tens ofkilogauss.Based on observations made with ESO telescopes at the La SillaObservatory under program ID 70.D-0339. Kinematics of the Open Cluster System in the GalaxyAbsolute proper motions and radial velocities of 202 open clusters inthe solar neighborhood, which can be used as tracers of the Galacticdisk, are used to investigate the kinematics of the Galaxy in the solarvicinity, including the mean heliocentric velocity components(u1,u2,u3) of the open cluster system,the characteristic velocity dispersions(σ1,σ2,σ3), Oortconstants (A,B) and the large-scale radial motion parameters (C,D) ofthe Galaxy. The results derived from the observational data of propermotions and radial velocities of a subgroup of 117 thin disk young openclusters by means of a maximum likelihood algorithm are:(u1,u2,u3) =(-16.1+/-1.0,-7.9+/-1.4,-10.4+/-1.5) km s-1,(σ1,σ2,σ3) =(17.0+/-0.7,12.2+/-0.9,8.0+/-1.3) km s-1,(A,B) =(14.8+/-1.0,-13.0+/-2.7) km s-1 kpc-1, and (C,D) =(1.5+/-0.7,-1.2+/-1.5) km s-1 k pc-1. A discussionon the results and comparisons with what was obtained by other authorsis given. The structure of planetary nebulae: theory vs. practiceContext.This paper is the first in a short series dedicated to thelong-standing astronomical problem of de-projecting the bi-dimensional,apparent morphology of a three-dimensional mass of gas. Aims.Wefocus on the density distribution in real planetary nebulae (and alltypes of expanding nebulae). Methods. We introduce some basictheoretical notions, discuss the observational methodology, and developan accurate procedure for determining the matter radial profile withinthe sharp portion of nebula in the plane of the sky identified by thezero-velocity-pixel-column (zvpc) of high-resolution spectral images.Results. The general and specific applications of the method (andsome caveats) are discussed. Moreover, we present a series of evolutivesnapshots, combining illustrative examples of both model and trueplanetary nebulae. Conclusions. The zvpc radial-densityreconstruction - added to tomography and 3D recovery developed at theAstronomical Observatory of Padua (Italy) - constitutes a very usefultool for looking more closely at the spatio-kinematics, physicalconditions, ionic structure, and evolution of expanding nebulae. Proper motion determination of open clusters based on the UCAC2 catalogueWe present the kinematics of hundreds of open clusters, based on theUCAC2 Catalogue positions and proper motions. Membership probabilitieswere obtained for the stars in the cluster fields by applying astatistical method uses stellar proper motions. All open clusters withknown distance were investigated, and for 75 clusters this is the firstdetermination of the mean proper motion. The results, including the DSSimages of the cluster's fields with the kinematic members marked, areincorporated in the Open Clusters Catalogue supported on line by ourgroup. The 3-D shaping of NGC 6741: A massive, fast-evolving Planetary Nebula at the recombination-reionization edgeWe infer the gas kinematics, diagnostics and ionic radial profiles,distance and central star parameters, nebular photo-ionization model,spatial structure and evolutionary phase of the Planetary Nebula NGC6741 by means of long-slit ESO NTT+EMMI high-resolution spectra at nineposition angles, reduced and analysed according to the tomographic and3-D methodologies developed at the Astronomical Observatory of Padua(Italy). NGC 6741 (distance≃2.0 kpc, age≃ 1400 yr, ionizedmass Mion≃ 0.06 Mȯ) is a dense(electron density up to 12 000 cm-3), high-excitation,almost-prolate ellipsoid (0.036 pc × 0.020 pc × 0.018 pc,major, intermediate and minor semi-axes, respectively), surrounded by asharp low-excitation skin (the ionization front), and embedded in aspherical (radius≃ 0.080 pc), almost-neutral, high-density (n(HI)≃ 7 ×103 atoms cm-3) halo containinga large fraction of the nebular mass (Mhalo≥ 0.20Mȯ). The kinematics, physical conditions and ionicstructure indicate that NGC 6741 is in a deep recombination phase,started about 200 years ago, and caused by the rapid luminosity drop ofthe massive (M*=0.66{-}0.68 Mȯ), hot (logT* ≃ 5.23) and faint (logL*/Lȯ ≃ 2.75) post-AGB star, which hasexhausted the hydrogen-shell nuclear burning and is moving along thewhite dwarf cooling sequence. The general expansion law of the ionizedgas in NGC 6741, Vexp(km s-1)=13 × R arcsec,fails in the innermost, highest-excitation layers, which move slowerthan expected. The observed deceleration is ascribable to the luminositydrop of the central star (the decreasing pressure of the hot-bubble nolonger balances the pressure of the ionized gas), and appears instriking contrast to recent reports inferring that acceleration is acommon property of the Planetary Nebulae innermost layers. A detailedcomparative analysis proves that the "U"-shaped expansion velocity fieldis a spurious, incorrect result due to a combination of: (a) simplisticassumptions (spherical shell hypothesis for the nebula); (b) unfitreduction method (emission profiles integrated along the slit); and (c)inappropriate diagnostic choice (λ4686 Å of He II, i.e. athirteen fine-structure components recombination line). Some generalimplications for the shaping mechanisms of Planetary Nebulae arediscussed. 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. Comparison of the Luminosity Functions of Open Clusters Based on USNO-A1 DataThe luminosity and mass functions of a group of Galactic open clustersare constructed by applying a statistical method to photometric datafrom the USNO-A1 catalog. Despite some limitations, this catalog can beused for statistical analyses in Galactic astronomy. Pairwisecomparisons of the derived cluster luminosity functions are performedfor five age intervals. The differences between the luminosity functionsof the open clusters are not statistically significant in most cases. Itis concluded that the luminosity functions are approximately universalthroughout a large volume in the solar neighborhood. Combined luminosityand mass functions are constructed for six age intervals. The slope ofthe mass spectrum may vary somewhat from cluster to cluster, and themean slope may be somewhat higher than the Salpetervalue. The distances of Type I planetary nebulaeThe distances D of planetary nebulae (PNe) are still extremelyuncertain. Although a variety of methods have been used to evaluate thisparameter, these are often in conflict, and subject to large random andsystematic errors. It is therefore important to evaluate D using as manyindependent procedures as possible. We outline here one further way inwhich this parameter may be assessed. It is noted that where the nebularmass range is narrow, then one might expect observed PNe radii to beroughly similar. This, where it occurs, would also result in acorrelation between their angular diameters Θ, and distances D.We find that just such a trend occurs for Type I nebulae, and we employthis to determine distances to a further 44 such outflows. Our meanvalues of D appear similar to those of Zhang [ApJS 98 (1995) 659],implying a relatively long PNe distance scale. 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. Metallicity distribution on the galactic diskDepending mainly on UBVCCD data, the metallicities of 91 open starclusters nearby the galactic disk have been estimated using Cameron's[A&A 147 (1985b) 39] method. The metallicity radial gradient alongthe galactic plane is found to be -0.09 dex/kpc; which is in a very goodagreement with Panagia and Tosi [A&A 96 (1981) 306] and Carraro etal. [MNRAS 296 (1998) 1045]. Vertically on the galactic disk, withinabout 800 pc, the metallicity gradient is found to be so trivial. Anaverage age-metallicity relation has been examined, which confirms theprevious suggestion that the metallicity of a cluster depending mainlyon its position on the galactic disk more than its age. 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} Angular dimensions of planetary nebulaeWe have measured angular dimensions of 312 planetary nebulae from theirimages obtained in Hα (or Hα + [NII]). We have appliedthree methods of measurements: direct measurements at the 10% level ofthe peak surface brightness, Gaussian deconvolution and second-momentdeconvolution. The results from the three methods are compared andanalysed. We propose a simple deconvolution of the 10% levelmeasurements which significantly improves the reliability of thesemeasurements for compact and partially resolved nebulae. Gaussiandeconvolution gives consistent but somewhat underestimated diameterscompared to the 10% measurements. Second-moment deconvolution givesresults in poor agreement with those from the other two methods,especially for poorly resolved nebulae. From the results of measurementsand using the conclusions of our analysis we derive the final nebulardiameters which should be free from systematic differences between small(partially resolved) and extended (well resolved) objects in our sample.Table 1 is only available in electronic form athttp://www.edpsciences.org The relation between elemental abundances and morphology in planetary nebulaeAn investigation of the variation of elemental abundances with planetarynebula morphology is of considerable interest, since it has a bearingupon how such sources are formed, and from which progenitors they areejected. Recent advances in morphological classification now enable usto assess such trends for a statistically significant number of sources.We find, as a result, that the distribution N[log(X/H)] of sources withrespect to elemental abundance (X/H) varies between the differingmorphologies. Circular sources tend to peak towards low abundancevalues, whilst bipolar nebulae (BPNe) peak towards somewhat highervalues. This applies for most elemental species, although it is perhapsleast apparent for oxygen. In contrast, elliptical sources appear todisplay much broader functions N[log(X/H)], which trespass upon thedomains of both circular and elliptical planetary nebulae (PNe).We take these trends to imply that circular sources derive fromlower-mass progenitors, bipolar sources from higher-mass stars, and thatelliptical nebulae derive from all masses of progenitor, high and low.Whilst such trends are also evident in values of mean abundance, they are much less clear. Only in the cases of He/H, N/H,Ne/H and perhaps Ar/H is there evidence for significant abundancedifferences.Certain BPNe appear to possess low abundance ratios He/H and Ar/H, andthis confirms that a few such outflows may arise from lower-massprogenitors. Similarly, we note that ratios are quite modestin elliptical planetary nebulae, and not much different from those forcircular and bipolar PNe; a result that conflicts with the expectationsof at least one model of shell formation. Planetary nebula carbon yields and the chemical evolution of the Galactic discTwo sets of observational carbon stellar yields for low-and-intermediate mass stars are computed based on planetary nebulaabundances derived from CIIλ4267 and CIIIλλ1906 +1909 lines, respectively. These observational yields are assumed inchemical evolution models for the solar vicinity and the Galactic disc.C/O values observed in stars in the solar vicinity and Galactic HIIregions are compared with those predicted by chemical evolution modelsfor the Galaxy. It is concluded that the C yields derived from permittedlines are in better agreement with the observational constraints thanare those derived from forbidden lines. On the Galactic Disk Metallicity Distribution from Open Clusters. I. New Catalogs and Abundance GradientWe have compiled two new open cluster catalogs. In the first one, thereare 119 objects with ages, distances, and metallicities available, whilein the second one, 144 objects have both absolute proper motion andradial velocity data, of which 45 clusters also have metallicity dataavailable. Taking advantage of the large number of objects included inour sample, we present an iron radial gradient of about -0.063+/-0.008dex kpc-1 from the first sample, which is quite consistentwith the most recent determination of the oxygen gradient from nebulaeand young stars, about -0.07 dex kpc-1. By dividing clustersinto age groups, we show that the iron gradient was steeper in the past,which is consistent with the recent result from Galactic planetarynebulae data, and also consistent with inside-out galactic diskformation scenarios. Based on the cluster sample, we also discuss themetallicity distribution, cluster kinematics, and space distribution. Adisk age-metallicity relation could be implied by those properties,although we cannot give conclusive result from the age- metallicitydiagram based on the current sample. More observations are needed formetal-poor clusters. From the second catalog, we have calculated thevelocity components in cylindrical coordinates with respect to theGalactic standard of rest for 144 open clusters. The velocitydispersions of the older clusters are larger than those of youngclusters, but they are all much smaller than that of the Galactic thickdisk stars. Proper Motions of Open Star Clusters and the Rotation Rate of the GalaxyThe mean proper motions of 167 Galactic open clusters withradial-velocity measurements are computed from the data of the Tycho-2catalog using kinematic and photometric cluster membership criteria. Theresulting catalog is compared to the results of other studies. The newproper motions are used to infer the Galactic rotation rate at the solarcircle, which is found to be ω0=+24.6±0.8 km s-1 kpc-1.Analysis of the dependence of the dispersion of ω0 estimates onheliocentric velocity showed that even the proper motions of clusterswith distances r>3 kpc contain enough useful information to be usedin kinematic studies demonstrating that the determination of propermotions is quite justified even for very distant clusters. Infrared Imaging and Optical Imaging and Spectroscopy of (mostly) typePlanetary Nebulae.Optical spectra, images in the main optical emission lines and IR imagesdirected towards the detection of molecular hydrogen, were obtained for14 planetary nebulae (PNe). Eleven of these comply with the two criteriadefining a typePNe (He-rich with a large N/O ratio). All are classifiedas bipolar, though two of them (A 24 and M 3-5) are either bipolars seenpole-on or elliptical. Shocked molecular hydrogen was detected in eightof these objects, but not in A 14, A 24, and M 1-57. KjPn 6 wastentatively classified as a typePN, though its helium abundance issmaller than the minimum prescribed value and the sum of its nitrogenand oxygen abundances is low. The object is about 800above the galacticplane, is bipolar and shows no trace of molecular hydrogen. The chemicalcomposition of A 24, A 79, M 1-57, and Sh 1-89 indicates that thirddredge-up episodes occured in their progenitor star. Shock waves arepropagating in the ionized component of A 79, M 1-28 and NGC 2818, andpossibly A 14, A 24, and K 3-46. An emission arc in the periphery ofWray 16-22, a typePN, may be interpreted as a bow shock produced by thesupersonic motion of this object in the interstellar medium. Imaging and Spectroscopy of Type I Planetary NebulaeNot Available Morphological analysis of open clusters' propertiesII. Relationships projected onto the galactic planeA morphological analysis study of open clusters' properties has beenachieved for a sample of 160 UBVCCD open star clusters of approximately128,000 stars near the galactic plane. The data was obtained and reducedfrom using the same reduction procedures, which makes this catalogue thelargest homogeneous source of open clusters' parameters. Integrated photometric characteristics of galactic open star clustersIntegrated UBVRI photometric parameters of 140 galactic open clustershave been computed. Integrated I(V-R)0 and I(V-I)0colours as well as integrated parameters for 71 star clusters have beenobtained for the first time. These, in combination with published data,altogether 352 objects, are used to study the integrated photometriccharacteristics of the galactic open clusters. The I(MV)values range from -9.0 to -1.0 mag corresponding to a range in totalmass of the star clusters from ~ 25 to 4*E4 Msun.The integrated colours have a relatively narrow range, e.g., I(B-V){_0}varies from -0.4 to 1.2 mag. The scatter in integrated colours at agiven integrated magnitude can be understood in terms of differences infraction of red giants/supergiants in the clusters. The observedintegrated magnitudes and colours agree with the synthetic ones, exceptthe dependences of I(V-R)0 and I(V-I)0 colours forclusters younger than ~ 100 Myrs and also of the integrated magnitudesof oldest clusters. The large sample provides the most accurate agedependence of integrated magnitudes and colours determined so far. Theluminosity function of the I(MV) has a peak around -3.5 magand its slope indicates that only ~ 1% of the open clusters in thegalactic disc are brighter than I(MV)=-11 mag. No variationhas been found of integrated magnitude with galactocentric distance andmetallicity. 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 The Extreme Type I Planetary Nebula M2-52High-resolution spectrophotometric data of the central zone of theBr-type planetary nebula M2-52 are presented. The nebula has a richspectrum, with high and low excitation lines. The chemical compositionderived from the spectra shows that He and N are very enhanced in M2-52.Thus, this object can be classified as an extreme Peimbert's Type I PN.The chemical composition of the ionized gas is: He/H = 0.165+/-0.010,O/H = (2.6+/-0.5)×10-4, N/O = 2.3+/-0.3, Ne/O =0.37+/-0.10, Ar/O = (9.2+/-2.0)×10-3 and S/O >2.0×10-3. The expansion velocity of the nebula is, onaverage, about 20+/-2 km s-1, but the low ionization species (N+ and S+)seem to systematically show slightly lower expansion velocities (18 kms-1) than O+ + and He+ which have v[exp] = 20 km s-1, whileH+ and He++ have v[exp] ~ 22 km s-1. This behavior couldindicate that the outer zones of the ionized gas are being deceleratedby the molecular ring located around the central star. A possible observational measure of evolution in bipolar nebulaeBipolar planetary nebulae (BPNe) possess a broad range of shapes,ranging from narrow-waisted butterfly-like structures, through to thosewhich are more nearly cylindrical. We point out that these morphologiesappear to be correlated with radio surface brightness Tb, inthe sense that higher values of Tb are associated withenvelopes having narrower waists. If one interprets the variation inTb as arising from shell evolution, as is usually assumed forother planetaries, this would then imply that shell morphology varieswith time in a manner which appears not to be consistent withevolutionary models. It also remains possible, however, that differentBPNe morphologies arise as a result of differing mechanisms offormation, and that the pre-collimation of high velocity central windsgives rise to narrower waists, and higher surface brightness nuclei. Ourresults, if this is true, may then imply that central winds possess abroad range of collimations. Membership, Binarity and Stellar Evolution of Red Giants in the Open Clusters NGC 2324, NGC 2818, NGC 3960 and NGC 6259Not Available Morphological analysis of open clusters' propertiesI. Properties' estimationsA sample of 160 UBVCCD observations of open star clusters near thegalactic plane has been studied, and a catalogue of their propertiesobtained. The main photometrical properties have been re-estimated selfconsistently and the results have been compared with those of Lynga[Lynga, G., 1987. Catalog of Open Cluster Data, 5th Edition, StellarData Centers, Observatoire de Strasbourg, France]. Red giants in open clusters. IX. NGC 2324, 2818, 3960 and 6259We present accurate radial velocities and photoelectric UBV photometryfor 73 and 57 red-giant candidates, respectively, in theintermediate-age open clusters NGC 2324, 2818, 3960 and 6259. These dataconfirm the membership of 47 stars, 12 of which (26%) are spectroscopicbinaries; three preliminary orbits have been determined in NGC 3960.From Washington photometry of 8 red giant members, the metallicity ofNGC 6259 is found to be [Fe/H] = 0.06 +/- 0.08. At the age of theseclusters, most of the red giants are observed in the core-helium (clump)burning phase, the general morphology of which is well reproduced bytheoretical models with convective overshooting. However, a number ofbona fide cluster giant members are found significantly to the red ofthe isochrones fitting the rest of the CMD of these and a few otherclusters. Some of these stars are binaries, but others seem to besingle. In either case, their red colours and/or low luminosities remainunexplained by current stellar evolution theory. Based on observationscollected with the Danish 1.54-m telescope at the European SouthernObservatory, La Silla (Chile) and at Cerro Tololo Inter-AmericanObservatory, National Optical Astronomy Observatories, operated by theAssociation of the Universities for Research in Astronomy, Inc., undercontract with the National Science Foundation. Bipolar Nebulae: The Missing PopulationWe have undertaken an analysis of observed aspect ratios for thecollimation disks in bipolar nebulae (BPNe). We find, as a result, thatmost such structures are likely to have a ratio of thickness to diameterb/a~0.23. Similarly, the paucity of disks having aspect ratios >=0.7implies that many of the sources are either unobserved or, what is moreprobably the case, misidentified. It seems likely, as a result, that thetrue population of BPNe is ~1.7 times greater than previously supposed.Similarly, we determine that as many as of 43% of annular or ringlikeplanetaries may represent bipolar sources oriented along the line ofsight. The low-excitation structures of planetary nebulaeThe low excitation properties of the planetary nebula (PN) NGC 6720 areknown to be unusual, and to imply large ring/core emission ratios. Wepoint out that such characteristics are by no means confined to thissource alone, and that high ratios may occur in a large fraction ofelliptical and circular PNe. Such trends may arise because of thepresence of thin low-excitation emission sheets `wrapped' within andaround the primary outflows. The widths of such shells are required tobe exceedingly small, and may (for certain cases) be of order<<10-2pc. Such a mechanism appears capable ofexplaining most of the observed emission properties, and may arisethrough shock interaction between differing envelopes. Alternativeexplanations in terms of bipolar or cylindrical outflows are shown to beimplausible.
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# Find the area of the region in the first quadrant enclosed by

Question:

Find the area of the region in the first quadrant enclosed by $x$-axis, line $x=\sqrt{3} y$ and the circle $x^{2}+y^{2}=4$

Solution:

The area of the region bounded by the circle, $x^{2}+y^{2}=4, x=\sqrt{3} y$, and the $x$-axis is the area $\mathrm{OAB}$.

The point of intersection of the line and the circle in the first quadrant is $(\sqrt{3}, 1)$.

Area $\mathrm{OAB}=$ Area $\triangle \mathrm{OCA}+$ Area $\mathrm{ACB}$

Area of $O A C=\frac{1}{2} \times O C \times A C=\frac{1}{2} \times \sqrt{3} \times 1=\frac{\sqrt{3}}{2}$         ...(1)

Area of $\mathrm{ABC}=\int_{\sqrt{3}}^{2} y d x$

$=\int_{\sqrt{3}}^{2} \sqrt{4-x^{2}} d x$

$=\left[\frac{x}{2} \sqrt{4-x^{2}}+\frac{4}{2} \sin ^{-1} \frac{x}{2}\right]_{\sqrt{3}}^{2}$

$=\left[2 \times \frac{\pi}{2}-\frac{\sqrt{3}}{2} \sqrt{4-3}-2 \sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)\right]$

$=\left[\pi-\frac{\sqrt{3} \pi}{2}-2\left(\frac{-}{3}\right)\right]$

$=\left[\pi-\frac{\sqrt{3}}{2}-\frac{2 \pi}{3}\right]$

$=\left[\frac{\pi}{3}-\frac{\sqrt{3}}{2}\right]$                            ...(2)

Therefore, required area enclosed $=\frac{\sqrt{3}}{2}+\frac{\pi}{3}-\frac{\sqrt{3}}{2}=\frac{\pi}{3}$ square units

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# AMS Sectional

## Structure Theory of CI-Groupoids of Bol-Moufang Type

A talk about groupoid classification at the Fall 2013 AMS Central Section Meeting.

## CSP for Commutative, Idempotent Groupoids

A talk about groupoids and the CSP at the Fall 2013 AMS Western Section Meeting.


text:
# I have a continious function f that is strictly increasing. And a continious function g

I have a continious function $f$ that is strictly increasing. And a continious function $g$ that is strictly decreasing. How to I rigorously prove that $f\left(x\right)=g\left(x\right)$ has a unique solution?

Intuitively, I understand that if I take limits to infinity, then $f$ grows really large and $g$ grows very small so the difference is less than zero. If I take the limits to negative infinity then the opposite happens. Using the intermediate value theorem, there must be an intersection, and since they are strictly increasing/decreasing, only one intersection happens.

My question is how do I apply the intermediate value theorem here? I don't have the interval to apply it on. I don't know when $f$ crosses $g$ and therefore can't take any interval. Or is there some sort of axiom applied here that I am missing?
You can still ask an expert for help

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Johnathan Morse
This is not necessarily true, take $f\left(x\right)={e}^{x}$ and $g\left(x\right)=-{e}^{x}$ then $f\left(x\right)=g\left(x\right)$ has no solution, if $f\left(x\right)=g\left(x\right)$ has a solution then it's unique!

Say $f\left(x\right)=g\left(x\right)$ has a solution

Consider the function $h\left(x\right)=f\left(x\right)-g\left(x\right)$, $h$ is continuous, and clearly $h$ is strictly incresing, so $h\left(x\right)=0$ cannot have more than 1 real roots as it would violate that it's strictly increasing.

(As if $h\left(x\right)=h\left(y\right)$ for $x\ne y$ then either $x>y$ or $y>x$ so either $f\left(x\right)>f\left(y\right)$ or $f\left(y\right)>f\left(x\right)$, anyway we get a contradiction! )

Hence if $f\left(x\right)=g\left(x\right)$ has a solution then it's unique!


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# Permutations and Combinations Questions

Q:

How many integers, greater than 999 but not greater than 4000, can be formed with the digits 0, 1, 2, 3 and 4, if repetition of digits is allowed?

 A) 376 B) 375 C) 500 D) 673

Explanation:

The smallest number in the series is 1000, a 4-digit number.

The largest number in the series is 4000, the only 4-digit number to start with 4.

The left most digit (thousands place) of each of the 4 digit numbers other than 4000 can take one of the 3 values 1 or 2 or 3.

The next 3 digits (hundreds, tens and units place) can take any of the 5 values 0 or 1 or 2 or 3 or 4.

Hence, there are 3 x 5 x 5 x 5 or 375 numbers from 1000 to 3999.

Including 4000, there will be 376 such numbers.

29 11823
Q:

Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?

 A) 25200 B) 52000 C) 120 D) 24400

Explanation:

Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4)

= $\inline&space;{\color{Black}(7C_3{}\times&space;4C_2{})}$

= 210.

Number of groups, each having 3 consonants and 2 vowels = 210.

Each group contains 5 letters.

Number of ways of arranging 5 letters among themselves

= 5!

= 120

Required number of ways = (210 x 120) = 25200.

11 11382
Q:

A committee of 5 persons is to be formed from 6 men and 4 women. In how many ways can this be done when at least 2 women are included ?

 A) 196 B) 186 C) 190 D) 200

Explanation:

When at least 2 women are included.

The committee may consist of 3 women, 2 men : It can be done in$\inline&space;\fn_jvn&space;{\color{Black}&space;4C_{3}\times&space;6C_{2}}$  ways

or, 4 women, 1 man : It can be done in $\inline&space;\fn_jvn&space;{\color{Black}&space;4C_{4}\times&space;6C_{1}}$ ways

or, 2 women, 3 men : It can be done in $\inline&space;\fn_jvn&space;{\color{Black}&space;4C_{2}\times&space;6C_{3}}$ ways.

Total number of ways of forming the committees

= $\inline&space;\fn_jvn&space;{\color{Black}&space;4C_{2}\times&space;6C_{3}+4C_{3}\times&space;6C_{2}+4C_{4}\times&space;6C_{1}}$

= 6 x 20 + 4 x 15 + 1x 6

= 120 + 60 + 6 =186

10 8573
Q:

How many 4-letter words with or without meaning, can be formed out of the letters of the word, 'LOGARITHMS', if repetition of letters is not allowed?

 A) 4050 B) 3600 C) 1200 D) 5040

Explanation:

'LOGARITHMS' contains 10 different letters.

Required number of words = Number of arrangements of 10 letters, taking 4 at a time.

= $\inline&space;{\color{Black}10P_4{}}$

= 5040.

17 7535
Q:

If the letters of the word SACHIN are arranged in all possible ways and these words are written out as in dictionary, then the word ‘SACHIN’ appears at serial number :

 A) 601 B) 600 C) 603 D) 602

Explanation:

If the word started with the letter A then the remaining 5 positions can be filled in  5! Ways.

If it started with c then the remaining 5 positions can be filled in 5! Ways.Similarly if it started with H,I,N the remaining 5 positions can be filled in 5! Ways.

If it started with S then the remaining position can be filled with A,C,H,I,N in alphabetical order as on dictionary.

The required word SACHIN can be obtained after the 5X5!=600 Ways i.e. SACHIN is the 601th letter.


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# showing that a regular language is regular after taking a letter off or after adding letters

I'll be happy to recieve help with this one:

Given the regular language $L$ defined over alphabet $\{a,b\}$, show that the following languages are also regular:

• $\{xy\mid xay\in L\}$ (meaning, taking one $a$ of the word in the middle)
• $\{x(a^*)y\mid xy\in L\}$ (meaning, adding in the middle of the word as much $a$'s as wanted)

-
If the question has been answered to your satisfaction, could you please indicate this by accepting an answer? (And preferably upvoting all answers you found helpful, as well.) –  Tara B Nov 17 '12 at 15:41

For the second problem, start with a right regular grammar for $L$. For each non-terminal symbol $X$ of $L$ add two new non-terminal symbols, $X'$ and $X''$. For each production of the form $X\to\alpha Y$ add the productions $X\to\alpha Y'$, $Y'\to aY'$, $Y'\to aY''$, and $X''\to\alpha Y''$. For each production of the form $X\to\alpha$ add the production $X''\to\alpha$. (Here $\alpha\in\{a,b\}$.) The idea is that the grammar starts out using the original non-terminals to generate the $x$ part of $xa^*y$, switches to one of the single-prime non-terminals to generate the $a^*$ part, and then generates the $y$ part using the double-prime non-terminals. It’s set up so that once it uses a primed variable, it can’t generate any of the original non-terminals, so it can generate an extra string of $a$’s only once.

You can use a similar idea for the first problem, though the details are a bit messier, and it’s easier to work with automata. Start with a DFA $M$ that accepts $L$; we’ll build an NFA $N$ that accepts the modified language. Start with two disjoint copies of $M$, say $M_1$ and $M_2$. The initial state of $N$ will be the initial state of $M_1$, and the acceptor states of $N$ will be the acceptor states of $M_2$. $M_1$ and $M_2$ will keep their original transitions. Finally, for every $a$-transition from a state $q_1$ to a state $q_2$ in $M_1$ we need an $\epsilon$-transition from $q_1$ in $M_1$ to the copy of $q_2$ in $M_2$. The idea is that we start in $M_1$ and proceed normally, except that at some point we follow one of those new $\epsilon$-transitions, simulating reading an $a$ and shifting over to $M_2$; then we continue just as if we were in the original $M$. The shift from $M_1$ to $M_2$ can happen only once, so we can delete only one $a$, and I’ve arranged matters so that we must delete one $a$ in order for the word to be accepted; how?

-
I don't study this material in english - what is a non-terminal symbol? thanks for the reply –  DanielY Nov 15 '12 at 21:49
@user1067083: The non-terminal symbols are the ones that are not part of the input alphabet. This article may help, even though it’s in English. –  Brian M. Scott Nov 15 '12 at 21:50
thanks alot for the answer for the first problem it was great!!! can you think of a NFA solution for the second as well? –  DanielY Nov 15 '12 at 21:52
add a state with a self $a$ transition in the middle of the $\varepsilon$-transitions you added –  Vincent Nivoliers Nov 15 '12 at 21:56
it's more than that I think, Vincent, because those a's can be added at any point in the word –  DanielY Nov 15 '12 at 21:59

Hint : imagine an automaton which recognizes the language L. Take a copy of this automaton, put it aside the original, add transitions between the two and select the right input and output nodes.

-

first part) Since L is a regular language, we can find a regular expression for it, and since we know that some of strings in L has symbol "a", the regular expression of L would be something like R1.a.R2+R3 where R1, R2 and R3 (R3 are the strings without "a") are regular expressions. since R1 and R2 are regular expressions, their concatenation would also be a regular expression. L(R1.R2)={xy∣xay∈L} so we could find a regular expression for {xy∣xay∈L}.

Second part can be answered similarly.

-
Wont work that easily. We might also want to insert a single $a$ in parts that are deep inside nested stars. –  Hendrik Jan Nov 16 '12 at 13:12


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# Determine Projectile Motion velocity

1. Oct 1, 2005

### dekoi

If the velocity of a projectile is known at t=1second and the gravitational force is unknown, how does one determine the velocity at t=0, 2, and 3 seconds.

The projectile does not start from rest.

If i use the formula: vfy = viy - gt, i can't solve for v, since the value for g is unknown.

Thank you .

2. Oct 1, 2005

### hotvette

Use the information you have. How can knowing the velocity at t=1 sec help? Hint: use your equation of motion.

3. Oct 2, 2005

### dekoi

Which equation for motion?

I still do not understand.

4. Oct 2, 2005

### Grogs

You're sure this is correct and you've listed everything that's been given to you (vy1)? If that's the case, you can't solve for vy0, vy2, etc. numerically. The best you could do is solve for them in terms of vy0 . For example:

$$v_{y1} = v_ {y0} + g't \Longrightarrow g' = \frac{v_{y1}-v_{y0}}{t}$$

You could then put vy2 and vy3 in terms of vy0, but that's as far as you could go. If you had one other piece of information you could solve for g' numerically and use it to solve for the other velocities as well.

5. Oct 2, 2005

### dekoi

The only problem with that is that in the second part of the question, I have to find the value of "g", so there must be a way to do it. Here is the exact question:

A physics student on Planet Exidor throws a ball, and it follows a parabolic trajectory. The ball's position is shown at 1 s intervals until t = 3 s. At t = 1 s, the ball's velocity is v = (2.0i + 2.0j) m/s. (i and j being unit vectors for x and y)
a) Determine the ball's velocity at t = 0s, 2s, and 3s.
b) What is the value of g on Planet Exidor?
c) What was the ball's launch angle?

6. Oct 2, 2005

### Päällikkö

Does the problem have a graph/figure ("The ball's position is shown" implies it does)? Can you see the height from that?

7. Oct 2, 2005

### dekoi

It does not land at 3 seconds; the graph only shows up until 3 seconds.

The graph is attached.

File size:
6.1 KB
Views:
475
8. Oct 2, 2005

### Päällikkö

$$v_y = v_{y0} - gt$$
What do you know at t = 1s and t = 2s?

9. Oct 2, 2005

### dekoi

At t = 1s, vx = 2.0 m/s and vy = 2.0 m/s.
At t = 2s, vx = 2.0 m/s and vy = (2.0 - g) m/s

10. Oct 2, 2005

### Fermat

From your graph, you can see that the total time of flight is 4s.
So the time to reach max height = 2s
You also know the vertical velocity at t = 1.
By symmetry the velocity at t = 3 will be the same as at t= 1, but in the opposite direction.
Now you can use your eqns of motion to solve for g, etc.

11. Oct 2, 2005

### Grogs

{quote snipped for clarity}

This answer is correct; however, by looking at your graph I can tell you the *exact* numerical value of v2y. Why is that?

12. Oct 2, 2005

### dekoi

Is this correct?

v2y= 0 m/s
therefore
0 m/s = 2.0 - g
g = 2.0 m/s2

v0 = (2.0i + 4.0j) m/s
v1 = (2.0i + 2.0j) m/s
v2 = (2.0i) m/s
v3 = (2.0i - 2.0j) m/s

v0 = sqrt(2.0^2 + 4.0^2) = sqrt(4 + 16) = sqrt20 m/s

angle = arccos(v0x / v0) = arccos(2 / sqrt20) = 1.1 degrees

I am not sure about the angle.

13. Oct 2, 2005

### dekoi

Any responses?

14. Oct 2, 2005


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# Homework Help: Measure space, null set

1. May 24, 2010

### complexnumber

1. The problem statement, all variables and given/known data

Let $$(X,\mathcal{A},\mu)$$ be a fixed measure space.

Let $$A_k \in \mathcal{A}$$ such that $$\displaystyle \sum^\infty_{k=1} \mu(A_k) < \infty$$. Prove that
\begin{align*} \{ x \in X | x \in A_k \text{ for infinitely many k} \} \end{align*}
is a null set.

2. Relevant equations

3. The attempt at a solution

Let $$S = \{ x \in X | x \in A_k \text{ for infinitely many k} \}$$.
Suppose $$\mu (S) > 0$$. Then $$\displaystyle \mu(\bigcap A_k) > 0, A_k \ni x, x \in S$$. Then $$\mu (A_k) > 0, A_k \ni x, x \in S$$ and hence
$$\displaystyle \sum^\infty_{k=1} \mu(A_k) = \infty$$, which


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How do I label images faster

I have around 1600 images extracted from videos shot at night time. I am labeling each image and trying to be as accurate as I can in assigning bounding boxes. I am labeling vehicles and traffic light/traffic signs. This is very time-consuming, I am wondering if someone have experience or done this before and can advise me on some automated methods of labeling night time images. The objects of interests are usually appear to be quite small. An example labelled image is given below

• use amazon mechanical turk perhaps – Van Peer May 26 '20 at 5:58

On the internet, LabelImg is pretty popular. It's a Python program that you can use to automatize drawing of bounding boxes. Alternatively there is HyperLabel, on their website they say it's free so it.s worth to give it a try.

The options available are so many I completely get lost while searching. I suggest you to read this nice review of tools first.

With these tools, labeling 1600 images is not going to be a problem IMHO. Good luck.

• I find all these tools are for manually labeling images, I have used LabelImg. You are right 1600 images are not that much of work, but I am getting more and more image data. Therefore, I do not want to stick to manual labeling, so specifically my question is how can I automate image labeling process? – Vendetta May 26 '20 at 7:02

If you're drawing bounding boxes around cars, you can use a pretrained model. Hundreds of pretrained networks are trained on the COCO dataset, which has a car category. Then you will get the labels as the output.

Then you can also double check with LabelImg by iterating through the folder with the keyboard arrows, and adjust the bounding boxes if they are off. But they shouldn't. Cars (and people) are maybe the most cropped category in the world. Use it to your advantage.

• Could you please give more information on "you can use a pretrained model"? How exactly I can use a pretrained model to obtain labels as the outputs. Can you please provide a link to some example where it has been done? – Vendetta May 26 '20 at 11:45
• check for transfer learning where you can use a pretrained model to get the labels for your own images. – ARAT Feb 23 at 19:20


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## Main result

We now prove that a left inverse of a square matrix is also a right inverse. In other words, we show the following:

Let $$A, N \in \mathbb{F}^{n\times n}$$ where $$\mathbb{F}$$ denotes a field. If $$NA = I$$, then $$AN = I$$.

Before we look at the proof, note that the above statement also establishes that a right inverse is also a left inverse because we can view $$A$$ as the right inverse of $$N$$ (as $$NA = I$$) and the conclusion asserts that $$A$$ is a left inverse of $$N$$ (as $$AN = I$$).

To prove the above statement, we first establish the claim that $$Ax = y$$ has a solution for all $$y \in \mathbb{R}^n$$.

The claim is not true if $$A$$ does not have a left inverse. We postpone the proof of this claim to the end. Let's see how we can use this claim to prove the main result.

Take an arbitrary element in $$\mathbb{F}^n$$ and call it $$y$$. Then the above result tells us that there is $$x' \in \mathbb{F}$$ such that $$Ax' = y$$. Multiplying both sides on the left by $$N$$, we get that $$N(Ax') = Ny$$, giving $$(NA)x' = Ny$$ by associativity of matrix multiplication. As $$NA = I$$, we have $$x' = Ny$$.

Hence, $$y = Ax' = A(Ny) = (AN)y$$. Let $$D$$ denote the product $$AN$$. So $$y = Dy$$. But $$y$$ is arbitrary. We must have $$D = I$$.

### Proof of claim

Suppose that there exists $$y' \in \mathbb{F}^n$$ such that $$Ax = y'$$ has no solution. Now, row reduce $$[A~y']$$ to $$[R~d]$$ where $$R$$ is in reduced row-echelon form.

As $$Ax = y'$$ has no solution, there must be an $$i$$ such that row $$i$$ of $$R$$ has all 0's and $$d_i \neq 0$$. Hence, because $$R$$ is a square matrix, not every column of $$R$$ can be a pivot column. So there is at least one free variable, implying that there is a nonzero $$\tilde{x} \in N(A)$$ (i.e., $$Ax' = 0_n$$). But $$\tilde{x} = I \tilde{x} = (NA)\tilde{x} = N(A\tilde{x}) = N 0_n = 0_n$$, contradicting that $$\tilde{x}$$ is nonzero! So the assumption that there exists $$y' \in \mathbb{F}^n$$ such that $$Ax = y'$$ has no solution cannot be true.

## Exercise

Let $$A = \begin{bmatrix} 2 & 0 \\ -1 & 0 \\ 1 & 1\end{bmatrix}$$ and let $$N = \begin{bmatrix} 1 & 1 & 0\\ -1 & -1 & 1 \end{bmatrix}$$. Compute the products $$NA$$ and $$AN$$. Do the answers contradict our main result? If not, why not?


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# botorch.test_functions¶

Synthetic test functions for testing and benchmarking purposes.

botorch.test_functions.neg_branin(X)[source]

Negative Branin test function.

Two-dimensional function (usually evaluated on [-5, 10] x [0, 15]):

B(x) = (x_2 - b x_1^2 + c x_1 - r)^2 + 10 (1-t) cos(x_1) + 10

Here b, c, r and t are constants where b = 5.1 / (4 * math.pi ** 2) c = 5 / math.pi, r = 6, t = 1 / (8 * math.pi) B has 3 minimizers for its global minimum at

z_1 = (-pi, 12.275), z_2 = (pi, 2.275), z_3 = (9.42478, 2.475)

with B(z_i) = 0.397887

Parameters

X (Tensor) – A Tensor of size 2 or k x 2 (k batch evaluations).

Return type

Tensor

Returns

-B(X), the negative value of the standard Branin function.

botorch.test_functions.neg_eggholder(X)[source]

Negative Eggholder test function.

Two-dimensional function (usually evaluated on [-512, 512]^2):

E(x) = (x_2 + 47) sin(R1(x)) - x_1 * sin(R2(x)) R1(x) = sqrt(|x_2 + x_1 / 2 + 47|) R2(x) = sqrt(|x_1 - (x_2 + 47)|)

Parameters

X (Tensor) – A Tensor of size 2 or k x 2 (k batch evaluations).

Return type

Tensor

Returns

-E(X), the negative value of the Eggholder function.

botorch.test_functions.neg_hartmann6(X)[source]

Negative Hartmann6 test function.

Six-dimensional function (typically evaluated on [0, 1]^6)

H(x) = - sum_{i=1}^4 ALPHA_i exp( - sum_{j=1}^6 A_ij (x_j - P_ij)**2 )

H has a 6 local minima and a global minimum at

z = (0.20169, 0.150011, 0.476874, 0.275332, 0.311652, 0.6573)

with H(z) = -3.32237

Parameters

X (Tensor) – A Tensor of size 6 or k x 6 (k batch evaluations).

Return type

Tensor

Returns

-H(X), the negative value of the standard Hartmann6 function.

botorch.test_functions.neg_holder_table(X)[source]

Negative Holder Table synthetic test function.

Two-dimensional function (typically evaluated on [0, 10] x [0, 10]):

H(x) = - | sin(x_1) * cos(x_2) * exp(| 1 - ||x|| / pi | ) |

H has 4 global minima with H(z_i) = -19.2085 at

z_1 = ( 8.05502, 9.66459) z_2 = (-8.05502, -9.66459) z_3 = (-8.05502, 9.66459) z_4 = ( 8.05502, -9.66459)

Parameters

X (Tensor) – A Tensor of size 2 or k x 2 (k batch evaluations).

Return type

Tensor

Returns

-H(X), the negative value of the standard Holder Table function.

botorch.test_functions.neg_michalewicz(X)[source]

Negative 10-dim Michalewicz test function.

10-dim function (usually evaluated on hypercube [0, pi]^10):

M(x) = sum_{i=1}^10 sin(x_i) (sin(i x_i^2 / pi)^20)

Parameters

X (Tensor) – A Tensor of size 10 or k x 10 (k batch evaluations).

Return type

Tensor

Returns

-M(X), the negative value of the Michalewicz function.

botorch.test_functions.neg_styblinski_tang(X)[source]

Negative Styblinski-Tang test function.

d-dimensional function (usually evaluated on the hypercube [-5, 5]^d):

H(x) = 0.5 * sum_{i=1}^d (x_i^4 - 16 * x_i^2 + 5 * x_i)

H has a single global mininimum H(z) = -39.166166 * d at z = [-2.903534]^d

Parameters

X (Tensor) – A Tensor of size d or k x d (k batch evaluations)

Return type

Tensor

Returns

-H(X), the negative value of the standard Styblinski-Tang function.

botorch.test_functions.neg_aug_branin(X)[source]

Negative augmented-Branin test function.

The last dimension of X is the fidelity. 3-dimensional function with domain [-5, 10] x [0, 15] * [0,1]:

B(x) = (x_2 - (b - 0.1 * (1 - x_3))x_1^2 + c x_1 - r)^2 +

10 (1-t) cos(x_1) + 10

Here b, c, r and t are constants where b = 5.1 / (4 * math.pi ** 2) c = 5 / math.pi, r = 6, t = 1 / (8 * math.pi). B has infinitely many minimizers with x_1 = -pi, pi, 3pi

and B_min = 0.397887

Parameters

X (Tensor) – A Tensor of size 3 or k x 3 (k batch evaluations).

Return type

Tensor

Returns

-B(X), the negative value of the augmented-Branin function.

botorch.test_functions.neg_aug_hartmann6(X)[source]

Negative augmented Hartmann6 test function.

The last dimension of X is the fidelity parameter. 7-dimensional function (typically evaluated on [0, 1]^7):

H(x) = -(ALPHA_1 - 0.1 * (1-x_7)) * exp(- sum_{j=1}^6 A_1j (x_j - P_1j) ** 2) -

sum_{i=2}^4 ALPHA_i exp( - sum_{j=1}^6 A_ij (x_j - P_ij) ** 2)

H has unique global minimizer x = [0.20169, 0.150011, 0.476874, 0.275332, 0.311652, 0.6573, 1]

with H_min = -3.32237

Parameters

X (Tensor) – A Tensor of size 7 or k x 7 (k batch evaluations).

Return type

Tensor

Returns

-H(X), the negative value of the augmented Hartmann6 function.

botorch.test_functions.neg_aug_rosenbrock(X)[source]

Augmented-Rosenbrock test function.

d-dimensional function (usually evaluated on [-5, 10]^(d-2) * [0, 1]^2), the last two dimensions are the fidelity parameters:

f(x) = sum_{i=1}^{d-1} (100 (x_{i+1} - x_i^2 + 0.1 * (1-x_{d-1}))^2 +

(x_i - 1 + 0.1 * (1 - x_d)^2)^2)

f has one minimizer for its global minimum at

z_1 = (1, 1, …, 1)

with f(z_i) = 0.0

Parameters

X (Tensor) – A Tensor of size d or k x d (k batch evaluations).

Returns

-f(X), the negative value of the Augmented-Rosenbrock function.


text:
# What is the term for this family of improper integrals?

What is the name of the integrals of this form?

$$\int_{0}^{\infty} \frac{\sin\left(\frac{x}{1}\right)\sin\left(\frac{x}{3}\right)\cdots\sin\left(\frac{x}{2n + 1}\right)}{\left(\frac{x}{1}\right)\left(\frac{x}{3}\right)\cdots\left(\frac{x}{2n + 1}\right)} \:dx$$

• @Travis - Thanks for the edit too :-) – user150203 Nov 14 '18 at 1:33

## 1 Answer

These are examples of Borwein integrals. See:

Borwein, D.; Borwein, J.M., "Some Remarkable Properties of $$\operatorname{Sinc}$$ and Related Integrals", Ramanujan J. 5 (2001), 73-89.

John Baez' Azimuth blog has an illuminating post discussing these integrals, and Greg Egan gave an intuitive explanation in terms of Fourier transforms for the pattern-breaking phenomenon often mentioned when these integrals come up.

• Thanks @Travis! – user150203 Nov 14 '18 at 0:42
• Cheers!$\!\!\!$ – Travis Nov 14 '18 at 0:47


text:
The suspense in the first chapter alone is enough to get hooked | BookishFirst

# The suspense in the first chapter alone is enough to get hooked

The title and cover of course made me think of Rapunzel, although I'm not sure this will be an outright retelling/reimagining of that fairy tale. The cover is simple but pleasing to the eye and captures your attention. The synopsis was interesting but didn't get me excited about the book. However, the first preview of the book had me hooked in chapter one. It is first person alternating view, which is something I tend to dislike but if there is a time it works it's with suspense. And this book promises suspense!


text:
### Home > CCA > Chapter 9 > Lesson 9.2.1 > Problem9-55

9-55.

For each parabola graphed below, visually estimate the $x$-intercepts. Then use the Quadratic Formula to confirm your estimates.

$\text{Quadratic Formula} =\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

1.  $y=x^2−5x+3$

$x\approx0.7$ or $4.3$

1.  $y=x^2+2x−6$

See part (a).


text:
# how to draw squares, circles and triangles [duplicate]

I'd like to explain the markers of a figure in the caption by drawing small marker symbols. I have seen in a previous question that you can do this with the tikz package.

I am using

\usepackage{tikz}

\newcommand{\square}[1]{\tikz{\filldraw[draw=#1,fill=#1] (0,0) rectangle (0.2cm,0.2cm);}}

and then \square{red} in the text when I want to draw a red square.

Could anyone show me how can I do the same for circles and triangles?

Thank you

-

## marked as duplicate by LaRiFaRi, Pouya, Thorsten, Masroor, T. VerronJul 8 '14 at 11:47

You can also use the picture environment. Or enhaced by pict2e package. – Manuel Jul 8 '14 at 11:12

If you just want something in same style, without using the useful node shapes of tikz then you can do the following. I have updated the code so it works in captions too, the replacement is to use \newrobustcmd from the etoolbox package instead of \newcommand:

\documentclass{article}

\usepackage{etoolbox}
\usepackage{tikz}

\newrobustcmd*{\square}[1]{\tikz{\filldraw[draw=#1,fill=#1] (0,0)
rectangle (0.2cm,0.2cm);}}

\newrobustcmd*{\mytriangle}[1]{\tikz{\filldraw[draw=#1,fill=#1] (0,0) --
(0.2cm,0) -- (0.1cm,0.2cm);}}

\begin{document}

\listoffigures

\section{Sample}
A square \square{red}, a circle \mycircle{green} and a triangle \mytriangle{blue}.

\begin{figure}[htp]
\centering
A figure
\caption{A square \square{red}, a circle \mycircle{green} and a triangle \mytriangle{blue}.}
\end{figure}

\end{document}

An alternative using node shape is as follows. Whilst there are more options to set, this is countered by more predefined shapes to choose between. See the pgf manual:

\documentclass{article}

\usepackage{tikz}
\usetikzlibrary{shapes}

\newcommand{\square}[1]{\tikz{\node[draw=#1,fill=#1,rectangle,minimum
width=0.2cm,minimum height=0.2cm,inner sep=0pt] at (0,0) {};}}

\newcommand{\mycircle}[1]{\tikz{\node[draw=#1,fill=#1,circle,minimum
width=0.2cm,minimum height=0.2cm,inner sep=0pt] at (0,0) {};}}

\newcommand{\mytriangle}[1]{\tikz{\node[draw=#1,fill=#1,isosceles
triangle,isosceles triangle stretches,shape border rotate=90,minimum
width=0.2cm,minimum height=0.2cm,inner sep=0pt] at (0,0) {};}}

\begin{document}

A square \square{red}, a circle \mycircle{green} and a triangle \mytriangle{blue}.

\end{document}

-
This works great in the main text, but it does not compile in the figure captions. Do you have a suggestion what the problem may be? Message: the use of \tikz@picture doesn't match it's definition. – Adam Gosztolai Jul 8 '14 at 11:51
@AdamGosztolai That is a robustness issue. I have updated my solution to take that in to account. – Andrew Swann Jul 8 '14 at 12:03
Thank you Andrew. The robustness issue is sorted now. I found that an alternative is to put an extra argument \caption[]{}. However the, \mycircle command still does not work. It says that it cannot parse this radius. It doesn't work for bigger radii either. – Adam Gosztolai Jul 8 '14 at 12:15
The extra argument, means that nothing will be written to the list of figures, which may be OK for you. The code above compiles fine for me - maybe you a have clashing definition of \mycircle lying around, the error messages can sometimes be rather cryptic; or may be you have slightly mistyped the code. – Andrew Swann Jul 8 '14 at 12:44

There are predefined mark symbols in the plotmarks library. Here are some examples

Code:

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{plotmarks}
\newcommand\marksymbol[2]{\tikz[#2,scale=1.2]\pgfuseplotmark{#1};}

\begin{document}
\foreach \symb/\col in {square/red,triangle/blue,diamond/green,oplus/blue!50!red}{
\symb: \marksymbol{\symb}{\col} and \marksymbol{\symb*}{\col}
\par}
%
circle: \marksymbol{o}{yellow} and \marksymbol{*}{yellow}
\end{document}

Edit

Inside captions you have to use \protect in front of this command.

Code:

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{plotmarks}
\newcommand\marksymbol[2]{\tikz[#2,scale=1.2]\pgfuseplotmark{#1};}

\begin{document}
\listoffigures

\section{Symbols}
\begin{figure}[ht]
\centering \fbox{figure}
\caption{\protect\marksymbol{heart}{blue} caption}
\end{figure}
\end{document}

If there is no \listoffigures you can alternativly use an empty optinal argument of \caption:

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{plotmarks}
\newcommand\marksymbol[2]{\tikz[#2,scale=1.2]\pgfuseplotmark{#1};}

\begin{document}
\section{Symbols}
\begin{figure}[ht]
\centering \fbox{figure}
\caption[]{\marksymbol{heart}{blue} caption}
\end{figure}
\end{document}

Or you can use \newrobustcmd from the etoolbox package like @Andrew Swann in his answer suggested.

-
This works great in the main text, but it does not work in the captions. It does not compile. – Adam Gosztolai Jul 8 '14 at 11:44

You can draw a circle in tikz using

\draw[<draw options>] (0,0) circle (1cm);

\newcommand{\circle}[1]{\tikz{\filldraw[draw=#1,fill=#1] (0,0) circle (0.2cm);}}

should do the job.

To draw a triangle, you'll have to define and connect 3 points like the following code does

\draw[<draw options>] (0,0) -- (0,2) -- (2,0)-- (0,0);

\newcommand{\triangle}[1]{\tikz{\filldraw[draw=#1,fill=#1] (0,0) -- (0,.2cm) -- (0.2cm,0) -- (0,0);}}

-

As these are to appear in text, the only change I would make to Andrew Swann's approach would be to make the size of circle, triangle and square dependent on the surrounding type. That can be done by using the em as the unit of measure:

\documentclass[12pt]{article}

\usepackage{tikz}

\newcommand{\square}[1]{\tikz{\filldraw[draw=#1,fill=#1] (0,0)
rectangle (0.6em,0.6em);}}


text:
# 1.4 Polynomials

 Page 1 / 15
In this section students will:
• Identify the degree and leading coefficient of polynomials.
• Multiply polynomials.
• Use FOIL to multiply binomials.
• Perform operations with polynomials of several variables.

Earl is building a doghouse, whose front is in the shape of a square topped with a triangle. There will be a rectangular door through which the dog can enter and exit the house. Earl wants to find the area of the front of the doghouse so that he can purchase the correct amount of paint. Using the measurements of the front of the house, shown in [link] , we can create an expression that combines several variable terms, allowing us to solve this problem and others like it.

First find the area of the square in square feet.

$\begin{array}{ccc}\hfill A& =& {s}^{2}\hfill \\ & =& {\left(2x\right)}^{2}\hfill \\ & =& 4{x}^{2}\hfill \end{array}$

Then find the area of the triangle in square feet.

Next find the area of the rectangular door in square feet.

$\begin{array}{ccc}\hfill A& =& lw\hfill \\ & =& x\cdot 1\hfill \\ \hfill & =& x\hfill \end{array}$

The area of the front of the doghouse can be found by adding the areas of the square and the triangle, and then subtracting the area of the rectangle. When we do this, we get $\text{\hspace{0.17em}}4{x}^{2}+\frac{3}{2}x-x\text{\hspace{0.17em}}{\text{ft}}^{2},$ or $\text{\hspace{0.17em}}4{x}^{2}+\frac{1}{2}x\text{\hspace{0.17em}}$ ft 2 .

In this section, we will examine expressions such as this one, which combine several variable terms.

## Identifying the degree and leading coefficient of polynomials

The formula just found is an example of a polynomial    , which is a sum of or difference of terms, each consisting of a variable raised to a nonnegative integer power. A number multiplied by a variable raised to an exponent, such as $\text{\hspace{0.17em}}384\pi ,$ is known as a coefficient    . Coefficients can be positive, negative, or zero, and can be whole numbers, decimals, or fractions. Each product $\text{\hspace{0.17em}}{a}_{i}{x}^{i},$ such as $\text{\hspace{0.17em}}384\pi w,$ is a term of a polynomial    . If a term does not contain a variable, it is called a constant .

A polynomial containing only one term, such as $\text{\hspace{0.17em}}5{x}^{4},$ is called a monomial    . A polynomial containing two terms, such as $\text{\hspace{0.17em}}2x-9,$ is called a binomial    . A polynomial containing three terms, such as $\text{\hspace{0.17em}}-3{x}^{2}+8x-7,$ is called a trinomial    .

We can find the degree    of a polynomial by identifying the highest power of the variable that occurs in the polynomial. The term with the highest degree is called the leading term    because it is usually written first. The coefficient of the leading term is called the leading coefficient    . When a polynomial is written so that the powers are descending, we say that it is in standard form.

## Polynomials

A polynomial    is an expression that can be written in the form

${a}_{n}{x}^{n}+...+{a}_{2}{x}^{2}+{a}_{1}x+{a}_{0}$

Each real number a i is called a coefficient    . The number $\text{\hspace{0.17em}}{a}_{0}\text{\hspace{0.17em}}$ that is not multiplied by a variable is called a constant . Each product $\text{\hspace{0.17em}}{a}_{i}{x}^{i}\text{\hspace{0.17em}}$ is a term of a polynomial    . The highest power of the variable that occurs in the polynomial is called the degree    of a polynomial. The leading term    is the term with the highest power, and its coefficient is called the leading coefficient    .

Given a polynomial expression, identify the degree and leading coefficient .

1. Find the highest power of x to determine the degree.
2. Identify the term containing the highest power of x to find the leading term.
3. Identify the coefficient of the leading term.

what is linear equation with one unknown 2x+5=3
-4
Joel
x=-4
Joel
x=-1
Joan
I was wrong. I didn't move all constants to the right of the equation.
Joel
x=-1
Cristian
what is the VA Ha D R X int Y int of f(x) =x²+4x+4/x+2 f(x) =x³-1/x-1
can I get help with this?
Wayne
Are they two separate problems or are the two functions a system?
Wilson
Also, is the first x squared in "x+4x+4"
Wilson
x^2+4x+4?
Wilson
thank you
Wilson
Wilson
f(x)=x square-root 2 +2x+1 how to solve this value
Wilson
what is algebra
The product of two is 32. Find a function that represents the sum of their squares.
Paul
if theta =30degree so COS2 theta = 1- 10 square theta upon 1 + tan squared theta
how to compute this 1. g(1-x) 2. f(x-2) 3. g (-x-/5) 4. f (x)- g (x)
hi
John
hi
Grace
what sup friend
John
not much For functions, there are two conditions for a function to be the inverse function:   1--- g(f(x)) = x for all x in the domain of f     2---f(g(x)) = x for all x in the domain of g Notice in both cases you will get back to the  element that you started with, namely, x.
Grace
sin theta=3/4.prove that sec square theta barabar 1 + tan square theta by cosec square theta minus cos square theta
acha se dhek ke bata sin theta ke value
Ajay
sin theta ke ja gha sin square theta hoga
Ajay
I want to know trigonometry but I can't understand it anyone who can help
Yh
Idowu
which part of trig?
Nyemba
functions
Siyabonga
trigonometry
Ganapathi
differentiation doubhts
Ganapathi
hi
Ganapathi
hello
Brittany
Prove that 4sin50-3tan 50=1
False statement so you cannot prove it
Wilson
f(x)= 1 x    f(x)=1x  is shifted down 4 units and to the right 3 units.
f (x) = −3x + 5 and g (x) = x − 5 /−3
Sebit
what are real numbers
I want to know partial fraction Decomposition.
classes of function in mathematics
divide y2_8y2+5y2/y2
wish i knew calculus to understand what's going on 🙂
@dashawn ... in simple terms, a derivative is the tangent line of the function. which gives the rate of change at that instant. to calculate. given f(x)==ax^n. then f'(x)=n*ax^n-1 . hope that help.
Christopher
thanks bro
Dashawn
maybe when i start calculus in a few months i won't be that lost 😎
Dashawn


text:
# zbMATH — the first resource for mathematics

Wake-induced relative motion of bubbles rising in line. (English) Zbl 1135.76453
Summary: The motion of nearly spherical air bubbles rising in a column in stagnant water is measured at Reynolds numbers ranging from $$Re_{D} = 0$$.2 to 35. The relative velocity is found to be dependent on the distance between bubbles and on their diameter. For the larger bubbles, the relative velocities increased with decreasing distance, reaching maximum values just prior to contact. For the smaller bubbles, the relative velocity decreased prior to coalescence. For the entire range of Reynolds number considered, the wake-induced relative motion results in collisions between bubbles. These collisions culminate in coalescence at the present levels of water purity and surface tension. In order to understand the basic features of the measured relative motion, a simple model is developed. It is based on the known flow field and viscous-wake structure around a single bubble, and examines how other bubbles move within this field. Oseen flow for $$Re \ll 1$$ and potential flow with a thin wake for $$Re \gg 1$$ are assumed. The approximations involved limit the validity of the model to distances larger than a few bubble diameters. The general agreement between the predictions and experimental results suggest that the model contains the most relevant mechanisms that govern the interaction, within its range of validity. Prior analyses for non-deformable bubbles that predicted an equilibrium due to balance between pressure gradients and wake-induced motion, are contradicted by the observed coalescence. A possible cause for the discrepancy is bubble deformation.

##### MSC:
 76Txx Multiphase and multicomponent flows
##### Keywords:
bubbles; coalescence; wake; bubbly flow
Full Text:


text:
# The Path to the Result

Input:

A grid of varying size only containing numbers in the range 00-99, the following four characters: ^; >; v; <, and spaces to fill (and new-lines of course).

Output:

The result of the sum (integer) when you follow the path.

How does this work?

You start at the arrow with the highest number, and follow the path the arrows point to.
For each of the four arrows we'll do a different arithmetic calculation:

• > = Addition (+);
• < = Subtraction (−);
• ^ = Multiplication (* or × or ·);
• v = Real / Calculator Division (/ or ÷);

(NOTE: The symbols between parenthesis are just added as clarification. If your programming language uses a completely different symbol for the arithmetic operation than the examples, then that is of course completely acceptable.)

Challenge rules:

• You can choose the input (to some extent). A string-array would be most logical I think, but a symbol/newline-separated single string is also fine. (Full functions as input that are simply executed in the actual code snippet aren't allowed!)
• Trailing spaces for the input are optional.
• Numbers 0-9 will have a leading 0 so all numbers plus arrow are of an equal three-character length.
The number & arrow will always have the following format: DDA (D being a digit; A being one of the four arrows).
• You use the current value with the previous arithmetic operation (the example below explains this better).
• The final arithmetic operation can be ignored because of the rule above.
• It is possible to have DDA's in the input-grid that aren't being used in the path to the resulting sum.
• There won't be any test-cases for arrows that form a loop.
• There won't be any test-cases for dividing by 0.
• There won't be any test-cases where dividing creates a decimal sum-result. You only have to worry about integers.
• EDIT: Since there weren't any answer yet, I decided to change the grids slightly to perhaps make it easier. I know this is usually not recommended to do with challenges, but since no answers were given yet, I think this will make it slightly easier, giving more (/any) answers: The spaces between de DDA's are random. There could even be no spaces in between at all. All DDA's and spaces are grouped per three. Also, it is possible to have no spaces between DDA's at all.

Example:

Input:

94>      66v
05v   08<
43>
03>   00^

Path it follows:

Path it follows   |   Current sum value
________________________________________
94>               |   94  (highest value)
66v               |   160 (+)
08<               |   20  (/)
05v               |   15  (-)
03>               |   5   (/)
00^               |   5   (+)
43>               |   215 (*)
'>' after 43 is ignored because it's the last number

Output: 215

General Rules:

• This is , so shortest answer in bytes wins.
Don't let code-golf languages discourage you from posting answers with non-codegolfing languages. Try to come up with an as short as possible answer for 'any' programming language.
• Standard rules apply for your answer, so you are allowed to use STDIN/STDOUT, functions/method with the proper parameters, full programs. Your call.
• Default Loopholes are forbidden.

Test cases:

94>      66v           ->     215
05v   08<
43>
03>   00^

77>   80v55^02<        ->     -19
20>   14^

22>11>15>80>14v        ->     -16
18^17>      10v
00^   90v43^02v
07^11<15<22v01<

99>98>97>96>95>94v     ->     102
93>92>   91v   03<

• I keep finding -12 for the third test case... – Paul Picard Sep 5 '16 at 14:45
• @PaulPicard Ah, I'm so sorry, there was a mistake in the grid.. 07< has been changed to 07^ to fix this. – Kevin Cruijssen Sep 5 '16 at 14:47
• Could somebody please tell me what I'm misunderstanding with the second one? As I see it, it is: 77+80/20+14*2-55=-11.3 without order of operations. – Daniel Sep 6 '16 at 3:36
• @Dopapp You start at the highest value, so it's 80/20+14*2-55=-19 – Kevin Cruijssen Sep 6 '16 at 6:15

# Python 2, 409390 377 bytes

def f(p):
p=map(lambda s:map(lambda d:d==3*(' ',)and(0,-1)or(int(d[0]+d[1]),'><^v'.index(d[2])),zip(s[::3],s[1::3],s[2::3])),p);x,y=max([(x,y)for x in range(len(p[0]))for y in range(len(p))],key=lambda(x,y):p[y][x][0]);r=o=0
try:
while 1:
1/(y*x>=0);d=p[y][x]
if-1<d[1]:r=eval('r%sd[0]'%'+-*/'[o]);o=d[1];a=(-(o==1),1)[o<1];b=(-(o==2),1)[o>2]
x+=a;y+=b
except:print r

The second and third indent levels are a raw tab and a raw tab plus a space, respectively. Markdown mangles the raw tabs, so they've been replaced by two spaces.

Ungolfed:

def follow(path):
# Split path's strs into DDAs ((0, -1) for blank, else (int, int))
path = map(
lambda s: map(
lambda dda: (0, -1) if dda == (' ', ' ', ' ') else (int(dda[0] + dda[1]), '><^v'.index(dda[2])),
zip(s[::3], s[1::3], s[2::3])
),
path
)

# Find highest DDA
x, y = max([(x, y) for x in range(len(path[0])) for y in range(len(path))], key=lambda (x, y): path[y][x][0])

result = 0
try:
# Keep following the path until we get an IndexError
while True:
if y < 0 or x < 0: raise IndexError

dda = path[y][x]
if dda[1] != -1:
# It's not empty - do the operation & change direction
const = dda[0]

result = eval('result %s const' % '+-*/'[op])

op = dda[1]

dx = 1 if op == 0 else -1 if op == 1 else 0
dy = 1 if op == 3 else -1 if op == 2 else 0

# Move to the next thing
x += dx
y += dy
except IndexError:
# We've exited the path - return
print result

Ideone it! (all test cases)

## Python 2, 449 bytes

i=input();p,d,a,o,l=[-1,0],">",0,"+",0
for y in range(len(i)):
for x in range(len(i[0])):
if i[y][x]in"><^v":
v=int(i[y][x-2]+i[y][x-1])
if v>l:l=v;p=[x-2,y]
while 1:
if d==">":p[0]+=1
if d=="<":p[0]-=1
if d=="^":p[1]-=1
if d=="v":p[1]+=1
try:
c=i[p[1]][p[0]]
if c in"><^v":
d=c
if p[0]<0 or p[1]<0:1/0
a=eval(str(a)+o+str(int(i[p[1]][p[0]-2]+i[p[1]][p[0]-1])));o={">":"+","<":"-","^":"*","v":"/"}[c]
except:print a;break

This takes an array of strings as input

Ungolfed:

pointer = [-1,0]
dir = ">"
output=0
operation="+"

largest = 0
for y in range(len(input)):
for x in range(len(input[0])):
if input[y][x] in "><^v":
value = int(input[y][x-2]+input[y][x-1])
if value > largest: largest = value; pointer = [x-2, y]

while 1:
if dir==">":pointer[0]+=1
if dir=="<":pointer[0]-=1
if dir=="^":pointer[1]-=1
if dir=="v":pointer[1]+=1
try:
current = input[pointer[1]][pointer[0]]
if current in "><^v":
dir = current
value = int(input[pointer[1]][pointer[0]-2]+input[pointer[1]][pointer[0]-1])

if pointer[0]<0 or pointer[1]<0:1/0
output=eval(str(output)+operation+str(value))

if current==">":operation="+"
if current=="<":operation="-"
if current=="^":operation="*"
if current=="v":operation="/"

except:
print output;break


text:
# Multiplying strings

Computer Science Level pending

Great! Now you want to print a dotted line below the message for Aditya, so that you can eventually print messages for several students on the same page, and easily cut along the dotted line to separate the messages. You're just doing your part to save the planet.

You write the following code:

 1 2 3 4 5 6 7 8 9 name = "Aditya" score = 98 liked = True message = "Hi %s. You achieved a score of %d." % (name, score) extra = "It's %s that I enjoyed having you as a student." % liked print message, extra print [---]

What do you use to replace [---] so that you get a row of 80 - with no spaces in between? Your output should like like this:

 1 2 Hi Aditya. You achieved a score of 98. It's True that I enjoyed having you as a student. --------------------------------------------------------------------------------
×


text:
# How do you solve 5x - 10 + 12 = 22?

Sep 27, 2015

color(blue)(x=4

#### Explanation:

 5x−10+12=22

$5 x + 2 = 22$

$5 x = 22 - 2$

$5 x = 20$

$x = \frac{20}{5}$

color(blue)(x=4

Aug 4, 2018

$x = 4$

#### Explanation:

We can combine the constants on the left to get

$5 x + 2 = 22$

Next, we can subtract $2$ from both sides to get

$5 x = 20$

Lastly, we can divide both sides by $5$ to get

$x = 4$

Hope this helps!


text:
Math Relationships Relationship

# Relationship

In mathematics as well as in real life one wants to assign certain interdependent variables to each other.

### Example

If we have the price of a kilo in the supermarket, we can easily extrapolate this.

• 1 kilogram of meat costs 4 euros.
• 2 kilos of meat cost 8 euros.
• 3 kilos of meat cost 12 euros.
• etc.

This calculation is based on a simple system. We can assign each value to another.

We can also represent this:

$\text{weight}\longmapsto\text{price}$

• $1 \longmapsto 4$
• $2 \longmapsto 8$
• $3 \longmapsto 12$
• ...
!

### Remember

A relationship assigns a different value to each value. One uses for the representation an arrow: $\longmapsto$.

## Presentation

There are various ways to represent relationships.

## Types

Typically, the school deals with two types of relationships as well as functions that also represent a relationship.


text:
# Distance Between Two Points 2D Formula

The distance between two points, in two dimensions, is given by solving pythagorean's theorem for the length the hypotenuse of the right triangle formed by the two points.


text:
# Central Tendency vs Dispersion

Hello everyone, welcome back! In this blog we will again focus into the some of widely used central tendency techniques and then measure of spread in the Statistical Analysis of the EDA part. If you are looking for a brief walk-through of a Statistical Data Analysis in Data Science please refer to this blog of mine.

## Central Tendency

Central tendency in statistics is the property of a data which tries to explain the central portion of the data. There are different types of the central tendency and each have their advantage and disadvantages. Lets explore them below.

We will calculate these values in the the titanic dataset.

### Mean

Mean is the most common measure of a central tendency where we divide the sum of all data points by the number of the data points. This is often called as an average and is denoted by Greek symbol $\mu$.

import numpy as np
import pandas as pd
import seaborn as sns

df.head()

PassengerId Survived Pclass Name Sex Age SibSp Parch Ticket Fare Cabin Embarked
0 1 0 3 Braund, Mr. Owen Harris male 22.0 1 0 A/5 21171 7.2500 NaN S
1 2 1 1 Cumings, Mrs. John Bradley (Florence Briggs Thayer) female 38.0 1 0 PC 17599 71.2833 C85 C
2 3 1 3 Heikkinen, Miss. Laina female 26.0 0 0 STON/O2. 3101282 7.9250 NaN S
3 4 1 1 Futrelle, Mrs. Jacques Heath (Lily May Peel) female 35.0 1 0 113803 53.1000 C123 S
4 5 0 3 Allen, Mr. William Henry male 35.0 0 0 373450 8.0500 NaN S
df.Age.mean()
29.69911764705882

The mean age of the passenger seems to be 29.6.

### Median

Median in statistics is the point of data which is exactly in the center of the data when it is sorted. Median should cut the data points into two halves. Median is often called as second quartile and is calculated as (where N is number of data point):

• When N is odd,
$$\text{median} = \left(\frac{N+1}{2} \right)^{th} \text{item}$$

• When N is even,

$$\text{median} = \frac{\left[ \left(\frac{N}{2} \right)^{th} \text{item} + \left(\frac{N}{2} +1 \right)^{th} \text{item} \right]}{2}$$

df.Age.median()
28.0

But the median age of the customer seems to be 28.

### Mode

Mode is the value which is most repeated in the data.

df.Age.mode()
0    24.0
dtype: float64

Age 24 seems to have repeated most. But is it true?

df.Age.value_counts().sort_values(ascending=False)
24.00    30
22.00    27
18.00    26
28.00    25
19.00    25
..
0.67      1
20.50     1
66.00     1
24.50     1
12.00     1
Name: Age, Length: 88, dtype: int64

Yes, it is. It seems that there are 30 passengers with age 24. Which is the most repeated.

### Quartiles

Besides mean, median and mode, quartiles are also sometimes taken along with median. Quartiles gives the values which lies in the 25, 50 and the 75 percentile of the data. But data has to be sorted first.

df.Age.describe()
count    714.000000
mean      29.699118
std       14.526497
min        0.420000
25%       20.125000
50%       28.000000
75%       38.000000
max       80.000000
Name: Age, dtype: float64

Looking over the quartile values above, we can see that the minimum value is 0.42, 25% is 20, median or 50% is 28, mean is 29 and 75% is 38 and max is 80. It seems that our data is slightly spread, as max value is far from the 3rd quartile.

### Box Plot

sns.boxplot(data=df, x="Age")

Box plot gives the values just like pandas describe. It seems that most of the data are within our 3rd quartile.

### Mid Range

This is not considered exactly as the central tendency but still it is useful to see how far is our average from the mid range. Mid Range gives the average of the highest and lowest value on the series. Which is simply an arithmetic mean.

(df.Age.min() + df.Age.max())/2
40.21

It seems that our mid range age is 40.21.

### When to use what?

• If the data is normally distributed, that is the mean is exactly in the center.
• We should use median rather than mean when there is huge outliers in the data. Because the mean gets skewed towards direction of outliers but not the median.

## Dispersion

Dispersion is a measure of the spread in the data. It is often used along with the central tendency and sometimes it depends on the central tendency too.

### Inter Quartile Range (IQR)

It is the difference between Q3 and Q1.

df.Age.quantile(0.75)-df.Age.quantile(0.25)
17.875

It seems that our two quartiles are not much far. Which represents that, majority of the data is around center and this is good in the sense of outliers.

### Range

It is the simplest measure of the dispersion, where we subtract smallest value from the largest value. This gives the idea of how huge is our data difference is.

df.Age.max()-df.Age.min()
79.58

It seems that the maximum difference between value is 79.58 but our IQR was only 17.875. There are certainly outliers in the data.

### Standard Deviation

It is the measure of the distance between each points of data with the mean data. This is most widely used measurement of the dispersion because this often gives the error in the data points.
$$s = \sqrt{\dfrac{\sum_{i=1}^{n}(x_i - \overline{x})^{2}}{n}}$$

df.Age.std()
14.526497332334044

### Z-Score

This answers the question How many standard deviation away is the data from the mean? It is calculated as:

$$z = \frac{x-\mu}{\sigma}$$

Where,

• x is the estimation of the mean or raw score
• 𝜇 is the population mean
• 𝜎 is the population standard deviation

Z Score is often used to compare the mean of two different samples and examine which is better than another. Lets take an example of comparing z score. A scored 172 in x and 50 on 37 on y. While mean and std for x and y are (151, 10) and (25.1, 6.4) resp. Z score for both is, 2.1, 1.86. We can conclude that, A scored relatively better in x than y. Lets do this in our Age by taking 500 random samples from our data and comparing it with our population z score.

pop = df.Age
sample1 = df.Age.sample(500)
sample2 = df.Age.sample(500)
sample1_z = (pop.mean()-sample1.mean())/pop.std()
sample2_z = (pop.mean()-sample2.mean())/pop.std()
sample1_z, sample2_z
(0.030374493169490115, 0.020496044621050025)

Looking over above two sample's z scores, we can say that the mean age of sample two is slightly nearer towards the population mean.

## When to use what?

• If we are about to compare means from two different population then we should be using z score.
• If we are about to look how huge data points are, then simple range should be used.
• For looking into how far are the mid points of two halves are, IQR should be used.
• For measuring how far is each data point is from the mean value, Standard Deviation is used.

## Shifting and Scaling of Mean and Standard Deviation

• Mean shifts by same amount as the data shift(add) amount.
• Mean increases by amount as the data was increased by amount.
• Median has property as same as of mean for shift and scale.
• Standard deviation does not shift but scales by amount.
• IQR has same property as of standard deviation.

Lets verify above properties. First calculate the describe of age and then another describe by multiplying age by 2.

df.Age.describe()
count    714.000000
mean      29.699118
std       14.526497
min        0.420000
25%       20.125000
50%       28.000000
75%       38.000000
max       80.000000
Name: Age, dtype: float64
(2*df.Age).describe()
count    714.000000
mean      59.398235
std       29.052995
min        0.840000
25%       40.250000
50%       56.000000
75%       76.000000
max      160.000000
Name: Age, dtype: float64

It seems that, mean and standard deviation has been increased by 2 times. But lets see what happens when we shift the data by 5 points.

(2+df.Age).describe()
count    714.000000
mean      31.699118
std       14.526497
min        2.420000
25%       22.125000
50%       30.000000
75%       40.000000
max       82.000000
Name: Age, dtype: float64

It seems that our means has been shifted but the standard deviation has not changed. Thus we can say that, the shift and the scaling property of dispersion and central tendency is different.

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# Converse to Bishop-Gromov Inequality

Is the converse to the Bishop-Gromov Inequality true?

In other words, if, for a complete $n$-dimensional Riemannian manifold $M$, there is $k \in \mathbb{R}$, such that defining $V_k(R)$ to be the size of a ball of radius $R$ in the standard space $S^n_k$, for any $p \in M$ and $R\geq 0$ we have that

$$vol_M(B_R(p)) \leq V_k(R)$$

then is it true that on $M$, $Ric \geq (n-1)k$?

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A very vague reason for this result not to hold is that, if it were true, this could have been used as a definition of Ricci curvature bounded form below for metric (or metric measured) spaces. In fact defining such bounds for Ricci curvature on metric spaces appears to need the machinery of optimal transport, as shown by Sturm, Lott and Villani, math.univ%2lyon1.fr/homes%2www/villani/Cedrif/P12.CIME.pdf and hal.archives-ouvertes.fr/docs/00/51/89/74/PDF/ohta.pdf are nice surveys on these questions. See also, arxiv.org/pdf/math/0701886. –  Thomas Richard Jan 20 '11 at 21:41
Thanks, this was my motivation for asking the question - I think that the B-G inequality is preserved under G-H convergence, so the space of metric spaces satisfying the B-H ineq. with a fixed n,k is compact, so I was confused as to why this didn't just solve the problem. –  Otis Chodosh Jan 21 '11 at 18:31

This is false for $n=3$, and for $k<0$. One can get a different pinching condition, $R(g)g-Ric \geq -4g$ which is weaker than $Ric \geq -2g$ ($R(g)$ is the scalar curvature), and which gives the same upper bound on volume as Gromov's inequality (for a manifold with negative curvature or no cut points). The idea is that in the proof of Gromov's inequality, one can integrate out the curvature tangent to the sphere of radius $r$ using the theorem egregrium and Gauss-Bonnet, giving a weaker curvature condition. There are metrics which satisfy the first curvature condition but not the Ricci lower bound. For example, there are homogeneous negatively curved metrics on $R^3$ which in one direction are foliated by totally geodesic hyperbolic planes of curvature $-K_1^2$, and in a perpendicular direction are foliated by totally geodesic hyperbolic planes of curvature $-K_2^2$ for $0 < K_1\neq K_2>0$ (this is a double warped product, and is Thurston's "ninth geometry" of the form $dr^2+e^{2K_1r}dx^2+e^{2K_2r}dy^2$). One obtains a counterexample for the choice of $K_1 =5/4,K_2=1/2$.

Added computations: The curvature operator of this metric may be normalized to have sectional curvatures $-K_1^2, -K_2^2, -K_1K_2$. If we assume that $K_1\geq K_2$, then the minimal curvature is $-K_1^2$. Under the same assumption, the minimal eigenvalue of the Ricci operator is $-K_1^2-K_1K_2$. The minimal eigenvalue of $Rg-Ric$ is $-2K_1^2-K_1K_2-K_2^2$. We then see that for $K_1=5/4, K_2=1/2$, the minimal eigenvalue of $Ric$ is $-35/16<-2$, and the minimal eigenvalue of $Rg-Ric$ is $-4$.

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Shouldn't it be even easier to construct counterexamples in higher dimensions, say $n = 4$, using a product metric? –  Deane Yang Jan 20 '11 at 20:16
It seems that your curvature condition survives under nice connected sum with thin necks. I.e. it can not give the same upper bound on volume as in Bishop inequality. It may hold locally and your counterexample seems to be OK, BUT you can not do global estimates with your curvature condition... –  Anton Petrunin Jan 20 '11 at 22:47
@ Deane: maybe, although I wouldn't expect a product to have as large volume growth, and I'm not sure how to estimate it. –  Ian Agol Jan 21 '11 at 0:59
@Anton: Nice observation - I wasn't sure if the estimate worked if there were cut points. How do you check that the curvature condition holds under connect sum? –  Ian Agol Jan 21 '11 at 1:50
@Agol - Thanks, this answers my question! @Deane Yang, could you provide such a counterexample if you know one? I'd be interested in the simplest possible such example. –  Otis Chodosh Jan 21 '11 at 18:34

If it is true for all $R,$ then the inequality follows from the infinitesimal calculation (as $R$ goes to $0.$)

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Igor, is this really right? It doesn't seem like there are enough variations to recover the full Ricci tensor at $p$. –  Deane Yang Jan 20 '11 at 15:44
It's a good calculation to check, but it sort-of suggests the opposite final answer. The first non-trivial derivative at $R = 0$ gives you the scalar curvature of the manifold, not the Ricci curvature. However, this is not a complete answer because we need a closed manifold by hypothesis. Possibly also the question is bad unless the manifold is simply connected. –  Greg Kuperberg Jan 20 '11 at 16:06