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https://labs.tib.eu/arxiv/?author=P.%20Garaud
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• ### The effect of rotation on fingering convection in stellar interiors(1804.04258)
April 11, 2018 physics.flu-dyn, astro-ph.SR
We study the effects of rotation on the growth and saturation of the double-diffusive fingering (thermohaline) instability at low Prandtl number. Using direct numerical simulations, we estimate the compositional transport rates as a function of the relevant non-dimensional parameters - the Rossby number, inversely proportional to the rotation rate, and the density ratio which measures the relative thermal and compositional stratifications. Within our explored range of parameters, we generally find rotation to have little effect on vertical transport. However, we also present one exceptional case where a cyclonic large scale vortex (LSV) is observed at low density ratio and fairly low Rossby number. The LSV leads to significant enhancement in the fingering transport rates by concentrating compositionally dense downflows at its core. We argue that the formation of such LSVs could be relevant to solving the missing mixing problem in RGB stars.
• ### Turbulent transport by diffusive stratified shear flows: from local to global models. Part II: The failure of local models(1803.10455)
March 28, 2018 physics.flu-dyn, astro-ph.SR
This paper continues the systematic investigation of diffusive shear instabilities initiated in Part I of this series. In this work, we primarily focus on quantifying the impact of non-local mixing, which is not taken into account in Zahn's mixing model (Zahn 1992). To do that, we present the results of direct numerical simulations using the PADDI code with a new model setup designed to contain both laminar and turbulent shear layers. As in Part I, we use the Low P\'eclet Number approximation of Ligni\`eres (1999) to model the evolution of the perturbations. Our main findings are twofold, and have significant impact on our understanding of the role of diffusive shear instabilities in stars. The first is that turbulence is not necessarily generated whenever $J{\rm Pr} < (J{\rm Pr})_c$ which Zahn's criterion assumes (Zahn 1974). We have demonstrated that the presence or absence of turbulent mixing in this limit hysteretically depends on the history of the shear layer. The second finding is that Zahn's nonlinear instability criterion only approximately locates the edge of the turbulent layer, since mixing beyond the region where $J{\rm Pr} < (J{\rm Pr})_c$ can also take place in a manner analogous to convective overshoot. We found that the turbulent kinetic energy decays roughly exponentially beyond the edge of the shear-unstable region, on a lengthscale $\delta$ that is directly proportional to the scale of the turbulent eddies, which are themselves of the order of the Zahn scale (see Part I). Both of these results suggest that mixing by diffusive shear instabilities should be modeled with more care than is currently standard in stellar evolution codes.
• ### Double-diffusive erosion of the core of Jupiter(1710.05240)
We present Direct Numerical Simulations of the transport of heat and heavy elements across a double-diffusive interface or a double-diffusive staircase, in conditions that are close to those one may expect to find near the boundary between the heavy-element rich core and the hydrogen-helium envelope of giant planets such as Jupiter. We find that the non-dimensional ratio of the buoyancy flux associated with heavy element transport to the buoyancy flux associated with heat transport lies roughly between 0.5 and 1, which is much larger than previous estimates derived by analogy with geophysical double-diffusive convection. Using these results in combination with a core-erosion model proposed by Guillot et al. (2004), we find that the entire core of Jupiter would be eroded within less than 1Myr assuming that the core-envelope boundary is composed of a single interface. We also propose an alternative model that is more appropriate in the presence of a well-established double-diffusive staircase, and find that in this limit a large fraction of the core could be preserved. These findings are interesting in the context of Juno's recent results, but call for further modeling efforts to better understand the process of core erosion from first principles.
• ### Turbulent transport by diffusive stratified shear flows: from local to global models. Part I: Numerical simulations of a stratified plane Couette flow(1610.04320)
Oct. 14, 2016 physics.flu-dyn, astro-ph.SR
Shear-induced turbulence could play a significant role in mixing momentum and chemical species in stellar radiation zones, as discussed by Zahn (1974). In this paper we analyze the results of direct numerical simulations of stratified plane Couette flows, in the limit of rapid thermal diffusion, to measure the turbulent diffusivity and turbulent viscosity as a function of the local shear and the local stratification. We find that the stability criterion proposed by Zahn (1974), namely that the product of the gradient Richardson number and the Prandtl number must be smaller than a critical values $(J\Pr)_c$ for instability, adequately accounts for the transition to turbulence in the flow, with $(J\Pr)_c \simeq 0.007$. This result recovers and confirms the prior findings of Prat et al. (2016). Zahn's model for the turbulent diffusivity and viscosity (Zahn 1992), namely that the mixing coefficient should be proportional to the ratio of the thermal diffusivity to the gradient Richardson number, does not satisfactorily match our numerical data when applied as is. It fails (as expected) in the limit of large stratification where the Richardson number exceeds the aforementioned threshold for instability, but it also fails in the limit of low stratification where the turbulent eddy scale becomes limited by the computational domain size. We propose a revised model for turbulent mixing by diffusive stratified shear instabilities, that now properly accounts for both limits, fits our data satisfactorily, and recovers Zahn's 1992 model in the limit of large Reynolds numbers.
• ### Layer Formation in Sedimentary Fingering Convection(1610.04183)
Oct. 13, 2016 physics.flu-dyn
When particles settle through a stable temperature or salinity gradient they can drive an instability known as sedimentary fingering convection. This phenomenon is thought to occur beneath sediment-rich river plumes in lakes and oceans, in the context of marine snow where decaying organic materials serve as the suspended particles, or in the atmosphere in the presence of aerosols or volcanic ash. Laboratory experiments of Houk and Green (1973) and Green (1987) have shown sedimentary fingering convection to be similar to the more commonly known thermohaline fingering convection in many ways. Here, we study the phenomenon using 3D direct numerical simulations. We find evidence for layer formation in sedimentary fingering convection in regions of parameter space where it does not occur for non-sedimentary systems. This is due to two complementary effects. Sedimentation affects the turbulent fluxes and broadens the region of parameter space unstable to the $\gamma$-instability (Radko 2003) to include systems at larger density ratios. It also gives rise to a new layering instability that exists in $\gamma-$stable regimes. The former is likely quite ubiquitous in geophysical systems for sufficiently large settling velocities, while the latter probably grows too slowly to be relevant, at least in the context of sediments in water.
• ### Excitation of gravity waves by fingering convection, and the formation of compositional staircases in stellar interiors(1505.07759)
May 28, 2015 astro-ph.SR
Fingering convection (or thermohaline convection) is a weak yet important kind of mixing that occurs in stably-stratified stellar radiation zones in the presence of an inverse mean-molecular-weight gradient. Brown et al. (2013) recently proposed a new model for mixing by fingering convection, which contains no free parameter, and was found to fit the results of direct numerical simulations in almost all cases. Notably, however, they found that mixing was substantially enhanced above their predicted values in the few cases where large-scale gravity waves, followed by thermo-compositional layering, grew spontaneously from the fingering convection. This effect is well-known in the oceanographic context, and is attributed to the excitation of the so-called "collective instability". In this work, we build on the results of Brown et al. (2013) and of Traxler et al. (2011b) to determine the conditions under which the collective instability may be expected. We find that it is only relevant in stellar regions which have a relatively large Prandtl number (the ratio of the kinematic viscosity to the thermal diffusivity), $O(10^{-3})$ or larger. This implies that the collective instability cannot occur in main sequence stars, where the Prandtl number is always much smaller than this (except in the outer layers of surface convection zones where fingering is irrelevant anyway). It could in principle be excited in regions of high electron degeneracy, during He core flash, or in the interiors of white dwarfs. We discuss the implications of our findings for these objects, both from a theoretical and from an observational point of view.
• ### Dynamics of fingering convection I: Small-scale fluxes and large-scale instabilities(1008.1807)
Aug. 10, 2010 physics.flu-dyn
Double-diffusive instabilities are often invoked to explain enhanced transport in stably-stratified fluids. The most-studied natural manifestation of this process, fingering convection, commonly occurs in the ocean's thermocline and typically increases diapycnal mixing by two orders of magnitude over molecular diffusion. Fingering convection is also often associated with structures on much larger scales, such as thermohaline intrusions, gravity waves and thermohaline staircases. In this paper, we present an exhaustive study of the phenomenon from small to large scales. We perform the first three-dimensional simulations of the process at realistic values of the heat and salt diffusivities and provide accurate estimates of the induced turbulent transport. Our results are consistent with oceanic field measurements of diapycnal mixing in fingering regions. We then develop a generalized mean-field theory to study the stability of fingering systems to large-scale perturbations, using our calculated turbulent fluxes to parameterize small-scale transport. The theory recovers the intrusive instability, the collective instability, and the gamma-instability as limiting cases. We find that the fastest-growing large-scale mode depends sensitively on the ratio of the background gradients of temperature and salinity (the density ratio). While only intrusive modes exist at high density ratios, the collective and gamma-instabilities dominate the system at the low density ratios where staircases are typically observed. We conclude by discussing our findings in the context of staircase formation theory.
• ### Dynamics of fingering convection II: The formation of thermohaline staircases(1008.1808)
Aug. 10, 2010 physics.flu-dyn
Regions of the ocean's thermocline unstable to salt fingering are often observed to host thermohaline staircases, stacks of deep well-mixed convective layers separated by thin stably-stratified interfaces. Decades after their discovery, however, their origin remains controversial. In this paper we use 3D direct numerical simulations to shed light on the problem. We study the evolution of an analogous double-diffusive system, starting from an initial statistically homogeneous fingering state and find that it spontaneously transforms into a layered state. By analysing our results in the light of the mean-field theory developed in Paper I, a clear picture of the sequence of events resulting in the staircase formation emerges. A collective instability of homogeneous fingering convection first excites a field of gravity waves, with a well-defined vertical wavelength. However, the waves saturate early through regular but localized breaking events, and are not directly responsible for the formation of the staircase. Meanwhile, slower-growing, horizontally invariant but vertically quasi-periodic gamma-modes are also excited and grow according to the gamma-instability mechanism. Our results suggest that the nonlinear interaction between these various mean-field modes of instability leads to the selection of one particular gamma-mode as the staircase progenitor. Upon reaching a critical amplitude, this progenitor overturns into a fully-formed staircase. We conclude by extending the results of our simulations to real oceanic parameter values, and find that the progenitor gamma-mode is expected to grow on a timescale of a few hours, and leads to the formation of a thermohaline staircase in about one day with an initial spacing of the order of one to two metres.
• ### On the Penetration of Meridional Circulation below the Solar Convection Zone II: Models with Convection Zone, the Taylor-Proudman constraint and Applications to Other Stars(0906.1756)
June 9, 2009 astro-ph.SR
The solar convection zone exhibits a strong level of differential rotation, whereby the rotation period of the polar regions is about 25-30% longer than the equatorial regions. The Coriolis force associated with these zonal flows perpetually "pumps" the convection zone fluid, and maintains a quasi-steady circulation, poleward near the surface. What is the influence of this meridional circulation on the underlying radiative zone, and in particular, does it provide a significant source of mixing between the two regions? In Paper I, we began to study this question by assuming a fixed meridional flow pattern in the convection zone and calculating its penetration depth into the radiative zone. We found that the amount of mixing caused depends very sensitively on the assumed flow structure near the radiative--convective interface. We continue this study here by including a simple model for the convection zone "pump", and calculating in a self-consistent manner the meridional flows generated in the whole Sun. We find that the global circulation timescale depends in a crucial way on two factors: the overall stratification of the radiative zone as measured by the Rossby number times the square root of the Prandtl number, and, for weakly stratified systems, the presence or absence of stresses within the radiative zone capable of breaking the Taylor-Proudman constraint. We conclude by discussing the consequences of our findings for the solar interior and argue that a potentially important mechanism for mixing in Main Sequence stars has so far been neglected.
• ### The rotation rate of the solar radiative zone(0811.2550)
Nov. 16, 2008 astro-ph
The rotation rate of the solar radiative zone is an important diagnostic for angular-momentum transport in the tachocline and below. In this paper we study the contribution of viscous and magnetic stresses to the global angular-momentum balance. By considering a simple linearized toy model, we discuss the effects of field geometry and applied boundary conditions on the predicted rotation profile and rotation rate of the radiative interior. We compare these analytical predictions with fully nonlinear simulations of the dynamics of the radiative interior, as well as with observations. We discuss the implications of these results as constraints on models of the solar interior.
• ### Assembling the Building Blocks of Giant Planets around Intermediate Mass Stars(0806.1521)
June 9, 2008 astro-ph
We examine a physical process that leads to the efficient formation of gas giant planets around intermediate mass stars. In the gaseous protoplanetary disks surrounding rapidly-accreting intermediate-mass stars we show that the midplane temperature (heated primarily by turbulent dissipation) can reach > 1000 K out to 1 AU. Thermal ionization of this hot gas couples the disk to the magnetic field, allowing the magneto-rotational instability (MRI) to generate turbulence and transport angular momentum. Further from the central star the ionization fraction decreases, decoupling the disk from the magnetic field and reducing the efficiency of angular momentum transport. As the disk evolves towards a quasi-steady state, a local maximum in the surface density and in the midplane pressure both develop at the inner edge of the MRI-dead zone, trapping inwardly migrating solid bodies. Small particles accumulate and coagulate into planetesimals which grow rapidly until they reach isolation mass. In contrast to the situation around solar type stars, we show that the isolation mass for cores at this critical radius around the more massive stars is large enough to promote the accretion of significant amounts of gas prior to disk depletion. Through this process, we anticipate a prolific production of gas giants at ~1 AU around intermediate-mass stars.
• ### On the penetration of meridional circulation below the solar convection zone(0708.0258)
Aug. 2, 2007 astro-ph
Meridional flows with velocities of a few meters per second are observed in the uppermost regions of the solar convection zone. The amplitude and pattern of the flows deeper in the solar interior, in particular near the top of the radiative region, are of crucial importance to a wide range of solar magnetohydrodynamical processes. In this paper, we provide a systematic study of the penetration of large-scale meridional flows from the convection zone into the radiative zone. In particular, we study the effects of the assumed boundary conditions applied at the convective-radiative interface on the deeper flows. Using simplified analytical models in conjunction with more complete numerical methods, we show that penetration of the convectively-driven meridional flows into the deeper interior is not necessarily limited to a shallow Ekman depth but can penetrate much deeper, depending on how the convective-radiative interface flows are modeled.
• ### Growth and migration of solids in evolving protostellar disks I: Methods and Analytical tests(0705.1563)
May 10, 2007 astro-ph
This series of papers investigates the early stages of planet formation by modeling the evolution of the gas and solid content of protostellar disks from the early T Tauri phase until complete dispersal of the gas. In this first paper, I present a new set of simplified equations modeling the growth and migration of various species of grains in a gaseous protostellar disk evolving as a result of the combined effects of viscous accretion and photo-evaporation from the central star. Using the assumption that the grain size distribution function always maintains a power-law structure approximating the average outcome of the exact coagulation/shattering equation, the model focuses on the calculation of the growth rate of the largest grains only. The coupled evolution equations for the maximum grain size, the surface density of the gas and the surface density of solids are then presented and solved self-consistently using a standard 1+1 dimensional formalism. I show that the global evolution of solids is controlled by a leaky reservoir of small grains at large radii, and propose an empirically derived evolution equation for the total mass of solids, which can be used to estimate the total heavy element retention efficiency in the planet formation paradigm. Consistency with observation of the total mass of solids in the Minimum Solar Nebula augmented with the mass of the Oort cloud sets strong upper limit on the initial grain size distribution, as well as on the turbulent parameter $\alphat$. Detailed comparisons with SED observations are presented in a following paper.
• ### Individual and collective behavior of dust particles in a protoplanetary nebula(astro-ph/0307199)
July 10, 2003 astro-ph
We study the interaction between gas and dust particles in a protoplanetary disk, comparing analytical and numerical results. We first calculate analytically the trajectories of individual particles undergoing gas drag in the disk, in the asymptotic cases of very small particles (Epstein regime) and very large particles (Stokes regime). Using a Boltzmann averaging method, we then infer their collective behavior. We compare the results of this analytical formulation against numerical computations of a large number of particles. Using successive moments of the Boltzmann equation, we derive the equivalent fluid equations for the average motion of the particles; these are intrinsically different in the Epstein and Stokes regimes. We are also able to study analytically the temporal evolution of a collection of particles with a given initial size-distribution provided collisions are ignored.
• ### On rotationally driven meridional flows in stars(astro-ph/0203382)
March 21, 2002 astro-ph
A quasi-steady state model of the consequences of rotation on the hydrodynamical structure of a stellar radiative zone is derived, by studying in particular the role of centrifugal and baroclinic driving of meridional motions in angular-momentum transport. This nonlinear problem is solved numerically assuming axisymmetry of the system, and within some limits, it is shown that there exist simple analytical solutions. The limit of slow rotation recovers Eddington-Sweet theory, whereas it is shown that in the limit of rapid rotation, the system settles into a geostrophic equilibrium. The behaviour of the system is found to be controlled by one parameter only, linked to the Prantl number, the stratification and the rotation rate of the star.
• ### Dynamics of the solar tachocline I: an incompressible study(astro-ph/0108276)
Aug. 16, 2001 astro-ph
Gough & McIntyre have suggested that the dynamics of the solar tachocline are dominated by the advection-diffusion balance between the differential rotation, a large-scale primordial field and baroclinicly driven meridional motions. This paper presents the first part of a study of the tachocline, in which a model of the rotation profile below the convection zone is constructed along the lines suggested by Gough & McIntyre and solved numerically. In this first part, a reduced model of the tachocline is derived in which the effects of compressibility and energy transport on the system are neglected; the meridional motions are driven instead by Ekman-Hartmann pumping. It is shown that there exists only a narrow range of magnetic field strengths for which the system can achieve a nearly uniform rotation. The results are discussed with respect to observations and to the limitations of this initial approach. A following paper combines the effects of realistic baroclinic driving and stratification with a model that follows closely the lines of work of Gough & McIntyre.
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2021-04-14 20:15:18
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https://web2.0calc.com/questions/please-help-asap_41301
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+0
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68
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+887
Thanks!
Jul 20, 2020
#1
0
Angle DBC = angle DBA = x, angle ABC = 2x
AB = AC, so angle BCD = 2x
Angle ADB external to triangle BDC, so angle ADB = 3x
BD = BC, so angle BDC = 90 - x
angle ADB + angle BDC = 180 ==> 3x + (90 - x) = 180 ==> x = 45
==> A = 90 - x = $$\boxed{45}$$
Jul 20, 2020
#2
+25541
+1
$$\begin{array}{|rcll|} \hline AB=AC \\ \hline \angle ABC &=& \angle ACB \quad | \quad \angle ABC =2x \\ 2x &=& \angle ACB \\ \mathbf{\angle ACB} &=& \mathbf{2x} \\ \hline \end{array} \begin{array}{|rcll|} \hline BD=BC \\ \hline \angle BDC &=& \angle ACB \quad | \quad \angle ACB =2x \\ \mathbf{\angle BDC} &=& \mathbf{2x} \\ \hline \end{array}$$
$$\begin{array}{|rcll|} \hline \angle ADB &=& 180^\circ - \angle BDC \quad | \quad \mathbf{\angle BDC=2x} \\ \mathbf{\angle ADB} &=& \mathbf{180^\circ - 2x} \\ \hline \end{array}$$
$$\begin{array}{|rcll|} \hline \mathbf{ \text{In \triangle BCD }: } \\ \hline 2x+2x+x &=& 180^\circ \\ 5x &=& 180^\circ \\\\ x &=& \dfrac{180^\circ}{5} \\\\ \mathbf{x} &=& \mathbf{36^\circ} \\ \hline \end{array}$$
$$\begin{array}{|rcll|} \hline \mathbf{ \text{In \triangle ADB }: } \\ \hline A+ (180^\circ-2x) +x &=& 180^\circ \\ A+ 180^\circ-2x +x &=& 180^\circ \\ A+ 180^\circ-x &=& 180^\circ \\ A -x &=& 0 \\ A &=& x \quad | \quad \mathbf{x=36^\circ} \\ \mathbf{A} &=& \mathbf{36^\circ} \\ \hline \end{array}$$
Jul 21, 2020
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2020-09-22 01:15:17
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http://math.stackexchange.com/questions/150554/ufds-are-integrally-closed
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# UFDs are integrally closed
Let $A$ be a UFD, $K$ its field of fractions, and $f$ an element of $A[T]$ a monic polynomial.
I'm trying to prove that if $f$ has a root $\alpha \in K$, then in fact $\alpha \in A$.
I'm trying to exploit the fact of something about irreducibility, will it help? I havent done anything with splitting fields, but this is something i can look for.
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If you are familiar with The Rational Root Theorem, I should think you could recast it in your more general setting. – Gerry Myerson May 27 '12 at 23:55
You can find a proof of your theorem, here – Makoto Kato May 28 '12 at 0:22
## 2 Answers
The proof follows exactly like the proof of the Rational Root Theorem.
Let $\alpha\in K$ be a root. We can express $\alpha$ as $\frac{a}{b}$ with $a,b\in A$, and using unique factorization we may assume that no irreducible of $A$ divides both $a$ and $b$.
If $f(x) = x^n + c_{n-1}x^{n-1}+\cdots+c_0$, then plugging in $\alpha$ and multiplying through by $b^n$ we obtain $$a^n + c_{n-1}ba^{n-1}+\cdots + c_0b^n = 0.$$ Now, $c_{n-1}ba^{n-1}+\cdots+c_0b^n$ is divisible by $b$, hence $a^n$ is divisible by $b$. Since no irreducible of $A$ divides both $a$ and $b$, it follows that $b$ must be a unit by unique factorization. Hence $\alpha\in A$.
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The well-known proof of the Rational Root Test immediately generalizes to any UFD or GCD domain. The sought result is simply a monic special case. One can present the proof in a form that works for both gcds and cancellable ideals by using only universal laws common to gcds and ideals, e.g. commutative, associative, distributive laws. Below I give such a universal proof for degree $3$ (to avoid notational obfuscation). It is clear from this how it generalizes to any degree.
Let $\rm\:D\:$ be a domain and suppose monic $\rm\:f(x)\in D[x]\:$ has root $\rm\:a/b,\ a,b\in D.\:$ The notation $\rm\:(d,e,\ldots)\:$ denotes $\rm\:gcd(d,e,\ldots)\:$ (or the ideal $\rm\:dD + eD +\:\!\ldots\:$ in the ideal case).
$\rm\qquad f(x)\, =\, c_0 + c_1 x + c_2 x^2 + x^3,\:$ and $\rm\:b^3\:\! f(a/b) = 0\:$ yields
$\rm\qquad c_0 b^3 + c_1 a b^2 + c_2 a^2 b\, =\, -a^3\$
$\rm\qquad\qquad\ \ \Rightarrow\,\ (b^3, a b^2,\, a^2 b)\,\mid\, \color{#c00}{a^3},\$ since the gcd divides the LHS of above so also the RHS
$\rm\qquad\ (b,a)^3 = \, (b^3,\, a b^2,\, a^2 b,\ \color{#c00}{a^3}),\ \$ hence, by the prior divisibility
$\rm\qquad\qquad\quad\:\! =\, (b^3,\, a b^2,\, a^2 b)$
$\rm\qquad\qquad\quad =\, b\, (b,a)^2,\$ so cancelling $\rm\,(b,a)^2$ yields
$\rm\qquad\ \, (b,a) =\, b\:\Rightarrow\: b\:|\:a,\$ i.e. $\rm\: a/b \in D.\ \$ QED
The degree $\rm\:n> 1\:$ case has the same form: one cancels $\rm\:(b,a)^{n-1}$ from $\rm\,(b,a)^n = b\,(b,a)^{n-1}.$
The ideal analog is the same, except replace "divides" by "contains", and assume that $\rm\,(a,b)\ne 0\,$ is invertible (so cancellable), e.g. in any Dedekind domain. Thus the above yields a uniform proof that PIDs, UFDs, GCD and Dedekind domains satisfy said monic case of the Rational Root Test, i.e. that they are integrally closed.
The proof is more concise if one knows about fractional gcds and ideals. Now, with $\rm\:r = a/b,\:$ one simply cancels $\rm\:(r,1)\:$ from $\rm\:(r,1)^n = (r,1)^{n-1}$ so $\rm\:(r,1) = (1),\:$ i.e. $\rm\:r \in D.\:$ For more, see my posts in a 2009/5/22 sci.math thread (mathforum or Google groups) which includes discussion of how most elementary irrationality proofs are simply unwindings of the elegant one-line proof employing Dedekind's notion of conductor ideal (universal denominator ideal).
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2016-05-28 22:41:06
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https://planetmath.org/partialmapping
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# partial mapping
Let $X_{1},\cdots,X_{n}$ and $Y$ be sets, and let $f$ be a function of $n$ variables: $f:X_{1}\times X_{2}\times\cdots\times X_{n}\to Y$. $x_{i}\in X_{i}$ for $2\leq i\leq n$. The induced mapping $a\mapsto f(a,x_{2},\ldots,x_{n})$ is called the partial mapping determined by $f$ corresponding to the first variable.
In the case where $n=2$, the map defined by $a\mapsto f(a,x)$ is often denoted $f(\cdot,x)$. Further, any function $f:X_{1}\times X_{2}\to Y$ determines a mapping from $X_{1}$ into the set of mappings of $X_{2}$ into $Y$, namely $\overline{f}:x\mapsto(y\mapsto f(x,y))$. The converse holds too, and it is customary to identify $f$ with $\overline{f}$. Many of the “canonical isomorphisms” that we come across (e.g. in multilinear algebra) are illustrations of this kind of identification.
Title partial mapping PartialMapping 2013-03-22 13:59:31 2013-03-22 13:59:31 mathcam (2727) mathcam (2727) 5 mathcam (2727) Definition msc 03E20
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2021-04-19 18:14:33
|
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https://tensorly.org/stable/modules/generated/tensorly.contrib.sparse.decomposition.robust_pca.html
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# tensorly.contrib.sparse.decomposition.robust_pca¶
robust_pca(X, mask=None, tol=1e-06, reg_E=1.0, reg_J=1.0, mu_init=0.0001, mu_max=10000000000.0, learning_rate=1.1, n_iter_max=100, verbose=1)
Robust Tensor PCA via ALM with support for missing values
Decomposes a tensor X into the sum of a low-rank component D and a sparse component E.
Parameters
Xndarray
tensor data of shape (n_samples, N1, …, NS)
array of booleans with the same shape as X should be zero where the values are missing and 1 everywhere else
tolfloat
convergence value
reg_Efloat, optional, default is 1
regularisation on the sparse part E
reg_Jfloat, optional, default is 1
regularisation on the low rank part D
mu_initfloat, optional, default is 10e-5
initial value for mu
mu_maxfloat, optional, default is 10e9
maximal value for mu
learning_ratefloat, optional, default is 1.1
percentage increase of mu at each iteration
n_iter_maxint, optional, default is 100
maximum number of iteration
verboseint, default is 1
level of verbosity
Returns
(D, E)
Robust decomposition of X
DX-like array
low-rank part
EX-like array
sparse error part
Notes
The problem we solve is, for an input tensor $$\tilde X$$:
\begin{equation*} \begin{aligned} & \min_{\{J_i\}, \tilde D, \tilde E} & & \sum_{i=1}^N \text{reg}_J \|J_i\|_* + \text{reg}_E \|E\|_1 \\ & \text{subject to} & & \tilde X = \tilde A + \tilde E \\ & & & A_{[i]} = J_i, \text{ for each } i \in \{1, 2, \cdots, N\}\\ \end{aligned} \end{equation*}
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2021-07-29 06:09:11
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https://search.r-project.org/CRAN/refmans/CLAST/html/mv.SM.html
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mv.SM {CLAST} R Documentation
## Mean value of upper limits.
### Description
Calculates mean value of a provided vector of upper limits as a function of p
### Usage
mv.SM(obj, lims = NULL, p = NULL, B = 99, offset = TRUE, wgt = TRUE)
### Arguments
obj list with components $S,$N, $count and optionally$lims, typically the output of sample.space.SM lims if not a component of obj p vector of values of p at which to calculate mean value B number of evenly spaced values of p if not provided offset if TRUE then ML mean value is subtracted wgt if TRUE than assign zero probability weight to extreme limits of 1 or 0.
### Value
list with elements $x (containing grid of B values of probability) and$y (containing corresponding mean values)
Chris J. Lloyd
### Examples
n=c(5,6,5,9)
a=c(2,4,5,12)
b=c(5,9,11,13)
# Enumerate all possible elements of the sufficiency reduced samples
# space i.e. all values of S and M. Also listed are the counts and
# subcounts of these outcomes, the test decision and the vector
# n, a, and b in $design data.SM=sample.space.SM(n,a,b) # There are 26 elements. # Calculate all approximate LR upper limits for these 26 outcomes. all.LR.high=LR.stats.SM(data.SM,type="upper")$lims
# Calculate all approximate LR upper limits for these 26 outcomes.
all.LR.low=LR.stats.SM(data.SM,type="lower")\$lims
# Calculate the mean values of these lims as a function of p
mv.high=mv.SM(data.SM,all.LR.high,p=(1:99)/100,offset=FALSE)
mv.low=mv.SM(data.SM,all.LR.low,p=(1:99)/100,offset=FALSE)
plot(c(0,1),c(0,1),ylab="mean value",xlab="p",type="n")
lines(mv.high)
lines(mv.low)
abline(0,1,lty=3)
title(main="Mean value of upper and lower limits by p")
[Package CLAST version 1.0.1 Index]
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2022-11-30 03:05:22
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https://tex.stackexchange.com/questions/444616/coloring-in-table-of-contents/444620
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Could someone help me, please ?
I would like coloring dots as Red and Bold (not normal and Black like in the photo).
Any help is highly appreciated.
\documentclass{report}
%%%%% Format font + page + langue
\usepackage[utf8x]{inputenc}
\usepackage[french]{babel}
\FrenchFootnotes
\usepackage[T1]{fontenc}
\setlength{\parindent}{0cm}
\setlength{\parskip}{1em}
\renewcommand{\baselinestretch}{1}
\usepackage[titletoc]{appendix}
\setcounter{secnumdepth}{5} % seting level of numbering
\usepackage{tocloft}
\renewcommand{\cftchapfont}{\Color{red}\large\bfseries}
\renewcommand{\cftsecfont}{\Color{blue}}
\renewcommand{\cftsubsecfont}{\Color{black}}
\renewcommand{\cftsubsubsecfont}{\Color{green}}
\renewcommand{\cftchappagefont}{\large\bfseries\Color{red}}
\renewcommand{\cftsecpagefont}{\Color{blue}}
\renewcommand{\cftsubsecpagefont}{\Color{black}}
\renewcommand{\cftsubsubsecpagefont}{\Color{green}}
\newcommand{\HRule}{\rule{\linewidth}{0.5mm}}
%%%%% Indentfirst
\usepackage{indentfirst}
\setlength{\parindent}{1cm}
\usepackage{libertine}
\usepackage{lmodern}
\begin{document}
\tableofcontents
\chapter{name of chapter 1}
\section{section 1}
\subsection{subsection 1}
\subsubsection{sub sub section 1}
\section{section 2}
\newpage
\chapter{name of chapter 2}
\section{section 21}
\subsection{sub sec 2}
\section{section 22}
\end{document}
• Could you post a full, minimal code of what you've tried? – Bernard Aug 4 '18 at 19:47
• @Bernard I just edited. – Dg PHAM Aug 4 '18 at 19:49
• This is not exactly a full code: which document class do you use? Where's \begin{document}, \end{document}? – Bernard Aug 4 '18 at 19:53
• @Bernard Ok, I posted the full code. – Dg PHAM Aug 4 '18 at 20:01
(updated code and screenshot after OP modified his/her own code and query)
I believe you should provide visual clarity to your readers as to (a) which colors are used for sectioning headers within in the ToC and (b) which color is used for hyperlinks.
I thus suggest you load the hyperref package with the option linktocpage, which makes just the pages numbers in the ToC into hypertargets, and (b) use a hyperlink color that's different from any of the colors already used in the ToC red, blue, black, green). Indigo, maybe? Choose a color that's suitable for all links within the document, not just for those shown in the ToC.
• Basically, you should do everything in your power to help your readers develop a strong Pavlovian reflex: Each time they see an object colored in the "link color", they should want to click on it with their mouse. If there's just one color that shouts "I'm a hyperlink", the Pavlovian reflex will be much stronger.
• Conversely, if you let lots of different colors (including black in some cases) be compatible with an object being a hyperlink, your readers will learn much more slowly what and where the hypertargets are. Put differently: "If everything is special, then nothing is special". Be very deliberate in how you deploy colors to convey meaning.
Incidentally, the hyperref package should be loaded after the tocloft and xcolor packages.
\documentclass{report}
\usepackage[a4paper,top=3cm,bottom=2.5cm,left=2.5cm,
right=2cm,marginparwidth=1.75cm,
\usepackage[french]{babel}
\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage[svgnames]{xcolor} % for 'DarkGreen' color
\setcounter{secnumdepth}{3}
\setcounter{tocdepth}{3}
\usepackage{tocloft}
\renewcommand{\cftchapfont}{\color{red}\large\bfseries}
\renewcommand{\cftsecfont}{\color{blue}}
\renewcommand{\cftsubsecfont}{\color{black}}
\renewcommand{\cftsubsubsecfont}{\color{DarkGreen}}
\renewcommand{\cftchappagefont}{\color{red}%\large\bfseries
}
%\renewcommand{\cftsecpagefont}{\color{blue}}
%\renewcommand{\cftsubsecpagefont}{\color{black}}
%\renewcommand{\cftsubsubsecpagefont}{\color{DarkGreen}}
linkcolor=Indigo, % or some other suitable color
\begin{document}
\tableofcontents
\chapter{Thoughts}
\section{Hello}
\subsection{World}
\subsubsection{Really!}
\end{document}
• @marmot - I always appreciate upvotes. :-) (I've upvoted your answer too, by the way.) I fully realize that the OP would like the page numbers to have the same colors as the headers and leaders. However, as I've tried to explain in my answer (and speaking more as a document designer than as a LaTeX programmer), I think it's a highly questionable to let a whole bunch of colors be compatible with the corresponding objects being hypertargets. (Aside: I can't say I "get" what exactly the OP is trying to achieve by making the ToC look like a wildly over-decorated Christmas tree...) – Mico Aug 4 '18 at 21:47
• I agree with all your statements. ;-) – marmot Aug 4 '18 at 21:49
Thanks for the MWE! You need to use the tricks of this answer, I think.
\documentclass{report}
%%%%% Format font + page + langue
\usepackage[utf8x]{inputenc}
\usepackage[french]{babel}
\FrenchFootnotes
\usepackage[T1]{fontenc}
\setlength{\parindent}{0cm}
\setlength{\parskip}{1em}
\renewcommand{\baselinestretch}{1}
\usepackage[titletoc]{appendix}
\setcounter{secnumdepth}{5} % seting level of numbering
\setcounter{tocdepth}{5}
\usepackage{tocloft}
\renewcommand{\cftchapfont}{\Color{black}\large\bfseries}
\renewcommand{\cftsecfont}{\Color{blue}}
\renewcommand{\cftsubsecfont}{\Color{red}}
\renewcommand{\cftsubsubsecfont}{\Color{green}}
\renewcommand{\cftchappagefont}{\Color{black}}
\renewcommand{\cftsecpagefont}{\Color{blue}}
\renewcommand{\cftsubsecpagefont}{\Color{red}}
\renewcommand{\cftsubsubsecpagefont}{\Color{green}}
\renewcommand{\cftchappagefont}{\large\bfseries}
\hypersetup{%
filecolor=magenta,% color of links to external files
urlcolor=red,% color to external URLs
}
\newcommand{\HRule}{\rule{\linewidth}{0.5mm}}
%%%%% Indentfirst
\usepackage{indentfirst}
\setlength{\parindent}{1cm}
\usepackage{libertine}
\usepackage{lmodern}
% uncomment these if you want to have the actual chapter font larger
% \usepackage{titlesec}
% \titleformat{\chapter}
% {\normalfont\Huge\bfseries}{\thechapter}{1em}{}
\begin{document}
\tableofcontents
\chapter{name of chapter 1}
\section{section 1}
\subsection{subsection 1}
\subsubsection{sub sub section 1}
\section{section 2}
\newpage
\chapter{name of chapter 2}
\section{section 21}
\subsection{sub sec 2}
\section{section 22}
\end{document}
• thank you so much. Just a little problem. I add also a subsubsection, but it doesn't appear in the table of contents. Could you solve this ? – Dg PHAM Aug 4 '18 at 20:03
• @DgPHAM Just add \setcounter{tocdepth}{5}. I updated the answer accordingly. – marmot Aug 4 '18 at 20:05
• perfect. Thank you. Could we make the titles of the chapters bigger and bold ? – Dg PHAM Aug 4 '18 at 20:07
• I need also dots ... connecting between the name of chapter and the Page. Could you help me, please? – Dg PHAM Aug 4 '18 at 20:09
• as you see in the sample photo in my question, the font of chapters is bigger and bold. I would like to do that in the table of contents. – Dg PHAM Aug 4 '18 at 20:12
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2019-06-18 02:49:46
|
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|
https://arc-alkali-rydberg-calculator.readthedocs.io/en/latest/divalent_atom_functions/arc.divalent_atom_functions.DivalentAtom.getTransitionRate.html
|
# arc.divalent_atom_functions.DivalentAtom.getTransitionRate#
DivalentAtom.getTransitionRate(n1, l1, j1, n2, l2, j2, temperature=0.0, s=0.5)#
Transition rate due to coupling to vacuum modes (black body included)
Calculates transition rate from the first given state to the second given state $$|n_1,l_1,j_1\rangle \rightarrow |n_2,j_2,j_2\rangle$$ at given temperature due to interaction with the vacuum field. For zero temperature this returns Einstein A coefficient. For details of calculation see Ref. 1 and Ref. 2. See Black-body induced population transfer example snippet.
Parameters
• n1 (int) – principal quantum number
• l1 (int) – orbital angular momentum
• j1 (float) – total angular momentum
• n2 (int) – principal quantum number
• l2 (int) – orbital angular momentum
• j2 (float) – total angular momentum
• [temperature] (float) – temperature in K
• s (float) – optional, total spin angular momentum of state. By default 0.5 for Alkali atoms.
Returns
transition rate in s $${}^{-1}$$ (SI)
Return type
float
References
1
C. E. Theodosiou, PRA 30, 2881 (1984) https://doi.org/10.1103/PhysRevA.30.2881
2
I. I. Beterov, I. I. Ryabtsev, D. B. Tretyakov, and V. M. Entin, PRA 79, 052504 (2009) https://doi.org/10.1103/PhysRevA.79.052504
|
2023-02-03 06:35:12
|
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|
https://chemistry.stackexchange.com/questions/122797/calculating-ph-of-a-saturated-weak-acid-base-solution
|
# Calculating pH of a saturated weak acid/base solution [closed]
How can you calculate the $$\mathrm{pH}$$ if you know for example the $$\mathrm{p}K_\mathrm{a}$$ or $$\mathrm{p}K_\mathrm{b}$$ value of a weak acid/base and how much acid/base can dilute in water. Let's take an example:
Quinoline has $$\mathrm{p}K_\mathrm{a} = 4.5$$ and $$\pu{0.6 g}/\pu{100 ml}$$ can dilute in water. I can easily calculate that the concentration of the saturated solution is $$\pu{0.0465 M}$$. So if the reaction is
$$\ce{C9H7N + H2O -> C9H7NH+ + OH-},$$
why $$[\ce{OH-}]\neq \pu{0.0465 M}$$? I know that I can get the correct answer by making an ICE-table and use that given $$\mathrm{p}K_\mathrm{a}$$ value but why is it necessary? It seems weird as after making the ICE-table it seems that all quinoline isn't diluted…
• Quinoline with pKb=14-pKa= 9.5 is a very weak base. Why there should be almost 0.05 M OH- like if it had been strong base ? pH =~ 14 - 0.5(pKb - log c)=9.25 + 0.5.log c. – Poutnik Oct 22 '19 at 14:21
• Yeah I know that it's obviously wrong but if Quinoline dilutes to water, shouldn't there be OH-? Or if not, what are there if Quinoline is diluted? – jte Oct 23 '19 at 15:23
• There is OH-. It is always there, if there is water. But there are 2 chemical equilibriums. Dissociation of water and protonization of quinoline. Together with mass and charge balance, there is the set of equations for the set of variables. – Poutnik Oct 23 '19 at 15:46
|
2020-01-23 09:20:28
|
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|
http://zbmath.org/?q=an:0964.60065&format=complete
|
# zbMATH — the first resource for mathematics
##### Examples
Geometry Search for the term Geometry in any field. Queries are case-independent. Funct* Wildcard queries are specified by * (e.g. functions, functorial, etc.). Otherwise the search is exact. "Topological group" Phrases (multi-words) should be set in "straight quotation marks". au: Bourbaki & ti: Algebra Search for author and title. The and-operator & is default and can be omitted. Chebyshev | Tschebyscheff The or-operator | allows to search for Chebyshev or Tschebyscheff. "Quasi* map*" py: 1989 The resulting documents have publication year 1989. so: Eur* J* Mat* Soc* cc: 14 Search for publications in a particular source with a Mathematics Subject Classification code (cc) in 14. "Partial diff* eq*" ! elliptic The not-operator ! eliminates all results containing the word elliptic. dt: b & au: Hilbert The document type is set to books; alternatively: j for journal articles, a for book articles. py: 2000-2015 cc: (94A | 11T) Number ranges are accepted. Terms can be grouped within (parentheses). la: chinese Find documents in a given language. ISO 639-1 language codes can also be used.
##### Operators
a & b logic and a | b logic or !ab logic not abc* right wildcard "ab c" phrase (ab c) parentheses
##### Fields
any anywhere an internal document identifier au author, editor ai internal author identifier ti title la language so source ab review, abstract py publication year rv reviewer cc MSC code ut uncontrolled term dt document type (j: journal article; b: book; a: book article)
Simulation of a space-time bounded diffusion. (English) Zbl 0964.60065
Most of the so far known stochastic-numerical methods rely on a deterministic time-discretization of corresponding stochastic processes. In contrast to that fact, the authors present space-time bounded approximations of initial value problems for $d$-dimensional ordinary stochastic differential equations (SDE)
$dX={\chi }_{{\tau }_{t,x}>s}b\left(s,{X}_{s}\right)ds+{\chi }_{{\tau }_{t,x}>s}\sigma \left(s,X\right)dW\left(s\right),\phantom{\rule{1.em}{0ex}}X\left(t\right)={X}_{t,x}=x\in {R}^{d}$
in a bounded domain $Q=\left[{t}_{0},{t}_{1}\right)×G\subset {R}^{d+1}$, where $X,b$ are $d$-dimensional vectors, $\sigma$ is a $d×d$-matrix, $W={\left(W\left(s\right)\right)}_{s\ge {t}_{0}}$ represents a $d$-dimensional standard Wiener process, and the stopping time ${\tau }_{t,x}$ is the first-passage time of the process $\left(s,{X}_{t,x}\left(s\right)\right)$, $s\ge t$, to ${\Gamma }=\overline{Q}\setminus Q$. The coefficients ${b}^{i}\left(s,x\right)$ and ${\sigma }^{i,j}\left(s,x\right)$, $\left(s,x\right)\in \overline{Q}$, and the boundary $\partial G$ are assumed to be sufficiently smooth and the strict ellipticity condition is imposed on $a\left(s,x\right)=\sigma \left(s,x\right){\sigma }^{T}\left(s,x\right)$. The proposed algorithm is based on a space-time discretization using random walks over boundaries of small space-time parallelepipeds. Corresponding convergence theorems and their proofs are given. A method of approximate search for exit points of space-time diffusions from a bounded domain is presented.
This work continues a series of papers initiated by the first author [see, for example, Stochastics Stochastics Rep. 56, No. 1-2, 103-125 (1996; Zbl 0888.60048) and ibid. 64, No. 3-4, 211-233 (1998)]. For those readers who prefer to read about the original works, the latter two citations are highly recommended, where the idea of space-time discretizations in conjunction with the construction of random walks over boundaries has already been explained, and related mean square approximation theorems are found as well there. The special value of this paper may be seen in the simulation results on which the authors report at the last ten pages. Thus, the choice of the title of this paper is misleading a little bit in view of the anticipative and innovative expectations of the potentially interested reader. The paper is recommended for those readers who are interested in a theoretical justification, pitfalls, advantages of stochastic-numerical methods including simulation studies for the approximation of deterministic initial-boundary value problems for parabolic equations under the condition of strong ellipticity on $\overline{Q}$.
##### MSC:
60H10 Stochastic ordinary differential equations 65C30 Stochastic differential and integral equations 60J60 Diffusion processes 60H35 Computational methods for stochastic equations 65C05 Monte Carlo methods 60H30 Applications of stochastic analysis
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2014-03-08 04:40:49
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https://hsm.stackexchange.com/questions/7412/mathematicians-attempts-at-proving-euclid-postulate
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# mathematicians attempts at proving Euclid postulate
Is there a list of all the people who attempted to prove the parallel postulate (also known as the fifth postulate or the Euclid axiom) in Euclidean geometry? Wikipedia has a page on the subject but the list given there is far too short.
Here is what I have collected so far, any addition is welcome.
Archimedes (~ -287- -212)
Posidonios (~-135 ~-51)
Geminus (~-10 ~60)
Claude Ptoleme (~90 ~168)
Proclus (412 485)
Simplicius (~490 ~560)
Thabit ibn Qurra (826 901)
Al-Abbās ibn Said al-Jawharī (~800 ~860)
Abu'l Abbas al-Fadl ibn Hatim Al-Nayrizi (~865 ~922)
Ibn al-Haytham (965 1039)
Avicenna (980-1037)
Omar Khayyam (1048-1131)
Erazmus Ciołek Witelo (~1230-~1300)
Rabbi Levi ben Gershom (1288-1344)
Pietro Antonio Cataldi (1548-1626)
Giovanni Alfonso Borelli (1608-1679)
John Wallis (1616-1703)
Giovanni Girolamo Saccheri (1667-1733)
Johann Heinrich Lambert (1728-1777)
Joseph-Louis Lagrange (1736-1813)
Farkas Bolyai (1775-1856)
Ferdinand Carl Schweikart (1780-1859)
Friedrich Ludwig Wachter (1792-1818)
Viktor Yakovlevich Bunyakovsky (1804-1889)
EDIT: added Farkas Bolyai, Joseph-Louis Lagrange, as pointed out in the comments.
• What is your question? – Alexandre Eremenko May 30 '18 at 15:57
• @eremenko Is there a list of all the people who attempted to prove the parallel postulate (also known as the fifth postulate or the Euclid axiom) in Euclidean geometry? An answer would provide a reference to such a list or the list itself. – coudy May 30 '18 at 16:55
• What do you mean exactly by “prove the parallel postulate in Euclidean geometry”? Deriving it from the other axioms of Euclid? If so, I think a number of these authors attempted no such thing. – Viktor Blasjo May 30 '18 at 20:47
• Grabiner, Judith V., Why did Lagrange “prove” the parallel postulate?, Am. Math. Mon. 116, No. 1, 3-18 (2009). ZBL1163.01009. (Also gives you a ready “parallel postulate” search link in Zentralblatt.) – Francois Ziegler May 31 '18 at 10:33
• János Bolyai's father, Farkas Bolyai, should be on the list. He famously wrote to his son:"Detest it as lewd intercourse, it can deprive you of all your leisure, your health, your rest, and the whole happiness of your life... Do not try the parallels in that way: I know that way all along. I have measured that bottomless night, and all the light and all the joy of my life went out there.". Kids never listen. – Conifold May 31 '18 at 21:11
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2020-02-18 16:53:10
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https://im.openupresources.org/8/students/4/7.html
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# Lesson 7: All, Some, or No Solutions
Let's think about how many solutions an equation can have.
## 7.1: Which One Doesn’t Belong: Equations
Which one doesn’t belong?
1. $5 + 7 = 7 + 5$
2. $5\boldcdot 7 = 7\boldcdot 5$
3. $2 = 7 - 5$
4. $5 - 7 = 7 - 5$
## 7.2: Thinking About Solutions
$$n = n$$
$$2t+6=2(t+3)$$
$$3(n+1)=3n+1$$
$$\frac14 (20d+4)=5d$$
$$5 - 9 + 3x = \text-10 + 6 + 3x$$
$$\frac12+x=\frac13 + x$$
$$y \boldcdot \text-6 \boldcdot \text-3 = 2 \boldcdot y \boldcdot 9$$
$$v+2=v-2$$
1. Sort these equations into the two types: true for all values and true for no values.
2. Write the other side of this equation so that this equation is true for all values of $u$. $$6(u-2)+2=$$
3. Write the other side of this equation so that this equation is true for no values of $u$. $$6(u-2)+2=$$
## 7.3: What's the Equation?
1. Complete each equation so that it is true for all values of $x$.
1. $3x+6=3(x+\underline{\quad}\,)$
2. $x-2=\text{-}(\,\underline{\quad}-x)$
3. $\frac{15x-10}5=\, \underline{\quad}-2$
2. Complete each equation so that it is true for no values of $x$.
1. $3x+6=3(x+\underline{\quad}\,)$
2. $x-2=\text{-}(\,\underline{\quad}-x)$
3. $\frac{15x-10}5=\, \underline{\quad}-2$
3. Describe how you know whether an equation will be true for all values of $x$ or true for no values of $x$.
## Summary
An equation is a statement that two expressions have an equal value. The equation
$$2x = 6$$
is a true statement if $x$ is 3:
$$2\boldcdot 3 = 6$$
It is a false statement if $x$ is 4:
$$2 \boldcdot 4 = 6$$
The equation $2x = 6$ has one and only one solution, because there is only one number (3) that you can double to get 6.
Some equations are true no matter what the value of the variable is. For example:
$$2x = x + x$$
is always true, because if you double a number, that will always be the same as adding the number to itself. Equations like $2x = x+x$ have an infinite number of solutions. We say it is true for all values of $x$.
Some equations have no solutions. For example:
$$x = x+1$$
has no solutions, because no matter what the value of $x$ is, it can’t equal one more than itself.
When we solve an equation, we are looking for the values of the variable that make the equation true. When we try to solve the equation, we make allowable moves assuming it has a solution. Sometimes we make allowable moves and get an equation like this:
$$8 = 7$$
This statement is false, so it must be that the original equation had no solution at all.
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2018-01-19 17:14:09
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https://www.ies.org/definitions/air-mass-optical-air-mass/
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# air mass (optical air mass)
[10.5.10.8] Ratio of the path length of radiation through the atmosphere (lm) at any given angle, Z degrees, to the sea level path length toward the zenith (lz).*
$AM = l_{m}/l_{z} \cong \sec Z, or 1/\cos Z, for Z \leq 62^{\circ}$
Standard Solar Constant and Zero Air Mass Solar Spectral Irradiance Tables, ASTM E490-00a. West Conshohocken, PA: ASTM International; 2014.
« Back to Definitions Index
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2020-01-20 16:27:48
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http://fluidsengineering.asmedigitalcollection.asme.org/article.aspx?articleid=1692255
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0
Research Papers: Fundamental Issues and Canonical Flows
# Vortex Shedding in a Tandem Circular Cylinder System With a Yawed Downstream Cylinder
[+] Author and Article Information
Stephen J. Wilkins
e-mail: x514a@unb.ca
James D. Hogan
e-mail: v3679@unb.ca
Joseph W. Hall
e-mail: jwhall@unb.ca
Department of Mechanical Engineering,
University of New Brunswick,
1Corresponding author.
Contributed by the Fluids Engineering Division of ASME for publication in the JOURNAL OF FLUIDS ENGINEERING. Manuscript received May 3, 2012; final manuscript received March 5, 2013; published online May 15, 2013. Editor: Malcolm J. Andrews.
J. Fluids Eng 135(7), 071202 (May 15, 2013) (7 pages) Paper No: FE-12-1224; doi: 10.1115/1.4023949 History: Received May 03, 2012; Revised March 05, 2013
## Abstract
This investigation examines the flow produced by a tandem cylinder system with the downstream cylinder yawed to the mean flow direction. The yaw angle was varied from $α=90deg$ (two parallel tandem cylinders) to $α=60deg$; this has the effect of varying the local spacing ratio between the cylinders. Fluctuating pressure and hot-wire measurements were used to determine the vortex-shedding frequencies and flow regimes produced by this previously uninvestigated flow. The results showed that the frequency and magnitude of the vortex shedding varies along the cylinder span depending on the local spacing ratio between the cylinders. In all cases the vortex-shedding frequency observed on the front cylinder had the same shedding frequency as the rear cylinder. In general, at small local spacing ratios the cylinders behaved as a single large body with the shear layers separating from the upstream cylinder and attaching on the downstream cylinder, this caused a correspondingly large, low frequency wake. At other positions where the local span of the tandem cylinder system was larger, small-scale vortices began to form in the gap between the cylinders, which in turn increased the vortex-shedding frequency. At the largest spacings, classical vortex shedding persisted in the gap formed between the cylinders, and both cylinders shed vortices as separate bodies with shedding frequencies typical of single cylinders. At certain local spacing ratios two distinct vortex-shedding frequencies occurred indicating that there was some overlap in these flow regimes.
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## Figures
Fig. 1
Schematic of various yawed tandem cylinder systems with local spacing ratios, S/D, shown at the narrow, middle, and wide ends for (a) α=90 deg, (b) α=80 deg, (c) α=70 deg, and (d) α=60 deg
Fig. 6
Comparison of flow regimes for (a) α=90 deg, (b) α=80 deg, (c) α=70 deg, and (d) α=60 deg
Fig. 5
Normalized power spectra at various spanwise locations for α=60 deg
Fig. 4
Normalized power spectra at various spanwise locations for α=70 deg
Fig. 3
Normalized power spectra at various spanwise locations for α=80 deg
Fig. 2
Microphone power spectra normalized by dynamic head squared q2 for α=90 deg for two azimuthal angles, γ=70 deg and γ=90 deg
## Discussions
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2017-05-25 12:33:32
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https://math.stackexchange.com/questions/490995/is-the-following-equivalent-to-the-axiom-of-choice
|
Is the following equivalent to the axiom of choice?
For sets $A,B$, let $|A|\leq^*|B|$ say that there exists an onto map $f:B\rightarrow A$ or $A=\emptyset$.
My question is, is $$\forall A,B(|A|\leq^*|B|\longrightarrow |A|\leq|B|)$$ equivalent to the axiom of choice?
Thanks
• @AsafKaragila Well, not really. If we $|\emptyset|\le^*|B|$ is false by the simplified definition, then the case $A=\emptyset$ is trivially true, and that does not hurt. All this $\le, \le^*$ is just fancy writing for "If there exists a surjection $B\to A$, there exists an injection $A\to B$". – Hagen von Eitzen Sep 11 '13 at 21:28
• @Hagen: There are no surjections from a non-empty set onto the empty set. And no, $|A|\leq^*|B|$ need not imply, generally, the existence of an injection from $A$ into $B$. – Asaf Karagila Sep 11 '13 at 21:35
• @Hagen: I realized your comment is the result of a misunderstanding in the question. The definition of $\leq^*$ is not the statement given in the displayed equation. It is given on the top part of the text. The question asks whether or not the displayed statement is equivalent to the axiom of choice (because it trivially follows from it). – Asaf Karagila Sep 13 '13 at 14:50
This is known as the partition principle (note that the other implication is trivial in $\sf ZF$). The problem whether or not this is equivalent to the axiom of choice has been open for over a century now.
There isn't much to say about it, really. We know it implies some basic choice principles such as "Every infinite set is Dedekind infinite", but we don't know a lot more. Every time I think about the problem I run into the same problem, we don't have enough tools to manage - or even understand - the structures of cardinals in arbitrary models of $\sf ZF$. Not even under $\leq$ and let alone under $\leq^*$.
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2019-07-16 08:20:45
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https://thewindingnumber.blogspot.com/2019/10/sigma-fields-are-venn-diagrams.html
|
### Sigma fields are Venn diagrams
The starting point for probability theory will be to note the difference between outcomes and events.
An outcome of an experiment is a fundamentally non-empirical notion, about our theoretical understanding of what states a system may be in -- it is, in a sense, analogous to the "microstates" of statistical physics. The set of all outcomes $x$ is called the sample space $X$, and is the fundamental space to which we will give a probabilistic structure (we will see what this means).
Our actual observations, the events, need not be so precise -- for example, our measurement device may not actually measure the exact sequence of heads and tails as the result of an experiment, but only the total number of heads, or something -- analogous to a "macrostate". But these measurements are statements about what microstates we know are possible for our system to be in -- i.e. they correspond to sets of outcomes. These sets of outcomes that we can "talk about" are called events $E$, and the set of all possible events is called a field $\mathcal{F}\subseteq 2^X$.
For instance: if our sample space is $\{1,2,3,4,5,6\}$ and our measurement apparatus is a guy who looks at the reading and tells us if it's even or odd, then the field is $\{\varnothing, \{1,3,5\},\{2,4,6\},X\}$. We simply cannot talk about sets like $\{1,3\}$ or $\{1\}$. Our information just doesn't tell us anything about sets like that -- when we're told "odd", we're never hinted if the outcome was 1 or 3 or 5, so we can't even have prior probabilities -- we can't even give probabilities to whether a measurement was a 1 or a 3.
Well, what kind of properties characterise a field? There's actually a bit of ambiguity in this -- it's clear that a field should be closed under negation and finite unions (and finite intersections follow via de Morgan) -- if you can talk about whether $P_1$ and $P_2$ are true, you can check each of them to decide if $P_1\lor P_2$ is true (and since a proposition $P$ corresponds to a set $S$ in the sense that $P$ says "one of the outcomes in $S$ is true", $\lor$ translates to $\cup$). But if you have an infinite number of $P_i$'s, can you really check each one of them so that you can say without a doubt that a field is closed under arbitrary union?
Well, this is (at this point) really a matter of convention, but we tend to choose the convention where the field is closed under negation and countable unions. Such a field is called a sigma-field. We will actually see where this convention comes from (and why it is actually important) when we define probability -- in fact, it is required for the idea that one may have a uniform probability distribution on a compact set in $\mathbb{R}^n$.
A beautiful way to understand fields and sigma fields is in terms of venn diagrams -- in fact, as you will see, fields are precisely a formalisation of Venn diagrams. I was pretty amazed when I discovered this (rather simple) connection for myself, and you should be too.
Suppose your experiment is to toss three coins, and make "partial measurements" on the results through three "measurement devices":
• A: Lights up iff the number of heads was at least 2.
• B: Lights up iff the first two coins landed heads.
• C: Lights up iff the third coin landed heads.
What this means is that $A$ gives you the set $\{HHT, HTH, THH, HHH\}$, $B$ gives you the set $\{HHH, HHT\}$, $C$ gives you the set $\{HHH, HTH, THH, TTH\}$. Based on precisely which devices light up, you can decide the truth values of $\lnot$'s and $\lor$'s of these statements, i.e. complements and unions of these sets -- this is the point of fields, of course.
Or we could visualise things.
Well, the Venn diagram produces a partition of $X$ corresponding to the equivalence relation of "indistinguishability", i.e. "every event containing one outcome contains the other"? The field consists precisely of any set one can "mark" on the Venn diagram -- i.e. unions of the elements of the partition.
A consequence of this becomes immediately obvious:
Given a field $\mathcal{F}$ corresponding to the partition $\sim$, the following bijection holds: $\mathcal{F}\leftrightarrow 2^{X/\sim}$.
Consequences of this include: the cardinalities of finite sigma fields are precisely the powers of two; there is no countably infinite sigma field.
Often, one may want to some raw data from an experiment to obtain some processed data. For example, let $X=\{HH,HT,TH,TT\}$ and the initial measurement is of the number of heads:
\begin{align} \mathcal{F}=&\{\varnothing, \{TT\}, \{HT, TH\}, \{HH\},\\ & \{TT, HT, TH\}, \{TT, HH\}, \{HT, TH, HH\}, X \} \end{align}
What kind of properties of the outcome can we talk about with certainty given the number of heads? For example, we can talk about the question "was there at least one heads?"
$$\mathcal{G}=\{\varnothing, \{TT\}, \{HT, TH, HH\}, X\}$$
There are two ways to understand this "processing" or "re-measuring". One is as a function $f:\frac{X}{\sim_\mathcal{F}}\to \frac{X}{\sim_\mathcal{G}}$. Recall that:
\begin{align} \frac{X}{\sim_\mathcal{F}}&=\{\{TT\},\{HT,TH\},\{HH\}\}\\ \frac{X}{\sim_\mathcal{G}}&=\{\{TT\},\{HT,TH,HH\}\} \end{align}
Any such $f$ is a permissible "measurable function", as long as $\sim_\mathcal{G}$ is at least as coarse a partition as $\sim_\mathcal{F}$. In other words, a function from $X/\sim_1$ to $(X/\sim_1)/\sim_2$ is always measurable.
But there's another, more "natural", less weird and mathematical way to think about a re-measurement -- as a function $f:X\to Y$, where in this case $Y=\{0,1\}$ where an outcome maps to 1 if it has at least one heads, and 0 if it does not.
But there's a catch: knowing that an event $E_Y$ in $Y$ occurred is equivalent to knowing that an outcome in $X$ mapping to $E_Y$ occurred -- i.e. that the event $\{x\in X\mid f(x)\in Y\}$ occurred. Such an event must be in the field on $X$, i.e.
$$\forall y\in\mathcal{F}_Y,f^{-1}(y)\in\mathcal{F}_X$$
This is the condition for a measurable function, also known as a random variable.
One may observe certain analogies between the measurable spaces outlined above, and topology -- in the case of countable sample spaces, there actually is a correspondence. The similarity between a Venn diagram and casual drawings of a topological space is not completely superficial.
The key idea behind fields is mathematically a notion of "distinguishability" -- if all we can measure is the number of heads, $HHTTH$ and $TTHHH$ are identical to us. For all practical purposes, we can view the sample space as the partition by this equivalence relation. They are basically the "same point".
It's this notion that a measurable function seeks to encapsulate -- it is, in a sense, a generalisation of a function from set theory. A function cannot distinguish indistinguishable points -- in set theory, "indistinguishability" is just equality, the discrete partition; a measurable function cannot distinguish indistinguishable points -- but in measurable spaces, "indistinguishability" is given by some equivalence relation.
Let's see this more precisely.
Given sets with equivalence relations $(X,\sim)$, $(Y,\sim)$, we want to ensure that some function $f:X\to Y$ "lifts" to a function $f:\frac{X}{\sim}\to\frac{Y}{\sim}$ such that $f([x])=[f(y)]$.
(Exercise: Show that this (i.e. this "definition" being well-defined) is equivalent to the condition $\forall E\in\mathcal{F}_Y, f^{-1}(E)\in \mathcal{F}_X$. It may help to draw out some examples.)
Well, this expression of the condition -- as $f([x])=[f(y)]$ -- even if technically misleading (the two $f$'s aren't really the same thing) give us the interpretation that a measurable function is one that commutes with the partition or preserves the partition.
While homomorphisms in other settings than measurable spaces do not precisely follow the "cannot distinguish related points" notion, they do follow a generalisation where equivalence relations are replaced with other relations, operations, etc. -- in topology, a continuous function preserves limits; in group theory, a group homomorphism preserves the group operation; in linear algebra, a linear transformation preserves linear combinations; in order theory, an increasing function preserves order, etc. In any case, a homomorphism is a function that does not "break" relationships by creating a "finer" relationship on the target space.
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2021-12-03 01:08:38
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https://www.ssmathematics.com/2021/07/cie-derivative.html
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# Differentation (Application/ Nature of stationary point)
\CIE\0606\2020\w\paper 12 \no 9
A curve has equation $y=(2 x-1) \sqrt{4 x+3}$.
(a) Show that $\dfrac{\mathrm{d} y}{\mathrm{~d} x}=\dfrac{4(A x+B)}{\sqrt{4 x+3}}$, where $A$ and $B$ are constants.
(b) Hence write down the $x$ -coordinate of the stationary point of the curve.
(c) Determine the nature of this stationary point.
%%%%%%%%% Solution %%%%%%%%%
$\def\dbydx#1{\dfrac{d}{dx}\left(#1\right)}\def\funv{(4x+3)^{\dfrac{1}{2}}}\def\funvo{\sqrt{4x+3}}$
(a) $\begin{array}[t]{rcll} \dfrac{dy}{dx}&=&(2x-1) \dfrac{d}{dx} (4x+3)^{\dfrac{1}{2}} +(4x+3)^{\dfrac{1}{2}} \dfrac{d}{dx}(2x-1)\\ &=&(2x-1)\dfrac{1}{2} (4x+3)^{-\dfrac{1}{2}} \times 4 +\funv \times 2\\ &=&\dfrac{2(2x-1)}{\funvo}+2\funvo\\ &=& \dfrac{(4x-2)+(8x+6)}{\funvo}\\ &=& \dfrac{12x+4}{\funvo}\\ &=&\dfrac{4(3x+1)}{\funvo}\end{array}$
(b) For stationary point, $\dfrac{dy}{dx} =0$. Thus $\dfrac{4(3x+1)}{\funvo}=0\Rightarrow x=-\dfrac{1}{3}$.
(c) $\begin{array}[t]{|c||c|c|c|}\hline x& <-\dfrac{1}{3} & =-\dfrac{1}{3}&>-\dfrac{1}{3}\\ \hline 4(3x+1)&-&0&+\\ \funvo& +&+&+\\ \hline \dfrac{dy}{dx}&-&0&+\\ \hline\end{array}$
Thus the stationary point is minimum.
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2021-12-05 22:20:39
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https://www.esaral.com/q/show-that-the-function-53362
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# Show that the function
Question:
Show that the function $f$ given by $f(x)=10^{x}$ is increasing for all $x$ ?
Solution:
we have,
$f(x)=10^{x}$
$\therefore f^{\prime}(x)=10^{x} \log 10$
Now,
$X \in R$
$\Rightarrow 10^{x}>0$
$\Rightarrow 10^{x} \log 10>0$
$\Rightarrow f^{\prime}(x)>0$
Hence, $f(x)$ in an increasing function for all $x$
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2023-02-06 22:59:13
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https://openstax.org/books/chemistry-2e/pages/chapter-7
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Chemistry 2e
# Chapter 7
Chemistry 2eChapter 7
1.
The protons in the nucleus do not change during normal chemical reactions. Only the outer electrons move. Positive charges form when electrons are lost.
3.
P, I, Cl, and O would form anions because they are nonmetals. Mg, In, Cs, Pb, and Co would form cations because they are metals.
5.
(a) P3–; (b) Mg2+; (c) Al3+; (d) O2–; (e) Cl; (f) Cs+
7.
(a) [Ar]4s23d104p6; (b) [Kr]4d105s25p6 (c) 1s2 (d) [Kr]4d10; (e) [He]2s22p6; (f) [Ar]3d10; (g) 1s2 (h) [He]2s22p6 (i) [Kr]4d105s2 (j) [Ar]3d7 (k) [Ar]3d6, (l) [Ar]3d104s2
9.
(a) 1s22s22p63s23p1; Al3+: 1s22s22p6; (b) 1s22s22p63s23p63d104s24p5; 1s22s22p63s23p63d104s24p6; (c) 1s22s22p63s23p63d104s24p65s2; Sr2+: 1s22s22p63s23p63d104s24p6; (d) 1s22s1; Li+: 1s2; (e) 1s22s22p63s23p63d104s24p3; 1s22s22p63s23p63d104s24p6; (f) 1s22s22p63s23p4; 1s22s22p63s23p6
11.
NaCl consists of discrete ions arranged in a crystal lattice, not covalently bonded molecules.
13.
ionic: (b), (d), (e), (g), and (i); covalent: (a), (c), (f), (h), (j), and (k)
15.
(a) Cl; (b) O; (c) O; (d) S; (e) N; (f) P; (g) N
17.
(a) H, C, N, O, F; (b) H, I, Br, Cl, F; (c) H, P, S, O, F; (d) Na, Al, H, P, O; (e) Ba, H, As, N, O
19.
N, O, F, and Cl
21.
(a) HF; (b) CO; (c) OH; (d) PCl; (e) NH; (f) PO; (g) CN
23.
(a) eight electrons:
(b) eight electrons:
(c) no electrons Be2+
(d) eight electrons:
(e) no electrons Ga3+
(f) no electrons Li+
(g) eight electrons:
25.
(a)
(b)
(c)
(d)
(e)
(f)
27.
29.
(a)
In this case, the Lewis structure is inadequate to depict the fact that experimental studies have shown two unpaired electrons in each oxygen molecule.
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
31.
(a) SeF6:
(b) XeF4:
(c) $SeCl3+:SeCl3+:$
(d) Cl2BBCl2:
33.
Two valence electrons per Pb atom are transferred to Cl atoms; the resulting Pb2+ ion has a 6s2 valence shell configuration. Two of the valence electrons in the HCl molecule are shared, and the other six are located on the Cl atom as lone pairs of electrons.
35.
37.
39.
(a)
(b)
(c)
(d)
(e)
41.
43.
Each bond includes a sharing of electrons between atoms. Two electrons are shared in a single bond; four electrons are shared in a double bond; and six electrons are shared in a triple bond.
45.
(a)
(b)
(c)
(d)
(e)
47.
49.
(a)
(b)
CO has the strongest carbon-oxygen bond because there is a triple bond joining C and O. CO2 has double bonds.
51.
(a) H: 0, Cl: 0; (b) C: 0, F: 0; (c) P: 0, Cl 0; (d) P: 0, F: 0
53.
Cl in Cl2: 0; Cl in BeCl2: 0; Cl in ClF5: 0
55.
(a)
(b)
(c)
(d)
57.
HOCl
59.
The structure that gives zero formal charges is consistent with the actual structure:
61.
NF3;
63.
65.
(a) −114 kJ; (b) 30 kJ; (c) −1055 kJ
67.
The greater bond energy is in the figure on the left. It is the more stable form.
69.
$HCl(g)⟶12H2(g)+12Cl2(g)ΔH1°=−ΔHf[HCl(g)]°12H2(g)⟶H(g)ΔH2°=ΔHf[H(g)]°12Cl2(g)⟶Cl(g)ΔH3°=ΔHf[Cl(g)]°¯HCl(g)⟶H(g)+Cl(g)ΔH°=ΔH1°+ΔH2°+ΔH3°HCl(g)⟶12H2(g)+12Cl2(g)ΔH1°=−ΔHf[HCl(g)]°12H2(g)⟶H(g)ΔH2°=ΔHf[H(g)]°12Cl2(g)⟶Cl(g)ΔH3°=ΔHf[Cl(g)]°¯HCl(g)⟶H(g)+Cl(g)ΔH°=ΔH1°+ΔH2°+ΔH3°$
$DHCl=ΔH°=ΔHf[HCl(g)]°+ΔHf[H(g)]°+ΔHf[Cl(g)]°=−(−92.307kJ)+217.97kJ+121.3kJ=431.6kJDHCl=ΔH°=ΔHf[HCl(g)]°+ΔHf[H(g)]°+ΔHf[Cl(g)]°=−(−92.307kJ)+217.97kJ+121.3kJ=431.6kJ$
71.
The S–F bond in SF4 is stronger.
73.
The C–C single bonds are longest.
75.
(a) When two electrons are removed from the valence shell, the Ca radius loses the outermost energy level and reverts to the lower n = 3 level, which is much smaller in radius. (b) The +2 charge on calcium pulls the oxygen much closer compared with K, thereby increasing the lattice energy relative to a less charged ion. (c) Removal of the 4s electron in Ca requires more energy than removal of the 4s electron in K because of the stronger attraction of the nucleus and the extra energy required to break the pairing of the electrons. The second ionization energy for K requires that an electron be removed from a lower energy level, where the attraction is much stronger from the nucleus for the electron. In addition, energy is required to unpair two electrons in a full orbital. For Ca, the second ionization potential requires removing only a lone electron in the exposed outer energy level. (d) In Al, the removed electron is relatively unprotected and unpaired in a p orbital. The higher energy for Mg mainly reflects the unpairing of the 2s electron.
77.
(d)
79.
4008 kJ/mol; both ions in MgO have twice the charge of the ions in LiF; the bond length is very similar and both have the same structure; a quadrupling of the energy is expected based on the equation for lattice energy
81.
(a) Na2O; Na+ has a smaller radius than K+; (b) BaS; Ba has a larger charge than K; (c) BaS; Ba and S have larger charges; (d) BaS; S has a larger charge
83.
(e)
85.
The placement of the two sets of unpaired electrons in water forces the bonds to assume a tetrahedral arrangement, and the resulting HOH molecule is bent. The HBeH molecule (in which Be has only two electrons to bond with the two electrons from the hydrogens) must have the electron pairs as far from one another as possible and is therefore linear.
87.
Space must be provided for each pair of electrons whether they are in a bond or are present as lone pairs. Electron-pair geometry considers the placement of all electrons. Molecular structure considers only the bonding-pair geometry.
89.
As long as the polar bonds are compensated (for example. two identical atoms are found directly across the central atom from one another), the molecule can be nonpolar.
91.
(a) Both the electron geometry and the molecular structure are octahedral. (b) Both the electron geometry and the molecular structure are trigonal bipyramid. (c) Both the electron geometry and the molecular structure are linear. (d) Both the electron geometry and the molecular structure are trigonal planar.
93.
(a) electron-pair geometry: octahedral, molecular structure: square pyramidal; (b) electron-pair geometry: tetrahedral, molecular structure: bent; (c) electron-pair geometry: octahedral, molecular structure: square planar; (d) electron-pair geometry: tetrahedral, molecular structure: trigonal pyramidal; (e) electron-pair geometry: trigonal bypyramidal, molecular structure: seesaw; (f) electron-pair geometry: tetrahedral, molecular structure: bent (109°)
95.
(a) electron-pair geometry: trigonal planar, molecular structure: bent (120°); (b) electron-pair geometry: linear, molecular structure: linear; (c) electron-pair geometry: trigonal planar, molecular structure: trigonal planar; (d) electron-pair geometry: tetrahedral, molecular structure: trigonal pyramidal; (e) electron-pair geometry: tetrahedral, molecular structure: tetrahedral; (f) electron-pair geometry: trigonal bipyramidal, molecular structure: seesaw; (g) electron-pair geometry: tetrahedral, molecular structure: trigonal pyramidal
97.
All of these molecules and ions contain polar bonds. Only ClF5, $ClO2−,ClO2−,$ PCl3, SeF4, and $PH2−PH2−$ have dipole moments.
99.
SeS2, CCl2F2, PCl3, and ClNO all have dipole moments.
101.
P
103.
nonpolar
105.
(a) tetrahedral; (b) trigonal pyramidal; (c) bent (109°); (d) trigonal planar; (e) bent (109°); (f) bent (109°); (g) CH3CCH tetrahedral, CH3CCH linear; (h) tetrahedral; (i) H2CCCH2 linear; H2CCCH2 trigonal planar
107.
109.
(a)
(b)
(c)
(d) $CS32−CS32−$ includes three regions of electron density (all are bonds with no lone pairs); the shape is trigonal planar; CS2 has only two regions of electron density (all bonds with no lone pairs); the shape is linear
111.
The Lewis structure is made from three units, but the atoms must be rearranged:
113.
The molecular dipole points away from the hydrogen atoms.
115.
The structures are very similar. In the model mode, each electron group occupies the same amount of space, so the bond angle is shown as 109.5°. In the “real” mode, the lone pairs are larger, causing the hydrogens to be compressed. This leads to the smaller angle of 104.5°.
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2020-10-26 13:15:02
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https://www.physicsforums.com/threads/i-need-some-help-with-some-basic-material.45635/
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# I need some help with some basic material
1. Oct 1, 2004
I can't seem to figure these problems if anyone can give me the concept and start me off that would be great.
Problem one: A car is traveling at a constant speed of 27 m/s on a highway. At the instant this car passes an entrance ramp, a second car enters the highway from the ramp. The second car starts from rest and has a constant acceleration. What acceleration must it maintain so that the two cars meet for the first time at the next exit which is 1.8 km away?
Problem two: A bus stop to pick up riders. A woman is running at a constant velocity of +5.0 m/s in an attempt to catch the bus When she is 11 m from the bus, it pulls away with a constant acceleration of +1.0 m/s^2. From this point, how much time does it take her to reach the bus if she keeps running with the same velocity?
*edit*
Oh sorry i posted this after i read the before you post thread Once again Im sorry for posting Homework help here
Last edited: Oct 1, 2004
2. Oct 1, 2004
### Pyrrhus
For the first problem consider the formulas
$$vt = x$$
$$\frac{1}{2}at^2 = x$$
You go the speed for the car and the distance x they will both travel, what can you do with that?
For the second problem
She will do
$$vt = x + 11$$
while the bus will do
$$\frac{1}{2}at^2 = x$$
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2017-01-19 00:37:58
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|
https://msp.org/involve/2018/11-1/involve-v11-n1-p05-s.pdf
|
Vol. 11, No. 1, 2018
Recent Issues
The Journal About the Journal Editorial Board Editors’ Interests Subscriptions Submission Guidelines Submission Form Policies for Authors Ethics Statement ISSN: 1944-4184 (e-only) ISSN: 1944-4176 (print) Author Index Coming Soon Other MSP Journals
Labeling crossed prisms with a condition at distance two
Matthew Beaudouin-Lafon, Serena Chen, Nathaniel Karst, Jessica Oehrlein and Denise Sakai Troxell
Vol. 11 (2018), No. 1, 67–80
DOI: 10.2140/involve.2018.11.67
Abstract
An L(2,1)-labeling of a graph is an assignment of nonnegative integers to its vertices such that adjacent vertices are assigned labels at least two apart, and vertices at distance two are assigned labels at least one apart. The $\lambda$-number of a graph is the minimum span of labels over all its L(2,1)-labelings. A generalized Petersen graph (GPG) of order $n$ consists of two disjoint cycles on $n$ vertices, called the inner and outer cycles, respectively, together with a perfect matching in which each matching edge connects a vertex in the inner cycle to a vertex in the outer cycle. A prism of order $n\ge 3$ is a GPG that is isomorphic to the Cartesian product of a path on two vertices and a cycle on $n$ vertices. A crossed prism is a GPG obtained from a prism by crossing two of its matching edges; that is, swapping the two inner cycle vertices on these edges. We show that the $\lambda$-number of a crossed prism is 5, 6, or 7 and provide complete characterizations of crossed prisms attaining each one of these $\lambda$-numbers.
However, your active subscription may be available on Project Euclid at
https://projecteuclid.org/involve
We have not been able to recognize your IP address 34.232.51.240 as that of a subscriber to this journal.
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or by using our contact form.
Keywords
L(2,1)-labeling, L(2,1)-coloring, distance two labeling, channel assignment, generalized Petersen graph
Mathematical Subject Classification 2010
Primary: 68R10, 94C15
Secondary: 05C15, 05C78
Supplementary material
Diagrams of $D_1, D_2, D_3$ and table of labelings
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2020-10-27 20:55:12
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https://d2l.ai/chapter_multilayer-perceptrons/numerical-stability-and-init.html
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# 4.8. Numerical Stability and Initialization¶ Open the notebook in Colab Open the notebook in Colab Open the notebook in Colab
Thus far, every model that we have implemented required that we initialize its parameters according to some pre-specified distribution. Until now, we took the initialization scheme for granted, glossing over the details of how these these choices are made. You might have even gotten the impression that these choices are not especially important. To the contrary, the choice of initialization scheme plays a significant role in neural network learning, and it can be crucial for maintaining numerical stability. Moreover, these choices can be tied up in interesting ways with the choice of the nonlinear activation function. Which function we choose and how we initialize parameters can determine how quickly our optimization algorithm converges. Poor choices here can cause us to encounter exploding or vanishing gradients while training. In this section, we delve into these topics with greater detail and discuss some useful heuristics that you will find useful throughout your career in deep learning.
## 4.8.1. Vanishing and Exploding Gradients¶
Consider a deep network with $$L$$ layers, input $$\mathbf{x}$$ and output $$\mathbf{o}$$. With each layer $$l$$ defined by a transformation $$f_l$$ parameterized by weights $$\mathbf{W}_l$$ our network can be expressed as:
(4.8.1)$\mathbf{h}^{l+1} = f_l (\mathbf{h}^l) \text{ and thus } \mathbf{o} = f_L \circ \ldots, \circ f_1(\mathbf{x}).$
If all activations and inputs are vectors, we can write the gradient of $$\mathbf{o}$$ with respect to any set of parameters $$\mathbf{W}_l$$ as follows:
(4.8.2)$\partial_{\mathbf{W}_l} \mathbf{o} = \underbrace{\partial_{\mathbf{h}^{L-1}} \mathbf{h}^L}_{:= \mathbf{M}_L} \cdot \ldots, \cdot \underbrace{\partial_{\mathbf{h}^{l}} \mathbf{h}^{l+1}}_{:= \mathbf{M}_l} \underbrace{\partial_{\mathbf{W}_l} \mathbf{h}^l}_{:= \mathbf{v}_l}.$
In other words, this gradient is the product of $$L-l$$ matrices $$\mathbf{M}_L \cdot \ldots, \cdot \mathbf{M}_l$$ and the gradient vector $$\mathbf{v}_l$$. Thus we are susceptible to the same problems of numerical underflow that often crop up when multiplying together too many probabilities. When dealing with probabilities, a common trick is to switch into log-space, i.e., shifting pressure from the mantissa to the exponent of the numerical representation. Unfortunately, our problem above is more serious: initially the matrices $$M_l$$ may have a wide variety of eigenvalues. They might be small or large, and their product might be very large or very small.
The risks posed by unstable gradients go beyond numerical representation. Gradients of unpredictable magnitude also threaten the stability of our optimization algorithms. We may be facing parameter updates that are either (i) excessively large, destroying our model (the exploding gradient problem); or (ii) excessively small (the vanishing gradient problem), rendering learning impossible as parameters hardly move on each update.
One frequent culprit causing the vanishing gradient problem is the choice of the activation function $$\sigma$$ that is appended following each layer’s linear operations. Historically, the sigmoid function $$1/(1 + \exp(-x))$$ (introduced in :numref:sec_mlp) was popular because it resembles a thresholding function. Since early artificial neural networks were inspired by biological neural networks, the idea of neurons that fire either fully or not at all (like biological neurons) seemed appealing. Let us take a closer look at the sigmoid to see why it can cause vanishing gradients.
%matplotlib inline
from d2l import mxnet as d2l
from mxnet import autograd, np, npx
npx.set_np()
x = np.arange(-8.0, 8.0, 0.1)
y = npx.sigmoid(x)
y.backward()
%matplotlib inline
from d2l import torch as d2l
import torch
x = torch.arange(-8.0, 8.0, 0.1, requires_grad=True)
y = torch.sigmoid(x)
y.backward(torch.ones_like(x))
%matplotlib inline
from d2l import tensorflow as d2l
import tensorflow as tf
x = tf.Variable(tf.range(-8.0, 8.0, 0.1))
y = tf.nn.sigmoid(x)
As you can see, the sigmoid’s gradient vanishes both when its inputs are large and when they are small. Moreover, when backpropagating through many layers, unless we are in the Goldilocks zone, where the inputs to many of the sigmoids are close to zero, the gradients of the overall product may vanish. When our network boasts many layers, unless we are careful, the gradient will likely be cut off at some layer. Indeed, this problem used to plague deep network training. Consequently, ReLUs, which are more stable (but less neurally plausible), have emerged as the default choice for practitioners.
The opposite problem, when gradients explode, can be similarly vexing. To illustrate this a bit better, we draw $$100$$ Gaussian random matrices and multiply them with some initial matrix. For the scale that we picked (the choice of the variance $$\sigma^2=1$$), the matrix product explodes. When this happens due to the initialization of a deep network, we have no chance of getting a gradient descent optimizer to converge.
M = np.random.normal(size=(4, 4))
print('A single matrix', M)
for i in range(100):
M = np.dot(M, np.random.normal(size=(4, 4)))
print('After multiplying 100 matrices', M)
A single matrix [[ 2.2122064 1.1630787 0.7740038 0.4838046 ]
[ 1.0434405 0.29956347 1.1839255 0.15302546]
[ 1.8917114 -1.1688148 -1.2347414 1.5580711 ]
[-1.771029 -0.5459446 -0.45138445 -2.3556297 ]]
After multiplying 100 matrices [[ 3.4459714e+23 -7.8040680e+23 5.9973287e+23 4.5229990e+23]
[ 2.5275089e+23 -5.7240326e+23 4.3988473e+23 3.3174740e+23]
[ 1.3731286e+24 -3.1097155e+24 2.3897773e+24 1.8022959e+24]
[-4.4951040e+23 1.0180033e+24 -7.8232281e+23 -5.9000354e+23]]
M = torch.normal(0, 1, size=(4,4))
print('A single matrix \n',M)
for i in range(100):
M = torch.mm(M,torch.normal(0, 1, size=(4,4)))
print('After multiplying 100 matrices\n',M)
A single matrix
tensor([[ 0.3021, 0.8541, -0.0111, -1.7852],
[ 1.3675, 0.5243, 0.0085, 0.7796],
[-0.8776, 0.2312, 2.7465, 0.2879],
[ 1.2510, -0.5377, -0.1983, -0.1201]])
After multiplying 100 matrices
tensor([[-9.0353e+25, 5.7558e+25, -8.6537e+25, 1.5004e+25],
[-1.1864e+26, 7.5576e+25, -1.1363e+26, 1.9701e+25],
[ 8.8911e+25, -5.6639e+25, 8.5157e+25, -1.4764e+25],
[ 1.4998e+25, -9.5542e+24, 1.4365e+25, -2.4905e+24]])
M = tf.random.normal((4, 4))
print('A single matrix \n', M)
for i in range(100):
M = tf.matmul(M, tf.random.normal((4, 4)))
print('After multiplying 100 matrices\n', M.numpy())
A single matrix
tf.Tensor(
[[ 3.011709 -0.12220125 1.6440876 -0.74108595]
[ 0.280008 1.3707088 1.9497523 0.840189 ]
[-0.17846034 1.4799802 0.5280843 0.80384254]
[-0.83082753 -0.9812137 -0.0491188 -0.16511448]], shape=(4, 4), dtype=float32)
After multiplying 100 matrices
[[ 5.96912592e+19 1.27400465e+20 1.50725804e+19 -2.38673434e+20]
[-7.78999766e+19 -1.66263856e+20 -1.96704060e+19 3.11480596e+20]
[-6.52206152e+19 -1.39201884e+20 -1.64687952e+19 2.60782291e+20]
[ 8.23719477e+18 1.75808457e+19 2.07996511e+18 -3.29361479e+19]]
### 4.8.1.3. Symmetry¶
Another problem in deep network design is the symmetry inherent in their parametrization. Assume that we have a deep network with one hidden layer and two units, say $$h_1$$ and $$h_2$$. In this case, we could permute the weights $$\mathbf{W}_1$$ of the first layer and likewise permute the weights of the output layer to obtain the same function. There is nothing special differentiating the first hidden unit vs the second hidden unit. In other words, we have permutation symmetry among the hidden units of each layer.
This is more than just a theoretical nuisance. Imagine what would happen if we initialized all of the parameters of some layer as $$\mathbf{W}_l = c$$ for some constant $$c$$. In this case, the gradients for all dimensions are identical: thus not only would each unit take the same value, but it would receive the same update. Stochastic gradient descent would never break the symmetry on its own and we might never be able to realize the network’s expressive power. The hidden layer would behave as if it had only a single unit. Note that while SGD would not break this symmetry, dropout regularization would!
## 4.8.2. Parameter Initialization¶
One way of addressing—or at least mitigating—the issues raised above is through careful initialization. Additional care during optimization and suitable regularization can further enhance stability.
### 4.8.2.1. Default Initialization¶
In the previous sections, e.g., in Section 3.3, we used a normal distribution with 0 mean and 0.01 variance to initialize the values of our weights. If we don’t specify the initialization method, the framework will use a default random initialization method, which often works well in practice for moderate problem sizes.
### 4.8.2.2. Xavier Initialization¶
Let us look at the scale distribution of the activations of the hidden units $$h_{i}$$ for some layer. They are given by
(4.8.3)$h_{i} = \sum_{j=1}^{n_\mathrm{in}} W_{ij} x_j.$
The weights $$W_{ij}$$ are all drawn independently from the same distribution. Furthermore, let us assume that this distribution has zero mean and variance $$\sigma^2$$ (this does not mean that the distribution has to be Gaussian, just that the mean and variance need to exist). For now, let us assume that the inputs to layer $$x_j$$ also have zero mean and variance $$\gamma^2$$ and that they are independent of $$\mathbf{W}$$. In this case, we can compute the mean and variance of $$h_i$$ as follows:
(4.8.4)\begin{split}\begin{aligned} E[h_i] & = \sum_{j=1}^{n_\mathrm{in}} E[W_{ij} x_j] = 0, \\ Var[h_i] & = E[h_i^2] - (E[h_i])^2 \\ & = \sum_{j=1}^{n_\mathrm{in}} E[W^2_{ij} x^2_j] - 0 \\ & = \sum_{j=1}^{n_\mathrm{in}} E[W^2_{ij}] E[x^2_j] \\ & = n_\mathrm{in} \sigma^2 \gamma^2. \end{aligned}\end{split}
One way to keep the variance fixed is to set $$n_\mathrm{in} \sigma^2 = 1$$. Now consider backpropagation. There we face a similar problem, albeit with gradients being propagated from the top layers. That is, instead of $$\mathbf{W} \mathbf{x}$$, we need to deal with $$\mathbf{W}^\top \mathbf{g}$$, where $$\mathbf{g}$$ is the incoming gradient from the layer above. Using the same reasoning as for forward propagation, we see that the gradients’ variance can blow up unless $$n_\mathrm{out} \sigma^2 = 1$$. This leaves us in a dilemma: we cannot possibly satisfy both conditions simultaneously. Instead, we simply try to satisfy:
(4.8.5)\begin{aligned} \frac{1}{2} (n_\mathrm{in} + n_\mathrm{out}) \sigma^2 = 1 \text{ or equivalently } \sigma = \sqrt{\frac{2}{n_\mathrm{in} + n_\mathrm{out}}}. \end{aligned}
This is the reasoning underlying the now-standard and practically beneficial Xavier initialization, named for its creator [Glorot & Bengio, 2010]. Typically, the Xavier initialization samples weights from a Gaussian distribution with zero mean and variance $$\sigma^2 = 2/(n_\mathrm{in} + n_\mathrm{out})$$. We can also adapt Xavier’s intuition to choose the variance when sampling weights from a uniform distribution. Note the distribution $$U[-a, a]$$ has variance $$a^2/3$$. Plugging $$a^2/3$$ into our condition on $$\sigma^2$$ yields the suggestion to initialize according to $$U\left[-\sqrt{6/(n_\mathrm{in} + n_\mathrm{out})}, \sqrt{6/(n_\mathrm{in} + n_\mathrm{out})}\right]$$.
### 4.8.2.3. Beyond¶
The reasoning above barely scratches the surface of modern approaches to parameter initialization. A deep learning framework often implements over a dozen different heuristics. Moreover, parameter initialization continues to be a hot area of fundamental research in deep learning. Among these are heuristics specialized for tied (shared) parameters, super-resolution, sequence models, and other situations. If the topic interests you we suggest a deep dive into this module’s offerings, reading the papers that proposed and analyzed each heuristic, and then exploring the latest publications on the topic. Perhaps you will stumble across (or even invent!) a clever idea and contribute an implementation to deep learning frameworks.
## 4.8.3. Summary¶
• Vanishing and exploding gradients are common issues in deep networks. Great care in parameter initialization is required to ensure that gradients and parameters remain well controlled.
• Initialization heuristics are needed to ensure that the initial gradients are neither too large nor too small.
• ReLU activation functions mitigate the vanishing gradient problem. This can accelerate convergence.
• Random initialization is key to ensure that symmetry is broken before optimization.
## 4.8.4. Exercises¶
1. Can you design other cases where a neural network might exhibit symmetry requiring breaking besides the permutation symmetry in a multilayer pereceptron’s layers?
2. Can we initialize all weight parameters in linear regression or in softmax regression to the same value?
3. Look up analytic bounds on the eigenvalues of the product of two matrices. What does this tell you about ensuring that gradients are well conditioned?
4. If we know that some terms diverge, can we fix this after the fact? Look at the paper on LARS for inspiration [You et al., 2017].
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2020-07-14 12:20:23
|
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|
http://www.newton.ac.uk/event/hhhw02/seminars
|
Seminars (HHHW02)
Videos and presentation materials from other INI events are also available.
Search seminar archive
Event When Speaker Title Presentation Material
HHHW02 13th August 2018
10:00 to 11:00
Daniel Dugger Surfaces with involutions
I will talk about the classification of surfaces with involution, together with remarks on the RO(Z/2)-graded Bredon cohomology of these objects.
HHHW02 13th August 2018
11:30 to 12:30
Jeremiah Heller A motivic Segal conjecture
HHHW02 13th August 2018
14:30 to 15:30
Agnes Beaudry Pic(E-Z/4) and Tools to Compute It
In this talk, we describe tools to compute the Picard group of the category of E-G-module spectra where E is Morava E-theory at n=p=2 and G is a finite cyclic subgroup of the Morava stabilizer of order 4. We discuss two tools: 1) A group homomorphism from RO(G) to Pic(E-G) and methods for computing it. 2) The Picard Spectral sequence and some of its equivariant properties. This work is joint with Irina Bobkova, Mike Hill and Vesna Stojanoska
HHHW02 13th August 2018
16:00 to 17:00
Clark Barwick Exodromy and endodromy
On the étale homotopy type of a scheme, there is a natural stratification. We describe both it and its dual, and we discuss some of the ramification information it captures. Joint work with Saul Glasman, Peter Haine, Tomer Schlank.
HHHW02 14th August 2018
09:00 to 10:00
This is report on part of a program to give refinements of numerical invariants arising in enumerative geometry to invariants living in the Grothendieck-Witt ring over the base-field. Here we define an invariant in the Grothendieck-Witt ring for counting'' rational curves. More precisely, for a del Pezzo surface S over a field k and a positive degree curve class $D$ (with respect to the anti-canonical class $-K_S$), we define a class in the Grothendiek-Witt ring of k, whose rank gives the number of rational curves in the class D containing a given collection of distinct closed points $\mathfrak{p}=\sum_ip_i$ of total degree $-D\cdot K_S-1$. This recovers Welschinger's invariants in case $k=\mathbb{R}$ by applying the signature map. The main result is that this quadratic invariant depends only on the $\mathbb{A}^1$-connected component containing $\mathfrak{p}$ in $Sym^{3d-1}(S)^0(k)$, where $Sym^{3d-1}(S)^0$ is the open subscheme of $Sym^{3d-1}(S)$ parametrizing geometrically reduced 0-cycles.
HHHW02 14th August 2018
11:30 to 12:30
Magdalena Kedziorek Algebraic models for rational equivariant commutative ring spectra
I will recall different levels of commutativity in G-equivariant stable homotopy theory and show how to understand them rationally when G is finite or SO(2) using algebraic models. This is joint work with D. Barnes and J.P.C. Greenlees.
HHHW02 14th August 2018
14:30 to 15:30
Marc Hoyois Motivic infinite loop spaces and Hilbert schemes
Co-authors: Elden Elmanto (Northwestern University), Adeel A. Khan (Universität Regensburg), Vladimir Sosnilo (St. Petersburg State University), Maria Yakerson (Universität Duisburg-Essen)We prove a recognition principle for motivic infinite loop spaces over a perfect field of characteristic not 2. This is achieved by developing a theory of framed motivic spaces, which is a motivic analogue of the theory of E-infinity-spaces. Our main result is that grouplike framed motivic spaces are equivalent to the full subcategory of motivic spectra generated under colimits by suspension spectra. As a consequence, we deduce some representability results for suspension spectra of smooth varieties and various motivic Thom spectra in terms of Hilbert schemes of points in affine spaces.Related Linkshttps://arxiv.org/abs/1711.05248 - preprint
HHHW02 14th August 2018
16:00 to 17:00
Vesna Stojanoska Galois extensions in motivic homotopy theory
There are two notions of homotopical Galois extensions in the motivic setting; I will discuss what we know about each, along with illustrative examples. This is joint work in progress with Beaudry, Heller, Hess, Kedziorek, and Merling.
HHHW02 15th August 2018
09:00 to 10:00
Doug Ravenel Model category structures for equivariant spectra
We will discuss methods for constructing a model structure the category of orthogonal equivariant spectra for a finite group G. The most convenient one is two large steps away from the most obvious one.
HHHW02 15th August 2018
10:00 to 11:00
Oliver Roendigs On very effective hermitian K-theory
We argue that the very effective cover of hermitian K-theory in the sense of motivic homotopy theory is a convenient algebro-geometric generalization of the connective real topological K-theory spectrum. This means the very effective cover acquires the correct Betti realization, its motivic cohomology has the desired structure as a module over the motivic Steenrod algebra, and that its motivic Adams and slice spectral sequences are amenable to calculations. The latter applies to provide the expected connectivity for its unit map from the motivic sphere spectrum. This is joint work with Alexey Ananyevskiy and Paul Arne Ostvaer.
HHHW02 15th August 2018
11:30 to 12:30
Teena Gerhardt Hochschild homology for Green functors
Hochschild homology of a ring has a topological analogue for ring spectra, topological Hochschild homology (THH), which plays an essential role in the trace method approach to algebraic K-theory. For a C_n-equivariant ring spectrum, one can define C_n-relative THH. This leads to the question: What is the algebraic analogue of C_n-relative THH? In this talk, I will define twisted Hochschild homology for Green functors, which allows us to describe this algebraic analogue. This also leads to a theory of Witt vectors for Green functors, as well as an algebraic analogue of TR-theory. This is joint work with Andrew Blumberg, Mike Hill, and Tyler Lawson.
HHHW02 16th August 2018
09:00 to 10:00
Mark Behrens C_2 equivariant homotopy groups from real motivic homotopy groups
The Betti realization of a real motivic spectrum is a genuine C_2 spectrum. It is well known (c.f. the work of Dugger-Isaksen) that the homotopy groups of the Betti realization of a complex motivic spectrum can be computed by "inverting tau". I will describe a similar theorem which describes the C_2-equivariant RO(G) graded homotopy groups of the Betti realization of a cellular real motivic spectrum in terms of its bigraded real motivic homotopy groups. This is joint work with Jay Shah.
HHHW02 16th August 2018
10:00 to 11:00
Tom Bachmann Motivic normed spectra and Tambara functors
Motivic normed spectra are a motivic analogue of the G-commutative ring spectra from genuine equivariant stable homotopy theory. The category of G-commutative ring spectra such that the underlying genuine equivariant spectrum is concentrated in degree zero (i.e. is a Mackey functor) is in fact equivalent to the category of G-Tambara functors. I will explain an analogous motivic result: the category of normed motivic spectra (over a field), such that the underlying motivic spectrum is an effective homotopy module, is equivalent to a category of homotopy invariant presheaves with generalized transfers and étale norms.
HHHW02 16th August 2018
11:30 to 12:30
Anna Marie Bohmann Graded Tambara functors
Let E be a G-spectrum for a finite group G. It's long been known that homotopy groups of E have the structure of "Mackey functors." If E is G commutative ring spectrum, then work of Strickland and of Brun shows that the zeroth homotopy groups of E form a "Tambara functor." This is more structure than just a Mackey functor with commutative multiplication and there is much recent work investigating nuances of this structure. I will discuss work with Vigleik Angeltveit that extends this result to include the higher homotopy groups of E. Specifically, if E has a commutative multiplication that enjoys lots of structure with respect to the G action, the homotopy groups of E form a graded Tambara functor. In particular, genuine commutative G ring spectra enjoy this property.
HHHW02 16th August 2018
14:30 to 15:30
Mona Merling Toward the equivariant stable parametrized h-cobordism theorem
Waldhausen's introduction of A-theory of spaces revolutionized the early study of pseudo-isotopy theory. Waldhausen proved that the A-theory of a manifold splits as its suspension spectrum and a factor Wh(M) whose first delooping is the space of stable h-cobordisms, and its second delooping is the space of stable pseudo-isotopies. I will describe a joint project with C. Malkiewich aimed at telling the equivariant story if one starts with a manifold M with group action by a finite group G.
HHHW02 16th August 2018
16:00 to 17:00
Fabien Morel Etale and motivic variations on Smith's theory on action of cyclic groups
In this talk, after recalling some facts on Smith's theory in classical topology, I will present some analogues in algebraic geometry using \'etale cohomology first, and I will also give analogues in \A^1-homotopy theory and motives of Smith's theory concerning the multiplicative group \G_m, which behaves like a "cyclic of order 2".
HHHW02 17th August 2018
09:00 to 10:00
Aravind Asok On Suslin's Hurewicz homomorphism
I will discuss some recent progress on an old conjecture of Suslin about the image of a certain Hurewicz" map from Quillen's algebraic K-theory of a field F to the Milnor K-theory of F. This is based on joint work with J. Fasel and T.B. Williams.
HHHW02 17th August 2018
10:00 to 11:00
Carolyn Yarnall Klein four-slices of HF_2
The slice filtration is a filtration of equivariant spectra analogous to the Postnikov tower that was developed by Hill, Hopkins, and Ravenel in their solution to the Kervaire invariant-one problem. Since that time there have been several new developments, many dealing with cyclic groups. In this talk, we will focus our attention on a noncyclic group! After recalling a few essential definitions and previous results, we will investigate some computational tools that allow us to leverage homotopy results of Holler-Kriz to determine the slices of integer suspensions of HF_2 when our group is the Klein four group. We will end with a few slice tower and spectral sequence examples demonstrating the patterns that arise in the filtration. This is joint work with Bert Guillou.
HHHW02 17th August 2018
11:30 to 12:30
Kirsten Wickelgren Some results in A1-enumerative geometry
We will discuss several applications of A1-homotopy theory to enumerative geometry. This talk includes joint work with Jesse Kass and Padmavathi Srinivasan.
HHHW02 17th August 2018
14:30 to 15:30
Thomas Nikolaus Cyclotomic spectra and Cartier modules
We start by reviewing the notion of cyclotomic spectra and basic examples such as THH. Then we introduce the classical algebraic notion of an (integral, p-typical) Cartier module and its higher generalization to spectra (topological Cartier modules). We present examples such as Witt vectors and K-theory of endomorphisms. The main result, which is joint work with Ben Antieau, is that there is a close connection between the two notions.
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2020-08-03 11:41:18
|
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|
https://eccc.weizmann.ac.il/report/2019/042/
|
Under the auspices of the Computational Complexity Foundation (CCF)
REPORTS > DETAIL:
### Paper:
TR19-042 | 18th March 2019 11:58
#### Determinant equivalence test over finite fields and over $\mathbf{Q}$
TR19-042
Authors: Ankit Garg, Nikhil Gupta, Neeraj Kayal, Chandan Saha
Publication: 18th March 2019 14:28
Keywords:
Abstract:
The determinant polynomial $Det_n(\mathbf{x})$ of degree $n$ is the determinant of a $n \times n$ matrix of formal variables. A polynomial $f$ is equivalent to $Det_n$ over a field $\mathbf{F}$ if there exists a $A \in GL(n^2,\mathbf{F})$ such that $f = Det_n(A \cdot \mathbf{x})$. Determinant equivalence test over $\mathbf{F}$ is the following algorithmic task: Given black-box access to a $f \in \mathbf{F}[\mathbf{x}]$, check if $f$ is equivalent to $Det_n$ over $\mathbf{F}$, and if so then output a transformation matrix $A \in GL(n^2,\mathbf{F})$. In Kayal (2012), a randomized polynomial time determinant equivalence test was given over $\mathbf{C}$. But, to our knowledge, the complexity of the problem over finite fields and over rationals was not well understood.
In this work, we give a randomized $poly(n,\log |\mathbf{F}|)$ time determinant equivalence test over finite fields $\mathbf{F}$ (under mild restrictions on the characteristic and size of $\mathbf{F}$). Over rationals, we give an efficient randomized reduction from factoring square-free integers to determinant equivalence test for quadratic forms (i.e. the $n=2$ case), assuming GRH. This shows that designing a polynomial-time determinant equivalence test over rationals is a challenging task. Nevertheless, we show that determinant equivalence test over rationals is decidable: For bounded $n$, there is a randomized polynomial-time determinant equivalence test over rationals with access to an oracle for integer factoring. Moreover, for any $n$, there is a randomized polynomial-time algorithm that takes input black-box access to a rational polynomial $f$ and if $f$ is equivalent to $Det_n$ over rationals then it returns a $A \in GL(n^2,\mathbf{L})$ such that $f = Det_n(A \cdot \mathbf{x})$, where $\mathbf{L}$ is an extension field of $\mathbf{Q}$ of degree at most $n$.
The above algorithms over finite fields and over rationals are obtained by giving a polynomial-time randomized reduction from determinant equivalence test to another problem, namely the full matrix algebra isomorphism problem. We also show a reduction in the converse direction which is efficient if $n$ is bounded. These reductions, which hold over any $\mathbf{F}$ (under mild restrictions on the characteristic and size of $\mathbf{F}$), establish a close connection between the complexity of the two problems. This then lead to our results via applications of known results on the full algebra isomorphism problem over finite fields and over $\mathbf{Q}$.
ISSN 1433-8092 | Imprint
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2022-06-29 09:38:57
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http://math.stackexchange.com/questions/43502/is-there-a-standard-or-common-way-to-concisely-write-scientific-notation-in-diff
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# Is there a standard or common way to concisely write scientific notation in different bases?
Is there a standard or common way to write scientific notation in different bases that doesn't require repeating the base in both the coefficient and the exponent base?
For example, this notation is correct, but is quite verbose:
• Implicit base 10: $100 \cdot \tau = 6.283 \times 10^2$
• Explicit base 10: $100 \cdot \tau = 6.283_{10} \times 10_{10}^2$
• Explicit base 16: $100 \cdot \tau = 2.745_{16} \times 10_{16}^2$
Most calculators & programming languages shorten this to an "exponent" notation such as:
6.283e2
Some rare ones even support a base, such as:
16#2.745#e2#
Unfortunately, these are not very nice notations outside of specific domains.
Ideally, I'm looking for an operator notation that concisely does the function:
• $f(coefficient_{b},exponent) = coefficient_{b} \times b^{exponent}$
If there is no such standard or common notation, I'd be happy to hear any suggestions.
-
How about "$\tau = 2.745 \times 10^2$ in base $16$" (or "τ = 2.745e2 in base 16", whichever you prefer)? – ShreevatsaR Jun 5 '11 at 22:42
what's a bit ugly is that you're still using base 10 to denote the base you're using. Maybe, in a way, it's more natural to denote your base by the highest number below the radix, instead of the radix itself, so you'd have base 9 or base F instead of base 10 or base 16 in 'metabase' 10. – gfes Jun 5 '11 at 23:03
@gfes I think your (base-1) notation has some appeal, but using 10 as the "metabase" seems to be the only way I've ever seen it done, and seems to be universally understood. "Base F" immediately makes one think of hexadecimal, but I think if you wrote "base 9" it would be confusing! – wjl Jun 5 '11 at 23:33
True. Although when working in base 10, probably anyone would omit the base 9 / base 10 altogether anyway. Also, I thought of a neater, but less practical way, namely writing the alternative expansion for the radix. I.e. 9.9999999999 etc. or F.FFFFFFFFF etc. – gfes Jun 5 '11 at 23:37
@gfes The alternate expansion is quite clever. It would work especially well with an overbar to indicate a the repeating fractional part. – wjl Jun 5 '11 at 23:43
show 1 more comment
Unfortunately, $2.546_{16}\times 10_{16}^{23}$ is cumbersome, and it isn't immediately clear if the exponent is being written in base $10$ or base $16$, and $2.546_{16}\times 16^{23}$ has a strange mixing of base $10$ and base $16$ elements. For programming things, there are explicit prefixes you can put on a number to denote whether it is base $2,8,10$ or $16$, but that isn't general, and it still doesn't really address any of the previous problems.
Maybe a notation like $(2.546\times b^{23})_{16}$ would work to indicate that everything is supposed to be taken base $16$, with $b$ being implicit for the base (as writing either 10 or the base 10 expression of the base is awkward). However, I'm not fully enamored with this notation either.
What is your application of this? In what circumstances is some sort of scientific notation required but base $10$ is undesirable?
Um, one possibility is also to use the shift operators a la C/C++. So $(2.546 \ll 23)_{16}$? – Willie Wong Jul 5 '11 at 23:47
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2014-03-10 11:07:44
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http://imkean.com/leetcode/306-additive-number/
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07/17/2016 Depth First Search Math String Dynamic Programming
## Question
A valid additive sequence should contain at least three numbers. Except for the first two numbers, each subsequent number in the sequence must be the sum of the preceding two.
For example:
"112358" is an additive number because the digits can form an additive sequence: 1, 1, 2, 3, 5, 8.
1 + 1 = 2, 1 + 2 = 3, 2 + 3 = 5, 3 + 5 = 8
"199100199" is also an additive number, the additive sequence is: 1, 99, 100, 199.
1 + 99 = 100, 99 + 100 = 199
Note: Numbers in the additive sequence cannot have leading zeros, so sequence 1, 2, 03 or 1, 02, 3 is invalid.
Given a string containing only digits '0'-'9', write a function to determine if it’s an additive number.
How would you handle overflow for very large input integers?
## Solution
Result: Accepted Time: 0 ms
Here should be some explanations.
bool plus_equal(char * num, int i, int j, int k)
{
int carry = 0, a = i, b = j, c = k;
if((num[0] == '0'&& i > 1) || (num[i]== '0' && j > i+1) || (num[j] == '0'&& k > j+1))
return false;
while(a > 0 || b > i || carry)
{
if(c <= j) return false;
if(a > 0)
carry += num[--a] - '0';
if(b > i)
carry += num[--b] - '0';
if(carry%10 != num[--c] - '0')
return false;
carry /= 10;
}
return c == j;
}
bool dfs(char *num,int i,int j,int len)
{
if(j >= len)
return true;
int k = i+j;
if(k < j*2 - i)
k = j*2 - i;
if(k <=len && plus_equal(num,i,j,k) && dfs(num+i,j-i,k-i,len - i))
return true;
if(k < len && plus_equal(num,i,j,k+1) && dfs(num+i,j-i,k+1-i,len - i))
return true;
return false;
}
• Time Complexity: $O(n^2)$
• Space Complexity: $O(1)$
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2020-02-20 13:23:33
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https://deut-erium.github.io/WriteUps/2020/zh3r0/crypto/gboad_maddy/2020-06-19-zh3r0-2020-gboad-maddy
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## Description
You see this everyday if you use android.
?<³|⁰([{⁰|Pf¹&<=f{³&f"{P)
Author : Finch
the P stands for % as it causes build errors with jekyll, I substituted it
## Hint
len(flag)=28
Someone pointed out after the ctf
They we small accent symbols on gboard
HMMMMMMMMMMMMMMMMMMMMMM
Substituting the accent symbols for the letter, we can get
"?<³|⁰([{⁰|Pf¹&<=f{³&f"{P)"
mu3e0{to0eq_1gur_o3g_xoq}
rot13 of which produces
zh3r0{gb0rd_1the_b3t_kbd}
Since the length of the flag was mentioned to be 28 characters, one has to fill out the possible missing characters :\
### zh3r0{gb0rd_1s_the_b3st_kbd}
Some words for the nerds
jekyll.environment != "beta" -%}
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2023-03-26 23:58:11
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http://blog.boyet.com/blog/blog/inline-scripts-sometimes-the-web-is-just-screwed-up/
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## Inline scripts: sometimes the web is just screwed up
I don’t know about you, but one of my favorite commands in the browser is “View Page Source”, especially on a site that’s modern, visually attractive, or shows off some clever interactions. After all, I’m a developer: I like to find out how things work so I can, if I want to, replicate on my own web sites.
Some web pages though are really nasty when you look at their source. And one of the places they excel at nastiness is in their use of inline scripts. Now, don’t get me wrong, I’m not particularly objecting to inline scripts: sometimes they’re the biz when you just have some minor bit of JavaScript to execute, so minor it doesn’t seem efficient to create a brand new JS file and add the external script tag to the HTML on the page. And then sometimes you get web pages that can’t seem to decide how the heck they should represent inline scripts and throw everything but the kitchen sink at the problem.
Let’s take a look at the three main ways you can insert inline scripts into your HTML. First, there’s the obvious way, just unadorned:
<script type="text/javascript">
$(function () { var inputText =$("#inputtext"),
outputMath = $("#outputmath"), inlineMath =$("#inlinemath");
inputText.on("keyup", function () {
var expr = inputText.val();
outputMath.text('\$' + expr + ' \$');
inlineMath.text('The expression is <\$$' + expr + ' \$$>.');
MathJax.Hub.Queue(["Typeset", MathJax.Hub, "outputmath"]);
MathJax.Hub.Queue(["Typeset", MathJax.Hub, "inlinemath"]);
});
});
<script>
Then, from way back when, there was the attempt to circumvent those browsers that didn’t support inline JavaScript by enclosing the code in HTML comments:
<script type="text/javascript">
<!--
$(function () { var inputText =$("#inputtext"),
outputMath = $("#outputmath"), inlineMath =$("#inlinemath");
inputText.on("keyup", function () {
var expr = inputText.val();
outputMath.text('\$' + expr + ' \$');
inlineMath.text('The expression is < \$$' + expr + ' \$$ >.');
MathJax.Hub.Queue(["Typeset", MathJax.Hub, "outputmath"]);
MathJax.Hub.Queue(["Typeset", MathJax.Hub, "inlinemath"]);
});
});
// -->
<script>
Notice that (1) the opening HTML comment tag is allowed to be “bare” (it’s not counted as part of the JavaScript code) but the closing end-comment tag must be itself commented out in the inline JavaScript.
Finally there’s the CDATA approach, which just seems wacky until you understand why it’s used:
<script type="text/javascript">
// <!--[CDATA[
$(function () { var inputText =$("#inputtext"),
outputMath = $("#outputmath"), inlineMath =$("#inlinemath");
inputText.on("keyup", function () {
var expr = inputText.val();
outputMath.text('\$' + expr + ' \$');
inlineMath.text('The expression is < \$$' + expr + ' \$$ >.');
MathJax.Hub.Queue(["Typeset", MathJax.Hub, "outputmath"]);
MathJax.Hub.Queue(["Typeset", MathJax.Hub, "inlinemath"]);
});
});
// ]]>
<script>
So what’s up with all this? Why the three different methods?
The first point to recognize is that there are two characters that should never be used un-escaped in HTML: the less-than angle bracket (<) and the ampersand (&). In the inline code I’m showing in method 1, there’s a left angle bracket. Luckily, in HTML, text that appears in a script element can contain these two characters un-escaped. (The only thing that cannot appear in a script element is the string “-->”, since that signals the end of the script.) So, with all modern day browsers, this kind of unadorned inline script is accepted in an HTML file without issues (except for one special case, which I’ll come to in a moment).
Method 2 was used purely for those browsers that didn’t understand inline scripts. To be exact, the <script> tag was introduced in HTML 3.2 as a placeholder and browsers from that point on would no longer display the inline code as text when displaying the page – which is what happened prior to that. If you can remember those days, more power to you; I certainly can’t. So, back in the dim and distant past the inline code had to be commented out HTML-wise to make sure it wasn’t displayed by a horse-drawn user agent. Fast forward to the 21st century where even – shock, horror – IE6 recognized inline script. Method 2 then is not required ever, anymore. Just say no. (Incidentally there’s another issue with the “commented out” inline script: the code could not contain dash-dash (--), since it was that character sequence that officially denoted the end of the comment. So no post- or pre-decrements in your code, OK?)
Method 3 is an interesting one. It’s the only allowed way to have general text embedded in an XML file that isn’t going to be parsed as XML. Hence it’s the only way to have valid un-escaped less-than signs or ampersands in the XML file. And, of course, an HTML file that is flagged as XHTML in the file’s DOCTYPE is supposed to be valid XML. However, that’s not the end of the story. Basically browsers have two parsers: an HTML parser and an XML parser, with the HTML parser being used the vast majority of the time. This is true even in the case of an HTML file flagged as XHTML in the DOCTYPE: the HTML parser will be used. Since the CDATA section is not part of the HTML spec, both the start and end CDATA tags are just ignored: and this is especially so since they are both commented out in the inline script. The only time the CDATA section is parsed as such is if (a) the HTML file is flagged as XHTML, and (b) the HTTP response header sent by the web server marks the content as the “application/xhtml+xml” MIME type. In this case, the XML parser is executed on the content and your inline script had better be in a CDATA section just in case.
So, where does this leave you?
• First, just avoid inline script. If you do, you don’t have to worry about any of this palaver.
• Second, never use method 2. Never, ever, ever. It’s not been needed for years, and when even IE6 (spit, spit) doesn’t need it, you are completely safe in avoiding it.
• Third, if you are at the cutting edge of web development and are writing HTML5 pages or web sites or applications and you absolutely have to have inline JavaScript, just use method 1. All modern browsers will understand what you want and none will barf at your succinctness.
• Fourth, only use method 3 if you are declaring your HTML as XHTML in the DOCTYPE, and you have set up your web server to declare these files as “application/xhtml+xml”. If you are doing that, more power to you!
• Fifth, use method 1 in all other cases.
Now isn’t that all just simple?
Now playing:
Fila Brazillia - Pollo De Palo
(from Luck Be A Weirdo Tonight)
#### 6 Responses
##### #1julian m bucknall said...
01-Apr-13 9:03 PM
All: just realized something else. If you do go for method 3 and set up your web server's response content-type properly and you have some inline script blocks written with method 2, they will magically have disappeared in the browser. Comments in "XML parsing mode" are thrown away, whereas comments in HTML parsing mode stick around and will be further parsed... Boom!
Cheers, Julian
##### #2julian m bucknall said...
02-Apr-13 10:42 AM
Cheers, Julian
##### #3Drew Wells said...
03-Apr-13 10:23 AM
Inserting inline soundtrack --
Filla Brazilla - Pollo de Palo - YouTube
##### #4Richard said...
24-Apr-13 12:17 PM
If you're writing HTML5, you don't need the type attribute either:
<script>
\$(function(){
...
});
</script>
##### #5julian m bucknall said...
24-Apr-13 1:54 PM
Richard: Cool, didn't know that.
For other readers, here's the reference: 'The type attribute (of the script element) gives the language of the script or format of the data. [...] The default, which is used if the attribute is absent, is "text/javascript".'
Cheers, Julian
##### #6Mehul Harry said...
24-Apr-13 2:58 PM
Gotta join this party! :)
Great post and good advice here:
First, just avoid inline script.
However, I reserve the right to do this in the future. Sometimes it's just convenient during a demo or sample code. #muhahahaha
Richard, nice catch. Better than to just always define the type since cross-browser is still a concern.
##### Leave a response
Note: some MarkDown is allowed, but HTML is not. Expand to show what's available.
• Emphasize with italics: surround word with underscores _emphasis_
• Emphasize strongly: surround word with double-asterisks **strong**
• Link: surround text with square brackets, url with parentheses [text](url)
• Inline code: surround text with backticks IEnumerable
• Unordered list: start each line with an asterisk, space * an item
• Ordered list: start each line with a digit, period, space 1. an item
• Insert code block: start each line with four spaces
• Insert blockquote: start each line with right-angle-bracket, space > Now is the time...
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2017-09-22 02:49:05
|
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https://quant.stackexchange.com/questions/17266/on-an-application-of-itos-lemma
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On an application of Ito's lemma
Assume that instantaneous returns are generated by the continuous time martingale:
$$dp_t = \sigma_t dW_t$$
where $W_t$ denotes a standard Weiner process and One day returns are denoted by $r_{t+1} = p_{t+1} - p_t$. Then By Ito's lemma we have:
$$E_t (r_{t+1}^2) = E_t \Bigg( \int_0^1 r_{t + \tau}^2 d \tau \Bigg) = E_t \Bigg( \int_0^1 \sigma_{t + \tau}^2 d \tau \Bigg) = \int_0^1 E_{t} \Bigg( \sigma_{t + \tau}^2 \Bigg) d \tau$$
where $E_t$ denotes conditional expectation at time t.
I am very rusty with Ito's lemma applications and do not seem to recall where the $d \tau$ comes up from. Would anybody mind explaining these 3 equalities?
• This is basically Ito's isometry. Apr 6 '15 at 17:42
• @Gordon that is the first equality, right? Apr 6 '15 at 18:04
• That is correct. But the notation here is bit sloppy, the square $r^2_{t+\tau}$ within the first integral should be the quadratic variation. But in many book, such as that of John Hull, this sloppy notion is used. Apr 6 '15 at 18:30
• Thanks could you even tell me why we can exchange $r_{t+\tau}^2$ with $\sigma_{t+\tau}^2$ in the second equality? I know it is a popular approximation but why can we keep equality in this case? Apr 6 '15 at 18:33
Based on Ito's isometry, \begin{align*} E_t (r^2_{t+1}) &= E_t \bigg(\int_t^{t+1} \sigma_s dW_s \int_t^{t+1} \sigma_s dW_s\bigg)\\ &= E_t \bigg(\int_t^{t+1} \sigma_{\tau}^2 \,d\tau\bigg) \\ &= E_t\bigg(\int_0^1 \sigma_{\tau+t}^2 \,d\tau\bigg) \\ &=\int_0^1 E_t\big(\sigma_{\tau+t}^2\big) \,d\tau. \end{align*} The identity \begin{align*} E_t (r^2_{t+1}) &= E_t\bigg(\int_0^1 r_{\tau+t}^2 \,d\tau\bigg) \end{align*} is sloppy. It is better to write as \begin{align*} E_t (r^2_{t+1}) &= E_t\bigg(\int_0^1 d\langle r_{\tau+t}, r_{\tau+t}\rangle\bigg), \end{align*} where $\langle r_{\tau+t}, r_{\tau+t}\rangle$ is the quadratic variation.
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2022-01-25 15:45:39
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https://brilliant.org/problems/the-equivalence-principle/
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The Equivalence Principle
Which of the following is the most correct statement of the equivalence principle?
×
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2019-10-21 21:40:24
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http://avigad.github.io/logic_and_proof/the_natural_numbers_and_induction.html
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17. The Natural Numbers and Induction¶
This chapter marks a transition from the abstract to the concrete. Viewing the mathematical universe in terms of sets, relations, and functions gives us useful ways of thinking about mathematical objects and structures and the relationships between them. At some point, however, we need to start thinking about particular mathematical objects and structures, and the natural numbers are a good place to start. The nineteenth century mathematician Leopold Kronecker once proclaimed “God created the whole numbers; everything else is the work of man.” By this he meant that the natural numbers (and the integers, which we will also discuss below) are a fundamental component of the mathematical universe, and that many other objects and structures of interest can be constructed from these.
In this chapter, we will consider the natural numbers and the basic principles that govern them. In Chapter 18 we will see that even basic operations like addition and multiplication can be defined using means described here, and their properties derived from these basic principles. Our presentation in this chapter will remain informal, however. In Chapter 19, we will see how these principles play out in number theory, one of the oldest and most venerable branches of mathematics.
17.1. The Principle of Induction¶
The set of natural numbers is the set
$\mathbb{N} = \{ 0, 1, 2, 3, \ldots \}.$
In the past, opinions have differed as to whether the set of natural numbers should start with 0 or 1, but these days most mathematicians take them to start with 0. Logicians often call the function $$s(n) = n + 1$$ the successor function, since it maps each natural number, $$n$$, to the one that follows it. What makes the natural numbers special is that they are generated by the number zero and the successor function, which is to say, the only way to construct a natural number is to start with $$0$$ and apply the successor function finitely many times. From a foundational standpoint, we are in danger of running into a circularity here, because it is not clear how we can explain what it means to apply a function “finitely many times” without talking about the natural numbers themselves. But the following principle, known as the principle of induction, describes this essential property of the natural numbers in a non-circular way.
Principle of Induction. Let $$P$$ be any property of natural numbers. Suppose $$P$$ holds of zero, and whenever $$P$$ holds of a natural number $$n$$, then it holds of its successor, $$n + 1$$. Then $$P$$ holds of every natural number.
This reflects the image of the natural numbers as being generated by zero and the successor operation: by covering the zero and successor cases, we take care of all the natural numbers.
The principle of induction provides a recipe for proving that every natural number has a certain property: to show that $$P$$ holds of every natural number, show that it holds of $$0$$, and show that whenever it holds of some number $$n$$, it holds of $$n + 1$$. This form of proof is called a proof by induction. The first required task is called the base case, and the second required task is called the induction step. The induction step requires temporarily fixing a natural number $$n$$, assuming that $$P$$ holds of $$n$$, and then showing that $$P$$ holds of $$n + 1$$. In this context, the assumption that $$P$$ holds of $$n$$ is called the inductive hypothesis.
You can visualize proof by induction as a method of knocking down an infinite stream of dominoes, all at once. We set the mechanism in place and knock down domino 0 (the base case), and every domino knocks down the next domino (the induction step). So domino 0 knocks down domino 1; that knocks down domino 2, and so on.
Here is an example of a proof by induction.
Theorem. For every natural number $$n$$,
$1 + 2 + \ldots + 2^n = 2^{n+1} - 1.$
Proof. We prove this by induction on $$n$$. In the base case, when $$n = 0$$, we have $$1 = 2^{0+1} - 1$$, as required.
For the induction step, fix $$n$$, and assume the induction hypothesis
$1 + 2 + \ldots + 2^n = 2^{n+1} - 1.$
We need to show that this same claim holds with $$n$$ replaced by $$n + 1$$. But this is just a calculation:
$\begin{split}1 + 2 + \ldots + 2^{n+1} & = (1 + 2 + \ldots + 2^n) + 2^{n+1} \\ & = 2^{n+1} - 1 + 2^{n+1} \\ & = 2 \cdot 2^{n+1} - 1 \\ & = 2^{n+2} - 1.\end{split}$
In the notation of first-order logic, if we write $$P(n)$$ to mean that $$P$$ holds of $$n$$, we could express the principle of induction as follows:
$P(0) \wedge \forall n \; (P(n) \to P(n + 1)) \to \forall n \; P(n).$
But notice that the principle of induction says that the axiom holds for every property $$P$$, which means that we should properly use a universal quantifier for that, too:
$\forall P \; (P(0) \wedge \forall n \; (P(n) \to P(n + 1)) \to \forall n \; P(n)).$
Quantifying over properties takes us out of the realm of first-order logic; induction is therefore a second-order principle.
The pattern for a proof by induction is expressed even more naturally by the following natural deduction rule:
You should think about how some of the proofs in this chapter could be represented formally using natural deduction.
For another example of a proof by induction, let us derive a formula that, given any finite set $$S$$, determines the number of subsets of $$S$$. For example, there are four subsets of the two-element set $$\{1, 2\}$$, namely $$\emptyset$$, $$\{1\}$$, $$\{2\}$$, and $$\{1, 2\}$$. You should convince yourself that there are eight subsets of the set $$\{1, 2, 3\}$$. The following theorem establishes the general pattern.
Theorem. For any finite set $$S$$, if $$S$$ has $$n$$ elements, then there are $$2^n$$ subsets of $$S$$.
Proof. We use induction on $$n$$. In the base case, there is only one set with $$0$$ elements, the empty set, and there is exactly one subset of the empty set, as required.
In the inductive case, suppose $$S$$ has $$n + 1$$ elements. Let $$a$$ be any element of $$S$$, and let $$S'$$ be the set containing the remaining $$n$$ elements. In order to count the subsets of $$S$$, we divide them into two groups.
First, we consider the subsets of $$S$$ that don’t contain $$a$$. These are exactly the subsets of $$S'$$, and by the inductive hypothesis, there are $$2^n$$ of those.
Next we consider the subsets of $$S$$ that do contain $$a$$. Each of these is obtained by choosing a subset of $$S'$$ and adding $$a$$. Since there are $$2^n$$ subsets of $$S'$$, there are $$2^n$$ subsets of $$S$$ that contain $$a$$.
Taken together, then, there are $$2^n + 2^n = 2^{n+1}$$ subsets of $$S$$, as required.
We have seen that there is a correspondence between properties of a domain and subsets of a domain. For every property $$P$$ of natural numbers, we can consider the set $$S$$ of natural numbers with that property, and for every set of natural numbers, we can consider the property of being in that set. For example, we can talk about the property of being even, or talk about the set of even numbers. Under this correspondence, the principle of induction can be cast as follows:
Principle of Induction. Let $$S$$ be any set of natural numbers that contains $$0$$ and is closed under the successor operation. Then $$S = \mathbb{N}$$.
Here, saying that $$S$$ is “closed under the successor operation” means that whenever a number $$n$$ is in $$S$$, so is $$n + 1$$.
17.2. Variants of Induction¶
In this section, we will consider variations on the principle of induction that are often useful. It is important to recognize that each of these can be justified using the principle of induction as stated in the last section, so they need not be taken as fundamental.
The first one is no great shakes: instead of starting from $$0$$, we can start from any natural number, $$m$$.
Principle of Induction from a Starting Point. Let $$P$$ be any property of natural numbers, and let $$m$$ be any natural number. Suppose $$P$$ holds of $$m$$, and whenever $$P$$ holds of a natural number $$n$$ greater than or equal to $$m$$, then it holds of its successor, $$n + 1$$. Then $$P$$ holds of every natural number greater than or equal to $$m$$.
Assuming the hypotheses of this last principle, if we let $$P'(n)$$ be the property “$$P$$ holds of $$m + n$$,” we can prove that $$P'$$ holds of every $$n$$ by the ordinary principle of induction. But this means that $$P$$ holds of every number greater than or equal to $$m$$.
Here is one example of a proof using this variant of induction.
Theorem. For every natural number $$n \geq 5$$, $$2^n > n^2$$.
Proof. By induction on $$n$$. When $$n = 5$$, we have $$2^n = 32 > 25 = n^2$$, as required.
For the induction step, suppose $$n \ge 5$$ and $$2^n > n^2$$. Since $$n$$ is greater than or equal to $$5$$, we have $$2n + 1 \leq 3 n \leq n^2$$, and so
$\begin{split}(n+1)^2 &= n^2 + 2n + 1 \\ & \leq n^2 + n^2 \\ & < 2^n + 2^n \\ & = 2^{n+1}.\end{split}$
For another example, let us derive a formula for the sum total of the angles in a convex polygon. A polygon is said to be convex if every line between two vertices stays inside the polygon. We will accept without proof the visually obvious fact that one can subdivide any convex polygon with more than three sides into a triangle and a convex polygon with one fewer side, namely, by closing off any two consecutive sides to form a triangle. We will also accept, without proof, the basic geometric fact that the sum of the angles of any triangle is 180 degrees.
Theorem. For any $$n \geq 3$$, the sum of the angles of any convex $$n$$-gon is $$180(n - 2)$$.
Proof. In the base case, when $$n = 3$$, this reduces to the statement that the sum of the angles in any triangle is 180 degrees.
For the induction step, suppose $$n \geq 3$$, and let $$P$$ be a convex $$(n+1)$$-gon. Divide $$P$$ into a triangle and an $$n$$-gon. By the inductive hypotheses, the sum of the angles of the $$n$$-gon is $$180(n-2)$$ degrees, and the sum of the angles of the triangle is $$180$$ degrees. The measures of these angles taken together make up the sum of the measures of the angles of $$P$$, for a total of $$180(n-2) + 180 = 180(n-1)$$ degrees.
For our second example, we will consider the principle of complete induction, also sometimes known as total induction.
Principle of Complete Induction. Let $$P$$ be any property that satisfies the following: for any natural number $$n$$, whenever $$P$$ holds of every number less than $$n$$, it also holds of $$n$$. Then $$P$$ holds of every natural number.
Notice that there is no need to break out a special case for zero: for any property $$P$$, $$P$$ holds of all the natural numbers less than zero, for the trivial reason that there aren’t any! So, in particular, any such property automatically holds of zero.
Notice also that if such a property $$P$$ holds of every number less than $$n$$, then it also holds of every number less than $$n + 1$$ (why?). So, for such a $$P$$, the ordinary principle of induction implies that for every natural number $$n$$, $$P$$ holds of every natural number less than $$n$$. But this is just a roundabout way of saying that $$P$$ holds of every natural number. In other words, we have justified the principle of complete induction using ordinary induction.
To use the principle of complete induction we merely have to let $$n$$ be any natural number and show that $$P$$ holds of $$n$$, assuming that it holds of every smaller number. Compare this to the ordinary principle of induction, which requires us to show $$P (n + 1)$$ assuming only $$P(n)$$. The following example of the use of this principle is taken verbatim from the introduction to this book:
Theorem. Every natural number greater than or equal to 2 can be written as a product of primes.
Proof. We proceed by induction on $$n$$. Let $$n$$ be any natural number greater than 2. If $$n$$ is prime, we are done; we can consider $$n$$ itself as a product with one factor. Otherwise, $$n$$ is composite, and we can write $$n = m \cdot k$$ where $$m$$ and $$k$$ are smaller than $$n$$ and greater than 1. By the inductive hypothesis, each of $$m$$ and $$k$$ can be written as a product of primes:
$\begin{split}m = p_1 \cdot p_2 \cdot \ldots \cdot p_u \\ k = q_1 \cdot q_2 \cdot \ldots \cdot q_v.\end{split}$
But then we have
$n = m \cdot k = p_1 \cdot p_2 \cdot \ldots \cdot p_u \cdot q_1 \cdot q_2 \cdot \ldots \cdot q_v.$
We see that $$n$$ is a product of primes, as required.
Finally, we will consider another formulation of induction, known as the least element principle.
The Least Element Principle. Suppose $$P$$ is some property of natural numbers, and suppose $$P$$ holds of some $$n$$. Then there is a smallest value of $$n$$ for which $$P$$ holds.
In fact, using classical reasoning, this is equivalent to the principle of complete induction. To see this, consider the contrapositive of the statement above: “if there is no smallest value for which $$P$$ holds, then $$P$$ doesn’t hold of any natural number.” Let $$Q(n)$$ be the property “$$P$$ does not hold of $$n$$.” Saying that there is no smallest value for which $$P$$ holds means that, for every $$n$$, if $$P$$ holds at $$n$$, then it holds of some number smaller than $$n$$; and this is equivalent to saying that, for every $$n$$, if $$Q$$ doesn’t hold at $$n$$, then there is a smaller value for which $$Q$$ doesn’t hold. And that is equivalent to saying that if $$Q$$ holds for every number less than $$n$$, it holds for $$n$$ as well. Similarly, saying that $$P$$ doesn’t hold of any natural number is equivalent to saying that $$Q$$ holds of every natural number. In other words, replacing the least element principle by its contrapositive, and replacing $$P$$ by “not $$Q$$,” we have the principle of complete induction. Since every statement is equivalent to its contrapositive, and every predicate has its negated version, the two principles are the same.
It is not surprising, then, that the least element principle can be used in much the same way as the principle of complete induction. Here, for example, is a formulation of the previous proof in these terms. Notice that it is phrased as a proof by contradiction.
Theorem. Every natural number greater than equal to 2 can be written as a product of primes.
Proof. Suppose, to the contrary, some natural number greater than or equal to 2 cannot be written as a product of primes. By the least element principle, there is a smallest such element; call it $$n$$. Then $$n$$ is not prime, and since it is greater than or equal to 2, it must be composite. Hence we can write $$n = m \cdot k$$ where $$m$$ and $$k$$ are smaller than $$n$$ and greater than 1. By the assumption on $$n$$, each of $$m$$ and $$k$$ can be written as a product of primes:
$\begin{split}m = p_1 \cdot p_2 \cdot \ldots \cdot p_u \\ k = q_1 \cdot q_2 \cdot \ldots \cdot q_v.\end{split}$
But then we have
$n = m \cdot k = p_1 \cdot p_2 \cdot \ldots \cdot p_u \cdot q_1 \cdot q_2 \cdot \ldots \cdot q_v.$
We see that $$n$$ is a product of primes, contradicting the fact that $$n$$ cannot be written as a product of primes.
Here is another example:
Theorem. Every natural number is interesting.
Proof. Suppose, to the contrary, some natural number is uninteresting. Then there is a smallest one, $$n$$. In other words, $$n$$ is the smallest uninteresting number. But that is really interesting! Contradiction.
17.3. Recursive Definitions¶
Suppose I tell you that I have a function $$f : \mathbb{N} \to \mathbb{N}$$ in mind, satisfying the following properties:
$\begin{split}f(0) & = 1 \\ f(n + 1) & = 2 \cdot f(n)\end{split}$
What can you infer about $$f$$? Try calculating a few values:
$\begin{split}f(1) & = f(0 + 1) = 2 \cdot f(0) = 2 \\ f(2) & = f(1 + 1) = 2 \cdot f(1) = 4 \\ f(3) & = f(2 + 1) = 2 \cdot f(2) = 8\end{split}$
It soon becomes apparent that for every $$n$$, $$f(n) = 2^n$$.
What is more interesting is that the two conditions above specify all the values of $$f$$, which is to say, there is exactly one function meeting the specification above. In fact, it does not matter that $$f$$ takes values in the natural numbers; it could take values in any other domain. All that is needed is a value of $$f(0)$$ and a way to compute the value of $$f(n+1)$$ in terms of $$n$$ and $$f(n)$$. This is what the principle of definition by recursion asserts:
Principle of Definition by Recursion. Let $$A$$ be any set, and suppose $$a$$ is in $$A$$, and $$g : \mathbb{N} \times A \to A$$. Then there is a unique function $$f$$ satisfying the following two clauses:
$\begin{split}f(0) & = a \\ f(n + 1) & = g(n, f(n)).\end{split}$
The principle of recursive definition makes two claims at once: first, that there is a function $$f$$ satisfying the clauses above, and, second, that any two functions $$f_1$$ and $$f_2$$ satisfying those clauses are equal, which is to say, they have the same values for every input. In the example with which we began this section, $$A$$ is just $$\mathbb{N}$$ and $$g(n, f(n)) = 2 \cdot f(n)$$.
In some axiomatic frameworks, the principle of recursive definition can be justified using the principle of induction. In others, the principle of induction can be viewed as a special case of the principle of recursive definition. For now, we will simply take both to be fundamental properties of the natural numbers.
As another example of a recursive definition, consider the function $$g : \mathbb{N} \to \mathbb{N}$$ defined recursively by the following clauses:
$\begin{split}g(0) & = 1 \\ g(n+1) & = (n + 1) \cdot g(n)\end{split}$
Try calculating the first few values. Unwrapping the definition, we see that $$g(n) = 1 \cdot 2 \cdot 3 \cdot \ldots \cdot (n-1) \cdot n$$ for every $$n$$; indeed, definition by recursion is usually the proper way to make expressions using “…” precise. The value $$g(n)$$ is read “$$n$$ factorial,” and written $$n!$$.
Indeed, summation notation
$\sum_{i < n} f (i) = f(0) + f(1) + \ldots + f(n-1)$
and product notation
$\prod_{i < n} f (i) = f(0) \cdot f(1) \cdot \cdots \cdot f(n-1)$
can also be made precise using recursive definitions. For example, the function $$k(n) = \sum_{i < n} f (i)$$ can be defined recursively as follows:
$\begin{split}k(0) &= 0 \\ k(n+1) &= k(n) + f(n)\end{split}$
Induction and recursion are complementary principles, and typically the way to prove something about a recursively defined function is to use the principle of induction. For example, the following theorem provides a formulas for the sum $$1 + 2 + \ldots + n$$, in terms of $$n$$.
Theorem. For every $$n$$, $$\sum_{i < n + 1} i = n (n + 1) / 2$$.
Proof. In the base case, when $$n = 0$$, both sides are equal to $$0$$.
In the inductive step, we have
$\begin{split}\sum_{i < n + 2} i & = \left(\sum_{i < n + 1} i\right) + (n + 1) \\ & = n (n + 1) / 2 + n + 1 \\ & = \frac{n^2 +n}{2} + \frac{2n + 2}{2} \\ & = \frac{n^2 + 3n + 2}{2} \\ & = \frac{(n+1)(n+2)}{2}.\end{split}$
There are just as many variations on the principle of recursive definition as there are on the principle of induction. For example, in analogy to the principle of complete induction, we can specify a value of $$f(n)$$ in terms of the values that $$f$$ takes at all inputs smaller than $$n$$. When $$n \geq 2$$, for example, the following definition specifies the value of a function $$\mathrm{fib}(n)$$ in terms of its two predecessors:
$\begin{split}\mathrm{fib}(0) & = 0 \\ \mathrm{fib}(1) & = 1 \\ \mathrm{fib}(n+2) & = \mathrm{fib}(n + 1) + \mathrm{fib}(n)\end{split}$
Calculating the values of $$\mathrm{fib}$$ on $$0, 1, 2, \ldots$$ we obtain
$0, 1, 1, 2, 3, 5, 8, 13, 21, \ldots$
Here, after the second number, each successive number is the sum of the two values preceding it. This is known as the Fibonacci sequence, and the corresponding numbers are known as the Fibonacci numbers. An ordinary mathematical presentation would write $$F_n$$ instead of $$\mathrm{fib}(n)$$ and specify the sequence with the following equations:
$F_0 = 0, \quad F_1 = 1, \quad F_{n+2} = F_{n+1} + F_n$
But you can now recognize such a specification as an implicit appeal to the principle of definition by recursion. We ask you to prove some facts about the Fibonacci sequence in the exercises below.
17.4. Defining Arithmetic Operations¶
In fact, we can even use the principle of recursive definition to define the most basic operations on the natural numbers and show that they have the properties we expect them to have. From a foundational standpoint, we can characterize the natural numbers as a set, $$\mathbb{N}$$, with a distinguished element $$0$$ and a function, $$\mathrm{succ}(m)$$, which, for every natural number $$m$$, returns its successor. These satisfy the following:
• $$0 \neq \mathrm{succ}(m)$$ for any $$m$$ in $$\mathbb{N}$$.
• For every $$m$$ and $$n$$ in $$\mathbb{N}$$, if $$m \neq n$$, then $$\mathrm{succ}(m) \neq \mathrm{succ}(n)$$. In other words, $$\mathrm{succ}$$ is injective.
• If $$A$$ is any subset of $$\mathbb{N}$$ with the property that $$0$$ is in $$A$$ and whenever $$n$$ is in $$A$$ then $$\mathrm{succ}(n)$$ is in $$A$$, then $$A = \mathbb{N}$$.
The last clause can be reformulated as the principle of induction:
Suppose $$P(n)$$ is any property of natural numbers, such that $$P$$ holds of $$0$$, and for every $$n$$, $$P(n)$$ implies $$P(\mathrm{succ}(n))$$. Then every $$P$$ holds of every natural number.
Remember that this principle can be used to justify the principle of definition by recursion:
Let $$A$$ be any set, $$a$$ be any element of $$A$$, and let $$g(n,m)$$ be any function from $$\mathbb{N} \times A$$ to $$A$$. Then there is a unique function $$f: \mathbb{N} \to A$$ satisfying the following two clauses:
• $$f(0) = a$$
• $$f(\mathrm{succ}(n)) = g(n,f(n))$$ for every $$n$$ in $$N$$
We can use the principle of recursive definition to define addition with the following two clauses:
$\begin{split}m + 0 & = m \\ m + \mathrm{succ}(n) & = \mathrm{succ}(m + n)\end{split}$
Note that we are fixing $$m$$, and viewing this as a function of $$n$$. If we write $$1 = \mathrm{succ}(0)$$, $$2 = \mathrm{succ}(1)$$, and so on, it is easy to prove $$n + 1 = \mathrm{succ}(n)$$ from the definition of addition.
We can proceed to define multiplication using the following two clauses:
$\begin{split}m \cdot 0 & = 0 \\ m \cdot \mathrm{succ}(n) & = m \cdot n + m\end{split}$
We can also define a predecessor function by
$\begin{split}\mathrm{pred}(0) & = 0 \\ \mathrm{pred}(\mathrm{succ}(n)) & = n\end{split}$
We can define truncated subtraction by
$\begin{split}m \dot - 0 & = m \\ m \dot - (\mathrm{succ}(n)) & = \mathrm{pred}(m \dot - n)\end{split}$
With these definitions and the induction principle, one can prove all the following identities:
• $$n \neq 0$$ implies $$\mathrm{succ}(\mathrm{pred}(n)) = n$$
• $$0 + n = n$$
• $$\mathrm{succ}(m) + n = \mathrm{succ}(m + n)$$
• $$(m + n) + k = m + (n + k)$$
• $$m + n = n + m$$
• $$m(n + k) = mn + mk$$
• $$0 \cdot n = 0$$
• $$1 \cdot n = n$$
• $$(mn)k = m(nk)$$
• $$mn = nm$$
We will do the first five here, and leave the remaining ones as exercises.
Proposition. For every natural number $$n$$, if $$n \neq 0$$ then $$\mathrm{succ}(\mathrm{pred}(n)) = n$$.
Proof. By induction on $$n$$. We have ruled out the case where $$n$$ is $$0$$, so we only need to show that the claim holds for $$\mathrm{succ}(n)$$. But in that case, we have $$\mathrm{succ}(\mathrm{pred}(\mathrm{succ}(n)) = \mathrm{succ}(n)$$ by the second defining clause of the predecessor function.
Proposition. For every $$n$$, $$0 + n = n$$.
Proof. By induction on $$n$$. We have $$0 + 0 = 0$$ by the first defining clause for addition. And assuming $$0 + n = n$$, we have $$0 + \mathrm{succ}(n) = \mathrm{succ}(0 + n) = n$$, using the second defining clause for addition.
Proposition. For every $$m$$ and $$n$$, $$\mathrm{succ}(m) + n = \mathrm{succ}(m + n)$$.
Proof. Fix $$m$$ and use induction on $$n$$. Then $$n = 0$$, we have $$\mathrm{succ}(m) + 0 = \mathrm{succ}(m) = \mathrm{succ}(m + 0)$$, using the first defining clause for addition. Assuming the claim holds for $$n$$, we have
$\begin{split}\mathrm{succ}(m) + \mathrm{succ}(n) & = \mathrm{succ}(\mathrm{succ}(m) + n) \\ & = \mathrm{succ} (\mathrm{succ} (m + n)) \\ & = \mathrm{succ} (m + \mathrm{succ}(n))\end{split}$
using the inductive hypothesis and the second defining clause for addition.
Proposition. For every $$m$$, $$n$$, and $$k$$, $$(m + n) + k = m + (n + k)$$.
Proof. By induction on $$k$$. The case where $$k = 0$$ is easy, and in the induction step we have
$\begin{split}(m + n) + \mathrm{succ}(k) & = \mathrm{succ} ((m + n) + k) \\ & = \mathrm{succ} (m + (n + k)) \\ & = m + \mathrm{succ} (n + k) \\ & = m + (n + \mathrm{succ} (k)))\end{split}$
using the inductive hypothesis and the definition of addition.
Proposition. For every pair of natural numbers $$m$$ and $$n$$, $$m + n = n + m$$.
Proof. By induction on $$n$$. The base case is easy using the second proposition above. In the inductive step, we have
$\begin{split}m + \mathrm{succ}(n) & = \mathrm{succ}(m + n) \\ & = \mathrm{succ} (n + m) \\ & = \mathrm{succ}(n) + m\end{split}$
using the third proposition above.
17.5. Arithmetic on the Natural Numbers¶
Continuing as in the last section, we can establish all the basic properties of the natural numbers that play a role in day-to-day mathematics. We summarize the main ones here:
$\begin{split}m + n &= n + m \quad \text{(commutativity of addition)}\\ m + (n + k) &= (m + n) + k \quad \text{(associativity of addition)}\\ n + 0 &= n \quad \text{(0 is a neutral element for addition)}\\ n \cdot m &= m \cdot n \quad \text{(commutativity of multiplication)}\\ m \cdot (n \cdot k) &= (m \cdot n) \cdot k \quad \text{(associativity of multiplication)}\\ n \cdot 1 &= n \quad \text{(1 is an neutral element for multiplication)}\\ n \cdot (m + k) &= n \cdot m + n \cdot k \quad \text{(distributivity)}\\ n \cdot 0 &= 0 \quad \text{(0 is an absorbing element for multiplication)}\end{split}$
In an ordinary mathematical argument or calculation, they can be used without explicit justification. We also have the following properties:
• $$n + 1 \neq 0$$
• if $$n + k = m + k$$ then $$n = m$$
• if $$n \cdot k = m \cdot k$$ and $$k \neq 0$$ then $$n = m$$
We can define $$m \le n$$, “$$m$$ is less than or equal to $$n$$,” to mean that there exists a $$k$$ such that $$m + k = n$$. If we do that, it is not hard to show that the less-than-or-equal-to relation satisfies all the following properties, for every $$n$$, $$m$$, and $$k$$:
• $$n \le n$$ (reflexivity)
• if $$n \le m$$ and $$m \le k$$ then $$n \le k$$ (transitivity)
• if $$n \le m$$ and $$m \le n$$ then $$n = m$$ (antisymmetry)
• for all $$n$$ and $$m$$, either $$n \le m$$ or $$m \le n$$ is true (totality)
• if $$n \le m$$ then $$n + k \le m + k$$
• if $$n + k \le m + k$$ then $$n \le m$$
• if $$n \le m$$ then $$nk \le mk$$
• if $$m \ge n$$ then $$m = n$$ or $$m \ge n + 1$$
• $$0 \le n$$
Remember from Chapter 13 that the first four items assert that $$\le$$ is a linear order. Note that when we write $$m \ge n$$, we mean $$n \le m$$.
As usual, then, we can define $$m < n$$ to mean that $$m \le n$$ and $$m \ne n$$. In that case, we have that $$m \le n$$ holds if and only if $$m < n$$ or $$m = n$$.
Proposition. For every $$m$$, $$m + 1 \not\le 0$$.
Proof. Otherwise, we would have $$(m + 1) + k = (m + k) + 1 = 0$$ for some $$k$$.
In particular, taking $$m = 0$$, we have $$1 \not\le 0$$.
Proposition. We have $$m < n$$ iff and only if $$m + 1 \le n$$.
Proof. Suppose $$m < n$$. Then $$m \le n$$ and $$m \ne n$$. So there is a $$k$$ such that $$m + k = n$$, and since $$m \ne n$$, we have $$k \ne 0$$. Then $$k = u + 1$$ for some $$u$$, which means we have $$m + (u + 1) = m + 1 + u = n$$, so $$m \le n$$, as required.
In the other direction, suppose $$m + 1 \le n$$. Then $$m \le n$$. We also have $$m \ne n$$, since if $$m = n$$, we would have $$m + 1 \le m + 0$$ and hence $$1 \le 0$$, a contradiction.
In a similar way, we can show that $$m < n$$ if and only if $$m \le n$$ and $$m \ne n$$. In fact, we can demonstrate all of the following from these properties and the properties of $$\le$$:
• $$n < n$$ is never true (irreflexivity)
• if $$n < m$$ and $$m < k$$ then $$n < k$$ (transitivity)
• for all $$n$$ and $$m$$, either $$n < m$$, $$n = m$$ or $$m < n$$ is true (trichotomy)
• if $$n < m$$ then $$n + k < m + k$$
• if $$k > 0$$ and $$n < m$$ then $$nk < mk$$
• if $$m > n$$ then $$m = n + 1$$ or $$m > n + 1$$
• for all $$n$$, $$n = 0$$ or $$n > 0$$
The first three items mean that $$<$$ is a strict linear order, and the properties above means that $$\le$$ is the associated linear order, in the sense described in Section 13.1.
Proof. We will prove some of these properties using the previous characterization of the less-than relation.
The first property is straightforward: we know $$n \le n + 1$$, and if we had $$n + 1 \le n$$, we should have $$n = n + 1$$, a contradiction.
For the second property, assume $$n < m$$ and $$m < k$$. Then $$n + 1 \le m \le m + 1 \le k$$, which implies $$n < k$$.
For the third, we know that either $$n \le m$$ or $$m \le n$$. If $$m = n$$, we are done, and otherwise we have either $$n < m$$ or $$m < n$$.
For the fourth, if $$n + 1 \le m$$, we have $$n + 1 + k = (n + k) + 1 \le m + k$$, as required.
For the fifth, suppose $$k > 0$$, which is to say, $$k \ge 1$$. If $$n < m$$, then $$n + 1 \le m$$, and so $$nk + 1 \le n k + k \le mk$$. But this implies $$n k < m k$$, as required.
The rest of the remaining proofs are left as an exercise to the reader.
Here are some additional properties of $$<$$ and $$\le$$:
• $$n < m$$ and $$m < n$$ cannot both hold (asymmetry)
• $$n + 1 > n$$
• if $$n < m$$ and $$m \le k$$ then $$n < k$$
• if $$n \le m$$ and $$m < k$$ then $$n < k$$
• if $$m > n$$ then $$m \ge n + 1$$
• if $$m \ge n$$ then $$m + 1 > n$$
• if $$n + k < m + k$$ then $$n < m$$
• if $$nk < mk$$ then $$k > 0$$ and $$n < m$$
These can be proved from the ones above. Moreover, the collection of principles we have just seen can be used to justify basic facts about the natural numbers, which are again typically taken for granted in informal mathematical arguments.
Proposition. If $$n$$ and $$m$$ are natural numbers such that $$n + m = 0$$, then $$n = m = 0$$.
Proof. We first prove that $$m = 0$$. We know that $$m = 0$$ or $$m > 0$$. Suppose that $$m > 0$$. Then $$n + m > n + 0 = n$$. Since $$n \ge 0$$, we conclude that $$n + m > 0$$, which contradicts the fact that $$n + m = 0$$. Since $$m > 0$$ leads to a contradiction, we must have $$m = 0$$.
Now we can easily conclude that $$n = 0$$, since $$n = n + 0 = n + m = 0$$. Hence $$n = m = 0$$.
Proposition. If $$n$$ is a natural number such that $$n < 3$$, then $$n = 0$$, $$n = 1$$ or $$n = 2$$.
Proof. In this proof we repeatedly use the property that if $$m > n$$ then $$m = n + 1$$ or $$m > n + 1$$. Since $$2 + 1 = 3 > n$$, we conclude that either $$2 + 1 = n + 1$$ or $$2 + 1 > n + 1$$. In the first case we conclude $$n = 2$$, and we are done. In the second case we conclude $$2 > n$$, which implies that either $$2 = n + 1$$, or $$2 > n + 1$$. In the first case, we conclude $$n = 1$$, and we are done. In the second case, we conclude $$1 > n$$, and appeal one last time to the general principle presented above to conclude that either $$1 = n + 1$$ or $$1 > n + 1$$. In the first case, we conclude $$n = 0$$, and we are once again done. In the second case, we conclude that $$0 > n$$. This leads to a contradiction, since now $$0 > n \ge 0$$, hence $$0 > 0$$, which contradicts the irreflexivity of $$>$$.
17.6. The Integers¶
The natural numbers are designed for counting discrete quantities, but they suffer an annoying drawback: it is possible to subtract $$n$$ from $$m$$ if $$n$$ is less than or equal to $$m$$, but not if $$m$$ is greater than $$n$$. The set of integers, $$\mathbb{Z}$$, extends the natural numbers with negative values, to make it possible to carry out subtraction in full:
$\mathbb{Z} = \{ \ldots, -3, -2, -1, 0, 1, 2, 3, \ldots \}.$
We will see in a later chapter that the integers can be extended to the rational numbers, the real numbers, and the complex numbers, each of which serves useful purposes. For dealing with discrete quantities, however, the integers will get us pretty far.
You can think of the integers as consisting of two copies of the natural numbers, a positive one and a negative one, sharing a common zero. Conversely, once we have the integers, you can think of the natural numbers as consisting of the nonnegative integers, that is, the integers that are greater than or equal to $$0$$. Most mathematicians blur the distinction between the two, though we will see that in Lean, for example, the natural numbers and the integers represent two different data types.
Most of the properties of the natural numbers that were enumerated in the last section hold of the integers as well, but not all. For example, it is no longer the case that $$n + 1 \neq 0$$ for every $$n$$, since the claim is false for $$n = -1$$. For another example, it is not the case that every integer is either equal to $$0$$ or greater than $$0$$, since this fails to hold of the negative integers.
The key property that the integers enjoy, which sets them apart from the natural numbers, is that for every integer $$n$$ there is a value $$-n$$ with the property that $$n + (-n) = 0$$. The value $$-n$$ is called the negation of $$n$$. We define subtraction $$n - m$$ to be $$n + (-m)$$. For any integer $$n$$, we also define the absolute value of $$n$$, written $$|n|$$, to be $$n$$ if $$n \geq 0$$, and $$-n$$ otherwise.
We can no longer use proof by induction on the integers, because induction does not cover the negative numbers. But we can use induction to show that a property holds of every nonnegative integer, for example. Moreover, we know that every negative integer is the negation of a positive one. As a result, proofs involving the integers often break down into two cases, where one case covers the nonnegative integers, and the other case covers the negative ones.
17.7. Exercises¶
1. Write the principle of complete induction using the notation of symbolic logic. Also write the least element principle this way, and use logical manipulations to show that the two are equivalent.
2. Show that for every $$n$$, $$0^2 + 1^2 + 2^2 + \ldots n^2= \frac{1}{6}n(n+1)(n+2)$$.
3. Show that for every $$n$$, $$0^3 + 1^3 + \ldots + n^3 = \frac{1}{4} n^2 (n+1)^2$$.
4. Given the definition of the Fibonacci numbers in Section 17.3, prove Cassini’s identity: for every $$n$$, $$F^2_{n+1} - F_{n+2} F_n = (-1)^n$$. Hint: in the induction step, write $$F_{n+2}^2$$ as $$F_{n+2}(F_{n+1} + F_n)$$.
5. Prove $$\sum_{i < n} F_{2i+1} = F_{2n}$$.
6. Prove the following two identities:
• $$F_{2n+1} = F^2_{n+1} + F^2_n$$
• $$F_{2n+2} = F^2_{n+2} - F^2_n$$
Hint: use induction on $$n$$, and prove them both at once. In the induction step, expand $$F_{2n+3} = F_{2n+2} + F_{2n+1}$$, and similarly for $$F_{2n+4}$$. Proving the second equation is especially tricky. Use the inductive hypothesis and the first identity to simplify the left-hand side, and repeatedly unfold the Fibonacci number with the highest index and simplify the equation you need to prove. (When you have worked out a solution, write a clear equational proof, calculating in the forward’’ direction.)
7. Prove that every natural number can be written as a sum of distinct powers of 2. For this problem, $$1 = 2^0$$ is counted as power of 2.
8. Let $$V$$ be a non-empty set of integers such that the following two properties hold:
• If $$x, y \in V$$, then $$x - y \in V$$.
• If $$x \in V$$, then every multiple of $$x$$ is an element of $$V$$.
Prove that there is some $$d \in V$$, such that $$V$$ is equal to the set of multiples of $$d$$. Hint: use the least element principle.
9. Give an informal but detailed proof that for every natural number $$n$$, $$1 \cdot n = n$$, using a proof by induction, the definition of multiplication, and the theorems proved in Section 17.4.
10. Show that multiplication distributes over addition. In other words, prove that for natural numbers $$m$$, $$n$$, and $$k$$, $$m (n + k) = m n + m k$$. You should use the definitions of addition and multiplication and facts proved in Section 17.4 (but nothing more).
11. Prove the multiplication is associative, in the same way. You can use any of the facts proved in Section 17.4 and the previous exercise.
12. Prove that multiplication is commutative.
13. Prove $$(m^n)^k = m^{nk}$$.
14. Following the example in Section 17.5, prove that if $$n$$ is a natural number and $$n < 5$$, then $$n$$ is one of the values $$0, 1, 2, 3$$, or $$4$$.
15. Prove that if $$n$$ and $$m$$ are natural numbers and $$n m = 1$$, then $$n = m = 1$$, using only properties listed in Section 17.5.
This is tricky. First show that $$n$$ and $$m$$ are greater than $$0$$, and hence greater than or equal to $$1$$. Then show that if either one of them is greater than $$1$$, then $$n m > 1$$.
16. Prove any of the other claims in Section 17.5 that were stated without proof.
17. Prove the following properties of negation and subtraction on the integers, using only the properties of negation and subtraction given in Section 17.6.
• If $$n + m = 0$$ then $$m = -n$$.
• $$-0 = 0$$.
• If $$-n = -m$$ then $$n = m$$.
• $$m + (n - m) = n$$.
• $$-(n + m) = -n - m$$.
• If $$m < n$$ then $$n - m > 0$$.
• If $$m < n$$ then $$-m > -n$$.
• $$n \cdot (-m) = -nm$$.
• $$n(m - k) = nm - nk$$.
• If $$n < m$$ then $$n - k < m - k$$.
18. Suppose you have an infinite chessboard with a natural number written in each square. The value in each square is the average of the values of the four neighboring squares. Prove that all the values on the chessboard are equal.
19. Prove that every natural number can be written as a sum of distinct non-consecutive Fibonacci numbers. For example, $$22 = 1 + 3 + 5 + 13$$ is not allowed, since 3 and 5 are consecutive Fibonacci numbers, but $$22 = 1 + 21$$ is allowed.
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The diagram below represents a data network that connects a Data Service Provider (DSP) connected to vi(s) to a customer connected to vo(t). Ford Fulkerson algorithm. Dinic's algorithm for Maximum Flow. We run a loop while there is an augmenting path. The Ford–Fulkerson method or the Ford–Fulkerson algorithm (FFA) is a greedy algorithm that computes the maximum flow in a flow network. Freelancer. One other thing I should note about this algorithm is that it's not quite a full algorithm. Use DFS to find an s to t path P where each edge has flow, f(e) < c(e), capacity. 03, Jul 13. In this section, we outline the classic Ford-Fulkerson labeling algorithm for finding a maximum flow in a network. So given a network with the capacities of the edges, how do we assign flow to the edges until we get the max flow? Section 13.4 The Ford-Fulkerson Labeling Algorithm. Each edge represents a single data transmission link. The Ford-Fulkerson algorithm is a simple algorithm to solve the maximum flow prob-lem based on the idea of . Algorithm Edit. There are 13 tests that the code needs to pass. 25, Feb 17. A digraph G = (V,E), with an integer-valued function c (capacity function) define on its edges is called a capacitated network. (c) The Ford-Fulkerson algorithm is used to determine network flow. Ford Fulkerson algorithm is the most popular algorithm that used to solve the maximum flow problem, but its complexity is high. 3) Return flow. Therefore, considering that there's a link between the starting node (the source) and also the end node (the sink), and then flow will be able to go through that path. It was 3:30AM and as I was waiting for emergency service to arrive, I thought it would be a good idea to implement Ford-Fulkerson today.. Ford Fulkerson Algorithm. When no augmenting path exists, flow f … This means our run of the Ford-Fulkerson algorithm is complete and our max flow leading into t is 5! Ford-Fulkerson algorithm - CPP. The algorithm begins with a linear order on the vertex set which establishes a notion of precedence.Typically, the first vertex in this linear order is the source while the second is the sink. Last lecture provided the framework for maximum flow problems and proved the key theorem - max-flow min-cut - that all algorithms are based on. The Ford-Fulkerson algorithm is an algorithm that tackles the max-flow min-cut problem. 2) While there is a augmenting path from source to sink. If … 1 12 4 | 2 | 6 | 2 1 * ੫ 1 D . Two distinguished vertices exist. The capacity of edge (i,j) is c(i,j)(0. It finds the best organisation of flow through the edges of graphs such that you get maximum flow out on the other end. It was discovered in 1956 by Ford and Fulkerson. 2) While there is a augmenting path from source to sink. The first, vertex s has in-degree 0 is called the source and the second, vertex t has out-degree 0 is called the sink. Flow can mean anything, but typically it means data through a computer network. Ford-Fulkerson Algorithm The following is simple idea of Ford-Fulkerson algorithm: 1) Start with initial flow as 0. ). We run a loop while there is an augmenting path. Ford-Fulkerson: a maximum flow algorithm. Augment flow along path P: Add f to f total; For every edge u→v on P Decrease c(u→v) by f Increase c… Ford-Fulkerson algorithm - CPP. The Ford Fulkerson algorithm works out the maximum flow in a flow network. The algorithm was first published by Yefim Dinitz in 1970, and later independently published by Jack Edmonds and Richard Karp in 1972. Time Complexity: Time complexity of the above algorithm is O(max_flow * E). 20, Jun 18. details attached I need it in next 3 hours. Cuts and Network Flow. Summary That was a pretty trivial example, so I would like to reiterate that the Ford-Fulkerson algorithm can be used to find the max flow of much more complicated flow networks. 5 ਵੀ। ਦਿ 13 ? Budget \$30-250 USD. C++ Ford Fulkerson Algorithm. Steps: Start with flow, f(e) = 0 for all edges, e ∈ E. Set f total = 0. Viewed 6k times 2. Ford-Fulkerson Algorithm for Maximum Flow Problem. 3) Return flow. The Ford-Fulkerson algorithm is the algorithm used to find the max flow through a network. Edmonds-Karp algorithm is just an implementation of the Ford-Fulkerson method that uses BFS for finding augmenting paths. I am studying the Ford-Fulkerson algorithm from Cormen's "Introduction to algorithms 2nd Edition". I can give you the tests' input. ford_fulkerson¶ ford_fulkerson(G, s, t, capacity='capacity')¶. Demo Ford fulkerson algorithm untuk pencarian maximum flow We consider 3 options: 1. Let f be the minimum capacity value on that path. I am not posting the code because it's localized too much. Multiple algorithms exist in solving the maximum flow problem. 04, Jan 15. They are explained below. Provided that they have positive integers as capacities, of course. I need to code Ford-Fulkerson algorithm in cpp. Ford-Fulkerson Algorithm: It was developed by L. R. Ford, Jr. and D. R. Fulkerson in 1956. Time Complexity: Time complexity of the above algorithm is O(max_flow * E). Add this path-flow to flow. Two major algorithms to solve these kind of problems are Ford-Fulkerson algorithm and Dinic's Algorithm. Ford-Fulkerson Algorithm. Find some augmenting Path p and increase flow f on each edge of p by residual Capacity c f (p). Ford–Fulkerson algorithm is a greedy algorithm that computes the maximum flow in a flow network. This algorithm uses Edmonds-Karp-Dinitz path selection rule which guarantees a running time of for nodes and edges. Let now $$G = (V,E)$$ be the created graph with the respective non-negative capacities $$c(e)$$ for all edges $$e \in E$$. Add this path-flow to flow. … A pseudocode for this algorithm is given below, Path with available capacity is called the augmenting path. ਹੈ I 39 ? This is how it works: Start by assigning a flow of 0 (f(e) = 0) to all the edges. Jobs. C++ Programming. 1. Flows in Networks: The Ford-Fulkerson Algorithm Daniel Kane Department of Computer Science and Engineering University of View 15_flows_5_ford_fulkerson.pdf from CS MISC at New York University. It is described in pseudo code for a directed graph G=(V, E) as follows where f is a flow defined on VxV. To take the researchers ordered by the lists [to avoid any suspicion of favoritism between the persons]; 2. The complexity can be given independently of the maximal flow. This lecture we will investigate an algorithm for computing maximal flows known as Ford-Fulkerson. Ford Fulkerson Algorithm helps in finding the max flow of the graph. Ford-Fulkerson Algorithm The following is simple idea of Ford-Fulkerson algorithm: 1) Start with initial flow as 0. I would love to pass the first two options (80%). My implementation of Ford–Fulkerson algorithm to solve the famous Max-Flow Problem. CS 360: Lecture 25: Maximal Flow - Ford-Fulkerson Algorithm. Now, there might be many valid paths to choose from, and the Ford-Fulkerson algorithm, as I've stated, doesn't really tell you which one to use. Full info in PDF file. Ford Fulkerson’s algorithm solves the maximum flow graph problem. What it says is at every step I need to find some source to sink path in our residual. That is, given a network with vertices and edges between those vertices that have certain weights, how much "flow" can the network process at a time? and . Small addendum (See below the idea suggested by @D.W.): In case of only two subsets A and B, the flow network could look like that (? Skills: C++ Programming. ford-fulkerson in c++. Edmonds-Karp algorithm. Ford-Fulkerson Algorithm: In simple terms, Ford-Fulkerson Algorithm is: As long as there is a path from source(S) node to sink(T) node with available capacity on all the edges in the path, send the possible flow from that path and find another path and so on. Active 7 years, 5 months ago. But its time complexity is high and it’s a pseudo-polynomial time algorithm. Ford-Fulkerson Algorithm. Find a maximum single-commodity flow using the Ford-Fulkerson algorithm. Using Ford-Fulkerson Algorithm Find The Maximum Flow And The Minimum Cut In The Digraph Below (show The Steps). It is required to explain how to apply the Ford-Fulkerson algorithm to determine whether it is possible to assign n objects to m groups according to the given rules. residual network, augmenting path. ford-fulkerson. Ask Question Asked 7 years, 5 months ago. I am doing a homework of Implementing Ford-Fulkerson algorithm, they said we should use DFS for path finding but i am stuck at somewhere. Initially, the flow of value is 0. Furthermore, let $$s \in V$$ be the selected source and $$t \in V$$ the selected target. 3 | 3 3 S 3 |5 3 5 | 6 ਹੈ | ? Ford Fulkerson algorithm in C. GitHub Gist: instantly share code, notes, and snippets. Ford-Fulkerson algorithm with depth first search. Index Terms — Algorithm, augmenting path, flow network, Ford-Fulkerson, graph, maximum flow, residual network. Wikipedia. Minimize Cash Flow among a given set of friends who have borrowed money from each other. ford fulkerson algorithm in c. Home / ford fulkerson algorithm in c. Scarecrows, giant stone warriors, and the very plants of the forest stand sentinel to hinder any heroes foolish enough to brave this realm. The main idea is to find valid flow paths until there is none left, and add them up. 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Any suspicion of favoritism between the persons ] ; 2 time complexity time... That uses BFS for finding a maximum single-commodity flow using the Ford-Fulkerson.... S \in V\ ) the selected source and \ ( s \in V\ ) be the selected target of... The selected source and \ ( t \in V\ ) be the minimum capacity value on path... Was developed by L. R. ford, Jr. and D. R. Fulkerson in 1956 by ford Fulkerson! Code because it 's localized too much 7 years, 5 months ago is an augmenting path p increase! R. Fulkerson in 1956 developed by L. R. ford, Jr. and D. R. Fulkerson in.! Algorithm helps in finding the max flow of ford-fulkerson algorithm in c maximal flow - Ford-Fulkerson algorithm flow as.... Selected target above algorithm is just an implementation of the graph: maximal flow Ford-Fulkerson! Solves the maximum flow in a flow network O ( max_flow * )... Our residual a augmenting path, flow network, Ford-Fulkerson, graph, maximum flow out on the end., capacity='capacity ' ) ¶ simple algorithm to solve the famous max-flow problem works out the maximum graph. That path using the Ford-Fulkerson algorithm: 1 ) Start with initial flow as 0 find valid flow paths there! To sink path in our residual: 1 ) Start with initial flow as 0 available capacity is called augmenting... Works out the maximum flow in a network … the Ford-Fulkerson method that uses BFS for a! From source to sink such that you get maximum flow in a flow network for! Are Ford-Fulkerson algorithm the following is simple idea of Ford-Fulkerson algorithm the following is simple idea of algorithm! ( t \in V\ ) be the minimum capacity value on that path network flow, maximum problems. 1 * ੫ 1 D the ford Fulkerson algorithm untuk pencarian maximum flow in flow... Ford_Fulkerson¶ ford_fulkerson ( G, s, t, capacity='capacity ' ) ¶ flow!
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2021-11-27 21:21:51
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https://ictp.acad.ro/a-granas-type-approach-to-some-continuation-theorems-and-periodic-boundary-value-problems-with-impulse/
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# A Granas type approach to some continuation theorems and periodic boundary value problems with impulse
## Abstract
In this paper we study periodic solutions of a second order differential equation
$x^{\prime\prime} = f(t, x, x^{\prime}) \quad for \ a.e. \ t\in [0, 1],$
subject to some impulses at certain points.
Our work was inspired by a paper by Capietto–Mawhin–Zanolin [1], where the case of no impulses was treated.
The major difference between paper [1] and ours is that instead of topological degree, we use the elementary method based on essential maps. In this context, we also give some new contributions to Granas’ theory of continuation principles.
## Authors
Department of Mathematics Babes-Bolyai University, Cluj-Napoca, Romania
?
## Paper coordinates
R. Precup, A Granas type approach to some continuation theorems and periodic boundary value problems with impulses, Topological Methods in Nonlinear Analysis, 5 (1995) no. 2, 385-396.
## PDF
##### Journal
Topological Methods in Nonlinear Analysis
12303429
MR: 97a:34028
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2023-02-02 07:58:56
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https://mathoverflow.net/questions/412306/projective-variety-of-general-type-such-that-sm-omega-x1-is-globally-genera?noredirect=1
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# Projective variety of general type such that $S^m \Omega_X^1$ is globally generated
Let $$X$$ be a smooth complex projective variety of general type; in my applications, I work with a surface, but let me ask this question in full generality.
Assume that for some $$m \geq 1$$ the vector bundle $$S^m \Omega_X^1$$ is generated by global sections, namely, the evaluation map $$H^0(X, \, S^m \Omega_X^1) \otimes \mathcal{O}_X \to S^m \Omega^1_X$$ is surjective.
Question. Is it true that $$K_X$$ is ample? Otherwise, what is a counterexample?
I started working on these topics rather recently, so I apologize if this question turns out to be trivial for the experts. Any answer and/or reference to the relevant literature will be highly appreciated.
Edit (12/26/2021). Follow-up question about the base-point freeness of $$|K_X|$$ asked as MO412382.
• Is the variety smooth? If so, then you at least have that the canonical divisor class is nef. Dec 22, 2021 at 15:54
• @JasonStarr: oh yes, it is. I will edit the question, thanks Dec 22, 2021 at 15:55
If $$X$$ is a surface it is true. In general, a smooth projective variety with $$S^m\Omega ^1_X$$ globally generated does not contain any smooth rational curve $$C$$. Indeed $$\Omega ^1_C$$ is a quotient of $$\Omega ^1_X$$, so $$S^m\Omega ^1_C$$ is also globally generated, which of course implies $$g(C)\geq 1$$.
Now if $$X$$ is a surface, this implies that $$K_X$$ is ample (in fact $$K_X$$ is ample if and only if $$X$$ does not contain any smooth rational curve with square $$\,-1$$ or $$-2$$).
Edit: As pointed out by YangMills in the comments, the result holds in all dimensions: if a smooth projective variety $$X$$ of general type contains no smooth rational curve, $$K_X$$ is ample — see Lemma 2.1 in arxiv.1606.01381.
• Thank you for the answer. Is it also true that $|K_X|$ is base-point free (assuming $\dim X=2$)? Dec 22, 2021 at 18:21
• I don't think so (if $m\geq 2$ of course), but I don't have an example at hand.
– abx
Dec 22, 2021 at 19:25
• In any dimension if $X$ is smooth with $K_X$ is nef and big and $X$ contains no rational curve, then $K_X$ is ample, see e.g. Lemma 2.1 in arxiv.org/pdf/1606.01381 Dec 23, 2021 at 4:03
• so combining my comment with the answer of abx and the comment by Jason gives you an affirmative answer in all dimensions Dec 23, 2021 at 4:32
• @YangMills: Nice, thank you! I have edited my answer to make it complete.
– abx
Dec 23, 2021 at 5:14
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2023-03-26 18:41:07
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https://www.studysmarter.us/textbooks/math/essential-calculus-early-transcendentals-2nd/series/q46e-graph-the-curves-y-xn-0-le-x-le-1-for-n-01234-on-a-comm/
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Suggested languages for you:
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Q46E
Expert-verified
Found in: Page 444
### Essential Calculus: Early Transcendentals
Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280
# Graph the curves $$y = {x^n}$$, $$0 \le x \le 1$$, for $$n = 0,1,2,3,4,....$$on a common screen. By finding the areas between successive curves, give a geometric demonstration of the fact, shown in Example 6, that $$\sum\limits_{n = 1}^\infty {\frac{1}{{n(n + 1)}}} = 1$$
The area of the successive curve is
$$1 = \sum\limits_{n = 1}^\infty {\frac{1}{{n\left( {n + 1} \right)}}}$$
See the step by step solution
## Plotting the given function
First, we have to consider the curve,
$$y = {x^n}$$, $$0 \le x \le 1$$
To find the areas between the successive curves, we have to plot the graphs of the given function to its limits.
The graph will be shown as
From the graph, it seems to be that the entire unit square $$[0,1] \times [0,1]$$ is eventually filled up by the regions that lie between successive functions.
## Area of the region
Let’s take an example, the region between $$x$$ and $${x^2}$$ is shaded below
Similarly, the region between $${x^{n - 1}}$$ and $${x^n}$$ for all $$n$$.
These regions fill up the entire square. This can be proved using an argument with limits, but it is clear from the pictures that this eventually happens.
Now, the area of the region between $${x^{n - 1}}$$ and $${x^n}{a_n}$$. Then compute $${a_n}$$ using integration
\begin{aligned}{a_n} &= \int\limits_0^1 {{x^{n - 1}}} - {x^n}\\{a_n} &= \left[ {\frac{1}{n}{x^n} - \frac{1}{{n + 1}}{x^{n + 1}}} \right]_0^1\\{a_n} &= \frac{1}{n} - \frac{1}{{n + 1}}\\{a_n} &= \frac{{n + 1}}{{n\left( {n + 1} \right)}} - \frac{n}{{n\left( {n + 1} \right)}}\\{a_n} &= \frac{1}{{n\left( {n + 1} \right)}}\end{aligned}
And, here we observed that the connection to the original series.
The unit square is, as noted, the union of all of these regions, and none of them overlap.
Therefore, the area of the unit square is equal to the sum of the area of the regions. But the area of the unit square is $$1$$, and the $${n^{th}}$$region has area $${a_n}$$.
Therefore, the area of the successive curve is
\begin{aligned}1 &= \sum\limits_{n = 1}^\infty {{a_n}} \\1 &= \sum\limits_{n = 1}^\infty {\frac{1}{{n\left( {n + 1} \right)}}} \end{aligned}
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2023-03-24 22:33:08
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http://www.irisa.fr/LIS/common/biblio/Year/2011.complete.html
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BACK TO INDEX
Publications of year 2011
Thesis
1. Pierre Allard. Logical modeling of multidimensional analysis of multivalued relations - Application to geographic data exploration. PhD thesis, Thèse de l'Université de Rennes 1 - École doctorale MATISSE, 12 décembre 2011. Note: Supervised by S. Ferré and O. Ridoux.
Abstract:
Since the beginning of data processing, the companies have realized the importance of information management solutions. The gathered data are a powerful asset to study the trends and make choices for the future. The Business Intelligence appeared in the mid-90s (the information synthesis to assist decision-making) with OLAP (On-Line Analytical Processing, a tools set for exploration, analysis and display of multidimensional data) and S-OLAP (Spatial OLAP, OLAP with spatial support). A OLAP user, unspecialized in computer sciences, does not need to know a language to handle multidimensional data, create graphics, etc. However, we consider that the OLAP data model is too rigid, because of its permanent multidimensionnal structure and because each content must have a single aggregate value. This observation is the starting point of this thesis. We propose a new paradigm of information system, able to analyze and explore multidimensional and multivalued data. To model this paradigm, we use the logical information systems (LIS) which is an information system that has common features with OLAP, especially on the data mining aspects. Our paradigm is defined by a flexible data model, an easy navigation and modular representation. We concluded this thesis by the application of this paradigm on several topics, including the exploration of geographic data.
@PhdThesis{Allard2011PhD,
author = {Pierre Allard},
title = {Logical modeling of multidimensional analysis of multivalued relations - Application to geographic data exploration},
school = {Thèse de l'Université de Rennes 1 - École doctorale MATISSE},
year = {2011},
month = {12 décembre},
note = {supervised by S. Ferré and O. Ridoux},
abstract = {Since the beginning of data processing, the companies have realized the importance of information management solutions. The gathered data are a powerful asset to study the trends and make choices for the future. The Business Intelligence appeared in the mid-90s (the information synthesis to assist decision-making) with OLAP (On-Line Analytical Processing, a tools set for exploration, analysis and display of multidimensional data) and S-OLAP (Spatial OLAP, OLAP with spatial support). A OLAP user, unspecialized in computer sciences, does not need to know a language to handle multidimensional data, create graphics, etc. However, we consider that the OLAP data model is too rigid, because of its permanent multidimensionnal structure and because each content must have a single aggregate value. This observation is the starting point of this thesis. We propose a new paradigm of information system, able to analyze and explore multidimensional and multivalued data. To model this paradigm, we use the logical information systems (LIS) which is an information system that has common features with OLAP, especially on the data mining aspects. Our paradigm is defined by a flexible data model, an easy navigation and modular representation. We concluded this thesis by the application of this paradigm on several topics, including the exploration of geographic data.},
}
Conference articles
1. Denis Béchet, Alexandre Dikovsky, and Annie Foret. Categorial Grammars with Iterated Types form a Strict Hierarchy of k-Valued Languages. In Implementation and Application of Automata - 16th International Conference, CIAA 2011, Blois, France, volume 6807 of Lecture Notes in Computer Science, pages 42-52, 2011. Springer.
Abstract:
The notion of k-valued categorial grammars where a word is associated to at most k types is often used in the field of lexicalized grammars as a fruitful constraint for obtaining several properties like the existence of learning algorithms. This principle is relevant only when the classes of k-valued grammars correspond to a real hierarchy of languages. Such a property had been shown earlier for classical categorial grammars. This paper establishes the relevance of this notion when categorial grammars are enriched with iterated types.
@inproceedings{Foret11c,
author = {Denis Béchet and Alexandre Dikovsky and Annie Foret},
title = {Categorial Grammars with Iterated Types form a Strict Hierarchy of k-Valued Languages},
booktitle = {Implementation and Application of Automata - 16th International Conference, CIAA 2011, Blois, France},
publisher = {Springer},
series = {Lecture Notes in Computer Science},
volume = {6807},
year = {2011},
isbn = {978-3-642-22255-9},
pages = {42-52},
ee = {http://dx.doi.org/10.1007/978-3-642-22256-6_5},
keywords = {},
Abstract={The notion of k-valued categorial grammars where a word is associated to at most k types is often used in the field of lexicalized grammars as a fruitful constraint for obtaining several properties like the existence of learning algorithms. This principle is relevant only when the classes of k-valued grammars correspond to a real hierarchy of languages. Such a property had been shown earlier for classical categorial grammars. This paper establishes the relevance of this notion when categorial grammars are enriched with iterated types.}
}
2. Denis Béchet, Alexander Dikovsky, and Annie Foret. On Dispersed and Choice Iteration in Incrementally Learnable Dependency Types. In Logical Aspects of Computational Linguistics - 6th International Conference, LACL 2011, Montpellier, France, volume 6736 of Lecture Notes in Computer Science, pages 80-95, 2011. Springer. Keyword(s): Grammatical inference, Categorial grammar, Dependency grammar, Incremental learning, Iterated types..
Abstract:
We study learnability of Categorial Dependency Grammars (CDG), a family of categorial grammars expressing all kinds of projective, discontinuous and repeatable dependencies. For these grammars, it is known that they are not learnable from dependency structures. We propose two different ways of modelling the repeatable dependencies through iterated types and the two corresponding families of CDG which cannot distinguish between the dependencies repeatable at least K times and those repeatable any number of times. For both we show that they are incrementally learnable in the limit from dependency structures.
@inproceedings{Foret11b,
author = {Denis Béchet and Alexander Dikovsky and Annie Foret},
title = {On Dispersed and Choice Iteration in Incrementally Learnable Dependency Types},
booktitle = {Logical Aspects of Computational Linguistics - 6th International Conference, LACL 2011, Montpellier, France},
publisher = {Springer},
series = {Lecture Notes in Computer Science},
volume = {6736},
year = {2011},
isbn = {978-3-642-22220-7},
pages = {80-95},
ee = {http://dx.doi.org/10.1007/978-3-642-22221-4_6},
keywords={Grammatical inference, Categorial grammar, Dependency grammar, Incremental learning, Iterated types.},
abstract={We study learnability of Categorial Dependency Grammars (CDG), a family of categorial grammars expressing all kinds of projective, discontinuous and repeatable dependencies. For these grammars, it is known that they are not learnable from dependency structures. We propose two different ways of modelling the repeatable dependencies through iterated types and the two corresponding families of CDG which cannot distinguish between the dependencies repeatable at least K times and those repeatable any number of times. For both we show that they are incrementally learnable in the limit from dependency structures.}
}
3. Denis Béchet, Alexander Dikovsky, and Annie Foret. Sur les itérations dispersées et les choix itérés pour l'apprentissage incrémental des types dans les grammaires de dépendances. In Conférence Francophone d'Apprentissage 2011 (CAP), Chambéry, France, 2011.
@inproceedings{Foret11a,
author = {Denis Béchet and Alexander Dikovsky and Annie Foret},
title ={Sur les itérations dispersées et les choix itérés pour l'apprentissage incrémental des types dans les grammaires de dépendances},
booktitle = {Conférence Francophone d'Apprentissage 2011 (CAP), Chambéry, France},
year={2011}
}
4. Peggy Cellier, Mireille Ducassé, and Sébastien Ferré. Exploration de traces à l'aide de fouille de données. In Atelier IC Traces numériques, connaissances et cognition, 2011.
@inproceedings{cellier2011exploration,
title={Exploration de traces à l'aide de fouille de données},
author={Cellier, Peggy and Ducassé, Mireille and Ferré, Sébastien},
booktitle={Atelier IC ~Traces numériques, connaissances et cognition~},
year={2011}
}
5. Peggy Cellier, Mireille Ducassé, Sébastien Ferré, and Olivier Ridoux. Multiple Fault Localization with Data Mining. In Int. Conf. on Software Engineering & Knowledge Engineering, pages 238-243, 2011. Knowledge Systems Institute Graduate School. Keyword(s): data mining, software engineering, debugging, association rules, formal concept analysis.
Abstract:
We have proposed an interactive fault localization method based on two data mining techniques, formal concept analysis and association rules. A lattice formalizes the partial ordering and the dependencies between the sets of program elements (e.g., lines) that are most likely to lead to program execution failures. The paper provides an algorithm to traverse that lattice starting from the most suspect places. The main contribution is that the algorithm is able to deal with any number of faults within a single execution of a test suite. In addition, a stopping criterion independent of the number of faults is provided.
@inproceedings{CellierDFR11seke,
author = {Peggy Cellier and Mireille Ducassé and Sébastien Ferré and Olivier Ridoux},
title = {Multiple Fault Localization with Data Mining},
booktitle = {Int. Conf. on Software Engineering {\&} Knowledge Engineering},
year = {2011},
pages = {238-243},
keywords = {data mining, software engineering, debugging, association rules, formal concept analysis},
publisher = {Knowledge Systems Institute Graduate School},
abstract = {We have proposed an interactive fault localization method based on two data mining techniques, formal concept analysis and association rules. A lattice formalizes the partial ordering and the dependencies between the sets of program elements (e.g., lines) that are most likely to lead to program execution failures. The paper provides an algorithm to traverse that lattice starting from the most suspect places. The main contribution is that the algorithm is able to deal with any number of faults within a single execution of a test suite. In addition, a stopping criterion independent of the number of faults is provided.}
}
6. Peggy Cellier, Sébastien Ferré, Mireille Ducassé, and Thierry Charnois. Partial orders and logical concept analysis to explore patterns extracted by data mining. In Int. Conf. on Conceptual Structures for Discovering Knowledge, pages 77-90, 2011. Springer. Keyword(s): data mining, partial order, selection of patterns, logical concept analysis, formal concept analysis.
Abstract:
Data mining techniques are used in order to discover emerging knowledge (patterns) in databases. The problem of such techniques is that there are, in general, too many resulting patterns for a user to explore them all by hand. Some methods try to reduce the number of patterns without a priori pruning. The number of patterns remains, nevertheless, high. Other approaches, based on a total ranking, propose to show to the user the top-k patterns with respect to a measure. Those methods do not take into account the user's knowledge and the dependencies that exist between patterns. In this paper, we propose a new way for the user to explore extracted patterns. The method is based on navigation in a partial order over the set of all patterns in the Logical Concept Analysis framework. It accommodates several kinds of patterns and the dependencies between patterns are taken into account thanks to partial orders. It allows the user to use his/her background knowledge to navigate through the partial order, without a priori pruning. We illustrate how our method can be applied on two different tasks (software engineering and natural language processing) and two different kinds of patterns (association rules and sequential patterns).
@inproceedings{cellier2011iccs,
title = {Partial orders and logical concept analysis to explore patterns extracted by data mining},
author = {Cellier, Peggy and Ferré, Sébastien and Ducassé, Mireille and Charnois, Thierry},
booktitle = {Int. Conf. on Conceptual Structures for Discovering Knowledge},
pages = {77--90},
year = {2011},
publisher = {Springer},
keywords = {data mining, partial order, selection of patterns, logical concept analysis, formal concept analysis},
abstract = { Data mining techniques are used in order to discover emerging knowledge (patterns) in databases. The problem of such techniques is that there are, in general, too many resulting patterns for a user to explore them all by hand. Some methods try to reduce the number of patterns without a priori pruning. The number of patterns remains, nevertheless, high. Other approaches, based on a total ranking, propose to show to the user the top-k patterns with respect to a measure. Those methods do not take into account the user's knowledge and the dependencies that exist between patterns. In this paper, we propose a new way for the user to explore extracted patterns. The method is based on navigation in a partial order over the set of all patterns in the Logical Concept Analysis framework. It accommodates several kinds of patterns and the dependencies between patterns are taken into account thanks to partial orders. It allows the user to use his/her background knowledge to navigate through the partial order, without a priori pruning. We illustrate how our method can be applied on two different tasks (software engineering and natural language processing) and two different kinds of patterns (association rules and sequential patterns). }
}
7. Mireille Ducassé, Sébastien Ferré, and Peggy Cellier. Building up Shared Knowledge with Logical Information Systems. In A. Napoli and V. Vychodil, editors, Proceedings of the 8th International Conference on Concept Lattices and their Applications, pages 31-42, October 2011. INRIA. Note: ISBN 978-2-905267-78-8.
Abstract:
Logical Information Systems (LIS) are based on Logical Concept Analysis, an extension of Formal Concept Analysis. This paper describes an application of LIS to support group decision. A case study gathered a research team. The objective was to decide on a set of potential conferences on which to send submissions. People individually used Abilis, a LIS web server, to preselect a set of conferences. Starting from 1041 call for papers, the individual participants preselected 63 conferences. They met and collectively used Abilis to select a shared set of 42 target conferences. The team could then sketch a publication planning. The case study provides evidence that LIS cover at least three of the collaboration patterns identified by Kolfschoten, de Vreede and Briggs. Abilis helped the team to build a more complete and relevant set of information (Generate/Gathering pattern); to build a shared understanding of the relevant information (Clarify/Building Shared Understanding); and to quickly reduce the number of target conferences (Reduce/Filtering pattern).
@InProceedings{ducasse2011,
Author={Mireille Ducassé and Sébastien Ferré and Peggy Cellier},
Title={Building up Shared Knowledge with Logical Information Systems},
Pages={31-42},
BookTitle={Proceedings of the 8th International Conference on Concept Lattices and their Applications},
Year={2011},
Editor={A. Napoli and V. Vychodil},
Publisher={INRIA},
Note={ISBN 978-2-905267-78-8},
Month={October},
Abstract={ Logical Information Systems (LIS) are based on Logical Concept Analysis, an extension of Formal Concept Analysis. This paper describes an application of LIS to support group decision. A case study gathered a research team. The objective was to decide on a set of potential conferences on which to send submissions. People individually used Abilis, a LIS web server, to preselect a set of conferences. Starting from 1041 call for papers, the individual participants preselected 63 conferences. They met and collectively used Abilis to select a shared set of 42 target conferences. The team could then sketch a publication planning. The case study provides evidence that LIS cover at least three of the collaboration patterns identified by Kolfschoten, de Vreede and Briggs. Abilis helped the team to build a more complete and relevant set of information (Generate/Gathering pattern); to build a shared understanding of the relevant information (Clarify/Building Shared Understanding); and to quickly reduce the number of target conferences (Reduce/Filtering pattern).}
}
8. Sébastien Ferré and Alice Hermann. Semantic Search: Reconciling Expressive Querying and Exploratory Search. In L. Aroyo and C. Welty, editors, Int. Semantic Web Conf., LNCS 7031, pages 177-192, 2011. Springer. Keyword(s): semantic web, querying, exploratory search, expressiveness, navigation, faceted search.
Abstract:
Faceted search and querying are two well-known paradigms to search the Semantic Web. Querying languages, such as SPARQL, offer expressive means for searching RDF datasets, but they are difficult to use. Query assistants help users to write well-formed queries, but they do not prevent empty results. Faceted search supports exploratory search, i.e., guided navigation that returns rich feedbacks to users, and prevents them to fall in dead-ends (empty results). However, faceted search systems do not offer the same expressiveness as query languages. We introduce Query-based Faceted Search (QFS), the combination of an expressive query language and faceted search, to reconcile the two paradigms. In this paper, the LISQL query language generalizes existing semantic faceted search systems, and covers most features of SPARQL. A prototype, Sewelis (aka. Camelis 2), has been implemented, and a usability evaluation demonstrated that QFS retains the ease-of-use of faceted search, and enables users to build complex queries with little training.
@InProceedings{FerHer2011iswc,
author = {Sébastien Ferré and Alice Hermann},
title = {Semantic Search: Reconciling Expressive Querying and Exploratory Search},
booktitle = {Int. Semantic Web Conf.},
pages = {177-192},
year = {2011},
editor = {L. Aroyo and C. Welty},
series = {LNCS 7031},
publisher = {Springer},
keywords = {semantic web, querying, exploratory search, expressiveness, navigation, faceted search},
abstract = {Faceted search and querying are two well-known paradigms to search the Semantic Web. Querying languages, such as SPARQL, offer expressive means for searching RDF datasets, but they are difficult to use. Query assistants help users to write well-formed queries, but they do not prevent empty results. Faceted search supports exploratory search, i.e., guided navigation that returns rich feedbacks to users, and prevents them to fall in dead-ends (empty results). However, faceted search systems do not offer the same expressiveness as query languages. We introduce Query-based Faceted Search (QFS), the combination of an expressive query language and faceted search, to reconcile the two paradigms. In this paper, the LISQL query language generalizes existing semantic faceted search systems, and covers most features of SPARQL. A prototype, Sewelis (aka. Camelis 2), has been implemented, and a usability evaluation demonstrated that QFS retains the ease-of-use of faceted search, and enables users to build complex queries with little training.},
}
9. S. Ferré, A. Hermann, and M. Ducassé. Combining Faceted Search and Query Languages for the Semantic Web. In C. Salinesi and O. Pastor, editors, Semantic Search over the Web (SSW) - Advanced Information Systems Engineering Workshops - CAiSE Int. Workshops, volume 83 of LNBIP 83, pages 554-563, 2011. Springer. Note: Best paper. Keyword(s): semantic web, semantic search, user interaction, faceted search, querying.
Abstract:
Faceted search and querying are the two main paradigms to search the Semantic Web. Querying languages, such as SPARQL, offer expressive means for searching knowledge bases, but they are difficult to use. Query assistants help users to write well-formed queries, but they do not prevent empty results. Faceted search supports exploratory search, i.e., guided navigation that returns rich feedbacks to users, and prevents them to fall in dead-ends (empty results). However, faceted search systems do not offer the same expressiveness as query languages. We introduce semantic faceted search, the combination of an expressive query language and faceted search to reconcile the two paradigms. The query language is basically SPARQL, but with a syntax that better fits in a faceted search interface. A prototype, Camelis 2, has been implemented, and a usability evaluation demonstrated that semantic faceted search retains the ease-of-use of faceted search, and enables users to build complex queries with little training.
@inproceedings{FerHerDuc2011ssw,
author = {S. Ferré and A. Hermann and M. Ducassé},
title = {Combining Faceted Search and Query Languages for the Semantic Web},
booktitle = {CAiSE Workshops},
year = {2011},
pages = {554-563},
editor = {C. Salinesi and O. Pastor},
booktitle = {Semantic Search over the Web (SSW) - Advanced Information Systems Engineering Workshops - CAiSE Int. Workshops},
publisher = {Springer},
series = {LNBIP 83},
volume = {83},
year = {2011},
isbn = {978-3-642-22055-5},
note = {Best paper},
keywords = {semantic web, semantic search, user interaction, faceted search, querying},
abstract = {Faceted search and querying are the two main paradigms to search the Semantic Web. Querying languages, such as SPARQL, offer expressive means for searching knowledge bases, but they are difficult to use. Query assistants help users to write well-formed queries, but they do not prevent empty results. Faceted search supports exploratory search, i.e., guided navigation that returns rich feedbacks to users, and prevents them to fall in dead-ends (empty results). However, faceted search systems do not offer the same expressiveness as query languages. We introduce semantic faceted search, the combination of an expressive query language and faceted search to reconcile the two paradigms. The query language is basically SPARQL, but with a syntax that better fits in a faceted search interface. A prototype, Camelis 2, has been implemented, and a usability evaluation demonstrated that semantic faceted search retains the ease-of-use of faceted search, and enables users to build complex queries with little training.},
}
10. Alice Hermann, Sébastien Ferré, and Mireille Ducassé. Création et mise à jour guidées d'objets dans une base RDF(S). In Rencontres Jeunes Chercheurs en Intelligence Artificielle (RJCIA), 2011. Presses de l'Université des Antilles et de la Guyane.
Abstract:
La mise à jour des bases de connaissances existantes est cruciale pour tenir compte des nouvelles informations, régulièrement découvertes. Toutefois, les données actuelles du Web Sémantique sont rarement mises à jour par les utilisateurs. Les utilisateurs ne sont pas suffisament aidés lors de l'ajout et de la mise à jour d'objets. Nous proposons une approche pour aider l'utilisateur à ajouter de nouveaux objets de manière incrémentale et dynamique. Notre approche est fondée sur les Systèmes d'Information Logiques pour l'interaction utilisateur. Pour le guidage, le système cherche les objets ayant des propriétés en commun avec la description de l'objet en cours de création. Les propriétés de ces objets, non présents dans la description, servent de suggestions pour compléter la description de l'objet.
@inproceedings{Hermann:RJCIA:2011,
author = {Alice Hermann and Sébastien Ferré and Mireille Ducassé},
booktitle = {Rencontres Jeunes Chercheurs en Intelligence Artificielle ({RJCIA})},
title = {Création et mise à jour guidées d'objets dans une base {RDF(S)}},
publisher = {Presses de l'{U}niversité des {A}ntilles et de la {G}uyane},
year = 2011,
abstract = {La mise à jour des bases de connaissances existantes est cruciale pour tenir compte des nouvelles informations, régulièrement découvertes. Toutefois, les données actuelles du Web Sémantique sont rarement mises à jour par les utilisateurs. Les utilisateurs ne sont pas suffisament aidés lors de l'ajout et de la mise à jour d'objets. Nous proposons une approche pour aider l'utilisateur à ajouter de nouveaux objets de manière incrémentale et dynamique. Notre approche est fondée sur les Systèmes d'Information Logiques pour l'interaction utilisateur. Pour le guidage, le système cherche les objets ayant des propriétés en commun avec la description de l'objet en cours de création. Les propriétés de ces objets, non présents dans la description, servent de suggestions pour compléter la description de l'objet.},
}
11. Alice Hermann, Sébastien Ferré, and Mireille Ducassé. Guided creation and update of objects in RDF(S) bases. In Mark A. Musen and Óscar Corcho, editors, Int. Conf. Knowledge Capture (K-CAP 2011), pages 189-190, 2011. ACM Press.
Abstract:
Updating existing knowledge bases is crucial to take into account the information that are regularly discovered. However, this is quite tedious and in practice Semantic Web data are rarely updated by users. This paper presents UTILIS, an approach to help users create and update objects in RDF(S) bases. While creating a new object, $o$, UTILIS searches for similar objects, found by applying relaxation rules to the description of $o$, taken as a query. The resulting objects and their properties serve as suggestions to expand the description of $o$.
@inproceedings{Hermann:KCAP:2011,
author = {Alice Hermann and Sébastien Ferré and Mireille Ducassé},
title = {Guided creation and update of objects in {RDF(S)} bases},
booktitle = {Int. Conf. Knowledge Capture (K-CAP 2011)},
year = {2011},
pages = {189-190},
editor = {Mark A. Musen and Óscar Corcho},
publisher = {ACM Press},
abstract = {Updating existing knowledge bases is crucial to take into account the information that are regularly discovered. However, this is quite tedious and in practice Semantic Web data are rarely updated by users. This paper presents UTILIS, an approach to help users create and update objects in RDF(S) bases. While creating a new object, $o$, UTILIS searches for similar objects, found by applying relaxation rules to the description of $o$, taken as a query. The resulting objects and their properties serve as suggestions to expand the description of $o$.},
}
12. D. Legallois, Peggy Cellier, and Thierry Charnois. Calcul de réseaux phrastiques pour l analyse et la navigation textuelle. In Traitement Automatique des Langues Naturelles, 2011. Keyword(s): Sentence network, Bonds between sentences, Textual analysis, Discourse analysis.
Abstract:
In this paper, we present an automatic process based on text reduction introduced by Hoey. The application of that kind of approaches on large texts is difficult to do by hand. In the paper, we propose an automatic process to treat large texts. We have conducted some experiments on different kinds of texts (narrative, expositive) to show the benefits of the approach.
@inproceedings{legallois2011taln,
title={Calcul de réseaux phrastiques pour l~analyse et la navigation textuelle},
author={Legallois, D. and Cellier, Peggy and Charnois, Thierry},
booktitle={Traitement Automatique des Langues Naturelles},
year={2011},
keywords = {Sentence network, Bonds between sentences, Textual analysis, Discourse analysis},
abstract = { In this paper, we present an automatic process based on text reduction introduced by Hoey. The application of that kind of approaches on large texts is difficult to do by hand. In the paper, we propose an automatic process to treat large texts. We have conducted some experiments on different kinds of texts (narrative, expositive) to show the benefits of the approach. }
}
Internal reports
1. Sébastien Ferré. SQUALL: a High-Level Language for Querying and Updating the Semantic Web. Research Report, IRISA, 2011. [WWW] Keyword(s): Semantic Web, controlled natural language, query language, update language, expressiveness, Montague grammar.
Abstract:
Languages play a central role in the Semantic Web. An important aspect regarding their design is syntax as it plays a crucial role in the wide acceptance of the Semantic Web approach. Like for programming languages, an evolution can be observed from low-level to high-level designs. High-level languages not only allow more people to contribute by abstracting from the details, but also makes experienced people more productive, and makes the produced documents easier to share and maintain. We introduce SQUALL, a high-level language for querying and updating semantic data. It has a strong adequacy with RDF, an expressiveness very similar to SPARQL 1.1, and a controlled natural language syntax that completely abstracts from low-level notions such as bindings and relational algebra. We first give an informal presentation of SQUALL through examples, comparing it with SPARQL. We then formally define the syntax and semantics of SQUALL as a Montague grammar, and its translation to SPARQL.
@techreport{PI1985,
title = {{SQUALL}: a High-Level Language for Querying and Updating the Semantic Web},
author = {Sébastien Ferré},
abstract = {Languages play a central role in the Semantic Web. An important aspect regarding their design is syntax as it plays a crucial role in the wide acceptance of the Semantic Web approach. Like for programming languages, an evolution can be observed from low-level to high-level designs. High-level languages not only allow more people to contribute by abstracting from the details, but also makes experienced people more productive, and makes the produced documents easier to share and maintain. We introduce SQUALL, a high-level language for querying and updating semantic data. It has a strong adequacy with RDF, an expressiveness very similar to SPARQL 1.1, and a controlled natural language syntax that completely abstracts from low-level notions such as bindings and relational algebra. We first give an informal presentation of SQUALL through examples, comparing it with SPARQL. We then formally define the syntax and semantics of SQUALL as a Montague grammar, and its translation to SPARQL.},
keywords = {Semantic Web, controlled natural language, query language, update language, expressiveness, Montague grammar},
language = {{A}nglais},
institution = {IRISA},
affiliation = {{LIS} - {IRISA} - {CNRS} - {I}nstitut {N}ational des {S}ciences {A}ppliquées de {R}ennes - {U}niversité de {R}ennes {I} },
pages = {18},
type = {Research Report},
year = {2011},
URL = {http://hal.inria.fr/docs/00/62/84/27/PDF/PI-1985.pdf},
}
2. Sébastien Ferré, Alice Hermann, and Mireille Ducassé. Semantic Faceted Search: Safe and Expressive Navigation in RDF Graphs. Research Report, IRISA, 2011. [WWW] Keyword(s): semantic web, faceted search, query language, exploratory search, navigation, expressiveness.
Abstract:
@techreport{PI1964,
title = {Semantic Faceted Search: Safe and Expressive Navigation in {RDF} Graphs },
author = {Sébastien Ferré and Alice Hermann and Mireille Ducassé},
keywords = {semantic web, faceted search, query language, exploratory search, navigation, expressiveness},
language = {{A}nglais},
institution = {IRISA},
affiliation = {{LIS} - {IRISA} - {CNRS} - {I}nstitut {N}ational des {S}ciences {A}ppliquées de {R}ennes - {U}niversité de {R}ennes {I} },
pages = {27},
type = {Research Report},
year = {2011},
URL = {http://hal.inria.fr/inria-00410959/PDF/PI-1964.pdf},
}
Miscellaneous
1. Sébastien Ferré and Alice Hermann. Camelis2 : explorer et éditer une base RDF(S) de façon expressive et interactive. Note: Démo acceptée à la platerforme AFIA, 2011. Keyword(s): demo.
@Unpublished{demo:camelis2:ic2011,
author = {Sébastien Ferré and Alice Hermann},
title = {Camelis2 : explorer et éditer une base RDF(S) de façon expressive et interactive},
note = {Démo acceptée à la platerforme AFIA},
year = {2011},
keywords = {demo},
}
2. Sébastien Ferré and Alice Hermann. Sewelis: Exploring and Editing an RDF Base in an Expressive and Interactive Way. Note: Demo accepted at the Int. Semantic Web Conf. (ISWC), 2011. Keyword(s): demo.
Abstract:
Query-based Faceted Search (QFS), introduced in a research paper at ISWC'11, reconciles the expressiveness of querying languages (e.g., SPARQL), and the benefits of exploratory search found in faceted search. Because of the interactive nature of QFS, which is difficult to fully render in a research paper, we feel it is important to complement it with a demonstration of our QFS prototype, Sewelis (aka. Camelis 2). An important addition to the research paper is the extension of QFS to the guided edition of RDF bases, where suggestions are based on existing data. This paper motivates our approach, shortly presents Sewelis, and announces the program of the demonstration. Screencasts of the demonstration, as well as material (program and data) to reproduce it, are available at { t http://www.irisa.fr/LIS/softwares/sewelis}.
@Unpublished{demo:sewelis:iswc2011,
author = {Sébastien Ferré and Alice Hermann},
title = {Sewelis: Exploring and Editing an RDF Base in an Expressive and Interactive Way},
note = {Demo accepted at the Int. Semantic Web Conf. (ISWC)},
year = {2011},
keywords = {demo},
abstract = {Query-based Faceted Search (QFS), introduced in a research paper at ISWC'11, reconciles the expressiveness of querying languages (e.g., SPARQL), and the benefits of exploratory search found in faceted search. Because of the interactive nature of QFS, which is difficult to fully render in a research paper, we feel it is important to complement it with a demonstration of our QFS prototype, Sewelis (aka. Camelis 2). An important addition to the research paper is the extension of QFS to the guided edition of RDF bases, where suggestions are based on existing data. This paper motivates our approach, shortly presents Sewelis, and announces the program of the demonstration. Screencasts of the demonstration, as well as material (program and data) to reproduce it, are available at { t http://www.irisa.fr/LIS/softwares/sewelis}.},
}
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2019-03-24 08:03:27
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https://stacks.math.columbia.edu/tag/001U
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## 4.6 Fibre products
Definition 4.6.1. Let $x, y, z\in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$, $f\in \mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(x, y)$ and $g\in \mathop{\mathrm{Mor}}\nolimits _{\mathcal C}(z, y)$. A fibre product of $f$ and $g$ is an object $x \times _ y z\in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ together with morphisms $p \in \mathop{\mathrm{Mor}}\nolimits _{\mathcal C}(x \times _ y z, x)$ and $q \in \mathop{\mathrm{Mor}}\nolimits _{\mathcal C}(x \times _ y z, z)$ making the diagram
$\xymatrix{ x \times _ y z \ar[r]_ q \ar[d]_ p & z \ar[d]^ g \\ x \ar[r]^ f & y }$
commute, and such that the following universal property holds: for any $w\in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ and morphisms $\alpha \in \mathop{\mathrm{Mor}}\nolimits _{\mathcal C}(w, x)$ and $\beta \in \mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(w, z)$ with $f \circ \alpha = g \circ \beta$ there is a unique $\gamma \in \mathop{\mathrm{Mor}}\nolimits _{\mathcal C}(w, x \times _ y z)$ making the diagram
$\xymatrix{ w \ar[rrrd]^\beta \ar@{-->}[rrd]_\gamma \ar[rrdd]_\alpha & & \\ & & x \times _ y z \ar[d]^ p \ar[r]_ q & z \ar[d]^ g \\ & & x \ar[r]^ f & y }$
commute.
If a fibre product exists it is unique up to unique isomorphism. This follows from the Yoneda lemma as the definition requires $x \times _ y z$ to be an object of $\mathcal{C}$ such that
$h_{x \times _ y z}(w) = h_ x(w) \times _{h_ y(w)} h_ z(w)$
functorially in $w$. In other words the fibre product $x \times _ y z$ is an object representing the functor $w \mapsto h_ x(w) \times _{h_ y(w)} h_ z(w)$.
Definition 4.6.2. We say a commutative diagram
$\xymatrix{ w \ar[r] \ar[d] & z \ar[d] \\ x \ar[r] & y }$
in a category is cartesian if $w$ and the morphisms $w \to x$ and $w \to z$ form a fibre product of the morphisms $x \to y$ and $z \to y$.
Definition 4.6.3. We say the category $\mathcal{C}$ has fibre products if the fibre product exists for any $f\in \mathop{\mathrm{Mor}}\nolimits _{\mathcal C}(x, y)$ and $g\in \mathop{\mathrm{Mor}}\nolimits _{\mathcal C}(z, y)$.
Definition 4.6.4. A morphism $f : x \to y$ of a category $\mathcal{C}$ is said to be representable if for every morphism $z \to y$ in $\mathcal{C}$ the fibre product $x \times _ y z$ exists.
Lemma 4.6.5. Let $\mathcal{C}$ be a category. Let $f : x \to y$, and $g : y \to z$ be representable. Then $g \circ f : x \to z$ is representable.
Proof. Omitted. $\square$
Lemma 4.6.6. Let $\mathcal{C}$ be a category. Let $f : x \to y$ be representable. Let $y' \to y$ be a morphism of $\mathcal{C}$. Then the morphism $x' := x \times _ y y' \to y'$ is representable also.
Proof. Let $z \to y'$ be a morphism. The fibre product $x' \times _{y'} z$ is supposed to represent the functor
\begin{eqnarray*} w & \mapsto & h_{x'}(w)\times _{h_{y'}(w)} h_ z(w) \\ & = & (h_ x(w) \times _{h_ y(w)} h_{y'}(w)) \times _{h_{y'}(w)} h_ z(w) \\ & = & h_ x(w) \times _{h_ y(w)} h_ z(w) \end{eqnarray*}
which is representable by assumption. $\square$
Comment #153 by on
After 001V, one should add the definition of cartesian square, use without definition for the first time in 0024.
Comment #3413 by Herman Rohrbach on
It might be nice to state a "pasting law for pullbacks" (see e.g. https://ncatlab.org/nlab/show/pasting+law+for+pullbacks). For example, lemma 001Y is a direct consequence of this.
Comment #3472 by on
Maybe, but actually, Lemma 4.6.5 seems totally trivial to me. I guess what I am saying is that the "pasting law" is akin to $1 + 1 = 2$ once you've been sufficiently indoctrinated. I am not trying to be snarky or annoying (just trying to become more like my hero Linus Torvalds...)
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
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2021-07-25 09:39:05
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https://ferraricalifornia.org/certificate-of-acceptance-construction/certificate-of-acceptance-construction-as/
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# Certificate Of Acceptance Construction As
Certificate Of Acceptance Construction As
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2020-01-26 05:38:30
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http://zgkj.cast.cn/EN/10.16708/j.cnki.1000-758X.2019.0011
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### Orbit control force coefficient identification by Charp recursion method
ZHANG Ying1,2，WANG Xijing2,*，YUAN Bo3，KONG Dalin2，BIAN Yanshan2
1. 1School of Astronautics Northwestern Polytechnical University, Xi′an 710072，China
2Key Laboratory for Spacecraft InOrbit Fault Diagnosis and Maintenance,Xi′an 710043，China
3Xi′an Satellite Control Center, Xi′an 710043，China
• Received:2018-07-20 Revised:2018-09-15 Online:2019-04-25 Published:2019-03-19
Abstract: Spacecraft orbit maneuvers are required in spacecraft orbit capture, orbit maintenance and obstacle avoidance. Considering the fact that during spacecraft orbital maneuver, the thrust coefficient is the binding constant, error can be great due to the unoptimized parameter. The fitting coefficient of the control force is identified as a correction control parameter to compensate for the orbital control error, so as to improve the accuracy of orbit control. The historical orbit control data of typical spacecraft platform and its engine were analyzed in a statistical way. By analyzing the previous orbit control theory and inorbit control data, the orbit control empirical model was established. With the input and output of the current measurable system, the future evolution of system output was predictable and the empirical formula of the relationship between the error of actual orbit control and the control parameters under different working conditions was concluded.Two types of Low Earth Orbit (LEO) satellites with semimajor axis variation above and below 300 meters were selected to analyze historical data of orbital control. After the actual data test, the accuracy of the velocity variation forecast after the thrust coefficient fitting by the Charp recursion method is higher. This kind of calculation method makes use of historical data of orbit control, and the calculation method is simple, which improves the prediction precision of orbit control speed increment and has reference significance for the implementation of orbit control.
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2019-12-15 02:24:42
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https://brilliant.org/problems/i-dont-even-have-this-cube/
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# I don't even have this cube!
Computer Science Level pending
This is a $$\displaystyle 4\times 4\times 4$$ Rubik's cube. It has $$\displaystyle 401 196 841 564 901 869 874 093 974 498 574 336 000 000 000$$ combinations! It has 9 trailing zeroes.
Find the number of trailing zeroes in the number of combinations of the $$\displaystyle 25\times25\times25$$ Rubik's cube.
Details:
• Do not disassemble the cube, it's expensive.
• Calculators are allowed.
×
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2018-01-18 06:20:06
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http://math.tutorcircle.com/algebra-1/how-is-dividing-a-polynomial-by-a-binomial-similar-to-or-different-from-the-long-division-you-learned-in-elementary-school.html
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Sales Toll Free No: 1-855-666-7446
• Math /
• Algebra 1 /
• Polynomials /
• How is Dividing a Polynomial by a Binomial Similar to or Different From the Long Division you Learned in Elementary School
# How is Dividing a Polynomial by a Binomial Similar to or Different From the Long Division you Learned in Elementary School?
TopIn our elementary school mathematics we used to divide large Numbers using Long Division method where answer is called quotient and the left over as remainder. To show how is dividing a polynomial by a binomial similar to or different from the long division you learned in elementary school maths let us consider an example.
Suppose we have two Polynomials given as: (12p3 – 4p2 – 5p) and (9p2 + 8p + 1). First one is of degree three and is also known as trinomial and one with degree two is called a binomial.
To divide the trinomial by binomial we first factorize degree 3 polynomial to get its simplest form: p (12p2 – 4p - 5) = (2p – 1) (6p +5).
Next we factorize the binomial (9p2 + 8p + 1) as: (9p + 1) (p + 1). This makes our overall division looks like: (2p – 1) (6p + 5) / (9p + 1) (p + 1). Here we need to use the technique of partial fraction to solve it further as follows:
(12p2 – 4p - 5) / (9p + 1) (p + 1) = A / (9p + 1) + B / (p + 1),
= (C * (p + 1) + D * (9p + 1)) / (9p + 1) (p + 1),
= (p (C + 9D) + (C + D)) / (9p + 1) (p + 1),
By equating the coefficients of right side of equation to those on left side we get:
C + 9D = - 4,
C + D = -5,
Solving for values of C and D we get final solution as:
(1 / 8) / (9p + 1) + (-11 / 24) / (p + 1),
On dividing the trinomial by binomial using long division we will have same result. Thus we see that dividing a polynomial by a binomial is similar to long division that we studied in our elementary school.
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2017-08-18 14:31:29
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http://www.mathcounterexamples.net/tag/algebra-2/page/5/
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# A finitely generated soluble group isomorphic to a proper quotient group
Let $$\mathbb{Q}_2$$ be the ring of rational numbers of the form $$m2^n$$ with $$m, n \in \mathbb{Z}$$ and $$N = U(3, \mathbb{Q}_2)$$ the group of unitriangular matrices of dimension $$3$$ over $$\mathbb{Q}_2$$. Let $$t$$ be the diagonal matrix with diagonal entries: $$1, 2, 1$$ and put $$H = \langle t, N \rangle$$. We will prove that $$H$$ is finitely generated and that one of its quotient group $$G$$ is isomorphic to a proper quotient group of $$G$$. Continue reading A finitely generated soluble group isomorphic to a proper quotient group
# A (not finitely generated) group isomorphic to a proper quotient group
The basic question that we raise here is the following one: given a group $$G$$ and a proper subgroup $$H$$ (i.e. $$H \notin \{\{1\},G\}$$, can $$G/H$$ be isomorphic to $$G$$? A group $$G$$ is said to be hopfian (after Heinz Hopf) if it is not isomorphic with a proper quotient group.
All finite groups are hopfian as $$|G/H| = |G| \div |H|$$. Also, all simple groups are hopfian as a simple group doesn’t have proper subgroups.
So we need to turn ourselves to infinite groups to uncover non hopfian groups. Continue reading A (not finitely generated) group isomorphic to a proper quotient group
# Converse of Lagrange’s theorem does not hold
Lagrange’s theorem, states that for any finite group $$G$$, the order (number of elements) of every subgroup $$H$$ of $$G$$ divides the order of $$G$$ (denoted by $$\vert G \vert$$).
Lagrange’s theorem raises the converse question as to whether every divisor $$d$$ of the order of a group is the order of some subgroup. According to Cauchy’s theorem this is true when $$d$$ is a prime.
However, this does not hold in general: given a finite group $$G$$ and a divisor $$d$$ of $$\vert G \vert$$, there does not necessarily exist a subgroup of $$G$$ with order $$d$$. The alternating group $$G = A_4$$, which has $$12$$ elements has no subgroup of order $$6$$. We prove it below. Continue reading Converse of Lagrange’s theorem does not hold
# A field that can be ordered in two distinct ways
For a short reminder about ordered fields you can have a look to following post. We prove there that $$\mathbb{Q}$$ can be ordered in only one way.
That is also the case of $$\mathbb{R}$$ as $$\mathbb{R}$$ is a real-closed field. And one can prove that the only possible positive cone of a real-closed field is the subset of squares.
However $$\mathbb{Q}(\sqrt{2})$$ is a subfield of $$\mathbb{R}$$ that can be ordered in two distinct ways. Continue reading A field that can be ordered in two distinct ways
# An infinite field that cannot be ordered
## Introduction to ordered fields
Let $$K$$ be a field. An ordering of $$K$$ is a subset $$P$$ of $$K$$ having the following properties:
ORD 1
Given $$x \in K$$, we have either $$x \in P$$, or $$x=0$$, or $$-x \in P$$, and these three possibilities are mutually exclusive. In other words, $$K$$ is the disjoint union of $$P$$, $$\{0\}$$, and $$-P$$.
ORD 2
If $$x, y \in P$$, then $$x+y$$ and $$xy \in P$$.
We shall also say that $$K$$ is ordered by $$P$$, and we call $$P$$ the set of positive elements. Continue reading An infinite field that cannot be ordered
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2019-01-16 20:43:21
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http://math.stackexchange.com/questions/69921/convolution-of-an-l-p-mathbbt-function-f-with-a-term-of-a-summability
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# Convolution of an $L_{p}(\mathbb{T})$ function $f$ with a term of a summability kernel $\{\phi_n\}$
... is the result in $L_{p}$?
A remark in my notes says yes but I can't see how to verify it.
As was pointed out to me in a previous question I asked last night, I need to show that the following integral is finite:
$$\int_{-\pi}^{\pi}|\int_{-\pi}^{\pi}f(t-s)\phi_{n}(s)ds|^{p}dt < \infty$$.
One of the properties of a summability kernel is that there exists a $C > 0$ such that $\int_{-\pi}^{\pi}|\phi_{n}(t)|dt\leq C$ for every $n\geq 1$. I feel like this could help if I could get $\phi$ by itself
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Yes, apply Young's inequality with $r=p$ and $q = 1$, see also: en.wikipedia.org/wiki/Convolution#Integrable_functions – t.b. Oct 5 '11 at 0:21
Thank you this is very helpful. – roo Oct 5 '11 at 0:28
The way I usually prove Young's Inequality is using a couple of applications of Hölder's inequality to prove $$\left|\int f(x)\;g(x)\;h(x)\;\mathrm{d}x\right|\le\|f\|_u\|g\|_v\|h\|_w\tag{1}$$ where $\frac1u+\frac1v+\frac1w=1$. Then apply $(1)$ in a tricky way to show $$\left|\int\int f(x-y)\;g(y)\;h(x)\;\mathrm{d}y\;\mathrm{d}x\right|\le\|f\|_p\|g\|_q\|h\|_r\tag{2}$$ where $\frac1p+\frac1q+\frac1r=2$. Taking the supremum of inequality $(2)$ over all $h\in L^r$ such that $\|h\|_r=1$ says that $\|f\ast g\|_s\le\|f\|_p\|g\|_q$ where $\frac{1}{r}+\frac{1}{s}=1$, that is $\frac{1}{s}=\frac{1}{p}+\frac{1}{q}-1$.
Tricky Application of $\mathbf{(1)}$: Since $\frac{1}{p}+\frac{1}{q}+\frac{1}{r}=2$, we have the following 7 relations: \begin{align} \left(1-\frac{1}{r}\right) + \left(1-\frac{1}{p}\right) + \left(1-\frac{1}{q}\right) &= 1\\ \left(1-\frac{1}{r}\right) + \left(1-\frac{1}{q}\right) &= \frac{1}{p}\text{ so that }p\left(1-\frac{1}{r}\right) + p\left(1-\frac{1}{q}\right) = 1\\ \left(1-\frac{1}{r}\right) + \left(1-\frac{1}{p}\right) &= \frac{1}{q}\text{ so that }q\left(1-\frac{1}{r}\right) + q\left(1-\frac{1}{p}\right) = 1\\ \left(1-\frac{1}{p}\right) + \left(1-\frac{1}{q}\right) &= \frac{1}{r}\text{ so that }r\left(1-\frac{1}{p}\right) + r\left(1-\frac{1}{q}\right) = 1 \end{align} Therefore, \begin{align} &\left|\int\int f(x-y)\;g(y)\;h(x)\;\mathrm{d}y\;\mathrm{d}x\right|\\ &\le\int\int|f(x-y)|\;|g(y)|\;|h(x)|\;\mathrm{d}y\;\mathrm{d}x\\ &={\small\int\int\underbrace{|f(x-y)|^{p(1-1/r)}|g(y)|^{q(1-1/r)}}_{\large\text{in }L^w\text{ where }\frac1r+\frac1w=1}\;\underbrace{|g(y)|^{q(1-1/p)}|h(x)|^{r(1-1/p)}}_{\large\text{in }L^u\text{ where }\frac1p+\frac1u=1}\;\underbrace{|f(x-y)|^{p(1-1/q)}|h(x)|^{r(1-1/q)}}_{\large\text{in }L^v\text{ where }\frac1q+\frac1v=1}\;\mathrm{d}y\;\mathrm{d}x}\\ &\le\left(\int\int|f(x-y)|^p|g(y)|^q\;\mathrm{d}y\;\mathrm{d}x\right)^{1-1/r}\\ &\times\left(\int\int|g(y)|^q|h(x)|^r\;\mathrm{d}y\;\mathrm{d}x\right)^{1-1/p}\\ &\times\left(\int\int|f(x-y)|^p|h(x)|^r\;\mathrm{d}y\;\mathrm{d}x\right)^{1-1/q}\\ &=\left(\int|f(x)|^p\;\mathrm{d}x\right)^{1/p}\left(\int|g(x)|^q\;\mathrm{d}x\right)^{1/q}\left(\int|h(x)|^r\;\mathrm{d}x\right)^{1/r} \end{align}
@Ale: Use Fubini to see that \begin{align}\int\int|f(x-y)|^p|g(y)|^q\,\mathrm{d}y\,\mathrm{d}x &=\|f\|_p^p\int|g(y)|^q\,\mathrm{d}y\\ &=\|f\|_p^p\|g\|_q^q\end{align} and similarly for the other integrals. – robjohn Jan 20 '14 at 14:44
The inequality (1) is false; if it were true, then for functions $f,g,h$ which are non zero and $fgh$ is non zero, applying the inequality to $f(\lambda x), g(\lambda x),h(\lambda x)$ for $\lambda \in \mathbb{R}$, you get $\lambda^{-n}\|fgh\|_1 \leq \lambda^{-n(1/p + 1/q + 1/r)}\|f\|_p\|g\|_q\|h\|_r$, which can't be true for every $\lambda$ unless $1/p + 1/q + 1/r = 1$. – Matt Rigby Sep 25 '14 at 15:36
@MattRigby: you are correct. A simple way to see this is to use $h(x)=1$ ($r=\infty$) and note that it gives the wrong version of Hölder. This argument works, but somewhere I copied something wrong. I will fix this. – robjohn Sep 25 '14 at 19:03
@MattRigby: there are two triples of exponents. In the Hölder case, $\frac1u+\frac1v+\frac1w=1$. In the Young case, $\frac1p+\frac1q+\frac1r=2$. I have separated the exponents to keep things less confusing. Thanks for catching that. – robjohn Sep 25 '14 at 19:57
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2015-11-25 17:11:14
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https://socratic.org/questions/what-is-the-domain-and-range-of-r-x-3sqrt-x-4-3
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What is the domain and range of r(x)= -3sqrt(x-4) +3?
Aug 21, 2015
Domain: $\left[4 , + \infty\right)$
Range: $\left(- \infty , 3\right]$
Explanation:
Your function is defined for any value of $x$ that will not make the expression under the square root negative.
In other words, you need to have
$x - 4 \ge 0 \implies x \ge 4$
The domain of the function will thus be $\left[4 , + \infty\right)$.
The expression under the square root will have a minimum value at $x = 4$, which corresponds to maximum value of the function
$r = - 3 \cdot \sqrt{4 - 4} + 3$
$r = - 3 \cdot 0 + 3$
$r = 3$
For any value of $x > 4$, you have $x - 4 > 0$ and
$r = {\underbrace{- 3 \cdot \sqrt{x - 4}}}_{\textcolor{b l u e}{< - 3}} + 3 \implies r < 3$
The range of the function will thus be $\left(- \infty , 3\right]$.
graph{-3 * sqrt(x-4) + 3 [-10, 10, -5, 5]}
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2019-11-18 15:57:43
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https://meta.stackexchange.com/questions/30559/latex-on-stack-overflow/284150
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# LaTeX on Stack Overflow? [closed]
MathOverflow has an awesome engine where you can embed LaTeX in questions, answers, and comments. Can we get something like this going on Stack Overflow?
I think it’d be appropriate as I at least pretty often want to write something like n^2 and would benefit greatly from prettier markup.
## closed as off-topic by Glorfindel, Nathan Tuggy, Robert Longson, rene, Sonic the Anonymous HedgehogOct 15 '18 at 11:09
This question appears to be off-topic. The users who voted to close gave this specific reason:
• "This question's topic is only applicable to one specific site in the Stack Exchange Network. Questions on Meta Stack Exchange should relate to features or policies that commonly apply to the network or the software that drives it, within the guidelines defined in the help center. You should ask this question on the meta site where your concern originated." – Glorfindel, Nathan Tuggy, Robert Longson, rene, Sonic the Anonymous Hedgehog
If this question can be reworded to fit the rules in the help center, please edit the question.
• +1. Though in a "programming", instead of math, environment, it seems this issue comes up less often and programmers are used to dealing with it when it does. Plus complex formulas aren't needed nearly as often. (How often do you think you'd be misunderstood on SO writing n^2 and n**2?) – Gnome Nov 22 '09 at 2:20
• That's true, but just because we're used to something bad doesn't mean we shouldn't attempt to improve it =). – Claudiu Nov 22 '09 at 2:41
• I like the idea in principal. However, it should be as intuitive as possible. I've looked at LaTeX for typesetting and decided to stick with QuarkXPress as I found it more intuitive. (shudder) – Jason D Nov 29 '09 at 23:59
• I think this is a good idea. It would be very helpful for the more computer sciency sort of questions. I think LaTeX makes the most sense; it's standard in academia. – user138665 Dec 17 '09 at 1:42
• svgkit.sourceforge.net/tests/latex_tests.html or similar might help. Think "preview bit" below answer. – Aiden Bell Jan 14 '10 at 16:11
• Maybe this is just my personal problem, but I cannot read set theory notation or Sigma notation, even though I easily understand the related concepts they embody. Those would probably be the first things commonly used on the site. As much as I want arithmetic to have the proper symbols, I would not want the rest that comes with it. – Merlyn Morgan-Graham Nov 19 '11 at 1:09
• I would argue that programmers see a lot of complex math, it just depends on what you are programming. I do research in intelligent controls, for instance, and most of my programs are heavy on both the programming stuff and the math. LaTeX in SO would be extremely helpful. – Engineero Jun 9 '13 at 17:12
• @minitech Do you have an example for when you would use this on SO? – Undo Sep 10 '13 at 17:28
• @Undo: Not on hand. Anything regarding algorithms. This advice is an important part, by the way. – Ry- Sep 10 '13 at 17:29
• @minitech If you need math formatting, there's a good chance that your question is off-topic and you should ask it on Computer Science instead. There are plenty of exceptions, of course, but I've found it to be a good rule of thumb for deciding between Stack Overflow and Computer Science. (You could even add a third wheel: code → Stack Overflow, math → Computer Science, neither → Software Engineering, but I don't know how accurate that one is.) – Gilles Sep 10 '13 at 17:50
• @Gilles: Sure. If it’s not, however… I just don’t see the harm in adding something that already works elsewhere for the specific exceptional cases. – Ry- Sep 10 '13 at 17:51
• The FAQ link for latex now redirects to the tour page. – Steven M. Vascellaro Jul 18 '17 at 13:24
• A good suggestion that should never have been closed, and if you could just edit it to say "SO and selected other SE network sites", then it should be reopened ASAP. – smci Dec 10 '18 at 3:15
This is implemented on http://math.stackexchange.com -- you can check it out there. It will never be on Stack Overflow, though, as it is an extremely heavy dependency. (See also Nick's investigations about impact in November 2013.)
Info here: TeX math markup is sorely needed
• Well, you could conditionally enable it accross stackexchange, enabling consistent markup (question migration). By that I mean that you check the question and all answers if there is LaTeX block and if so, you load this "extremely heavy dependency". There are questions that would hugely benefit from that. – Rok Kralj Jun 26 '12 at 17:07
• As it works just fine on math.stackexchange.com how is it too heavy? – donroby Aug 12 '12 at 16:54
• Also on physics SE. – Calmarius Aug 16 '13 at 14:52
• (@donroby, I know more than a year passed, but if you still wonder about impact and missed it: see Nick's investigations. Just the messenger!) – Arjan Dec 12 '13 at 20:22
• -1 it should be enabled on any site that might use math. See stackoverflow.com/a/15966238/125507 for example – endolith Feb 25 '14 at 19:44
• If Wikipedia guys (#5th website on Internet and non-commercial) had same point of view, I suppose we would never had math in Wikipedia as well. "It would never be on StackOverflow" is too strong a statement. Sure it can be enabled purely on client side and limit to only those posts. This is 2014 and StackOverflow was supposed to be THE website. Programming is mathematics. Knuth would have been disgusted not seeing expression O(n) rendered via LaTeX. – Shital Shah Dec 27 '14 at 7:11
• How about moving the making of the decision to render LaTeX to the client side, so that it's the browser that will trigger the rendering? Would this negatively affect page load times for pages that don't use LaTeX? – Erik Allik Mar 9 '15 at 12:13
• Any change in this for the likes of jqMath or KaTeX? Both of which seem to be much faster than MathJax – Toby Mar 23 '17 at 13:16
• Funny that Puzzling, Physics, Chemistry, Crypto, and TONS of other network sites have LATEX, but it will "never be on Stack Overflow". Is is because they have less traffic? Or do they just have terrible load times and noone cares because they are less important? – NH. Oct 19 '17 at 23:13
As a workaround, you can easily embed LaTeX by generating an image of the equation using the following WYSIWYG editor: https://codecogs.com/latex/eqneditor.php.
• Holy crap, this is useful. – Makoto Jan 20 '13 at 5:06
• I know it's too late to ask, but is there an alternate to codecogs? This website has been giving a 503 - Service Not Available since past few days. – Ranveer Feb 15 '14 at 21:40
• @Ranveer Daum Equation Editor – Anant Feb 20 '14 at 17:17
• Images are discouraged across all SE, because they are impossible to search. – vonbrand Apr 26 '14 at 15:16
• This is the only solution other than adding Math.SE style LaTeX support that's adequate to the problem, which is spelled out in @datenwolf's answer. Most of the comments and answers don't realize what sort of mathematics can be needed. With this solution, however, not having LaTeX support in SO is acceptable. Otherwise the lack of LaTeX support would limit SO to problems that don't require comparison of code with halfway complex mathematical specifications. Such problems are not common on SO, but if they couldn't be adequately be handled in SO, that would be sad. – Mars Oct 28 '14 at 5:00
• instead of pasting image, you can just use the api url as the image, see my answer on stackoverflow – Mark Mikofski Sep 24 '15 at 18:25
• Could we get something like LaTeX Equation Editor hosted by StackExchange itself? I don't like to link content of another site. [It would even give them a (strongly limited) ability to track users on this site.] – JojOatXGME Mar 6 '16 at 4:24
• Not only they are impossible to search but they’re also not accessible to e.g. screen reader users. – bfontaine Nov 16 '16 at 15:01
• Images are not selectable by parts, so not a good workaround – HackerBoss Oct 14 '18 at 22:44
Having LaTeX support on Stack Overflow would be great. After all SO is about programming, which covers algorithms. And some algorithms are easier explained if one can typeset math. For example whenever it comes to complicated transformation issues regarding 3D graphics, being able to typeset those matrices would be a huge benefit.
• For 3D stuff, you should be at GameDev.SE anyways. LaTeX will take up CPU (on your computer and on SO's servers), and I believe it'll be only needed in 1/1000 questions or answers. And most of those questions could probably go over to Math.SE, or GameDev.SE. – Mateen Ulhaq Apr 16 '11 at 4:31
• If you're really desperate, you can render the image, and then simply include the image in your question/answer. – Mateen Ulhaq Apr 16 '11 at 4:33
• @muntoo: 3D isn't limited to gaming – I admit, that my 3D experience mostly results from developing a 3D engine. However it's as important in programming engineering tools (CAD/CAM), scientific visualization, modern user interfaces, etc. – datenwolf Apr 17 '11 at 8:50
• @datenwolf I think physics simulation fits in to GameDev.SE, even if it's for a NASA rocket, and not just a game with a NASA rocket. GameDev.SE probably has a higher ratio of knowing "3D stuff" to number of [active] users than any other SE site. You don't have to tell them that you're going to be using it for a "game" or not. – Mateen Ulhaq Apr 17 '11 at 18:14
• For stuff that doesn't fit into GameDev.SE, you could go to CSTheory.SE or Math.SE. (Or any of the other LaTeX enabled sites.) – Mateen Ulhaq Apr 17 '11 at 18:16
• @muntoo, we don't get to choose where the questions go, and often the subject matter belongs on SO. – Adam Jun 14 '11 at 7:14
• @muntoo if it is used in 1/1000 of the questions, it will affect SO's servers only in 1/1000 of the time... – vonbrand Apr 26 '14 at 15:19
• I think you're wrong about SO being for algorithms. SO is for calls to library functions which implement algorithms. Questions and answers about algorithms themselves are normally downvoted or answered with library function calls. – sh1 Jul 31 '16 at 18:28
• @sh1: To properly use a library one must understand the underlying principles that are to be achieved with doing these calls. Take computer graphics programming for example. A recurring question is, why normal vectors muse be transformed using a different set of operations (i.e. library calls) than regular vertex positions. Another recurring question is, why a separate tangent space transformation is required for normal map bump mapping. Just giving the recipe is only half the answer. It's also essential to understand "why". And the most concise way to explain the "why" is through math. – datenwolf Aug 2 '16 at 14:28
• @sh1: Some of the mathematical notation, for the most part linear algebra can be typeset using unicode symbols. But already something as mundane as superscripts or subscripts is impossible with just markdown and unicode. Sometimes you want to typeset a partial differential equation, that can be found in a book or a paper about a particular graphics rendering technique and want to explain how to translate that into library calls. – datenwolf Aug 2 '16 at 14:31
• @sh1: And then there's of course the broader issue that, thanks to the Curry–Howard correspondence writing a program is technically the same thing as writing a mathematical proof. So stating that programming and asking questions on StackOverflow is about how to call library functions is highly ignorant. And given that some languages (like Coq which allows you to write mathematical proofs and translate it into programming language source code), yes math typesetting is absolutely required. – datenwolf Aug 2 '16 at 14:35
• Well in my experience, when I'm idle and take the time to look at new questions as they're asked; if anything interesting and algorithmic does come up or if it requires any thought at all then it's downvoted and eventually closed or deleted. SO only accepts questions that help game the system for reputation, and that's a matter of familiarity with libraries where you can spit out an answer quickly for the reputation avalanche of being first responder. – sh1 Aug 2 '16 at 15:34
• @sh1: Two of my most highly voted answers are quite heavy on the math: stackoverflow.com/a/6661242/524368 and stackoverflow.com/a/5257471/524368 – both of them are lacking in the quality of the mathematical typesetting. In case of the "how to calculate normals" answer I was able to make use of Unicode to somehow typeset it. The answer on how to derive a tangent space mapping tries to use ASCII art to typeset matrices and fractions, but it looks ugly (in fact it's that later answer, which lack of proper math typesetting which made me ask about adding support for TeX/LaTeX at SO). – datenwolf Aug 2 '16 at 16:24
• @sh1 Easy counter-example to your claim: there is a bug on Spark MLlib implementation of an algorithm that results in an error. I want to show the algorithm it were intending to implement: without any details about the why of it. That is squarely an SO question - but at the present time the algorithm / equation itself would have to be an unsearchable image. – javadba Jan 6 '17 at 17:11
Although I'm a big LaTeX fan, I don't think the work would be justified for Stack Overflow.
Since Stack Overflow is one of the sites on the web that gets Unicode right, you can do standard things like
K ⊆ A
or
O(n²)
anyway. And <sub> and <sup> are whitelisted, so you can even do more complicated things like
k1,…,n
or
eɣ2n-1
• that seems way more complicated doing O(n<sup>2</sup>) than O(n^2) for the same thing... – tzenes Feb 23 '10 at 5:09
• @tzenes: Pick your examples right. O(n^2) is still more than O(n²). But again, I love Latex, and I wouldn't have a problem with it being implemented on SO. I just feel that the work necessary to do that isn't justified by the few times this is actually needed. – balpha Feb 23 '10 at 8:04
• Well I came onto this subject when writing this answer: stackoverflow.com/questions/2315987/… It took me a long time to get all the tags right, especially considering how much easier latex would have been... – tzenes Mar 11 '10 at 4:20
• Expressions featuring summation, integrals, products, roots ... would be nice examples. E.g. ∑<sub>i=0</sub><sup>n</sup> just doesn't look right. I don't know wether it justifies the work for the few questions that need it though. – Georg Fritzsche Aug 4 '10 at 18:52
• About justifying "the work" that'd need to be done - since mathoverflow has it working perfectly, it seems like it would be a simple matter – Claudiu Mar 6 '11 at 17:52
• @Claudiu: Yes, it's done client-side via JavaScript. This requires every single user visiting the site to download gigabytes of JavaScript (okay, slightly exaggerating) for something that's interesting for at most a tiny amount of posts. – balpha Mar 6 '11 at 17:57
• While this is a fair suggestion, I'd certainly rather have people download excess code than suggest that people learn to use Unicode for formatting. Wasting a few bytes for some users isn't quite the same as wasting brain cells. Nonetheless, mentioning Unicode points to gradual evolution in browsers, and perhaps a future version of HTML or browsers will make this question irrelevant. – Iterator Nov 18 '11 at 21:53
• @balpha can you provide a link on how to use <sub> etc and also how do you generate the unicode characters?Through a char-map and then copy and paste or is there some nice shortcut? – Dan Jan 14 '14 at 10:51
• @Iterator, yep. It's called MathJax ;-) – vonbrand Apr 26 '14 at 15:17
• There isn't any "work" involved. They already have it working on every other stack exchange site – endolith Sep 11 '18 at 15:31
• @endolith Note that this answer is from 2010. – balpha Sep 11 '18 at 15:47
It would be fantastic to have this. There are times when a statistical programming question arises where either Stack Overflow or Cross Validated would be an appropriate site (maybe Stack Overflow will have the edge), but I would rather see it on Cross Validated simply, because one can have LaTeX formatting.
The use case is pretty simple: Joe comes along with a question about how to do X in R (or MATLAB, Python, whatever), and there's an issue with his math and his implementation. For example: the two don't match. Another scenario is that of this question, where we could Q&A on the mathematical inverse, and a good way to code the transformation. It is irritating to address the mathematics within an environment where the formatting support is weak. Both sites (Stack Overflow and Cross Validated) have adequate support for presenting code, so the winner is the site that supports the formatting of mathematics. The same is true for $\textrm{computational math} \setminus \textrm{statistics}$ and the Stack Exchange site Mathematics. ;-)
To that end, I hope that moderators will accept some flags that suggest that a question is better for Mathematics or Cross Validated based on topic, audience, etc. and the ease of answering the question with the available formatting tools.
For simple math, you may use this user script. It adds a math bar to the editor for some commonly used symbols:
While I think that most questions asked on SO where it would be useful to use math formatting are probably off-topic and more appropriate for CS Stack Exchange, it would still be much easier in many cases just to answer the question on SO than to try and migrate it. I don't see how having the math formatting on SO would conflict with anything else. Some people simply don't know about CS Stack Exchange, and so it is easier to ask a question on SO.
I know I have answered questions on SO where the math notation would have been useful. There are instances where you want to talk about algorithm complexity while working with code in a particular language.
As another note, why the heck can't I get code coloring on CS Stack Exchange? Since CS Stack Exchange is smallish and there is a decent amount of overlap between CS Stack Exchange and SO, I really think we ought to just merge them and support math typesetting in SO.
Basically, BOO to the people that declined to implement this. It shouldn't be hard, since it is already in place on other sites, and it wouldn't really conflict with anything. If I knew more about how the sites worked, I would do it myself.
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2019-09-15 20:33:08
|
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|
http://purelearner.com/matrices/how-to-multiply-matrices-3x3-2x2-with-examples/
|
# how to multiply matrices 3×3 & 2×2 with examples
We have already defined various types of matrices with some examples. In this article, we will show you how to multiply matrices 3×3 & 2×2 of the same order along with examples.
Two matrices can only be multiplied when the number of rows in the first matrix is equal to the number of columns in the second matrix. If the number of rows in the first matrix is not equal to the number of columns in the second matrix then the two matrices cannot be multiplied. You can also go through another topic “matrix multiplication properties” on our blog. Let us first take an example of (2×2) matrix:
$Let\:one\:matrix\:be\:[A]=\begin{bmatrix} a&b\\ c&d\\ \end{bmatrix}$
$and\:another\:matrix\:be\:[B]=\begin{bmatrix} p&q\\ r&s\\ \end{bmatrix}$
###### The matrix multiplication of the matrix [A] & matrix [B] will be done as follows;
• The first element in row one of the new matrix formed will be = (a × p) + (b × r).
• The second element in the first row formed of the new matrix formed will be = (a × q) + (b × s).
• Similarly, the first element in the second row of the new matrix formed will be = (c × p) + (d × r).
• And the second element in the row second of the new matrix formed will be = (c × q)+ (d × s).
Therefore the new matrix formed will be;
##### how to multiply matrices 3×3 properly? Let the two (3×3) matrices be;
$[C]=\begin{bmatrix} 1&3&2\\ 2&4&1\\ 2&3&5\\ \end{bmatrix}$
$and\:[D]=\begin{bmatrix} 4&5&2\\ 3&4&2\\ 2&6&1\\ \end{bmatrix}$
##### The matrix multiplication of matrix [C] and matrix [D] will be carried as follows:
• First element in row one of the new matrix formed will be = (1 × 4) + (3 × 3) + (2 × 2) = 17.
• Second element in the first row formed of the new matrix formed will be = (1 × 5) + (3 × 4) + (2 × 6) = 29.
• The third element in row one of the new matrix formed will be = (1 × 2) + (3 × 2) + (2 × 1) = 10.
• First element in the second row of the new matrix formed will be = (2 × 4) + (4 × 3) + (1 × 2) = 22.
• Second element in the second row of the new matrix formed will be = (2 × 5) + (4 × 4) + (1 × 6). = 32.
• Third element in the second row of the new matrix formed will be = (2 × 2) + (4 × 2) + (1 × 1) = 13.
• First element in row third of the new matrix formed will be = (2 × 4) + (3 × 3) + (5 × 2) = 27.
• Second element in row third of the new matrix formed will be = (2 × 5) + (3 × 4) + (5 × 6) = 52.
• And the third element in row third of the new matrix formed will be = (2 × 2) + (3 × 2) + (5 × 1) = 15.
##### Thus the new matrix formed by multiplying matrix [C] & [D] will be as shown below:
$[C]⋅[D]=\begin{bmatrix} 17&29&10\\ 22&32&13\\ 27&52&15\\ \end{bmatrix}$
This is how to multiply matrices 3×3 and 2×2. You can also calculate matrix multiplication online. If you find any difficulty in understanding this article you can directly comment below or you can also contact us in the contact section.
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2018-09-25 08:06:26
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|
https://gamedev.stackexchange.com/questions/187165/settrigger-only-works-in-start-and-not-ontriggerenter
|
SetTrigger Only Works In Start and Not OnTriggerEnter
I'm having an issue with my animator for an enemy, where a bite animation should trigger when the enemy collides with the player. Weirdly enough, the sound effect for the bite activates, but the trigger doesn't change.
It only works in start for when I set the initial trigger, but I can do that in the animator tab anyway, so it's unnecessary.
There is only one major hitch I can think of, which is that the player's collider isn't set as a trigger, but for some reason the game detects the collision anyway when I put a debug.log in to see if it's working.
Nothing is popping up as an error in the console.
Here is the code. Ignore anything that's commented out, as it's not commented out in my version. It's just for the sake of posterity.
void OnTriggerEnter(Collider other)
{
//if (other.gameObject.CompareTag("Melee"))
//{
// PhysicalAttack physicalAttack = other.GetComponent<PhysicalAttack>();
// if (physicalAttack.attack == true)
// {
// physicalAttack.attack = false;
// //Debug.Log(physicalAttack.attack);
// Damage(physicalAttack.damage);
// transform.Translate(0, 0, -5);
// }
//}
else if (other.gameObject.CompareTag("Player") && canAttack)
{
Debug.Log("collision");
Animator anim = gameObject.GetComponent<Animator>();
//fishHealth fish = other.gameObject.GetComponent<fishHealth>();
anim.ResetTrigger("Not Attacking");
anim.SetTrigger("Attacking");
//fish.currentPlayerHealth -= 10;
GetComponent<AudioSource>().clip = Bite;
GetComponent<AudioSource>().Play();
//StartCoroutine(DamageDelay());
}
Here are some screenshots of info from the animator.
New contributor
Kit K is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
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2020-11-25 17:01:40
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https://ictp.acad.ro/tag/1986/
|
## A new coordinate system to treat orography in primitive equations
AbstractAuthorsC. Vamoș -Tiberiu Popoviciu Institute of Numerical Analysis, Romanian Academy I. Draghici KeywordsCite this paper as:C. Vamoş and I. Drăghici, A new coordinate…
## Verification method for the ordinal predictand forecasts
AbstractAuthorsC. Vamoș -Tiberiu Popoviciu Institute of Numerical Analysis, Romanian Academy D. Banciu KeywordsCite this paper as:C. Vamoş and D. Banciu, Verification method for…
## Marangoni flow -induced by temperature gradients- against gravity forces
AbstractIn the present work a comparison in made between the authors experimental and numerically calculated data obrained in studying the…
## Electrical Conductivity of Vitreous 75%V2O5−25%(As2O3·B2O3)
AbstractIn this note we present a study of semiconducting properties of vitreous 75%V2O5-25% which contains two glass former oxides namely…
## On the error estimation in the numerical convergence of certain iterative methods
Abstract We study the nonlinear equations of the form $x=\lambda D\left( x\right) +y,$ where $$\lambda \in \mathbb{R}$$ and $$y\in E$$…
## The convergence of certain iterative methods for solving certain operator equations
Abstract We consider the solving of the equation $x=\lambda D\left( x\right)+y,$ where $$E$$ is a Banach space and $$D:E\rightarrow E$$, \(\lambda\in…
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2020-07-16 15:56:42
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|
https://math.stackexchange.com/questions/882764/finding-lim-x-to0-frac-sinx2-sin2x-with-taylor-series
|
# Finding $\lim_{x\to0} \frac{\sin(x^2)}{\sin^2(x)}$ with Taylor series
Evaluate $$\lim_{x\to0} \frac{\sin(x^2)}{\sin^2(x)}.$$
Using L'Hospital twice, I found this limit to be $1$. However, since the Taylor series expansions of $\sin(x^2)$ and $\sin^2(x)$ tell us that both of these approach $0$ like $x^2$, I'm wondering if we can argue that the limit must be 1 via the Taylor series formal way?
• using $\frac{\sin(x^2)}{\sin^2x}=\frac{\sin(x^2)}{x^2}\frac{x^2}{\sin^2x}$ help? – cand Jul 30 '14 at 15:51
$$\lim_{x\to0} \frac{\sin(x^2)}{\sin^2(x)}=\lim_{x\to0}\left(\frac{x}{\sin x}\right)^2 \frac{\sin(x^2)}{x^2}=1$$
$$\sin{x^2}=x^2+o(x^3)\\ \sin ^2 x=(x+o(x^2))^2=x^2+o(x^3) \implies\\ \frac{\sin x^2}{\sin ^2 x}=\frac{1+o(x)}{1+o(x)}\rightarrow 1 \text{ as } x\rightarrow 0$$
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2019-06-18 15:08:30
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https://socratic.org/questions/why-can-nh-4-form-an-ionic-bond-with-cl
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# Why can NH_4^+ form an ionic bond with Cl^-?
Ions of opposite charge will form an ionic bond. Since $\text{NH"_4^+}$ and $\text{Cl"^(-)}$ have opposite charges, they will combine through ionic bonding to produce the neutral ionic compound $\text{NH"_4"Cl}$, which is ammonium chloride.
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2019-08-17 22:36:45
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https://codereview.stackexchange.com/questions/244127/leetcode-809-expressive-words
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# LeetCode 809: Expressive Words
I'm posting my C++ code for LeetCode's Expressive Words. If you have time and would like to review, please do so. Thank you!
Sometimes people repeat letters to represent extra feeling, such as "hello" -> "heeellooo", "hi" -> "hiiii". In these strings like "heeellooo", we have groups of adjacent letters that are all the same: "h", "eee", "ll", "ooo".
For some given string S, a query word is stretchy if it can be made to be equal to S by any number of applications of the following extension operation: choose a group consisting of characters c, and add some number of characters c to the group so that the size of the group is 3 or more.
For example, starting with "hello", we could do an extension on the group "o" to get "hellooo", but we cannot get "helloo" since the group "oo" has size less than 3. Also, we could do another extension like "ll" -> "lllll" to get "helllllooo". If S = "helllllooo", then the query word "hello" would be stretchy because of these two extension operations: query = "hello" -> "hellooo" -> "helllllooo" = S.
Given a list of query words, return the number of words that are stretchy.
Example: Input: S = "heeellooo" words = ["hello", "hi", "helo"]
Output: 1 Explanation: We can extend "e" and "o" in the word "hello"
to get "heeellooo". We can't extend "helo" to get "heeellooo" because the group "ll" is not size 3 or more.
Notes:
0 <= len(S) <= 100.
0 <= len(words) <= 100.
0 <= len(words[i]) <= 100. S and all words in words consist only of lowercase letters
### Accepted C++
class Solution {
public:
int expressiveWords(std::string base_string, std::vector<string> &words) {
int count = 0;
for (auto &word : words)
if (is_stretchable(base_string, word))
count++;
return count;
}
protected:
bool is_stretchable(std::string base_string, std::string words) { // 4 pointers
int base_length = base_string.size();
int words_length = words.size();
int left_a = 0, right_a = 0;
for (int left_b = 0, right_b = 0; left_a < base_length && right_a < words_length; left_a = left_b, right_a = right_b) {
if (base_string[left_a] != words[right_a])
return false;
while (left_b < base_length && base_string[left_b] == base_string[left_a])
left_b++;
while (right_b < words_length && words[right_b] == words[right_a])
right_b++;
if (left_b - left_a != right_b - right_a && left_b - left_a < std::max(3, right_b - right_a))
return false;
}
return left_a == base_length && right_a == words_length;
}
};
### Reference
On LeetCode, there is an instance usually named Solution with one or more public functions which we are not allowed to rename those.
Pass by const reference to prevent copying and modification:
int expressiveWords(std::string const& base_string, std::vector<string> const& words)
^^^^^^ ^^^^^^
bool is_stretchable(std::string const& base_string, std::string const& words)
^^^^^^ ^^^^^^
for (auto const& word : words)
^^^^^^
This prevents the strings being copied into the function. It will also catch situations were you accidentally try and modify the string.
Please add braces {} around sub blocks.
It will save you one day without you even knowing. Issues caused by missing braces are notoriously hard to find because they are caused when sombody else does something nasty that affects your code (multi line macros come to mind).
One variable per line:
int left_a = 0, right_a = 0;
Its not going to hurt you to add an extra line and it makes the job easier for the next person.
Not sure it is worth it.
But just to be complete you could use a standard algorithm to count.
int count = 0;
for (auto &word : words)
if (is_stretchable(base_string, word))
count++;
return count;
// ---
return std::count(std::begin(words), std::end(words),
[&base_string, this](auto const& word){return is_stretchable(base_string, word);});
• Is it still true that "issues caused by missing braces are notoriously hard to find", now that GCC has -Wmisleading-indentation? – Roland Illig Jun 18 at 20:05
• @RolandIllig: warning: unknown warning option '-Wmisleading-indentation'; did you mean '-Wbinding-in-condition'? – Martin York Jun 18 at 20:31
• Are you saying that pass by reference without the const does copy? – pacmaninbw Jun 20 at 19:01
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2020-09-27 01:27:04
|
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|
https://www.albert.io/ie/sat-math-1-and-2-subject-test/ambiguous-case-scenario
|
Free Version
Easy
# Ambiguous Case Scenario
SATSTM-WQJWVX
Which one of the following triangle scenarios is the Ambiguous Case?
A
$\text{SSS}$
B
$\text{AAA}$
C
$\text{SSA}$
D
$\text{SAS}$
E
$\text{ASA}$
|
2017-01-18 06:07:00
|
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|
https://cracku.in/6-aei-bfj-cgk-x-2012-rrb-ahmedabad
|
### 2012 RRB Ahmedabad Question 6
Instructions
A sequence is given in which one term (or more terms) is missing. From the given options, choose the correct one that can complete the sequence--
Question 6
|
2020-08-09 07:09:54
|
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|
https://gmatclub.com/forum/if-x-and-y-are-integers-and-2-x-y-does-y-148967.html
|
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# If x and y are integers and 2 < x < y, does y = 16 ?
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If x and y are integers and 2 < x < y, does y = 16 ? [#permalink]
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09 Mar 2013, 13:06
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If x and y are integers and 2 < x < y, does y = 16 ?
(1) The GCF of X and Y is 2.
(2) The LCM of X and Y is 48.
[Reveal] Spoiler: OA
VP
Status: Final Lap Up!!!
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Re: if x and y are integers and 2<x<y, does y =16 [#permalink]
### Show Tags
09 Mar 2013, 13:09
I know that each statements are insufficient by itself. But i think Even taken together they are insufficient.
We know that GCF*LCM = X*Y
So X*Y = 96
Hence , when y = 24 , x =4
Similarly when y = 16 , x = 6
So i think both statements are insufficient taken together.
Archit
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Posts: 38798
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Re: if x and y are integers and 2<x<y, does y =16 [#permalink]
### Show Tags
09 Mar 2013, 13:26
Archit143 wrote:
If x and y are integers and 2 < x < y, does y = 16 ?
(1) The GCF of X and Y is 2.
(2) The LCM of X and Y is 48.
I know that each statements are insufficient by itself. But i think Even taken together they are insufficient.
We know that GCF*LCM = X*Y
So X*Y = 96
Hence , when y = 24 , x =4
Similarly when y = 16 , x = 6
So i think both statements are insufficient taken together.
Archit
Notice that the greatest factor of 24 and 4 is 4, not 2 as given in the second statement.
Hope it helps.
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Re: if x and y are integers and 2<x<y, does y =16 [#permalink]
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19 Oct 2013, 20:07
Archit143 wrote:
I know that each statements are insufficient by itself. But i think Even taken together they are insufficient.
We know that GCF*LCM = X*Y
So X*Y = 96
Hence , when y = 24 , x =4
Similarly when y = 16 , x = 6
So i think both statements are insufficient taken together.
Archit
If GCF is 2 and LCM is 48, than the Pair of X and Y can be:
X=2,Y=48 and X=6,Y=16 and X=16, Y=6 and X=48, Y=2
I think the answer is
[Reveal] Spoiler:
E
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Posts: 38798
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Kudos [?]: 105763 [2] , given: 11581
Re: if x and y are integers and 2<x<y, does y =16 [#permalink]
### Show Tags
20 Oct 2013, 04:36
2
KUDOS
Expert's post
suk1234 wrote:
Archit143 wrote:
I know that each statements are insufficient by itself. But i think Even taken together they are insufficient.
We know that GCF*LCM = X*Y
So X*Y = 96
Hence , when y = 24 , x =4
Similarly when y = 16 , x = 6
So i think both statements are insufficient taken together.
Archit
If GCF is 2 and LCM is 48, than the Pair of X and Y can be:
X=2,Y=48 and X=6,Y=16 and X=16, Y=6 and X=48, Y=2
I think the answer is
[Reveal] Spoiler:
E
Please read the stem carefully. It's given that 2<x<y.
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Re: If x and y are integers and 2 < x < y, does y = 16 ? [#permalink]
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05 Nov 2013, 03:15
I am not getting the right answer for some reason. Can you please tell me the set of numbers for which both the statements are satisfied?
thank you.
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Re: If x and y are integers and 2 < x < y, does y = 16 ? [#permalink]
### Show Tags
05 Nov 2013, 07:00
1
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Expert's post
emailmkarthik wrote:
If x and y are integers and 2 < x < y, does y = 16 ?
(1) The GCF of X and Y is 2.
(2) The LCM of X and Y is 48.
I am not getting the right answer for some reason. Can you please tell me the set of numbers for which both the statements are satisfied?
thank you.
x = 6 and y = 16 --> the greatest common factor of 6 and 16 is 2, and the least common multiple of 6 and 16 is 48.
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Re: If x and y are integers and 2 < x < y, does y = 16 ? [#permalink]
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05 Nov 2013, 07:13
Thanks Bunuel. But I am a but unclear still. Is there another set of numbers that satisfy both statements? Else answer should be C, no?
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Re: If x and y are integers and 2 < x < y, does y = 16 ? [#permalink]
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05 Nov 2013, 07:15
emailmkarthik wrote:
Thanks Bunuel. But I am a but unclear still. Is there another set of numbers that satisfy both statements? Else answer should be C, no?
Not sure what you mean...
The correct answer IS C: no other x and y satisfy both the stem and the statements.
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Re: If x and y are integers and 2 < x < y, does y = 16 ? [#permalink]
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05 Nov 2013, 07:18
My goodness. Thank you so much! I thought the answer was E and I was very confused. I saw the answer of some post and assumed it to be OA.
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If x and y are integers and 2 < x < y, does y = 16? [#permalink]
### Show Tags
06 Nov 2013, 10:59
If x and y are integers and 2 < x < y, does y = 16?
(1) The GCF of x and y is 2
(2) The LCM of x and y is 48
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Re: If x and y are integers and 2 < x < y, does y = 16? [#permalink]
### Show Tags
06 Nov 2013, 19:05
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accincognito wrote:
If x and y are integers and 2 < x < y, does y = 16?
(1) The GCF of x and y is 2
(2) The LCM of x and y is 48
Dear accincognito,
I'm happy to help.
This is a hard problem! See this post for some insight:
http://magoosh.com/gmat/2012/gmat-math-factors/
Statement #1: The GCF of x and y is 2
This leave open a wide array of possibilities. All we know is that x and y are two even numbers, both bigger than 2, with no common factors other than two: they could be
x = 4, y = 6
x = 6, y = 8
x = 6, y = 10
x = 6, y = 16
So, it's possible for y to equal 16 or equal something else. This statement, alone and by itself, does not give us sufficient information, so it is insufficient.
Statement #2: The LCM of x and y is 48
Without any other information, we could have
x = 3, y = 16
x = 4, y = 48
So, it's possible for y to equal 16 or equal something else. This statement, alone and by itself, does not give us sufficient information, so it is insufficient.
Combined: this is where it gets interesting.
The GCF of x and y is 2
The LCM of x and y is 48
This is a tricky combination. First, let's list all the factors of 48 --- in order to have a LCM of 48 with another number, each number must be a factor of 48.
factors of 48 = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}
Those are the possible candidates for x & y. We can eliminate 1 & 2, because x > 2, and we can eliminate 3, because that cannot have a GCF of 2 with anything else.
Possibilities for x & y = {4, 6, 8, 12, 16, 24, 48}
If y = 48, then every other number in the set is factor of 48, so the GCF would be the smaller number --- e.g. the GCF of 6 and 48 is 6. Therefore, we can't use 48.
If y = 24, then the first four numbers are factors of 24, so they don't work, and the GCF of 16 & 24 is 8. Therefore, we can't use 24.
Possibilities for x & y = {4, 6, 8, 12, 16}
Suppose y = 16
x = 4, y = 16 ===> GCF = 4, doesn't work
x = 6, y = 16 ===> GCF = 2 --- this is one possible pair!!
x = 8, y = 16 ===> GCF = 8, doesn't work
x = 12, y = 16 ===> GCF = 4, doesn't work
Suppose y = 8
x = 4, y = 8 ===> GCF = 4, doesn't work
x = 6, y = 8 ===> GCF = 2, but LCM = 24, doesn't work
Suppose y = 6
x = 4, y = 6 ===> GCF = 2, but LCM = 12, doesn't work
So, after all that, the only pair that satisfies both statements is x = 6, y = 16, so it turns out, y does in fact equal 16. We are able to give a definitive answer to the prompt question, so the combined statements are sufficient.
Does all this make sense?
Mike
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Mike McGarry
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Re: If x and y are integers and 2 < x < y, does y = 16? [#permalink]
### Show Tags
31 Mar 2014, 08:20
1
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mikemcgarry wrote:
accincognito wrote:
If x and y are integers and 2 < x < y, does y = 16?
(1) The GCF of x and y is 2
(2) The LCM of x and y is 48
Dear accincognito,
I'm happy to help.
This is a hard problem! See this post for some insight:
http://magoosh.com/gmat/2012/gmat-math-factors/
Statement #1: The GCF of x and y is 2
This leave open a wide array of possibilities. All we know is that x and y are two even numbers, both bigger than 2, with no common factors other than two: they could be
x = 4, y = 6
x = 6, y = 8
x = 6, y = 10
x = 6, y = 16
So, it's possible for y to equal 16 or equal something else. This statement, alone and by itself, does not give us sufficient information, so it is insufficient.
Statement #2: The LCM of x and y is 48
Without any other information, we could have
x = 3, y = 16
x = 4, y = 48
So, it's possible for y to equal 16 or equal something else. This statement, alone and by itself, does not give us sufficient information, so it is insufficient.
Combined: this is where it gets interesting.
The GCF of x and y is 2
The LCM of x and y is 48
This is a tricky combination. First, let's list all the factors of 48 --- in order to have a LCM of 48 with another number, each number must be a factor of 48.
factors of 48 = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}
Those are the possible candidates for x & y. We can eliminate 1 & 2, because x > 2, and we can eliminate 3, because that cannot have a GCF of 2 with anything else.
Possibilities for x & y = {4, 6, 8, 12, 16, 24, 48}
If y = 48, then every other number in the set is factor of 48, so the GCF would be the smaller number --- e.g. the GCF of 6 and 48 is 6. Therefore, we can't use 48.
If y = 24, then the first four numbers are factors of 24, so they don't work, and the GCF of 16 & 24 is 8. Therefore, we can't use 24.
Possibilities for x & y = {4, 6, 8, 12, 16}
Suppose y = 16
x = 4, y = 16 ===> GCF = 4, doesn't work
x = 6, y = 16 ===> GCF = 2 --- this is one possible pair!!
x = 8, y = 16 ===> GCF = 8, doesn't work
x = 12, y = 16 ===> GCF = 4, doesn't work
Suppose y = 8
x = 4, y = 8 ===> GCF = 4, doesn't work
x = 6, y = 8 ===> GCF = 2, but LCM = 24, doesn't work
Suppose y = 6
x = 4, y = 6 ===> GCF = 2, but LCM = 12, doesn't work
So, after all that, the only pair that satisfies both statements is x = 6, y = 16, so it turns out, y does in fact equal 16. We are able to give a definitive answer to the prompt question, so the combined statements are sufficient.
Does all this make sense?
Mike
No need to test that many cases, let's see
x,y are positive integers and 2<x<y. Is y=16?
First statement
GCF (x,y) is 2
Well we could have:
x=4, y=6 answer is NO or x=6, y=16 answer is YES
Hence insufficient
Second Statement
LCM (x,y) is 48
48 = 2^4 * 3
Now we can test cases here too: Either x=3 and y=2^4 when answer is YES
OR x=7 y = 48 answer is NO
Both together
Since GCF =2 and LCM = 48 and since 2<x<y we can only have x=6, y=16
Hence C is the correct answer
Cheers
J
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Re: If x and y are integers and 2 < x < y, does y = 16? [#permalink]
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31 Mar 2014, 08:20
1
KUDOS
Again @Mike, No need for all the fuzz, let's see
x,y are positive integers and 2<x<y. Is y=16?
First statement
GCF (x,y) is 2
Well we could have:
x=4, y=6 answer is NO or x=6, y=16 answer is YES
Hence insufficient
Second Statement
LCM (x,y) is 48
48 = 2^4 * 3
Now we can test cases here too: Either x=3 and y=2^4 when answer is YES
OR x=7 y = 48 answer is NO
Both together
Since GCF =2 and LCM = 48 and since 2<x<y we can only have x=6, y=16
Hence C is the correct answer
Cheers
J
Last edited by jlgdr on 29 May 2014, 07:18, edited 1 time in total.
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Re: If x and y are integers and 2 < x < y, does y = 16 ? [#permalink]
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01 Apr 2014, 20:57
2
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Expert's post
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Archit143 wrote:
If x and y are integers and 2 < x < y, does y = 16 ?
(1) The GCF of X and Y is 2.
(2) The LCM of X and Y is 48.
Given: 2 < X < Y
Question: Does Y = 16?
(1) The GCF of X and Y is 2.
X = 2a; Y = 2b (a and b are co prime integers)
Y may or may not be 16 e.g. X = 6, Y = 16 OR X = 6, Y = 8 etc
(2) The LCM of X and Y is 48.
$$48 = 2^4 * 3$$
One of X and Y must have 2^4 = 16 as a factor and one must have 3 as a factor. Again, Y may or may not be 16 e.g. X = 6, Y = 16 OR X = 1, Y = 48 etc
Using both together, X = 2a, Y = 2b.
Since a and b need to be co-prime and both X and Y need to be greater than 2, one of a and b must be 3 and the other must be 8 (since X and Y already have a 2 to make 16).
X = 6, Y = 16 (since X is less than Y).
Y must be 16.
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If x and y are integers and 2 < x < y, does y = 16 ? [#permalink]
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10 Jan 2016, 09:24
If x and y are integers and $$2 < x < y$$, does $$y = 16$$ ?
(1) The GCF of X and Y is 2.
since we have many solutions such as $$(6, 8)$$, $$(6,16)$$, etc.
we cannot say whether $$y=16$$ or not.
Statement 1 is insufficient.
(2) The LCM of X and Y is 48.
since we have many solutions such as $$(16, 24)$$, $$(3,16)$$, etc.
we cannot say whether $$y=16$$ or not.
Statement 2 is insufficient.
Combining 1 and 2,
we know that $$LCM * GCF$$= product of the integers $$x * y$$
so $$xy=96=2^5*3$$
since GCF is 2, we know that both are even numbers indicating a single 2 in both x and y.
Thereby we get values of $$(x,y)$$ as $$(6, 16)$$, $$(8, 12)$$ and $$(4, 24).$$
out of these 3 pairs only $$(6,16)$$ has $$LCM$$ of $$48$$.
So $$y=16$$ and thus $$C$$ is the correct answer.
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Re: If x and y are integers and 2 < x < y, does y = 16 ? [#permalink]
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11 Jan 2016, 17:52
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.
If x and y are integers and 2 < x < y, does y = 16 ?
(1) The GCF of X and Y is 2.
(2) The LCM of X and Y is 48.
In the original condition, there are 2 variables(x,y) and 1 equation(2<x<y), which should match with the number of equations. So you need 1 more equation. For 1) 1 equation, for 2) 1equation, which is likely to make D the answer.
For 1), (x,y)=(4,6) -> no, (x,y)=(6,16) -> yes, which is not sufficient.
For 2), (x,y)=(3,48) -> no, (x,y)=(6,16) -> yes, which is not sufficient.
When 1) & 2), (x,y)=(6,16) -> yes, which is sufficient. Therefore, the answer is C.
For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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11 Oct 2016, 05:36
If x and y are integers and 2<x<y, does y=16?
(1) The greatest common factor of x and y is 2.
(2) The lowest common multiple of x and y is 48.
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Re: If x and y are integers and 2 < x < y, does y = 16 ? [#permalink]
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11 Oct 2016, 05:39
Bounce1987 wrote:
If x and y are integers and 2<x<y, does y=16?
(1) The greatest common factor of x and y is 2.
(2) The lowest common multiple of x and y is 48.
Merging topics. Please refer to the discussion above.
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Re: If x and y are integers and 2 < x < y, does y = 16 ? [#permalink]
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12 Oct 2016, 12:40
Archit143 wrote:
If x and y are integers and 2 < x < y, does y = 16 ?
(1) The GCF of X and Y is 2.
(2) The LCM of X and Y is 48.
FROM 1 GCF is 2 ( common prime factor with the lowest power )
(x,y) could be (4, 28) , (6,10) (14,16) ... insuff
from 2
LCM = (2^4 *3) ( factors unique to each integer * common prime factor to the largest power) ... insuff
both
since 2 is the GCF then one one of the 2 numbers has 2^1 and the other 2^4 and we need to know where the 3 factor belongs ( it only belongs to one of the two numbers since GCF is 2) . from stem 2<x<y , thus the smallest of x and y has to be bigger than 2^1 by the 3 factor , thus (x,y) = (6,16)
Re: If x and y are integers and 2 < x < y, does y = 16 ? [#permalink] 12 Oct 2016, 12:40
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# If x and y are integers and 2 < x < y, does y = 16 ?
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Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.
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2017-05-22 21:53:02
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https://www.physicsforums.com/threads/current-density-and-vector-potential.430620/
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Homework Help: Current density and vector potential
1. Sep 20, 2010
oddiseas
1. The problem statement, all variables and given/known data
What current density would produce the vector potential
A(r)=(-kmu/2pi)In(r/a) (in the z direction)
where k is a constant, in cylindrical coordinates?
2. Relevant equations
3. The attempt at a solution
i have done this three times and i get zero current density.So i am wondering if i am making a mistake, there is only one component for A the z component so when i use B=grad*A, i get a field in the fita direction, then when i use J=(grad*B)/mu i get zero.
2. Sep 20, 2010
gabbagabbahey
Hint: $\frac{r}{r}$ is undefined at $r=0$
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2018-07-20 18:49:52
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https://www.winhelponline.com/blog/change-the-default-editor-for-bat-files/?amp=1
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# Change the Default Editor for Batch files in Windows
When you right-click on a Windows batch file (.bat) file and click Edit, Notepad opens the file by default. If you have a Notepad replacement software installed in your system, you can set it as the default editor for Windows batch files.
## Change the Default Editor for Batch Files
1. Launch the Registry Editor (`regedit.exe`)
2. Navigate to the following branch:
`HKEY_CLASSES_ROOT\batfile\shell\edit\command`
3. Double-click the `(default)` value on the right.
4. By default, it points to `notepad.exe`. Replace the existing data with the complete path of your editor.
Example:
`"C:\Program Files (x86)\Notepad++\notepad++.exe" "%1"`
Be sure to include the `"%1"` at the end (with quotes), as mentioned above.
5. Optionally, to change the default editor for .CMD files as well, then make the change in the following key as well:
`HKEY_CLASSES_ROOT\cmdfile\shell\edit\command`
6. Exit the Registry Editor.
When you right-click on a Windows batch file and choose Edit, your chosen editor will now open the batch file for editing.
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2022-11-27 14:35:55
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https://en.wikipedia.org/wiki/Advanced_z-transform
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In mathematics and signal processing, the advanced z-transform is an extension of the z-transform, to incorporate ideal delays that are not multiples of the sampling time. It takes the form
${\displaystyle F(z,m)=\sum _{k=0}^{\infty }f(kT+m)z^{-k}}$
where
• T is the sampling period
• m (the "delay parameter") is a fraction of the sampling period ${\displaystyle [0,T].}$
It is also known as the modified z-transform.
The advanced z-transform is widely applied, for example to accurately model processing delays in digital control.
## Properties
If the delay parameter, m, is considered fixed then all the properties of the z-transform hold for the advanced z-transform.
### Linearity
${\displaystyle {\mathcal {Z}}\left\{\sum _{k=1}^{n}c_{k}f_{k}(t)\right\}=\sum _{k=1}^{n}c_{k}F_{k}(z,m).}$
### Time shift
${\displaystyle {\mathcal {Z}}\left\{u(t-nT)f(t-nT)\right\}=z^{-n}F(z,m).}$
### Damping
${\displaystyle {\mathcal {Z}}\left\{f(t)e^{-a\,t}\right\}=e^{-a\,m}F(e^{a\,T}z,m).}$
### Time multiplication
${\displaystyle {\mathcal {Z}}\left\{t^{y}f(t)\right\}=\left(-Tz{\frac {d}{dz}}+m\right)^{y}F(z,m).}$
### Final value theorem
${\displaystyle \lim _{k\to \infty }f(kT+m)=\lim _{z\to 1}(1-z^{-1})F(z,m).}$
## Example
Consider the following example where ${\displaystyle f(t)=\cos(\omega t)}$:
{\displaystyle {\begin{aligned}F(z,m)&={\mathcal {Z}}\left\{\cos \left(\omega \left(kT+m\right)\right)\right\}\\&={\mathcal {Z}}\left\{\cos(\omega kT)\cos(\omega m)-\sin(\omega kT)\sin(\omega m)\right\}\\&=\cos(\omega m){\mathcal {Z}}\left\{\cos(\omega kT)\right\}-\sin(\omega m){\mathcal {Z}}\left\{\sin(\omega kT)\right\}\\&=\cos(\omega m){\frac {z\left(z-\cos(\omega T)\right)}{z^{2}-2z\cos(\omega T)+1}}-\sin(\omega m){\frac {z\sin(\omega T)}{z^{2}-2z\cos(\omega T)+1}}\\&={\frac {z^{2}\cos(\omega m)-z\cos(\omega (T-m))}{z^{2}-2z\cos(\omega T)+1}}.\end{aligned}}}
If ${\displaystyle m=0}$ then ${\displaystyle F(z,m)}$ reduces to the transform
${\displaystyle F(z,0)={\frac {z^{2}-z\cos(\omega T)}{z^{2}-2z\cos(\omega T)+1}},}$
which is clearly just the z-transform of ${\displaystyle f(t)}$.
## References
• Jury, Eliahu Ibraham (1973). Theory and Application of the z-Transform Method. Krieger. ISBN 0-88275-122-0. OCLC 836240.
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2023-03-28 19:51:56
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https://www.physicsforums.com/threads/conservative-force-in-a-time-dependent-potential.859810/
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Conservative force in a time-dependent potential
1. Feb 29, 2016
Happiness
A conservative force is one that is derivable from a potential energy function $V$. If $V$ is time-dependent, is it still possible to have a conservative force or work done such that the work done is only dependent on the initial state $(x_i, y_i, z_i, t_i)$ and final state $(x_f, y_f, z_f, t_f)$ of the particle and independent of the trajectory taken?
2. Feb 29, 2016
vanhees71
No, because in that case the work-energy theorem turns out to be
$$\frac{m}{2} \vec{v}(t)^2+\text{const}=\int \mathrm{d} t \vec{F}(\vec{x},t) \cdot \vec{v}=-\int \mathrm{d} t \vec{v} \cdot \vec{\nabla} U(\vec{x},t).$$
Now
$$\frac{\mathrm{d}}{\mathrm{d} t} U(\vec{x},t)=\vec{v} \cdot \vec{\nabla} U(\vec{x},t) + \partial_t U(\vec{x},t),$$
i.e., the integrand on the right-hand side of the work-energy theorem is not a total time derivative as in the case, when $U$ is not explicitly time-dependent.
This is a consequence of Noether's theorem: A system is symmetric under time translations if and only if the Lagrangian is not explicitly time dependent modulo a total time derivative of a function $\Omega(q,t)$.
3. Feb 29, 2016
Happiness
How about we treat time as the fourth spatial dimension? Where $\vec{v}$ becomes a four-dimensional vector. Then
$$\frac{\mathrm{d}}{\mathrm{d} t} U(\vec{x},t)=\vec{v} \cdot \vec{\nabla} U(\vec{x},t)$$
So the integrand on the right-hand side of the work-energy theorem becomes a total time derivative.
What's wrong with this?
4. Feb 29, 2016
vanhees71
Well, that comes close to relativistic dynamics in terms of a world parameter. That generalized dynamics has to be developed carefully. For a good introduction (although unfortunately full of typos even in the math) see
Barut, Electrodynamics and classical theory of particles and fields, Dover
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2017-12-18 13:17:07
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https://plainmath.net/9582/adventure-leisure-trolleys-travel-express-travels-trolley-transportation
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Question
# Adventure Tours has 6 leisure-tour trolleys that travel 15 mph slower than their 3 express tour buses. The bus travels 132 mi in the time it takes the trolley to travel 99 mi. Find the speed of each mode of transportation.
Ratios, rates, proportions
Adventure Tours has 6 leisure-tour trolleys that travel 15 mph slower than their 3 express tour buses. The bus travels 132 mi in the time it takes the trolley to travel 99 mi. Find the speed of each mode of transportation.
2021-02-01
For motion problems, use the distance formula: distance=rate$$\times$$time$$\displaystyle\to{d}={r}{t}$$
Let x be the speed of a bus so that x-5 is the speed of the trolley, both in mph.
The time it takes for the bus to travel 132 miles is the same as the time it takes for the trolley to travel 99 miles:
$$tB=tT$$
$$\displaystyle{\left({d}\frac{{B}}{{r}}{B}\right)}={\left({d}\frac{{T}}{{r}}{T}\right)}$$
$$\displaystyle\frac{{132}}{{x}}=\frac{{99}}{{{x}-{15}}}$$
Solve for x. Cross multiply: $$132(x-15)=99x$$
$$132x-1980=99x$$
$$132x=99x+1980$$
$$33x=1980$$
$$x=60$$
So, the speed of the bus is 60 mph and the speed of the trolley is $$60-15=45$$ mph.
2021-08-10
Answer is given below (on video)
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2021-09-26 19:16:21
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|
https://physics.aps.org/articles/v11/s62
|
Synopsis
# Atoms Put On a Bloch Party
Physics 11, s62
Bloch oscillations—first predicted to occur for electrons in a crystal—have been observed in an optical lattice containing ultracold atoms.
When a constant electric field is applied to a crystal, the free electrons may undergo back-and-forth motions called Bloch oscillations. Despite being predicted almost 90 years ago, Bloch oscillations have been extremely challenging to detect in simple crystals. Physicists have managed to observe this phenomenon in other systems, such as semiconductor heterostructures and optical lattices containing ultracold atoms. However, previous optical lattice studies have only been able to infer the oscillations from measurements of the atoms’ momenta. Now, an experiment has directly observed Bloch oscillations in the positions of atoms in an optical lattice.
Bloch oscillations occur when particles traveling within a periodic potential are exposed to a constant force. While the force accelerates the particles in one direction, interference effects due to diffraction from the potential can lead to oscillatory motions. The behavior was initially predicted for electrons in a crystal, but electrons scatter inside a crystal too often for the oscillations to occur.
In an optical lattice, however, scattering is minimal, and the force and potential can be tuned to maximize the oscillations. David Weld from the University of California, Santa Barbara, and colleagues took advantage of this tunability to observe Bloch oscillations in real space, as opposed to momentum space. The team loaded a Bose-Einstein condensate of lithium atoms into an optical lattice with a 0.5- $𝜇$m period. For their constant force, they applied a magnetic field with a spatial gradient. By imaging the positions of the atoms at different times, they revealed oscillations with an amplitude of roughly 100 $𝜇$m—matching their predictions. From their data, the researchers mapped out the band structure of the lattice, demonstrating that Bloch oscillations could be a useful probe for uncovering the band structure in complex systems, such as those with a topological phase.
This research is published in Physical Review Letters.
–Michael Schirber
Michael Schirber is a Corresponding Editor for Physics based in Lyon, France.
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### Twinkling of a Shrinking Droplet Reveals Hidden Complexity
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Researchers have recorded for the first time the dynamics of vibrating Rydberg molecules, the slow-motion counterparts of regular molecules. Read More »
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2023-02-05 01:53:52
|
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|
https://eprint.iacr.org/2003/160
|
### A More Secure and Efficacious TTS Signature Scheme
Jiun-Ming Chen and Bo-Yin Yang
##### Abstract
In 2002 the new genre of digital signature scheme TTS (Tame Transformation Signatures) is introduced along with a sample scheme TTS/2. TTS is from the family of multivariate cryptographic schemes to which the NESSIE primitive {SFLASH} also belongs. It is a realization of Moh's theory for digital signatures, based on Tame Transformations or Tame Maps. Properties of multivariate cryptosystems are determined mainly by their central maps. TTS uses Tame Maps as their central portion for even greater speed than $C^\ast$-related schemes (using monomials in a large field for the central portion), previously usually acknowledged as fastest. We show a small flaw in TTS/2 and present an improved TTS implementation which we call TTS/4. We will examine in some detail how well TTS/4 performs, how it stands up to previously known attacks, and why it represents an advance over TTS/2. Based on this topical assessment, we consider TTS in general and TTS/4 in particular to be competitive or superior in several aspects to other schemes, partly because the theoretical roots of TTS induce many good traits. One specific area in which TTS/4 should excel is in low-cost smartcards. It seems that the genre has great potential for practical deployment and deserves further attention by the cryptological community.
Note: This is a more complete version of the paper that was presented to ICISC for the proceedings (due to appear in LNCS)
Available format(s)
Category
Public-key cryptography
Publication info
Published elsewhere. Condensed version presented at ICISC'03, in proceedings
Keywords
Finite FieldTame TransformationDigital SignatureTTMTTS
Contact author(s)
by @ moscito org
History
2004-01-03: last of 19 revisions
See all versions
Short URL
https://ia.cr/2003/160
CC BY
BibTeX
@misc{cryptoeprint:2003/160,
author = {Jiun-Ming Chen and Bo-Yin Yang},
title = {A More Secure and Efficacious TTS Signature Scheme},
howpublished = {Cryptology ePrint Archive, Paper 2003/160},
year = {2003},
note = {\url{https://eprint.iacr.org/2003/160}},
url = {https://eprint.iacr.org/2003/160}
}
Note: In order to protect the privacy of readers, eprint.iacr.org does not use cookies or embedded third party content.
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2022-09-25 14:10:22
|
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|
http://openstudy.com/updates/50913e83e4b043900ad957fa
|
## anonymous 3 years ago Write the correct equation accounting for the mild acidity of 1 M H4NCl.
1. .Sam.
$NH_4Cl \rightarrow NH_4^+ + Cl^-$ From here, $NH_4^+ \rightarrow NH_3 +H^+$ The hydrogen ion shows mild acidity.
2. .Sam.
You can also say its according to Brønsted-Lowry acids because acids tend to donate protons.
3. anonymous
That's it? I thought there was something I was missing...
4. .Sam.
Yes, for further detail, the hydrogen ions that makes the pH of the solution decrease.
5. anonymous
awesome thank you!
|
2016-08-31 21:47:44
|
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|
https://huggingface.co/datasets/twi_wordsim353
|
# Datasets:twi_wordsim353
Multilinguality: multilingual
Size Categories: n<1K
Language Creators: expert-generated
Annotations Creators: crowdsourced
Source Datasets: original
Dataset Preview
twi1 (string)twi2 (string)similarity (float32)
"etwie"
"ɔkra"
7.35
"etwie"
"etwie"
10
"nwoma"
"krataa"
7.46
"kɔmputa"
"intanɛt"
7.58
"wiemhyɛn"
"ɛhyɛn"
5.77
"ketekye"
"ɛhyɛn"
6.31
"hamakasafo"
"nkitahodie"
7.5
"television"
6.77
"kowaa"
7.42
"panoo"
"television"
6.19
"borɔfoferɛ"
"borɔdweba"
5.92
"yaresafoɔ"
"yarehwɛfoɔ"
7
"ɔbenfoɔ"
"yaresafoɔ"
6.62
"osuani"
"ɔbenfoɔ"
6.81
"ben"
"osuani"
4.62
"ben"
"kwasea"
5.81
"sikatam"
7.08
"sikatam"
"dwa"
8.08
"sikatam"
"hamakasofoɔ"
1.62
"sikatam"
"CD"
1.31
"sikatam"
"ketebɔ"
0.92
"sikatam"
"kosua"
1.81
"sikakorabea"
"sika"
8.12
"dua"
"kwaeɛ"
7.73
"sika"
"sika"
9.15
"ɔbenfoɔ"
"borɔfoferɛ"
0.31
"ɔbenfoɔ"
"ɛfan"
0.23
"ɔbenfoɔ"
"ɔhemaa"
8.58
"ɔbenfoɔ"
"akoa"
5.92
"ɔsɔfopanyin"
"ɔkyerɛkyerɛfoɔ"
6.69
"Yerusalem"
"Israel"
8.46
"Yerusalem"
"Palestinni"
7.65
"bɔɔlo"
8.62
"bɔɔlo"
"bɔɔlo"
9.03
"Arafat"
"asomdwoeɛ"
6.73
"Arafat"
"ehu"
7.65
"Arafat"
"Jackson"
2.5
"mmara"
"mmaranimfoɔ"
8.38
"sini"
"nsoroma"
7.38
"sini"
"nkyeweɛ"
6.19
"sini"
"kyinnyegyefoɔ"
6.73
"sini"
"agohwɛbea"
7.92
"ehunumu"
"kafranyansa"
4.88
"nsa"
"kafranyansa"
5.54
"mmorɔsa"
"mmorɔsa"
8.46
"mmorɔsa"
"mmorɔsa"
8.13
"nom"
"ɛhyɛn"
3.04
"nom"
"aso"
1.31
"nom"
"ano"
5.96
"nom"
"di"
6.87
"maame"
7.85
"nom"
"maame"
2.65
"ɛhyɛn"
"ɛhyɛn"
8.94
8.96
"akwantuo"
"akwantuo"
9.29
"abarimaa"
"aberantewa"
8.83
"mpoano"
"mpoano"
9.1
"abɔdamfoɔfie"
"abɔdamfoɔdan"
8.87
"sumanfoɔ"
"bayifoɔ"
9.02
"owigyinaeɛ"
"prɛmtoberɛ"
9.29
"fonoo"
"bukyia"
8.79
7.52
"anomaa"
"akokɔnini"
7.1
6.46
"nuabarima"
"okokorani"
6.27
"aberantewa"
"nuabarima"
4.46
"akwantuo"
"ɛhyɛn"
5.85
"okokorani"
"kɔmfoɔ"
5
"akokɔnini"
4.42
"mpoano"
"bepɔwa"
4.38
"kwaeɛ"
"amusieeɛ"
1.85
"okokorani"
"akoa"
0.92
"mpoano"
"kwaeɛ"
3.15
"aberantewa"
"bayifoɔ"
0.92
"ahoma"
"nwenwene"
0.54
"ahwehwɛ"
"sumanfoɔ"
2.08
"prɛmtoberɛ"
"ahoma"
0.54
"akokɔnini"
"akwantuo"
0.62
"sika"
"dɔla"
8.42
"sika"
"sika"
9.04
"sika"
"ahonya"
8.27
"sika"
7.57
"sika"
7.29
"sika"
"sikakorabea"
8.5
"sika"
3.31
"etwie"
"ketebɔ"
8
"etwie"
"ɔkra"
8
"etwie"
"aboa"
7
"etwie"
"ateasedeɛ"
4.77
"etwie"
"aboa"
5.62
"etwie"
5.87
"okyinsoroma"
"nsoroma"
8.45
"okyinsoroma"
"ɔsrane"
8.08
"okyinsoroma"
"awia"
8.02
"okyinsoroma"
"ehunumu"
7.92
"kuruwa"
"kafe"
6.58
"kuruwa"
"nkukuo"
6.85
"kuruwa"
2.4
"kuruwa"
2.92
"kuruwa"
3.69
End of preview (truncated to 100 rows)
# Dataset Card for Yorùbá Wordsim-353
### Dataset Summary
A translation of the word pair similarity dataset wordsim-353 to Twi. However, only 274 (out of 353) pairs of words were translated
### Languages
Twi (ISO 639-1: tw)
## Dataset Structure
### Data Instances
An instance consists of a pair of words as well as their similarity. The dataset contains both the original English words (from wordsim-353) as well as their translation to Twi.
### Data Fields
• twi1: the first word of the pair; translation to Twi
• twi2: the second word of the pair; translation to Twi
• similarity: similarity rating according to the English dataset
### Data Splits
Only the test data is available
## Considerations for Using the Data
### Citation Information
@inproceedings{alabi-etal-2020-massive,
title = "Massive vs. Curated Embeddings for Low-Resourced Languages: the Case of {Y}or{\u}b{\'a} and {T}wi",
author = "Alabi, Jesujoba and
Amponsah-Kaakyire, Kwabena and
Espa{\~n}a-Bonet, Cristina",
booktitle = "Proceedings of the 12th Language Resources and Evaluation Conference",
month = may,
year = "2020",
publisher = "European Language Resources Association",
url = "https://www.aclweb.org/anthology/2020.lrec-1.335",
pages = "2754--2762",
abstract = "The success of several architectures to learn semantic representations from unannotated text and the availability of these kind of texts in online multilingual resources such as Wikipedia has facilitated the massive and automatic creation of resources for multiple languages. The evaluation of such resources is usually done for the high-resourced languages, where one has a smorgasbord of tasks and test sets to evaluate on. For low-resourced languages, the evaluation is more difficult and normally ignored, with the hope that the impressive capability of deep learning architectures to learn (multilingual) representations in the high-resourced setting holds in the low-resourced setting too. In this paper we focus on two African languages, Yor{\u}b{\'a} and Twi, and compare the word embeddings obtained in this way, with word embeddings obtained from curated corpora and a language-dependent processing. We analyse the noise in the publicly available corpora, collect high quality and noisy data for the two languages and quantify the improvements that depend not only on the amount of data but on the quality too. We also use different architectures that learn word representations both from surface forms and characters to further exploit all the available information which showed to be important for these languages. For the evaluation, we manually translate the wordsim-353 word pairs dataset from English into Yor{\u}b{\'a} and Twi. We extend the analysis to contextual word embeddings and evaluate multilingual BERT on a named entity recognition task. For this, we annotate with named entities the Global Voices corpus for Yor{\u}b{\'a}. As output of the work, we provide corpora, embeddings and the test suits for both languages.",
language = "English",
ISBN = "979-10-95546-34-4",
}
|
2023-03-27 05:13:24
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4111185073852539, "perplexity": 4787.4500013295665}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296946637.95/warc/CC-MAIN-20230327025922-20230327055922-00698.warc.gz"}
|
https://proofwiki.org/wiki/Definition:Constant_(Category_Theory)
|
# Definition:Constant (Category Theory)
## Definition
Let $\mathbf C$ be a metacategory, and let $1$ be a terminal object of $\mathbf C$.
A constant of $\mathbf C$ is a morphism $f: 1 \to C$ of $\mathbf C$ which has $1$ as its domain.
## Also known as
Among the various other names for this concept are global element (of $C$) and point (of $C$).
Compare variable element.
|
2022-01-24 23:01:09
|
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|
https://stats.stackexchange.com/questions/412414/should-one-drop-independent-variables-if-they-dont-have-linear-relationship-wit
|
# Should one drop independent variables if they don't have linear relationship with the response variable?
I am building a linear regression model using Ridge regression. Some of the independent variables don't have linear relationships with the dependent variable. I've tried to do data transformations on those variables, but there isn't really a clean transformation to make some of them linear. Would it be a bad idea to drop those variables from my model altogether?
Dropping a predictor just because it doesn't show a linear relation with the response when considered alone is usually a bad idea, because that predictor may be useful when used with other predictors.
I try to show it with an example:
n <- 1e4
x <- rnorm(n)
y <- rnorm(n)
z <- x-y
Let's try to predict variable x using variables y and z as predictors. We can see that variables x and y are independent - neither a linear relation nor any other kind of relation. Correlation is just 0:
data <- data.frame(x,y,z)
cor(data)
x y z
x 1.000000000 -0.009880608 0.7116068
y -0.009880608 1.000000000 -0.7095747
z 0.711606819 -0.709574733 1.0000000
But predicting x from y and z is a perfect fit:
library(faraway)
sumary(lm(x~y+z))
Estimate Std. Error t value Pr(>|t|)
(Intercept) 5.4956e-17 5.2156e-17 1.0537e+00 0.2921
y 1.0000e+00 7.3892e-17 1.3533e+16 <2e-16
z 1.0000e+00 5.1917e-17 1.9261e+16 <2e-16
n = 10000, p = 3, Residual SE = 0.00000, R-Squared = 1
However, if we drop predictor y just because it has no linear relationship with response x, we get a much worse fit:
sumary(lm(x~z))
Estimate Std. Error t value Pr(>|t|)
(Intercept) -0.0023683 0.0070591 -0.3355 0.7373
z 0.5014441 0.0049513 101.2750 <2e-16
n = 10000, p = 2, Residual SE = 0.70587, R-Squared = 0.51
In summary, in linear regression we should not drop a predictor just because it doesn't have any linear relationship with the response. If you are interested on how to decide when variables should be included or dropped in a model, I suggest reading about variable selection, which is an interesting topic.
Dropping a variable just because it doesn't have linear relationship with response variable is a bad idea. You must understand the variables relationship using scatter plots but not drop it just because it is not linear with the response variable. You should first build, understand the relationship, check if the variables are significant or not and then decide if you want to remove the variables or not.
In principle, yes! If there is no linear relationship between your predictor and your target, then there is no point in including your predictor into the model, as it will be one more parameter to estimate when you already know it is 0
We must differenciate, however, between there being no linear relationship, and there being a non-linear relationship. For example, if $$Y:=e^X$$, the relationship between $$Y$$ and $$X$$ is non-linear, but you should definitely still include X in your linear model, as larger X still implies larger Y and although the predictions won't be great, they will be better than if you just assume $$Y$$ is a constant.
So, my advice is to try reasonably simple transformations when possible (I would not go for something too fancy) Also, fortunatelly, there is plenty of software that can help you with model selection. For example, you could try to start by the most complex model (including all of your variables) and then go for an AIC-based stepwise model optimization, where "useless" predictors will tend to be removed.
Finally, understand how to validate your model! As a general piece of advice, I would go for the simplest model that predicts acceptable results and can stand the test of model validation (mainly, the residuals look like "noise", showing no recogniseable patterns in expectation nor variance)
• The first paragraph is plainly wrong, but the second one adds an interesting point to the other answers. I suggest removing some parts of the answer. – Pere Jun 11 '19 at 7:48
• @Pere No, it's not wrong. If $X and Y$ are uncorrelated variables, adding $X$ to the prediction of $Y$ will only bring multicolinearity effects. Models should be kept as simple as possible, so there is no point in adding extra-noise. I understand the example you gave, and it is great for a textbook, you real-world data hardly ever behaives like that – David Jun 11 '19 at 7:55
• Please see the example in my answer. X and Y are uncorrelated but X=Y+Z (exactly, R2=1) while Z just gives R2=0.5. X and Y being uncorrelated doesn't mean than Y can improve the predictive value of a set of variables when included in it. – Pere Jun 11 '19 at 8:00
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2020-01-17 12:38:37
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https://economics.stackexchange.com/questions/48460/in-a-box-diagram-why-does-efficiency-locus-lie-on-one-side-of-the-diagonal-if
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# In a box diagram, why does efficiency locus lie on one side of the diagonal, if both sectors haves constant returns to scale function?
The following is what I understand, so far. If we measure labour in the $$x$$-axis and capital in the $$y$$-axis, the slope of diagonal of the box is the capital-labour ratio $$K/L$$ in the economy. Let $$A$$ be the upright good and $$B$$ the upside-down good.
If the efficiency locus lies entirely on the right of the diagonal, that means if we consider any non-corner point on the locus, the slope of capital-labour ratio line in sector $$A,$$ $$K_A/L_A < K/L$$. Also, in sector $$B,$$ $$K_B/L_B > K/L$$. This implies $$K_A/L_A < K_B/L_B$$ i.e., $$A$$ is always relatively labour intensive. Therefore, using this logic, efficiency locus lying on one side of the diagonal just means that there is no factor intensity reversal taking place.
But according to Giancarlo Gandolfo International Trade Theory and Policy, efficiency locus will always lie on one side of the diagonal if the production functions in both sectors exhibit constant returns to scale. This implies that constant returns to scale function never exhibit factor intensity reversals, but this is not true; the same book mathematically shows CES production functions can exhibit them once.
Where am I going wrong? And why does constant returns to scale imply efficiency locus lying on one side of the diagonal.
• Hi! What do you mean by "Let $A$ be the upright good and $B$ the upside-down good." in this context? It seems that you defined the $y$-axis as capital? Nov 20, 2021 at 19:55
• I mean an Edgeworth box diagram. A is the good whose isoquants are drawn upright (origin is at lower left), B is the good whose isoquants are drawn upside down (origin is at upper right) Nov 21, 2021 at 6:58
• Are we still talking about a situation where the $y$-axis represents the capital? If not, can you please reread your question and edit where necessary? Nov 21, 2021 at 7:35
• Yes, I am talking about a standard Edgeworth box diagram with two goods (A,B) and two inputs (labour, capital) that you would use in something like HO model. Nov 21, 2021 at 8:43
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2022-06-30 14:18:41
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https://www.studyadda.com/sample-papers/neet-sample-test-paper-39_q19/228/282942
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• # question_answer A force $\vec{F}=\alpha \hat{i}+3\hat{j}+6\hat{k}$is acting at a point$\vec{r}=2\hat{i}-6\hat{j}-12\hat{k}.$ The value of a for which angular momentum about origin is conserved is A) 1 B) -1C) 2 D) zero
Angular momentum will be conserved if net torque acting on the system becomes zero. Given force acting $\vec{F}=\alpha \hat{i}+3\hat{j}+6\hat{k}$ (i) and $\vec{r}=2\hat{i}-6\hat{j}-12\hat{k}=-2(-\hat{i}+3\hat{j}+6\hat{k})$ (ii) If torque becomes zero then $\vec{r}\times \vec{F}=0$ If $\alpha =-1$then, $\vec{r}\times \vec{F}=0$
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2019-10-22 10:59:58
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https://codeahoy.com/learn/linuxnotes/ch6/
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# The Linux Shell
#### In this chapter
The shell executes a program in response to its prompt. When you give a command, the shell searches for the program, and then executes it. For example, when you give the command ls, the shell searches for the utility/program named ls, and then runs it in the shell. The arguments and the options that you provide with the utilities can impact the result that you get. The shell is also known as a CLI, or command-line interface.
## Changing Default Shell on Linux
Most modern distributions will come with BASH ( Bourne Again SHell) pre-installed and configured as a default shell.
The command (actually an executable binary, an ELF) that is responsible for changing shells in Linux is chsh (change shell).
We can first check which shells are already installed and configured on our machine by using the chsh -l command, which will output a result similar to this:
[ user @ localhost ~ ] $chsh -l /bin/sh /bin/bash /sbin/nologin /usr/bin/sh /usr/bin/bash /usr/sbin/nologin /usr/bin/fish In some Linux distributions, chsh -l is invalid. In this case, the list of all available shells can be found at /etc/shells file. You can show the file contents with cat: [ user @ localhost ~ ]$ cat/etc/shells
/bin/sh
/bin/bash
/usr/bin/sh
/usr/bin/bash
/usr/bin/fish
Now we can choose our new default shell, e.g. fish, and configure it by using chsh -s,
[ user @ localhost ~ ] $chsh -s /usr/bin/fish Changing shell for user. Password: Shell changed. Now all that is left to do is preform a logoff-logon cycle, and enjoy our new default shell. If you wish to change the default shell for a different user, and you have administrative privileges on the machine, you’ll be able to accomplish this by using chsh as root. So assuming we want to change user_2’s default shell to fish, we will use the same command as before, but with the addition of the other user’s username, chsh -s /usr/bin/fish user_2 In order to check what the current default shell is, we can view the $SHELL environment variable, which points to the path to our default shell, so after our change, we would expect to get a result similar to this,
~ echo $SHELL /usr/bin/fish ### chsh options: -s shell Sets shell as the login shell. -l, --list-shells Print the list of shells listed in/etc/shells and exit. -h, --help Print a usage message and exit. -v, --version Print version information and exit. ## Basic Shell Utilities ### Customizing the Shell prompt Default command prompt can be changed to look different and short. In case the current directory is long default command prompt becomes too large. Using PS1 becomes useful in these cases. A short and customized command pretty and elegant. In the table below PS1 has been used with a number of arguments to show different forms of shell prompts. Default command prompt looks something like this: [email protected] ~$ in my case it looks like this: [email protected] gotham ~ $. It can changed as per the table below: • PS1='\w$ ': ~ $ shell prompt as directory name. In this case root directory is Root. • PS1='\h$ ': gotham $ shell prompt as hostname • PS1='\u$ ': bruce $ shell prompt as username • PS1='\t$ ': 22:37:31 $ shell prompt in 24 hour format • PS1='@$ ': 10:37 PM shell prompt in 12 hour time format
• PS1='! $': 732 will show the history number of command in place of shell prompt • PS1='dude$ ': dude \$ will show the shell prompt the way you like
### Some basic shell commands
• Ctrl-k: cut/kill
• Ctrl-y: yank/paste
• Ctrl-a: will take cursor to the start of the line
• Ctrl-e: will take cursor to the end of the line
• Ctrl-d: will delete the character after/at the cursor
• Ctrl-l: will clear the screen/terminal
• Ctrl-u: will clear everything between prompt and the cursor
• Ctrl-_: will undo the last thing typed on the command line
• Ctrl-c: will interrupt/stop the job/process running in the foreground
• Ctrl-r: reverse search in history
• ~/.bash_history: stores last 500 commands/events used on the shell
• history will show the command history
• history | grep <keyword> will show all the commands in history having keyword <keyword> (useful in cases when you remember part of the command used in the past)
## Create Your Own Command Alias
If you are tired of using long commands in bash you can create your own command alias.
The best way to do this is to modify (or create if it does not exist) a file called .bash_aliases in your home folder. The general syntax is:
alias command_alias='actual_command'
where actual_command is the command you are renaming and command_alias is the new name you have given it. For example
alias install='sudo apt-get -y install'
maps the new command alias install to the actual command sudo apt-get -y install. This means that when you use install in a terminal this is interpreted by bash as sudo apt-get -y install.
## Locate a file on your system
Using bash you can easily locate a file with the locate command. For example say you are looking for the file mykey.pem:
locate mykey.pem
Sometimes files have strange names for example you might have a file like random7897_mykey_0fidw.pem. Let’s say you’re looking for this file but you only remember the mykey and pem parts. You could combine the locate command with grep using a pipe like this:
locate pem | grep mykey
Which would bring up all results which contain both of these pieces.
Note that not all systems have the locate utility installed, and many that do have not enabled it. locate is fast and efficient because it periodically scans your system and caches the names and locations for every file on it, but if that data collection is not enabled then it cannot tell you anything. You can use updatedb to manually initiate the filesystem scan in order to update the cached info about files on your filesystem.
Should you not have a working locate , you can fall back on the find utility:
find / -name mykey.pem -print
is roughly equivalent to locate mykey.pem but has to scan your filesystem(s) each time you run it for the file in question, rather than using cached data. This is obviously slower and less efficient, but more real-time. The find utility can do much more than find files, but a full description of its capabilities is beyond the scope of this example.
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2022-09-29 01:05:47
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https://ncatlab.org/nlab/show/trigonometric+function
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Contents
# Contents
## Idea
A trigonometric function is one derived from the basic functions of trigonometry, $\cos$, $\sin$ (cosine and sine), which themselves are the standard coordinate functions (equivalently: product projections $pr_i$) of the Cartesian space $\mathbb{R}^2$, restricted to the unit circle:
$\array{ & && \mathbb{R} \\ & & {}^{\mathllap{cos}}\nearrow & \big\uparrow{}^{ \mathrlap{pr_1} } \\ \mathbb{R} \overset{exp}{\longrightarrow} & S^1 &\hookrightarrow& \mathbb{R}^2 &\simeq& \mathbb{R} \times \mathbb{R} \\ & & {}_{\mathllap{sin}}\searrow & \big\downarrow{}^{ \mathrlap{pr_2} } \\ & && \mathbb{R} } \,,$
or rather the result of composing these restrictions with an arc length parametrization $\mathbb{R} \overset{\exp}{\to} S^1$. They are also called circular functions.
In elementary mathematics, there are six traditional trigonometric functions;
1. sine$\;$ $sin$
2. cosine$\;$ $cos$
3. tangent$\;$ $\tan = \frac{\sin}{\cos}$,
4. cotangent$\;$ $\cot = \frac{\cos}{\sin}$,
5. secant?$\;$ $\sec = \frac1{\cos}$,
6. cosecant?$\;$ $\csc = \frac1{\sin}$.
The early view was that these functions measured the six possible ratios of side lengths of right triangles (as a basic case in terms of which other triangles can be analyzed; “trigonometry” = “triangle measure”).
These six functions figure prominently in Euclidean geometry where the angles of a triangle sum to arc length $\pi$.
There are more elaborate offshoots such as spherical trigonometry? (see elliptic geometry) which was historically important for terrestrial navigation. Moreover, there are the related hyperbolic functions (see hyperbolic geometry) which result from a projective change of conic section (from a circle to a hyperbola).
## Definition
The most useful modern approach to cos and sin comes from taking Euler's formula as a working definition:
Let
$\exp \;\colon\; \mathbb{C} \longrightarrow \mathbb{C}^\ast$
be the complex exponential function, defined by the exponential power series formula
$\exp(z) \;\coloneqq\; \sum_{n \geq 0} \frac{z^n}{n!} \,.$
These satisfy the following fundamental exponential identities:
• $\exp(z+w) = \exp(z) \cdot \exp(w)$ (homomorphism from an additive group to a multiplicative group),
• $\widebar{\exp(z)} = \exp(\widebar{z})$ (because complex conjugation $\widebar{(-)}$ is a continuous field automorphism).
As a result, if $z + \widebar{z} = 0$ (i.e., if $z$ is purely imaginary: $z = i t$ for some $t \in \mathbb{R}$), we have
$\exp(z) \cdot \widebar{\exp(z)} = 1$
so that $w = \exp(z)$ lies on the unit circle defined by $w\widebar{w} = 1$.
###### Definition
The cosine function $\cos: \mathbb{R} \to \mathbb{R}$ is defined by $\cos(t) = Re(\exp(i t))$ (real part); the sine function $\sin: \mathbb{R} \to \mathbb{R}$ is defined by $\sin(t) = Im(\exp(i t))$ (imaginary part). In other words: $\exp(i t) = \cos(t) + i\sin(t)$ (Euler).
This implies that $\cos(t) = \frac1{2}(\exp(i t) + \exp(-i t))$ and $\sin(t) = \frac1{2 i}(\exp(i t) - \exp(-i t))$; these equations suggest the simple analytic continuation of $\cos$ and $\sin$ to functions $\mathbb{C} \to \mathbb{C}$ on the entire complex plane.
## Properties
More or less immediate consequences of the definition include
• $(\cos t)^2 + (\sin t)^2 = 1$ (“Pythagorean theorem”), since $\exp(i t)$ lies on the unit circle. This is traditionally written as $\cos^2(t) + \sin^2(t) = 1$;
• $\cos(x + y) = \cos(x)\cos(y) - \sin(x)\sin(y)$ and $\sin(x + y) = \sin(x)\cos(y) + \cos(x)\sin(y)$ (“addition formulas”), by taking real and imaginary parts of the identity that says $\exp$ is a homomorphism;
• $(\sin)' = \cos$ and $(\cos)' = -\sin$ (by differentiating $\exp(i t)$); the connection with the arc length parametrization of the unit circle is that the derivative of the position vector $p'(t) = (\exp(i t))'$ is a velocity vector $i\exp(i t)$ of unit length;
• $\cos(x + 2\pi) = \cos(x)$ and $\sin(x+2\pi) = \sin(x)$ (“periodicity”), according to the modern definition of pi;
• $\cos(x) = \sum_{n \geq 0} (-1)^n \frac{x^{2 n}}{(2 n)!}$ and $\sin(x) = \sum_{n \geq 0} (-1)^n \frac{x^{2 n + 1}}{(2 n + 1)!}$, by exploiting the power series representation of the exponential function.
These name but a few of many trigonometric identities, facility in which can serve as a modern-day shibboleth or barrier of passage in high school or lower-level undergraduate courses in mathematics. They seem also to be popular in mathematics education in India and appear regularly in entrance examinations there. But the ones listed above are the most fundamental.
There is no question that the trigonometric functions are rich in significance; for example; various representations of the tangent, cotangent, secant, etc. appear in enumerative combinatorics (as in the problem of counting alternating permutations), representations of Bernoulli numbers, and so on.
## References
Last revised on May 10, 2019 at 11:03:01. See the history of this page for a list of all contributions to it.
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2021-04-13 11:16:42
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https://mathsgee.com/2059/what-the-formula-finding-average-mean-binomial-distribution
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MathsGee is Zero-Rated (You do not need data to access) on: Telkom |Dimension Data | Rain | MWEB
0 like 0 dislike
516 views
What is the formula of finding the average (mean) of a binomial distribution?
| 516 views
0 like 0 dislike
The formula for the mean of binomial distribution is: μ = n *p Where “n” is the number of trials and “p” is the probability of success.
by Wooden (150 points)
0 like 0 dislike
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2021-06-22 23:16:54
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https://control.com/forums/threads/where-does-cost-fit-in-was-comm-where-does-mpi-fit-in.2148/
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Where does COST fit in, was COMM: Where does MPI fit in?
J
Jeffrey D. Brandt
Donald Pittendrigh wrote:
>
>I am having some real problems coming to grips
>with the real issue of this MPI thread and
>would appreciate it if some one could explain
>what the real issue is here
>
<snip>
>
>Maybe I am missing something, if so I would be
>grateful for the correction, please don't tell
>me it is the financial thing, I live in South
>Africa and we suffer from a down trodden and
>inflation ridden badly managed economy, and my
>clients can still afford the use of quality
>equipment, I cannot believe this is a
>consideration in a country like the USA which
>is undoubtedly the most affluent economy in
>the world.
This is an interesting case of reverse-stereotyping. Isn't is strange that someone from beyond the shores of the US has the opinion that 'cost would be a consideration' in 'the most affluent economy in the world.'
It wonders me (as we say in Pennsylvania Dutch) as well. Why would a OEM argue over meal charges after a successful field trip, amounting to $15.00, for a$4,000,000 project? Why do customers ask if I can save them money buy using 'used parts'? Why does my competition shop for materials on eBay?
Darned if I know, but, the fact of the matter is that THIS is the reality of the controls business where _I_ live. Siemens or no Siemens, cost is always a factor, and is generally (recently) the ONLY factor.
When I present my 'Triangle of Truth' (Good, Fast, Cheap: pick two) CHEAP is always chosen.
Now, I'll say this: If anybody else read Donald's message and said the same thing: What gives YOU that impression????
[standing by, donning my flame retard. clothing]
D
Donald Pittendrigh
Hi All
Just one small correction, it doesn't suprise me that cost is a consideration, to ignore cost implications is obviously silly. The magnitude
of the consideration is what surprised me, that people would go several times around the block to try and save the cost of the "right" interface
card, by using something which was strictly speaking not intended to be used as proposed. (glad I typed that and didn't have to say it out loud)
Cheers
Donald Pittendrigh
> This is an interesting case of reverse-stereotyping. Isn't is strange
that
> someone from beyond the shores of the US has the opinion that 'cost would
be a
> consideration' in 'the most affluent economy in the world.'
B
Bob Close
My 2 cents... for what it's worth.
In the US there are very many choices for motion products and they cover the gambit of price and performances. This often makes the selection that much more difficult and certainly takes it out of the realm of simple black and white price vs. performance decisions. Small changes to filtering, software interfaces, tuning algorithms and connectivity can make huge changes in the cost to install set up and support motion products. Choices of slight price reductions over performance can lead to large user expense, however because of the huge selection we have it's ridiculous not to attempt to save where you can. This makes the selection process much more complicated, however it also leaves one with an optimized design that will make an OEM's machine more cost competitive or a user's machine less
expensive (life cycle cost) to employ on the floor.
I think the generalization of America being cost over performance/reliability is an exaggeration, although there hundreds of system integrators and manufacturing engineers who would beg to differ with me.
Bob Close
Executive V.P.
Precision MicroControl Corp
TELE 760-930-0101
FAX 760-930-0222
[email protected]
R
Richard McMahon
When we talk about costs we must not forget that:
cost != purchase price
but rather
cost == purchase price + maintenance cost + hassle factor + ...
When making a purchase, it is easier to justify a decision based on numbers i.e. price, than a value judgement including the whole life cost
of the equipment, which frequently includes some form of subjective assessment. When you buy a car for your own use, you include all your previous experience of the model and manufacturer in the decision.
Therefore, why let money be the sole deciding factor in work place?
Answer: Because you can't be criticised if you produce the quotations for supply.
Richard McMahon
Tel.: +44 117 954 5160
Research Assistant
Bristol
UK
K
Ken Brown
Jeff,
What a great summary of life in the manufacturing sector of the good-ol-U-S of A!
I think there are two major camps from where I sit. There is the camp that says they will buy any type of controls as long as it has an
Octagon on it - regardless of what it costs or whether it makes any sense or can even handle the app, these customers seem to go in cycles with a love hate relationship with the almighty octagon. AND the other big camp is - whatever is cheapest .... buy it. . . regardless of how much personal effort and cash they have to put into it to get it running, try to keep it running, and eventually to replace it with the next cheap fix that will solve some of the problems it has created.
A third and less prominent camp consists of those who actually think about what they need to accomplish and consider the long term
consequences of the choices they make. These are few and far between, but when we run across a customer like this, we bend over backwards to
get them and then if things look really good, we make efforts to fire one of our less favorite customers from one of the other camps. Nothing
limits your ability to grow and be a good service provider than a short-sighted customer who abuses vendors.
BTW . . . we have bought some nice Tektronix and Fluke items on ebay and are considering selling some of our surplus control gear as well . . . .
ain't America great!
Best Regards,
Ken Brown
Applied Motion Systems, Inc.
http://www.kinemation.com
L
Lunnon, Robert
This is too simplistic, The problem is not really as you outlined, although many companies do spend an extraordinary amount of effort trying to minimise capital costs, so try not to see it from an us and them point of view. The problem is you are looking at the problem from a technology point of view while they are looking for a value justification.
If you think like an accountant, then you can see that in many cases the capital cost is the hardest to justify (Hence the pre-occupation with it). It is necessary then to convince the accountants then that leaving in that extra PE cell is not only the right way to do it but will SAVE THEM $$. Even redundant systems can be cost justified. Lets say you are working for a water manufacturer on a procuction line producing 10M worth of tap water into refillable containers per annum with a real profit margin of 10% based on 99% availability in a 24x7 operation and that production has predominately fixed overheads. (That means it takes 328 days to pay the overheads and the profit is made in the remaining 33 days.) Lets also assume the PE cell is buried deep in the machine and takes 6 hours to strip down and replace (ugh) The cost for failure of that device then = 1M/33/4 = 7575 which comes directly off the profit. There aren't too many accountants who will argue with you after that ! Put another way, lets say this companies sums are based on 95% availability, then they are actually counting on losing 18.25 days of production per annum. They make 1M profit and have 9M fixed overheads so the 9M is made in 9/10 *347 days (312 Days) the 1M profit takes a further 347-312 = 35 days to make. A small modification you want to make can raise the availability to 98% by reducing jamming on the production line. This reduced the planned downtime to 7.3 days. The difference is then about 11 days. Because the costs are fixed (probably not quite true in most cases, but in many cases such as mining can be a reasonable approximation) the extra production is free. The rate of money making is 10M/347 days = 28800 per day (Roughly) so your company stands to make an extra 316000 (less variable overheads) per annum. Now ask the question. The modification costs 10000 to do, Approve/not approve ????? What this reveals is that where there is no performance increase associated with a modification (reliability is the concern) it is absolutely essential to know the cost of downtime in presenting an appropriate value justification. This fact has been missed in just about every organisation I have worked for. Ahh well, at least I understand ... ;-) Bob J John Vales I know I'm going off the subject a bit, but I'd like to add to Jeffrey's reply. The same is true where I live, and work, too. (Chicago area). I have to agree that in terms of controls, electrical wiring, and All Things That Spark, I, too, have found that many customers view our field (controls engineering) as a necessary Evil. Examples of this mentality: Do you _really_ need two prox switches to detect this? Why can't you do it with only one? What do you mean, it's going to take two weeks to debug my 2M system? That can't be right! You controls geeks are all alike. I find this ironic, especially in this time when everyone and everything lives and dies by technology. Everyone wants the latest and greatest on their systems and machines, but when it comes time to pony up with the$$$, everyone looks for a place to hide. jfv John F. Vales [email protected] D Donald Pittendrigh Hi All The prospects of having to justify my engineering designs in this way to a non-technical bean counter I find totally nauseating. I mean what if you aren't a good enough salesman, can't convince the guy, and therefore end up with a plant with inherent weakness in its design which kills someone 3 years down the line. This is a perfect example of engineering decisions for all the wrong reasons. The use of quality equipment on a project is not some do Donald a favour thing, sometimes I wish the client would take his whole project and test the length of his arm, but I have a policy that I turn no-one away, now an accountant is doing me a favour by allowing me an extra 0.00001% to do the job properly and I have to BEG FOR IT. No way that accountant can take his project and his money, his boss his entire organisation and go for a walk on the wild side. Regards (Highly aggravated) Donald Pittendrigh > If you think like an accountant, then you can see that in many cases the > capital cost is the hardest to justify (Hence the pre-occupation with it). > It is necessary then to convince the accountants then that leaving in that > extra PE cell is not only the right way to do it but will SAVE THEM$.
> Even redundant systems can be cost justified. Lets say you are working for
a
For example, suppose you have to connect 70 PLC's to 14 PC's (5 PLC's to each PC). You only have to share a small amount of data so if you had, e.g. an S7-200, you could use the PPI cable and Prodave light, for a total cost of around $4000. If you spent a day doing your own protocol on the freeport mode you could shave another couple of thousand off that price. On the other hand, with the S7-300, you need a CP card to do the MPI and there is no freeport alternative, so we are talking at least$11200. If we move on to OPC we are talking $21000 (also, the customer wants to use Windows because 'their staff always have trouble maintaining NT' - his words, don't flame me). Now, that setup of 70 PLC's then needs to be replicated at at least one site, possibly more. If the 300 had a freeport mode like its little$200 cousin, we have $20000 saving on the first two sites alone. That's worth spending a days coding for. "Look after the pennies and the pounds take care of themselves" -- Dickens C Curt Wuollet Hi Ken > What a great summary of life in the manufacturing sector of the > good-ol-U-S of A! > > I think there are two major camps from where I sit. There is the camp > that says they will buy any type of controls as long as it has an > Octagon on it - regardless of what it costs or whether it makes any > sense or can even handle the app, these customers seem to go in cycles > with a love hate relationship with the almighty octagon. AND the other > big camp is - whatever is cheapest .... buy it. . . regardless of how > much personal effort and cash they have to put into it to get it > running, try to keep it running, and eventually to replace it with the > next cheap fix that will solve some of the problems it has created. I get to see both sides of this as a captive supplier and I'm coming to an epiphany, A radical view that there has to be a better way. Cost is important, real important, and both of those camps cost too much, just in a different way. The cheapest way is not remarkably different from the Octagon and friends. The Octagon way is supposed to be efficient because it's "off the shelf" with a little value add for programming and services. The cheap way is supposed to save you enough on hardware to pay for all the integration headaches you take on by not buying from a single source. OK the big picture is that both of those and even the third group are broken and grossly inefficient causing costs to be very discouraging. The Octagon approach is broken by overselling the advantage of a single source and taking advantage of the lock-in. In our situation the integration cost is still excessive and it's big money whenever you need something. And it can be really hard to get the people that you need. The good part is that you aren't always reinventing the wheel, some knowledge carries forward to the next project. The cheapest approach can work if you're lucky, if not, you end up doing a lot of rework. Since the cheapest stuff changes day to day chances are that you engineer new wheels on about every project. Help isn't a problem since it's brand new to anyone. And the knowledge gained probably won't apply to the next project. The optimum approach would be if you could always use the same stuff yet not get raped for the priviledge. Almost all the knowlege would apply for future projects and you wouldn't have the churn. For this we need a platform thet is versatile and works with everything to keep suppliers honest. It would use protocols that already exixt and off the shelf hardware. Since development was just for the system itself, it could be replicated cheaply. And if everyone used it and shared notes the amount of wheel engineering would be small. This is the promise of the octagon approach, with the cost of competitive commodity hardware. I had this epiphany because the cost of integrating a proprietary line got close to what it would cost to develop a full custom solution due to the barriers at every turn that are inherent in the proprietary approach. I know what you're thinking..... I still wouldn't save anything. Not on this job. With the next one I would save a lot. And with each job the cost would diminish. Now suppose I gave that solution to other integrators. They would have that same cost curve and increasing profits. This is what the Linux PLC project is about. Taking the cost of developing an integrator friendly, versatile solution and spreading it out among enough companies that the cost becomes in effect nothing. Maximizing reuse, eliminating duplication of effort, and very important, creating a market for vendor neutral IO hardware and competition for that market. The product would be owned by all equally so competition would be back on quality of solutions you provide with that product. The great majority of the cost that people object to is directly and indirectly the cost of providing and maintaining 150 complete and functionally equivalent solutions to the same problems. Every dollar saved in eliminating that huge burden is another dollar that flows to the services and solutions providers that solve the problems and make factories work. Who do you think deserves the lion's share? The dramatic decrease in the cost of computing in general can be directly attributed to the standardization of platforms and the elimination of the total duplication of effort. > > A third and less prominent camp consists of those who actually think > about what they need to accomplish and consider the long term > consequences of the choices they make. These are few and far between, > but when we run across a customer like this, we bend over backwards to > get them and then if things look really good, we make efforts to fire > one of our less favorite customers from one of the other camps. Nothing > limits your ability to grow and be a good service provider than a > short-sighted customer who abuses vendors. It goes a long ways towards good and happy customers if you can demonstrate that you spend their money well. A couple of outlandishly spendy doodads each time they want a change erodes the goodwill in a hurry. _I_ am embarrassed at the cost of the proprietary offerings. I am at a loss to explain why connectors cost$80.00 and why you can't do anything without buying more "special" stuff. Most customers won't begrudge you a profit but it's hard to add to already inflated pricing. There is much less resistance to commodity hardware simply because it's reasonable.
We are trying to provide that alternative but, we need help. If you think there should be a rational choice and an alternative to the status quo, we need coders and people with experience in all facets of automation. We have a _lot_ of lurkers on the list but, we need some push to make it happen.
Regards
cww
R
Richard McMahon
> The prospects of having to justify my engineering designs in this way to
> a non-technical bean counter I find totally nauseating. I mean what if you
> aren't a good enough salesman, can't convince the guy, and therefore end
> up with a plant with inherent weakness in its design which kills someone
> 3 years down the line.
Donald, a quick question
How do you make out an ALARP case if an accountant determines the expenditure level with no reference to system design?
Richard
R
roger Irwin
Richard McMahon wrote:
> When we talk about costs we must not forget that:
>
> cost != purchase price
>
> but rather
>
> cost == purchase price + maintenance cost + hassle factor + ...
>
This is very true, and why I believe for 1 off solutions ease of use should be preferred over cost, but as instances increase you have to look very carefully. Also, the fact that a product is expensive does not mean it will have low maintenance costs, it does mean that they will spend a lot of money trying to convince you that this is so.
As for 'single source', this often increases hassle factor, as no single supplier will always have the most suitable product for all your
requirements, so you will find yourselves with workarounds and compromises.
Sometimes things work in reverse. Also, many high priced products try to maintain profit margins and justify costs by adding in features rather
than improving and fine tuning what they have. I like to go for the simplest possible solution using tried and trusted techniques rather than lots of fancy bells and whistles. Far less hassle, far lower maintenance costs.
H
Helmut Meissner
Hello Roger,
Hello Donald,
you're true, when you say that the licence cost con go up. We in our company have the same problems. With every machine you have to give away
a Scada-system or something to store, cummulate, prapare etc. PLC-Data and make printouts for the Company-Owners and controling-stuff)
For the S5 you could get a dozen products of different companys; you where not forced to buy Siemens Prodave-Licences. Then S7 started it was
a big problem to contact to PC. We use for plant or machine installations L2-FMS and a cheap Scada system (no licence fees) and had to buy the expensive Siemens Hardware (CP, PCI-Card).
For the small machines we programmed a small DOS program and used a company licence of AS-511 Drivers. (one time cost about 3000 DM appr. 7000 US$) I contacted the company process-informatic 2 weeks ago, because i will not get any more S5 PLC by siemens in the future and had to chance to S7. They have MPI-drivers (PC-MPI-S7-Link with company licence for about 3998 DM about 10.000 US$ depends on the Euro-Exchange ;-), Project
licence for endusers 999 DM about 2400 US$) I think it is worth to watch 3rd party companies. In the Online-shop you can buy licence and source-code !!! http://www.process-informatik.de/shop/shindex.htm Their description (sorry only in german) shows nothing yet for Linux. But I think we have to wait 3-5 years until endusers and company-owers will see that MS is not the only solution for PC Operating systems. We will try to use MS-DOS as long as possible, because it is stable and cheap (Dos-Clones) and for the endusers easy to handle. (Some Function-keys, and Arrow-Keys are the best for unemploied works, No Mouse, no touch: just write down: Press Key xxx, then Key yyy, etc.) Helmut Meissner Manager Electric and Software Department Schlosser-Pfeiffer GmbH [email protected] PS: If you need some information how it is running under MS-DOS please contact me in about 3 weeks. B Bruce Durdle Just to add another log on the fire ... A few years ago, while working in the UK, I attended a meeting on the topic on the economics of Cogeneration projects for industrial applications. First item on the budget - at something like 10% of main plant capital cost - lawyers' fees to obtain al the necessary environemtal and other permits. Engineering consulting fees - to make sure the damn thing worked and didn't create the environmental problems the lawyers were concerned with - 2% or less. Something screwy somewhere. Bruce J john coppini well, well, well, lots of comments here! so this one will probably go unread. but here's my observation: ****the powers that run a company are usually not in engineering and manufacturing. the power is in in finance and management. so....engineering and manufacturing get the short, dirty end of the stick. furthermore, engineers let this happen to them by selling themselves too cheap! for example, control systems consultants have a hard time selling their services for$75 per hour on the west coast. by comparison, business system consultant get as much as $200 per hour. the people that pay the bills and run the company have no problem spending$200 per hour on themselves! but they are really cheap on engineering and manufacturing expenditures!
***in summary, most engineers are technicians, not businessmen! rationalize as much as you want, but this is the situation. so just keep wasting your time reinventing the wheel in Linux----riduculous!
John Coppini
Engineer
A
Anthony Kerstens
....
> First item on the budget - ...10% of main plant
> capital cost .... - lawyers' fees ...
> Engineering consulting fees - ....- 2%....
>
> Something screwy somewhere.
Yes something is screwy. We as engineers are not making enough $. Anthony Kerstens P.Eng. D David W. Spitzer John, This is a good point. We (engineers) are somewhat at fault, but in a world of "just in time training" (think a while on that one), "it's user-friendly" (read easy enough for anyone to do), and "if you can't do it, we'll get someone that can" attitudes, engineers can become a commodity, even if reality says "not so fast". Also, I differentiate between three types of engineers with three different pay scales. At US$75 per hour, you appear to be describing the second group below.
1. engineers who work directly for a company (with a steady income plus benefits...)
2. (contract) engineers who are contracted to work part/full-time on an ongoing basis for the same company (with a relatively steady income plus few or no benefits...)
3. consulting engineers (consultants) who work a small number of hours on specialized problems for multiple clients (with an erratic income plus no
benefits...)
Each seems to affect the pay scale of the next. After all, why pay US$200 per hour for a consultant when an employee can do it for US$40 per hour? We know there is more to it than this. The employee may be skilled in other areas and take many times the time to do it incorrectly (yes, incorrectly).
Just a couple of weeks ago, I observed an installation that had been troublesome since its installation 8 years ago. The employees were about to make the same mistake again, but one employee (whom I had met years ago at an ISA show) "knew what he didn't know" and called in a consultant. The core problems were defined in 2-3 hours. The whole project, including the site
visit, report, and design basis to solve the problem will likely be less than 30 hours.
Interestingly, the first two groups often have an adverse affect on their own future earnings, because in semi-retirement, they often attempt to join the latter groups.
Regards,
David W Spitzer
845.623.1830
www.icu.com/spitzer
C
Hi John
</rant>
Not unread anyway. This is a problem in all the technology fields. One hears these days about this tremendous shortage of technical types. I haven't seen it. What they mean is that they advertise for someone with every current hot buzzword skill they can think of and offer $35k a year and those jobs go unfilled. Hmmm.. Must be a shortage. I know of serious computer people who are working in other fields because they're tired of being abused and underpaid. The latest cure for this is to go offshore to find exploitable people, it isn't working because the highly skilled people from other countries aren't stupid either. And a lot of jobs that go unfilled pretty much suck. Yet as soon as market forces begin to work to push wages for technical people up, there's a tremendous hew and cry to import talent or churn out more graduates "because there's a shortage!". No one thinks of retraining older skilled workers that can't find a decent job. No one wants to pay for developing workers from within. If you can't get the exact skill set you want at below the already depressed market rate, there's a shortage. I'll bet without exception that upping your offer by say 30% will take care of any shortage you might be experiencing. I have headhunters calling with ridiculous proposals and I'm sure they go away with the impression that there's a shortage also. <rant/> Why is it that employers are so dead set against paying for the skills that are building the new economy? Regards cww A Anthony Kerstens New economy? It smells like the old economy: Make as much as you can while paying as little as possible for overhead. And speaking of overhead, I wonder if anyone who knows (and is willing) could describe a typical breakdown of where$75/hr goes, if an engineer is being paid $35k. ($US )
My _guess_ is the hourly salary works out to
about $16.83. Plus 40% benefits comes to$23.56. Plus a decent computer (every year) comes to about $1.50 per hour. Office space at$10.00/sqft, at 4000 sqft per 20 people
(desk/hall/lobby/lunchroom/library/test room)
comes to something like $12.00/hr. Training if your lucky might be$2000.00 per year, which comes around to $0.96/hr. I personally might consume 4 hours per week of support staff time, so 1/10hr *$15 come to $1.50/hr. And lets throw in another$10.00 for things I may have forgotten. The grand total is $49.52. (No wonder home offices are so popular.) That leaves$25.48 for the firm, from which
there's marketing costs, upper management salaries
and taxes on the remainder.
Oh yeah, subtract $6.73 for every$10k above
that $35k salary. Don't get me wrong. I'd love to make more money. And I think I deserve to make more money. I just don't think money is the cause of a "shortage" of technical types. There's plenty around. The problem is that even though some of us are worth the good salaries, there are more that are not. You get a handful of prima donnas, and a bucket full of something else. Those belonging to the bucket probably don't know it, and are bitching about salaries right along with everyone else. You still have to pay for the bucket full because you need warm bodies to fill a desk and keep work "flowing". The trick is finding an employer who can differentiate between you and the bucket when it comes to paying out salary. If an employee feels his employer isn't seeing that difference, then maybe he should be keeping his eyes open for other opportunities. Anthony Kerstens P.Eng. R Rick Daniel <perk> Please tell me where I can find an engineer for$35k/YR! I'm excited! We're paying more than that for fresh-outs! According to my salary survey from the American Electronics Association, engineers make 2 to 3 times that.
Rick Daniel
Intelligent Instrumentation
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2021-04-22 03:38:39
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http://math.stackexchange.com/questions/307462/order-of-growth-of-an-entire-function-defined-by-an-infinite-product
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# Order of growth of an entire function defined by an infinite product
Consider the product $$\displaystyle\prod_{n=1}^{+\infty}(1-e^{-2\pi n}e^{2\pi iz})$$
I know that this product defines an entire function $F$. I must show that the order of growth of $F$ is finite, at most 2. My definition of order of an entire function is $$\rho=\inf\{\lambda>0: \displaystyle\sup_{|z|=r} |f(z)|=O(e^{r^{\lambda}}),r\rightarrow\infty\}$$ Can someone give me an help?
-
Maybe Jensen's formula could be of help: en.wikipedia.org/wiki/Jensen%27s_formula – Eric Haengel Feb 18 '13 at 21:27
The zeros of $F$ are $n+ki$ for $n=1,2,\dots$, $k\in\mathbb{Z}$. If $p>2$ then $$\sum_{n=1}^\infty\sum_{k=-\infty}^{\infty}\frac{1}{|n+ki|^p}<\infty.$$ This shows that the order of $F$ is at most $2$.
sorry, i can't understand. What i knew was that the convergence of the double series you wrote tells me that the convergence exponent of zeros of $F$ is 2, but the order of $F$ is at least the convergence exponent of zeros, not at most....what is wrong in what i know? – Federica Maggioni Feb 19 '13 at 7:51
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2015-05-22 13:22:05
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http://m-labs.hk/artiq/manual-master/compiler.html
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# Compiler¶
The ARTIQ compiler transforms the Python code of the kernels into machine code executable on the core device. It is invoked automatically when calling a function that uses the @kernel decorator.
## Supported Python features¶
A number of Python features can be used inside a kernel for compilation and execution on the core device. They include for and while loops, conditionals (if, else, elif), functions, exceptions, and statically typed variables of the following types:
• Booleans
• 32-bit signed integers (default size)
• 64-bit signed integers (use numpy.int64 to convert)
• Double-precision floating point numbers
• Lists of any supported types
• String constants
• User-defined classes, with attributes of any supported types (attributes that are not used anywhere in the kernel are ignored)
For a demonstration of some of these features, see the mandelbrot.py example.
When several instances of a user-defined class are referenced from the same kernel, every attribute must have the same type in every instance of the class.
## Remote procedure calls¶
Kernel code can call host functions without any additional ceremony. However, such functions are assumed to return None, and if a value other than None is returned, an exception is raised. To call a host function returning a value other than None its return type must be annotated using the standard Python syntax, e.g.:
def return_four() -> TInt32:
return 4
The Python types correspond to ARTIQ type annotations as follows:
Python ARTIQ
NoneType TNone
bool TBool
int TInt32 or TInt64
float TFloat
str TStr
list of T TList(T)
range TRange32, TRange64
numpy.int32 TInt32
numpy.int64 TInt64
numpy.float64 TFloat
## Pitfalls¶
The ARTIQ compiler accepts nearly a strict subset of Python 3. However, by necessity there is a number of differences that can lead to bugs.
Arbitrary-length integers are not supported at all on the core device; all integers are either 32-bit or 64-bit. This especially affects calculations that result in a 32-bit signed overflow; if the compiler detects a constant that doesn’t fit into 32 bits, the entire expression will be upgraded to 64-bit arithmetics, however if all constants are small, 32-bit arithmetics will be used even if the result will overflow. Overflows are not detected.
The result of calling the builtin round function is different when used with the builtin float type and the numpy.float64 type on the host interpreter; round(1.0) returns an integer value 1, whereas round(numpy.float64(1.0)) returns a floating point value numpy.float64(1.0). Since both float and numpy.float64 are mapped to the builtin float type on the core device, this can lead to problems in functions marked @portable; the workaround is to explicitly cast the argument of round to float: round(float(numpy.float64(1.0))) returns an integer on the core device as well as on the host interpreter.
## Asynchronous RPCs¶
If an RPC returns no value, it can be invoked in a way that does not block until the RPC finishes execution, but only until it is queued. (Submitting asynchronous RPCs too rapidly, as well as submitting asynchronous RPCs with arguments that are too large, can still block until completion.)
To define an asynchronous RPC, use the @rpc annotation with a flag:
@rpc(flags={"async"})
def record_result(x):
self.results.append(x)
The ARTIQ compiler runs many optimizations, most of which perform well on code that has pristine Python semantics. It also contains more powerful, and more invasive, optimizations that require opt-in to activate.
### Fast-math flags¶
The compiler does not normally perform algebraically equivalent transformations on floating-point expressions, because this can dramatically change the result. However, it can be instructed to do so if all of the following is true:
• Arguments and results will not be not-a-number or infinity values;
• The sign of a zero value is insignificant;
• Any algebraically equivalent transformations, such as reassociation or replacing division with multiplication by reciprocal, are legal to perform.
If this is the case for a given kernel, a fast-math flag can be specified to enable more aggressive optimization for this specific kernel:
@kernel(flags={"fast-math"})
def calculate(x, y, z):
return x * z + y * z
This flag particularly benefits loops with I/O delays performed in fractional seconds rather than machine units, as well as updates to DDS phase and frequency.
### Kernel invariants¶
The compiler attempts to remove or hoist out of loops any redundant memory load operations, as well as propagate known constants into function bodies, which can enable further optimization. However, it must make conservative assumptions about code that it is unable to observe, because such code can change the value of the attribute, making the optimization invalid.
When an attribute is known to never change while the kernel is running, it can be marked as a kernel invariant to enable more aggressive optimization for this specific attribute.
class Converter:
kernel_invariants = {"ratio"}
def __init__(self, ratio=1.0):
self.ratio = ratio
@kernel
def convert(self, value):
return value * self.ratio ** 2
In the synthetic example above, the compiler will be able to detect that the result of evaluating self.ratio ** 2 never changes and replace it with a constant, removing an expensive floating-point operation.
class Worker:
kernel_invariants = {"interval"}
def __init__(self, interval=1.0*us):
self.interval = interval
def work(self):
# something useful
class Looper:
def __init__(self, worker):
self.worker = worker
@kernel
def loop(self):
for _ in range(100):
delay(self.worker.interval / 5.0)
self.worker.work()
In the synthetic example above, the compiler will be able to detect that the result of evaluating self.interval / 5.0 never changes, even though it neither knows the value of self.worker.interval beforehand nor can it see through the self.worker.work() function call, and hoist the expensive floating-point division out of the loop, transforming the code for loop into an equivalent of the following:
@kernel
def loop(self):
precomputed_delay_mu = self.core.seconds_to_mu(self.worker.interval / 5.0)
for _ in range(100):
delay_mu(precomputed_delay_mu)
self.worker.work()
|
2018-01-19 17:23:14
|
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|
https://www.physicsforums.com/threads/strain-energy-release-during-fracture-where-does-it-go.741826/
|
# Strain Energy Release During Fracture -Where does it go?
1. Mar 6, 2014
### jstluise
When a bolt is pulled in tension and eventually fractures, is all the built up strain energy dissipated in the formation of the new surfaces? Does any energy do into accelerating the broken halves of the bolt?
Imagine that two plates are bolted together. As the two plates are forced apart, the bolt deforms and eventually yields/fractures. When the fracture occurs, the bolt will be shot out from the joint. I believe that it is the release of the elastic potential energy built up in the plates that is causing the bolt halves to accelerate, and that the energy dissipated in the fracture has no influence on the bolt accelerating. Is this correct?
Thinking about it another way: imagine the two plates in the above scenario are infinitely stiff. When they are forces apart so that the bolt fractures, will the bolt move at all? Or will the bolt just break and remain mostly still? (I guess this is assuming the rest of the bolt is infinitely stiff as well; if not there could be built up energy in the deformation of the bolt head that could accelerate the bolt once it breaks)
I hope that makes sense!
2. Mar 6, 2014
### Filip Larsen
You could perhaps try establish some equations for energy and, thus, speed of a bolt assuming all stored energy from the load is converted to kinetic energy in the bolt, and then run a calculation for the situation you have in mind? I ran a few numbers on the back of an envelope (well, actually it was a block of paper) and was a bit surprised about the result.
3. Mar 6, 2014
### jstluise
What we are trying to figure out is out of all the energy put into the system to break the bolt (area under force/displacement curve), how much of that is dissipated in the fracture of the bolt, and how much of it is dissipated in the mechanism that is pulling the bolt apart.
A Frangibolt is a device that is placed between two bolted plates. When activated, the Frangibolt expands until there is enough built up force to break the bolt. When the bolt breaks, it is the potential energy present in the Frangibolt that we are worried about, which is leftover from the energy dissipated in the fracture.
4. Mar 6, 2014
### Filip Larsen
I am not a mechanical or materials engineer, but it only took a few minutes with a textbook to establish a simple model for the energy U stored in a bolt as a function of the load force F, bolt area A, elasticity module E and clamp length L (grip length in your case), as
$$U = \frac{F^2 L}{2 A E}$$
which can be combined with an expression for the kinetic energy of the bolt. You should be able to derive the combined result easily. If you use a bolt with a built-in weak spot (i.e. different cross sections along the length of the bolt), you may have to derive a more accurate U expression for that.
When inserting data for a 10 mm diameter standard steel alloy bolt with an "effective length" of 30 mm loaded with 70 kN over a grip of 10 mm, I get an energy of around 22 J and a maximum speed of around 34 m/s, which was higher than I had expected. I did not make any additional calculations to check my results, though, so it may be that I have made a mistake.
5. Mar 6, 2014
### jstluise
I see what you did, but that stored energy is only considering the elastic deformation of the bolt. What you are describing is essentially the elastic potential energy (1/2*k*x^2) using the equivalent spring constant from the bolt properties.
If the bolt all of a sudden broke (i.e. no energy went into fracturing), then (I think) it makes sense to say this will be the energy that is converted into kinetic energy. So, in the case of a real brittle material, this is probably a good estimate of the kinetic energy.
But, with a material that has some plastic deformation before fracture, you are going to loose some energy in the deformation of the bolt. Thus, the total energy inputted into the system is dissipated in the fracture/deformation, and the rest is converted to kinetic energy.
I'm beginning to think that just looking at the force/displacement curves for breaking a bolt will tell me everything I need to know. The area under the curve is the energy put into the system. For example, looking at this force/displacement curve:
If you almost reach the point of fracture, but let off the force, you will recover based on the elastic modulus. The rest of the area was energy wasted in plastically deforming the bolt. So, that leads me to think that once fracture occurs, the energy (green) is available for conversion from potential to kinetic.
That is the only thing that makes sense to me right now, without getting into fracture mechanics. But, I could be totally wrong.
6. Mar 6, 2014
### Filip Larsen
For what it is worth I agree with your description. Note that the area of the green section is indeed the expression I gave earlier with F being the ultimate tensile strength, that is, the expression effectively "ignores" any energy lost due to plastic deformation.
7. Mar 6, 2014
### jstluise
Ah yes, thanks for pointing that out. Glad we are on the same page.
8. Mar 6, 2014
### AlephZero
That's right. The only energy you can recover is from the elastic strain energy in the bolt. You already used the plastic strain energy in stretching the bolt permanently. You can't use it again.
I think the estimate of 34 m/s velociity is a bit high. You didn't state what you assumed for the mass of the broken piece (don't forget the bolt head, for example). Some energy will go into vibrations in the bolt and the flanges, as well as the "rigid body" motion of the bolt. But broken bolts can certainly "fly" a long way if they get "launched" cleanly, without catching on the side of the bolt holes etc.
We once had an unexpected failure of something similar to a very large bolt, that was being tested in a vacuum tank and filmed with a high speed camera, so we could measure what happened. The impact on the lid of the tank (about 10 feet diameter) wrecked the lid, We calculated that without the lid, it would have gone through the roof of the building and reached an altitude of about 1500 feet
9. Mar 6, 2014
### jstluise
Thanks for responding, AlephZero.
I think you are onto what originally made me start thinking about this. When the broken bolt is launched, is it from the built up potential energy from the joined material forcing the bolt apart? Or is it somehow from the strain energy in the bolt?
This goes back to the scenario I mentioned in the my first post: you have two infinitely stiff plates being joined by a bolt. The two plates are forced apart, stretching the bolt to failure. When the bolt breaks, there will be no stored energy in the plates (right?), so will the bolt be launched out? Where does the elastic strain energy go?
Scary!
10. Mar 6, 2014
### Cowpoke
I would imagine the resulting force is a result of both the bolt's and the plate's potential energies being released simultaneously. The material of which the plate was constructed, and the tensile strength of the bolt could possibly give you the numbers needed to calculate how much each contributed to the bolt's acceleration.
Jim
11. Mar 6, 2014
### jstluise
I can totally understand how the potential energy in the plate could cause the bolt to accelerate and shoot out; the plate is essentially a spring in compression that is all of a sudden released when the bolt breaks.
But, when the bolt is being pulled apart, it is always in tension like an extension spring. So why would the energy in the bolt cause it to shoot outwards? The only thing I can think of would be the reaction happening between the bolt head and the plate; as the broken piece of the bolt contracts toward the head of the bolt and it is this momentum that accelerates the bolt. Think of shooting a rubber band off the tip of your finger; your finger is acting as the plate, the loop of rubber band around your finger is acting as the bolt head, and your release hand is acting as the bolt fracturing.
12. Mar 6, 2014
### Cowpoke
aah yes it seems I was thinking unidirectionally. Now this is bugging me as well.
13. Mar 6, 2014
### jstluise
The analogy I made with the rubber band shooting makes the most sense to me thus far. Assuming that the each half of the broken bolt are the same mass, and the plates are infinitely stiff, then the elastic potential energy in the bolt divided in half and each half of the bolt will be slingshotted out away from eachother.
Now considering the stiffness of the plates, and I get a bit confused. Imagine the point right before the bolt is going to break...you have the stored elastic potential energy in the bolt, but now you have stored elastic potential energy in the plates. So that means there is more energy in the system? i.e. you did more work to both stretch the bolt AND compress the plate? The dynamics get a bit more messy, but I imagine a model could be made with just masses and springs prior to release (fracture).
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2017-08-16 13:45:01
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https://ltwork.net/the-hcf-of-smallest-2-digit-composite-number-and-smallest--7775187
|
# The HCF of smallest 2 digit composite number and smallest 3 digit prime number is
###### Question:
The HCF of smallest 2 digit composite number and smallest 3 digit prime number is
### The organized list below shows all possible outcomes when four fair coins are flipped.Each coin lands
The organized list below shows all possible outcomes when four fair coins are flipped. Each coin lands facing either heads up (H) or tails up (T). H HT HTHH HHTT HHTH HTHT HTTH HT T TH THTT TTHH TTHT THTH THHT TH What is the probability that at least three coins land facing tails up? Give the probab...
### BIOLOGY HOMEWORK HELP ASAP
BIOLOGY HOMEWORK HELP ASAP $BIOLOGY HOMEWORK HELP ASAP$...
### HELP ME AND THANK YOU IF YOU HELP ME
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### Out of 32 students in a class five said they ride their bikes to school based on these results how many
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### What is the area of a regular triangle with a side length of 17 and an apothem of 4? If necessary round
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### An expression is shown below: 2x3y + 18xy - 10x2y - 90y part a: rewrite the expression so that the gcf
An expression is shown below: 2x3y + 18xy - 10x2y - 90y part a: rewrite the expression so that the gcf is factored completely. part b: rewrite the expression completely factored. show the steps of your work. part c: if the two middle terms were switched so that the expression became 2x3y - 10x2...
### Select the response that best answers the question. which of the following is a word borrowed from the
Select the response that best answers the question. which of the following is a word borrowed from the spanish language and used in the english language? el poncho el sándwich el champú el zíper which of the following is a word borrowed from the english language and used in the spanish language?...
### Why does the question 'what are the causes and effects of natural disasters' relate to the article called planning an explosive
Why does the question "what are the causes and effects of natural disasters" relate to the article called planning an explosive study of mount st. helens...
### You and your friends are planning a party. you are in charge of delegating all responsibilities to the group. use formal
You and your friends are planning a party. you are in charge of delegating all responsibilities to the group. use formal ustedes commands to tell everyone what to do in a minimum of 5 sentences. you must include five of the following verbs in the formal ustedes command form: ir, poner, traer, decor...
### Guess the song:Are you one of them girls?That peels off the Bud-Light labelJust might run a pool tableRoll your eyes if I call you
Guess the song: Are you one of them girls? That peels off the Bud-Light label Just might run a pool table Roll your eyes if I call you an angel...
### In a group 40% like football ,70% like cricket and 30% like both games
In a group 40% like football ,70% like cricket and 30% like both games...
### Write a Java program that can maintain information about a personalbookshelf. The program asks for information about a new book suchthe title, author
Write a Java program that can maintain information about a personal bookshelf. The program asks for information about a new book suchthe title, author and price. Use a class to hold book information, and twomethods one to get information from the user, second one to printinformation in the text fil...
### Of mice and men candy character favorite things?
Of mice and men candy character favorite things?...
### February revolution
February revolution...
### What is trepanning, bloodletting and cauterisationPLS HELP IT'S DUE TMRW
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### If the garbage truck and minivan in Part A get into an accident with each other, how can safety restraints in a car can save a life? Explain
If the garbage truck and minivan in Part A get into an accident with each other, how can safety restraints in a car can save a life? Explain your response using one of Newton’s laws. Which of Newton’s laws of motion act upon the vehicles at the point of impact? Explain your answer....
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2022-12-08 16:31:34
|
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|
https://leanprover-community.github.io/archive/stream/113488-general/topic/rec_on.20works.20in.20tactic.20mode.2C.20but.20fails.20in.20a.20term.html
|
## Stream: general
### Topic: rec_on works in tactic mode, but fails in a term
#### Chris Wong (Jul 23 2020 at 11:01):
The following code works:
import data.list.basic
variables {α : Type*}
open list
inductive palindrome : list α → Prop
| zero : palindrome []
| one : ∀ a, palindrome [a]
| many : ∀ a {l}, palindrome l → palindrome (a :: (l ++ [a]))
theorem to_reverse_eq {l : list α} (p : palindrome l) : reverse l = l :=
begin
refine p.rec_on _ _ _,
{ refl },
{ intros a, refl },
{ intros a l p h, simp [h] },
end
But when I replace the begin ... end block with:
p.rec_on rfl (λ a, rfl) (λ a l p h, by simp [h])
I get this error:
type mismatch at application
p.rec_on rfl (λ (a : α), rfl)
term
λ (a : α), rfl
has type
∀ (a : α), ?m_2[a] = ?m_2[a]
but is expected to have type
∀ (a : α), nil = [a]
What's the reason for this discrepancy?
#### Shing Tak Lam (Jul 23 2020 at 11:04):
Change it to not use projection notation, so palindrome.rec_on p instead of p.rec_on then it works. There's an explanation somewhere, I'll try to find it
theorem to_reverse_eq {l : list α} (p : palindrome l) : reverse l = l :=
palindrome.rec_on p rfl (λ a, rfl) (λ a l p h, by simp [h])
#### Shing Tak Lam (Jul 23 2020 at 11:15):
Found it
You can't use projection notation with @[elab_as_eliminator] functions
https://leanprover.zulipchat.com/#narrow/stream/113489-new-members/topic/Natural.20Numbers.20Game/near/199965854
#### Chris Wong (Jul 23 2020 at 11:25):
Thanks! That does seem rather obscure.
I wonder if we can special case this – maybe warn when projection notation is used with an @[elab_as_eliminator]? Or even outright ban them, if it's never a good idea to call them this way.
#### Mario Carneiro (Jul 23 2020 at 11:32):
I think it's a bug in lean somewhere. There isn't any principled reason for it to be like that AFAICT
Last updated: May 10 2021 at 00:31 UTC
|
2021-05-10 02:33:56
|
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|
https://or.stackexchange.com/questions/9789/solver-for-convex-optimization-with-exponent-in-the-objective-function
|
# Solver for convex optimization with exponent in the objective function
My question is related to a previous one: Dedicated solver for convex problems
To minimize a convex function of the form $$~f(x_i) = \left[ C + mx_i + \frac{s}{x_i+t} \right]^p$$
with various parameters $$~C$$, $$m$$, $$s$$, $$t~$$ and $$~p$$, $$~$$where $$~0 \le x_i \le 1~$$ and $$~p \ge 1$$.
(By generating a graph of the function, I ensure that the function $$f(x_i)$$ is convex. $$~$$The constraints are all linear.)
Edit (5:30am GMT January 25): Apologies for not bringing this up earlier, since I assumed that it wouldn't make a difference. The variable actually is a vector $$X =$$ ($$x_1$$, $$x_2$$, $$\cdots$$, $$x_n$$).. The objective function is:
Minimise $$~Z(X) = \sum_{i=1}^n Z_i = \sum_{i=1}^n f(x_i)$$.
If each $$f(x_i)$$ is convex, then their sum $$Z(X)$$ is also convex.
My question is, which solvers are well-suited to minimize such a convex function (subject to linear constraints)?
Of all solvers at NEOS, only MINOS seems to come close, and even this is not perfect (sometimes returns errors for instances for which manually I'm able to obtain optimal feasible solutions).
• Could you please clarify: (1) x is the only variable, all the other letters are parameters; (2) Is x a single number? If so, you can forget about the exponent, let g(x) = C + mx + s/(x+t), and solve directly the equation g'(x) = 0, i.e., x = -t +-sqrt(s/m).
– Stef
Jan 24 at 14:26
• If x is not a real number, but a vector, then I don't know what s/(x + t) means. But if x is a real number, then I don't know what you mean when you say "the constraints are linear", as linear constraints in one dimension are nothing more than just setting an interval.
– Stef
Jan 24 at 14:35
• To Stef -- good point! Let me be honest! There are several $x_i$ (that is, $~x_i$, $~$1 $\le i \le n$).. The objective function is actually (Minimise) $Z$ $= \sum_i Z_i$ $= \sum_i (~C + mx_i + s/(x_i + t) ~)^p$.. So I have linear constraints that connect different $x$'s, such as $x_i$ + $x_j$ $\le K_{i,j}$, etc.. If each $Z_i$ is convex, then the entire sum $Z$ is also convex. (I assume that the answers provided so far don't change with this extra info?) Jan 24 at 20:35
• Ah, it makes a lot more sense then! I suggest editing your question to include what you just said in a comment. As for Henrik Friberg's answer below, the way I understand it, it looks like it assumed that the objective function was f(x) and x was a real number and one of the variables, but there might be other variables that were involved in the constraints without being involved in the objective function. Note that the answer emits a reserve: "Hence, if f(x) really is all there is to your objective function".
– Stef
Jan 24 at 23:05
• The answers of Oscar Dowson and Sutanu now don't make much sense in the new light that x is a vector and the objective function is a sum of functions of the x_i.
– Stef
Jan 24 at 23:05
## 3 Answers
You are missing some details for the claimed convexity of $$f(x)$$. I will assume that you meant to say convex on the domain $$C+mx+\frac{s}{x+t} \geq 0$$ and $$x+t \geq 0$$. In this case the epigraph of $$f(x)$$ has conic normal form: $$\begin{array}{ll} &f \geq \left(C+mx+\frac{s}{x+t}\right)^p\\\Leftrightarrow &f \geq \left(C+mx+y\right)^p,\quad y \geq \frac{s}{x+t},\\\Leftrightarrow &f \geq z^p,\quad z=C+mx+y,\quad y \geq \frac{s}{x+t},\\\Leftrightarrow &f \geq |z|^p,\quad z=C+mx+y \geq 0,\quad y (x+t) \geq s. \end{array}$$
where the last deduction adds in the missing convexity details. You can now input the representation in any Disciplined Convex Programming language using one power cone for $$f \geq |z|^p$$ (see, e.g, https://docs.mosek.com/modeling-cookbook/powo.html#powers), and one rotated quadratic cone for $$y (x+t) \geq s$$ (see, e.g., https://docs.mosek.com/modeling-cookbook/cqo.html#rotated-quadratic-cones). These two cones are so simple that their presence will hardly affect the interior-point algorithm from a complexity point of view.
As a last remark, note that minimizing $$\left(C+mx+\frac{s}{x+t}\right)^p$$ on $$p\geq 1$$ is equivalent to just minimizing $$C+mx+\frac{s}{x+t}$$. Hence, if $$f(x)$$ really is all there is to your objective function, you will get the same optimal solution by just minimizing $$z$$. This will allow you to get rid of the epigraph variable $$f$$ and the power cone that constrain it.
As to the question of which solver to use with Disciplined Convex Programming, I refer to this answer to your previous question: https://or.stackexchange.com/a/2727.
• Thanks.. Yes, you're correct -- certainly we need $~\left(C + mx + \frac{s}{x+t} \right) \ge 0~$.. Just to make sure -- your $t$ (on the left side), is it the same as my $t$ (on the right side)? $~$Also, any reason why $~y \ge \frac{s}{x+t}$, and not $~y = \frac{s}{x+t}~$? Jan 24 at 4:08
• @ShuxueJiaoshou (1) Yes that variable should not have been called $t$. I changed it so the function $f(x)$ is now represented by the epigraph variable $f$. (2) You want $y=\frac{s}{x+t}$ but that is nonconvex. Fortunately, the convex relaxtion $y \geq \frac{s}{x+t}$ is exact because $(C+mx+y)^p$ is nondecreasing in $y$ (see docs.mosek.com/modeling-cookbook/…). (3) Great achievement, the DCP rules are indeed powerful. I can highly recommend MOSEK. Jan 24 at 8:25
• @HenrikAlsingFriberg, would you please, what's happened for parameter $t$ if we would replace it with the new variable $t'$? We should simply change the parameter to the variables!!? Jan 25 at 10:20
• @A.Omidi: In my reformulation it doesn't matter if $t$ is a parameter or a variable as long as the convexity requirement, $x+t \geq 0$, is valid for the problem you are trying to solve. Jan 25 at 13:09
• @HenrikAlsingFriberg, thanks for your explanation. 🙏 Jan 26 at 9:43
Use Ipopt and a modeling language. For example, JuMP: https://jump.dev/JuMP.jl/stable/tutorials/nonlinear/introduction/
julia> using JuMP, Ipopt
julia> model = Model(Ipopt.Optimizer)
A JuMP Model
Feasibility problem with:
Variables: 0
Model mode: AUTOMATIC
CachingOptimizer state: EMPTY_OPTIMIZER
Solver name: Ipopt
julia> set_silent(model)
julia> @variable(model, 0 <= x <= 1)
x
julia> C, m, s, t, p = 1, 1, 1, 0.1, 3
(1, 1, 1, 0.1, 3)
julia> @NLobjective(model, Min, (C + m * x + s / (x + t))^p)
julia> optimize!(model)
julia> value(x)
0.8999994186940709
julia> objective_value(model)
24.389000000008522
julia> (C + m * value(x) + s / (value(x) + t))^p
24.389000000008522
Best way is to apply log to the function as optimal solution to log of a function is the same for the function.
$$\log f = p[\log (C(x+t)+ mx(x+t)+s)-\log(x+t)]$$
Then you have to get the actual solution by the exp of $$\log$$ base used (e, 2 or 10).
However log functions need to be transformed into auxillary variables. You can also use the exp with $$p$$ but these also have to expressed as additional variables using addition constraints. Here is the link to suggested piecewise solution by Gurobi. Gurobi provides constraints as model methods like addGenConstrLog() and addGenConstrExp(). These are available at Gurobi's reference guide. Solvers won't have a problem with 2nd degree $$x^2$$
• Doesn't it require log(f(x)) to be convex on the domain for this to work? Jan 23 at 9:53
• @HenrikAlsingFriberg: log preserves convexity Jan 23 at 16:26
• Thanks, but one worry is that taking $\log$ weakens the degree of convexity.. However as Henrik said in his answer, the exponent $p$ may be unnecessary for the minimisation. Jan 23 at 23:00
• I don't think log weakens convexity, otherwise max loglikelihood wouldn't have used log to optimize in logistic/probit regression. Also not sure if $f(x)=x$ and $f(x)=x^2$ will have same solution for $x=[-2,2]$ Jan 23 at 23:38
• @Sutanu: You are able to use log-transformation of the likelihood objective because $\log(x)$ is strictly increasing (doesn't change optimal solution set), and you prefer to do it because log-likelihood is concave and likelihood alone isn't (you want concave objectives when maximizing, and convex objectives when minimizing). Please post a new question on this forum if you need more details as the comment section isn't suited for proper explanation. Jan 24 at 8:59
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2023-03-20 13:29:34
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https://www.gamedev.net/forums/topic/693477-need-help-with-rastertek-tutorial-making-a-2d-font-overlay/
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# DX11 Need help with Rastertek tutorial - making a 2D font overlay
## Recommended Posts
Hi guys, anyone experienced with DX11 could look at my graphics.cpp class? I got fonts rendering correctly
with painters algorithm - painting over the other 3d stuff each frame, however, whenever I turn the camera left or right,
the fonts get smushed narrower and narrower, then disappear completely.
It seems like the fix must be a very small change, untying their rendering from the cam direction,
but I just can't figure out how to do it under all this rendering complexity. Any tips would be helpful, thanks.
Edited by mister345
##### Share on other sites
When displaying text, you are not suppose to use the viewMatrix. Only the orthoProjection is needed but I think in this tutorial you will also need the basic world Matrix. The problem is that you are storing rotation inside your viewMatrix or any other type of modelMatrix and this cause to rotate your quad containing your text. It produces your bad effects, get narrow ...
Hope it helps
Edited by Thibault Ober
##### Share on other sites
On 11/12/2017 at 10:20 PM, Thibault Ober said:
When displaying text, you are not suppose to use the viewMatrix. Only the orthoProjection is needed but I think in this tutorial you will also need the basic world Matrix. The problem is that you are storing rotation inside your viewMatrix or any other type of modelMatrix and this cause to rotate your quad containing your text. It produces your bad effects, get narrow ...
Hope it helps
So you're saying in the following function, dont pass in the view matrix right? What matrix should I pass in instead, just identity?
SetShaderParameters(deviceContext, worldMatrix, identity /*viewMatrix*/, projectionMatrix, texture);
##### Share on other sites
I don't know how you handle your mouse. But you should put the resulting rotation somewhere (in a matrix). In order to draw your quad properly (in front) you shouldn't try to use them.
I will recommend to take in to account only the orthoProjection Matrix you have.
SetShaderParameters(deviceContext, identity, identity /*viewMatrix*/, projectionMatrix, texture);
projectionMatrix i think is computed with an orthoProjection. Try the code above and tell us your result.
good luck
Edited by Thibault Ober
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• By owenjr
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• Hello guys,
I've released my game for the first time. I'm very excited about it and I hope you'll enjoy the game - Beer Ranger. It's a retro-like puzzle-platfromer which makes you think a lot or die trying. You have a squad of skilled dwarfs with special powers and your goal is tasty beer. There is a lot of traps as well as many solutions how to endure them - it is up to your choice how to complete the level!
Link to the project: Project site
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Some screens:
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2018-02-25 00:20:30
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https://socratic.org/questions/590a40087c01491a7ff53dfe
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# What is the equation of a parabola, whose vertex is (0,0) and directrix is x=-3?
Equation of parabola is ${y}^{2} = 12 x$
As directrix is $x = - 3$ and vertex is $\left(0 , 0\right)$, the equation is of type ${y}^{2} = 4 a x$
In such an equation of parabola, directrix is $x = - a$. Hence $a = 3$ and equation of parabola is ${y}^{2} = 12 x$
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2019-09-16 20:39:47
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https://www.biostars.org/p/44378/
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Filtering A Sam File For Quality Scores
2
5
Entering edit mode
10.0 years ago
Varun Gupta ★ 1.2k
Hi I have a sam file. The 5th column of the sam file tells us the quality score. In my sam alignment file there are reads which have quality score of 2. I would like to remove them from my sam file because i think these reads which have a low quality score are not that important in my analysis. I want to keep the headers of sam file too.
Any code or unix one liner would be appreciated.
Thanks for the help
Regards Varun
sam • 39k views
12
Entering edit mode
10.0 years ago
You can use -q paramter, from the man page and provide a custom threshold to it
-q INT Skip alignments with MAPQ smaller than INT [0]
So, for a bam file the code would be samtools view -bq 2 file.bam > filtered.bam and for a sam, it is
output in bam
samtools view -bSq 2 file.sam > filtered.bam
output in sam
samtools view -Sq 2 file.sam > filtered.sam
Cheers
2
Entering edit mode
Is it not possible get rid of the chaining for the quality filtering?
samtools view -bSq 2 file.sam > filtered.bam
0
Entering edit mode
I tried the code in a jiffy, and for sure missed a parameter, so had to put the pipe. Of course, it works, I'll edit again :)
0
Entering edit mode
thnks sukhdeep works smoothly cheers
0
Entering edit mode
No worries, could you also edit the question to something like Filtering a Sam file for quality scores Cheers
3
Entering edit mode
10.0 years ago
I wouldn't worry about removing them. If you are variant calling they will be filtered.
You can also use samtools view -f 3 to only include reads that mapped and have their mate. This will reduce the size of the SAM file if that is a concern.
if you really want to remove them.
awk '$5 > x {print}' your.sam where x is the minimum value. ADD COMMENT 3 Entering edit mode the print is superflous, you can also do awk '$5 > x || \$1 ~ /^@/'
0
Entering edit mode
Thanks. I didn't know you could skip the print statement.
0
Entering edit mode
Hi I tried that, but using that command does not give the headers(does it??). I need the headers(starting with @) in my output as well
Regards
3
Entering edit mode
Yes, using awk or grep or something like that will usually strip the headers, though they can be added on again with samtools reheader. It's usually best to use samtools view where possible, and samtools view can filter on mapping quality, as you want it to.
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2022-05-22 13:49:16
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https://math.stackexchange.com/questions/1258515/mordell-weil-rank-in-elliptic-surfaces
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Mordell-Weil rank in elliptic surfaces
Suppose that an elliptic smooth K3 surface $X$ defined over a number field $k$ has arithmetic Picard rank $r$ and assume that it is equipped with a $k$ fibration over $\mathbb{P}^1$ that has a section over $k$, i.e. $f:X\to \mathbb{P}^1$ is a dominant map with the generic fiber $f^{-1}(\eta)$ being an elliptic curve. Is there a relation between the Mordell-Weil rank of the generic fiber and $r$ ?
• Just to be clear, do you know the answer in case $k$ is instead something algebraically closed? – user64687 Apr 30 '15 at 5:25
• No, not really:( – Captain Darling Apr 30 '15 at 12:41
• OK, let me try to write something that applies in that case, and then maybe you can see how to adapt it to your setting... – user64687 May 1 '15 at 16:49
• Maybe just an explicit example would be ok ? – Captain Darling May 1 '15 at 19:13
Note: let $k$ be a field and let $C/k$ be a smooth projective curve defined over the field $k$ and has genus $g$. The function field of $C/k$ will be denoted by $K=k(C)$. An elliptic surface $\mathcal{E}$ over the curve $C$ is, by definition, a two-dimensional projective variety together with (i) a morphism $\pi: \mathcal{E} \to C$ such that for all but finitely many points $t\in C(\overline{k})$, the fiber $\mathcal{E}_t=\pi^{-1}(t)$ is a non-singular curve of genus $1$, and (ii) a section to $\pi$ (the zero section) $\sigma_0: C \to \mathcal{E}$. With this definition, the generic fibre of $\pi$, $E/K$, may be regarded as an elliptic curve over the field $K$.
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2019-10-19 19:47:19
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http://www.linuxjournal.com/article/7657?page=0,1&quicktabs_1=2
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# At the Sounding Edge: LilyPond, Part 1
An introduction to this music notation software for Linux.
Putting in the Pond
LP's developers state that building the program from source code is a complicated process, and they advise installing a pre-built package, conveniently providing a list of available packages on the LilyPond Web site. Packages currently are available for Red Hat, Debian, Mandrakelinux and Slackware systems. The common source tarball also is available. However you install it, once installation is complete you can start using LP.
The following examples assume LP in version 2.2.0 or higher. LP is a moving target with a brisk development pace; your mileage may vary if you run the examples with another version.
Jumping In
The LilyPond documentation includes a good tutorial guide to the basics of using the language interface. The following code is a fairly simple example of a LilyPond source file:
% This is a comment.
\score {
\notes { c'4 e' g' b'}
}
In this example, the \score element prepares a default 5-line staff with treble clef and common (4/4) time signature. Its braces hold the data to be represented in the score. In this example, the score includes a sequence of notes that starts from middle C and arpeggiates a Cmaj7 chord in quarter notes. Pitch elements include pitch name (c), an octave specifier (') and a duration value (4). If no duration value is given for the next element, it takes its duration from the preceding note.
To create a PostScript graphics file from this code, simply process it with the LP compiler, as follows:
lilypond lj-ex-1.ly
Figure 1 shows the output file, lj-ex-1.ps, as displayed by GhostView.
Figure 1. Basic LilyPond
Now we add a few more language elements to create something a little more complete:
title = "Simple"
composer = "DLP 2004"
}
\score {
\notes { c'4 e' g'8 a'8 b'4 c''1 \bar "|."}
\midi {\tempo 4=132}
\paper { }
}
The new elements include a header block, further specifications for rhythm and pitch, a specifier for a double-bar and directives for MIDI output (with tempocontrol) and LilyPond's default print-ready formats. Figure 2 displays the compiled output, again in the GhostView PostScript file viewer.
Figure 2. More Symbols and Signs
Notation-savvy musicians may notice that I did not specify a bar line expected between b' and c''. As advertised, LilyPond automates many aspects of the score layout but always allows the possibility of customizing the output to virtually any degree.
As a last code example, here's a complex fragment in two staves with considerably more complicated rhythms:
\version "2.2.0"
title = "Toccata Vivace"
subtitle = "For Flute and Bassoon"
composer = "DLP 2004"
}
Flute = \notes \context Voice = Flute {
\set Staff.instrument = "Flute"
\set Staff.midiInstrument = "flute"
\key c \major
\clef treble
\time 3/4 \partial4 r4 | r8. g''16 -\staccato ges'' -\staccato f'' -\staccato b'-\mf -\accent fis''-\accent r e'-\sf r8 |
r4 \times 2/3 { ees'16 -\staccato d'' -\staccato des'' } f''8\< ~ f''4\! \bar "||"
}
Bassoon = \notes \context Voice = Bassoon {
\set Staff.instrument = "Bassoon"
\set Staff.midiInstrument = "bassoon"
\key c \major
\clef bass
\time 3/4 \partial4 r16 cis'-\mf -\staccato c' -\staccato d' -\staccato |
f-\accent e'-\accent r a,-\sf r4 r8 \times 2/3 { a,16 -\staccato g -\staccato cis'\< ~ } |
cis'4\! \>~ \times 2/3 { cis'8\! ( b bes, } d,4-\accent) \bar "||"
}
FluteStaff = \context Staff = FluteStaff <<
\Flute
>>
BassoonStaff = \context Staff = BassoonStaff <<
\Bassoon
>>
\score {
<<
\FluteStaff
\BassoonStaff
>>
\paper { }
\midi {\tempo 4 = 160}
}
Figure 3 displays the PostScript output. With a little thought you should be able to figure out the logic of the code. One point of assistance: LP uses 'is' to indicate a sharp, 'es' to indicate a flat. The other new elements are fairly self-explanatory, especially with reference to Figure 3, and I refer the interested reader to LilyPond's documentation to clarify any remaining obscurity.
Figure 3. A More Complicated Score Engraved by LP
______________________
Similis sum folio de quo ludunt venti.
## Comment viewing options
### Re: At the Sounding Edge: LilyPond, Part 1
Musical Notation has not only evolved over centuries, it is also facing numerous dilemmas: Instruction to the performer vs. compositional concept, difficulties with tuning systems, and so on. I own about 6 kilograms of books about the subject, so I am not surprised that there can be fiery discussions about notation programs.
My favourite tools for notating music will always be some sheets of paper and a 4b pen, and I will never touch an eraser when composing or arranging.
I agree, music notation has become a wildly tangled garden, and it is indeed a difficult thing to come up with a program that could satisfy all possible demands.
Btw, your last comment reminded me of Morton Feldman's statement that he always composed with pen & ink. He claimed it made him really think about whether he should write down what he was considering...
Best,
dp
### Re: At the Sounding Edge: LilyPond, Part 1
An archive of folk tunes in LilyPond format is at http://sniff.numachi.com/~rickheit/dtrad/.
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2017-04-27 06:52:58
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http://moestuininfo.com/diy-twin-mdsar/in-the-polynomial-function-the-coefficient-of-is-92a35c
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The leading term of this polynomial 5x 3 − 4x 2 + 7x − 8 is 5x 3. For the function $f\left(x\right)$, the highest power of $x$ is $3$, so the degree is $3$. Learn how to find the degree and the leading coefficient of a polynomial expression. The highest power of the variable that occurs in the polynomial is called the degree of a polynomial. Degree, Leading Term, and Leading Coefficient of a Polynomial Function. there, done. 1. For Example: For the polynomial we could rewrite it in descending order of exponents, to get which makes clear that as the ‘leading coefficient’ of . I'm trying to write a function that takes as input a list of coefficients (a0, a1, a2, a3.....a n) of a polynomial p(x) and the value x. Coefficient of x: If we refer to a specific variable when talking about a coefficient, we are treating everything else besides that variable (and its exponent) as part of the coefficient. Share. Coefficient of x in 14x 3 y is 14y. A polynomial in one variable is a function . what is the polynomial function of the lowest degree with lead coefficient 1 and roots 1 and 1+i? For the function $g\left(x\right)$, the highest power of $x$ is $6$, so the degree is $6$. The Degree of a Polynomial. . Show that the coefficient of $[x^nu^m]$ in the bivariate generating function $\\dfrac{1}{1-2x+x^2-ux^2}$ is ${n+1\\choose n-2m}.$ I tried to do this by using the … A polynomial containing three terms, such as $-3{x}^{2}+8x - 7$, is called a trinomial. Ask Question Asked 4 years, 9 months ago. The leading term is the term containing that degree, $-4{x}^{3}$. Polynomials. 3 8 4 π. Here is a typical polynomial: Notice the exponents (that is, the powers) on each of the three terms. e. The term 3 cos x is a trigonometric expression and is not a valid term in polynomial function, so n(x) is not a polynomial function. Functions are a specific type of relation in which each input value has one and only one output value. A family of nth degree polynomial functions that share the same x-intercepts can be defined by f(x) = — — a2) (x — an) where k is the leading coefficient, k e [R, k 0 and al, a2,a3, , zeros of the function. A polynomial with one variable is in standard form when its terms are written in descending order by degree. The first two functions are examples of polynomial functions because they contain powers that are non-negative integers and the coefficients are real numbers. About It Sketch the graph of a fifth-degree polynomial function whose leading coefficient is positive and that has a zero at x=3 of multiplicity 2. Use the degree of the function, as well as the sign of the leading coefficient to determine the behavior. Here, is the th coefficient and . always. In the latter case, the variables appearing in the coefficients are often called parameters, and must be clearly distinguished from the other variables. f (x) = x4 - 3x2 - 4 f (x) = x3 + x2 - 4x - 4 Which second degree polynomial function has a leading coefficient of - 1 and root 4 with multiplicity 2? Often, the leading coefficient of a polynomial will be equal to 1. The leading term is the term containing that degree, $6{x}^{2}$. The highest power of $x$ is $2$, so the degree is $2$. 10x: the coefficient is 10. It tells us that the number of positive real zeroes in a polynomial function f(x) is the same or less than by an even numbers as the number of changes in the sign of the coefficients. The degree of the polynomial is the power of x in the leading term. In the following video, you will see additional examples of how to identify a polynomial function using the definition. e. The term 3 cos x is a trigonometric expression and is not a valid term in polynomial function, so n(x) is not a polynomial function. Generally, unless … In any polynomial, the degree of the leading term tells you the degree of the whole polynomial, so the polynomial above is a "second-degree polynomial", or a "degree-two polynomial". We call the term containing the highest power of x (i.e. In mathematics, a coefficient is a multiplicative factor in some term of a polynomial, a series, or any expression; it is usually a number, but may be any expression (including variables such as a, b and c). It is often helpful to know how to identify the degree and leading coefficient of a polynomial function. Note that the second function can be written as $g\left(x\right)=-x^3+\dfrac{2}{5}x$ after applying the distributive property. Find all coefficients of a polynomial, including coefficients that are 0, by specifying the option 'All'. 9. How many turning points can it have? Find an answer to your question “In the polynomial function below what is the leading coefficient f (x) = 1/4x^5+8x-5x^4-19 ...” in Mathematics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.“In the polynomial function below what Coefficient. This means that m(x) is not a polynomial function. Each product ${a}_{i}{x}^{i}$ is a term of a polynomial. Identify the degree, leading term, and leading coefficient of the polynomial $4{x}^{2}-{x}^{6}+2x - 6$. a. f(x) = 3x 3 + 2x 2 – 12x – 16. b. g(x) = -5xy 2 + 5xy 4 – 10x 3 y 5 + 15x 8 y 3 To learn more about polynomials, terms, and coefficients, review the lesson titled Terminology of Polynomial Functions, which covers the following objectives: Define polynomials … When a polynomial is written so that the powers are descending, we say that it is in standard form. Notice that these quartic functions (left) have up to three turning points. The degree of a polynomial in one variable is the largest exponent in the polynomial. Possible degrees for this graph include: Negative 1 4 and 6. In other words roots of a polynomial function is the number, when you will plug into the polynomial, it will make the polynomial zero. Or you could view each term as a monomial, as a polynomial with only one term in it. What is sought is a theorem that says something to the effect that the coefficient sum of a function of a polynomial is the value of that function evaluated with the base of the polynomial set equal to the multiplicative identity. Because of the form of a polynomial function, we can see an infinite variety in the number of terms and the powers of the variables. Polynomial functions are sums of terms consisting of a numerical coefficient multiplied by a unique power of the independent variable. Viewed 3k times 10. \displaystyle 384\pi 384π, is known as a coefficient. Coefficients can be positive, negative, or zero, and can be whole numbers, … For the following polynomials, identify the degree, the leading term, and the leading coefficient. Four or less. The degree of a polynomial is the degree of the leading term. 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. Polynomials are algebraic expressions that are created by adding or subtracting monomial terms, such as $-3x^2$, where the exponents are only non-negative integers. Leading Coefficient (of a polynomial) The leading coefficient of a polynomial is the coefficient of the leading term. Polynomial function whose general form is f (x) = A x 2 + B x + C, where A ≠ 0 and A, B, C ∈ R. A second-degree polynomial function in which all the coefficients of the terms with a degree less than 2 are zeros is called a quadratic function. Polynomial functions are the addition of terms consisting of a numerical coefficient multiplied by a unique power of the independent variables. Coefficient[expr, form, n] gives the coefficient of form^n in expr. A polynomial function is made up of terms called monomials; If the expression has exactly two monomials it’s called a binomial.The terms can be: Constants, like 3 or 523.. Variables, like a, x, or z, A combination of numbers and variables like 88x or 7xyz. A polynomial function is a function that can be expressed in the form of a polynomial. Which of the following are polynomial functions? What is the polynomial function of lowest degree with lead coefficient 1 and roots i, - 2, and 2? R. = QQ[] List1= [x^(2), y^(2),z^(2)] List2= [x^(2)+y^(2)+z^(2), 3*x^(2),4*y^(2)] List3=[] For example if I do List2[0].coefficient(List1[0]), Sage immediately outputs 1. Which of the following are polynomial functions? Fill in the blanks. Cost Function of Polynomial Regression. positive or zero) integer and $$a$$ is a real number and is called the coefficient of the term. Because there i… Definition. Solved: Find the nth degree polynomial function having the following : n = 4, 2i, 7 and -7 are zeros; leading coefficient is 1. Recall that a polynomial is an expression of the form ax^n + bx^(n-1) + . The degree of this polynomial 5x 3 − 4x 2 + 7x − 8 is 3. So, in standard form, the degree of the first term indicates the degree of the polynomial, and the leading coefficient is the coefficient of the first term. The first term has an exponent of 2; the second term has an \"understood\" exponent of 1 (which customarily is not included); and the last term doesn't have any variable at all, so exponents aren't an issue. The degree of a polynomial is given by the term with the greatest degree. If the leading coefficient of a polynomial function is negative, then the left end of the graph ____ points down. The Linear Factorization Theorem tells us that a polynomial function will have the same number of factors as its degree, and that each factor will be in the form $$(x−c)$$, where c is a complex number. We can call this function like any other function: for x in [-1, 0, 2, 3.4]: print (x, p (x))-1 -6 0 0 2 6 3.4 97.59359999999998 import numpy as np import matplotlib.pyplot as plt X = np. Hello so I am using the .coefficient function to extract the coefficient of a monomial given some polynomial. Polynomials in one variable are algebraic expressions that consist of terms in the form $$a{x^n}$$ where $$n$$ is a non-negative (i.e. Terms. The degree of a polynomial in one variable is the largest exponent in the polynomial. If it is, write the function in standard form and state its degree, type and leading coefficient. The required Monic polynomial say p(x) has three zeros ; 1, (1+i) & (1-i). A polynomial is an expression that can be written in the form. Find all coefficients of 3x 2. ... Gradient descent is an optimization algorithm used to find the values of parameters (coefficients) of a function that minimizes a cost function … The result for the graphs of polynomial functions of even degree is that their ends point in the same direction for large | x |: up when the coefficient of the leading term is positive, down when the coefficient is negative. We generally represent polynomial functions in decreasing order of the power of the variables i.e. Finding the coefficient of the x² term in a Maclaurin polynomial, given the formula for the value of any derivative at x=0. Factors And Coefficients Of A Polynomial Factor: When numbers (constants) and variables are multiplied to form a term, then each quantity multiplied is called a factor of the term. If f is a polynomial function with real coefficients, and a+bi is an imaginary solution of f,then a-bi is also a zero of f. Descartes' Rule of Signs. A number multiplied by a variable raised to an exponent, such as. It's called a polynomial. Coefficient[expr, form] gives the coefficient of form in the polynomial expr. Since the leading coefficient is positive, the graph rises to the right. Coefficient[expr, form] gives the coefficient of form in the polynomial expr. . Watch the next video for more examples of how to identify the degree, leading term and leading coefficient of a polynomial function. Coefficients can be positive, negative, or zero, and can be whole numbers, decimals, or fractions. Summary. Just as we identified the degree of a polynomial, we can identify the degree of a polynomial function. 0. We can find the degree of a polynomial by identifying the highest power of the variable that occurs in the polynomial. In mathematics, a coefficient is a multiplicative factor in some term of a polynomial, a series, or any expression; it is usually a number, but may be any expression (including variables such as a, b … Learn how to write the equation of a polynomial when given complex zeros. Polynomial functions contain powers that are non-negative integers and the coefficients are real numbers. To do this, follow these suggestions: $\begin{array}{ccc}f\left(x\right)=5x^7+4\hfill \\ g\left(x\right)=-x^2\left(x-\dfrac{2}{5}\right)\hfill \\ h\left(x\right)=\dfrac{1}{2}x^2+\sqrt{x}+2\hfill \end{array}$, http://cnx.org/contents/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1/Preface, Determine if a given function is a polynomial function, Determine the degree and leading coefficient of a polynomial function, Identify the term containing the highest power of. The returned coefficients are ordered from the highest degree to the lowest degree. We can now find the equation using the general cubic function, y = ax3 + bx2 + cx+ d, and determining the values of a, b, c, and d. We can find the value of the leading coefficient, a, by using our constant difference formula. Solution for Write a polynomial function f of least degree that has rational coefficients, a leading coefficient of 1, and the zeros 1,3, and 2−i. The function is not a polynomial function because the term 3 x does not have a variable base and an … So those are the terms. I don't want to use the Coefficient[] function in Mathematica, I just want to understand how it is done. x 3 − 3x 2 + 4x + 10. If the highest exponent of a polynomial function is odd, then the range of the function is ____ all real numbers. More precisely, a function f of one argument from a given domain is a polynomial function if there exists a polynomial + − − + ⋯ + + + that evaluates to () for all x in the domain of f (here, n is a non-negative integer and a 0, a 1, a 2, ..., a n are constant coefficients). A polynomial is generally represented as P(x). The function will return p(x), which is the value of the polynomial when evaluated at x. A polynomial function with degree n and leading coefficient a_{n} is a function of the form f(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\\cdots+a_{2} x… Active 4 years, 8 months ago. x 3. The leading coefficient in the polynomial function ¾(4x⁵-2x)+2x³+3 is - 30035759 A polynomial containing two terms, such as $2x - 9$, is called a binomial. A constant factor is called a numerical factor while a variable factor is called a literal factor. This graph has _____turning point(s). The leading coefficient here is 3. Let $$f$$ be a polynomial function with real coefficients, and suppose $$a +bi$$, $$b≠0$$, is a zero of $$f(x)$$. In a polynomial function, the leading coefficient (LC) is in the term with the highest power of x (called the leading term). Since all the coefficients of the polynomials equal $1$ or $-1$ except for the polynomial expanded in $(3)$, we have as our coefficient $$\binom{21+3-1}{21} - \binom{6+3-1}{6} - \binom{5+3-1}{5} = 204$$ Note: I hadn't seen Andre's solution prior to typing this. Now let's think about the coefficients of each of the terms. Polynomial, In algebra, an expression consisting of numbers and variables grouped according to certain patterns.Specifically, polynomials are sums of monomials of the form ax n, where a (the coefficient) can be any real number and n (the degree) must be a whole number. Polynomial functions have all of these characteristics as well as a domain and range, and corresponding graphs. The Coefficient Sum of a Function of a Polynomial. Example 7. The leading coefficient in a polynomial is the coefficient of the leading term. A polynomial in the variable x is a function that can be written in the form,. To review: the degree of the polynomial is the highest power of the variable that occurs in the polynomial; the leading term is the term containing the highest power of the variable or the term with the highest degree. A function is a fifth-degree polynomial. If the coefficients of a polynomial are all integers, and a root of the polynomial is rational (it can be expressed as a fraction in lowest terms), the Rational Root Theorem states that the numerator of the root is a factor of a0 and the denominator of the root … 1. Simple enough. Positive. 15x 2 y: the coefficient is 15. 1. 8. sometimes. List all possible rational zeros of f(x)=2 x 4 −5 x 3 + x 2 … Coefficient of polynomials is the number multiplied to the variable. The leading coefficient of that polynomial is 5. Listing All Possible Rational Zeros. The sign of the leading coefficient for the polynomial equation of the graph is . ... Get Coefficient of polynomial excluding variables. In the first example, we will identify some basic characteristics of polynomial functions. Roots of second degree polynomial=4,4 because multiplicity 2 means roots are repeated two times . Polynomials in one variable are algebraic expressions that consist of terms in the form $$a{x^n}$$ where $$n$$ is a non-negative (i.e. Determine the degree of the following polynomials. The third function is not a polynomial function because the variable is under a square root in the middle term, therefore the function contains an exponent that is not a non-negative integer. All Coefficients of Polynomial. we will define a class to define polynomials. For real-valued polynomials, the general form is: p (x) = p n x n + p n-1 x n-1 + … + p 1 x + p 0. Root of a polynomial also known as zero of polynomial which means to find the root of polynomial we can set up the polynomial equal to zero to get the value ( root) of the variable. A polynomial’s degree is that of its monomial of highest degree. Determine if a Function is a Polynomial Function. The largest exponent is the degree of the polynomial. A polynomial function consists of either zero or the sum of a finite number of non-zero terms, each of which is a product of a number, called the coefficient of the term, and a variable raised to a non-negative integer power. where a n, a n-1, ..., a 2, a 1, a 0 are constants. a. f(x) = 3x 3 + 2x 2 – 12x – 16. b. g(x) = -5xy 2 + 5xy 4 – 10x 3 y 5 + 15x 8 y 3 We have introduced polynomials and functions, so now we will combine these ideas to describe polynomial functions. It is often helpful to know how to identify the degree and leading coefficient of a polynomial function. A number multiplied by a variable raised to an exponent, such as $384\pi$, is known as a coefficient. Polynomial can be employed to model different scenarios, like in the stock market to observe the way and manner price is changing over time. Which is the polynomial function of lowest degree with rational real coefficients, a leading coefficient of 3 and roots StartRoot 5 EndRoot and 2? Introduction. The coefficient is what's multiplying the power of x or what's multiplying in the x part of the term. The leading coefficient is the coefficient of that term, $6$. General equation of second degree polynomial is given by Descartes' rule of sign is used to determine the number of real zeros of a polynomial function. The leading term in a polynomial is the term with the highest degree . For Example: (i) 7, x and 7x are factors […] Poly, it has many terms. Decide whether the function is a polynomial function. Each product ${a}_{i}{x}^{i}$, such as $384\pi w$, is a term of a polynomial. $\begin{array}{lll} f\left(x\right)=5{x}^{2}+7-4{x}^{3} \\ g\left(x\right)=9x-{x}^{6}-3{x}^{4}\\ h\left(x\right)=6\left(x^2-x\right)+11\end{array}$. (image is √3) 2 See answers jdoe0001 jdoe0001 Reload the page, if you don't see above yet hmmmmm shoot, lemme fix something, is off a bit. The leading coefficient of a polynomial is the coefficient of the leading term. The term with the highest degree is called the leading term because it is usually written first. I have written an algorithm that given a list of words, must check each unique combination of four words in that list of words (regardless of order). Degree of a polynomial function is very important as it tells us about the behaviour of the function P(x) when x becomes v… For polynomial. This means that m(x) is not a polynomial function. We generally write these terms in decreasing order of the power of the variable, from left to right * . polynomials. A polynomial function is a function that can be defined by evaluating a polynomial. an are the We can use this general equation to find the equation of a family of polynomial functions with a given set of zeros. The leading term is the term containing that degree, $-{x}^{6}$. Example 6. Give the degree of the polynomial, and give the values of the leading coefficient and constant term, if any, of the following polynomial: 2x 5 – 5x 3 – 10x + 9 Leading coefficient is called the coefficient is what 's multiplying the power of the rises... Is often helpful to know how to identify a polynomial we generally write these terms in decreasing order of variable! 0 are constants which is the coefficient of a polynomial is the power of the leading term is coefficient! As P ( x ) is 3 functions, in the polynomial function the coefficient of is now we will identify and evaluate functions. Terms consisting of a polynomial ’ s degree is that of its monomial of highest degree to the variable it... Expressed in the polynomial is the polynomial when evaluated at x a n-1,,. Just as we identified the degree of a function is negative, or fractions repeated two times term does contain! Degree to the variable of P ( x ), which is coefficient. Coefficients that are non-negative integers and the coefficients in the polynomial function the coefficient of is ordered from the highest degree is called the of! Be expressed in the polynomial function is a function that can be written in decreasing order of variable... Exponents ( that is, write the equation of a polynomial function with leading coefficient in this case, say! [ latex ] -4 { x } ^ { 2 } [ /latex ] bx^ ( n-1 ) + degree! Because multiplicity 2 4 −5 x 3 + x 2 … polynomials represented as P ( x has..., which is the polynomial function is odd, then the left end of the leading coefficient of a by! And functions, so now we will combine these ideas to describe polynomial functions a numerical factor while a raised... To know how to identify a polynomial when given complex zeros turning points form in the following,! Input value has one and only one term in it multiplied to the right to variable! Will return P ( x ) has three zeros ; 1, ( 1+i ) (... Which each input value has one and only one term in a polynomial is written that... By using this website, you will see additional examples of how to identify the of., ( 1+i ) & ( 1-i ) x 3 − 4x 2 + 7x − 8 is 3. Return P ( x ) is a fifth-degree polynomial independent variable } ^ { 6 [., one takes some terms and adds ( and subtracts ) them together order by degree ) is a number. ) ) is not a polynomial equation of the three terms n ] gives the coefficient of term! A specific type of relation in which each input value has one and one. Monic polynomial say P ( x ) form ax^n + bx^ ( )! Coefficient 1 and roots I, - 2, and the coefficients are real numbers, one takes some and... Generally represent polynomial functions multiplying the power of x or what 's multiplying in the polynomial function the coefficient of is the following polynomials, identify degree! 5X 3 − 4x 2 + 7x − 8 is 3 3 } [ /latex.... 'S think about the coefficients are real numbers now we will combine these ideas to describe functions. Used to determine the number in the polynomial function the coefficient of is by a unique power of x, the leading coefficient in a is. Functions, so now we will identify and evaluate polynomial functions because they contain that. Call the term with the highest power, and we call the term variable is standard! Variable is the term n, a n-1,..., a n-1,..., a 0 are.! Given by the term with the highest power of the term containing that degree leading. One output value 384π, is called the leading term is the polynomial definition of a numerical coefficient multiplied a! ( that is, the graph is that it is usually written in the first example, we will and. 2, and corresponding graphs months ago ^ { 2 } [ /latex ] 7x 8! Know how to identify the degree of this polynomial 5x 3 is called a.. In descending order by degree factor is called a literal factor polynomial functions constant is. These quartic functions ( left ) have up to three turning points,! A cubic represented as P ( x ) =2 x 4 −5 x +. X 3 − 3x 2 + 7x − 8 is 3 monomial some... Term, [ latex ] –4 [ /latex ] have introduced polynomials and functions so. ) has three zeros ; 1, ( 1+i ) & ( 1-i ) well. Ax^N + bx^ ( n-1 ) + power, and we call term... That is, write the function will return P ( x ), leading term is the with! Are descending, we can identify the degree of a polynomial in one variable is term. Turning points be … it 's called a binomial does not contain a variable raised to an,. Specific type of relation in which each input value has one and only one in. If in the polynomial function the coefficient of is term does not contain a variable raised to an exponent such. Number multiplied by a unique power of x ( i.e [ /latex ] n the leading term is the containing! By degree be written in descending order by degree in other words, the nonzero coefficient of x in 3! From the highest exponent of a polynomial equation x or what 's multiplying the power the. Integers and the coefficients are ordered from the highest power of x in the form ax^n + (. … it 's called a binomial second degree polynomial=4,4 because multiplicity 2 means roots are repeated two.. And evaluate polynomial functions because they contain powers that are 0, by the. Polynomial equation is, the leading term have a monic polynomial ) on each of the polynomial expr turning.! Coefficient in the polynomial is an expression of the terms function of lowest degree with. The graph ____ points down polynomial in one variable is in standard form state. Multiplicity 2 terms with the highest power of the leading term x ( i.e, then the range of three! Words, the nonzero coefficient of the leading coefficient is the term an exponent, as! 30035759 a function that can be derived from the definition of a function negative. Addition of terms with the stated coefficient { 3 } [ /latex ] have. Characteristics of polynomial functions ; 1, a 1, a 1, 0. Of polynomials is the term left ) have up to three turning.... Two terms, such as [ latex ] -1 [ /latex ] s degree is called the leading of. Cookie Policy two functions are sums of terms consisting of a polynomial, we can find the second polynomial=4,4! Are ordered from the highest exponent of a polynomial, one takes some terms and (., so now we will identify some basic characteristics of polynomial functions are the addition of terms with highest... 2, a 2, and we call a n x n ) the leading term is the containing. Exponent, such as [ latex ] –4 [ /latex ] n-1 +... At x 3 − 4x 2 + 7x − 8 is 5x 3 factor..., and we call the term containing that degree, [ latex ] 2x - 9 [ /latex ] a... Characteristics of polynomial functions a number multiplied to the right number and called! Because they contain powers that are non-negative integers and the coefficients are real numbers these quartic functions ( left have... Are constants will combine these ideas to describe polynomial functions in decreasing order of the variables i.e degrees this... Its terms are written in descending order by degree it 's called a coefficient. Is that in the polynomial function the coefficient of is its monomial of highest degree is called a binomial use degree... Function that can be derived from the definition can be expressed in the polynomial is 5x 3 4x... The term only one term in a polynomial, one takes some terms and adds ( and )! The next video for more examples of how to identify the degree of a polynomial with one variable is term! Find the degree of the term exponent is the degree of a that! To the variable of P ( x ) has three zeros ; 1 a. Website, you agree to our Cookie Policy three terms Asked 4 years 9! Of x, the polynomial a n-1,..., a 1, a 1, a,! Means that m ( x ) =2 x 4 −5 x 3 − 4x +! Expression of the independent variables term is the coefficient of a polynomial function leading! We can identify the degree of a polynomial by identifying the highest power the. Polynomial 5x 3 − 4x 2 + 7x − 8 is 3 in standard form its! One variable is the term containing that degree, leading term graph include negative. 'S called a numerical coefficient multiplied by a variable, it is usually written in decreasing order of the x. Polynomials and functions, so now we will identify and evaluate polynomial functions the )... By degree 14x 3 y is 14y coefficient to determine the number multiplied by a variable factor called... Numbers, polynomials may be … it 's called a polynomial in one variable is in standard form graph. One output value a typical polynomial: Notice the exponents ( that,... More examples of how to identify a polynomial, including coefficients that are 0 by..., is called a polynomial is the number of real zeros of (... S degree is called a numerical coefficient multiplied by a unique power the... Website, you agree to our Cookie Policy the leading term is the number multiplied by a power...
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2022-12-06 06:36:29
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http://www.algebra.com/cgi-bin/show-question-source.mpl?solution=223851
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Question 313034
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Step 1: Determine the slope of the desired line. In this case, you can simply do this by inspection. That is because parallel lines have equal slopes and your given line is in slope-intercept form, namely: *[tex \Large y\ =\ mx\ +\ b]
Hence, the slope of the given line, which is the same as the slope we need for the desired line is just the coefficient on *[tex \Large x]
Step 2: Use the point-slope form of the equation of a line to write an equation of the desired line:
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ -\ y_1\ =\ m(x\ -\ x_1) ]
where *[tex \Large \left(x_1,y_1\right)] are the coordinates of the given point and *[tex \Large m] is the slope of the desired line. All you need do is plug in the numbers:
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ -\ (-1)\ =\ (-25)(x\ -\ (-1))]
Step 3: The only thing left to do is to put this equation into slope-intercept form so that it matches the answer you gave.
Distribute across the parentheses in the RHS:
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ +\ 1\ =\ -25x\ -\ 25]
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ -25x\ -\ 26]
Note that you cannot answer the question you asked exactly. That is because you wrote: "Write <b><i>the</i></b> equation of the line that is parallel..."
The reason you can't do that is because there is no such thing as <b><i>the</i></b> equation of a line. There are an infinite number of representations of any given line, so all you can do is to write <b><i>an</i></b> equation of the line you want.
Also, since you didn't specify the form of the answer, *[tex \LARGE y\ =\ -25x\ -\ 26] is not the only valid answer to this question. *[tex \LARGE y\ +\ 1\ =\ (-25)(x\ +\ 1)] and *[tex \LARGE 25x\ +\ y\ =\ -26] are both equally valid answers to the question <i>as you posed it.</i>
John
*[tex \LARGE e^{i\pi} + 1 = 0]
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2013-05-19 06:51:05
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https://handwiki.org/wiki/Linear_least_squares
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# Linear least squares
Linear least squares (LLS) is the least squares approximation of linear functions to data. It is a set of formulations for solving statistical problems involved in linear regression, including variants for ordinary (unweighted), weighted, and generalized (correlated) residuals. Numerical methods for linear least squares include inverting the matrix of the normal equations and orthogonal decomposition methods.
## Main formulations
The three main linear least squares formulations are:
• Ordinary least squares (OLS) is the most common estimator. OLS estimates are commonly used to analyze both experimental and observational data.
The OLS method minimizes the sum of squared residuals, and leads to a closed-form expression for the estimated value of the unknown parameter vector β: $\displaystyle{ \hat{\boldsymbol\beta} = (\mathbf{X}^\mathsf{T}\mathbf{X})^{-1} \mathbf{X}^\mathsf{T} \mathbf{y}, }$ where $\displaystyle{ \mathbf{y} }$ is a vector whose ith element is the ith observation of the dependent variable, and $\displaystyle{ \mathbf{X} }$ is a matrix whose ij element is the ith observation of the jth independent variable. The estimator is unbiased and consistent if the errors have finite variance and are uncorrelated with the regressors:[1] $\displaystyle{ \operatorname{E}[\,\mathbf{x}_i\varepsilon_i\,] = 0, }$ where $\displaystyle{ \mathbf{x}_i }$ is the transpose of row i of the matrix $\displaystyle{ \mathbf{X}. }$ It is also efficient under the assumption that the errors have finite variance and are homoscedastic, meaning that E[εi2|xi] does not depend on i. The condition that the errors are uncorrelated with the regressors will generally be satisfied in an experiment, but in the case of observational data, it is difficult to exclude the possibility of an omitted covariate z that is related to both the observed covariates and the response variable. The existence of such a covariate will generally lead to a correlation between the regressors and the response variable, and hence to an inconsistent estimator of β. The condition of homoscedasticity can fail with either experimental or observational data. If the goal is either inference or predictive modeling, the performance of OLS estimates can be poor if multicollinearity is present, unless the sample size is large.
• Weighted least squares (WLS) are used when heteroscedasticity is present in the error terms of the model.
• Generalized least squares (GLS) is an extension of the OLS method, that allows efficient estimation of β when either heteroscedasticity, or correlations, or both are present among the error terms of the model, as long as the form of heteroscedasticity and correlation is known independently of the data. To handle heteroscedasticity when the error terms are uncorrelated with each other, GLS minimizes a weighted analogue to the sum of squared residuals from OLS regression, where the weight for the ith case is inversely proportional to var(εi). This special case of GLS is called "weighted least squares". The GLS solution to an estimation problem is $\displaystyle{ \hat{\boldsymbol\beta} = (\mathbf{X}^\mathsf{T} \boldsymbol\Omega^{-1} \mathbf{X})^{-1}\mathbf{X}^\mathsf{T}\boldsymbol\Omega^{-1}\mathbf{y}, }$ where Ω is the covariance matrix of the errors. GLS can be viewed as applying a linear transformation to the data so that the assumptions of OLS are met for the transformed data. For GLS to be applied, the covariance structure of the errors must be known up to a multiplicative constant.
## Alternative formulations
Other formulations include:
• Iteratively reweighted least squares (IRLS) is used when heteroscedasticity, or correlations, or both are present among the error terms of the model, but where little is known about the covariance structure of the errors independently of the data.[2] In the first iteration, OLS, or GLS with a provisional covariance structure is carried out, and the residuals are obtained from the fit. Based on the residuals, an improved estimate of the covariance structure of the errors can usually be obtained. A subsequent GLS iteration is then performed using this estimate of the error structure to define the weights. The process can be iterated to convergence, but in many cases, only one iteration is sufficient to achieve an efficient estimate of β.[3][4]
• Instrumental variables regression (IV) can be performed when the regressors are correlated with the errors. In this case, we need the existence of some auxiliary instrumental variables zi such that E[ziεi] = 0. If Z is the matrix of instruments, then the estimator can be given in closed form as $\displaystyle{ \hat{\boldsymbol\beta} = (\mathbf{X}^\mathsf{T}\mathbf{Z}(\mathbf{Z}^\mathsf{T}\mathbf{Z})^{-1}\mathbf{Z}^\mathsf{T}\mathbf{X})^{-1}\mathbf{X}^\mathsf{T}\mathbf{Z}(\mathbf{Z}^\mathsf{T}\mathbf{Z})^{-1}\mathbf{Z}^\mathsf{T}\mathbf{y}. }$ Optimal instruments regression is an extension of classical IV regression to the situation where E[εi | zi] = 0.
• Total least squares (TLS)[5] is an approach to least squares estimation of the linear regression model that treats the covariates and response variable in a more geometrically symmetric manner than OLS. It is one approach to handling the "errors in variables" problem, and is also sometimes used even when the covariates are assumed to be error-free.
In addition, percentage least squares focuses on reducing percentage errors, which is useful in the field of forecasting or time series analysis. It is also useful in situations where the dependent variable has a wide range without constant variance, as here the larger residuals at the upper end of the range would dominate if OLS were used. When the percentage or relative error is normally distributed, least squares percentage regression provides maximum likelihood estimates. Percentage regression is linked to a multiplicative error model, whereas OLS is linked to models containing an additive error term.[6]
In constrained least squares, one is interested in solving a linear least squares problem with an additional constraint on the solution.
## Objective function
In OLS (i.e., assuming unweighted observations), the optimal value of the objective function is found by substituting the optimal expression for the coefficient vector: $\displaystyle{ S=\mathbf y^\mathsf{T} (\mathbf{I} - \mathbf{H})^\mathsf{T} (\mathbf{I} - \mathbf{H}) \mathbf y = \mathbf y^\mathsf{T} (\mathbf{I} - \mathbf{H}) \mathbf y, }$ where $\displaystyle{ \mathbf{H}=\mathbf{X}(\mathbf{X}^\mathsf{T}\mathbf{X})^{-1} \mathbf{X}^\mathsf{T} }$, the latter equality holding since $\displaystyle{ (\mathbf{I} - \mathbf{H}) }$ is symmetric and idempotent. It can be shown from this[7] that under an appropriate assignment of weights the expected value of S is m − n. If instead unit weights are assumed, the expected value of S is $\displaystyle{ (m - n)\sigma^2 }$, where $\displaystyle{ \sigma^2 }$ is the variance of each observation.
If it is assumed that the residuals belong to a normal distribution, the objective function, being a sum of weighted squared residuals, will belong to a chi-squared ($\displaystyle{ \chi ^2 }$) distribution with m − n degrees of freedom. Some illustrative percentile values of $\displaystyle{ \chi ^2 }$ are given in the following table.[8]
$\displaystyle{ m - n }$ $\displaystyle{ \chi ^2_{0.50} }$ $\displaystyle{ \chi ^2 _{0.95} }$ $\displaystyle{ \chi ^2 _{0.99} }$
10 9.34 18.3 23.2
25 24.3 37.7 44.3
100 99.3 124 136
These values can be used for a statistical criterion as to the goodness of fit. When unit weights are used, the numbers should be divided by the variance of an observation.
For WLS, the ordinary objective function above is replaced for a weighted average of residuals.
## Discussion
In statistics and mathematics, linear least squares is an approach to fitting a mathematical or statistical model to data in cases where the idealized value provided by the model for any data point is expressed linearly in terms of the unknown parameters of the model. The resulting fitted model can be used to summarize the data, to predict unobserved values from the same system, and to understand the mechanisms that may underlie the system.
Mathematically, linear least squares is the problem of approximately solving an overdetermined system of linear equations A x = b, where b is not an element of the column space of the matrix A. The approximate solution is realized as an exact solution to A x = b', where b' is the projection of b onto the column space of A. The best approximation is then that which minimizes the sum of squared differences between the data values and their corresponding modeled values. The approach is called linear least squares since the assumed function is linear in the parameters to be estimated. Linear least squares problems are convex and have a closed-form solution that is unique, provided that the number of data points used for fitting equals or exceeds the number of unknown parameters, except in special degenerate situations. In contrast, non-linear least squares problems generally must be solved by an iterative procedure, and the problems can be non-convex with multiple optima for the objective function. If prior distributions are available, then even an underdetermined system can be solved using the Bayesian MMSE estimator.
In statistics, linear least squares problems correspond to a particularly important type of statistical model called linear regression which arises as a particular form of regression analysis. One basic form of such a model is an ordinary least squares model. The present article concentrates on the mathematical aspects of linear least squares problems, with discussion of the formulation and interpretation of statistical regression models and statistical inferences related to these being dealt with in the articles just mentioned. See outline of regression analysis for an outline of the topic.
## Properties
If the experimental errors, $\displaystyle{ \varepsilon }$, are uncorrelated, have a mean of zero and a constant variance, $\displaystyle{ \sigma }$, the Gauss–Markov theorem states that the least-squares estimator, $\displaystyle{ \hat{\boldsymbol{\beta}} }$, has the minimum variance of all estimators that are linear combinations of the observations. In this sense it is the best, or optimal, estimator of the parameters. Note particularly that this property is independent of the statistical distribution function of the errors. In other words, the distribution function of the errors need not be a normal distribution. However, for some probability distributions, there is no guarantee that the least-squares solution is even possible given the observations; still, in such cases it is the best estimator that is both linear and unbiased.
For example, it is easy to show that the arithmetic mean of a set of measurements of a quantity is the least-squares estimator of the value of that quantity. If the conditions of the Gauss–Markov theorem apply, the arithmetic mean is optimal, whatever the distribution of errors of the measurements might be.
However, in the case that the experimental errors do belong to a normal distribution, the least-squares estimator is also a maximum likelihood estimator.[9]
These properties underpin the use of the method of least squares for all types of data fitting, even when the assumptions are not strictly valid.
### Limitations
An assumption underlying the treatment given above is that the independent variable, x, is free of error. In practice, the errors on the measurements of the independent variable are usually much smaller than the errors on the dependent variable and can therefore be ignored. When this is not the case, total least squares or more generally errors-in-variables models, or rigorous least squares, should be used. This can be done by adjusting the weighting scheme to take into account errors on both the dependent and independent variables and then following the standard procedure.[10][11]
In some cases the (weighted) normal equations matrix XTX is ill-conditioned. When fitting polynomials the normal equations matrix is a Vandermonde matrix. Vandermonde matrices become increasingly ill-conditioned as the order of the matrix increases. In these cases, the least squares estimate amplifies the measurement noise and may be grossly inaccurate. Various regularization techniques can be applied in such cases, the most common of which is called ridge regression. If further information about the parameters is known, for example, a range of possible values of $\displaystyle{ \mathbf{\hat{\boldsymbol{\beta}}} }$, then various techniques can be used to increase the stability of the solution. For example, see constrained least squares.
Another drawback of the least squares estimator is the fact that the norm of the residuals, $\displaystyle{ \| \mathbf y - X\hat{\boldsymbol{\beta}} \| }$ is minimized, whereas in some cases one is truly interested in obtaining small error in the parameter $\displaystyle{ \mathbf{\hat{\boldsymbol{\beta}}} }$, e.g., a small value of $\displaystyle{ \|{\boldsymbol{\beta}}-\hat{\boldsymbol{\beta}}\| }$. However, since the true parameter $\displaystyle{ {\boldsymbol{\beta}} }$ is necessarily unknown, this quantity cannot be directly minimized. If a prior probability on $\displaystyle{ \hat{\boldsymbol{\beta}} }$ is known, then a Bayes estimator can be used to minimize the mean squared error, $\displaystyle{ E \left\{ \| {\boldsymbol{\beta}} - \hat{\boldsymbol{\beta}} \|^2 \right\} }$. The least squares method is often applied when no prior is known. Surprisingly, when several parameters are being estimated jointly, better estimators can be constructed, an effect known as Stein's phenomenon. For example, if the measurement error is Gaussian, several estimators are known which dominate, or outperform, the least squares technique; the best known of these is the James–Stein estimator. This is an example of more general shrinkage estimators that have been applied to regression problems.
## Applications
### Uses in data fitting
The primary application of linear least squares is in data fitting. Given a set of m data points $\displaystyle{ y_1, y_2,\dots, y_m, }$ consisting of experimentally measured values taken at m values $\displaystyle{ x_1, x_2,\dots, x_m }$ of an independent variable ($\displaystyle{ x_i }$ may be scalar or vector quantities), and given a model function $\displaystyle{ y=f(x, \boldsymbol \beta), }$ with $\displaystyle{ \boldsymbol \beta = (\beta_1, \beta_2, \dots, \beta_n), }$ it is desired to find the parameters $\displaystyle{ \beta_j }$ such that the model function "best" fits the data. In linear least squares, linearity is meant to be with respect to parameters $\displaystyle{ \beta_j, }$ so $\displaystyle{ f(x, \boldsymbol \beta) = \sum_{j=1}^{n} \beta_j \varphi_j(x). }$
Here, the functions $\displaystyle{ \varphi_j }$ may be nonlinear with respect to the variable x.
Ideally, the model function fits the data exactly, so $\displaystyle{ y_i = f(x_i, \boldsymbol \beta) }$ for all $\displaystyle{ i=1, 2, \dots, m. }$ This is usually not possible in practice, as there are more data points than there are parameters to be determined. The approach chosen then is to find the minimal possible value of the sum of squares of the residuals $\displaystyle{ r_i(\boldsymbol \beta)= y_i - f(x_i, \boldsymbol \beta),\ (i=1, 2, \dots, m) }$ so to minimize the function $\displaystyle{ S(\boldsymbol \beta)=\sum_{i=1}^{m}r_i^2(\boldsymbol \beta). }$
After substituting for $\displaystyle{ r_i }$ and then for $\displaystyle{ f }$, this minimization problem becomes the quadratic minimization problem above with $\displaystyle{ X_{ij} = \varphi_j(x_i), }$ and the best fit can be found by solving the normal equations.
## Example
A plot of the data points (in red), the least squares line of best fit (in blue), and the residuals (in green)
As a result of an experiment, four $\displaystyle{ (x, y) }$ data points were obtained, $\displaystyle{ (1, 6), }$ $\displaystyle{ (2, 5), }$ $\displaystyle{ (3, 7), }$ and $\displaystyle{ (4, 10) }$ (shown in red in the diagram on the right). We hope to find a line $\displaystyle{ y=\beta_1+\beta_2 x }$ that best fits these four points. In other words, we would like to find the numbers $\displaystyle{ \beta_1 }$ and $\displaystyle{ \beta_2 }$ that approximately solve the overdetermined linear system: \displaystyle{ \begin{alignat}{3} \beta_1 + 1\beta_2 + r_1 &&\; = \;&& 6 & \\ \beta_1 + 2\beta_2 + r_2 &&\; = \;&& 5 & \\ \beta_1 + 3\beta_2 + r_3 &&\; = \;&& 7 & \\ \beta_1 + 4\beta_2 + r_4 &&\; = \;&& 10 & \\ \end{alignat} } of four equations in two unknowns in some "best" sense.
$\displaystyle{ r }$ represents the residual, at each point, between the curve fit and the data: \displaystyle{ \begin{alignat}{3} r_1 &&\; = \;&& 6 - (\beta_1 + 1\beta_2) & \\ r_2 &&\; = \;&& 5 - (\beta_1 + 2\beta_2) & \\ r_3 &&\; = \;&& 7 - (\beta_1 + 3\beta_2) & \\ r_4 &&\; = \;&& 10 - (\beta_1 + 4\beta_2) & \\ \end{alignat} }
The least squares approach to solving this problem is to try to make the sum of the squares of these residuals as small as possible; that is, to find the minimum of the function: \displaystyle{ \begin{align} S(\beta_1, \beta_2) &= r_1^2 + r_2^2 + r_3^2 + r_4^2 \\[6pt] &= [6-(\beta_1+1\beta_2)]^2 + [5-(\beta_1+2\beta_2)]^2 + [7-(\beta_1+3\beta_2)]^2 + [10-(\beta_1+4\beta_2)]^2 \\[6pt] &= 4\beta_1^2 + 30\beta_2^2 + 20\beta_1\beta_2 - 56\beta_1 - 154\beta_2 + 210 \\[6pt] \end{align} }
The minimum is determined by calculating the partial derivatives of $\displaystyle{ S(\beta_1, \beta_2) }$ with respect to $\displaystyle{ \beta_1 }$ and $\displaystyle{ \beta_2 }$ and setting them to zero: $\displaystyle{ \frac{\partial S}{\partial \beta_1}=0=8\beta_1 + 20\beta_2 -56 }$ $\displaystyle{ \frac{\partial S}{\partial \beta_2}=0=20\beta_1 + 60\beta_2 -154. }$
This results in a system of two equations in two unknowns, called the normal equations, which when solved give: $\displaystyle{ \beta_1=3.5 }$ $\displaystyle{ \beta_2=1.4 }$ and the equation $\displaystyle{ y = 3.5 + 1.4x }$ is the line of best fit. The residuals, that is, the differences between the $\displaystyle{ y }$ values from the observations and the $\displaystyle{ y }$ predicated variables by using the line of best fit, are then found to be $\displaystyle{ 1.1, }$ $\displaystyle{ -1.3, }$ $\displaystyle{ -0.7, }$ and $\displaystyle{ 0.9 }$ (see the diagram on the right). The minimum value of the sum of squares of the residuals is $\displaystyle{ S(3.5, 1.4)=1.1^2+(-1.3)^2+(-0.7)^2+0.9^2=4.2. }$
More generally, one can have $\displaystyle{ n }$ regressors $\displaystyle{ x_j }$, and a linear model $\displaystyle{ y = \beta_0 + \sum_{j=1}^{n} \beta_{j} x_{j}. }$
The result of fitting a quadratic function $\displaystyle{ y=\beta_1+\beta_2x+\beta_3x^2\, }$ (in blue) through a set of data points $\displaystyle{ (x_i, y_i) }$ (in red). In linear least squares the function need not be linear in the argument $\displaystyle{ x, }$ but only in the parameters $\displaystyle{ \beta_j }$ that are determined to give the best fit.
Importantly, in "linear least squares", we are not restricted to using a line as the model as in the above example. For instance, we could have chosen the restricted quadratic model $\displaystyle{ y=\beta_1 x^2 }$. This model is still linear in the $\displaystyle{ \beta_1 }$ parameter, so we can still perform the same analysis, constructing a system of equations from the data points: \displaystyle{ \begin{alignat}{2} 6 &&\; = \beta_1 (1)^2 + r_1 \\ 5 &&\; = \beta_1 (2)^2 + r_2 \\ 7 &&\; = \beta_1 (3)^2 + r_3 \\ 10 &&\; = \beta_1 (4)^2 + r_4 \\ \end{alignat} }
The partial derivatives with respect to the parameters (this time there is only one) are again computed and set to 0: $\displaystyle{ \frac{\partial S}{\partial \beta_1} = 0 = 708 \beta_1 - 498 }$ and solved $\displaystyle{ \beta_1 = 0.703 }$ leading to the resulting best fit model $\displaystyle{ y = 0.703 x^2. }$
## References
1. Lai, T.L.; Robbins, H.; Wei, C.Z. (1978). "Strong consistency of least squares estimates in multiple regression". PNAS 75 (7): 3034–3036. doi:10.1073/pnas.75.7.3034. PMID 16592540. Bibcode1978PNAS...75.3034L.
2. del Pino, Guido (1989). "The Unifying Role of Iterative Generalized Least Squares in Statistical Algorithms". Statistical Science 4 (4): 394–403. doi:10.1214/ss/1177012408.
3. Carroll, Raymond J. (1982). "Adapting for Heteroscedasticity in Linear Models". The Annals of Statistics 10 (4): 1224–1233. doi:10.1214/aos/1176345987.
4. Cohen, Michael; Dalal, Siddhartha R.; Tukey, John W. (1993). "Robust, Smoothly Heterogeneous Variance Regression". Journal of the Royal Statistical Society, Series C 42 (2): 339–353.
5. Nievergelt, Yves (1994). "Total Least Squares: State-of-the-Art Regression in Numerical Analysis". SIAM Review 36 (2): 258–264. doi:10.1137/1036055.
6. Tofallis, C (2009). "Least Squares Percentage Regression". Journal of Modern Applied Statistical Methods 7: 526–534. doi:10.2139/ssrn.1406472.
7. Hamilton, W. C. (1964). Statistics in Physical Science. New York: Ronald Press.
8. Spiegel, Murray R. (1975). Schaum's outline of theory and problems of probability and statistics. New York: McGraw-Hill. ISBN 978-0-585-26739-5.
9. Margenau, Henry; Murphy, George Moseley (1956). The Mathematics of Physics and Chemistry. Princeton: Van Nostrand.
10. Gans, Peter (1992). Data fitting in the Chemical Sciences. New York: Wiley. ISBN 978-0-471-93412-7.
11. Deming, W. E. (1943). Statistical adjustment of Data. New York: Wiley.
12. Acton, F. S. (1959). Analysis of Straight-Line Data. New York: Wiley.
13. Guest, P. G. (1961). Numerical Methods of Curve Fitting. Cambridge: Cambridge University Press.
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2022-08-14 04:12:13
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https://www.gradesaver.com/textbooks/math/precalculus/precalculus-6th-edition-blitzer/chapter-6-review-exercises-page-798/44
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Precalculus (6th Edition) Blitzer
$(x-1.5)^2+y^2=2.25$ See graph.
Step 1. Multiply the equation with $r$; we have $r^2=3r\ cos\theta$. Step 2. Using $r^2=x^2+y^2$ and $r\ cos\theta=x$, we have $x^2+y^2=3x$ which gives $x^2+y^2-3x=0$ and $(x-1.5)^2+y^2=1.5^2$ or $(x-1.5)^2+y^2=2.25$ Step 3. We can identify the above equation as a circle with center $(1.5,0)$ and radius $r=1.5$ Step 4. See graph.
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2021-06-24 18:40:59
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https://www.physicsforums.com/threads/non-integrable-functions.57349/
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# Non integrable functions
hai
i heard that the function x^x doesnt have any indefinite integral and hence one cant find definite integral by normal methods .. So one has to go for numerical methods , i havent tried this ...
just curious to know if there exist more fuctions in the same class ...
regards
mahesh
## Answers and Replies
Well, the integral "exists" of course, but it's not the expressible in elementary functions.
There are many such integrals of this nature.
Thanx wolfe, can u tell me where i can find such functions
Mahesh
dextercioby
Science Advisor
Homework Helper
DeadWolfe said:
Well, the integral "exists" of course, but it's not the expressible in elementary functions.
There are many such integrals of this nature.
Yap,just about elliptic integrals.Basically most of the type $\sqrt{P(x)}$,where P(x) is a polynomial with real coeffcients of degree larger of equal with 3,get the "chance" of not having a "cute" antiderivative.Mathematicians invented the famous syntagma "nonelementary function",referring to this sort of functions which come up when searching for antiderivatives.They couldn't come up with a decent definition for this "nonelementary". :tongue2:
Anyway,when you spot something wrong,i.e.u can't find an antiderivative,try for other tools.Numerical analysis works,but only in the case on definite integral,where the result is a number.Sometimes,u can expand the integrand in Taylor series (though the ray may be small) or express it terms on tabulated "nonelementary functions".The books by Abramowitz & Stegun and Gradsteyn & Rytzhik may turn out to be handy.
Daniel.
PS.If the antiderivatives exist,but cannot be expressed in terms of "elementary" functions,then the function which makes up the integrand is integrable.
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2021-06-13 00:45:28
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http://stackoverflow.com/questions/9267584/when-documenting-in-roxygen-how-do-i-make-an-itemized-list-in-details
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# When documenting in Roxygen: How do I make an itemized list in @details?
What is the appropriate syntax to add an itemized list to roxygen2, for instance, in the @details section? Can I create a latex list environment?
It seems that line brakes are simply ignored, i.e.
#' @details text describing parameter inputs in more detail
#'
#' parameter 1: stuff
#'
#' parameter 2: stuff
thanks!
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do \describe{}, \itemize{} as in the R extensions manual work ... ? – Ben Bolker Feb 13 '12 at 20:41
@Ben yes they do – hadley Feb 14 '12 at 1:16
Here is a roxygen2 example following your problem formulation.
##'
##' @details text describing parameter inputs in more detail.
##' \itemize{
##' \item{"parameter 1"}{Stuff}
##' \item{"parameter 2"}{Stuff}
##' }
##'
This will allow you to use itemize in details section. You can also use it in the @param sections.
Hope this helps.
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Can you point me to the documentation where you found this? I'm having a hard time figuring out the syntax for some of the Latex features in Roxygen. – Jeff Allen Jul 17 '12 at 15:56
@JeffAllen I just looked through the cran.r-project.org/doc/manuals/R-exts.html#Lists-and-tables section of the "Writing R Extensions" manual and tried it in Roxygen, which works fine for me. Is there a specific problem you're having? – Dr. Mike Jul 30 '12 at 9:31
@JeffAllen Note that this is just the standard R doc (.Rd) style for lists, which is just the LaTeX style. Trouble with both roxygen and .Rd is that it can be hard to predict when something that works in tex will work in .Rd, and when what works in .Rd will work in roxygen... – cboettig Nov 13 '13 at 3:54
This works for me, except that the args for \item are just written out as-is without formatting. Note: no space between \itemize and "{". – Roger May 23 '15 at 1:50
I think you mean \describe{...}, not \itemize{...}. \describe takes a named parameter and a definition, \itemize is just a bulleted list. – Ken Williams Jul 24 '15 at 22:13
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2016-05-26 09:08:22
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{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9109259843826294, "perplexity": 3250.061804358032}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-22/segments/1464049275764.90/warc/CC-MAIN-20160524002115-00140-ip-10-185-217-139.ec2.internal.warc.gz"}
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http://semantic-portal.net/concept:1794
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# The ::selection
//the following example makes the selected text red on a yellow background:
::selection {
color: red;
background: yellow;
}
Matches the portion of an element that is selected by a user.
The following CSS properties can be applied to ::selection: color, background, cursor, and outline.
The ::selection
## The ::selection — Structure map
Clickable & Draggable!
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2021-09-17 10:51:18
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{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3852325677871704, "perplexity": 13935.79402931942}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780055632.65/warc/CC-MAIN-20210917090202-20210917120202-00417.warc.gz"}
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http://centralcareers.org/f1z004va/numerical-solution-of-first-order-differential-equations-8bb23d
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Higher order ODEs can be solved using the same methods, with the higher order equations first having to be reformulated as a system of first order equations. To fully specify a particular solution, we require two additional conditions. 0000002412 00000 n Courses 0000044201 00000 n 0000031273 00000 n As a result, we need to resort to using numerical methods for solving such DEs. syms y (t) [V] = odeToVectorField (diff (y, 2) == (1 - y^2)*diff (y) - y) V =. Home 0000031432 00000 n 0000045893 00000 n 0000032603 00000 n Solutions to Linear First Order ODE’s 1. The general solution to the differential equation is given by. They are "First Order" when there is only dy dx, not d 2 y dx 2 or d 3 y dx 3 etc. We give an in depth overview of the process used to solve this type of differential equation as well as a derivation of the formula needed for the integrating factor used in the solution process. Differential Equations 0000059998 00000 n Modify, remix, and reuse (just remember to cite OCW as the source. Your use of the MIT OpenCourseWare site and materials is subject to our Creative Commons License and other terms of use. We then get two differential equations. » Flash and JavaScript are required for this feature. Integrating factors. 0000045823 00000 n In the previous session the computer used numerical methods to draw the integral curves. 0000033201 00000 n L �s^d�����9���Ie9��-[�"�#I��M-lB����%C8�ʾ>a���o������WB��B%�5��%L Download files for later. �����HX�8 ,Ǩ�ѳJE � ��((�?���������XIIU�QPPPH)-�C)�����K��8 [�������F��д4t�0�PJ��q�K mĞŖ|Ll���X�%XF. Any differential equation of the first order and first degree can be written in the form. differential equations in the form $$y' + p(t) y = g(t)$$. 0000028617 00000 n Send to friends and colleagues. Matlab has facilities for the numerical solution of ordinary differential equations (ODEs) of any order. There are many ways to solve ordinary differential equations (ordinary differential equations are those with one independent variable; we will assume this variable is time, t). For many of the differential equations we need to solve in the real world, there is no "nice" algebraic solution. In this paper, a method was proposed based on RBF for numerical solution of first-order differential equations with initial values that are valued by Z -numbers. >�d�����S Let and such that differentiating both equations we obtain a system of first-order differential equations. If we stepped by 0.0001 we would get even closer and closer and closer. For these DE's we can use numerical methods to get approximate solutions. We first express the differential equation as ′= ( , )=4 0.8 −0.5 and then express it as an Euler’s iterative formula, (+1)= ()+ℎ(4 0.8 ( 0+ Þℎ)−0.5 ()) With 0=0 and ℎ=1, we obtain (+1)= ()+4 0.8 Þ−0.5 ()=0.5 ()+4 0.8 Þ. Initialization: (0)=2. Find materials for this course in the pages linked along the left. 0000029218 00000 n Here we will look at solving a special class of Differential Equations called First Order Linear Differential Equations. The numerical algorithm for solving “first-order linear differential equation in fuzzy environment” is discussed. The Euler method is the simplest algorithm for numerical solution of a differential equation. 0000069568 00000 n In this paper, a novel iterative method is proposed to obtain approximate-analytical solutions for the linear systems of first-order fuzzy differential equations (FDEs) with fuzzy constant coefficients (FCCs) while avoiding the complexities of eigen-value computations. Mathematics Since we obtained the solution by integration, there will always be a constant of integration that remains to be specified. dy dx + P(x)y = Q(x). x�bf}�����/� �� @1v� Use OCW to guide your own life-long learning, or to teach others. Massachusetts Institute of Technology. In this document we first consider the solution of a first order ODE. A first-order differential equation is an Initial value problem (IVP) of the form, The formula for Euler's method defines a recursive sequence: where for each . 0000049934 00000 n 0000014784 00000 n If we ch… 0000043601 00000 n 0000061617 00000 n 0000057010 00000 n … It follows, by the application of Theorem 4.5, that the solution of any noncommensurate multi-order fractional differential equation may be arbitrarily closely approximated over any finite time interval [0,T] by solutions of equations of rational order (which may in turn be solved by conversion to a system of equations of low order). Many differential equations cannot be solved exactly. Hence, yn+1 = yn +0.05{yn −xn +[yn +0.1(yn −xn)]−xn+1}. Finite difference solution for the second order ordinary differential equations. Then v'(t)=y''(t). The first is easy 1.10 Numerical Solution to First-Order Differential Equations 95 Solution: Taking h = 0.1 and f(x,y)= y −x in the modified Euler method yields y∗ n+1 = yn +0.1(yn −xn), yn+1 = yn +0.05(yn −xn +y ∗ n+1 −xn+1). There's no signup, and no start or end dates. We will start with Euler's method. 0000050365 00000 n Knowledge is your reward. 0000015447 00000 n The simplest numerical method for approximating solutions of differential equations is Euler's method. This equation is called a first-order differential equation because it contains a The techniques discussed in these pages approximate the solution of first order ordinary differential equations (with initial conditions) of the form In other words, problems where the derivative of our solution at time t, y(t), is dependent on that solution and t (i.e., y'(t)=f(y(t),t)). 0000058223 00000 n FIRST ORDER SYSTEMS 3 which finally can be written as !.10 (1.6) You can check that this answer satisfies the equation by substituting the solution back into the original equation. This is actually how most differential equations or techniques that are derived from this or that are based on numerical methods similar to this are how most differential equations gets solved. 0000057397 00000 n ), Learn more at Get Started with MIT OpenCourseWare, MIT OpenCourseWare makes the materials used in the teaching of almost all of MIT's subjects available on the Web, free of charge. Use the tangent line to approximate at a small time step : where . 0000062862 00000 n Many differential equations cannot be solved exactly. In the previous session the computer used numerical methods to draw the integral curves. 0000069965 00000 n Use Runge-Kutta Method of Order 4 to solve the following, using a step size of h=0.1\displaystyle{h}={0.1}h=0.1 for 0≤x≤1\displaystyle{0}\le{x}\le{1}0≤x≤1. So there's a bunch of interesting things here. trailer <<4B691525AB324A9496D13AA176D7112E>]>> startxref 0 %%EOF 115 0 obj <>stream It we assume that M = M 0 at t = 0, then M 0 = A e 0 which gives A = M 0 The solution may be written as follows M(t) = M 0 e - k t 0000014336 00000 n 0000006840 00000 n The differential equation. Solve the above first order differential equation to obtain M(t) = A e - k t where A is non zero constant. 0000034709 00000 n » Freely browse and use OCW materials at your own pace. 0000045099 00000 n Example. can also be written as. How to use a previous numerical solution to solve a differential equation numerically? That is, we can't solve it using the techniques we have met in this chapter (separation of variables, integrable combinations, or using an integrating factor), or other similar means. Hot Network Questions AWS recommend 54 t2.nano EC2 instances instead one m5.xlarge The ddex1 example shows how to solve the system of differential equations y 1 ' ( t ) = y 1 ( t - 1 ) y 2 ' ( t ) = y 1 ( t - 1 ) + y 2 ( t - 0 . dy dt = f (t,y) y(t0) =y0 (1) (1) d y d t = f ( t, y) y ( t 0) = y 0. We will also discuss more sophisticated methods that give better approximations. N���ػM�Pfj���1h8��5Qbc���V'S�yY�Fᔓ� /O�o��\�N�b�|G-��F��%^���fnr��7���b�~���Cİ0���ĦQ������.��@k���:�=�YpЉY�S�%5P�!���劻+9_���T���p1뮆@k{���_h:�� h\$=:�+�Qɤ�;٢���EZ�� �� The differential equations we consider in most of the book are of the form Y′(t) = f(t,Y(t)), where Y(t) is an unknown function that is being sought. The numerical solutions are compared with (i)-gH and (ii)-gH differential (exact solutions concepts) system. solution and its numerical approximation. In most of these methods, we replace the di erential equation by a di erence equation … 0000002869 00000 n Consider the differential equation: The first step is to convert the above second-order ode into two first-order ode. 2. 0000059172 00000 n Systems of first-order equations and characteristic surfaces. 0000052745 00000 n 0000051866 00000 n For these DE's we can use numerical methods to get approximate solutions. The first part has stated the amount of limitation of the fragmentation solution, while the second part has described the assurance of the first part. In order to select 0000070325 00000 n If you're seeing this message, it means we're having trouble loading external resources on our website. The proposed method consists of two parts. 0000025843 00000 n Solution. 0000002580 00000 n This is one of over 2,200 courses on OCW. 0000002144 00000 n The study on numerical methods for solving partial differential equation will be of immense benefit to the entire mathematics department and other researchers that desire to carry out similar research on the above topic because the study will provide an explicit solution to partial differential equations using numerical methods. 3. • y=g(t) is a solution of the first order differential equation means • i) y(t) is differentiable • ii) Substitution of y(t) and y’(t) in equation satisfies the differential equation identically Differential equations of the first order and first degree. » 0000050727 00000 n 0000007909 00000 n The classification of partial differential equations can be extended to systems of first-order equations, where the unknown u is now a vector with m components, and the coefficient matrices A ν are m by m matrices for ν = 1, 2,… n. The partial differential equation takes the form 0000025489 00000 n Differential equations with only first derivatives. 0000035725 00000 n If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. Consider a first order differential equation with an initial condition: The procedure for Euler's method is as follows: 1. You can represent these equations with … 0000032007 00000 n Made for sharing. 0000007272 00000 n » 58 0 obj <> endobj xref 58 58 0000000016 00000 n 0000051500 00000 n 0000007623 00000 n Linear. A scheme, namely, “Runge-Kutta-Fehlberg method,” is described in detail for solving the said differential equation. 0000045610 00000 n 0000030177 00000 n This is a standard operation. A first order differential equation is linear when it can be made to look like this:. This method was originally devised by Euler and is called, oddly enough, Euler’s Method. The given function f(t,y) of two variables defines the differential equation, and exam ples are given in Chapter 1. Unit I: First Order Differential Equations, Unit II: Second Order Constant Coefficient Linear Equations, Unit III: Fourier Series and Laplace Transform, Motivation and Implementation of Euler's Method (PDF). 0000024570 00000 n Learn more », © 2001–2018 Numerical Solution of Ordinary Di erential Equations of First Order Let us consider the rst order di erential equation dy dx = f(x;y) given y(x 0) = y 0 (1) to study the various numerical methods of solving such equations. Let’s start with a general first order IVP. Advanced Math Solutions – Ordinary Differential Equations Calculator, Separable ODE Last post, we talked about linear first order differential equations. using a change of variables. Construct the tangent line at the point and repeat. Existence of a solution. 0000030266 00000 n 0000053769 00000 n where d M / d t is the first derivative of M, k > 0 and t is the time. %PDF-1.6 %���� Bernoulli’s equation. In this section we shall be concerned with the construction and the analysis of numerical methods for first-order differential equations of the form y′ = f(x,y) (1) for the real-valued function yof the real variable x, where y′ ≡ dy/dx. We will start with Euler's method. method, a basic numerical method for solving initial value problems. 0000015145 00000 n (x - 3y)dx + (x - 2y)dy = 0. Linear Equations – In this section we solve linear first order differential equations, i.e. 0000044616 00000 n 2 ) y 3 ' ( t ) = y 2 ( t ) . Contruct the equation of the tangent line to the unknown function at :where is the slope of at . MIT OpenCourseWare is a free & open publication of material from thousands of MIT courses, covering the entire MIT curriculum. 0000029673 00000 n Numerical Methods. No enrollment or registration. Unit I: First Order Differential Equations > Download from Internet Archive (MP4 - 97MB), > Download from Internet Archive (MP4 - 10MB), > Download from Internet Archive (MP4 - 23MB). First Order. 0000025058 00000 n Module: 5 Numerical Solution of Ordinary Differential Equations 8 hours First and second order differential equations - Fourth order Runge – Kutta method. Let v(t)=y'(t). The concept is similar to the numerical approaches we saw in an earlier integration chapter (Trapezoidal Rule, Simpson's Rule and Riemann Su… 0000002207 00000 n Adams-Bashforth-Moulton predictor-corrector methods. We don't offer credit or certification for using OCW. 0000060793 00000 n » 0000033831 00000 n \begin{equation*}y = C_1\sin(3x) + C_2\cos(3x)\text{,}\end{equation*} where $$C_1$$ and $$C_2$$ are arbitrary constants. 0000062329 00000 n 0000001456 00000 n This is the simplest numerical method, akin to approximating integrals using rectangles, but it contains the basic idea common to all the numerical methods we will look at. It usually gives the least accurate results but provides a basis for understanding more sophisticated methods. First Order Linear Equations In the previous session we learned that a first order linear inhomogeneous ODE for the unknown function x = x(t), has the standard form x … We are going to look at one of the oldest and easiest to use here. With more than 2,400 courses available, OCW is delivering on the promise of open sharing of knowledge. Line at the point and repeat and repeat: the procedure for Euler 's method is as follows:.... Module: 5 numerical solution of a differential equation with an initial condition: the procedure for Euler 's is! Use OCW to guide your own pace offer credit or certification for using OCW fully specify particular! P ( x ) y 3 ' ( t ) M / t... Order ode devised by Euler and is called a first-order differential equation numerically with more than courses! ( ii ) -gH and ( ii ) -gH differential ( exact solutions concepts ) system matlab facilities. Module: 5 numerical solution of a differential equation is given by step is to convert the above second-order into. Web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked any order order.. Opencourseware is a free & open publication of material from thousands of MIT courses, covering the entire MIT.... Or to teach others p ( x - 3y ) dx + ( x ) y 3 ' t... » Unit numerical solution of first order differential equations: first order differential equations in the form \ ( y ' + (... A first order and first degree of differential equations ( ODEs ) of any order solutions. Is Euler 's method to cite OCW as the source let and such that differentiating both equations obtain. » numerical methods y = g ( t ) the least accurate results but provides a basis for understanding sophisticated... We obtained the solution by integration, there will always be a of. Numerical algorithm for numerical solution of a differential equation with an initial condition: the procedure Euler... Or certification for using OCW to cite OCW as the source for approximating solutions differential! To our Creative Commons License and other terms of use usually gives the least results., and no start or end dates free & open publication of from! The numerical solutions are compared with ( i ) -gH and ( ii ) -gH and ( ii -gH... Site and materials is subject to our Creative Commons License and other terms of use differential of! Numerical methods to draw the integral curves method was originally devised by Euler and is called, oddly enough Euler. You 're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are.., yn+1 = yn +0.05 { yn −xn + [ yn +0.1 yn. A result, we need to resort to using numerical methods to get approximate.. Of open sharing of knowledge the above second-order ode into two first-order ode Euler method is follows! General first order differential equation with an initial condition: the procedure for Euler 's defines. And *.kasandbox.org are unblocked −xn ) ] −xn+1 } a constant of integration that remains to be specified the! We obtained the solution by integration, there will always be a of! 'S method defines a recursive sequence: where 8 hours first and second order differential equation first-order differential is. Procedure for Euler 's method is the time y ' + p ( x - 2y ) dy =.... Gives the least accurate results but provides a basis for understanding more sophisticated methods give. At your own life-long learning, or to teach others use numerical methods to get approximate solutions provides a for. The solution by integration, there will always be a constant of integration remains! Basic numerical method for solving such DEs both equations we obtain a system of first-order equations! Fuzzy environment numerical solution of first order differential equations is described in detail for solving such DEs / d t is the first differential... Differential ( exact solutions concepts numerical solution of first order differential equations system can be made to look this! Are compared with ( i ) -gH differential ( exact solutions concepts ).... *.kastatic.org and *.kasandbox.org are unblocked condition: the procedure for Euler 's method the time where is slope! - 2y ) dy = 0 open publication of material from thousands of MIT courses, covering the entire curriculum. First step is to convert the above second-order ode into two first-order ode written in the pages along! ( t ) ’ s start with a general first order and first degree 're... To be specified d M / d t is the simplest algorithm for numerical of... The promise of open sharing of knowledge for each methods for solving such DEs differential ( exact solutions concepts system... ( ii ) -gH and ( ii ) -gH and ( ii ) -gH and ii! With more than 2,400 courses available, OCW is delivering on the promise open! First-Order differential equations » Unit i: first order differential equations » Unit i: first order.... 'S we can use numerical methods to draw the integral curves value problems use OCW to guide own! For solving such DEs equation of the first order and first degree can be made to look like this.. Methods to draw the integral curves remix, and no start or end dates of knowledge p x! We solve linear first order differential equations, i.e we obtained the solution of a order. Remix, and reuse ( just remember to cite OCW as the source y 3 ' ( t y. Concepts ) system using numerical methods linear differential equation: the first ode..., please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked we get. Numerical solution of a first order differential equations » Unit i: first order and degree... Opencourseware site and materials is subject to our Creative Commons License and other terms of use and and... 8 hours first and second order ordinary differential equations to convert the above second-order ode into two first-order...., it means we 're having trouble loading external resources on our website ( exact solutions concepts ) system first! As the source when it can be written in the form \ ( y ' + p ( )... Method, a basic numerical method for approximating solutions of differential equations ODEs. Can use numerical methods for solving the said differential equation: the order. Is the simplest algorithm for solving the numerical solution of first order differential equations differential equation because it a., we require two additional conditions for the numerical solution to solve a differential.. Solving “ first-order linear differential equation is called a first-order differential equation in fuzzy environment ” is described in for! Basis for understanding more sophisticated methods that give better approximations with more than courses... Of open sharing of knowledge we first consider the solution by integration, there always... ) system = y 2 ( t ) =y ' ( t.! This is one of over 2,200 courses on OCW +0.05 { yn +. Discuss more sophisticated methods this message, it means we 're having trouble loading external resources on our website ii... Two first-order ode, OCW is delivering on the promise of open sharing of.... A constant of integration that remains to be specified courses available, OCW is delivering the... Or end dates be specified the general solution to solve a differential equation linear. 'S we can use numerical methods to get approximate solutions equations – in this we... Finite difference solution for the second order differential equation with an initial condition: first... Y ' + p ( t ) y 3 ' ( t ) solutions concepts ) system Massachusetts Institute Technology. Other terms of use - 3y ) dx + ( x - 2y ) dy = 0 form \ y! For solving “ first-order linear differential equation because it contains a Integrating factors 2. 0 and t is the time linear first order differential equations of the first of! Or end dates 's no signup, and no start or end.... Own pace things here OpenCourseWare is a free & open numerical solution of first order differential equations of material thousands! Offer credit or certification for using OCW “ first-order linear differential equation the. ( exact solutions concepts ) system more », © 2001–2018 Massachusetts of! 'Re having trouble loading external resources on our website a small time step: where is the algorithm... A free & open publication of material from thousands of MIT courses, covering entire. ( ii ) -gH and ( ii ) -gH and ( ii ) -gH and ( ii -gH. Usually gives the least accurate results but provides a basis for understanding more methods. Massachusetts Institute of Technology of any order credit or certification for using OCW yn! The simplest numerical method for approximating solutions of differential equations - Fourth order –... Solution of a differential equation of the MIT OpenCourseWare is a free & open publication of material thousands! Of MIT courses, covering the entire MIT curriculum system of first-order differential equations of the MIT OpenCourseWare a... Then v ' ( t ) =y ' ( t ) = y (! Point and repeat please make sure that the domains *.kastatic.org and *.kasandbox.org unblocked... Trouble loading external resources on our website previous session the computer used numerical methods to the! The left materials is subject to our Creative Commons License and other terms of use the previous session the used! Two first-order ode equation of the MIT OpenCourseWare site and materials is subject to Creative. A basis for understanding more sophisticated methods to teach others there 's signup... 0 and t is the simplest algorithm for numerical solution of a differential equation in fuzzy environment ” discussed. Approximating solutions of differential equations is Euler 's method is as follows: 1 the general solution the! Is one of over 2,200 courses on OCW interesting things here cite OCW as the source course in previous... Is discussed there 's a bunch of interesting things here 2,400 courses available, OCW is delivering the...
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2022-01-16 11:00:51
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https://space.stackexchange.com/questions/30261/angular-momentum-in-non-inertial-frame
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# Angular momentum in non inertial frame [closed]
How to express the angular momentum in non inertial Earth frame like Earth Fixed frame so that angular momentum is conserved ?
Is H = R x mV still valid when R and V are taken in the rotating frame ?
## closed as off-topic by Phiteros, peterh says reinstate Monica, Heopps, Rob, Organic MarbleAug 24 '18 at 20:21
This question appears to be off-topic. The users who voted to close gave this specific reason:
• "This question is about other space sciences (physics, weather, astronomy, etc), and does not directly pertain to space exploration as outlined in the help center." – Phiteros, peterh says reinstate Monica, Heopps
If this question can be reworded to fit the rules in the help center, please edit the question.
• Welcome on the Space SE! Your question is not really about the space exploration, I suggest to try physics.stackexchange.com . – peterh says reinstate Monica Aug 24 '18 at 15:50
• This question is better suited to Physics.SE, and off topic for this site. – Rob Aug 24 '18 at 16:12
• It's a good question for sure. Before you ask there, I'd recommend you look there for existing answers. – uhoh Aug 24 '18 at 17:22
Yes, $\vec H = \vec r \times (m \vec v)$ is still valid in a rotating frame. That's the definition, after all. Whether this definition has any meaning is a different question. Angular momentum is not necessarily conserved in a rotating frame. Consider, for example, a spacecraft in a geosynchronous orbit that has a non-zero inclination with respect to the equator and a non-zero eccentricity. The angular velocity vector as defined by $\vec r \times (m \vec v)$ points all over the place.
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2019-11-12 18:55:31
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https://www.physicsforums.com/threads/jet-engine-thermodynamic-non-sequitur.176000/
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# Jet engine thermodynamic non sequitur.
1. Jul 5, 2007
### parsec
Hi all,
This has been bugging me all day, so I thought I'd seek the help and advice of the internet.
When a plane is stationary, you’d have to say that its jet engines are doing no useful work. The engines apply a static force on the airframe and could be said to be running at zero efficiency (ignoring the fact that they generate shaft power and compressed air for the plane’s auxiliary systems). I guess this means that the hot exhaust gases ejected from the rear of the engine lose all of their energy through heating up the surrounding atmosphere (entirely plausible).
Now consider a plane taxiing at a slow velocity at the same engine speed (rpm). The engines are doing useful work, and the propulsive power it is producing could be said to be the product of the plane’s velocity and frictional force the plane’s airframe sees; P=FV (in the tyre-tarmac interface and drag).
What causes this dramatic increase in efficiency? Mechanically, the same scenario is easy to visualize and rationalize in a car. When the engine is decoupled from the drivetrain (in neutral), the car revs freely, and not much air and fuel (accelerator pedal excursion) is required to redline the car. Here, the work done by the engine is used purely to overcome piston friction and other mechanical losses. When the car is coupled to the drivetrain, much more air and fuel is needed to speed the engine up as the energy is consumed by accelerating the vehicle.
From the jet engine’s point of view, the only thing that has changed is the velocity of the intake air. When this intake air is no longer stationary, the engine seems to do useful work. It is hard to visualize how this works.
A simpler related problem involves the propulsion of an inflated balloon. If the balloon is held static while it deflates, all of its energy is lost in heat and noise. By simply allowing the balloon to be unrestrained, and hence achieve a non zero velocity relative to the surrounding air, the balloon seems to do useful work in the form of giving itself kinetic energy. How can simply allowing the balloon to be unrestrained so dramatically increase its propulsive efficiency?
Imagine the scenario from the jet engine's or balloon's point of view. All that has changed is the ability to propel yourself forward, but you are still applying the same force you were applying when restrained.
The implication is that when unrestrained, the environment or surroundings aren't as hot as a result of your jet engine or balloon exhaust, because some of this energy has gone into the plane's/balloon's inertia. I can accept this, but I don't see how restraint (or lack thereof) can cause the exhaust to reconfigure itself magically in this fashion.
It would seem at first glance that this lack of restraint allows the jet stream to create less turbulence and more ordered work in the form of propulsive thrust.
2. Jul 5, 2007
### Staff: Mentor
For the purpose of efficiency, the work done by a jet engine is measured from the mass flow rate of air out the exhaust. The plane may not utilize that work, but it is still being generated.
edit: also, it is true that idling engines are typically pretty inefficient. In an idling car engine, for example, all of the energy generated is lost to friction. This isn't any big deal.
Last edited: Jul 5, 2007
3. Jul 5, 2007
### K.J.Healey
Whats your "propulsive efficiency" equation that you're talking about? Is it a specific quantity?
Lets assume an inflated ballon. It has potential energy U. The work done by the system when the balloon is held in place while it deflates would be:
W = W_air + W_balloon
Well W_balloon = F*d, since no displacement == 0.
The AIR on the otherhand, is moving at a a velocity defined such that
W_air = U =>> Turns into complete kinetic energy of the air. The air moves really fast. All the balloon's potential becomes kinetic in the air.
Now DON'T restrain the balloon. What happens?
W = W_air + W_balloon
Now W_balloon isn't zero. Which means that :
U = W_air + W_balloon
W_air = U - W_balloon
As you can see, the work done on the AIR is much less. And this makes sense if you think about it relative to the balloon. If the balloon is sitting still its air displacement can be measure from either an observer or the balloon and its the same. But when the balloon gets moving, the balloon still sees the air moving out just as fast as before, BUT a stationary OBSERVER sees the air move much slower relative to the initial position of the balloon.
Think about the possibility when the velocity of the air leaving the balloon is equal to the balloon's speed.
A stationalry observer would see the air stay still, while the balloon flies off at 2*V, where V is the velocity of the air as witnessed by the balloon.
Work done is always the same as the potential energy created by the system.
It just depends on what you ask the work is being done on. (oh man thats so grammatically errored)
4. Jul 5, 2007
### rcgldr
"Useful work" depends on what you consider as the work output of a jet engine. Even when "staionary", jet engines are performing work by accelerating air and burnt fuel. The result of this work is in increase in kinetic energy (relative to the surrounding air), and the thrust produced as a result of the acceleration of air and burnt fuel (there's also a big increase in temperature).
Taxing on the ground isn't going to be much more efficient that standing still. However a jet engine's efficiency does improve with speed and altitude. At speed, ram air effect supplies the intake with higher pressure air. At higher altitudes, air temperature and density lowers, which also helps.
Most of the losses are heat related, not friction. The heat losses go into heating the cooling systems, and the heated air out of the exhaust.
As I mentioned, the velocity of intake air produces a ram air effect that help, but your point is about useful work done by the thrust of a jet engine. In this case, thrust (force) times distance is the work done. Power would be thrust times air speed. Note that as speeds increase, power increases, as maximum thrust doesn't decrease with speed (within reason). This is just an issue of what you consider to be the work done by a jet engine, the acceleration of air and burnt fuel, or the thrust times the distance moved.
Last edited: Jul 5, 2007
5. Jul 5, 2007
### FredGarvin
The unrestrained scenario simply allows you to use the definition of work with respect to what is outside of the engine's control volume. The engine is always producing work. Whether you utilize that work does not have any bearing on it.
Technically in static conditions, even the definition of thrust horsepower causes one to conclude that there is no work being done. However, we know that there is always the $$\tau \dot{\theta}$$ power being produced. I find it analogous to the person pushing on a wall; they are exerting and using energy, even to the point of exhaustion, but doing no work.
6. Jul 5, 2007
### parsec
Everything everyone has posted makes sense so far. I'd like to quote individual posts but it'd take too much time.
The lengthy post involving the balloon makes sense, but what about at the instant the balloon is released, where the balloon has no forward momentum, and for all intensive purposes is doing the same amount of work accelerating the air as when it was restrained.
Also, does that mean a propulsive device is most efficient when its forward velocity equals the exit velocity of its air/jet stream?
I agree with this sentiment, however I can't get my head around the jet stream changing its properties due to the plane being unrestrained.
consider the following:
Say the engine is supplied with so much fuel with a thermal content of P joules/sec or watts.
In the static scenario, ignoring friction in the engine and the auxillary systems, the jet stream exhausted out of the rear of the engine delivers Q watts of heat to the outside atmosphere. A lot of this is in heat, the rest is noise and the high speed jet stream shearing against the motionless surrounding air. We can say P=Q in this scenario, and all of the power put into the engine results in heating the atmosphere.
Now release the plane's brakes. Some of the power P must be invested in building up the plane's inertia, so Q can no longer equal P. P = Q + U (power invested in moving the plane) From the jet engine's point of view, its jet stream must be losing less energy to the atmosphere. It is still being supplied with P watts of fuel, but now its jet stream has suddenly reconfigured itself so that P - U energy is exhausted into the atmosphere.
Sorry if that's a little scatterbrained. Hope it makes some sense. Clearly I'm slightly confused :)
7. Jul 5, 2007
### rcgldr
Which means work is peformed on the air that is accelerated (ignoring the heat aspect).
I think you're missing the point of my post. Aircraft engines perform work on air, regardless of whether the aircraft is moving with respect to the air or not. So even in a "static" situation, the power of a jet engine, based on the rate of work peformed on the air can be determined, and it's not zero.
The comparason to pushing on a wall is not a good one, because the net result of the forces is to compress the wall and what ever is pushing the wall a small amount, and there's no movement (except for the initial compression reaction to the application of forces). In the case of a jet engine, a force is applied to a mass of air over a distance as the air accelerates, and that translates into real work being done, and in turn the cacpacity to accelerate the air can be converted into power.
Last edited: Jul 5, 2007
8. Jul 5, 2007
### parsec
sure, work is being performed on the air. i agree with this, but surely when the plane is allowed to move, less work is performed on the air as some of the work must be used to actually move the plane.
i don't understand what part or the jet engine or its interaction with the air changes to allow some of this work to be invested in the plane's inertia.
9. Jul 6, 2007
### rcgldr
The jet stream output of a "static" "typical" military type jet engine is around 1300mph. Assuming it remains at around 1300mph, then as the aircraft speed increases, then the acceleration of air to 1300mph represents less work done and there would be less thrust. However military jet engines have nozzles, there's a ram air effect that increases intake air pressure, and at altitude the air is much colder, so a high rate of thrust can be maintained at high speeds and high altitudes. There are some jets that can reach mach 1.5 at 200 feet above sea level (F-111). The SR71 uses very high speed jet exhaust, uses a movable cone in front of the intake to allow the supersonic shock wave to help compress intake air, and also uses partial compressor stage bypass tubes to feed air directly to the combustion chamber to run at Mach 3 (2000+mph), at about 30% "throttle" (top speed is restricted the temperature the aircraft can take due to air friction). Obviously, the engines on a SR-71 have an exhaust speed well over 2000mph.
The point of all of this is that jet engines have the ability to maintain thrust (and work done on the air) at higher speeds because the configurations can be change dynamically (nozzles, inlet valves, bypass pipes, moving cones, ...).
Another wierdness of jet engines, is that the rate of fuel consumed verus distance traveled remains about the same regardless of aircraft speed.
A jet engine accelerates air backwards, and the air responds to this backwards acceleration with a forwards reaction force that is the "thrust" that moves or accelerates the aircraft forwards.
Last edited: Jul 6, 2007
10. Jul 6, 2007
### FredGarvin
That is somewhat true. Actually it is a combination of a little less work performed plus less wasted kinetic energy of the jet. That is why, if you look at a plot of propulsive efficiency vs. airspeed, it pretty much increases up to around 85% (depending on altitude).
I guess all I can say to this is that energy is not being wasted restraining and thus wasting the energy provided by the jet. At zero forward speed, the propulsive efficiency is zero because of the definition of the efficiency. Remember, propulsive efficiency is the ratio of the useful propulsive power divided by the sum of that energy PLUS the wasted or unused kinetic energy of the jet. BY DEFINITION, if the aircraft is at a standstill, you are wasting 100% of the provided kinetic energy of the jet and thus your propulsive efficiency is 0. The jet doesn't really care one way or another what is going on around it. It's going to grab a gulp of air, compress it and throw it out the back. How much of that work we grab and how much is wasted gets rolled into that efficiency number.
$$\eta_{propulsive} = \frac{2}{1+\frac{V_{exhaust}}{V_{aircraft}}}$$ or if you prefer the long winded version...
$$\eta_{propulsive} = \frac{m V_a(V_j-V_a)}{m\frac{\left[V_a(V_j-V_a)+(V_j-V_a)^2\right]}{2}}$$
You can think about it this way when it comes to the propulsive efficiency:
- $$F$$ is maximum when $$V_a =0$$ however $$\eta_p =0$$
- $$\eta_p$$ is maximum at $$\frac{V_j}{V_a}=1$$ but then $$F = 0$$
Last edited: Jul 6, 2007
11. Jul 12, 2007
### parsec
Thanks. I think I understand what I was misunderstanding. The concept seems more intuitive if I think about the useful thrust being generated through a momentum exchange mechanism. I guess that means that when the aircraft is travelling at the same speed as the exhaust, the air remains relatively undisturbed and all of the work goes into propelling the aircraft.
Presumably your propulsive efficiency equation does not apply to the case where the exhaust is slower than the aircraft? (I guess no useful thrust can be produced in this scenario).
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2018-02-25 06:46:28
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https://www.physicsoverflow.org/42930/what-the-difference-between-product-states-entangled-states
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# what is the difference between product states and entangled states?
+ 0 like - 0 dislike
853 views
now , i know that product states can be factorisable into two separate superpositions.
e:g (a l0> + b l1>) * (c l0> + d l1>) = ac l00> + ad l01> + bc l10> + bd l11> how does this differ from entangled states which also has 4 coefficients (e:g) a l00> + b l01> + c l10> + d l11>.
i know that entangled states aren't factorizable but how do they differ from product states if they end up with the same number of coefficients. like why does entangled states give us 2^n computational states while superrpositions without entanglement give us 2n although as written above its the same number of coefficents?
if we add a third qubit entangled states will have 8 coefficients and product states 6, though both will give all possible combinations of the three qubit system (e:g 000, 001, 011, .....) when we multiply out the product states made of three separate superpositions. how does that make any difference?
recategorized May 17, 2020
+ 0 like - 0 dislike
Take $(a\left|0\right>+b\left|1\right>)(c\left|0\right>+d\left|1\right>)$.
If $a\neq0$, $c\neq 0$, $ac(\left|0\right>+\frac{b}{a}\left|1\right>)(\left|0\right>+\frac{d}{c}\left|1\right>)$, so you have 3 degrees of freedom: $ac$, $\frac{b}{a}$ and $\frac{d}{c}$.
Take $a\left|00\right>+b\left|10\right>+c\left|01\right>+d\left|11\right>$.
If $a\neq0$, $c\neq 0$, $a\left|00\right>+b\left|10\right>+c\left|01\right>+d\left|11\right>$ and there are 4 degrees of freedom.
answered May 19, 2020 by (30 points)
edited May 19, 2020 by Iliod
Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor) Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOverfl$\varnothing$wThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.
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2023-02-04 03:36:54
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https://www.lmfdb.org/Variety/Abelian/Fq/3/23/ax_jh_aceq
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# Properties
Label 3.23.ax_jh_aceq Base Field $\F_{23}$ Dimension $3$ Ordinary No $p$-rank $2$ Principally polarizable Yes Contains a Jacobian No
## Invariants
Base field: $\F_{23}$ Dimension: $3$ L-polynomial: $( 1 - 9 x + 23 x^{2} )( 1 - 14 x + 92 x^{2} - 322 x^{3} + 529 x^{4} )$ Frobenius angles: $\pm0.112386341891$, $\pm0.135789939707$, $\pm0.314924263207$ Angle rank: $3$ (numerical)
This isogeny class is not simple.
## Newton polygon
$p$-rank: $2$ Slopes: $[0, 0, 1/2, 1/2, 1, 1]$
## Point counts
This isogeny class is principally polarizable, but does not contain a Jacobian.
$r$ 1 2 3 4 5 6 7 8 9 10 $A(\F_{q^r})$ 4290 135624060 1808039324520 21969744191881200 266739205154465707950 3244372546491988002336960 39472897837403617193938415370 480257541194117605634752487193600 5843233974769971834562226813683119960 71094398476475424486958657923029006922300
$r$ 1 2 3 4 5 6 7 8 9 10 $C(\F_{q^r})$ 1 483 12214 280543 6438851 148046004 3404938769 78312090383 1801159729318 41426540165163
## Decomposition and endomorphism algebra
Endomorphism algebra over $\F_{23}$
The isogeny class factors as 1.23.aj $\times$ 2.23.ao_do and its endomorphism algebra is a direct product of the endomorphism algebras for each isotypic factor. The endomorphism algebra for each factor is:
All geometric endomorphisms are defined over $\F_{23}$.
## Base change
This is a primitive isogeny class.
## Twists
Below is a list of all twists of this isogeny class.
Twist Extension Degree Common base change 3.23.af_al_hc $2$ (not in LMFDB) 3.23.f_al_ahc $2$ (not in LMFDB) 3.23.x_jh_ceq $2$ (not in LMFDB)
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2020-06-01 09:08:23
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http://www.reference.com/browse/Cumulative_incidence
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Definitions
# Cumulative incidence
Cumulative incidence is a measure of frequency, as in epidemiology, where it is a measure of disease frequency during a period of time.
Cumulative incidence is the incidence calculated using a period of time during which all of the individuals in the population are considered to be at risk for the outcome. It is sometimes referred to as the incidence proportion or the attack rate.
Cumulative incidence is calculated by the number of new cases during a period divided by the number of people at risk in the population at the beginning of the study.
It may also be calculated by the incidence rate multiplied by duration.
$CI\left(t\right)=1-e^\left\{-IR\left(t\right) * D\right\},$
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2013-06-20 07:01:30
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https://www.physicsforums.com/threads/variation-of-lagrangian-w-r-to-canonical-momenta.864477/
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# Variation of Lagrangian w/r to canonical momenta
• A
muscaria
Hi,
I've been working through Cornelius Lanczos book "The Variational Principles of Mechanics" and there's something I'm having difficulty understanding on page 168 of the Dover edition (which is attached). After introducing the Legendre transformation and transforming the Lagrangian equations of motion into their canonical form, he is now trying to motivate a ("extended") variational principle which gives rise directly to Hamilton's equations of motion, in which the action integral is varied w/r to the 2n positional phase space variables ##(q,p)##. Basically, I don't understand the passage from equation (64.1) to (64.2), where he considers the variation of $$L=\sum_{i=1}^n p_i\dot{q}_i - H,$$ with respect to the ##p_i##, which he writes as $$\delta L = \sum_{i=1}^n\left(\dot{q}_i - \frac{\partial H}{\partial p_i}\right)\delta p_i.$$ My problem is aren't the ##\dot{q}_i## a function of ##(q,p)##, so that there should be an extra term?
My attempt at understanding what is going on:
Because of the duality of the Legendre transform, we can write $$L= \sum_{i=1}^n p_i\frac{\partial H}{\partial p_i} - H,$$ so that the variation of ##L## w/r to the ##p_i## in this form, is what we had above but with an extra piece: $$\delta L = \sum_{i=1}^n\left(\dot{q}_i - \frac{\partial H}{\partial p_i}\right)\delta p_i + \sum_{i,j}p_i\frac{\partial^2H}{\partial p_j\partial p_i}\delta p_j.$$ The first term vanishes on account of the duality of the Legendre transformation but what's the meaning of the 2nd term, if there is any? If we follow the assumption that there is no coupling between different canonical momenta (which seems reasonable otherwise conservation of momentum would be violated, right?), then we will only have the diagonal terms of the double sum:##\sum_{i,j} p_i\frac{\partial^2H}{\partial p_j\partial p_i}\delta p_j = \sum_i p_i \frac{\partial^2 H}{\partial p_i^2}\delta p_i##, i.e $$\delta L = \sum_{i=1}^n\left(\dot{q}_i - \frac{\partial H}{\partial p_i}\right)\delta p_i + \frac{1}{2}\sum_i \frac{\partial^2 H}{\partial p_i^2}\delta (p_i)^2.$$ Is this suggesting in some way that a variation of the Lagrangian w/r to the ##p_i## only enters as a second order effect and can be put to 0 given that we are looking for stationary values of the action? Or is there something I'm missing or misinterpreting which will make all of this obvious?
As a side note, you can get to the result if you consider the variation of the action integral w/r to the ##p_i## (as opposed to variation of the bare Lagrangian) then integration by parts shows that there is no variation of the action from arbitrary variations of the ##p_i##, so all fine, you can still form the ##2n## differential equations in the way he does on page 169, but still, I'd like to understand how he can just go between those 2 equations.. To me it seems like what he is getting at only works if the variation is integrated out w/r to time, we need that integration by parts.
Thanks for your help and I hope this message reaches you in good spirits!
#### Attachments
• Canonical Integral_Lanczos.pdf
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secur
To compute the coefficient of the variation of p you use the partial derivative NOT the total derivative. The qdot's have no explicit dependence on p, so they're just constants. True, in the physical space there is a dependence - which would have to be taken into account if, for some reason, you were taking a total derivative here (which actually wouldn't make much sense). But for the partial, qdots are constants multiplying p so you get simply qdot for that first term in 64.2, coming from the first term in 64.1.
As for what you tried to do in your "attempt at understanding what is going on" - it may work out somehow; I didn't examine it. Certainly not the right way to go, though you get points for trying!
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When I first learned about Lagrangian mechanics and that q and qdot were to be treated as formally independent variables, I was similarly confused. How can q and qdot be independent if qdot is fully determined by q? It finally made sense to me when I realized that the Lagrangian (and Hamiltonian) are defined not just along the path of the motion, but on all possible potential paths. In the case of Hamiltonian, H is a function of p and q, and p and q are just coordinates, not the actual position and momentum. You can ask, what is the energy at this instant if the position is q and the momentum is p, regardless of if the system will actually reach that point in phase space. Therefore, p and q are independent. The advantage of this formulism is that you can use it for quantum mechanics, where there is no precise trajectory.
muscaria
Hi guys, thank you for your responses and apologies for the rather late reply!
When I first learned about Lagrangian mechanics and that q and qdot were to be treated as formally independent variables, I was similarly confused. How can q and qdot be independent if qdot is fully determined by q?
This was not what I was confused about, ##\dot{q}## is only determined by ##q(t)##, i.e once you have solved the equations of motion.. they're independent dynamical variables which need to be specified independently of each other in order for a given solution trajectory to be taken - I get that. Without meaning to be presumptuous and I appreciate the response, from your reply I think you may have misunderstood/overlooked what I was asking. My question was basically, if the "new" dynamical space is the set of independent dynamical variables ##(q,p)##, then any function ##f## of the motion (be it actual or tentative) is a function of these dynamical variables -> ##f(q,p)##. The generalised velocities are no different in this respect, i.e. for each point in phase space there is a definite set of associated ##\dot{q}_i=\dot{q}_i(q_1,\cdots,q_n,p_1,\cdots,p_n)##; so my line of thinking was that OK well, a variation of ##p_i## will induce a variation of ##\dot{q}_i##. And so in considering ##\delta L## resulting from ##\delta p_i##, why does ##\delta\dot{q}_i## not play any part?
To compute the coefficient of the variation of p you use the partial derivative NOT the total derivative.
I don't see how I was computing a total derivative..? I computed a partial just in the same way as you would for the following situation say: Given some ##f(x,y)=xg(x,y)## (meant to represent ##p_i\dot{q}_i##), the variation ##\delta f## which results from varying ##x## is simply $$\delta f =\frac{\partial\left(xg(x,y)\right)}{\partial x}\delta x$$
But for the partial, qdots are constants multiplying p so you get simply qdot for that first term in 64.2, coming from the first term in 64.1.
But not because
The qdot's have no explicit dependence on p, so they're just constants
I think I've just finally got it actually, or just about.. the point seems to be that the ##\dot{q}_i## are slopes of a given function in ##(q,p)## space, namely ##\frac{\partial H(q,p)}{\partial p_i}##, so although ##\dot{q}_i=\dot{q}_i(q_1,\cdots,q_n,p_1,\cdots,p_n)## is true, there's a deeper structure which follows from the Legendre transform... It's not simply a point transformation we're carrying out on our dynamical space in going from Lagrangian to Hamiltonian, we're substituting a set of variables by the slope of a given function in this dynamical space. But given that the "new" set of variables correspond to (1st order) slopes in the "old" set of variables,a variation of these will only manifest as a second order effect in the reciprocal space, through the duality of the Legendre Transform and the E-L equations of motion. And the reciprocity is due to the structure of how the E-L e.o.m : ##\dot{p}_i =\frac{\partial L}{\partial q_i}## and the Legendre transform combine. Does this line of reasoning sound reasonable? I guess that's essentially what the calculation that came out in the original post is saying also, because I don't see how it is wrong..
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2023-02-03 10:29:43
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http://mathoverflow.net/questions/88356/comparing-blow-ups-with-comparable-centres
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# Comparing blow-ups with comparable centres
Let $X$ a variety. Say $X=\operatorname{Spec} A$.
Consider two ideals of $A$, say $I$ and $J$, with equal radical ; and consider the blow-ups of X with centre $I$ and $J$, say $Y_I$ and $Y_J$. How can I decide if the blow-up map $Y_I \to X$ factors through $Y_J$ ?
For example, if $X = \operatorname{Spec}k[x,y]$ is the affine plane. Then we can consider the blow-ups with centre $I_1 = (x,y)$, $I_2 = (x,y^2)$, $I_3=(x^2,y^2)$ and $I_4=(x^2,xy,y^2)=I_1^2$. We have $I_4\subset I_3\subset I_2\subset I_1$. Let $Y_i$ denote the corresponding blow-ups.
It is well-known that $Y_1$ and $Y_4$ are isomorphic. However $Y_2$ and $Y_3$ are two other different varieties. Using the universal property of the blow-up we see that $Y_1\to X$ factors through $Y_3$ but not through $Y_2$.
What kind of property should I look at to predict this kind of factorization, without computing the actual blow-up ?
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There are two conditions I can think of.
1. Perhaps $I = J \cdot J'$ for some other ideal $J'$. Indeed, blowing up a product of ideals is the same as blowing up one ideal (say $J$) and then blowing up the total transform of the other (say $J'$).
2. The other is requiring that $I = \overline{J}$, here this is integral closure of ideals (more generally $I$ and $J$ have the same integral closure and $I \subseteq J$). Indeed, two ideals have the same integral closure if and only if they have the same normalized blow-up and if their total transforms on the common normalized blow-up agree.
Let me give an alternate explanation of 2., suppose for example that $I = \overline{J}$. Then the blow-up of $I$ and the blow-up of $J$ have the same normalization and $Y_I$ is just a partial normalization of $Y_J$.
So this explains some of your examples. Indeed, the integral closure of $I_3$ is just $I_4$. On the other hand, $I_4 = I_1^2$. So by 1., $I_1$ and $I_4$ have the same blowup but by 2., the blow up of $I_4$ factors through the blowup of $I_3$. However, if you consider $I_1 \cdot I_2$ or $I_3 \cdot I_2$, you will find that their blowups factor through $I_2$.
For smooth surfaces, there is a theory of factorization of complete (integrally closed) ideals due to Zariski. This should allow you to answer these sorts of questions even more precisely.
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I think this will help me a lot! The point 2 is really great, I was not aware of it. – Lierre Feb 14 '12 at 8:19
The universal property of the blowup $\pi:Bl_I X\to X$ of an ideal $I$ in $X$ is that the sheaf of ideals $\pi^{-1}I\cdot O_{Bl_I X}$ is invertible. So, if you want to find out whether $Bl_I$ factors through $Bl_J$ you just have to check whether $\pi^{-1}J\cdot O_{Bl_I X}$ is invertible.
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2015-11-29 14:18:08
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http://dataspace.princeton.edu/jspui/handle/88435/dsp01vh53wz05d
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Please use this identifier to cite or link to this item: http://arks.princeton.edu/ark:/88435/dsp01vh53wz05d
Title: A Contemporary Theory of Responsibility: What Aristotle Can Teach us Authors: Di Rosa, Elena Advisors: Lorenz, Hendrik Contributors: Cooper, John M. Department: Philosophy Class Year: 2015 Abstract: In my thesis, I advance a theory of ethical responsibility based on Aristotle’s discussion of the voluntary in the Nicomachean Ethics. I argue that people are ethically responsible for their actions when they have sufficiently developed the capacity to deliberate and act on decision, in the Aristotelian sense, and as such, can reasonably be held ethically responsible for their voluntary actions. After all, once they have developed the capacity for deliberation and decision, their actions, for the most part, can be said to reflect their relatively stable characters for which they are responsible. Extent: 42 pages URI: http://arks.princeton.edu/ark:/88435/dsp01vh53wz05d Type of Material: Princeton University Senior Theses Language: en_US Appears in Collections: Philosophy, 1924-2016
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2016-10-27 11:00:08
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https://git.rtems.org/rtems/log/tools/update/310_to_320_list?id=e013fe3cd07bd9e1025d10c7d3d3034f8439446d&showmsg=1
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summaryrefslogtreecommitdiffstats log msg author committer range
path: root/tools/update/310_to_320_list (follow)
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* Ralf Corsepius reported a number of missing CVS Id's:Joel Sherrill1998-01-161-0/+2
| | | | | | | | | | > RTEMS is under CVS control and has been since rtems 3.1.16 which was > around May 1995. So I just to add the \$Id\$. If you notice other files > with missing \$Id\$'s let me know. I try to keep w\up with it. Now that you have asked -- I'll attach a list of files lacking an RCS-Id to this mail. This list has been generated by a little sh-script I'll also enclose.
* Initial revisionJoel Sherrill1995-05-111-0/+543
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2022-10-03 18:53:16
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https://stacks.math.columbia.edu/tag/02XJ
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## 4.32 Fibred categories
A very brief discussion of fibred categories is warranted.
Let $p : \mathcal{S} \to \mathcal{C}$ be a category over $\mathcal{C}$. Given an object $x \in \mathcal{S}$ with $p(x) = U$, and given a morphism $f : V \to U$, we can try to take some kind of “fibre product $V \times _ U x$” (or a base change of $x$ via $V \to U$). Namely, a morphism from an object $z \in \mathcal{S}$ into “$V \times _ U x$” should be given by a pair $(\varphi , g)$, where $\varphi : z \to x$, $g : p(z) \to V$ such that $p(\varphi ) = f \circ g$. Pictorially:
$\xymatrix{ z \ar@{~>}[d]^ p \ar@{-}[r] & ? \ar[r] \ar@{~>}[d]^ p & x \ar@{~>}[d]^ p \\ p(z) \ar[r] & V \ar[r]^ f & U }$
If such a morphism $V \times _ U x \to x$ exists then it is called a strongly cartesian morphism.
Definition 4.32.1. Let $\mathcal{C}$ be a category. Let $p : \mathcal{S} \to \mathcal{C}$ be a category over $\mathcal{C}$. A strongly cartesian morphism, or more precisely a strongly $\mathcal{C}$-cartesian morphism is a morphism $\varphi : y \to x$ of $\mathcal{S}$ such that for every $z \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S})$ the map
$\mathop{Mor}\nolimits _\mathcal {S}(z, y) \longrightarrow \mathop{Mor}\nolimits _\mathcal {S}(z, x) \times _{\mathop{Mor}\nolimits _\mathcal {C}(p(z), p(x))} \mathop{Mor}\nolimits _\mathcal {C}(p(z), p(y)),$
given by $\psi \longmapsto (\varphi \circ \psi , p(\psi ))$ is bijective.
Note that by the Yoneda Lemma 4.3.5, given $x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S})$ lying over $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ and the morphism $f : V \to U$ of $\mathcal{C}$, if there is a strongly cartesian morphism $\varphi : y \to x$ with $p(\varphi ) = f$, then $(y, \varphi )$ is unique up to unique isomorphism. This is clear from the definition above, as the functor
$z \longmapsto \mathop{Mor}\nolimits _\mathcal {S}(z, x) \times _{\mathop{Mor}\nolimits _\mathcal {C}(p(z), U)} \mathop{Mor}\nolimits _\mathcal {C}(p(z), V)$
only depends on the data $(x, U, f : V \to U)$. Hence we will sometimes use $V \times _ U x \to x$ or $f^*x \to x$ to denote a strongly cartesian morphism which is a lift of $f$.
Lemma 4.32.2. Let $\mathcal{C}$ be a category. Let $p : \mathcal{S} \to \mathcal{C}$ be a category over $\mathcal{C}$.
1. The composition of two strongly cartesian morphisms is strongly cartesian.
2. Any isomorphism of $\mathcal{S}$ is strongly cartesian.
3. Any strongly cartesian morphism $\varphi$ such that $p(\varphi )$ is an isomorphism, is an isomorphism.
Proof. Proof of (1). Let $\varphi : y \to x$ and $\psi : z \to y$ be strongly cartesian. Let $t$ be an arbitrary object of $\mathcal{S}$. Then we have
\begin{align*} & \mathop{Mor}\nolimits _\mathcal {S}(t, z) \\ & = \mathop{Mor}\nolimits _\mathcal {S}(t, y) \times _{\mathop{Mor}\nolimits _\mathcal {C}(p(t), p(y))} \mathop{Mor}\nolimits _\mathcal {C}(p(t), p(z)) \\ & = \mathop{Mor}\nolimits _\mathcal {S}(t, x) \times _{\mathop{Mor}\nolimits _\mathcal {C}(p(t), p(x))} \mathop{Mor}\nolimits _\mathcal {C}(p(t), p(y)) \times _{\mathop{Mor}\nolimits _\mathcal {C}(p(t), p(y))} \mathop{Mor}\nolimits _\mathcal {C}(p(t), p(z)) \\ & = \mathop{Mor}\nolimits _\mathcal {S}(t, x) \times _{\mathop{Mor}\nolimits _\mathcal {C}(p(t), p(x))} \mathop{Mor}\nolimits _\mathcal {C}(p(t), p(z)) \end{align*}
hence $z \to x$ is strongly cartesian.
Proof of (2). Let $y \to x$ be an isomorphism. Then $p(y) \to p(x)$ is an isomorphism too. Hence $\mathop{Mor}\nolimits _\mathcal {C}(p(z), p(y)) \to \mathop{Mor}\nolimits _\mathcal {C}(p(z), p(x))$ is a bijection. Hence $\mathop{Mor}\nolimits _\mathcal {S}(z, x) \times _{\mathop{Mor}\nolimits _\mathcal {C}(p(z), p(x))} \mathop{Mor}\nolimits _\mathcal {C}(p(z), p(y))$ is bijective to $\mathop{Mor}\nolimits _\mathcal {S}(z, x)$. Hence the displayed map of Definition 4.32.1 is a bijection as $y \to x$ is an isomorphism, and we conclude that $y \to x$ is strongly cartesian.
Proof of (3). Assume $\varphi : y \to x$ is strongly cartesian with $p(\varphi ) : p(y) \to p(x)$ an isomorphism. Applying the definition with $z = x$ shows that $(\text{id}_ x, p(\varphi )^{-1})$ comes from a unique morphism $\chi : x \to y$. We omit the verification that $\chi$ is the inverse of $\varphi$. $\square$
Lemma 4.32.3. Let $F : \mathcal{A} \to \mathcal{B}$ and $G : \mathcal{B} \to \mathcal{C}$ be composable functors between categories. Let $x \to y$ be a morphism of $\mathcal{A}$. If $x \to y$ is strongly $\mathcal{B}$-cartesian and $F(x) \to F(y)$ is strongly $\mathcal{C}$-cartesian, then $x \to y$ is strongly $\mathcal{C}$-cartesian.
Proof. This follows directly from the definition. $\square$
Lemma 4.32.4. Let $\mathcal{C}$ be a category. Let $p : \mathcal{S} \to \mathcal{C}$ be a category over $\mathcal{C}$. Let $x \to y$ and $z \to y$ be morphisms of $\mathcal{S}$. Assume
1. $x \to y$ is strongly cartesian,
2. $p(x) \times _{p(y)} p(z)$ exists, and
3. there exists a strongly cartesian morphism $a : w \to z$ in $\mathcal{S}$ with $p(w) = p(x) \times _{p(y)} p(z)$ and $p(a) = \text{pr}_2 : p(x) \times _{p(y)} p(z) \to p(z)$.
Then the fibre product $x \times _ y z$ exists and is isomorphic to $w$.
Proof. Since $x \to y$ is strongly cartesian there exists a unique morphism $b : w \to x$ such that $p(b) = \text{pr}_1$. To see that $w$ is the fibre product we compute
\begin{align*} & \mathop{Mor}\nolimits _\mathcal {S}(t, w) \\ & = \mathop{Mor}\nolimits _\mathcal {S}(t, z) \times _{\mathop{Mor}\nolimits _\mathcal {C}(p(t), p(z))} \mathop{Mor}\nolimits _\mathcal {C}(p(t), p(w)) \\ & = \mathop{Mor}\nolimits _\mathcal {S}(t, z) \times _{\mathop{Mor}\nolimits _\mathcal {C}(p(t), p(z))} (\mathop{Mor}\nolimits _\mathcal {C}(p(t), p(x)) \times _{\mathop{Mor}\nolimits _\mathcal {C}(p(t), p(y))} \mathop{Mor}\nolimits _\mathcal {C}(p(t), p(z))) \\ & = \mathop{Mor}\nolimits _\mathcal {S}(t, z) \times _{\mathop{Mor}\nolimits _\mathcal {C}(p(t), p(y))} \mathop{Mor}\nolimits _\mathcal {C}(p(t), p(x)) \\ & = \mathop{Mor}\nolimits _\mathcal {S}(t, z) \times _{\mathop{Mor}\nolimits _\mathcal {S}(t, y)} \mathop{Mor}\nolimits _\mathcal {S}(t, y) \times _{\mathop{Mor}\nolimits _\mathcal {C}(p(t), p(y))} \mathop{Mor}\nolimits _\mathcal {C}(p(t), p(x)) \\ & = \mathop{Mor}\nolimits _\mathcal {S}(t, z) \times _{\mathop{Mor}\nolimits _\mathcal {S}(t, y)} \mathop{Mor}\nolimits _\mathcal {S}(t, x) \end{align*}
as desired. The first equality holds because $a : w \to z$ is strongly cartesian and the last equality holds because $x \to y$ is strongly cartesian. $\square$
Definition 4.32.5. Let $\mathcal{C}$ be a category. Let $p : \mathcal{S} \to \mathcal{C}$ be a category over $\mathcal{C}$. We say $\mathcal{S}$ is a fibred category over $\mathcal{C}$ if given any $x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S})$ lying over $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ and any morphism $f : V \to U$ of $\mathcal{C}$, there exists a strongly cartesian morphism $f^*x \to x$ lying over $f$.
Assume $p : \mathcal{S} \to \mathcal{C}$ is a fibred category. For every $f : V \to U$ and $x\in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}_ U)$ as in the definition we may choose a strongly cartesian morphism $f^\ast x \to x$ lying over $f$. By the axiom of choice we may choose $f^*x \to x$ for all $f: V \to U = p(x)$ simultaneously. We claim that for every morphism $\phi : x \to x'$ in $\mathcal{S}_ U$ and $f : V \to U$ there is a unique morphism $f^\ast \phi : f^\ast x \to f^\ast x'$ in $\mathcal{S}_ V$ such that
$\xymatrix{ f^\ast x \ar[r]_{f^\ast \phi } \ar[d] & f^\ast x' \ar[d] \\ x \ar[r]^{\phi } & x' }$
commutes. Namely, the arrow exists and is unique because $f^*x' \to x'$ is strongly cartesian. The uniqueness of this arrow guarantees that $f^\ast$ (now also defined on morphisms) is a functor $f^\ast : \mathcal{S}_ U \to \mathcal{S}_ V$.
Definition 4.32.6. Assume $p : \mathcal{S} \to \mathcal{C}$ is a fibred category.
1. A choice of pullbacks1 for $p : \mathcal{S} \to \mathcal{C}$ is given by a choice of a strongly cartesian morphism $f^\ast x \to x$ lying over $f$ for any morphism $f: V \to U$ of $\mathcal{C}$ and any $x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}_ U)$.
2. Given a choice of pullbacks, for any morphism $f : V \to U$ of $\mathcal{C}$ the functor $f^* : \mathcal{S}_ U \to \mathcal{S}_ V$ described above is called a pullback functor (associated to the choices $f^*x \to x$ made above).
Of course we may always assume our choice of pullbacks has the property that $\text{id}_ U^*x = x$, although in practice this is a useless property without imposing further assumptions on the pullbacks.
Lemma 4.32.7. Assume $p : \mathcal{S} \to \mathcal{C}$ is a fibred category. Assume given a choice of pullbacks for $p : \mathcal{S} \to \mathcal{C}$.
1. For any pair of composable morphisms $f : V \to U$, $g : W \to V$ there is a unique isomorphism
$\alpha _{g, f} : (f \circ g)^\ast \longrightarrow g^\ast \circ f^\ast$
as functors $\mathcal{S}_ U \to \mathcal{S}_ W$ such that for every $y\in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}_ U)$ the following diagram commutes
$\xymatrix{ g^\ast f^\ast y \ar[r] & f^\ast y \ar[d] \\ (f \circ g)^\ast y \ar[r] \ar[u]^{(\alpha _{g, f})_ y} & y }$
2. If $f = \text{id}_ U$, then there is a canonical isomorphism $\alpha _ U : \text{id} \to (\text{id}_ U)^*$ as functors $\mathcal{S}_ U \to \mathcal{S}_ U$.
3. The quadruple $(U \mapsto \mathcal{S}_ U, f \mapsto f^*, \alpha _{g, f}, \alpha _ U)$ defines a pseudo functor from $\mathcal{C}^{opp}$ to the $(2, 1)$-category of categories, see Definition 4.28.5.
Proof. In fact, it is clear that the commutative diagram of part (1) uniquely determines the morphism $(\alpha _{g, f})_ y$ in the fibre category $\mathcal{S}_ W$. It is an isomorphism since both the morphism $(f \circ g)^*y \to y$ and the composition $g^*f^*y \to f^*y \to y$ are strongly cartesian morphisms lifting $f \circ g$ (see discussion following Definition 4.32.1 and Lemma 4.32.2). In the same way, since $\text{id}_ x : x \to x$ is clearly strongly cartesian over $\text{id}_ U$ (with $U = p(x)$) we see that there exists an isomorphism $(\alpha _ U)_ x : x \to (\text{id}_ U)^*x$. (Of course we could have assumed beforehand that $f^*x = x$ whenever $f$ is an identity morphism, but it is better for the sake of generality not to assume this.) We omit the verification that $\alpha _{g, f}$ and $\alpha _ U$ so obtained are transformations of functors. We also omit the verification of (3). $\square$
Lemma 4.32.8. Let $\mathcal{C}$ be a category. Let $\mathcal{S}_1$, $\mathcal{S}_2$ be categories over $\mathcal{C}$. Suppose that $\mathcal{S}_1$ and $\mathcal{S}_2$ are equivalent as categories over $\mathcal{C}$. Then $\mathcal{S}_1$ is fibred over $\mathcal{C}$ if and only if $\mathcal{S}_2$ is fibred over $\mathcal{C}$.
Proof. Denote $p_ i : \mathcal{S}_ i \to \mathcal{C}$ the given functors. Let $F : \mathcal{S}_1 \to \mathcal{S}_2$, $G : \mathcal{S}_2 \to \mathcal{S}_1$ be functors over $\mathcal{C}$, and let $i : F \circ G \to \text{id}_{\mathcal{S}_2}$, $j : G \circ F \to \text{id}_{\mathcal{S}_1}$ be isomorphisms of functors over $\mathcal{C}$. We claim that in this case $F$ maps strongly cartesian morphisms to strongly cartesian morphisms. Namely, suppose that $\varphi : y \to x$ is strongly cartesian in $\mathcal{S}_1$. Set $f : V \to U$ equal to $p_1(\varphi )$. Suppose that $z' \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}_2)$, with $W = p_2(z')$, and we are given $g : W \to V$ and $\psi ' : z' \to F(x)$ such that $p_2(\psi ') = f \circ g$. Then
$\psi = j \circ G(\psi ') : G(z') \to G(F(x)) \to x$
is a morphism in $\mathcal{S}_1$ with $p_1(\psi ) = f \circ g$. Hence by assumption there exists a unique morphism $\xi : G(z') \to y$ lying over $g$ such that $\psi = \varphi \circ \xi$. This in turn gives a morphism
$\xi ' = F(\xi ) \circ i^{-1} : z' \to F(G(z')) \to F(y)$
lying over $g$ with $\psi ' = F(\varphi ) \circ \xi '$. We omit the verification that $\xi '$ is unique. $\square$
The conclusion from Lemma 4.32.8 is that equivalences map strongly cartesian morphisms to strongly cartesian morphisms. But this may not be the case for an arbitrary functor between fibred categories over $\mathcal{C}$. Hence we define the $2$-category of fibred categories as follows.
Definition 4.32.9. Let $\mathcal{C}$ be a category. The $2$-category of fibred categories over $\mathcal{C}$ is the sub $2$-category of the $2$-category of categories over $\mathcal{C}$ (see Definition 4.31.1) defined as follows:
1. Its objects will be fibred categories $p : \mathcal{S} \to \mathcal{C}$.
2. Its $1$-morphisms $(\mathcal{S}, p) \to (\mathcal{S}', p')$ will be functors $G : \mathcal{S} \to \mathcal{S}'$ such that $p' \circ G = p$ and such that $G$ maps strongly cartesian morphisms to strongly cartesian morphisms.
3. Its $2$-morphisms $t : G \to H$ for $G, H : (\mathcal{S}, p) \to (\mathcal{S}', p')$ will be morphisms of functors such that $p'(t_ x) = \text{id}_{p(x)}$ for all $x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S})$.
In this situation we will denote
$\mathop{Mor}\nolimits _{\textit{Fib}/\mathcal{C}}(\mathcal{S}, \mathcal{S}')$
the category of $1$-morphisms between $(\mathcal{S}, p)$ and $(\mathcal{S}', p')$
Note the condition on $1$-morphisms. Note also that this is a true $2$-category and not a $(2, 1)$-category. Hence when taking $2$-fibre products we first pass to the associated $(2, 1)$-category.
Lemma 4.32.10. Let $\mathcal{C}$ be a category. The $(2, 1)$-category of fibred categories over $\mathcal{C}$ has 2-fibre products, and they are described as in Lemma 4.31.3.
Proof. Basically what one has to show here is that given $F : \mathcal{X} \to \mathcal{S}$ and $G : \mathcal{Y} \to \mathcal{S}$ morphisms of fibred categories over $\mathcal{C}$, then the category $\mathcal{X} \times _\mathcal {S} \mathcal{Y}$ described in Lemma 4.31.3 is fibred. Let us show that $\mathcal{X} \times _\mathcal {S} \mathcal{Y}$ has plenty of strongly cartesian morphisms. Namely, suppose we have $(U, x, y, \phi )$ an object of $\mathcal{X} \times _\mathcal {S} \mathcal{Y}$. And suppose $f : V \to U$ is a morphism in $\mathcal{C}$. Choose strongly cartesian morphisms $a : f^*x \to x$ in $\mathcal{X}$ lying over $f$ and $b : f^*y \to y$ in $\mathcal{Y}$ lying over $f$. By assumption $F(a)$ and $G(b)$ are strongly cartesian. Since $\phi : F(x) \to G(y)$ is an isomorphism, by the uniqueness of strongly cartesian morphisms we find a unique isomorphism $f^*\phi : F(f^*x) \to G(f^*y)$ such that $G(b) \circ f^*\phi = \phi \circ F(a)$. In other words $(G(a), G(b)) : (V, f^*x, f^*y, f^*\phi ) \to (U, x, y, \phi )$ is a morphism in $\mathcal{X} \times _\mathcal {S} \mathcal{Y}$. We omit the verification that this is a strongly cartesian morphism (and that these are in fact the only strongly cartesian morphisms). $\square$
Lemma 4.32.11. Let $\mathcal{C}$ be a category. Let $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$. If $p : \mathcal{S} \to \mathcal{C}$ is a fibred category and $p$ factors through $p' : \mathcal{S} \to \mathcal{C}/U$ then $p' : \mathcal{S} \to \mathcal{C}/U$ is a fibred category.
Proof. Suppose that $\varphi : x' \to x$ is strongly cartesian with respect to $p$. We claim that $\varphi$ is strongly cartesian with respect to $p'$ also. Set $g = p'(\varphi )$, so that $g : V'/U \to V/U$ for some morphisms $f : V \to U$ and $f' : V' \to U$. Let $z \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S})$. Set $p'(z) = (W \to U)$. To show that $\varphi$ is strongly cartesian for $p'$ we have to show
$\mathop{Mor}\nolimits _\mathcal {S}(z, x') \longrightarrow \mathop{Mor}\nolimits _\mathcal {S}(z, x) \times _{\mathop{Mor}\nolimits _{\mathcal{C}/U}(W/U, V/U)} \mathop{Mor}\nolimits _{\mathcal{C}/U}(W/U, V'/U),$
given by $\psi ' \longmapsto (\varphi \circ \psi ', p'(\psi '))$ is bijective. Suppose given an element $(\psi , h)$ of the right hand side, then in particular $g \circ h = p(\psi )$, and by the condition that $\varphi$ is strongly cartesian we get a unique morphism $\psi ' : z \to x'$ with $\psi = \varphi \circ \psi '$ and $p(\psi ') = h$. OK, and now $p'(\psi ') : W/U \to V/U$ is a morphism whose corresponding map $W \to V$ is $h$, hence equal to $h$ as a morphism in $\mathcal{C}/U$. Thus $\psi '$ is a unique morphism $z \to x'$ which maps to the given pair $(\psi , h)$. This proves the claim.
Finally, suppose given $g : V'/U \to V/U$ and $x$ with $p'(x) = V/U$. Since $p : \mathcal{S} \to \mathcal{C}$ is a fibred category we see there exists a strongly cartesian morphism $\varphi : x' \to x$ with $p(\varphi ) = g$. By the same argument as above it follows that $p'(\varphi ) = g : V'/U \to V/U$. And as seen above the morphism $\varphi$ is strongly cartesian. Thus the conditions of Definition 4.32.5 are satisfied and we win. $\square$
Lemma 4.32.12. Let $\mathcal{A} \to \mathcal{B} \to \mathcal{C}$ be functors between categories. If $\mathcal{A}$ is fibred over $\mathcal{B}$ and $\mathcal{B}$ is fibred over $\mathcal{C}$, then $\mathcal{A}$ is fibred over $\mathcal{C}$.
Proof. This follows from the definitions and Lemma 4.32.3. $\square$
Lemma 4.32.13. Let $p : \mathcal{S} \to \mathcal{C}$ be a fibred category. Let $x \to y$ and $z \to y$ be morphisms of $\mathcal{S}$ with $x \to y$ strongly cartesian. If $p(x) \times _{p(y)} p(z)$ exists, then $x \times _ y z$ exists, $p(x \times _ y z) = p(x) \times _{p(y)} p(z)$, and $x \times _ y z \to z$ is strongly cartesian.
Proof. Pick a strongly cartesian morphism $\text{pr}_2^*z \to z$ lying over $\text{pr}_2 : p(x) \times _{p(y)} p(z) \to p(z)$. Then $\text{pr}_2^*z = x \times _ y z$ by Lemma 4.32.4. $\square$
Lemma 4.32.14. Let $\mathcal{C}$ be a category. Let $F : \mathcal{X} \to \mathcal{Y}$ be a $1$-morphism of fibred categories over $\mathcal{C}$. There exist $1$-morphisms of fibred categories over $\mathcal{C}$
$\xymatrix{ \mathcal{X} \ar@<1ex>[r]^ u & \mathcal{X}' \ar[r]^ v \ar@<1ex>[l]^ w & \mathcal{Y} }$
such that $F = v \circ u$ and such that
1. $u : \mathcal{X} \to \mathcal{X}'$ is fully faithful,
2. $w$ is left adjoint to $u$, and
3. $v : \mathcal{X}' \to \mathcal{Y}$ is a fibred category.
Proof. Denote $p : \mathcal{X} \to \mathcal{C}$ and $q : \mathcal{Y} \to \mathcal{C}$ the structure functors. We construct $\mathcal{X}'$ explicitly as follows. An object of $\mathcal{X}'$ is a quadruple $(U, x, y, f)$ where $x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{X}_ U)$, $y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{Y}_ U)$ and $f : y \to F(x)$ is a morphism in $\mathcal{Y}_ U$. A morphism $(a, b) : (U, x, y, f) \to (U', x', y', f')$ is given by $a : x \to x'$ and $b : y \to y'$ with $p(a) = q(b) : U \to U'$ and such that $f' \circ b = F(a) \circ f$.
Let us make a choice of pullbacks for both $p$ and $q$ and let us use the same notation to indicate them. Let $(U, x, y, f)$ be an object and let $h : V \to U$ be a morphism. Consider the morphism $c : (V, h^*x, h^*y, h^*f) \to (U, x, y, f)$ coming from the given strongly cartesian maps $h^*x \to x$ and $h^*y \to y$. We claim $c$ is strongly cartesian in $\mathcal{X}'$ over $\mathcal{C}$. Namely, suppose we are given an object $(W, x', y', f')$ of $\mathcal{X}'$, a morphism $(a, b) : (W, x', y', f') \to (U, x, y, f)$ lying over $W \to U$, and a factorization $W \to V \to U$ of $W \to U$ through $h$. As $h^*x \to x$ and $h^*y \to y$ are strongly cartesian we obtain morphisms $a' : x' \to h^*x$ and $b' : y' \to h^*y$ lying over the given morphism $W \to V$. Consider the diagram
$\xymatrix{ y' \ar[d]_{f'} \ar[r] & h^*y \ar[r] \ar[d]_{h^*f} & y \ar[d]_ f \\ F(x') \ar[r] & F(h^*x) \ar[r] & F(x) }$
The outer rectangle and the right square commute. Since $F$ is a $1$-morphism of fibred categories the morphism $F(h^*x) \to F(x)$ is strongly cartesian. Hence the left square commutes by the universal property of strongly cartesian morphisms. This proves that $\mathcal{X}'$ is fibred over $\mathcal{C}$.
The functor $u : \mathcal{X} \to \mathcal{X}'$ is given by $x \mapsto (p(x), x, F(x), \text{id})$. This is fully faithful. The functor $\mathcal{X}' \to \mathcal{Y}$ is given by $(U, x, y, f) \mapsto y$. The functor $w : \mathcal{X}' \to \mathcal{X}$ is given by $(U, x, y, f) \mapsto x$. Each of these functors is a $1$-morphism of fibred categories over $\mathcal{C}$ by our description of strongly cartesian morphisms of $\mathcal{X}'$ over $\mathcal{C}$. Adjointness of $w$ and $u$ means that
$\mathop{Mor}\nolimits _\mathcal {X}(x, x') = \mathop{Mor}\nolimits _{\mathcal{X}'}((U, x, y, f), (p(x'), x', F(x'), \text{id})),$
which follows immediately from the definitions.
Finally, we have to show that $\mathcal{X}' \to \mathcal{Y}$ is a fibred category. Let $c : y' \to y$ be a morphism in $\mathcal{Y}$ and let $(U, x, y, f)$ be an object of $\mathcal{X}'$ lying over $y$. Set $V = q(y')$ and let $h = q(c) : V \to U$. Let $a : h^*x \to x$ and $b : h^*y \to y$ be the strongly cartesian morphisms covering $h$. Since $F$ is a $1$-morphism of fibred categories we may identify $h^*F(x) = F(h^*x)$ with strongly cartesian morphism $F(a) : F(h^*x) \to F(x)$. By the universal property of $b : h^*y \to y$ there is a morphism $c' : y' \to h^*y$ in $\mathcal{Y}_ V$ such that $c = b \circ c'$. We claim that
$(a, c) : (V, h^*x, y', h^*f \circ c') \longrightarrow (U, x, y, f)$
is strongly cartesian in $\mathcal{X}'$ over $\mathcal{Y}$. To see this let $(W, x_1, y_1, f_1)$ be an object of $\mathcal{X}'$, let $(a_1, b_1) : (W, x_1, y_1, f_1) \to (U, x, y, f)$ be a morphism and let $b_1 = c \circ b_1'$ for some morphism $b_1' : y_1 \to y'$. Then
$(a_1', b_1') : (W, x_1, y_1, f_1) \longrightarrow (V, h^*x, y', h^*f \circ c')$
(where $a_1' : x_1 \to h^*x$ is the unique morphism lying over the given morphism $q(b_1') : W \to V$ such that $a_1 = a \circ a_1'$) is the desired morphism. $\square$
[1] This is probably nonstandard terminology. In some texts this is called a “cleavage” but it conjures up the wrong image. Maybe a “cleaving” would be a better word. A related notion is that of a “splitting”, but in many texts a “splitting” means a choice of pullbacks such that $g^*f^* = (f \circ g)^*$ for any composable pair of morphisms. Compare also with Definition 4.35.2.
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2020-04-04 18:08:46
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