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Numerical scheme for treatment of Uehling-Uhlenbeck equation for two-particle interactions in relativistic plasma
Numerical scheme for treatment of Uehling-Uhlenbeck equation for two-particle interactions in relativistic plasma
Abstract
We present a new efficient method to compute Uehling-Uhlenbeck collision integral for all two-particle interactions in relativistic plasma with drastic improvement in computation time with respect to existing methods. Plasma is assumed isotropic in momentum space. The set of reactions consists of: Moeller and Bhabha scattering, Compton scattering, two-photon pair annihilation, and two-photon pair production, which are described by QED matrix elements. In our method exact energy and particle number conservation laws are satisfied. Reaction rates are compared, where possible, with the corresponding analytical expressions and convergence of numerical rates is demonstrated.
keywords:
Uehling-Uhlenbeck equations, collision integral, binary interactions, relativistic plasma.
\cortext
[cor1]Corresponding author
1 Introduction
Relativistic plasma, for which , where is Boltzmann constant, is speed of light, is electron mass, is temperature, is relevant in different branches of astrophysics. In the early universe ultrarelativistic electron-positron pairs contribute to the matter contents of the Universe [1]. X-ray and gamma-ray radiation from numerous astrophysical sources such as gamma-ray bursts [2]; [3]; [4], active galactic nuclei [5]; [6], and X-ray binaries [7] points out to existence of relativistic electron-positron plasma in these objects. The upcoming high-energy laser facilities aiming at generation of femtosecond laser pulses with intensity more than aim generation of relativistic plasma by interacting laser pulses. At present relativistic electron-positron jets are generated by interaction of laser pulses with condensed matter [8]; [9]; [10]; [11].
Solving the Boltzmann equations with collision integral containing a quantum cross-section represents the most general and complete method to describe a behavior of relativistic plasma [12]; [13]; [14]. The one-particle distribution function (DF) is defined on a seven dimensional space, three dimensions for the physical space and three dimensions for the momentum space, and one dimension for the time. Thus one has a multidimensional problem which is a real challenge from the computational point of view. Beside the dimensionality problem, there are other difficulties which are related to kinetic equations in general [15]; [16]. Our main goal in this paper is to tackle the challenge associated with the calculation of the collision integral, dealing with two key issues. First, the computational cost related to the evaluation of the collision operator involving multidimensional integrals which should be solved in each point of the coordinate space. Second, the presence of multiple scales requires the development of adapted numerical schemes capable of solving stiff dynamics. Different deterministic approaches are used to tackle collision integral from a numerical point of view: finite volume, semi-Lagrangian and spectral schemes [16]; [17]; [18]; [19]; [20]. While the deterministic methods could normally reach high order of accuracy, the probabilistic ones, such as Monte-Carlo (MC) method, are often faster.
MC methods are traditionally used to model Coulomb interactions in non-relativistic plasma[21]; [22]; [23]; [24]. As a rule MC techniques are based on the random pairing of particles in close vicinity and the calculation of a scattering angle due to the interaction. Small-angle Coulomb collisions which allow small energy and momentum transfer are often described in diffusion approximation by the Fokker-Planck equation [25]. The principal feature of relativistic plasma is a presence of pair creation and pair annihilation processes, which are often included in MC based models [9]; [10]. However Fokker-Planck approximation is no longer valid in relativistic plasma [26].
Classical Boltzmann equation does not take into account quantum statistics of particles. The generalization of classical Boltzmann equation including quantum corrections is Uehling-Uhlenbeck (U-U) equation, which contains additional Pauli blocking and Bose enhancement multipliers that give rise to equilibrium solution with Bose-Einstein and Fermi-Dirac distributions [27]; [28]. The main problem of the MC methods in application to U-U equations is that total reaction rate is unknown as distribution function is unknown too. Compensation methods include smoothing of the delta-function distribution of MC-particles over cells in the phase space, but it suffers from a large number of simulation particles and cells needed to reproduce Bose-Einstein steady state distribution. Spectral methods based on the Fourier transformation of the velocity distribution function require very dense computational grid to reach high accuracy [29]; [30]; [31]; [32]; [33]. Process-oriented approach to the U-U collision integral presented in this work allows one to get high accuracy results with low computational cost.
In this paper we further develop the method first used in the work [34]. This method was successfully applied to follow the thermalization of relativistic plasma [35]; [36]; [26]; [37] and to investigate thermalization timescales for an electron-positron plasma [38]. In section 2 we recall Boltzmann and UU equations and present usual scheme of their analytic treatment. Section 3 is devoted to the description of our numerical scheme, while section 4 shows comparison between our code results and known analytic formulae for non-degenerate case. Conclusion follows.
2 Formulation
The Boltzmann equation governs an evolution of one-particle distribution function . We assume that plasma is homogeneous and isotropic in coordinate space and isotropic in momentum space, thus distribution function depends on absolute value of momentum (energy) and time. DF is normalized on particles concentration, so that .
Consider an interaction of two initial particles of type I and II which are in states 1 and 2, correspondingly, and creation of two final particles of type III and IV which are in states 3 and 4, correspondingly. Let us image the process by the following scheme:
I1+II2→III3+IV4. (1)
The corresponding inverse process is:
III3+IV4→I1+II2. (2)
If every particle has momentum , which lies in interval , then a number of interactions in unit time value and unit space volume is:
w(p3,p4;p1,p2)fIfIId3p1d3p2d3p3d3p4, (3)
function is called a transition rate for a given reaction.
An effective cross-section of a given interaction is defined by the formula:
dσ=wureld3p3d3p4, (4)
where is a relative velocity of initial particles.
In quantum field theory an expression for interaction cross-section is:
dσ=ℏ2c6(2π)21urel|Mif|216ϵ1ϵ2ϵ3ϵ4δ(ϵ1+ϵ2−ϵ3−ϵ4)δ(p1+p2−p3−p4)d3p3d3p4, (5)
where are a matrix elements calculated with a methods of quantum field theory.
Comparing two last formulas one can derive the following expression for transition rate:
w(p3,p4;p1,p2)=ℏ2c6(2π)2|Mif|216ϵ1ϵ2ϵ3ϵ4δ(ϵ1+ϵ2−ϵ3−ϵ4)δ(p1+p2−p3−p4), (6)
Let us write the Boltzmann equation for DF of particle I for a given process:
˙fI=∫d3p2d3p3d3p4[w(p3,p4;p1,p2)fIII(p3,t)fIV(p4,t)−w(p3,p4;p1,p2)fI(p1,t)fII(p2,t)], (7)
equations for particle DFs of remaining types can be derived by the corresponding replacement of indices.
Specifically, for a scattering with and the inverse process is the same as the direct one since pairs of indices and can be interchanged. The relation holds for all processes listed in Table1. The right hand side of the Boltzmann equation is a collision integral denoted as St. The first term in collision integral describes outgoing particles which leave a given element of the phase space (denoted ) and the second term describes incoming particles which arrive to a given element of the phase space (denoted ).
For the particle in the state U-U equation has the following form:
˙fI=∫d3p2d3p3d3p4 ×[w(p3,p4;p1,p2)fIII(p3,t)fIV(p4,t)(1+υfI(p1,t)2h−3)(1+υfII(p2,t)2h−3)−w(p1,p2;p3,p4)fI(p1,t)fII(p2,t)(1+υfIII(p3,t)2h−3)(1+υfIV(p4,t)2h−3)], (8)
where is defined through
υ=⎧⎨⎩+1,for Bose-Einstein statistic,−1,for Fermi-Dirac statistic,0,for Maxwell-Boltzmann statistic. (9)
When incoming or outgoing particles coincide ( and/or ) quantum indistinguishability gives the term in front of the corresponding terms, see e.g. [39], [14].
For numerical evaluation phase space is divided into zones, in calculations we approximate continuous DF by its average over each zone (see Eq. (22) below). For this purpose we add an integral over in U-U equation (8). The right-hand side of resulting equation will have the same form for each particle type differing only by sign and its integration limits:
±∫d3p1d3p2d3p3d3p4 ×[w(p3,p4;p1,p2)fIII(p3,t)fIV(p4,t)(1+υfI(p1,t)2h−3)(1+υfII(p2,t)2h−3)−w(p1,p2;p3,p4)fI(p1,t)fII(p2,t)(1+υfIII(p3,t)2h−3)(1+υfIV(p4,t)2h−3)], (10)
where the upper sign corresponds to particles type I and II and the lower sign corresponds to particles type III and IV.
Particle in a given momentum state may participate in different reactions, which enter collision integrals multiple times. The essence of the reaction-oriented approach consists in evaluation of expression (10) to be used for calculation of collision integrals for all reactions simultaneously. In this way a given reaction with fixed particle state is considered only once, allowing considerable reduction in the computation time.
In this paper we deal with all two-particle QED processes in relativistic plasma, which are collected in Table 2. The exact QED matrix elements for given processes can be found in the standard textbooks, e.g. [40]; [41].
Let us make a remark regarding conservation laws for interacting particles. Energy and momentum conservations read [26]
^ε=ε1+ε2=ε3+ε4,^p=p1+p2=p3+p4. (11)
There are 4 delta-functions in Eq. (6) representing conservation of energy and momentum (11). Three integrations over momentum of particle can be performed immediately
∫dp3δ3(p1+p2−p3−p4)⟶1. (12)
In the integration over energy of particle it is necessary to take into account that is now a function of energy and angles of particles and , as well as angles of particle , so we have
∫dε4δ(ε1+ε2−ε3−ε4)⟶11−(β3/β4)n3⋅n4, (13)
where is the unit vector in the direction of particle momentum, is the absolute value of particle momentum, , and a dot denotes scalar product of 3-vectors.
We use spherical coordinates in momentum space: , , where is the particle energy, and and are polar and azimuthal angles, respectively. Then energy and angles of particle and energy of particle follow from energy and momentum conservations (11) and relativistic energy-momentum relation, namely
ε4=c√p24+m2IVc2,p4=p4n4, (14) ε3=^ε−ε4,p3=^p−p4, n3=p3p3,n4=p4p4, (15) n3=(√1−μ32cosϕ3,√1−μ32sinϕ3,μ3), (16) n4=(√1−μ24cosϕ4,√1−μ24sinϕ4,μ4), (17) p4=AB±√A2+4m2IVc2(B2−1)2(B2−1), (18) A=c^ε[^p2+(m2III−m2IV)c2]−^εc,B=c^εn4⋅^p. n3⋅n4=μ3μ4+√(1−μ23)(1−μ24)cos(ϕ3−ϕ4). (19)
Then we introduce these relations into collision integral (10). We also use spherical symmetry in momentum space to fix angles of the particle : , and to perform the integration over azimuthal angle of particle : , setting in the remaining integrals. Then final expression for collision integral is
StfI=ℏ232π∫dε2dμ2 dμ4dϕ4p2p4|Mfi|2ε1ε3[1−(β3/β4)n3⋅n4] ×[fIII(ε3,t)fIV(ε4,t)(1+υfI(ε1,t)2h−3)(1+υfII(ε2,t)2h−3)−fI(ε1,t)fII(ε2,t)(1+υfIII(ε3,t)2h−3)(1+υfIV(ε4,t)2h−3)]. (20)
For numerical integration, however, another expression is proved useful
StfI=ℏ232π[∫dε3 dε4dμ4 dμ2dϕ2δ(ε1−ε)p2p4|Mfi|2ε1ε3[1−(β1/β2)n1⋅n2]×fIII(ε3,t)fIV(ε4,t)(1+υfI(ε1,t)2h−3)(1+υfII(ε2,t)2h−3)−∫dε2dμ2 dμ4dϕ4p2p4|Mfi|2εε3[1−(β3/β4)n3⋅n4] ×fI(ε,t)fII(ε2,t)(1+υfIII(ε3,t)2h−3)(1+υfIV(ε4,t)2h−3)], (21)
where the first term is expressed in the form ready for replacement by the sum over incoming particles and . In this term are given by relations (18) with indices exchange , , , .
This collision integral of any of two-particle processes is a four-dimensional integral in momentum space. In Sec. 3 we show how such integral is computed numerically on a finite grid.
Here we note that in the case of homogeneous and isotropic pair plasma one has to satisfy only two conservation laws, namely energy and particle conservation laws. Momentum conservation law should be added for anisotropic in momentum space DF, see e.g. [42]. In our method electric charge is conserved due to conservation of particles because we use between cell interpolation for the same kind of particles.
3 Numerical Scheme
The phase space is divided in zones. The zone for particle specie corresponds to energy , cosine of polar angle and azimuthal angle , where indices run in the following ranges , , and . The zone boundaries are , , . The length of the -th energy zone is . On finite grid does not depend on and , and number density of particle in zone is
Yαa(t)=4π∫εa+1/2εa−1/2c−3ε√ε2−m2αc4fα(ε,t)dε=4πc−3εa√ε2a−m2αc4fα(εa,t)Δεa. (22)
In this variables discretized U-U equation for particle and energy zone reads
dYαa(t)dt=∑[St+YIa+St−YIa], (23)
where the sum is taken over all processes involving particle . Coefficients of particles income and outcome on the grid are obtained by integration of (21) for two-particle processes over the zone. The corresponding integrals are replaced by sums on the grid. For instance, coefficient of particles income in two-particle process (1) is
St+YIa=ℏ2c48(4π)2∑b,j,s,kΔμIIjΔμIVsΔϕIVk|Mfi|2p4ε3[1−(β3/β4)n3⋅n4]××YIa(t)εIaYIIb(t)εIIb×[1+υYIIIc(t)¯YIIIc][1+υYIVd(t)¯YIVd]. (24)
Coefficient of particles outcome in process (2) from integration of (21) is
Extra open brace or missing close brace (25)
and
Ca(ε1)=⎧⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪⎩εa−ε1εa−εa−1,εa−1<ε1<εa,εa+1−ε1εa+1−εa,εa<ε1<εa+1,0,otherwise. (26)
In integration of (21) over the zone one can integrate out the -function . However, when energies of incoming particles are fixed on the grid, the energies of outgoing particles are not on the grid. Hence an interpolation (26) is adopted, which enforces the exact number of particles and energy conservation in each two-particle process due to redistribution of outgoing particle with energy over two energy zones with .
The redistribution of final particles should also satisfy requirements of quantum statistics. Therefore if a process occurs, when final particle should be distributed over the quantum states which are fully occupied, such process should be forbidden. Thus we introduce the Bose enhancement/Pauli blocking coefficients in (24) and (25) as
[1+υYαa(t)¯Yαa]=min(1+υYαn(t)¯Yαn,1+υYαn+1(t)¯Yαn+1). (27)
The sum over angles can be found once and for all at the beginning of the calculations. We then store in the program for each set of the incoming and outgoing particles the corresponding terms and redistribution coefficients given by Eq. (26).
Representation of discretized collisional integral for particle and energy zone in processes (1), (2), is
˙YIa=∑Pabcd×YIIIc(t)YIVd(t)×[1+υYIa(t)¯YIa][1+υYIIb(t)¯YIIb]−∑Rabcd×YIa(t)YIIb(t)×[1+υYIIIc(t)¯YIIIc][1+υYIVd(t)¯YIVd], (28)
where constant coefficients are obtained from the summation over angles in the sums (24), (25). In the nondegenerate case of Boltzmann equation the indices in the first sum and in the second sum become dummy, equation (28) can be partially summed and takes the following form:
˙YIa=∑Pacd×YIIIc(t)YIVd(t)−∑Rab×YIa(t)YIIb(t), (29)
where . The last quantity is essentially reaction rate usually used for description of binary processes simply connected to the total cross section. The full U-U equation (23) contains similar sums for all processes from Table 2. Each individual term in these sums appears in the system of discretized equations four times in emission and absorption coefficients for each particle entering a given process. Then each term can be computed only once and added to all corresponding sums, that is the essence of our “reaction-oriented” approach [43]; [44].
We point out that unlike classical Boltzmann equation for binary interactions such as scattering, more general interactions are typically described by four collision integrals for each particle that appears both among incoming and outgoing particles.
4 Numerical results
The results of numerical calculations are presented below. As all known analytical expressions for reaction rates in relativistic plasma concern nondegenerate case, we present our results for nondegenerate plasma. We do not present numerical results for Coulomb scattering due to absence of the corresponding analytical results. Notice that for Coulomb scattering we have implemented a cutoff scheme based on minimal scattering angle [26]; [45]. We consider mildly relativistic plasma with
0.01≲e≲100, (30)
where is particle kinetic energy divided by electron rest energy, this range contains both relativistic and non-relativistic domains. The upper limit is chosen to avoid thermal production of other particles such as neutrinos and muons, while the lower limit is required to have sufficient pair density.
We introduce logarithmic energy grid with nodes for all calculations and different homogeneous grids for angular variables, -grid is 2 time denser then -grid (typically -grid contains nodes). We use Coppi&Blandford [46] analytical expressions (2.3),(3.2),(4.3) for , which corresponds to . Belmont [47] formula (10), Svensson [48] formula (55), Peer&Waxman [49] formula (28) are used for quantity , which corresponds to .
To compare numerical results with analytical one, we introduce the following quantity for each given process
Q=1a2max∑a,b|Rab/¯¯¯¯¯¯vσ(ea,eb)−1|, (31)
expressing relative deviations of numerical results from analytical one for all energy grid nodes, where is analytical expression for reaction rate. Table 2 presents values of for selected number of angular grid nodes. It is evident that the relative error decreases with increasing of number of angular grid nodes reaching about 1 % with 128 nodes. This demonstrates convergence of numerical results to the corresponding analytical ones.
Below we present several plots for the reaction rates of Compton scattering and creation/annihilation of pairs. Energy is measured in electron rest energy units. Presented results reproduce both nonrelativistic and relativistic energy cases.
Compton scattering presents well-known challenge for numerical treatment as all the analytical formulas for scattering rate behave badly numerically in different parameter areas, see e.g. [49]; [47]. We easily bypass this difficulty as we numerically integrate well-behaved differential cross-section, as one can see for non-relativistic regime in Fig.1 and for relativistic regime in Fig.2. Figure 1 presents analytic photon spectrum for the reaction as solid line and our numerical results shown by dots. Overall there is good agreement between numerical and analytical results. Small deviations in high-energy of the spectrum arise from leakage of the particles to kinematically forbidden area at the boundary of energy zones.
Figure 2 shows the total reaction rate of the same process. Again there is good agreement between numerical and analytical results. Small discrepancy arises from truncation of reactions where final particles get out of the grid to higher or lower energies. As a result numerical reaction rates are systematically lower than analytic ones.
Annihilation photon spectrum for reaction is illustrated in Fig.3 and total reaction rate in this process in Fig.4. Figure 3 shows that the method is able to accurately reproduce the spectrum of annihilation photons in the range of more than two orders of magnitude. Reaction truncation errors, hardly seen at Fig.4, are much lower for annihilation as low-energy photons are rare in this process.
Balance between pair creation and annihilation represent an independent test for the numerical scheme, as it is not automatically satisfied due to different numerical treatment of incoming and outgoing particles in the reactions. Pair creation spectra for reaction are reproduced well, see Fig. 5, as well as total reaction rates, see Fig. 6. Numerical balance can be checked by the form of particle distributions in numerical equilibrium, that was verified to be within 5 % of corresponding Boltzmann distributions.
5 Conclusions
In this paper, we propose a new numerical method to accurately calculate Uehling\unichar8211Uhlenbeck collision integral for two-particle interactions in relativistic plasma. After calculation of collision integral discretized Uehling\unichar8211Uhlenbeck equations transforms into system of ODEs, which can be treated by various methods suitable to solve stiff ODEs. The method admits parallelization on GPU/CPU. Improvement in computation time with respect to previous work is achieved. Our reaction-oriented approach can be easily applied to any other types of particles and any other binary interactions, for instanse, weak interactions of neutrinos or electromagnetic ones of protons. Generalization of the proposed method for triple interactions is straightforward.
Our results show that reaction rates in relativistic plasma are well reproduced with moderate number of grid nodes in energy and angles (see Figures and Table 2) both for non-relativistic and relativistic particle energies. This allows development of an efficient method of solution for relativistic Uehling\unichar8211Uhlenbeck equation.
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44. I. A. Siutsou, A. G. Aksenov, and G. V. Vereshchagin. On thermalization of electron-positron-photon plasma. In American Institute of Physics Conference Series, volume 1693 of American Institute of Physics Conference Series, page 070007, December 2015.
45. E. Haug. Energy loss and mean free path of electrons in a hot thermal plasma. Astronomy and Astrophysics, 191:181–185, February 1988.
46. P. S. Coppi and R. D. Blandford. Reaction rates and energy distributions for elementary processes in relativistic pair plasmas. MNRAS, 245:453–507, August 1990.
47. R. Belmont. Numerical computation of isotropic Compton scattering. Astronomy and Astrophysics, 506:589–599, November 2009.
48. R. Svensson. The pair annihilation process in relativistic plasmas. Astrophysical Journal, 258:321–334, July 1982.
49. A. Pe’er and E. Waxman. Time-dependent Numerical Model for the Emission of Radiation from Relativistic Plasma. Astrophysical Journal, 628:857–866, August 2005.
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[Tex/LaTex] Bibliography style with elsarticle
elsarticle
I have trouble getting the form of citations I need. Once I cite I want to see authors' names and not numbers. The unconventional thing is that I have to use the documentclass elsarticle (This is editor's request), which always produce numbers. Here is it what I have
\documentclass{elsarticle}
\usepackage{natbib}
\usepackage[colorlinks=true,linkcolor=black, citecolor=blue, urlcolor=blue]{hyperref}
\begin{document}
\bibliographystyle{plainnat}
\bibliography{references}
\end{document}
Can someone help me on this?
Here's an expanded MWE.
\documentclass{elsarticle}
\usepackage{amssymb}
\usepackage{graphicx}
\usepackage{amsmath, mathtools}
\usepackage{natbib}
\usepackage[colorlinks=true,linkcolor=black, citecolor=blue, urlcolor=blue]{hyperref}
\usepackage{changes}
\usepackage{appendix}
\usepackage{filecontents}
\usepackage{multirow}
\usepackage{booktabs}
\begin{document}
\section{The Model}
The main environment modifies that of \cite{Diamond1965}, into two main dimensions
\newpage
\bibliographystyle{elsarticle-harv}
\bibliography{references}
\end{document}
Best Answer
You should specify authoryear in the class options.
\begin{filecontents*}{\jobname.bib}
@article{Diamond1965,
author={J. A. L. Diamond},
title={Some Like It Hot},
journal={J. High Level Cinema},
year={1965},
}
\end{filecontents*}
\documentclass[authoryear]{elsarticle}
\usepackage[colorlinks=true,linkcolor=black, citecolor=blue, urlcolor=blue]{hyperref}
\begin{document}
\section{The Model}
The main environment modifies that of \cite{Diamond1965}, into two main dimensions
\bibliographystyle{plainnat}
\bibliography{\jobname}
\end{document}
The filecontents* environment is just to make the example selfcontained, you don't need it; change the \bibliography line to have the name of your .bib file.
Yes, I know that “Some Like It Hot” wasn't shot in 1965. ;-)
Adding the authoryear to natbib is not a solution, because the elsarticle class already loads it, so having
\usepackage[authoryear]{natbib}
in the preamble will only cause an error about conflicting options.
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Miami Pedestrian Bridge, Part II55
Miami Pedestrian Bridge, Part II
(OP)
Site management has requested that we limit the length of the other thread by forming a new one. This subject may require III, IV....
RE: Miami Pedestrian Bridge, Part II
Quote (BARetired)
The end diagonal, column and canopy slab still seem to form a triangle after collapse, although the connection between the end diagonal and column cannot be seen.
That triangle is the remaining parts of the members. The initial triangle is bigger.
RE: Miami Pedestrian Bridge, Part II
Quote (hoikie66)
All the photos show me that the failure was a joint failure rather that a member failure. Both ends of Member 10 developed hinges at the same time.
Can you points out these hinges in #10? #10 appears intact and aligned with its nodes until after collapse from what I can see. #11 deforms and loses alignment with the top node from the outset of collapse. My money is on the top of #11.
RE: Miami Pedestrian Bridge, Part II
For the benefit of PART II of this discussion, and so we are all talking about the same member #'s, here is a reference graphic:
RE: Miami Pedestrian Bridge, Part II
It should be interesting to know where the cracking occurred.
Dik
RE: Miami Pedestrian Bridge, Part II
When bridge consultants design cable stayed bridges for state DOTs...typically there are scale models of the bridge constructed...where the structure is put in a wind tunnel and wind analysis is performed...(you can't look up in ASCE-7 and get wind loads on something like this...I wonder if a scale model of the bridge was created (maybe at FIU?)..not only for the wind loads in Miami...but also for testing of the gravity loads... As Hokie has stated from the beginning of this thread (as he knows his stuff!) a concrete truss was a bad idea...I had never even heard of a concrete truss until this hit the news...I don't think I've ever seen one in person.
RE: Miami Pedestrian Bridge, Part II
I think this really opens up a can for State DOT's as well...I would say in most of the US, that pedestrian bridges are not designed by bridge engineers, but are designed by building engineers (or at least in my state). A client (hospital or school) will hire a building engineer and the building engineer submits plans to have a bridge over a DOTs right-of-way to the state. Typically, there is no secondary check from a different consultant required as there was in this case (I'm guessing that was required because federal funds were involved).
But from a liability perspective, if a state DOT looks at the plans prepared by the consultant, how much liability does he take on? Even though the state DOT is not likely not an expert in the structure being designed (and most bridge engineers aren't even familiar with ASCE 7 loading)...if a State DOT PE reviews the plans and then approves it...IMO he is taking on some liability with the structure.
Sorry...bit of a tangent :)
RE: Miami Pedestrian Bridge, Part II
(OP)
Tomfh,
I don't really know which node failed first. They were probably all failing. But in the dashcam video, both the top and bottom chord hinged at the ends of member 10.
RE: Miami Pedestrian Bridge, Part II
Quote (Structuralengr89)
.IMO he is taking on some liability with the structure.
Not a tangent, but, the next phase... I'm surprised a lawsuit has not already been filed.
Moreso in the US, than in other parts of the world, the lawyer will use a 'scattergun' approach and name all parties. If the lawyer misses one that he should have included, he can be sued.
FDOT, the city, the University, the consultants, the constructor, and any of the sub-trades are fair targets...
Dik
RE: Miami Pedestrian Bridge, Part II
Quote (Hokie)
But in the dashcam video, both the top and bottom chord hinged at the ends of member 10.
I couldn't tell if the failure was at the ends of the member or at the panel point the members framed in to. Could you?
Dik
RE: Miami Pedestrian Bridge, Part II
Quote (dik)
FDOT, the city, the University, the consultants, the constructor, and any of the sub-trades are fair targets...
Well, targets certainly; 'fair' is debatable. Probably include the concrete testing lab, special inspectors, material suppliers, etc, to the list too.
RE: Miami Pedestrian Bridge, Part II
(OP)
Just my opinion, but I think the truss was doomed to fail at the joints. At some stage, I hope we see some further information about the details, not just how the PT was arranged and anchored, but what was supposed to reinforce the joints. Based on the reporting which the NTSB released after the I35W collapse, we will eventually find out. It is ironic that Figg was the designer of the I35W bridge replacement.
RE: Miami Pedestrian Bridge, Part II
Quote (hokie66)
They were probably all failing. But in the dashcam video, both the top and bottom chord hinged at the ends of member 10.
Can you post a screen grab of the phenomena you are talking about?
RE: Miami Pedestrian Bridge, Part II
Quote (dik)
I'm surprised a lawsuit has not already been filed.
The insurers would all be notified already, and are no doubt working out their respective positions regarding the inevitable claims.
RE: Miami Pedestrian Bridge, Part II
Quote (Hokie)
Just my opinion, but I think the truss was doomed to fail at the joints. At some stage, I hope we see some further information about the details, not just how the PT was arranged and anchored, but what was supposed to reinforce the joints.
Just realised that I couldn't find any reference to confinement reinforcing in the anchorage zones... those high compressive loads from the post-tensioning would give rise to some high tensile stresses in the vicinity of the anchor.
Dik
RE: Miami Pedestrian Bridge, Part II
Quote (dik)
Just realised that I couldn't find any reference to confinement reinforcing in the anchorage zones... those high compressive loads from the post-tensioning would give rise to some high tensile stresses in the vicinity of the anchor.
Yes the stresses would be fairly intense. The node region would be working real hard. I was wondering if the temporary rod was acting as confinement in some way, and when it broke or was released it led to the overall rupture.
RE: Miami Pedestrian Bridge, Part II
Quote (hokie66)
I don't really know which node failed first. They were probably all failing. But in the dashcam video, both the top and bottom chord hinged at the ends of member 10.
hokie66....I agree that hinges formed but it looks like those occurred first at the top of members 11/10, then at the bottom of 11 and then at the bottom of 10. But....sometimes hard to tell failure sequence in progressive collapses.
My best guess, and I assure you that's all it is, is that there was buckling or punching shear at the top of 11, rotational failure at the bottom of 11, then tension failure at the bottom of 10.
Issues such at these usually get solved by the numerous ideas posed in such conversation as we are all having. Hopefully these discussions will lead to someone reaching a logical conclusion of the failure, but as a group we don't have the benefit of the cumulative information from design and construction that will inevitably be forthcoming...either through determined analysis or the litigation process. The resolution of such failures is important to the future success of innovative design and construction, it's just unfortunate that such a high price had to be paid for this lesson.
RE: Miami Pedestrian Bridge, Part II
Just in case anyone is interested in a rough idea of the tension / compression loading of the trusses when:
A-In transit supported underneath in 2 places mid span
B-In Situ supported at either end of the span
These are not to scale or necessarily accurate, but give a general idea. Draw your own conclusions here.
RE: Miami Pedestrian Bridge, Part II
Quote (Ingenuity)
Well, targets certainly; 'fair' is debatable. Probably include the concrete testing lab, special inspectors, material suppliers, etc, to the list too.
Concur... had lumped the others under sub-contractors... Wonder if VSL were involved with stressing the cables... They've been involved with post-tensioning for decades that I'm aware of, and are very good...
Quote (tomfh)
The insurers would all be notified already, and are no doubt working out their respective positions regarding the inevitable claims.
Shortly after it hit the ground...
Dik
RE: Miami Pedestrian Bridge, Part II
Member 11 may have been a tension member during construction but it was a compression member after the temporary supports were removed. Assuming a total weight of 950 tons, the bridge reaction at each end would have been in the order of 950 kips under dead load only. Member 11 appears to be oriented at about 35o to the horizontal, so it would have been loaded to about 1650 kips, a compressive stress of 3,300 psi on a 24" x 21" section under dead load only. It did not need any prestress at that stage.
BA
RE: Miami Pedestrian Bridge, Part II
Nobodyimportant:
Your arrows are drawn to show the forces in the members and not the forces at the panel points; is that correct? Nice drawings...
And add to that the forces induced by the 'fake' cable stays, it gets more interesting. These additional forces should be included in the design for the blocks at the tensioning cable intersections.
Just because the cable stays are fake, doesn't mean they don't take load or influence forces in other members.
Dik
RE: Miami Pedestrian Bridge, Part II
(OP)
Tomfh,
I was just referring to the deflected shape of the truss as it fell. You posted it. Both the top and bottom chord kinked at the ends of Member 10. But I don't know where a failure first occurred.
RE: Miami Pedestrian Bridge, Part II
Just a rough average spread of weight across the span to give a rough view of either tension or compression in both load scenarios.
RE: Miami Pedestrian Bridge, Part II
Quote (dik)
Wonder if VSL were involved with stressing the cables...
Yes, VSL (USA licensee is STRUCTURAL Group) did the supply, install and stressing of the PT - $440K subcontract sum. Source: Link And one of their field crew members died in the collapse - RIP. RE: Miami Pedestrian Bridge, Part II (OP) If member 11 failed first, why not member 2, which would have had more force and was longer? RE: Miami Pedestrian Bridge, Part II Ingenuity: Thanks for the info... and, as you noted in an earlier post, one of their employees lost his life in the collapse. RIP, too. Dik RE: Miami Pedestrian Bridge, Part II Ron: Were you able to determine if the failure occurred at the ends of the web members or at the panel point? Dik RE: Miami Pedestrian Bridge, Part II In an ideal world, the DOT would take no responsibility if it was neither the designer or the client (ie the commissioner of the bridge design and construction). Its review would be reviewing the road with a new obstruction/hazard built by others. Eg is the bridge high enough for legal overheight trucks, allows for future resurfacing, doesn't block planned widening, piers are protected if within the traffic hazard zone. In reality, it has deep pockets and Florida probably has joint and several liability (haven't checked), so it will be drawn in and may pay up if the insurance of the real culprit doesn't. RE: Miami Pedestrian Bridge, Part II (OP) dik, As to your question of Ron, what's the difference? The ends of the web members and the panel points are the same thing. RE: Miami Pedestrian Bridge, Part II Quote (Tomfh) I was wondering if the temporary rod was acting as confinement in some way If it were acting in compression, it would be aggravating the problem, I would think. Dik RE: Miami Pedestrian Bridge, Part II Quote (Hokie) The ends of the web members and the panel points are the same thing. A failure at the end of a member may be different than the panel point 'crushing' or 'exploding'. Dik RE: Miami Pedestrian Bridge, Part II Does anyone else not like the V shaped profile under the deck, and the SPMT using quite narrow wedge shaped load spreaders on top the shoring? Hope they loaded it in the right places too, that truss arrangement has some significant load complexities. RE: Miami Pedestrian Bridge, Part II From FOX news (I'm not a subscriber), "The two firms responsible for building Florida International University’s "instant bridge," which suddenly collapsed Thursday and left six people dead, are coming under increased scrutiny as details emerge of past engineering failures and inspection fines -- including a recent accusation that one hired “unskilled” and “careless” workers." Dik RE: Miami Pedestrian Bridge, Part II Nobody important - This was one of my concerns, particularly after re-watching the the full length bridge move & the pause, that took place. Driving over the center divider did tilt the bridge. Reading up on the SPMT vehicles, some of them are built with telescopic axles to accommodate uneven ground. Still, it has to be done correctly. RE: Miami Pedestrian Bridge, Part II Okay, this has been really eating me up... #11 is clearly a compression member in the configuration at failure. So, why was it not detensioned immediately after/during placement? Or if it was, why was it (presumably) being tensioned after the fact? This is the single biggest mystery to me. RE: Miami Pedestrian Bridge, Part II Does anybody have experience with the capability of the self-propelled modular transporters to keep the bridge lifted at a more or less constant plane, avoiding transferring ground irregularities, like cross slope of the road and road separator, to the structure? It seems to me that the I-beam shape of the bridge made it especially vulnerable to twisting loads and deformations, which could deteriorate the nodes of the web. "Where the spirit does not work with the hand, there is no art." - Leonardo da Vinci RE: Miami Pedestrian Bridge, Part II Not knowing, but wouldn't the jack on the rod have to work up to the load in the rod and a little more so the bracket on the rod cold either be taken off permanently or at least until some of the load was released and then re-set. Maybe it got away with them on the pull to come up to the load in the rod, just enuf extra to tear the end anchor loose. Or sufficient pull to cause a compression failure at a joint. RE: Miami Pedestrian Bridge, Part II Re: The blue box. It appears that hydraulic lines are leading to the box and there appears to be a pressure gauge attached to the box. I'm betting that it is the hydraulic pump for the tensioning jack. Bill -------------------- "Why not the best?" Jimmy Carter RE: Miami Pedestrian Bridge, Part II Quote (hokie66) If member 11 failed first, why not member 2, which would have had more force and was longer? It's a larger member RE: Miami Pedestrian Bridge, Part II Quote (oldestguy) Not knowing, but wouldn't the jack on the rod have to work up to the load in the rod and a little more so the bracket on the rod cold either be taken off permanently or at least until some of the load was released and then re-set. Yes, to achieve 'lift-off', whereby the coarse-thread nut "just" achieves 'daylight' from the bearing plate, the stressing load is a small increment above the pre-existing PT bar load. For manual systems, usually the 'stressing stool' has an opening on one face enabling a technician to have a open wrench on the nut with a human-effort to verify that lift has been achieved (when the nut part-turns). Other systems use a displacement gauge to verify lift-off. For more sophisticated systems, the stressing stool has a in-built, chain-driven wrench, however, the principle is the same. Keep mind that the hyd pump operator is monitoring the applied force via gauge pressure. I have often used a load cell 'in-line' to verify applied loads with a bit more accuracy - especially true in proof or performance testing of rock/soil anchors. RE: Miami Pedestrian Bridge, Part II Quote (dik) If it were acting in compression, it would be aggravating the problem, I would think. What I meant was, maybe that additional compressive load was stopping it from failing in some other way, as opposed to a compressive failure. There would be tensile stresses trying to leak out in all directions... RE: Miami Pedestrian Bridge, Part II Quote (Lnewqban) It seems to me that the I-beam shape of the bridge made it especially vulnerable to twisting loads and deformations, which could deteriorate the nodes of the web. Agree. The truss members have to stabilize the top. RE: Miami Pedestrian Bridge, Part II Quote (hokie66) Both the top and bottom chord kinked at the ends of Member 10. But I don't know where a failure first occurred. Yeah the top and bottom chord kinked, but it appears to be beyond the panel points. I can't see rotation of the ends of #10 or the respective panel points themselves. That's why I was asking... RE: Miami Pedestrian Bridge, Part II Regardless of the stresses, the canopy portion of the bridge seems to be the weak link compared to the walkway. It appears it failed here first RE: Miami Pedestrian Bridge, Part II Lnewqban - you raise two points I was interested in. It is essentially an I beam with holes poked in the web, which is not a great recipe for torsional stiffness, and the loads while installing or constructing a bridge have always fascinated me. In this case trundling down the highway with the thing on a couple of platforms seems like a design case nobody could really pin down without measuring the road profile etc. As an expansion, if the I beam was twisted at some point, would that stress the tendon anchorages in ways that would not normally be considered in the design process? Cheers Greg Locock New here? Try reading these, they might help FAQ731-376: Eng-Tips.com Forum Policies http://eng-tips.com/market.cfm? RE: Miami Pedestrian Bridge, Part II Quote (Tomfh) Agree. The truss members have to stabilize the top. There were some sets of top flange/canopy stabilizer cables installed at truss ends, and at transporter support points: RE: Miami Pedestrian Bridge, Part II I think #11 failed because one of the tensioning rods ruptured, leaving an asymmetric preload. It may not have helped that the bottom end of the rod blew out some concrete. One thing that isn't clear to me is the way loads are transferred from the tendons in the deck to members #2 and #11. RE: Miami Pedestrian Bridge, Part II Quote (3DDave) One thing that isn't clear to me is the way loads are transferred from the tendons in the deck to members #2 and #11. Not a lot of detail/s on this, but there was a thickened end-diaphragm where the bottom chord PT terminates and where the diagonals intersect. RE: Miami Pedestrian Bridge, Part II 2 @ Lnewqban I have a bit of experience with SPMTs. They are really a wonderfully designed piece of equipment, there are several manufacturers now but they all operate the same way. All of the axle lines during a normal transport are on the same hydraulic circuit which allows each axle to stroke up and down independent of adjacent axle lines while maintaining a constant bearing, so crossing normal bumps and slight elevation changes are no problem. But the axles only have 1ft of up/down stroke so there are limits. Also they talk about them in terms of the number of axles but there really are no axles, they have independent, two tire hubs on either side of an axle line. And just because I think they are so cool, here's some more. Any hub can be isolated and raised, for instance if a tire blows out. They can turn the hubs on either side of the axle line independent from each other so you can walk the trailer nearly 90 deg from its long axis, or almost pivot in place, wonderful things. RE: Miami Pedestrian Bridge, Part II It's pretty clear from the pictures they had a hydraulic jack attached to the PT rod of a compression member. I understand why it was needed for transport, but once it was placed on the permanent supports it went from a tension member to a compression member. They were jacking a PT rod in a compression member because in order to de-tension it you have to actually increase the tension first? Isn't that just putting more compression force into a compression member? If,during post tensioning, you damage the member or the member's connection to the rest of the truss you're screwed because there isn't any redundancy. How many times have you heard of something getting "blown out" during a post tensioning job? It seems to me it would take a pretty darn sophisticated structural model to account for the affect of all these post tension forces in addition to the forces in the truss to due self weight. Build a steel truss, if you need extra weight use an oversized concrete deck. Sometimes stuff seems really cool on paper, but it sucks in real life. RE: Miami Pedestrian Bridge, Part II waross, Re: The blue box. Yes, you are correct. That has already been established in part I of this thread. Ingenuity even identified the manufacturer and model of the pump. He/she seems to have experience with it's use. RE: Miami Pedestrian Bridge, Part II Quote (Tomfh) Agree. The truss members have to stabilize the top. and because of the length, they would also have to stabilise the unbalanced load from the bottom flange, too or at least a large portion of it. Dik RE: Miami Pedestrian Bridge, Part II OSUCivlEng, the point you make about the PT in #11 probably being intended for the transport phase and then detensioning for the permanent phase makes sense and has been assumed by several of us here. The timing seems odd to me, if that was the intent of the design and the construction sequencing, wouldn't they have detensioned #11 before Thursday if the bridge was set on Saturday. There is a video of someone from the construction company mentioning detensioning 2 cables on top of the bridge after the bridge was set. I think the video is from the day the bridge was set (Saturday). It is not clear whether he was referring to the 2 PT rods in #11 or 2 of the longitudinal tendons in the canopy/top chord, but either way it seems like it would have been done before Thursday. Does that mean that the work being done Thursday was not the normal design/construction sequence, but was some sort of remediation for another problem, maybe the cracking that has been reported? Any thoughts? RE: Miami Pedestrian Bridge, Part II Following on from my last post... if the intent was to detension #11 then wouldn't #2 probably get the same treatment? The construction company video mentioned detensioning 2 cables, but we think that #11 and #2 had 2 rods each, 4 total, so did he mean 2 members, or was he referring to 2 longitudinal tendons in the canopy/top chord and not the rods in the diagonals? A lot of unknowns. It will be interesting to learn the details when they come out. RE: Miami Pedestrian Bridge, Part II Here is a photo of a closer view of the north-end failure, from NTSB video: And another showing spalling to underside of member #11 - the same side that the PT bar that was being stressed/de-stressed was located: NTSB B-roll video here: Link RE: Miami Pedestrian Bridge, Part II Correct me if I'm wrong, but, with a prestressed compression member, the prestressing does not increase the compression loading... The member is loaded in compression until the prestressed value is reached before the stress increases in the section. Is that not correct? Dik RE: Miami Pedestrian Bridge, Part II Quote (Ingenuity) And another showing spalling to underside of member #11 - the same side that the PT bar that was being stressed/de-stressed was located: So there were two bars in #11? Not a central bar? RE: Miami Pedestrian Bridge, Part II Quote (dik) Correct me if I'm wrong, but, with a prestressed compression member, the prestressing does not increase the compression loading... The member is loaded in compression until the prestressed value is reached before the stress increases in the section. Is that not correct? The loads are additive. RE: Miami Pedestrian Bridge, Part II Nobody important (Computer) Quote (Nobody Important) 18 Mar 18 23:32 Just in case anyone is interested in a rough idea of the tension / compression loading of the trusses when: A-In transit supported underneath in 2 places mid span B-In Situ supported at either end of the span These are not to scale or necessarily accurate, but give a general idea. Draw your own conclusions here. That (rough) compression-tension FEA sketch is logical, and explains how the top members are in great compression across greater length/limited thickness elements = classic buckling failure at the odd-angled connections of each member that is in compression. But - Why would the "missing" center vertical temporary vertical pier NOT be installed? Does the analysis/design team truly think that a cable-stayed bridge only needs the cable stays and central pier for "looks"? RE: Miami Pedestrian Bridge, Part II (OP) That's what they thought. It wasn't a cable stayed bridge, it was a truss, unfortunately made of the wrong material. RE: Miami Pedestrian Bridge, Part II Quote (Tomfh) The loads are additive Thanks for the clarification... I seemed to recall from Lin's book that the precompression was 'absorbed' until it was exceeded by the axial load, was also thinking of preloaded A325 bolts not increasing in tensile stress until the preload was attained... never done work with prestressed compression members. Dik RE: Miami Pedestrian Bridge, Part II No dik, that is not correct. If a member is prestressed, applying external compression will relieve some of the prestress (because of the change in strain) but not all of it. If a member is compressed by externally applied load and then it is prestressed, the two effects will be additive. BA RE: Miami Pedestrian Bridge, Part II Thanks BART... I'll have to look into it... it's been 40+ years since I read Lin's book... read it starting Friday till Monday night, day and night and had about 1" of notes... can't do that now. When the load is equal to the prestressed load, what is the stress in the column? Dik RE: Miami Pedestrian Bridge, Part II Quote (dik) When the load is equal to the prestressed load, what is the stress in the column? If the load is applied first, then the prestressing, the stress is 2P/A where P is the applied load. If the prestressing is applied first, then the load, the stress is less than 2P/A but close to it. BA RE: Miami Pedestrian Bridge, Part II Quote (BARetired) If the prestressing is applied first, then the load, the stress is less than 2P/A but close to it. ...and then greater when they repull the rod. RE: Miami Pedestrian Bridge, Part II Quote (Tomfh) So there were two bars in #11? Not a central bar? Correct, as per the attached photo: RE: Miami Pedestrian Bridge, Part II It seems clear that the bottom chord was not attached to the pier, so it relied on attachment to the end diagonal and would have a tension equal to the horizontal component of the end diagonal (#11) excluding any prestressing force. The collapse is consistent with the bottom chord pulling away from Member #11. BA RE: Miami Pedestrian Bridge, Part II This from the commercial press: Munilla Construction Management (MCM), the South Miami-based firm that designed the FIU foot bridge, has been sued multiple times for unsafe practices in the past. In early March MCM was sued by a construction worker who was severely injured when MCM’s “makeshift bridge” at Miami’s International Airport collapsed. MCM is a Cuban-American, family-owned Miami company founded in 1983 that employs more than 1,000 people in several states. The company is a federal military contractor for the U.S. Army and Navy. MCM was awarded the$14.2 million minority contract to design and build the cable-stayed bridge. The company is well-connected in Miami politics and it promotes inclusion and diversity in the workforce.
Employees at MCM | Munilla Construction Management
Melanie Rowan, P.E.
Senior Project Manager
Nelson Gomez Jr
Project Engineer
Maira Suarez
Alex Suarez
Senior Project Manager
Nelson Nunez
Fleet Manager
RE: Miami Pedestrian Bridge, Part II
Quote (racookpe)
This from the commercial press:
Munilla Construction Management (MCM), the South Miami-based firm that designed the FIU foot bridge, has been sued multiple times for unsafe practices in the past.
I thought it was FIGG who designed the bridge.
BA
RE: Miami Pedestrian Bridge, Part II
Quote (BAretired)
If a member is prestressed, applying external compression will relieve some of the prestress (because of the change in strain) but not all of it. If a member is compressed by externally applied load and then it is prestressed, the two effects will be additive.
Is this not the likely culprit?
I know it's too soon to know, but:
• Say #11 was prestressed to resist the tension caused by the temporary transport load (the end of the truss was basically being hung by #11). No numbers here -- just call that Tension Force A.
• The bridge gets placed on the abutment. Load reversal city as the temporary supports get pulled away.
• Now say #11 is carrying 50% of the bridge self-weight in compression. And, at that steep angle. Call this Compressive Load B.
Wouldn't those PT rods go "slack" in that case?
If so, wouldn't re-tightening them result in Compressive Load B + Tension Load A to occur at that node?
That'd far exceed any other conceivable design load demand there, right?
"We shape our buildings, thereafter they shape us." -WSC
RE: Miami Pedestrian Bridge, Part II
I know one early NTSB member was quoted with the claim that the cable stays were "decorative only" - but is he correct? They are after all, traffic. Not bridge/truss design.
The entire "truss" design was intended to transmit the visual and real loads from each cable stay down through the angled truss members to the lower bridge walkway. They could not be built and tensioned without assuming the load.
RE: Miami Pedestrian Bridge, Part II
I thought it was FIGG who designed the bridge.
Munilla Construction Management could be a sub-contractor to FIGG as prime, or the reverse: Put MCM as the "bidder" to the government/political people to get the minority preference (visible) contract, with the real work being done by the FIGG people. Or, the story I found could be dead wrong.
RE: Miami Pedestrian Bridge, Part II
Quote (MJB315)
Wouldn't those PT rods go "slack" in that case?
No, because the rods are far more highly strained (i.e. stretched) than the amount the column shortens under the self weight loading. PT rods/strands stretch a lot.
Quote (MJB)
That'd far exceed any other conceivable design load demand there, right?
You'd include all the loads in the design.
RE: Miami Pedestrian Bridge, Part II
It was design/build. MCM is the prime contractor and Figg is an engineering subconsultant. I think.
RE: Miami Pedestrian Bridge, Part II
Here is a video showing the bridge move from the pylon side of the bridge: Link
RE: Miami Pedestrian Bridge, Part II
Quote (MJB315)
Is this not the likely culprit?
I know it's too soon to know, but:
Say #11 was prestressed to resist the tension caused by the temporary transport load (the end of the truss was basically being hung by #11). No numbers here -- just call that Tension Force A.
The bridge gets placed on the abutment. Load reversal city as the temporary supports get pulled away.
Now say #11 is carrying 50% of the bridge self-weight in compression. And, at that steep angle. Call this Compressive Load B.
Wouldn't those PT rods go "slack" in that case?
If so, wouldn't re-tightening them result in Compressive Load B + Tension Load A to occur at that node?
That'd far exceed any other conceivable design load demand there, right?
This seems somewhat plausible. It is a rookie mistake all round but entirely plausible.
RE: Miami Pedestrian Bridge, Part II
Quote (OCUCivlEng)
First lawsuit to be filed Monday.
http://www.wesh.com/article/orlando-based-attorney...
...and that, along all the many more to follow, will have FIGG's attorneys and insurers issuing FIGG with a immediate cautionary notice (demand?) to talk to no one!
Quote (FIGG BRIDGE)
“We are stunned by the tragic collapse of a pedestrian bridge that was under construction over Southwest Eighth Street in Miami. Our deepest sympathies are with all those affected by this accident. We will fully cooperate with every appropriate authority in reviewing what happened and why. In our 40-year history, nothing like this has ever happened before. Our entire team mourns the loss of life and injuries associated with this devastating tragedy, and our prayers go out to all involved.”
Speaking from past experience (a past project of remarkable similarity: bridge, post-tensioning, collapse, deaths...and its following lawsuits) these things get real ugly. The technical part of you wishes to assist the investigators, but you get 'muzzled' by the powers-that-be: i.e. lawyers/insurers/corporate/management.
RE: Miami Pedestrian Bridge, Part II
The relatively small amount of load that would have been applied by the stressing of one prestress bar in a diagonal should not be sufficient to cause the collapse if the overall design was ok. With normal factors of safety etc, the amount of force we are talking about compared to the overstrength required to be built into the design is insignificant.
There has to be some other cause.
RE: Miami Pedestrian Bridge, Part II
Given the uniqueness of this bridge configuration, I wonder if the design engineers [FIGG] gave any thought to embedding some structural monitoring equipment (e.g. strain or vibrating wire gauges to rebar, concrete, strand, PT bar etc)?
A couple of 16 channel datalogger, a multiplexer, some software, and gauges etc, are readily affordable today, with very good reliability and ease-of-use.
Damn, it would have been a good research project for a select few FIU engineering students.
RE: Miami Pedestrian Bridge, Part II
Quote (Ingenuity)
Damn, it would have been a good research project for a select few FIU engineering students.
Could still be an educational endeavour... Careful you don't do this...
Dik
RE: Miami Pedestrian Bridge, Part II
Re the compression effect on the force in the prestress, remember we are dealing with (in Metric) concrete strengths of 30 - 60MPa and prestress steel strengths of about 1800MPa. And Modulus Es / Ec of about 7-8.
So if there is 20MPa compression in the concrete (about the maximum allowed at .5Fc), then the reduced tension in the prestress will be about 20 * 8 = 160MPa, so about 10% of the ultimate prestress force, or up to about 15% of the force after stressing.
RE: Miami Pedestrian Bridge, Part II
Has anyone worked out the load/stress in #11?
RE: Miami Pedestrian Bridge, Part II
Quote (rapt)
The relatively small amount of load that would have been applied by the stressing of one prestress bar in a diagonal should not be sufficient to cause the collapse if the overall design was ok. With normal factors of safety etc, the amount of force we are talking about compared to the overstrength required to be built into the design is insignificant.
I think the idea is that it may have pushed an already deficient (or otherwise compromised) structure over the edge. The proverbial straw that broke the camel's back.
RE: Miami Pedestrian Bridge, Part II
Member #11, his joints and member #12 got awarded with blue X by NTSB.
Detail footage on member #11 from NTSB > https://youtu.be/aeJKqojmHgY?t=1m25s
RE: Miami Pedestrian Bridge, Part II
Quote (Tomf)
I think the idea is that it may have pushed an already deficient (or otherwise compromised) structure over the edge. The proverbial straw that broke the camel's back.
Exactly. What is pretty much clear is that the design and/or construction was defective. What is also almost certain is that the activities on the bridge by workers at the time were the trigger of the actual collapse.
My gut feeling is we are still largely talking about punching failure around the top connection. The cracking that was observed prior was likely this already happening. The works at the time was a supposed remedy. You have a thin top cord with high and concentrated punching loads from the web connections. It seems a fragile design to begin with. You want to get the reinforcing around those members done correctly.
RE: Miami Pedestrian Bridge, Part II
TomH,
I think that is what I was saying! It was on the point of collapse at the time that the diagonal stressing (or it could have been destressing!) was being done.
The final straw may have been the works they were doing, but it should not have taken 1 straw to cause a collapse. In my experience a collapse normally requires bad design errors + bad construction errors + bad luck. Unfortunately they succeeded.
RE: Miami Pedestrian Bridge, Part II
Quote (Tomfh)
Has anyone worked out the load/stress in #11?
BA did a quick back-of-the-envelope calc, above [in 'pagan' units] as follows:
Quote (BAretired)
Member 11 may have been a tension member during construction but it was a compression member after the temporary supports were removed. Assuming a total weight of 950 tons, the bridge reaction at each end would have been in the order of 950 kips under dead load only. Member 11 appears to be oriented at about 35o to the horizontal, so it would have been loaded to about 1650 kips, a compressive stress of 3,300 psi on a 24" x 21" section under dead load only. It did not need any prestress at that stage.
BA
About 23 MPa, in 'gods' units!
RE: Miami Pedestrian Bridge, Part II
rapt, yes I think we agree. Clearly it was unhappy already.
RE: Miami Pedestrian Bridge, Part II
Looking a the collapse, I think it started as a failure in the bottom slab immediately on the outer side of the point where diagonal 10 connects (towards the support).
The movement of 11 happens slightly after the the slab to the outside of the connection at 10/11 disintegrates on the 10 side.
RE: Miami Pedestrian Bridge, Part II
Probably the both anchors of member #11 tensioning rods.
One of them looks still attached on deck.
RE: Miami Pedestrian Bridge, Part II
The more I think about this the more concerns I have around the length of the main deck and lack of a proper 'I' beam.
This is a 50m concrete structure supported at either end with very little useful stiffening to stop longitudinal deflection.
You have a handful of asymmetrical concrete trusses and 12 longitudinal tendons stopping the deck from wanting to deflect down and buckle.
Add to that any structural weirdness that may have occurred during transport. Did the main deck longitudinal tendons start to fail, we assume they were PT'd and grouted.
Maybe the strange behavior around #11 pulled the trigger, but by way the largest forces in the bridge is the main deck in tension.
Did something fail around #11 that had a significant effect on the main deck tendons, which were already on their way out?
If you take away the compressive force of the longitudinal tendons at one end, you get a very similar failure pattern that is seen in the dashcams
RE: Miami Pedestrian Bridge, Part II
If the longitudinal tendons fail, they would be unlikely to fail at the same time, so the deck would have twisted when that side failed first. If they had all failed at the same time, the deck would have been pushed apart by the compression load in the canopy.
The fallen deck is shattered, but continuous and it doesn't twist in a notable way on the way down, so it seems unlikely that the initial failure was due to the longitudinals. Since the deck is still tied together I expect the longitudinals are still continuous.
I don't know how much twist would happen if only one of the longitudinals near the center line failed, but it seems likely that there would have been a cascade of failure, again yielding either twist or separation.
Since there was no twist and the deck and canopy are still tied together without notable separations, the failure is somewhere else. The only other place is the truss-like members and their attachment to the deck and canopy.
RE: Miami Pedestrian Bridge, Part II
(OP)
3DDave,
If you watch the video, there is a pronounced twisting of the upper deck as the truss fell.
Were the longitudinal tendons grouted? I thought not, someone said they were, but I am not sure.
RE: Miami Pedestrian Bridge, Part II
Does it bother anyone else that during transportation on the failed side one the transporter looks like it is in the middle of a span and not on a node?
RE: Miami Pedestrian Bridge, Part II
Quote (gte447f (Structural)19 Mar 18 03:15 OSUCivlEng, the point you make about the PT in #11 probably being intended for the transport phase and then detensioning for the permanent phase makes sense and has been assumed by several of us here. The timing seems odd to me, if that was the intent of the design and the construction sequencing, wouldn't they have detensioned #11 before Thursday if the bridge was set on Saturday. There is a video of someone from the construction company mentioning detensioning 2 cables on top of the bridge after the bridge was set. I think the video is from the day the bridge was set (Saturday). It is not clear whether he was referring to the 2 PT rods in #11 or 2 of the longitudinal tendons in the canopy/top chord, but either way it seems like it would have been done before Thursday. Does that mean that the work being done Thursday was not the normal design/construction sequence, but was some sort of remediation for another problem, maybe the cracking that has been reported? Any thoughts?)
I would imagine the 2 cables to be detensioned would be in the top chord or "canopy" of the bridge since it would be a compression member once the span was set in place. Let's also not forget that the span had been in place for 5 days before it collapsed. I don't think what they were doing on Thursday was part of the design/construction sequence, it was a remedy for the cracking that had occured.
My hunch is the adjustment to the PT rod in member #11 had to do with the cracking that was found, and was the subject of the voicemail by the FIGG engineer to the FDOT engineer. They had a 2 hour meeting about the cracking a few hours before the collapse. Engineers for FIGG "delivered a technical presentation" about the crack. http://www.tampabay.com/news/FIU-Firm-had-meeting-...
The crack was on the north end and they were messing with the rod on the north end, I don't think that is a coincidence. Like others have said, the tensioning/detensioning action was probably the straw that broke the camel's back.
By now I'm sure there are people who know exactly what happened, but can't or won't talk because of legal fears and orders from their lawyers or employer.
RE: Miami Pedestrian Bridge, Part II
The 'cable stays' have no bearing on the collapse, but, may have merit if the installation of the bridge required 'some' support at their locations. No one has provided a procedure for the installation of the bridge, and, this may come out in court if for no other reason than to cloud issues.
The cable stays, although decorative and not in place, would have an influence on the overall behaviour of the bridge. 8 - 1-1/2dia bolts (grade and anchorage unknown) would have a significant impact on load transfer. The stay, although on an angle, is in axial tension and has the axial stiffness of a steel member, whereas the bridge is concrete and in flexure.
Because they are above the shear center, they would also provide some stability.
Dik
RE: Miami Pedestrian Bridge, Part II
3
The stressing of #11 was due to cracking that was observed and was an attempt to fix an existing issue. And I do not believe they were destressing it, but were stressing the tendon...Though #11 may have caused the collapse...the collapse was due to a design failure of inadequate shear capacity.
Lets's assume Hokie's numbers are correct and that there is 1650K of compression in #11...Though we want to consider #11 with pinned ends..it is not..and is acting like a beam. The vertical shear at the bottom end of #11 would be about 950k...therefore Vu would have been about 1330k!...If f'c=6ksi...PhiVc would only be about 60k...there is not enough room in #11 (24"x21") to provide enough stirrups to handle the vertical shear in that member.
I believe that prior to failure #11 had some upward bending (concave up) and had cracking on the bottom side of #11. To address this, the engineer attempted to tighted the lower strand. The rupture first occurred at the bottom of #11..and with #11 removed..hinges were formed in the bottom of the slab near #10 and the top of the slab adjacent to #11.
When I step back and look at #11...It just doesn't not look big enough to handle the shear loads that would be imposed.
RE: Miami Pedestrian Bridge, Part II
nothing so far has nailed down the triggering cause of the collapse....I still maintain that hokie66's comments on the conc truss and it's potential problems especially @ the connections where the possibility of development of incidental moments could occur may well hold the clue to the collapse..also no mention so far of mild stl reinforcing @ the joints...frankly, the magnitude of these loads would scare me as I have never done any PT design...
RE: Miami Pedestrian Bridge, Part II
Structuralengr89,
I agree with your assessment. And you brought up another point which I had not thought of, namely that in addition to its poor performance in tension, the high E of concrete makes it a terrible material for use in a truss. Idealized trusses use pinned connections at the joints, steel trusses have enough flexibility (and redundancy in the material) that even though our nodes and connections are stiff, it still behaves mostly like an ideal truss. Concrete does not have that luxury, being such a rigid material it cannot accommodate the small rotations required at the points of intersection, and once it cracks it has lost all of its shear strength. If the design accounted for the shear strength of concrete + rebar (I'm making a leap here), when the PT rod let go and the member 11 lost the compression holding it together..........
Can someone who designs reinforced concrete please tell me how you typically design for shear in reinforced concrete?
Just Rebar or Concrete + Rebar?
RE: Miami Pedestrian Bridge, Part II
Attached is a research paper about testing of a 160' prestressed roof truss. During the testing it was noted that the first crack was a horizontal crack at the first diagonal. To me, it supports hokie66's theory of a node failure. In looking at Sheet B-17 of the DB proposal, in Detail A, note that the centroids of the roof member (top chord) and two diagonals do not coincide but that may not mean anything.
In the attached paper, the truss that was investigated had HS rods as verticals; they also pointed out the need for confinement reinforcement.
Quote (By now I'm sure there are people who know exactly what happened, but can't or won't talk because of legal fears and orders from their lawyers or employer. )
You can take that to the bank.
RE: Miami Pedestrian Bridge, Part II
In the pic (screen shot from NTSB investigation video) posted above by Meerkat007, the PT rod anchor plates from Member #11 are visible. The lower one, which is the one that was being stressed and still has the hydraulic ram attached appears to have moved a good ways toward the upper end of the #11 as the the PT rod erupted out of the top chord with the ram still attached. Ingenuity proposed in part 1 of this thread that the rod had not ruptured, but had erupted out the top after a failure in the dead end anchorage zone in the node at the far (lower) end of #11. That appears to be the case. There is also the spalling off of the entire bottom face of #11. Don't know if that happened prior to, simultaneously to, or after the dead anchorage zone/lower node failure.
RE: Miami Pedestrian Bridge, Part II
EPCI: The typical shear design includes plain concrete shear capacity combined with steel shear capacity. Prestressing adds a third element; compression. Compressing the member provides additional shear capacity. You are correct that severe cracking reduces or eliminates the plain concrete shear capacity, for this reason high-seismic concrete design cannot take credit for any plain concrete shear capacity due to the extreme spalling and cracking expected in a high-seismic loading event.
If shear was indeed the failure then a sudden loss of compression in the member and/or any severe cracking would reduce a significant amount of the shear capacity of the member.
Professional Engineer (ME, NH, MA) Structural Engineer (IL)
American Concrete Industries
RE: Miami Pedestrian Bridge, Part II
Bridgebuster,
the title of that article is "Truss-Girder"...that is not a truss but a beam with holes near the center where vert. shear is low. It is solid webbed as it approaches the reactions where vert. shear capacity is needed...We don't have that with this truss
RE: Miami Pedestrian Bridge, Part II
But the diagonal truss members whose ends are visible show a "clean" finish: No rebar coming out of the exposed ends, no cracking, no distortion. Just smooth concrete surfaces as if the ends of the angled trusses were wedged into notches in the upper and lower surfaces at both ends. The PT cables continue through into what is now open air, but no rebar is visible.
That makes little sense.
RE: Miami Pedestrian Bridge, Part II
Quote (gte447f)
There is also the spalling off of the entire bottom face of #11. Don't know if that happened prior to, simultaneously to, or after the dead anchorage zone/lower node failure.
Question for the more concrete experienced:
Is it possible that this extreme spalling on the under side of this member is indicative of a massive internal moment generated by near-instantaneous removal of one of the PT tendons and its associated compression load?
In other words:
-#11 starts out with 2 tendons which are apparently symmetrically placed about the beam's neutral axis
-Tendon A fails, leaving Tendon B intact and leaving an eccentric load in the member
-#11 is subjected to a giant internal moment and a bunch of material spalls off of the tension side
It also looks like there's a fair amount of steel inside those members beyond PT elements. Has no one found a reinforcement schedule yet?
RE: Miami Pedestrian Bridge, Part II
EPCI: Further to TME's comment. Some jurisdictions do not allow the concrete to take any shear if the stress is beyond a certain limit.
Dik
RE: Miami Pedestrian Bridge, Part II
Just speculating, a couple of scenarios:
The original procedure was that #11 tensioned rod was to be de-tensioned, once in place and was forgotten. A crack developed and reminded the contractor that #11 was to be de-tensioned because it was a major compression member and didn't need the prestress. In process of detentioning, coincidentally or due to a change in the loading regime; the collapse of the walkway was precipitated.
alternatively, if the stressing of the rod in #11 was increased and the rod failed, the sudden release of this energy may have caused vibrations that precipitated the collapse.
I still cannot wrap my ears around the 'fake' stays... If not intended, I would have designed a connection that allowed movement. It doesn't make sense that they had no purpose.
I did renovations to Polo Park shopping centre about 30 years ago... and when we were demolishing one of the small structures that had a huge sign connected to it, I discovered the 'connection' of the sign to the building was simply a pipe sleeved in a larger pipe. I talked to the engineer that did the original building and he explained that it was a city requirement that a sign of that size had to be attached to a building and he knew better than to attach a huge sign to a little building, so, he provided a slip connection that looked like an attachment.
Dik.
RE: Miami Pedestrian Bridge, Part II
Quote (jgKRI)
Has no one found a reinforcement schedule yet?
Not yet, need shop drawings to see what should have been placed.
Dik
RE: Miami Pedestrian Bridge, Part II
jgKRI - the massive spalling is from the rod being torn out the side during collapse. The retaining rings of rebar are severed as the rod zippered down the beam. The B-Roll of the NTSB, as mentioned above shows the rod still attached to the lower end. Here's a link to the investigation of that area: https://youtu.be/aeJKqojmHgY?t=130
I think this means the lower member #11 PT bar did not fail because it would not be sufficiently anchored to act like a pull string and rip out the bottom of the beam.
RE: Miami Pedestrian Bridge, Part II
@structuralengr89 - what we have at FIU isn't a truss either but a beam with web cutouts.
RE: Miami Pedestrian Bridge, Part II
I’m not a bridge guy, but while we’re speculating about the design (which might not have been the problem, we’ll have to see)...
I don’t know how you go about designing those nodes. Strut and Tie? Probably not a lot of textbook examples for STM models resembling this situation. The fact that there are multiple bars in each truss diagonal makes it even more complex, as others have noted earlier.
You’ve got a highly stressed, determinate structure whose local behavior is complex and analyzed using a lower-bound method. If the mild steel around those nodes isn't sized and oriented just right...
I still can’t believe how fast this thing failed, though. From the video of the collapse, it almost looks like the bottom flange fails in tension. Maybe the nodal failure was at the support. Maybe the temporary PT in member 11 was helping confine the joint by adding a vertical clamping force across a horizontal plane just above the longitudinal PT anchorage. Think shear-friction design. The base of Member 11 simply shears across a horizontal plane and punche out the end of the structure. The problem with this theory is that the shear plane I'm thinking of is way too large.
RE: Miami Pedestrian Bridge, Part II
In answer to dik, No, I'm not structural. I have spent most of my working career in the field, on the ground.
I have seen a lot of stuff go wrong. Today, everything is planned so that nothing breaks.
When I started, a lot of equipment was stressed to the breaking point.
"If something breaks, where will it end up?"
"Don't be there!"
I always had a keen interest in the root causes of any jobsite failure as my future safety may depend on it.
Example, not a failure but an illustration;
We had a lot of anchor bolts to set in concrete for the structures of a substation.
The bolts were inserted into the holding templates and the nuts were spun on by hand to position the bolts vertically.
The templates were positioned on the forms and the concrete was poured.
A worker was tasked with removing the nuts in preparation to erecting the columns.
Why is it taking so long?
It only took half this long to place the nuts and the worker is only half finished.
When the concrete was poured, very small splashes of concrete were deposited on the threads of each bolt.
Now instead of spinning the nuts off by hand, there is just enough drag that each nut has to turned all the way off with a wrench. The worker was working hard but taking off now takes about 4 or 5 times as long as putting on.
What's the point?
The Florida crew has probably been installing the nuts for the most part by hand. They have probably done most of the post tensioning on this project.
It is possible that a spec of concrete or a grain of sand or other material became lodged in the threads and this nut could not be removed by hand.
I have seen too many crews that would keep increasing the tension trying to free the bolt to disregard this possibility.
If the excess tension pulled the anchor on the other end of the PT rod through the concrete, that may explain the failure of the strut and the spalling on the bottom of the strut.
This may not be the reason for the failure but it is too possible to be discounted offhand.
I have seen too many times concrete that did not meet the design strength. (Too much water makes it easier to work with.)
I have seen too many concrete samples that did not reflect the strength of the installed concrete. (Don't add the water until the inspector leaves with his sample.)
Stuff happens in the field.
Sometimes more stuff happens with design build projects.
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Miami Pedestrian Bridge, Part II
SocklessJ, BAretired has hit on this also, a failure of the tension chord at or near the lower joint of diagonal #11. I think this is very plausible, and I think the direction and manner of the movement of the bottom PT rod in #11 as the collapse happened could support this theory.
RE: Miami Pedestrian Bridge, Part II
Am I the only one who is concerned about the mismatch of diagonal supports? The members 2,4,6,8 and 10 all point more or less the same angle one way resisted only by nos 11 and 9 pointing the other.
Does this not induce a force which is essentially trying to push the top section to the right (North end) heaping even more misery on the no 11 support? (see Ingenuities diagram on 18 mar 20:55). Also this is where the initial collapse was focused, but may have been due to other issues.
I've read all the posts and watched the videos lots of times and the speed of collapse is just frightening and seems almost impossible to say what actually broke first.
Certainly fiddling about with the PT bars in that overworked no 11 strut was clearly not a good idea, but is this simply a flawed design from the get go?
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
RE: Miami Pedestrian Bridge, Part II
Well, ALL of the angled truss members are "pointed" back vertically to align with the "not needed" (missing) cable stays/pipes going back up to the "not needed" (missing) vertical pier.
RE: Miami Pedestrian Bridge, Part II
VolsCE84 I think you're on the right track but just note, some of the prestress force would be lost because of compressive strain in the concrete diagonal.
RE: Miami Pedestrian Bridge, Part II
The more I watch the slo-mo video, the less I like my previous theory. I'm gonna go back to thinking that differential bar tensions in member 11 caused an eccentricity that led to local, and then complete compression failure in member 11.
Once the bridge started deflecting, the angle of the compression member decreased , causing the load to further increase and cause collapse. Could be as simple as that.
RE: Miami Pedestrian Bridge, Part II
waross... was just joking... we structural guys often have the same list you posted...
Dik
RE: Miami Pedestrian Bridge, Part II
Quote (Ingenuity)
About 23 MPa, in 'gods' units!
cubits?
RE: Miami Pedestrian Bridge, Part II
Was it here that it was mentioned that the founder of the bridge company died a while back. I think this is described in Petrovski's Engineer's of Dreams, where a successful company fails to maintain continuity of expertise, allowing the reputation to exceed the capability. The founder worked his way from smaller projects and probably encountered smaller failures which were overcome and thereafter avoided. The newer management didn't get that experience.
It would have been so easy to take this bridge, lift it at the same locations and set it back down on the construction site with simulated pylons, perhaps only a few feet above the ground, and do actual load testing and so forth before placing it. They could have made multiple engineering class projects out of it; measuring deflection and comparing it to envelope calculations and sophisticated computer models over a couple of semesters before finally placing it. It's supposed to be in place 100 years; 8 months is very little to ask.
RE: Miami Pedestrian Bridge, Part II
I'm ready to make a small bet. By the time his gets in court and resolved, all the time and energy will be spent on who gets sued and how much. All these engineering theories will never be mentioned in court OR WHO IS AT FAULT. SPEAK FROM EXPERIENCE.
OG: concur
RE: Miami Pedestrian Bridge, Part II
@hokie66 - you're going to have to start Part III pretty soon.
BTW, does anyone know where KootK has been? I'd be interested in his perspective.
RE: Miami Pedestrian Bridge, Part II
2
Quote (KootK)
BTW, does anyone know where KootK has been? I'd be interested in his perspective.
Knowing him, he's likely read the whole thing.
Perhaps he's doing like I am and not guessing until we have more factual information? Not disparaging any for their current discussion; only that I personally prefer to wait until more information is available in things like this before I discuss it.
Professional Engineer (ME, NH, MA) Structural Engineer (IL)
American Concrete Industries
RE: Miami Pedestrian Bridge, Part II
Quote (structuralengr89)
Snopes says it is. I found this by Googling "florida bridge women". I Googled the title of the Sandra Rose article, and I found no web or blog sites that I know to be reliable.
I have corrected this to link to Snopes, as I originally intended. There is lots of anti-snopes stuff in the comments to the Sandra Rose blog. I observe that the post I was replying to has been deleted.
--
JHG
RE: Miami Pedestrian Bridge, Part II
Quote (LittleInch)
Am I the only one who is concerned about the mismatch of diagonal supports? The members 2,4,6,8 and 10 all point more or less the same angle one way resisted only by nos 11 and 9 pointing the other.
Does this not induce a force which is essentially trying to push the top section to the right (North end) heaping even more misery on the no 11 support? (see Ingenuities diagram on 18 mar 20:55). Also this is where the initial collapse was focused, but may have been due to other issues.
I think this is correct. The reverse slope of diagonals 3 and (almost) 5 would have concentrated the compressive and bending stresses in the canopy towards the side of the bridge that failed.
It's possible/likely the asymmetric design of the truss, executed in service of making the bridge look like a cable-stayed structure, will prove to be an important factor in creating stresses the bridge design may not have fully anticipated.
RE: Miami Pedestrian Bridge, Part II
Quote (LittleInch )
Certainly fiddling about with the PT bars in that overworked no 11 strut was clearly not a good idea, but is this simply a flawed design from the get go?
Like many people I suspect that's what it boils down to. Tensioning and snapping a bar in a compression strut shouldn't bring the thing down.
I'd put my money on #11 struggling and failure being triggered by the PT in some way or another.
We won't know for sure until we find out where the crack was and what exactly the guys were doing up there. It's a public enough disaster that I suspect we'll find out. At least I hope so.
RE: Miami Pedestrian Bridge, Part II
Quote (3DDave)
8 months is very little to ask
I find people get angry enough if you cost them a day.
RE: Miami Pedestrian Bridge, Part II
Quote (mibro)
the asymmetric design of the truss, executed in service of making the bridge look like a cable-stayed structure
That's the strangest part of all this. The bridge was a intended to be fake. Decorative tower and cable stays?! What a bizarre concept.
RE: Miami Pedestrian Bridge, Part II
Quote (Tomfh)
That's the strangest part of all this. The bridge was a intended to be fake. Decorative tower and cable stays?! What a bizarre concept.
Yup, it gets stranger the more you think about it. What an odd series of design decisions when an actual cable-stayed structure would arguably have made a lot more sense. At the least it would have removed a hard-working concrete truss from the design.
RE: Miami Pedestrian Bridge, Part II
Yep, and it would have been able to be built out on a cantilever basis. Maybe shut down a lane or two over time, depending on the sequencing. Or, alternatively, use a steel truss for their accelerated construction concept.
Speaking generally, the default mode of concrete joints is fixed while the default mode of steel joints is pinned. Each can be made to do the other, but only deliberately and with (sometimes) great effort. Meanwhile my statics textbook, with one of this firm's (or it's predecessor's) bridges on the cover, tells me that a truss has pinned joints, by definition. Someone forgot the basics.
RE: Miami Pedestrian Bridge, Part II
3
I'm wondering why anyone would even post a link to white supremacist/sexist propaganda. That garbage was on twitter on day 1. If you use your engineering brain, you could figure out that it is inflammatory propaganda that shouldn't even be allowed to stay on this forum. The EOR is W. Denny Pate. Google him. He's a well-respected Caucasian male in his 60s with many bridges under his belt.
Even if it were designed by a woman, why would anyone in their right mind blame what happened on that. For shame on this well-respected forum.
RE: Miami Pedestrian Bridge, Part II
Quote (Tomfh)
That's the strangest part of all this. The bridge was a intended to be fake. Decorative tower and cable stays?! What a bizarre concept.
It is a fairly common concept across the world. It is a given for buildings and is common in bridges. I would almost suggest that a majority of pedestrian bridges have decorative features. Please lets not go back 50 years to purely function buildings devoid of character and life.
(Even in steelwork in industrial production we occasionally make some choices that are about looking nice rather than purely engineering. Only occasionally though mostly things are simply about function in my industry.)
[img https://c2.staticflickr.com/4/3138/3036688576_f65e...]
https://en.wikipedia.org/wiki/Bolte_Bridge
A simple concrete cantilever bridge which has fake towers simply to look more interesting.
RE: Miami Pedestrian Bridge, Part II
Human909,
Yes, don’t get me wrong, I know decoration is common. Our Sydney Harbour bride for example has decorative pylons.
In this instance I’m taking about the actual structure itself. The structure itself - the member orientation etc - was designed as decorative adornment to the phoney towers and cables.
That’s completely different to when the architect sticks a couple of fins on the building.
RE: Miami Pedestrian Bridge, Part II
True. True.
Quote (Archie264)
Yep, and it would have been able to be built out on a cantilever basis. Maybe shut down a lane or two over time, depending on the sequencing. Or, alternatively, use a steel truss for their accelerated construction concept.
Speaking generally, the default mode of concrete joints is fixed while the default mode of steel joints is pinned. Each can be made to do the other, but only deliberately and with (sometimes) great effort. Meanwhile my statics textbook, with one of this firm's (or it's predecessor's) bridges on the cover, tells me that a truss has pinned joints, by definition. Someone forgot the basics.
Forgetting the basics of pinned vs fixed seems surprisingly common in my observations. I'd hope it isn't common in bridge design but in more simple structure people seem to get lazy.
I was once involved in a construction where there were issues with columns bending on a simple single story structure. The structural engineer was blaming the fabrication/erection for it. I casually asked whether the column to beam joins were modeled as pinned connections to which the the engineer yes of course. I then pointed to the connection and asked does that look like a pinned connection to you? The columns were bending because the connection allowed significant moment transfer from the beam to the column.
RE: Miami Pedestrian Bridge, Part II
I think the problems with this bridge begin with the boots on the ground.
RE: Miami Pedestrian Bridge, Part II
Just can't resist getting my 2 cents in on this one.
Question, was this concrete bridge cast monolithic, or are there some construction joints somewhere. I'm guessing each of the truss diagonals are individual members who's ends don't fit PERFECTLY together, maybe leaving small voids inviting crushing at the bearing points between members. Not sure about getting accurate calculation of shear across a construction joint either.
Also, I'm having a hard time believing an Engineer would CHOOSE to have the last diagonal in a concrete truss in compression, with a tendency to push it's way off the end if something failed.
Just watching.
LonnieP
RE: Miami Pedestrian Bridge, Part II
LonnieP, the member orientation is to make them look like hangers aligned with the phoney cables above..
RE: Miami Pedestrian Bridge, Part II
Tom, got that from the side view rendering. Failure occurred at the end diagonal with the lessor internal load. curious.
RE: Miami Pedestrian Bridge, Part II
The dash cam video has a crane obstructing the joint at the top of member 11. But the short “CCTV” video posted by Tomfh on 18 Mar 18 01:39 appears to show the space between members 10 and 11 diminishing before the collapse. This looks like a shear failure at the top of member 11.
RE: Miami Pedestrian Bridge, Part II
If you look at the 5 second video posted by MattOhio in the first thread, you can see the initial collapse point was at the bottom chord on the canal side of the bridge (the node where 9 & 10 intersect in the diagram near the beginning of this thread). If the failure was due to bending the collapse would have occurred near mid-span. This clearly indicates that the collapse was due to a shear failure in a global sense. The failure probably occurred in one of the web members at this joint. The web(s) could have failed in either shear, tension or compression. Initially member 11 was in tension during transport since the end was cantilevered beyond the support trolley. After the truss was lowered onto the piers the stress in members 10 and 11 would have been reversed. Proper procedure would have probably required the tendons in 10 to be stressed and some or all of the strands in 11 to be released on the day the bridge was placed. Apparently some or all of the strands were not done on Saturday because after the job meeting on Thursday morning, workers went up on top of the truss to do what ever was needed. In that video posted by MattOHIO it appears that there were 2 or 3 workers on the roof of the bridge above the node where 10 & 11 intersect.
RE: Miami Pedestrian Bridge, Part II
3
I have read many, but not all, of the comments here and just for a pause would like to go back to the basics of the structure. The bridge is basically a simple span girder/truss with a fairly narrow top flange and extremely wide bottom flange. As a girder/truss it is a torsionally weak section which is supported on each end on the bottom edge and open to possible lateral torsional failure and sever shear lag problems in the wide bottom flange. The top flange is composed of a two thin curved sections presumably with solid concrete diaphragms at the termination of the diagonals. The truss members are concrete which is in itself regardless of strength a relatively brittle material with low tensile and shear strength unless reinforced and/or pre/post tensioned.
The top flange when exposed to bright sunlight will partially or fully shade the bottom flange and web members resulting in an average temperature gradient between the top flange and the bottom flange and web members. Because the web consists of concrete members the joints must be assumed to be fixed. The support of the entire bridge on the bottom of the section requires that all torsional stability is provided by the web members of the girder. This results in members that ultimately are subject to tension and compression plus biaxial bending based on rigid joints and extra stresses imposed by temperature differentials.
It is my opinion that the truss is much more like a beam than a truss mainly because from looking at the truss joints is does not appear to me that the streses from the tension and compression diagonals at the top or the bottom can be transferred directly to each other but require shear flow transfer to occur into the top and bottom flange. The top flange is hollow and I assume with diaphragms at the junction of truss diagonals. There does not appear to be any significant stirrup arrangement in the diaphragms or in the bottom slab so in effect it is then plain concrete in shear. The bottom flange is thin and wide and only a small fraction would be effective in transferring shear flow due to major shear lag effects in the bottom flange.
In summary I believe that the structure is much more complex than a "normal" truss but may have been designed as simply a truss and that the joint design was not suitable to transfer all of the loads between truss diagonals without complex stresses at the joints which were not designed to fully transfer the loads without involving substantial local effects on the top and bottom flange. Substantial shear stresses in the top and bottom chords and additional tensile and compressive stresses in the diagonals led to failure and with the maximum shear in the truss panel where member 11 lived due both to the end bearing reaction, culmination of any temperature gradients and maximum torsional stability stresses. This would also be the location of maximum shear flow if we look at the bridge as a girder where shear flow = Vq/I. The complexity which I believe is there plus the lower load factors taken for dead load do not allow in my opinion for making any simplifying assumptions.
Sorry for going on so long but for me to get this thing straight I need to logically go through this thing to conclude that it is an ill conceived design which is much too costly and in my opinion led to inevitable failure of the structure.
appster
RE: Miami Pedestrian Bridge, Part II
I noticed I can no longer find any of the vidio of the bridge being turned 90 degrees prior to placement, it has all been removed. It appeared to me in that vidio they were basically skidding a transporter with 450 tons around 90 degrees, the pivot transporter wheels did not appear to be following a circular radius although the outboard transporter appeared to be on radius. I wonder if using the bridge as a giant lever was included in the design requirements as obviously the spreader beams were not set up to handle the kind of twist that would be generated if this were the case. It also appears the walkway initially collapsed near that piviot point. Just observations.
RE: Miami Pedestrian Bridge, Part II
Nittanyray,
Agreed, that is the location that I suggested. But I think the failure is actually a compression failure in the concrete in the slab at the connection. You have the very concentrated high compression stress in the concrete in the slab at the node from 9 combined with the precompression from the prestress tendons in the bottom slab that is also relatively high.
At the 3 - 4 frame in the video you see something like crushing in this area with an explosion of material upwards from the top (not as in caused by explosives). The diagonals still seem to be intact at this point. Next frame, the top drops very quickly and you get the rotations in the 10/11 node at the top that have been noted previously as the bottom slab at this location crumbles.
RE: Miami Pedestrian Bridge, Part II
Quote (LonnieP)
Also, I'm having a hard time believing an Engineer would CHOOSE to have the last diagonal in a concrete truss in compression, with a tendency to push it's way off the end if something failed.
If all is detailed correctly... that's what concrete is best at... compression is good.
Dik
RE: Miami Pedestrian Bridge, Part II
Quote (tomfh)
LonnieP, the member orientation is to make them look like hangers aligned with the phoney cables above..
Those phoney cables are connected with 8 - 1-1/2"dia bolts (grade unknown) That could have a capacity of 400K or 500K. Not at all a slip connection.
Dik
RE: Miami Pedestrian Bridge, Part II
Quote (dik)
Those phoney cables are connected with 8 - 1-1/2"dia bolts (grade unknown) That could have a capacity of 400K or 500K. Not at all a slip connection.
Even with the dreaded APPENDIX D!!!!!!!????? I'm not so sure but would have to plug it into Profis to see.
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RE: Miami Pedestrian Bridge, Part II
FIGG Bridge Engineers, Inc. is the designer of the bridge, working for MCM. The project is part of a \$19.4 million U.S. Department of Transportation TIGER grant, and was bid competitively using design-build procurement by Florida International University.
http://www.figgbridge.com/FIU_Statement.html
https://heavy.com/news/2018/03/florida-internation...
https://heavy.com/news/2018/03/denney-denny-pate-f...
General Plan and Elevation
https://heavy.com/news/2018/03/munilla-constructio...
RE: Miami Pedestrian Bridge, Part II
Quote (EPCI-Steel)
@ Lnewqban
I have a bit of experience with SPMTs. They are really a wonderfully designed piece of equipment, there are several manufacturers now but they all operate the same way. All of the axle lines during a normal transport are on the same hydraulic circuit which allows each axle to stroke up and down independent of adjacent axle lines while maintaining a constant bearing, so crossing normal bumps and slight elevation changes are no problem. But the axles only have 1ft of up/down stroke so there are limits. Also they talk about them in terms of the number of axles but there really are no axles, they have independent, two tire hubs on either side of an axle line.
And just because I think they are so cool, here's some more. Any hub can be isolated and raised, for instance if a tire blows out. They can turn the hubs on either side of the axle line independent from each other so you can walk the trailer nearly 90 deg from its long axis, or almost pivot in place, wonderful things.
Thank you very much for the detailed explanation, EPCI-Steel.
1) Can the plastic flow of the steel, prior snapping, be seen in the gauge or be informed to the operator by some kind of alarm?
2) If so, does the hydraulic machine have any automatic device that prevents it from applying additional tension to the cable/tendon/tensioner beyond its yield point?
3) Can the machine fail in a way that excessive force is applied (failing switch, inaccurate gauge)?
"Where the spirit does not work with the hand, there is no art." - Leonardo da Vinci
RE: Miami Pedestrian Bridge, Part II
Quote (NITTANYRAY)
you can see the initial collapse point was at the bottom chord on the canal side of the bridge
My money is on the top of #11.
RE: Miami Pedestrian Bridge, Part II
Quote (JAE)
Even with the dreaded APPENDIX D!!!!!!!????? I'm not so sure but would have to plug it into Profis to see.
I don't know how anything is anchored or what grade the steel anchor is or if there is any confinement... the attachment is anything but a 'non-load' transferring connection, and, with this capacity, it has the potential for throwing a real wrench into the analysis of the walkway.
I've never used a 1-1/2 dia Kwikbolt...
Dik
RE: Miami Pedestrian Bridge, Part II
The owner is holding back payments...
I don't know how this is undertaken contractually... whether a Certificate of Payment can be negated due to on site damage, or what. I've never encountered this. I don't know how to 'claw' back money once it has been released. I've withheld funds because work has been rejected (one of the few ways payment can be withheld, and, still honour the contract.
Dik
RE: Miami Pedestrian Bridge, Part II
2
Quote (Lnewqban)
Can the plastic flow of the steel, prior snapping, be seen in the gauge or be informed to the operator by some kind of alarm?
It's likely that the rod failed at a thread; there would be little ductility or indication of failure before the rod 'broke'.
Quote (Lnewqban)
2) If so, does the hydraulic machine have any automatic device that prevents it from applying additional tension to the cable/tendon/tensioner beyond its yield point?
With yielding materials, hydraulic machines are very good for relaxing load and the load falls off quickly.
Quote (Lnewqban)
3) Can the machine fail in a way that excessive force is applied (failing switch, inaccurate gauge)?
Yes, but the equipment is generally very reliable. The devices are robust and simple, and nothing can go wrong, go wrong, go wrong, go wrong, go wrong, go wrong, go wrong...
Dik
RE: Miami Pedestrian Bridge, Part II
Quote (dik)
The owner is holding back payments...
No, really?
Quote (dik)
I don't know how this is undertaken contractually... whether a Certificate of Payment can be negated due to on site damage, or what. I've never encountered this. I don't know how to 'claw' back money once it has been released.
That would depend on the contract. But you're in a better position if the money's in your bank account than theirs. D&C probably helps too since it doesn't matter whether it was a design or a construction error, it all 'your' problem from the client's perspective.
RE: Miami Pedestrian Bridge, Part II
Quote (steveh49)
No, really?
Nothing quite like Aussie sarcasm!
RE: Miami Pedestrian Bridge, Part II
@Lnewqban,
To further Dik's replies to your questions.
I have used these type of hydraulic power units before to operate other devices, not a tensioning ram but other types of cylinders.
1) Can the plastic flow of the steel, prior snapping, be seen in the gauge or be informed to the operator by some kind of alarm?
No. These are very simple devices. The power pack consists of a pump, mounted onto a steel box which serves as a reservoir for the hydraulic fluid required, an inline pressure gauge, and a valve. The ones I have used are for operating a double acting cylinder so the valve has three positions, Open to line 1, closed, and open to line 2. One line will extend the cylinder, one line will retract the cylinder. If you want to derive useful force information, you must know the area of the piston where you are applying pressure (retraction is less because you must subtract the rod area) and multiply that by the hydraulic pressure from the gauge. From an untrained or inexperienced operator's perspective the only thing he will see once the steel is fully plastic is that the pressure on the gauge in not increasing.
2) If so, does the hydraulic machine have any automatic device that prevents it from applying additional tension to the cable/tendon/tensioner beyond its yield point?
No, the only limit is the ability of the pump to apply pressure and the individual components of the system. Usually the pump is the limting factor in the system and the cylinder is pressure rated to at least the pump pressure. 3000 and 10000 psi are the normal maximum operating pressures I have seen for these types of units.
3) Can the machine fail in a way that excessive force is applied (failing switch, inaccurate gauge)?
Everything can fail, in almost any conceivable combination, but a runaway pump scenario is very unlikely. Gauges can be inaccurate but they are supposed to be tested annually and given a certificate of calibration. Human error is much more likely.
RE: Miami Pedestrian Bridge, Part II
House mover: Here's a re-post of the video.
Quote (hokie66)
hokie66 (Structural)
16 Mar 18 01:03
An article from heavy.com...there is a time lapse of the bridge move. Hopefully, some video at the time of the collapse will turn up.
https://heavy.com/news/2018/03/florida-internation...
I can't see the position of the transporter wheels.
They have the ability to castor individually or be steered individually.
I doubt that the wheels would be dragging sideays.
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Miami Pedestrian Bridge, Part II
Quote (BAretired)
Member 11 may have been a tension member during construction but it was a compression member after the temporary supports were removed. Assuming a total weight of 950 tons, the bridge reaction at each end would have been in the order of 950 kips under dead load only. Member 11 appears to be oriented at about 35o to the horizontal, so it would have been loaded to about 1650 kips, a compressive stress of 3,300 psi on a 24" x 21" section under dead load only. It did not need any prestress at that stage.
At that cross section and length, how stable would that member be against buckling?
RE: Miami Pedestrian Bridge, Part II
(OP)
We should remember that the drawings we have seen are preliminary or proposal drawings. The construction drawings may well differ. The pipe stays are one thing that could have been changed to allow movement. The cast in bolts could have bolted down a bracket, which then had a slip connection to the pipe. But that is conjecture, and I agree if built the way the preliminary drawings show, the stays would have changed the way the final structure worked dramatically.
Lonnie P asked if the truss/frame was monolithically cast. I think it was, more or less. There must have been some construction joints, but I don't think it was a matter of precast elements being connected together.
RE: Miami Pedestrian Bridge, Part II
Quote (Lnewqban)
1) Can the plastic flow of the steel, prior snapping, be seen in the gauge or be informed to the operator by some kind of alarm?
2) If so, does the hydraulic machine have any automatic device that prevents it from applying additional tension to the cable/tendon/tensioner beyond its yield point?
3) Can the machine fail in a way that excessive force is applied (failing switch, inaccurate gauge)?
The stressing equipment used on this project at the time of collapse was very simple in design and use. Basically a 10,000 psi MAX pressure hydraulic pump (electrically operated, with a wired remote), with a 4-way valve (controls fluid directional flow), a gauge, a set of hyd hoses that connect to a center-hole, double-acting ram/jack, that sat upon a big-astressing stool/chair whereby the 'stressor' can turn the nut. This hyd design/setup is about 40 years old - tried-and-true. Simple. Stressing equipment gets calibrated, typically every 6-12 months, and sometimes right before significant use. It is usually calibrated as a 'system' - the pump, jack, gauge - using a load-cell, and traceable back to NIST standards. If you have a 200 ton jack setup on a 150 ton capacity PT bar, then it is possible to fail the bar during stressing. BUT this did not happen on this project. The PT bar did not undergo tensile failure. The answer to all three of your questions is a general 'NO'. There is equipment that is more sophisticated, and incorporates displacement gauges and in-line load cells, etc., but not particularly well suited to typical field operations. A more specific answer to your question #3 is more to do with operator error (or stupidity?) than equipment malfunction. A hyd pump and a 4-way valve is very simple in operation. Provided the operator is trained and experienced in stressing, knows what the stressing force (gauge pressure) is, then seldom (very seldom) do bars get overstressed. There are also pressure relief valves on pumps and rams to avoid overload above 10,000 psi (700 bar). Keep in mind that the 1-3/4" grade 150 ksi bar was stressed to 280 kips (max), about 70% of UTS. The bar has a ultimate tensile strength of 390 kips - so for an operator to go way beyond 280 kips - all the way close to 390 kips (40% more) - would take a huge screw-up. RE: Miami Pedestrian Bridge, Part II Cover of Benaim's text of PSC bridges... RE: Miami Pedestrian Bridge, Part II One possible explaination..., RE: Miami Pedestrian Bridge, Part II Why would a crack form in an area that is under considerable compression both from the PT tendons and the overhanging load? The distance between the canopy and the deck remains constant at #10, while outboard of #10 they both hinge together. RE: Miami Pedestrian Bridge, Part II Quote (steveh49) That would depend on the contract. But you're in a better position if the money's in your bank account than theirs. D&C probably helps too since it doesn't matter whether it was a design or a construction error, it all 'your' problem from the client's perspective. I've never had a project that 'stopped' so abruptly... had a couple where the contractor went bankrupt and one where the contractor was terminated with questionable cause. I would think that the contractor should be given the opportunity to make good, or, be sued for non-performance, but, I'm not sure. It gets real cloudy if you decide you no longer want the 'bridge'. Unless the contractor is terminated, he has every contractual right to complete the bridge (not likely he would). As far as withholding the funds, there is a 'legal' out. Since the last draw, the work has been rejected for reasons... a legal contractual out. I occasionally have had clients that don't like something and want to withhold funds, and, pointed out contractually unless the work is rejected as non-compliant there is no easy manner of legally withholding funds. Contractually, it's usually black or white, with no allowance for grey. Dik RE: Miami Pedestrian Bridge, Part II (OP) They killed 6 people. They have more to worry about than getting paid. RE: Miami Pedestrian Bridge, Part II The Governor of Florida suspended all remaining Federal payments to MCM. Since it's Federal money it passes through FDOT first. http://www.miamiherald.com/news/local/community/mi... FDOT is trying to get out in front of this and distance themselves as much as possible. I haven't heard or read of anyone pointing any fingers at them. Also interesting that FDOT says it didn't know of any "stress testing" (????) that was going on before the collapse. "Over the weekend, FIU said the cracks were the subject of the two-hour Thursday meeting with FDOT, in which a FIGG engineer “concluded that there were no safety concerns and the crack did not compromise the structural integrity of the bridge,” according to a statement. FDOT then responded by saying its consultant who attended was acting in an administrative capacity only to ensure that the project was on time and still qualified for federal funds." Really, the DOT sends a non engineer consultant to a meeting about a bridge being constructed over Florida SH-41? I know a little bit about a pedestrian bridge that was built over an interstate in the state where I live. The DOT here was all over every detail of the thing like stink on a pig's butt. Maybe hindsight is causing folks at FDOT to realize they should have exercised a bit more oversight over a bridge being built over one of their highways? RE: Miami Pedestrian Bridge, Part II Quote (Hokie) They killed 6 people. They have more to worry about than getting paid. Agreed... and, there could be criminal charges forthcoming, and how to prevent this from occurring again... something has to be learned from this terrible situation, but, someone is likely looking into how to 'wrap up' the project and what to do next. Was the bridge really necessary? will they construct another one? how are finances recovered? Lots of little unresolved questions. Dik RE: Miami Pedestrian Bridge, Part II Quote (OSUCivlEng) FDOT then responded by saying its consultant who attended was acting in an administrative capacity only to ensure that the project was on time and still qualified for federal funds." They can say what they want. That won't stop them from being included... he may have been party to the cause/harbinger of the failure and he was likely an engineer... and FDOT has money... It would be interesting to see what was presented at the meeting and see the minutes of that meeting if they had time to prepare them. RE: Miami Pedestrian Bridge, Part II Hi, all. Our version of the story was published. Let me know what you think. Thanks to many here for their insightful comments. I have quoted a few of you in the story. https://www.engineering.com/BIM/ArticleID/16670/A-... Roopinder Tara Director of Content ENGINEERING.com RE: Miami Pedestrian Bridge, Part II So just out of curiosity, who here has ever "stress tested" an newly built structure? (I know its media BS, please don't think I'm that naive). I have weighed things, load tested things that need certification, and proof loaded structures that have a life saving function, but have never seen a "stress test" outside of a university laboratory. RE: Miami Pedestrian Bridge, Part II Another reason I haven't joined the discussion here. Premature discussion here without all the facts may run against section 2.5 of the engineering code of ethics. Whether eng-tips posts are "public" is debatable (I'd argue that it's not) but regardless they are available to the public and could be quoted by media. Just keep this in mind when you post here. Professional Engineer (ME, NH, MA) Structural Engineer (IL) American Concrete Industries https://www.facebook.com/AmericanConcrete/ RE: Miami Pedestrian Bridge, Part II EPCI: We stress test overhead cranes and custom lifting devices, it's mandated by OSHA and CMAA. Of course, it's not done over people or (hopefully) in an unsafe manner but for cranes it's often done in the final installed location. Professional Engineer (ME, NH, MA) Structural Engineer (IL) American Concrete Industries https://www.facebook.com/AmericanConcrete/ RE: Miami Pedestrian Bridge, Part II Thats a load test RE: Miami Pedestrian Bridge, Part II EPCI: Oh, I misread your question. You're referring to specifically putting strain gauges on a structure (or similar)? Only one I know of off the top of my head outside of university was the Penobscot Narrows cable stayed bridge near me (interestingly enough also designed by FIGG) was installed with strain gauges on a few specialized carbon fiber cables to measure they're performance over time. I don't know if this is common or not to put strain gauges on modern cable stayed bridges but I remember them touting the fact that the DOT can monitor the stresses and loading on the bridge remotely via these gauges. Professional Engineer (ME, NH, MA) Structural Engineer (IL) American Concrete Industries https://www.facebook.com/AmericanConcrete/ RE: Miami Pedestrian Bridge, Part II Now that is interesting. I forgot about this. I was at Texas Tech for high wind design class many years ago and they showed us how they build full scale buildings, instrumented them up with strain gauges and pitot tubes and used borrowed military aircraft to wind load them until failure. We only go the see them shoot 2x4's through a wall, but they showed us the videos and results of their testing. I would have liked to have been involved in that. RE: Miami Pedestrian Bridge, Part II Quote (dik) We missed you... You just have to be careful about how you reply... Thanks. I'm hoping we can get some idea of the joint detail that went into the frame before I start contributing to this discussion. I suspect we'll find that this bridge works more like a Vierendeel bridge than a truss bridge but between the angled web members, the post-tensioning, and all the unknowns I don't want to dive into it much further until we have more factual details. Just way too many things that could change whether this was a design error or not. Professional Engineer (ME, NH, MA) Structural Engineer (IL) American Concrete Industries https://www.facebook.com/AmericanConcrete/ RE: Miami Pedestrian Bridge, Part II 3 A (hopefully somewhat gentle) reminder to all that: 1. This is a public forum. As such, it should be fair to quote anything posted here. Don't want to be quoted? Don't post. 2. Most participants aren't posting under their own name and those who do so have chosen to. 3. The title of this subforum is, "Engineering Failures and Disasters." If it shouldn't exist that subject should be discussed with the owners or moderators. 4. Would we all prefer that this discussion be left to the nonengineers? 5. There is far too much speech suppression in the world today as it is. I suspect this post might earn me some ill-will but so be it. RE: Miami Pedestrian Bridge, Part II If this were analyzed with FEA instead of as a truss and some fixity was assumed across the joint, would the diagonal pass a unity check? RE: Miami Pedestrian Bridge, Part II Roads and Bridges magazine has already sent out 2 surveys speculating as to the cause of the collapse. Congratulations hokie66. RE: Miami Pedestrian Bridge, Part II Quote (Phil1934) If this were analyzed with FEA instead of as a truss and some fixity was assumed across the joint, would the diagonal pass a unity check? With a 'brittle' concrete structure, the analysis would have to include for the fixity at the panel points and reinforced accordingly. If a joint rotates and reinforcing is not provided, it will likely crack... it may not have any issue with overall strength, but, may have an adverse impact on esthetics. Dik RE: Miami Pedestrian Bridge, Part II Quote (bridgebuster) Roads and Bridges magazine has already sent out 2 surveys speculating as to the cause of the collapse. Interesting how FDOT is trying to distance themselves. It may not be that simple. Being a resource with money may make them a 'target'. https://www.roadsbridges.com/fiu-broke-agreement-f... I guess one has to ask the questions: Were they aware the construction was on going? Were they aware of the agreement/requirement with FIU? Did they have anyone on staff visit the site during construction? Why did they not intervene? Dik RE: Miami Pedestrian Bridge, Part II Quote (Archie) ...I suspect this post might earn me some ill-will but so be it. On the contrary, nice summary. As I said before, I don't disparage anyone for commenting and discussing this failure. As you noted, I'm one of those people who chooses to attach identifiable information to my posts. This is part of the reason for my choice to not participate in the discussion yet. For those who post anonymously that's obviously not a concern. In short, please keep discussing this; I just caution people keep in mind who may be reading it (public, media, etc.), I'm sure one or two FIGG engineers are watching this discussion and probably tearing their hair out that they can't join in and provide further details. Professional Engineer (ME, NH, MA) Structural Engineer (IL) American Concrete Industries https://www.facebook.com/AmericanConcrete/ RE: Miami Pedestrian Bridge, Part II 2 I do not find cable-stayed brdiges (of any size) very attractive, but, my opinion about somebody's else's tastes don't matter. (That my tax dollars ARE taken to pay for somebody's (lack of) taste IS irritating, but I can do nothing about that right now.) Regardless, I read more and more often that the large tower, the cable-stayed members themselves (hollow 8 inch pipes ?) and their connection fittings (8 inch pipe flanges and bolts ?) and the cable-stayed connections into the angled truss members are "decorative. " If so, then these "decorations" are directly to blame, are they not? The "truss" cannot be symmetrical because every truss inclined member must align visually to its matching cable-unstay diagonal member coming up from the bridge lower walkway. Unsymmetrical members equal greater stress in some member connections than others, but all connections need to be "visually" identical = some are overbuilt = too expensive, take too long to build properly. Regardless of intent, the 8 inch pipes are very long, and will sag under gravity: This creates an elegant appearance when they are installed in a traditional suspension bridge, but an ugly distraction when hung basically sideways in this case. Thus, whether intended or not, the cable un-stays MUST be under tension to be visually straight under their natural gravity load at the extreme angles of their design,(and must be heavier (stronger) to resist that sag if NOT under their proper function as true cable stays. (A true cable stay cable will be straight BECAUSE) it is in tension holding the bridge up.) These decorative cable un-stays would need to be even heavier and thicker-walled to resist the inevitable bending (sagging and droop) than needed to be in "real tension" as a true cable stay. Further, a true cable-stay holding the original lighter weight of the bridge if it were a true cable-stayed bridge, would be a smaller diameter, lighter cable: Holding "up" the canal side of the bridge, thrusting down on the tower and foundation, and holding "up" the traffic side of the bridge. The smaller diameter of a true cable under simple tension creates less hurricane wind force under live load conditions. 100-150-200 feet of 8 inch pipe at 20 to 23 pounds per foot = added dead weight on each connection point in each truss, rather than a simple true cable stay that is lifting the walkway at the same point. The pipe (fake) cable un-stays are themselves heavy and have mass and wind resistance. As long round objects in the unpredictable gusty winds of a hurricane, this round shape creates a near-maximum turbulence and whipping load in a hurricane, and this extra load must be added to their gravity load under wind conditions = heavier, more expensive bridge = greater profits for the brddige design team, right? So, regardless of claim as decorative devices, there are gravity loads, extra stress, and extra un-symmetric stresses added to each connection point of every member. Forget the cost and time of the claimed "decorative" tower and its cable un-stays: The tower itself and its cable un-stays add dead weight to the walkway, add hurricane wind loadings to the walkway at each connectino point, and create even worse cases of the very conditions a true cable-stayed bridge is intended to reduce and support. RE: Miami Pedestrian Bridge, Part II There can be no question whether or not a title including the term "Unsafe" is appropriate. The bridge failed catastrophically under conditions of essentially "no load" (no foot traffic, no winds, no snow (obviously!) or rain, and with innocents underneath. RE: Miami Pedestrian Bridge, Part II Quote: The stressing equipment used on this project at the time of collapse was very simple in design and use. Basically a 10,000 psi MAX pressure hydraulic pump (electrically operated, with a wired remote), with a 4-way valve (controls fluid directional flow), a gauge, a set of hyd hoses that connect to a center-hole, double-acting ram/jack, that sat upon a big-a stressing stool/chair whereby the 'stressor' can turn the nut.
This hyd design/setup is about 40 years old - tried-and-true. Simple.
Interesting. Hydraulic fastener stretchers are very common in manufacturing and typically require redundant load cells for reasonable accuracy and repeatability, a hydraulic pressure gauge isnt even included on many.
RE: Miami Pedestrian Bridge, Part II
Quote (latexman)
Engineers have vastly improved living conditions, chemicals, medical equipment, prescription drugs, transportation, petrochemicals, utilities, agriculture, food, etc., etc., etc., i.e. almost everything, to the point that our score card is nowhere close to being in the red.
Some of these items may not have improved living conditions... in addition, there may be other engineered products that have definitely harmed mankind. I tend not to be quite so smug about our accomplishments.
Dik
RE: Miami Pedestrian Bridge, Part II
2
A defense of the design:
To the structurals here, who hasn't, at the behest of an architect, worked their tail off to make something superflous work? Is that not part of our job? Ok, so the "cable stays" really aren't. So?
Any of us who has come into a project halfway through has had the thought, "This, this, and this look really strange and I hope that this, this and this were accounted for." Same situation here - we're all coming in after the fact and slowly learning what the design really is. I, for one, am willing to give Figg some slack here on all the design issues: the un-symmetry, the not-really cable stays, the non-redundancy, the kinda-a-truss but kinda-a-beam-with-holes issues. That firm isn't just 2 people in a garage somewhere. They have designed many, many bridges throughout the United States (including the I-35W replacement in Minneapolis).
Concerning the article title, to say that the design would have only "looked safe" is a tremendous extrapolation of the speculative comments here.
Maybe, when it's all said and done, we'll know one way of the other if the final design was safe (and maybe not if everything gets settled and everyone signs an NDA). Until then, all we're doing is throwing thoughts around. Yes, it failed, but there's only a couple hundred other factors that could have ultimately caused the collapse.
RE: Miami Pedestrian Bridge, Part II
Quote (dik)
With a 'brittle' concrete structure, the analysis would have to include for the fixity at the panel points and reinforced accordingly. If a joint rotates and reinforcing is not provided, it will likely crack... it may not have any issue with overall strength, but, may have an adverse impact on esthetics.
what program would this be used in? I'm familiar with using ETABS although I assume this would be in SAP or some other program. (I'm a buildings structures guy). I ask because when modeling in ETABS and to a large extent in RISA as well, most FEA objects just end up being line objects that connect at specific points rather than an object that you can see how stress varies across the thickness of the section etc.
RE: Miami Pedestrian Bridge, Part II
Quote (Latexman)
"Time will tell who failed whom, but, if it is discovered that an Engineer(s) did fail here, I am very comfortable that Engineers have vastly improved living conditions, chemicals, medical equipment, prescription drugs, transportation, petrochemicals, utilities, agriculture, food, etc., etc., etc., i.e. almost everything, to the point that our score card is nowhere close to being in the red. We just need to drill down to the facts in this case, so it never happens again."
In school if we got a 95/100 on an exam it was considered very good by most people (even the PE exam only requires a 70% to pass) but in the real world one miscalculation out of hundreds or thousands on a project can become a life and death issue. Very few professions require as much near perfection as ours. The fact that these types of tragedies are rare speaks well for our profession.
RE: Miami Pedestrian Bridge, Part II
This is speculation of course but a theory of the failure I've not seen yet in this thread is sheer failure at the interface between 11/12 and the deck. In the dash cam footage there appears to be motion of 12 spanward as hinging starts to develop. But the base of 12 remains on the pylon so it might have been spalling? The last segment of the deck fell straight down and the upper flange initially moved downward rapidly before pivoting. One of the PT rods in 11, the inboard one, zippered to stay with the moving deck, the other stayed in the base of 12. The triangle consisting of 11,12 and the last segment of the upper flange retains substantially its original geometry post-failure. That sheer happened there is clear. Could it have been the point of initial failure and the joint failures at 9/10 and 10/11 secondary?
RE: Miami Pedestrian Bridge, Part II
Quote (ke27on)
what program would this be used in?
Any of those programs you mentioned would be able to do it... About 45 years back, I wrote a 2D Frame Analysis program that would run in 64K and swapped in and out of a 5-1/4 disk... By adding a couple of joints within a foot or so of the panel point I could increase the stiffness of the 'short' member to provide a better model of fixity. Almost any frame program can treat the joints as fixed and will generate a moment at the member-panel point junction.
Dik
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Electronic Theses and Dissertations
# The Effect of Throwing Intensity on Overhand Throwing Mechanics
2021-12-07
## Author
Wasserberger, Kyle
PhD Dissertation
## Department
Kinesiology
Overhead throwing is a dynamic and demanding movement that places great amounts of stress on the throwing arm. Current consensus attributes most throwing-related injuries to repeated microtrauma of the musculoskeletal structures surrounding the shoulder and elbow. After sustaining an injury to the shoulder or elbow, overhead athletes complete an interval throwing program as part of the rehabilitation process. Interval throwing programs gradually increase throwing intensity over several weeks by manipulating throwing volume and distance to produce successively more throws at successively longer distances. Recently, there has been an increased call for objective methods of quantifying how and when to progress through the interval throwing program, including monitoring throwing intensity with a calibrated radar gun. Although objective methods of throwing intensity quantification are beneficial to the rehabilitation process, how the demands placed on the throwing arm change throughout the intensity range typically seen in throwing rehabilitation programs is not well understood. Therefore, the purpose of this research was to model changes in throwing arm joint loads as pitchers progressed from low to high intensity throwing. Thirty-two skilled throwing athletes were recruited to participate ($21 \pm 2$ yrs; $1.86 \pm 0.08$ m; $89.0 \pm 10.2$ kg). Once participants completed their typical non-throwing warm up, 50 reflective markers were placed at relevant anatomical locations and participants had their throwing mechanics recorded during their throwing warm up using a passive optical motion capture system. The primary findings of this project were two fold: first, the intra-participant relationship between throwing arm joint loading and throwing speed appears to be nonlinear in form. Specifically, the form of this relationship is quadratic linear concave up indicating that, as throwing arm joint loading increases, corresponding increases in throwing speed become successively smaller and smaller. Second, as body mass and height increase, the slope estimating the relationship between throwing arm joint loading and throwing speed decreases. In addition to providing novel insight into the intra and inter-participant relationships between throwing arm joint loading and throwing speed, these findings also serve as an initial exploration of, and proof-of-concept for, the use of multilevel modeling strategies in sports biomechanics research.
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1. ## What are the answers to these problems?
Solve equation
(t+1)^2 - 2(t-1)^2 = 6t-5
x^2-a(3x-2a+b)-b =0
I got no solution for both, is it right?
2. Originally Posted by yitsongg
Solve equation
(t+1)^2 - 2(t-1)^2 = 6t-5
x^2-a(3x-2a+b)-b =0
I got no solution for both, is it right?
No.
$
t^2+2t+1 - 2(t^2-2t+1) - 6t + 5 = 0$
$-t^2 + 4 = 0$
(use the difference of two squares to solve)
-----------------
$x^2 - 3xa + 2a^2 - ab - b = 0$
Assuming $a \, , \, b \neq \, f(x)$
$x^2 - 3xa + (2a^2-ab-b) = 0$
(Use the quadratic formula to solve in terms of a and b)
3. Got it
first one is t= plus or minus 2
second one is
a=3
b=4
x=6
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# fourteen definition
• noun:
• The cardinal quantity corresponding to 13 + 1.
• The 14th in a set or series.
• One thing having 14 parts, products, or members.
• The cardinal number happening after thirteen and before fifteen, represented in Roman numerals as XIV plus Arabic numerals as 14.
• The sum of ten and four; forteen devices or things.
• emblematic representing fourteen, as 14 or xiv.
• The sum of ten and four, or thirteen and something.
• symbolic representing fourteen units, as 14, XIV, or xiv.
• the cardinal quantity that is the sum of thirteen and one
• The cardinal quantity equal to 13 + 1.
• The 14th in a group or series.
• some thing having 14 parts, devices, or users.
• The cardinal number occurring after thirteen and before fifteen, represented in Roman numerals as XIV as well as in Arabic numerals as 14.
• the sum of the ten and four; forteen products or items.
• emblematic representing fourteen, as 14 or xiv.
• the sum ten and four, or thirteen plus one.
• symbolic representing fourteen units, as 14, XIV, or xiv.
• the cardinal number that's the sum of thirteen plus one
• The cardinal quantity equal to 13 + 1.
• The 14th in a collection or series.
• anything having 14 components, units, or people.
• The cardinal number happening after thirteen and before fifteen, represented in Roman numerals as XIV and in Arabic numerals as 14.
• the sum ten and four; forteen units or items.
• emblematic representing fourteen, as 14 or xiv.
• The cardinal quantity add up to 13 + 1.
• The 14th in a collection or series.
• anything having 14 components, units, or people.
• The cardinal number happening after thirteen and before fifteen, represented in Roman numerals as XIV as well as in Arabic numerals as 14.
• The sum of ten and four; forteen devices or items.
• the sum of the ten and four, or thirteen plus one.
• The cardinal quantity equal to 13 + 1.
• The cardinal quantity equal to 13 + 1.
• The 14th in a collection or series.
• symbolic representing fourteen products, as 14, XIV, or xiv.
• The 14th in a collection or sequence.
• Something having 14 parts, units, or people.
• the cardinal quantity that is the sum of thirteen plus one
• some thing having 14 components, products, or people.
• emblematic representing fourteen, since 14 or xiv.
• The cardinal number occurring after thirteen and before fifteen, represented in Roman numerals as XIV and in Arabic numerals as 14.
• The cardinal quantity occurring after thirteen and before fifteen, represented in Roman numerals as XIV plus in Arabic numerals as 14.
• the sum of the ten and four; forteen units or objects.
• The cardinal quantity add up to 13 + 1.
• the sum of the ten and four, or thirteen and something.
• symbolic representing fourteen, because 14 or xiv.
• The 14th in a collection or sequence.
• the sum of the ten and four; forteen devices or objects.
• A symbol representing fourteen products, as 14, XIV, or xiv.
• Something having 14 components, products, or members.
• A symbol representing fourteen, because 14 or xiv.
• the cardinal number that is the sum of thirteen and something
• the sum ten and four, or thirteen and another.
• The cardinal quantity occurring after thirteen and before fifteen, represented in Roman numerals as XIV as well as in Arabic numerals as 14.
• The cardinal number corresponding to 13 + 1.
• The cardinal quantity corresponding to 13 + 1.
• The cardinal quantity equal to 13 + 1.
• The 14th in a group or series.
• The 14th in a group or series.
• symbolic representing fourteen products, as 14, XIV, or xiv.
• the sum of the ten and four; forteen products or items.
• the cardinal quantity that's the sum of thirteen plus one
• symbolic representing fourteen, as 14 or xiv.
• The cardinal quantity add up to 13 + 1.
• the sum of the ten and four, or thirteen and something.
• The 14th in a group or sequence.
• The 14th in a set or sequence.
• emblematic representing fourteen products, as 14, XIV, or xiv.
• Something having 14 parts, devices, or users.
• some thing having 14 parts, units, or users.
• One thing having 14 components, units, or users.
• Something having 14 components, devices, or members.
• The cardinal number add up to 13 + 1.
• the cardinal number that's the sum of thirteen and one
• The cardinal number happening after thirteen and before fifteen, represented in Roman numerals as XIV and in Arabic numerals as 14.
• The 14th in a collection or sequence.
• anything having 14 components, devices, or people.
• The cardinal quantity occurring after thirteen and before fifteen, represented in Roman numerals as XIV plus Arabic numerals as 14.
• The cardinal number occurring after thirteen and before fifteen, represented in Roman numerals as XIV plus Arabic numerals as 14.
• The cardinal quantity occurring after thirteen and before fifteen, represented in Roman numerals as XIV and in Arabic numerals as 14.
• The sum of ten and four; forteen units or objects.
• The sum of ten and four; forteen products or things.
• the sum ten and four; forteen units or things.
• The sum of ten and four; forteen products or things.
• symbolic representing fourteen, since 14 or xiv.
• symbolic representing fourteen, since 14 or xiv.
• symbolic representing fourteen, as 14 or xiv.
• symbolic representing fourteen, since 14 or xiv.
• The sum of ten and four, or thirteen and another.
• the sum ten and four, or thirteen and one.
• the sum ten and four, or thirteen plus one.
• the sum ten and four, or thirteen and one.
• emblematic representing fourteen products, as 14, XIV, or xiv.
• emblematic representing fourteen products, as 14, XIV, or xiv.
• the cardinal number that is the sum of thirteen plus one
• symbolic representing fourteen devices, as 14, XIV, or xiv.
• emblematic representing fourteen units, as 14, XIV, or xiv.
• The cardinal quantity corresponding to 13 + 1.
• the cardinal number that is the amount of thirteen and another
• The 14th in a set or sequence.
• the cardinal number this is the amount of thirteen plus one
• the cardinal number that is the amount of thirteen and another
• One thing having 14 components, devices, or people.
• The cardinal number occurring after thirteen and before fifteen, represented in Roman numerals as XIV plus Arabic numerals as 14.
• the sum ten and four; forteen devices or items.
• A symbol representing fourteen, as 14 or xiv.
• the sum ten and four, or thirteen and one.
• emblematic representing fourteen products, as 14, XIV, or xiv.
• the cardinal quantity that is the sum of thirteen and another
• The cardinal number corresponding to 13 + 1.
• The 14th in a set or sequence.
• Something having 14 parts, products, or users.
• The cardinal number happening after thirteen and before fifteen, represented in Roman numerals as XIV plus Arabic numerals as 14.
• the sum of the ten and four; forteen units or objects.
• A symbol representing fourteen, since 14 or xiv.
• the sum of the ten and four, or thirteen and something.
• emblematic representing fourteen products, as 14, XIV, or xiv.
• the cardinal quantity that is the amount of thirteen and something
• The cardinal quantity occurring after thirteen and before fifteen, represented in Roman numerals as XIV plus in Arabic numerals as 14.
• The sum of ten and four, or thirteen and another.
• A symbol representing fourteen units, as 14, XIV, or xiv.
• the cardinal number this is the sum of thirteen and another
• The sum of ten and four; forteen products or items.
• A symbol representing fourteen, since 14 or xiv.
• The sum of ten and four, or thirteen and one.
• emblematic representing fourteen units, as 14, XIV, or xiv.
• the cardinal quantity this is the sum of thirteen and something
• Four and ten more; two times seven.
• becoming another than thirteen
• Four and ten more; twice seven.
• being one more than thirteen
• Four and ten more; twice seven.
• Four and ten more; two times seven.
• becoming one more than thirteen
• Four and ten more; two times seven.
• Four and ten more; two times seven.
• being yet another than thirteen
• Four and ten more; twice seven.
• becoming one more than thirteen
• Four and ten even more; twice seven.
• being another than thirteen
• Four and ten more; two times seven.
• Four and ten even more; twice seven.
• Four and ten more; twice seven.
• being an additional than thirteen
• being yet another than thirteen
• becoming yet another than thirteen
• Four and ten even more; two times seven.
• becoming one more than thirteen
• Four and ten more; two times seven.
• being yet another than thirteen
• Four and ten even more; twice seven.
• becoming another than thirteen
• being another than thirteen
• becoming an additional than thirteen
• others:
• Four more than ten, or yet another than thirteen: a cardinal numeral.
• Four significantly more than ten, or another than thirteen: a cardinal numeral.
• Four over ten, or one more than thirteen: a cardinal numeral.
• Four significantly more than ten, or yet another than thirteen: a cardinal numeral.
• Four significantly more than ten, or another than thirteen: a cardinal numeral.
• Four significantly more than ten, or yet another than thirteen: a cardinal numeral.
• Four over ten, or one more than thirteen: a cardinal numeral.
• Four more than ten, or an additional than thirteen: a cardinal numeral.
• Four above ten, or one more than thirteen: a cardinal numeral.
• Four above ten, or an additional than thirteen: a cardinal numeral.
• Four above ten, or another than thirteen: a cardinal numeral.
• Four significantly more than ten, or an additional than thirteen: a cardinal numeral.
• Four more than ten, or one more than thirteen: a cardinal numeral.
• Four over ten, or one more than thirteen: a cardinal numeral.
## Related Sources
• Definition for "fourteen"
• The cardinal quantity corresponding to 13 + 1.
• Sentence for "fourteen"
• If you won the title fourteen
• Quotes for "fourteen"
• "Where I live, nobody who's fourteen…"
• Phrases for "fourteen"
• fourteen inches
• Rhyme for "fourteen"
• Aileen
• Hypernym for "fourteen"
• large integer
• Equivalent for "fourteen"
• cardinal
• Etymologically Related for "fourteen"
• fourteenth
• Form for "fourteen"
• fourteener
• Urban Dictionary for "fourteen"
• a gay guy, pansy.a ineffective age
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# Thread: could anyone try to solve this question..
1. ## could anyone try to solve this question..
Y=(a-b*r1-c*P1)*(m-1+(1+r1)^(T-1)*(p1^T+(1-l0)*(1-p1^T)))/(1+r3)^T
Z=(a-b*r2-c*P1)*(m-1+(1+r2)^(T-1)*(p1^T+(1-l0)*(1-p1^T)))/(1+r3)^T
Y=Z,then p1=?
I wanna find the value of p1 in terms of other variable..may be you can use mathematica software or maple since it involved a lot of variables.thank you
2. Are these your two equations (what you have is a little difficult to read and I don't know if what I have here is any better). Also, we have $P_1$ and $p_1$ - are these the same or different?
$Y=\dfrac{(a-b r_1-c P_1) \{m-1+(1+r_1)^{T-1} [p_1^T+(1-l_0) (1-p_1^T)]\}}{(1+r_3)^T}$
$Z=\dfrac{(a-b r_2-c P_1) \{m-1+(1+r_2)^{T-1} [p_1^T+(1-l_0) (1-p_1^T)]\}}{(1+r_3)^T}$
3. sorry for the mistake..p1 and P1 are same actually..typing error.thanks for the reply..I really need the answer.I tried use the mathematica software but couldnt get the answer.
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In this post, understanding the fundamental concepts consist of matter, particles and multiverses are described.
$$P$$ stands for Polynomial time while $$NP$$ stands for Non-Polynomial time. The idea behind the concept of putting P Vs NP is to pose a quesiton – If there is a problem whose solution can be verified in polynomial time, then can we also solve that problem in polynomial time. As a result, $$P \eq NP$$ or $$P \neq NP$$
P-Vs-NP
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ISSN 1002-1027 CN 11-2952/G2
A New Solid-Phase Microextraction with Activated Charcoal Coated Fiber
WANG Yonghua
1. College of Environmental Sciences, Peking University, Beijing, 100871, E-mail: yhwang@urban.pku.edu.cn
• Received:2005-02-16 Online:2006-03-20 Published:2006-03-20
Abstract: A new solid phase microextraction (SPME) fiber coated with porous layer activated charcoal has been developed for the quantification of benzene in gas. There is a linear relationship between the amount of benzene extracted by the SPME coated with porous layer activated charcoal and gas volume (Vg) at ≤100mL of gas phase volume. The linear equation is A=25.815Vg+36.462 with correlation coefficient (R2) of 0.999. The linear equation between amounts of benzene extracted by the SPME coated with porous layer activated charcoal and concentration of benzene (C) in gas phase is A=140.09C+16.24 with correlation coefficient (R2) of 0.996, ranging of concentration from 0.0879 to 52ng·mL-1. The linear equation between the reciprocal of amounts of benzene extracted (1/A) and the reciprocal of gas volume (1/Vg) is 1/A=0.11805/Vg+0.00011, with correlation coefficient (R2) of 0.9999. The extraction recovery and sorption coefficient (K)for benzene are 92.47% and 2.15×106, respectively. The relative standard deviation (RSD) and relative error (RE) are 6.98% and 7.50% for benzene, respectively. Detection limit of 5.86×10-3ng·mL-1 and linear range of 10.4 are measured. The maximum amount of the adsorbed benzene is 2636ng. The results obtained by use of this porous layer activated charcoal fiber are also compared with results obtained by use of a polydimethylsiloxane(PDMS) coated fiber(SUPELCO, USA).
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Albion College Mathematics and Computer Science Colloquium
Title: Combinatoria Poetica: Counting and Visualizing Rhyme Patterns in Sonnets Speaker: Hartmut F.W. Höft Professor Computer Science Eastern Michigan University Ypsilanti, Michigan Abstract: I will give a brief overview of sonnets, citing some examples, and describe a notation for the grouping structure and end rhyme pattern types of individual poems and sonnet sequences. Then I construct the sets of rhyme patterns of poems with even rhymes and compute their counts. Cascading this construction over rhyme groupings leads to counts for a variety of sonnet forms. The structure and counts for two types of sonnets are visualized as trees. I then visualize end rhyme patterns as color bands for two types of patterns: (1) systematically generated rhyme patterns for sections of sonnets, and (2) sonnet sequences from the literature. These bands provide a holistic visual overview that can give insight into the structure of poem sequences that may span hundreds of lines. Different color assignments can also be used to exhibit and enhance the visual beauty inherent in rhyme patterns. Mathematica 7 is used to create counts, summary tables and images of end rhyme patterns. Location: Palenske 227 Date: 1/28/2010 Time: 3:10 pm
@abstract{MCS:Colloquium:HartmutFWHöft:2010:1:28,
author = "{Hartmut F.W. Höft}",
title = "{Combinatoria Poetica: Counting and Visualizing Rhyme Patterns in Sonnets}",
address = "{Albion College Mathematics and Computer Science Colloquium}",
month = "{28 January}",
year = "{2010}"
}
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### Symmetry, Integrability and Geometry: Methods and Applications (SIGMA)
SIGMA 12 (2016), 049, 15 pages arXiv:1512.07104 https://doi.org/10.3842/SIGMA.2016.049
Contribution to the Special Issue on Orthogonal Polynomials, Special Functions and Applications
### Shell Polynomials and Dual Birth-Death Processes
Erik A. van Doorn
Department of Applied Mathematics, University of Twente, P.O. Box 217, 7500 AE Enschede, The Netherlands
Received January 02, 2016, in final form May 14, 2016; Published online May 18, 2016
Abstract
This paper aims to clarify certain aspects of the relations between birth-death processes, measures solving a Stieltjes moment problem, and sets of parameters defining polynomial sequences that are orthogonal with respect to such a measure. Besides giving an overview of the basic features of these relations, revealed to a large extent by Karlin and McGregor, we investigate a duality concept for birth-death processes introduced by Karlin and McGregor and its interpretation in the context of shell polynomials and the corresponding orthogonal polynomials. This interpretation leads to increased insight in duality, while it suggests a modification of the concept of similarity for birth-death processes.
Key words: orthogonal polynomials; birth-death processes; Stieltjes moment problem; shell polynomials; dual birth-death processes; similar birth-death processes.
pdf (364 kb) tex (21 kb)
References
1. Anderson W.J., Continuous-time Markov chains. An applications-oriented approach, Springer Series in Statistics: Probability and its Applications, Springer-Verlag, New York, 1991.
2. Berg C., Christensen J.P.R., Density questions in the classical theory of moments, Ann. Inst. Fourier (Grenoble) 31 (1981), 99-114.
3. Berg C., Christiansen J.S., A question by T.S. Chihara about shell polynomials and indeterminate moment problems, J. Approx. Theory 163 (2011), 1449-1464, arXiv:1102.2723.
4. Berg C., Thill M., A density index for the Stieltjes moment problem, in Orthogonal Polynomials and their Applications (Erice, 1990), IMACS Ann. Comput. Appl. Math., Vol. 9, Baltzer, Basel, 1991, 185-188.
5. Berg C., Thill M., Rotation invariant moment problems, Acta Math. 167 (1991), 207-227.
6. Berg C., Valent G., The Nevanlinna parametrization for some indeterminate Stieltjes moment problems associated with birth and death processes, Methods Appl. Anal. 1 (1994), 169-209.
7. Berg C., Valent G., Nevanlinna extremal measures for some orthogonal polynomials related to birth and death processes, J. Comput. Appl. Math. 57 (1995), 29-43.
8. Chihara T.S., Chain sequences and orthogonal polynomials, Trans. Amer. Math. Soc. 104 (1962), 1-16.
9. Chihara T.S., On determinate Hamburger moment problems, Pacific J. Math. 27 (1968), 475-484.
10. Chihara T.S., An introduction to orthogonal polynomials, Mathematics and its Applications, Vol. 13, Gordon and Breach Science Publishers, New York - London - Paris, 1978.
11. Chihara T.S., Indeterminate symmetric moment problems, J. Math. Anal. Appl. 85 (1982), 331-346.
12. Chihara T.S., The parameters of a chain sequence, Proc. Amer. Math. Soc. 108 (1990), 775-780.
13. Chihara T.S., Shell polynomials and indeterminate moment problems, J. Comput. Appl. Math. 133 (2001), 680-681.
14. Coolen-Schrijner P., van Doorn E.A., Orthogonal polynomials on ${\mathbb R}^+$ and birth-death processes with killing, in Difference Equations, Special Functions and Orthogonal Polynomials, Editors S. Elaydi, J. Cushing, R. Lasser, A. Ruffing, V. Papageorgiou, W. Van Assche, World Sci. Publ., Hackensack, NJ, 2007, 726-740.
15. Fralix B., When are two Markov chains similar?, Statist. Probab. Lett. 107 (2015), 199-203.
16. Karlin S., McGregor J.L., The differential equations of birth-and-death processes, and the Stieltjes moment problem, Trans. Amer. Math. Soc. 85 (1957), 489-546.
17. Karlin S., McGregor J., The classification of birth and death processes, Trans. Amer. Math. Soc. 86 (1957), 366-400.
18. Lenin R.B., Parthasarathy P.R., Scheinhardt W.R.W., van Doorn E.A., Families of birth-death processes with similar time-dependent behaviour, J. Appl. Probab. 37 (2000), 835-849.
19. Pedersen H.L., Stieltjes moment problems and the Friedrichs extension of a positive definite operator, J. Approx. Theory 83 (1995), 289-307.
20. Pollett P.K., Similar Markov chains, J. Appl. Probab. 38A (2001), 53-65.
21. Shohat J.A., Tamarkin J.D., The problem of moments, Math. Surveys Monogr., Vol. 1, Amer. Math. Soc., New York, 1943.
22. van Doorn E.A., The indeterminate rate problem for birth-death processes, Pacific J. Math. 130 (1987), 379-393.
23. van Doorn E.A., Representations for the decay parameter of a birth-death process based on the Courant-Fischer theorem, J. Appl. Probab. 52 (2015), 278-289.
24. van Doorn E.A., Spectral properties of birth-death polynomials, J. Comput. Appl. Math. 284 (2015), 251-258.
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# Why is the reaction H2+Cl2 = 2HCl exothermic?
## A: Energy involved in the bonds breaking is greater than that of the bonds forming. B: Energy involved in the bonds forming is greater than that of the bonds breaking. C: More bonds are broken than are formed. D: More bonds are formed than are broken. I know that Exothermic reactions are bond-forming and the energy involved in bond forming is greater than that of bond breaking. According to my book, the answer is B, but I want to know why D can't be the proper answer. Thanks
Nov 7, 2016
#### Answer:
Well, $D$ is not even true.
#### Explanation:
${H}_{2} \left(g\right) + C {l}_{2} \left(g\right) \rightarrow 2 H C l \left(g\right)$
In this reaction, $2 \times H - C l$ bonds are formed, and 2 elemental bonds are broken (i.e. $1 \times H - H$, and $1 \times C l - C l$). And so the number of bonds formed versus bonds broken are EQUAL.
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main-content
## Über dieses Buch
This book constitutes revised selected papers from the 22nd International Conference on Information Security and Cryptology, ICISC 2019, held in Seoul, South Korea, in December 2019.
The total of 18 papers presented in this volume were carefully reviewed and selected from 43 submissions. The papers were organized in topical sections named: public-key encryption and implementation; homomorphic encryption; secure multiparty computation; post-quantum cryptography; secret sharing and searchable encryption; storage security and information retrieval; and attacks and software security.
## Inhaltsverzeichnis
### Revised Version of Block Cipher CHAM
Abstract
CHAM is a family of lightweight block ciphers published in 2017 [22]. The CHAM family consists of three ciphers, CHAM-64/128, CHAM-128/128, and CHAM-128/256. CHAM can be implemented with a remarkably low area in hardware compared to other lightweight block ciphers, and it also performs well on software. We found new (related-key) differential characteristics and differentials of CHAM using a SAT solver. Although attacks using the new characteristics are limited to the reduced rounds of CHAM, it is preferable to increase the number of rounds to ensure a sufficient security margin. The numbers of rounds of CHAM-64/128, CHAM-128/128, and CHAM-128/256 are increased from 80 to 88, 80 to 112, and 96 to 120, respectively. We provide strong evidence that CHAM with these new numbers of rounds is secure enough against (related-key) differential cryptanalysis. Because increasing the number of rounds does not affect the area in low-area hardware implementations, the revised CHAM is still excellent in lightweight hardware implementations. In software, the revised CHAM is still comparable to SPECK, one of the top-ranked algorithms in software.
Dongyoung Roh, Bonwook Koo, Younghoon Jung, Il Woong Jeong, Dong-Geon Lee, Daesung Kwon, Woo-Hwan Kim
### Systematic Construction of Nonlinear Product Attacks on Block Ciphers
Abstract
A major open problem in block cipher cryptanalysis is discovery of new invariant properties of complex type. Recent papers show that this can be achieved for SCREAM, Midori64, MANTIS-4, T-310 or for DES with modified S-boxes. Until now such attacks are hard to find and seem to happen by some sort of incredible coincidence. In this paper we abstract the attack from any particular block cipher. We study these attacks in terms of transformations on multivariate polynomials. We shall demonstrate how numerous variables including key variables may sometimes be eliminated and at the end two very complex Boolean polynomials will become equal. We present a general construction of an attack where multiply all the polynomials lying on one or several cycles. Then under suitable conditions the non-linear functions involved will be eliminated totally. We obtain a periodic invariant property holding for any number of rounds. A major difficulty with invariant attacks is that they typically work only for some keys. In T-310 our attack works for any key and also in spite of the presence of round constants.
Nicolas T. Courtois, Matteo Abbondati, Hamy Ratoanina, Marek Grajek
### Authenticated Encryption Based on Lesamnta-LW Hashing Mode
Abstract
Authenticated encryption refers to symmetric cryptography providing both privacy and authenticity. It is most common to construct it as a block-cipher mode of operation. Another promising approach is to construct it based on cryptographic hashing. This paper proposes a nonce-based authenticated encryption scheme based on the Lesamnta-LW hashing mode. Lesamnta-LW is a block-cipher-based iterated hash function, which is specified in the ISO/IEC 29192-5 lightweight hash-function standard. This paper also shows that the proposed scheme is secure if the underlying block cipher is a pseudorandom permutation. Both of the other ISO/IEC 29192-5 mechanisms, PHOTON and SPONGENT, are hardware-oriented sponge-based hash functions, and nonce-based authenticated encryption schemes can also be constructed based on them. On the other hand, Lesamnta-LW is a software-oriented Merkle-Damgård hash function. Thus, the proposed scheme is a new option for authenticated encryption based on lightweight cryptographic hashing.
Shoichi Hirose, Hidenori Kuwakado, Hirotaka Yoshida
### All the HIGHT You Need on Cortex–M4
Abstract
In this paper, we present high-speed and secure implementations of HIGHT block cipher on 32-bit ARM Cortex-M4 microcontrollers. We utilized both data parallelism and task parallelism to reduce the execution timing. In particular, we used the 32-bit wise ARM–SIMD instruction sets to perform the parallel computations in efficient way. Since the HIGHT block cipher is constructed upon 8-bit word, four 8-bit operations are performed in the 32-bit wise ARM–SIMD instruction of ARM Cortex-M4 microcontrollers. We also presented a novel countermeasure against fault attack on target microcontrollers. The method achieved the fault attack resistance with intra-instruction redundancy feature with reasonable performance. Finally, the proposed HIGHT implementation achieved much better performance and security level than previous works.
Hwajeong Seo, Zhe Liu
### Fast AES Implementation Using ARMv8 ASIMD Without Cryptography Extension
Abstract
While the ARMv8-A ISA allows for hardware accelerated cryptographic instructions, such extension is not available for every device, being added at the discretion of the CPU manufacturer. Prime examples of ARMv8 devices without this support are the low cost Raspberry Pi 3B/3B+/4 single board computers. This work presents an optimized AES implementation targeting CPUs without Cryptography Extension instructions, relying only on ASIMD operations. We show a new implementation that processes four blocks at the same time, which requires block permutations and modified versions of the main layers. In particular, we provide a new efficient formula for computing the MixColumns layer. The time performance our AES implementation outperforms the current ASIMD implementation found in the Linux Kernel by about 5%.
Hayato Fujii, Félix Carvalho Rodrigues, Julio López
### FACE–LIGHT: Fast AES–CTR Mode Encryption for Low-End Microcontrollers
Abstract
In this paper, we revisited the previous Fast AES–CTR mode Encryption (FACE) method for high-end processors and tailored the method to the microcontrollers, namely FACE–LIGHT. We targeted the 32-bit counter mode of operation for AES in constant timing. This optimized technique pre-computes the 2 Add-RoundKey, 2 Sub-Bytes, 2 Shift-Rows and 1 Mix-Columns operations. The FACE–LIGHT is implemented on the representative low-end microcontrollers (e.g. 8-bit AVR). The execution timing of AES–CTR algorithm for 128-bit and 256-bit security levels achieved the 2,218 and 3,184 clock cycles, respectively. This is faster than previous works by 22 % for 128-bit security level. The FACE–LIGHT can be used to extend the FACE to round 3. The AES is also implemented to be secure against the CPA (Correlation Power Analysis).
Kyungho Kim, Seungju Choi, Hyeokdong Kwon, Zhe Liu, Hwajeong Seo
### Sum It Up: Verifiable Additive Homomorphic Secret Sharing
Abstract
In many situations, clients (e.g., researchers, companies, hospitals) need to outsource joint computations based on joint inputs to external cloud servers in order to provide useful results. Often clients want to guarantee that the results are correct and thus, an output that can be publicly verified is required. However, important security and privacy challenges are raised, since clients may hold sensitive information and the cloud servers can be untrusted. Our goal is to allow the clients to protect their secret data, while providing public verifiability i.e., everyone should be able to verify the correctness of the computed result.
In this paper, we propose three concrete constructions of verifiable additive homomorphic secret sharing (VAHSS) to solve this problem. Our instantiations combine an additive homomorphic secret sharing (HSS) scheme, which relies on Shamir’s secret sharing scheme over a finite field $$\mathbb {F}$$, for computing the sum of the clients’ secret inputs, and three different methods for achieving public verifiability. More precisely, we employ: (i) homomorphic collision-resistant hash functions; (ii) linear homomorphic signatures; as well as (iii) a threshold RSA signature scheme. In all three cases we provide a detailed correctness, security and verifiability analysis and discuss their efficiency.
Georgia Tsaloli, Aikaterini Mitrokotsa
### There Is Always an Exception: Controlling Partial Information Leakage in Secure Computation
Abstract
Private Function Evaluation (PFE) enables two parties to jointly execute a computation such that one of them provides the input while the other chooses the function to compute. According to the traditional security requirements, a PFE protocol should leak no more information, neither about the function nor the input, than what is revealed by the output of the computation. Existing PFE protocols inherently restrict the scope of computable functions to a certain function class with given output size, thus ruling out the direct evaluation of such problematic functions as the identity map, which would entirely undermine the input privacy requirement. We observe that when not only the input x is confidential but certain partial information g(x) of it as well, standard PFE fails to provide meaningful input privacy if g and the function f to be computed fall into the same function class.
Our work investigates the question whether it is possible to achieve a reasonable level of input and function privacy simultaneously even in the above cases. We propose the notion of Controlled PFE (CPFE) with different flavours of security and answer the question affirmatively by showing simple, generic realizations of the new notions. Our main construction, based on functional encryption (FE), also enjoys strong reusability properties enabling, e.g. fast computation of the same function on different inputs. To demonstrate the applicability of our approach, we show a concrete instantiation of the FE-based protocol for inner product computation that enables secure statistical analysis (and more) under the standard Decisional Diffie–Hellman assumption.
Máté Horváth, Levente Buttyán, Gábor Székely, Dóra Neubrandt
### An Automated Security Analysis Framework and Implementation for MTD Techniques on Cloud
Abstract
Cloud service providers offer their customers with on-demand and cost-effective services, scalable computing, and network infrastructures. Enterprises migrate their services to the cloud to utilize the benefit of cloud computing such as eliminating the capital expense of their computing need. There are security vulnerabilities and threats in the cloud. Many researches have been proposed to analyze the cloud security using Graphical Security Models (GSMs) and security metrics. In addition, it has been widely researched in finding appropriate defensive strategies for the security of the cloud. Moving Target Defense (MTD) techniques can utilize the cloud elasticity features to change the attack surface and confuse attackers. Most of the previous work incorporating MTDs into the GSMs are theoretical and the performance was evaluated based on the simulation. In this paper, we realized the previous framework and designed, implemented and tested a cloud security assessment tool in a real cloud platform named UniteCloud. Our security solution can (1) monitor cloud computing in real-time, (2) automate the security modeling and analysis and visualize the GSMs using a Graphical User Interface via a web application, and (3) deploy three MTD techniques including Diversity, Redundancy, and Shuffle on the real cloud infrastructure. We analyzed the automation process using the APIs and showed the practicality and feasibility of automation of deploying all the three MTD techniques on the UniteCloud.
Hooman Alavizadeh, Hootan Alavizadeh, Dong Seong Kim, Julian Jang-Jaccard, Masood Niazi Torshiz
### Security Analysis of Group Action Inverse Problem with Auxiliary Inputs with Application to CSIDH Parameters
Abstract
In this paper, we consider the security of a problem called Group Action Inverse Problem with Auxiliary Inputs (GAIPwAI). The Group Action Inverse Problem (GAIP) plays an important role in the security of several isogeny-based cryptosystems, such as CSIDH, SeaSign and CSI-FiSh.
Briefly speaking, given two isogenous supersingular curves E and $$E'$$ over $$\mathbb F_p$$, where $$E'$$ is defined by an ideal $$\mathfrak a$$ in the $$\mathbb F_p$$-endomorphism ring of E and denoted by $$E' = [\mathfrak a]*E$$, GAIP requires finding $$\mathfrak a \subset {\text {End}}_{\mathbb F_p}(E)$$. Its best classical algorithm is based on the baby-step-giant-step method and it runs in time $$O(p^{1/4})$$.
In this paper, we show that if E and $$E'$$ are given together with $$[\mathfrak a^d]*E$$ for a positive divisor d that divides the order of the class group of $${\mathbb Z}[\sqrt{-p}]$$, then $$\mathfrak a$$ can be computed in $$O\big ( ( p^{1/2} /d)^{1/2} + d^{1/2} \big )$$ time complexity. In particular, when $$d \approx p^{1/4}$$, it can be solved in time $$O( p^{1/8} )$$ which is significantly less than $$O( p^{1/4} )$$.
Applying the idea to CSIDH-512 parameters, we show that, if an additional isogenous curve $$[\mathfrak a^d] * E$$ is given, the security level of this cryptosystem reduces to 68-bit security instead of 128-bit security as originally believed.
Taechan Kim
### Secure Key Encapsulation Mechanism with Compact Ciphertext and Public Key from Generalized Srivastava Code
Abstract
Code-based public key cryptosystems have been found to be an interesting option in the area of Post-Quantum Cryptography. In this work, we present a key encapsulation mechanism (KEM) using a parity check matrix of the Generalized Srivastava code as the public key matrix. Generalized Srivastava codes are privileged with the decoding technique of Alternant codes as they belong to the family of Alternant codes. We exploit the dyadic structure of the parity check matrix to reduce the storage of the public key. Our encapsulation leads to a shorter ciphertext as compared to DAGS proposed by Banegas et al. in Journal of Mathematical Cryptology which also uses Generalized Srivastava code. Our KEM provides IND-CCA security in the random oracle model. Also, our scheme can be shown to achieve post-quantum security in the quantum random oracle model.
Jayashree Dey, Ratna Dutta
### Improvement of Binary and Non Binary Statistical Decoding Algorithm
Abstract
The security of McEliece’s cryptosystem relies heavily on the hardness of decoding a random linear code. The best known generic decoding algorithms are derived from the Information-Set Decoding (ISD) algorithm. This was first proposed in 1962 by Prange and subsequently improved in 1989 by Stern and later in 1991 by Dumer. In 2001 Al Jabri introduced a new decoding algorithm for general linear block codes which does not belong to this family, called Statistical Decoding (SD). Since then, like for the Information Set Decoding algorithm, there have been numerous work done to improve and generalize the SD algorithm. In this paper, we improve the SD algorithm using the notion of bases lists in binary case. Then, we give a non binary version of this improvement. Finally, we have computed complexity analysis and have made a complexity comparison of our results with that of recent results on SD algorithm and complexity of classic ISD algorithm.
Pierre-Louis Cayrel, Cheikh Thiécoumba Gueye, Junaid Ahmad Khan, Jean Belo Klamti, Edoardo Persichetti
### LizarMong: Excellent Key Encapsulation Mechanism Based on RLWE and RLWR
Abstract
The RLWE family algorithms submitted to the NIST post-quantum cryptography standardization process have each merit in terms of security, correctness, performance, and bandwidth. However, there is no splendid algorithm in all respects. Besides, various recent studies have been published that affect security and correctness, such as side-channel attacks and error dependencies. To date, though, no algorithm has fully considered all the aspects. We propose a novel Key Encapsulation Mechanism scheme called LizarMong, which is based on RLizard. LizarMong combines the merit of each algorithm and state-of-the-art studies. As a result, it achieves up to 85% smaller bandwidth and 3.3 times faster performance compared to RLizard. Compared to the NIST’s candidate algorithms with a similar security, the bandwidth is about 5–42% smaller, and the performance is about 1.2-4.1 times faster. Also, our scheme resists the known side-channel attacks.
Chi-Gon Jung, JongHyeok Lee, Youngjin Ju, Yong-Been Kwon, Seong-Woo Kim, Yunheung Paek
### Efficient Identity-Based Encryption from LWR
Abstract
The Learning with Rounding (LWR) problem is a deterministic variant of the classical Learning with Errors (LWE) problem, for which sampling an instance does not involve discrete Gaussian sampling. We propose the first probabilistic Identity-Based Encryption (IBE) from the LWR problem which is secure in the standard model. The encryption of our IBE scheme does not require discrete Gaussian sampling as it is based on the LWR problem, and hence it is simpler and faster than that of LWE-based IBEs such as ABB scheme. We also present an efficient instantiation employing algebraic ring structure and MP12 trapdoor sampling algorithms with an implementation result. With our proposed parameter sets, the ciphertext sizes can be reduced in a large extent compared to the ABB scheme with the same security level.
Jung Hee Cheon, Haejin Cho, Jaewook Jung, Joohee Lee, Keewoo Lee
### Faster Bootstrapping of FHE over the Integers
Abstract
In FHE over the integers, decryption function is simplified by sparse subset subset sum problem (SSSP) assumption, which is introduced by Dijk et al. (Eurocrypt 2010), so that bootstrapping can be achieved successfully. Later, Nuida and Kurowasa (Eurocrypt 2015) proposed an advanced method of which the degree is very low and the message space is non-binary. These previous methods require low degree but more than $$O(\lambda ^4)$$ homomorphic multiplications which make them very slow. For a general bootstrapping method in FHE over the integers, the number of homomorphic multiplications and the degree of decryption function are important factors for the efficiency of bootstrapping procedure.
In this paper, we propose a new bootstrapping method for FHE over the integers requiring only $$O(\log ^2 \lambda )$$ homomorphic multiplications which is significantly lower than previous methods. Implementing our bootstrapping method on the scale-invariant FHE over the integers called CLT scheme, it takes 6 s for 500-bit message space and 80-bit security on a desktop. We also apply our bootstrapping method to the homomorphic evaluation of AES-128 circuit: It takes about 8 s per 128-bit block and is faster than the previous results of homomorphic AES evaluation using FHEs over the integers without bootstrapping.
Jung Hee Cheon, Kyoohyung Han, Duhyeong Kim
### Complete Addition Law for Montgomery Curves
Abstract
Montgomery curves allow efficient and side-channel resistant computation of ECDH using the Montgomery ladder. But the addition law of a Montgomery curve derived from the chord-tangent method is less efficient than other curve models such as a short Weierstrass curve and an Edwards curve. So, the usage of a Montgomery curve is strictly limited to ECDH only, such as $$\mathsf {X25519}$$ and $$\mathsf {X448}$$ functions in IETF RFC 7748. For other operations including fixed-base and multiple scalar multiplications, their birationally-equivalent (twisted) Edwards curves are recommended for use since the conversions between Montgomery curves and their Edwards equivalents are simple. This conversion enables the use of the efficient complete addition law of the Edwards curve that works for all pairs of input points with no exceptional cases. As a result, the combination allows secure and exception-free implementations, but at the expense of additional storage for the two curve parameters and for the conversion between them. However, smart devices in IoT environments that mainly operate ECDH (for example, RawPublicKey mode of IETF RFC 7250) do not need to implement such a conversion if a complete addition law does exist for the Montgomery curves.
To make such implementations possible, we provide a complete addition law on Montgomery curves. The explicit formulas for the complete addition law are not as efficient as those of Edwards curves, but they can make the Montgomery curve addition operation more efficient compared to using the conversion to the (twisted) Edwards equivalent. We also confirmed the validity of the comparison by implementing such two methods of realizing the addition operation on $$\text {Curve25519}$$.
Jaeheon Kim, Je Hong Park, Dong-Chan Kim, Woo-Hwan Kim
### Improved CRT-RSA Secret Key Recovery Method from Sliding Window Leakage
Abstract
In this paper, we discuss side-channel attacks on the CRT-RSA scheme (RSA scheme with Chinese Remainder Theorem) implemented by the left-to-right sliding window method. This method calculates exponentiations by repeating squaring and multiplication. In CHES 2017, Bernstein et al. proposed side-channel attacks on the CRT-RSA signature scheme implemented by the left-to-right sliding window method. We can obtain square-and-multiply sequences by their side-channel attacks, but cannot calculate CRT-RSA secret keys because there are multiple candidates of multiplications. Then, Bernstein et al. calculated CRT-RSA secret keys by using two methods. First, they recovered CRT-RSA secret keys partially and calculated all secret key bits by using the Heninger–Shacham method. Second, they applied the Heninger–Shacham method to square-and-multiply sequences directly. They showed that we can calculate CRT-RSA secret keys more efficiently when we use square-and-multiply sequences directly. They also showed that we can recover CRT-RSA secret keys in polynomial time when $$w \le 4$$. Moreover, they experimentally showed that we can recover secret keys of 2048-bit CRT-RSA scheme when $$w=5$$. However, their latter method is simple and has room for improvement. Here, we study bit recovery more profoundly to improve their method. First, we calculate the exact rate of all knowable bits. Next, we propose a new method for calculating the proportion of each bit 0 or 1 in each nonrecovery bit. Finally, we propose a new method for calculating CRT-RSA secret key using this bit information. In our proposed algorithm, we extend Bernstein et al.’s method in combination with Kunihiro et al.’s method. We calculate more secret keys when $$w=5$$ by our proposed method compared to Bernstein et al.’s method.
Kento Oonishi, Xiaoxuan Huang, Noboru Kunihiro
### Differential Random Fault Attacks on Certain CAESAR Stream Ciphers
Abstract
We show that a particular class of stream ciphers – namely those in which the output function contains a bitwise AND operation – are susceptible to a differential fault attack using random faults. Several finalists and other candidates from the recent CAESAR competition fall into this category, including the AEGIS variants, Tiaoxin and the MORUS family. Attack outcomes range from key or full state recovery for Tiaoxin, to full state recovery for the AEGIS family and partial state recovery for MORUS. We present attack requirements and success probabilities on these ciphers, along with design considerations to mitigate against this attack.
Kenneth Koon-Ho Wong, Harry Bartlett, Leonie Simpson, Ed Dawson
### Backmatter
Weitere Informationen
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# Milling PCBs?
#### DNA Robotics
Joined Jun 13, 2014
603
I just finished my second board yesterday. It is for PIC 16F877 & 18F4550 tutorials & development. Designed & layout in Eagle 6.4.0 and used pcb-gcode-3.6.0.4 to generate single pass G-code to engrave & drill with Mach3Mill.
I used a .2mm, .0078" 30 degree engraving cutter and a .8mm, .032" mill/drill at 10,000 RPM and 10 IPM feed. In the future, I will use larger VIA pads.
Nice clean cuts. I didn't even have to sand the edges.
Joined Jul 18, 2013
23,393
I just finished my second board yesterday. It is for PIC 16F877 & 18F4550 tutorials & development. Designed & layout in Eagle 6.4.0 and used pcb-gcode-3.6.0.4 to generate single pass G-code to engrave & drill with Mach3Mill.
What machine make/model?
Max.
#### spinnaker
Joined Oct 29, 2009
7,835
Nice. Is it me or are some of your holes off center? Looks like some are oblong too. Was that intentional?
#### ISB123
Joined May 21, 2014
1,236
Nice. Is it me or are some of your holes off center? Looks like some are oblong too. Was that intentional?
Those holes are still better centered than I could possibly do.
Last edited:
#### DNA Robotics
Joined Jun 13, 2014
603
What machine make/model?
I used my 75 year old milling machine that I converted to CNC using large stepper motors. The free Mach3 program compensates for the 0.030” backlash nicely.
I made a separate 10,000 RPM spindle for engraving.
#### DNA Robotics
Joined Jun 13, 2014
603
Nice. Is it me or are some of your holes off center? Looks like some are oblong too. Was that intentional?
That was a little accumulated error. I'm glad it wasn't any more. The pcb-gcode drill pattern zigzagged back and forth from one side of the board to the other instead of going down a row of IC pins or header pins.
What might look like oblonged holes, there is a 40 pin IC socket inserted but not soldered in that first board picture.
Last edited:
#### DNA Robotics
Joined Jun 13, 2014
603
Those holes are still better centered than I could possibly do.
The pcb-gcode has an option to engrave the center of each hole after cutting the traces. So when drilling in the next step, that tiny drill falls right into the center mark without wandering or breaking.
Last edited:
#### spinnaker
Joined Oct 29, 2009
7,835
That is one heck of a machine! Does the table move? You must have huge stepper motors to move it? How do you think it would have faired at 8k rpm? That is the max on the cheap I am looking. I actually was considering converting a compound table but the one I was looking had only about an inch of travel on the y axis. How much travel do you have?
#### atferrari
Joined Jan 6, 2004
4,389
Hola DNA
How do you hold the board in position?
#### ISB123
Joined May 21, 2014
1,236
Could be double-sided tape.
#### Lestraveled
Joined May 19, 2014
1,946
#### DNA Robotics
Joined Jun 13, 2014
603
The table moves 24" in the X direction & 8" in Y.
I believe in overkill & didn't want any question about whether the motors would move all that iron so I got these motors http://www.mpja.com/Stepper-Motor-NEMA-34-78V/productinfo/17457 MS and these drivers http://www.mpja.com/Stepper-Motor-Driver-5A-Max-Wide-Range-CW250/productinfo/17452 MS
This looks like my interface
DIY Engraving Machine MACH3 USB CNC 4-axis Interface Breakout Board Adapter $15.00 You would have to use slower feed at 8K RPMs. There is a formula to determine Inches per minute to cut 2 or 3 thousandths of an inch per revolution of the cutter. WT 3-Way Mill & Drill Table$125.20
Thanks guys.
Yes it is held in place with double sided carpet tape. I flexed the board a little to take the bow out of it. When in place there was only .002" variation across the board.
#### DNA Robotics
Joined Jun 13, 2014
603
I actually was considering converting a compound table but the one I was looking had only about an inch of travel on the y axis.
Maybe keep watching eBay & craigslist for a compound table. You may even find one cheap with motors on it. Places that sell used machine tools & tooling may have just what you want. I'll bet there are plenty in an industrial area like Pittsburgh.
Joined Jul 18, 2013
23,393
The main thing with CNC is the leadscrews, most manual machines and tables come with an Acme screw, to convert to CNC it is customary to swap them out for Ball screw, even if the cheaper rolled type rather than the better ground versions.
Also the B. screw nut should be pre-loaded for zero back lash.
Max..
#### spinnaker
Joined Oct 29, 2009
7,835
Yes it is held in place with double sided carpet tape. I flexed the board a little to take the bow out of it. When in place there was only .002" variation across the board.
Would you mind detailing how this is done in Mach3? I have yet to use it so I am clueless.
Have you tried pc-probe?
http://www.instructables.com/id/Mak...-desktop-CNC-m/step3/Make-G-Code-for-a-Board/
I am not sure what the heck he is trying to describe. The article is not very clear.
I see you are milling on a piece of Plexiglas. Do you drill on that too? Or do you swap out a spoil board? If you swap out the board how do you line things back up?
Maybe you can write a blog on a generic milling procedure? I have yet to see one that is easy to understand. I would be happy to work with you on it. If I could understand it then you would have created a guide that anyone could follow.
#### DNA Robotics
Joined Jun 13, 2014
603
It had been over a year since I made my first board & I even had to reinstall Eagle 6.4.0 and pcb-gcode-3.6.0.4 so I had to read the "How to" again.
The Eagle "tuorial-en.pdf " is good & quick with sample projects. These are also good but I didn't need them this time.
https://learn.sparkfun.com/tutorials/using-eagle-schematic/introduction
https://learn.sparkfun.com/tutorials/using-eagle-board-layout
The pcbgcode.pdf User's manual is in their \docs folder. It was pretty much strait forward step by step also. Mine was single sided so I just told it to etch the bottom in one pass & drill from the bottom. Don't check the preview box. You probably don't have the JAVAW.EXE file that it needs.
I am milling on an aluminum plate that was milled flat. I drill .070" into a .062" board so it does drill into the aluminum but using tape, I'm not worried about it. A dial indicator said the board was flat within .002" & I cut .010" deep so I didn't use pcb-probe.
Mach3Mill requires a lot of setup. Their step by step documentation is pretty good. You start by wiring your parallel port cable to your interface / breakout board. DIY Engraving Machine MACH3 USB CNC 4-axis Interface Breakout Board Adapter Some wires you do their way and others, you tell the setup how they are wired. (active high or low etc.) Then set up the machine by telling it how many steps per inch or MM for each axis, then acceleration rate for the stepper motors and maximum speed. Then tell it how much backlash is in each axis. Mach3 compensates for backlash by adding that many steps every time it changes direction. They say it is a last resort but it works nicely on my old machine.
I put a pen in the chuck & had the machine draw that circuit board on paper until it was perfect before cutting the real board.
#### spinnaker
Joined Oct 29, 2009
7,835
Thanks for the tips DNA. I don't use Eagle Cad but I guess that PDF should still be helpful. It is just a guide on milling PCBs correct?
That breakout board you posted, is it all that is needed for the controller? Because when I see break out board I think of the easy to use boards you might use to interface to a mcu.
Will the USB port on that board work with Mach3 as well?
#### DNA Robotics
Joined Jun 13, 2014
603
This documentation on the interface board that I got wasn’t very clear. When I got it sorted out, I labeled everything on the photo of it. Wherever they use P, as in P10 it refers to that Pin on the parallel port. The USB is only to supply 5V interface power to the board, not data. This board sends step & direction signals to your stepper motor drivers. It does not drive the stepper motors directly.
Product Name: MACH3 interface board BV2
1, new products and that the appearance and size is similar to the
original function has only added to the
5V and 12V output.
This paragraph 2, the upgraded new constant prices, better
features, there are three relays can be used, you can use up to
six axes.
First, the scope
For numerical control software the MACH3 the host computer and
peripheral power control signal transfer and processing.
Second, the characteristics
1, The output and input models by Philips dedicated high-speed chip processing, signal integer, zoom, in addition to the unnecessary interference signals. Operation of the machine is more stable.
2, USB powered with external power supply. Power supply isolation. Computer and machining data is more secure
3, All of the input signal all the optical coupling isolation, of which P11, P12 port for high-speed optical coupling can be used to pick the spindle encoder, or handwheel A equivalent, strictly signal.
4, The ad hoc one off power relay of the power supply automatically when the computer shuts down, turn off all power of the engraving machine equipment.
5, Has a relay switch coolant supply, support MACH3, G code programming (parallel port P3, G code M7 open, M9 off), switch lights on the relay. Convenient to determine the power supply off.
6, Has a spindle-powered relay, MACH3, G code programming (parallel port P2, G code M3 open, M5 off), switch lights on the relay. Convenient to determine the power supply off
7, For the parallel port on the spindle speed control: ad hoc high-speed optical coupling isolation, you can output the PWM signal, the output port P1.
8, This interface board has a PWM converted to 0-10V digital-analog conversion circuit. Output 0-10V level to control there is only analog speed control inverter to control the spindle speed. The output for the parallel port P1.
9, The interface board can be connected to a common cathode or positive output to the stepper drives, servo drives. The output of the 5V level.
10, Wide supply voltage can be 12V or 16-35V two groups choose a
set of power supply. Power input the anti-reverse design.
Third, the Function Description
Input ports
P10 Emergency stop
P11 Alternate
P12 Alternate
P13 3-axis limit
P15 The knife
Output ports
P4 X pulse
P5 X direction
P6 Y Pulse
P7 Y direction
P8 Z pulse
P9 Z direction
P16 4 pulse
P17 4 directions
P2 Relay port
P3 Relay port
P14 An alternate port
The board needs power input in two ways. One is 16-32V input
(the design of anti-reverse function)
There is also a group of 5V is taken from the USB port, which means that
the interface board in this paragraph must be connected a USB cable to make it
work. (Our ship is equipped with a USB cable ). Because to do isolation
protection function, so these two parts of the supply voltage must
be with the USBpower supply voltage is relatively independent of the job
CNCMaker.com E-mail:CNCMaker@hotmail.com
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# Software:gretl
Developer(s) Screenshot of gretl the gretl team 31 January 2000; 22 years ago Through git C Cross-platform Multilingual (11) Statistical software GNU GPLv3 gretl.sourceforge.net
gretl is an open-source statistical package, mainly for econometrics. The name is an acronym for Gnu Regression, Econometrics and Time-series Library.
It has both a graphical user interface (GUI) and a command-line interface. It is written in C, uses GTK+ as widget toolkit for creating its GUI, and calls gnuplot for generating graphs. The native scripting language of gretl is known as hansl (see below); it can also be used together with TRAMO/SEATS, R, Stata, Python, Octave, Ox and Julia.
It includes natively all the basic statistical techniques employed in contemporary Econometrics and Time-Series Analysis. Additional estimators and tests are available via user-contributed function packages, which are written in hansl.[1] gretl can output models as LaTeX files.
Besides English, gretl is also available in Albanian, Basque, Bulgarian, Catalan, Chinese, Czech, French, Galician, German, Greek, Italian, Polish, Portuguese (both varieties), Romanian, Russian, Spanish, Turkish and Ukrainian.
Gretl has been reviewed several times in the Journal of Applied Econometrics[2][3][4] and, more recently, in the Australian Economic Review.[5]
A review also appeared in the Journal of Statistical Software[6] in 2008. Since then, the journal has featured several articles in which gretl is used to implement various statistical techniques.
## Supported data formats
gretl offers its own fully documented, XML-based data format.
It can also import ASCII, CSV, databank, EViews, Excel, Gnumeric, GNU Octave, JMulTi, OpenDocument spreadsheets, PcGive, RATS 4, SAS xport, SPSS, and Stata files. Since version 2020c, the GeoJSON and Shapefile formats are also supported, for thematic map creation.
It can export to Stata, GNU Octave, R, CSV, JMulTi, and PcGive file formats.
## hansl
Gretl has its own scripting language, called hansl (which is a recursive acronym for Hansl's A Neat Scripting Language).
Hansl is a Turing-complete, interpreted programming language, featuring loops, conditionals, user-defined functions and complex data structures.[7] It can be considered a domain-specific language for econometrics.[8] Like other scientifically oriented programming languages, such as MATLAB and Julia, matrices are supported natively as a primitive variable type.
A simple example of hansl:
matrix A = {1, 2 ; 3, 4}
matrix B = inv(A)
matrix C = A*B
print A B C
loop i=-3..3
printf "Phi(%d) = %7.3f\n", i, cdf(N, i)
endloop
Running the above code produces
A (2 x 2)
1 2
3 4
B (2 x 2)
-2 1
1.5 -0.5
C (2 x 2)
1.0000 0.0000
8.8818e-16 1.0000
Phi(-3) = 0.001
Phi(-2) = 0.023
Phi(-1) = 0.159
Phi( 0) = 0.500
Phi( 1) = 0.841
Phi( 2) = 0.977
Phi( 3) = 0.999
## Random Number Generation
Random Number Generation (RNG) in gretl has been examined and tested in Yalta & Schreiber (2012).[9] The authors conclude "Our results show that the RNG related procedures in gretl are implemented soundly and perform well in the three crush test suites of the TestU01".
## Gretl as a teaching tool
Due to its libre nature and the breadth of econometric techniques it contains, gretl is widely used for teaching econometrics, from the undergraduate level onwards. Datasets in gretl format are available for several popular textbooks.
The following is a list of textbooks that use gretl as their software of choice:
• Dougherty, Christopher Introduction to Econometrics (Oxford University Press)
• Kufel, Tadeusz Ekonometria (Wydawnictwo Naukowe PWN); in Polish (Russian version also available)
In addition, a free supplement[10] to Hill, Griffiths and Lim Principles of Econometrics (Wiley) is available.
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heegard diagram - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T03:14:28Z http://mathoverflow.net/feeds/question/109261 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109261/heegard-diagram heegard diagram mark 2012-10-09T23:50:02Z 2012-10-11T14:30:21Z <p>It seems like there is an algorithm to find the Heegard diagram of a 3 manifold obtained by surgery on a link. Also someone told me I can find it in the Gompf and Stipciz's book. But I could not find it. Can anyone help? </p> http://mathoverflow.net/questions/109261/heegard-diagram/109295#109295 Answer by Daniel Moskovich for heegard diagram Daniel Moskovich 2012-10-10T13:04:44Z 2012-10-11T08:30:48Z <p>I don't know about Gompf and Stipsicz, but the algorithm to obtain a Heegaard diagram from a surgery presentation may be found for example in <a href="http://www.jstor.org/discover/10.2307/2154000?uid=3738992&uid=2129&uid=2&uid=70&uid=4&sid=21101302227937" rel="nofollow">A Simple Proof of the Fundamental Theorem of Kirby Calculus on Links</a> by Ning Lu, Trans. Amer. Math. Soc. Vol 331(1) pp. 143-156 (1992). It's not difficult, and it's worth knowing.</p> http://mathoverflow.net/questions/109261/heegard-diagram/109307#109307 Answer by Kevin Walker for heegard diagram Kevin Walker 2012-10-10T16:05:40Z 2012-10-11T14:30:21Z <p>(1) Choose a planar presentation of your link, approximately in the plane of the blackboard. Let $N$ be a tubular neighborhood of the link in this position.</p> <p>(2) For each crossing $c$ of the presentation, add a 1-handle $T_c$ to $N$ which is perpendicular to the blackboard and connects the upper and lower parts of the crossing. Let $H$ be the union of $N$ and all the 1-handles $T_c$.</p> <p>(3) $S^3 \setminus H$ is a handlebody. The corresponding set of Heegaard curves on $\partial H$ bijects with the complementary regions of the (flattened) planar diagram of the link.</p> <p>(4) $H$ is, of course, also a handlebody. For the corresponding Heegaard curves take the surgery curves on $N$ (this involves a choice to make them disjoint from the attaching disks of the $T_c$) union the obvious small, disk-bounding curves on the boundary of each $T_c$. (Correction added later: Actually, if there are $k$ components of the link then one should omit $k-1$ of the $T_c$ curves. The omitted crossings should be minimal with respect to connecting the components of the link. Also, I'm assuming that the planar projection is a connected graph.)</p>
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How can I use LaTeX-commands inside of a listing?
I'm using the listings package in a LaTeX document. It works great, but it doesn't allow me to put LaTeX commands inside of the listing. I would like to use a counter to enumerate some of the interesting parts in the source code.
Is there a way to enable LaTeX commands in source listings?
OK, found the answer. You can use escapechar=<char> to set a character that will escape you to the latex mode. See The Listings Package, 4.3.13 Escaping to LaTeX.
• Here is an example: if you set escapechar=\%, you could use it inside the listings as follows: \begin{lstlisting} t %\leftarrow% 0 \end{lstlisting}. – thiagowfx Jul 9 '17 at 23:28
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# Past Papers’ Solutions | Assessment & Qualification Alliance (AQA) | AS & A level | Mathematics 6360 | Pure Core 1 (6360-MPC1) | Year 2007 | June | Q#4
Question A model helicopter takes off from a point O at time t=0 and moves vertically so that its height, y cm, above O after time t seconds is given by , a. Find: i. ii. b. Verify that y has a stationary value when and determine whether this stationary value is a maximum value or a […]
# Past Papers’ Solutions | Assessment & Qualification Alliance (AQA) | AS & A level | Mathematics 6360 | Pure Core 1 (6360-MPC1) | Year 2007 | January | Q#6
Question The curve with equation is sketched below. The curve cuts the x-axis at the point A (-1, 0) and cuts the y-axis at the point B. a. i. State the coordinates of the point B and hence find the area of the triangle AOB, where O is the origin. […]
# Past Papers’ Solutions | Assessment & Qualification Alliance (AQA) | AS & A level | Mathematics 6360 | Pure Core 1 (6360-MPC1) | Year 2007 | January | Q#5
Question The diagram shows an open-topped water tank with a horizontal rectangular base and four vertical faces. The base has width metres and length metres, and the height of the tank is metres. The combined internal surface area of the base and four vertical faces is 54m2. a. i. Show that . ii. Hence express in terms of . iii. Hence show […]
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# Diagonal
Can a rhombus have the same length diagonal and side?
Result
#### Solution:
$\alpha =6{0}^{\circ }$
We would be very happy if you find an error in the example, spelling mistakes, or inaccuracies, and please send it to us. We thank you!
Tips to related online calculators
#### You need to know the following knowledge to solve this word math problem:
We encourage you to watch this tutorial video on this math problem:
## Next similar math problems:
• Rhombus diagonals
In the rhombus ABCD are given the sizes of diagonals e = 24 cm; f = 10 cm. Calculate the side length of the diamond and the size of the angles, calculate the content of the diamond
The regular quadrilateral pyramid has a base edge a = 1.56 dm and a height h = 2.05 dm. Calculate: a) the deviation angle of the sidewall plane from the base plane b) deviation angle of the side edge from the plane of the base
• Railway embankment
The section of the railway embankment is an isosceles trapezoid, the sizes of the bases of which are in the ratio 5: 3. The arms have a length of 5 m and the height of the embankment is 4.8 m. Calculates the size of the embankment section area.
• Right triangle
A right triangle ABC is given, c is a hypotenuse. Find the length of the sides a, b, the angle beta if c = 5 and angle alfa = A = 35 degrees.
• The staircase
The staircase has a total height of 3.6 m and forms an angle of 26° with the horizontal. Calculate the length of the whole staircase.
• Triangular pyramid
Calculate the volume of a regular triangular pyramid with edge length a = 12cm and pyramid height v = 20cm.
• Five circles
On the line segment CD = 6 there are 5 circles with radius one at regular intervals. Find the lengths of the lines AD, AF, AG, BD, and CE
• A cliff
A line from the top of a cliff to the ground passes just over the top of a pole 5 ft high and meets the ground at a point 8 ft from the base of the pole. If the point is 93 ft from the base of the cliff, how high is the cliff?
• Construction
Construction the triangle ABC, if you know: the size of the side AC is 6 cm, the size of the angle ACB is 60° and the distance of the center of gravity T from the vertex A is 4 cm. (Sketch, analysis, notation of construction, construction)
• There
There is a triangle ABC: A (-2,3), B (4, -1), C (2,5). Determine the general equations of the lines on which they lie: a) AB side, b) height to side c, c) Axis of the AB side, d) median ta to side a
• Surface and volume
Find the surface and volume of the rotating cone if the circumference of its base is 62.8 m and the side is 25 m long.
• Digging a pit
The pit has the shape of a regular quadrilateral truncated pyramid. The edges of the bases are 14m and 10m long. The sidewalls form an angle of 135° with a smaller base. Determine how many m3 of soil were excavated when digging the pit?
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# Unit Vector (G Dataflow)
Version:
Finds the norm of a vector and normalizes the vector with the norm.
## input vector
Input vector.
This input can be an array of double-precision floating-point numbers or an array of complex double-precision floating-point numbers.
If input vector is an empty array, unit vector is also an empty array, and norm is NaN.
## norm type
Type of norm this node uses to compute the norm.
Name Value Description
1-norm 1 Uses 1-norm.
2-norm 2 Uses 2-norm.
Inf-norm 3 Uses infinity-norm.
-Inf-norm 4 Uses -infinity-norm.
User Defined 5 Uses user defined norm as the norm type.
Algorithm for Calculating norm with Each norm type
This node calculates norm using the following equations:
1-norm $‖x‖=|{x}_{0}|+|{x}_{1}|+\dots +|{x}_{n-1}|$ 2-norm $‖x‖=\surd \left({|{x}_{0}|}^{2}+{|{x}_{1}|}^{2}+\dots +{|{x}_{n-1}|}^{2}\right)$ Inf-norm $‖x‖={\mathrm{max}}_{i}\left(|{x}_{i}|\right)$ -Inf-norm $‖x‖={\mathrm{min}}_{i}\left(|{x}_{i}|\right)$ User Defined $‖x‖={\left({|{x}_{0}|}^{y}+{|{x}_{1}|}^{y}+\dots +{|{x}_{n-1}|}^{y}\right)}^{\frac{1}{y}}$
where
• x is input vector
• y is user defined norm
• ||x|| is norm
Default: 2-norm
## user defined norm
User-defined norm type.
This node uses user defined norm as the norm type only if you set norm type to User Defined. user defined norm must be nonzero.
Default: -1
## error in
Error conditions that occur before this node runs. The node responds to this input according to standard error behavior.
Default: No error
## unit vector
Output normalized vector.
## norm
Norm of the input vector.
## error out
Error information. The node produces this output according to standard error behavior.
## Algorithm for Calculating unit vector
This node calculates unit vector using the following equation:
$U=\frac{X}{‖X‖}$
where U is unit vector.
Where This Node Can Run:
Desktop OS: Windows
FPGA: Not supported
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# Terminology for expressing a graph as a sum of cliques (mod 2)
I am interested in the problem of expressing the edges of a given (undirected, simple) graph as the sum of edge sets of cliques modulo $$2$$.
To be more concrete, given a graph $$G=(V,E)$$, I am seeking to find cliques $$C_1=(V_1,E_1)$$, $$\dots$$, $$C_k=(V_1,E_k)$$ so that $$V_i\subseteq V$$ for all $$i$$, each edge set $$E_i$$ consists of all edges between pairs of vertices of $$V_i$$, and (most importantly), every edge $$e\in E$$ lies in an odd number of $$E_i$$ and every non-edge $$e\notin E$$ lies in an even number of $$E_i$$.
For example, in this sense we can express the claw as the sum (modulo $$2$$) of $$K_4$$ and $$K_3$$: (And of course any graph $$G=(V,E)$$ is the sum of $$|E|$$ copies of $$K_2$$, but many graphs have less trivial expressions as sums.)
Have you seen this sort of question asked in the literature? I have not been able to find any terminology for this question, or any literature on it, so I would appreciate almost anything MathOverflow users could share with me about it.
• I'm not sure if this is worthy of a full answer, but there is a similar question about expressing a graph in terms of the symmetric difference of complete bipartite graphs here: mathoverflow.net/questions/76043/… Dec 31, 2018 at 17:05
The closest notion to this that I have found in the literature is that of subgraph complementation, introduced by Kamiński, Lozin, and Milanič, in which the edges of a given induced subgraph of a graph $$G$$ are complemented. Then we may phrase your problem as building $$G$$ by taking successive subgraph complements, starting with the empty graph on $$V(G)$$.
In fact, your question inspired a research project amongst myself and the two others who responded to this post, the product of which is available here: arXiv:2101.06180.
We provide upper bounds on the minimum number of cliques in an expression of $$G$$ as you describe in terms of the number of vertices, the number of edges, and the size of a minimum vertex cover. We relate this problem to the minimum rank problem over the field of order $$2$$, enabling us in some cases to find the minimum size of such an expression. We also show that, similar to the minimum rank problem over $$\mathbb{F}_2$$, the class of graphs which may be expressed as a sum of $$k$$ cliques is hereditary and finitely defined for any positive integer $$k$$.
By considering an incidence matrix corresponding to a collection of cliques in an expression, we see that your problem is equivalent to that of finding a faithful orthogonal representation of a graph over $$\mathbb{F}_2$$; that is, an assignment of vectors over $$\mathbb{F}_2$$ to the vertices of $$G$$ so that two vectors are orthogonal if and only if they represent non-adjacent vertices. In the study of minimizing the dimension of such a representation, Lozin and Alekseev use an early variant of the subgraph complement (see the proof of Theorem 3).
I have never come across anything like this in the literature, but it is a fun question. It is reminiscent of the Lights-Out puzzle on general graphs.
Here is a greedy approach that gives $$\leq |V(G)|-1$$ cliques. Suppose we have some graph $$G$$ that we want to express as a sum of cliques.
Start with an empty graph $$H$$ with $$V(H)=V(G)$$. While $$H$$ is not isomorphic to $$G$$, choose a vertex $$v$$ that has the highest number of "incorrect" (non-)edges, i.e. the highest number of other vertices $$w$$ such that $$vw \in E(H)\triangle E(G)$$ (symmetric difference). Add a clique on the set of vertices $$\{v\} \cup \{ w |vw \in E(H)\triangle E(G)\}$$. Now all of $$v$$'s edges are "correct", and $$v$$ will never be chosen again to appear in a clique.
Note that this is not always optimal if the goal is to express as a sum of as few cliques as possible. Take, for example, the bowtie graph (two triangles that share a vertex). The above greedy algorithm uses 3 cliques, when it is easy to see that this graph is expressible as a sum of 2 cliques.
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# [OS X TeX] Re: List of feature requests on this list
Victor Ivrii vivrii at gmail.com
Mon Jul 7 19:58:27 EDT 2008
Well, the real problem is (at least for a mathematician) the line
Comparing equations (8.1), (7.9),(5.3) and (1.4),...
With AR or TS etc you click on the link (8.1), select page area
using corresponding tool:
rectangle tool in TS and Skim or snapshot tool in AR (looks like
photo camera; remember in AR you can customize toolbar), copy it (in
TS you press Cmd-C, in AR it is done automatically), paste it as
graphics in TextEdit (or any graphics editor) window, go to previous
view (thus returning to the page you are reading) and repeat the
procedure for each of equations. So you end-up with the TextEdit
window containing all these equations and can actually compare them.
Beside equations it could be theorems etc
This is not extremely convenient but (1) it is far more convenient
than for a printed paper where your snapshot tool is a pencil and a
paper (2) Adobe' people think that they already provided us with all
tools needed.
I definitely believe that just going back-and-forth without copy-paste
makes comparison rather superficial unless there are very few very
short equations. This would be true and for another window displaying
at any moment just one of the equations you want to compare.
Surely if you are the author you can always write: Comparing equations
(\ref{A}), (\ref{B}),...
$$............... \tag{\ref{A}}$$
$$............... \tag{\ref{b}}$$
but using this trick frequently would increase the number of pages
with grave consequences for printed article; even pdf article will be
overburdened.
Luckily Ross demonstrated a solution in a different mailing list
(works only in AR since only it understands Javascript). Hopefully he
will make a package.
Right now beside of asking him 'pretty-pretty-please" one can achieve
similar effect but not as nice looking (and actually requiring some
cumbersome work from the author) using fancytooltips package by Robert
Marik. What is worse: with Ross solution viewers different from AR
just follow the link, while with solution based on fancytooltips
package it does not work
While this definitely is not endemic to MacOSX, I think that asking
Ross for a really correct tool should be made in any occasion. I would
prefer, however, if (say) control-clicking on link would actually
follow it instead of pop-up (so the reader can chose between pop-uping
and going).
Victor
--
========================
Victor Ivrii, Professor, Department of Mathematics, University of Toronto
http://www.math.toronto.edu/ivrii
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My Math Forum Extremum search
Calculus Calculus Math Forum
March 5th, 2011, 01:51 PM #1 Newbie Joined: Mar 2011 Posts: 22 Thanks: 0 Extremum search Hello, I have the following exercise: $f(x,y)= x^2 - 2xy + 2y^2$ I want to search the extremum with help of a matrix. $\frac{d}{dx} = 2x - 4y \frac{d}{dy} = -2x + 4y$ So I get the following matrix: $\begin{array}{cc} \frac{d^2f}{d^2x} & \frac{d^2f}{dxdy}\\ \frac{d^2f}{dydx} & \frac{d^2f}{d^2y}\\ \end{array} = \begin{array}{cc} 2 & -2 \\ -2 & 4\\ \end{array}$ In this example is it a local minimum because of -2. But I dont understand this very well, can someone explain this to me. What I have to to is take the derivatives like above, fill in (0,0) and look at the numbers that come out, ok I got that BUT wy can you say it is a local minimum because of the -2? In another exercise I have (longer) I get the following matrix: $\begin{array}{cc} 2 & 0 \\ 0 & 2 \\ \end{array}$ Because of eigenvalue 2 we have a local minimum. If you want I will post the entire exercise but because of the length I didn't post it in the first place, just ask. Thank you
March 5th, 2011, 03:32 PM #2 Newbie Joined: Mar 2011 Posts: 22 Thanks: 0 Re: Extremum search Anyone an idea? Or didn't I explain my question good enough?
March 5th, 2011, 04:46 PM #3 Senior Member Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: Extremum search f(x,y) = x²-2xy+2y² Check your work. $\frac{\partial f}{\partial x}=2x-2y,\quad\frac{\partial f}{\partial y}=-2x+4y$ but the matrix is correct and the partial derivatives exist everywhere. For a critical point, you need to solve 2x-2y=0 and -2x+4y=0; these are both lines and they intersect at (0,0). The determinant of the matrix is (2)(4)-(-2)(-2) = 8-4 = 4, which is positive, therefore the fact that $\frac{\partial^2f}{\partial x^2}$ is positive shows that (0,0) is a minimum. Also, the reason you don't get an immediate reply is that this is a forum rather than a chatroom.
March 5th, 2011, 05:05 PM #4 Newbie Joined: Mar 2011 Posts: 22 Thanks: 0 Re: Extremum search The error you noticed was due to copying, Thank you for noticing. So what you are saying is: To check if it is a critical point, I have to solve: partials derivatives = 0 If they intersect at (0,0) check: If the determinant of the matrix is positive it is a minimum, else if the determinant is negative it is a positive. but what if the partials derivatives intersect at minus something, or plus something?
March 5th, 2011, 05:46 PM #5 Newbie Joined: Mar 2011 Posts: 22 Thanks: 0 Re: Extremum search Do you fill in the matrix the values found by solving the derivates = 0 if needed?
March 5th, 2011, 06:27 PM #6 Senior Member Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: Extremum search The function f needs to have partial derivatives everywhere in the region. Then, critical points are points where $\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}=0$. Second partial derivative test If the determinant is positive, then the critical point is a maximum (if $\frac{\partial^2 f}{\partial x^2}<0$) or a minimum (if $\frac{\partial^2 f}{\partial x^2}>0$). If the determinant is negative, the critical point is a saddle point. If the determinant is zero, this test fails.
March 5th, 2011, 06:41 PM #7 Newbie Joined: Mar 2011 Posts: 22 Thanks: 0 Re: Extremum search Ok, I understand it now. Thank you very much for explaining this
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## Tuesday, 2 April 2013
### Using Powershell to perform automatic Hg Bisect
The process of bisection in the source control system Mercurial or Hg is done with the command Hg Bisect. The Hg Bisect command is usually run manually and consists of manual steps done by the user where the developer running the hg bisect command looks for a certain condition should hold true and mark each revision as either good or bad. This is done in a binary search result manner, which quickly narrows down the number of revisions to look at by half for each iteration. The goal is to find the first revision containing a bad feature. An automatic function for finding such bad features will now be presented. Here is the Powershell cmdlet or function followed by an example of a call to this function:
function Get-HgChangeSet(
){
param(
[Parameter(Mandatory=$true)] [scriptblock]$test,
[Parameter()]
$good = 0, [Parameter()]$bad = 'tip'
)
hg bisect --reset;
hg bisect --bad $bad; (hg bisect --good$good) | out-null;
while ($output -notmatch 'The first bad revision is'){$result = & $test; if ($result){
$output = (hg bisect --good ) | out-string; } else {$output = (hg bisect --bad) | out-string;
}
}
$output; } Push-Location 'C:\toaurs-he\demorepo\' Get-HgChangeSet { !(Test-Path 'test.txt') -or (Get-Content test.txt) -notmatch 'ultrabad' } Pop-Location In this example, a simple demo repository has a file called test.txt. I want to find the first revision where the text ultrabad was inserted. The truth condition then, is that either the file does not exist (yet) in a revision, or that the file exists and does not match ultrabad. This is a simple example, but it shows how one can search for a given text or source code by specifying this in the script block passed to the function or cmdlet Get-HgChangeSet. When the script is run, it finds the first occurence of the text ultrabad (which is bad) at revision number 8. PS C:\toaurs-he\demorepo[ default ]> Hit Line breakpoint on 'C:\Users\Tore Aurstad\Documents\WindowsPowerShell\Scripts\Hg\BisectTool.ps1:22' PS C:\toaurs-he\demorepo[ default ]> The first bad revision is: changeset: 8:25be1d61e90d user: Tore Aurstad date: Tue Apr 02 22:09:15 2013 +0200 summary: foo 4 So there you have it, an automatic way via a Powershell script to find an introduction of a bad feature in a Hg repository with the aid of a Powershell function or cmdlet. To use this script, it is important to understand that the passed in first argument is a truth condition. It will usually always be necessary to accept if the file to test does not exist yet combined with and -or condition and then specify -notmatch and the text or source code which is not desired. This will usually be more precise than running hg bisect manually, which is kind of tedious still. Also note that the second and third parameter is set to$good equals 0 and $bad equals 'tip'. In Mercurial or Hg, 'tip' is the newest revision. Usually, for a repository with many revisions, it will be quicker if the values for$good and $bad is specified in a well-known range, say revision 5000 to 5500 or what have you.. This is just another example of how scripting can lighten our daily workload as developers. #### 2 comments: 1. Isn't that what "hg bisect -c$CMD" is for?
2. Hello Peter, I have not tried that command. If it is a one-liner command, obviously this Powershell script is unecessary. Thanks for the tip.
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# How to calculate density of states (DOS) from 8 energy eigenvalues of a Quantum model calculated by exact diagonalization?
• A
## Summary:
I have energy spectrum of a single sample. I want to calculate and plot DOS. Since DOS is dN/dE number of states available in unit interval of energy . How I can calculate this unit interval? If I plot histogram and select bins of equal size, it doesn't give me unit interval.Data = np.array([-1.61032636, -1.23577245, -0.50587484, -0.28348457,
-0.18748945, 0.4537447, 1.2338455, 2.13535718])
## Main Question or Discussion Point
Data = np.array([-1.61032636, -1.23577245, -0.50587484, -0.28348457, -0.18748945, 0.4537447, 1.2338455, 2.13535718])
print("Data is: ", Data)
print(Data.shape)
n,bins,patches = plt.hist(Data,bins=4)
print("n: ",n)
print("bins: ",bins)
plt.savefig("./DOS")
## Answers and Replies
Related Atomic and Condensed Matter News on Phys.org
ZapperZ
Staff Emeritus
Summary: I have energy spectrum of a single sample. I want to calculate and plot DOS. Since DOS is dN/dE number of states available in unit interval of energy . How I can calculate this unit interval? If I plot histogram and select bins of equal size, it doesn't give me unit interval.Data = np.array([-1.61032636, -1.23577245, -0.50587484, -0.28348457,
-0.18748945, 0.4537447, 1.2338455, 2.13535718])
Data = np.array([-1.61032636, -1.23577245, -0.50587484, -0.28348457, -0.18748945, 0.4537447, 1.2338455, 2.13535718])
print("Data is: ", Data)
print(Data.shape)
n,bins,patches = plt.hist(Data,bins=4)
print("n: ",n)
print("bins: ",bins)
plt.savefig("./DOS")
The phrase "energy spectrum" is vague. WHAT energy spectrum? Is this out of an experiment? If so, what TYPE of experiment.
For example, I could have an "energy spectrum" corresponding to emission levels. But this depends on how these excitations were obtained. You can't always use the intensity height to indicate the density of such a state, because it depends on the cross-section of absorption to excite that state, and you also do not have a clear calibration of the photon flux that was used to excite that state.
This is why we do not often calculate the DOS from such a spectrum unless we know a few other factors, something that you did not provide.
Zz.
The phrase "energy spectrum" is vague. WHAT energy spectrum? Is this out of an experiment? If so, what TYPE of experiment.
For example, I could have an "energy spectrum" corresponding to emission levels. But this depends on how these excitations were obtained. You can't always use the intensity height to indicate the density of such a state, because it depends on the cross-section of absorption to excite that state, and you also do not have a clear calibration of the photon flux that was used to excite that state.
This is why we do not often calculate the DOS from such a spectrum unless we know a few other factors, something that you did not provide.
Zz.
I have calculated the energy spectrum of a Quantum model using exact denationalization method using system size L=3 so there are 2^L means (8) energy levels. Now I want to plot DOS for these 8 levels.
I assume you have also determined the degeneracy of the energy levels. With that in hand it you should be ready to determine the DOS by the standard prescription.
There is no degeneracy among energy levels. If I have values [-1.5,-1.2,-0.5,-0.2,-0.1,0.5,1.0]. How we can calculate DOS and then plot it with standard prescription? DOS =dN/dE, where dN= N(E2)-N(E1) and dE= E2-E1 i.e, difference in the number of states the difference in the energy b/w states. If I plot histogram y-axis gives be counts(how many values in each bin) and x-axis is the bins. If the first bin has width 1.5 then I can say there are 3 engery eigenvalues in this bin. Then how to calculate dN= N(E2)-N(E1)? Moreover DOS has the units of 1/E so how to get these units in histogram?
Last edited:
If there is no degeneracy then clearly ##dN = 1##, but if you need more detail then here it is.
Your density of states is give by:
$$D(E_{n}) = \frac{dN}{dE}$$
where ##N(E_{n})## is the total number of states up to energy ##E_{n}##:
$$N(E_{n}) = \sum_{k=0}^n g_{k}$$
As you stated there is no level degeneracy so ##g_{k} = 1##
For a discreet setting determine ##dN## and ##dE##
$$dN = N(E_{n+1}) - N(E_{n}) = (n+1) - (n) = 1$$
I leave it to you to workout the pointwise determination of ##dE## from the 1x8 energy list you provided above.
member659127
That's what it should look like. I only showed 4, but you have 8. Ideally you have to put dirac delta functions at your energy values. However this is not very convenient since you can't plot them. Here I used very narrow triangles. If I were to code I could use gaussians or lorentzians, etc... that's what typical quantum chemistry software does actually (but they have hundreds of energy values so overlapping functions give some sort of different shapes.). Don't get yourself confused by the equations... This is what it should look like in the end, I never use histograms for such a purpose, a gaussian centered on energy value with a pre-determined arbitrary width of your choice does the job.
I have tried to calculate DOS which seem to be incorrect. I used formula DOS=dN/dE, where dE is simply energy difference b/w consecutive eigenvalue, So I simply put 1/dE, So I got 7 values where as originally they were 8. where x-axis is simply energy level. I'm dealing with 1D spin chain. Most of the explanation about DOS is for particle moving in 1D box. I'm really confused about this. Can someone please tell me where I'm wrong and what should I correct.
with histogram I can calculate DOS. we select some interval and then count states in that interval. But I want to plot graph b/w E vs DOS not using histogram
In my last post I said you needed to determine the interval ##dE## that matches ##dN=1##. If you are sitting at an energy interval in between two levels it should be abundantly clear that ##dN=0##. So for ##dE<(E_{n+1} - E_{n-1})##, would you agree with the following statement?
$$dN = \begin{cases} 1 & \text{if the interval } dE \text{ includes } E_{n} \\ 0 & \text{otherwise} \end{cases}$$
This holds even as ##dE \rightarrow 0##. What function matches this? If you read erbahar's message he told you:
Ideally you have to put dirac delta functions at your energy values.
So the functional representation of ##D(E)## should look like this:
$$f(x) = \alpha\delta(x-x_0)$$
What do you think ##\alpha## and ##x_0## should be? Hint: one is related to the degeneracy (which I alluded to in my first post) and one is related to the positioning of the energy levels (as I alluded to in my second post.)
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# michal-h21/mathdimen
Vertical positioning and scaling of (not only) math images with tex4ht
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# Mathdimen
With tex4ht, it is possible to convert LaTeX files to various output formats. Certain elements with difficult formating, typically math content, can be converted to images. Unfortunately, when these images are used in the running text, they can have wrong vertical placement or wrong size.
This package enables to precisely place these converted images, in terms of the vertical alignment and correct setting of the dimensions.
## Usage
There are four steps needed for correct use of this package
## ToDo
• add more lowlevel details into the documentation
• do more testing
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Algebraic & Geometric Topology
Bordism groups of solutions to differential relations
Abstract
In terms of category theory, the Gromov homotopy principle for a set valued functor $F$ asserts that the functor $F$ can be induced from a homotopy functor. Similarly, we say that the bordism principle for an abelian group valued functor $F$ holds if the functor $F$ can be induced from a (co)homology functor.
We examine the bordism principle in the case of functors given by (co)bordism groups of maps with prescribed singularities. Our main result implies that if a family $J$ of prescribed singularity types satisfies certain mild conditions, then there exists an infinite loop space $Ω∞BJ$ such that for each smooth manifold $W$ the cobordism group of maps into $W$ with only $J$–singularities is isomorphic to the group of homotopy classes of maps $[W,Ω∞BJ]$. The spaces $Ω∞BJ$ are relatively simple, which makes explicit computations possible even in the case where the dimension of the source manifold is bigger than the dimension of the target manifold.
Article information
Source
Algebr. Geom. Topol., Volume 9, Number 4 (2009), 2311-2347.
Dates
Revised: 18 May 2009
Accepted: 19 May 2009
First available in Project Euclid: 20 December 2017
https://projecteuclid.org/euclid.agt/1513797085
Digital Object Identifier
doi:10.2140/agt.2009.9.2311
Mathematical Reviews number (MathSciNet)
MR2558312
Zentralblatt MATH identifier
1179.57044
Citation
Sadykov, Rustam. Bordism groups of solutions to differential relations. Algebr. Geom. Topol. 9 (2009), no. 4, 2311--2347. doi:10.2140/agt.2009.9.2311. https://projecteuclid.org/euclid.agt/1513797085
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# self-reproduction
This topic is 4258 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
If you intended to correct an error in the post then please contact us.
## Recommended Posts
I made a program where there's a bunch of flying orbs linked with lines to other orbs: http://i90.photobucket.com/albums/k276/krims2006/orbs.gif http://i90.photobucket.com/albums/k276/krims2006/orbs.gif http://i90.photobucket.com/albums/k276/krims2006/orbs.gif What I'm going to do with these orbs is to make them into a 3-dimensional representation of an algorithm. I will assign a certain function to each orb and its relation to other functions will be determined by its links. For example, here's a simple algorithm represented in C++:
a = 3 + 4;
In my 3d system I will break each of these pieces into orbs and link them in a tree form:
(a)
|
(+)
/ \
(3) (4)
Once I've done this, I want to create a 3d algorithm that reproduces itself, so that the end result is like a biological cell reproducing itself. The problem is that I can't think of a way for how to implement this self-reproduction. I could make a function-orb that takes all the orbs nearby and reproduces them exactly, but that ruins the whole point of what I'm trying to do. And what I'm trying to do is to create a 3d evolving system where even the means by which the "organisms" reproduce can be mutated and evolved. So what I need is to break down self-reproduction into little steps that I will represent with orbs. I could store all the information on how to reconstruct an organism in some orbs, but then that is not really self-reproduction but having a really complex organism make smaller organisms... Is that understandable? It's very abstract, but I happen to have become interested in this sort of thing because I have a lot of free time to waste. :D So any ideas on how I can use very simple rules to create a self-reproducing organism in this 3d system?
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Here is a way that I've thought about how to do the self-reproduction:
There will be a section of the organism called the "reader" that will traverse through all the orbs in the organism, and send a command to a "copier" section to copy each orb with a link attached to the original (like an umbilical cord). The next phase is to copy all the links in the copied orbs the same way as they are linked in the original organism. The final step is to cut the "umbilical cords" and animate the new organism.
The problem with this is that it is too complicated. I tried for a long time to try to do a schematic of this organism on paper, but it was just mind-boggling trying to figure how to put it on paper.
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Reminds me of L-systems and Conway's Game of Life. Perhaps those can give you some inspiration. Both have a simple ruleset and especially the latter can result in very interesting situations.
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I'm a little confused about what exactly your problem is.
Are you asking:
"how do I create a copy of an arbitrary graph (collection of nodes and vertices)"; or,
"how do I make new functions from old functions that are still valid functions (given your constraints on your initial function space)"; or,
"what kinds of functions are functions of themselves"?
I'll try rereading your post a few more times to see if it helps... but perhaps you could give us a little more detail about the specific problem you have.
Cheers,
Timkin
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You are evolving a genetic algorithm.
Quote:
The problem is that I can't think of a way for how to implement this self-reproduction. I could make a function-orb that takes all the orbs nearby and reproduces them exactly, but that ruins the whole point of what I'm trying to do. And what I'm trying to do is to create a 3d evolving system where even the means by which the "organisms" reproduce can be mutated and evolved.
You could specify a fitness function with orbs, then your orb tree will mutate until one of acceptable fitness is created.
Using this method you could mutate the mutators and the mutator mutator mutators, etc :)
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I'm not sure if I agree with the 3D space and orb methodology, but it is an interesting idea. I think you'll end up spending as much or more time managing the 3D simulation and lose computational power on the real question at hand. I'd be interested in the results, regardless.
You may get some good information from the following papers.
http://www.sq3.org.uk/papers/evselfreps.pdf
Hutton, T. J. (2002). "Evolvable self-replicating molecules in an artificial chemistry." Artificial Life 8: 341-356.
http://citeseer.comp.nus.edu.sg/cache/papers/cs/27414/http:zSzzSzcogprints.ecs.soton.ac.ukzSzarchivezSz00002888zSz01zSzNRC-44969.pdf/smith03selfreplicating.pdf
http://johnnyvon.sourceforge.net/applet/2.0/
Smith, A., P. Turney, et al. (2003). "Self-replicating machines in continuous space with virtual physics." Artif Life 9(1): 21-40.
(Very good paper showing that with the proper enabling environment and rules, you can get self-replication. The applet lets you see it in action.)
http://www.nature.com/nature/journal/v437/n7059/abs/437636a.html
Griffith, S., D. Goldwater, et al. (2005). "Robotics: self-replication from random parts." Nature 437(7059): 636. (Very similar to the work by Hutton, but these folks take it into the real world. Here, instead of simulating the process in a virtual world, they've done it with small robots on an air table. Excellent video under Supplementary info.)
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Thanks for the links, I'll take a look later.
Here is what is the problem I'm having: I don't know how to make a self-reproducing algorithm in my 3d system.
Before I know what functions I will assign to orbs, I have to know how it will reproduce itself, and then I will figure out what functions I need.
This self-reproducing algorithm has to read itself in 3D SPACE and reconstruct the exact same structure of orbs and links. So try to imagine this - it is an algorithm, represented in 3d network form, that is reproducing a copy of itself, while morphing and shape-shifting because of the things that it's doing.
That's a very complicated task... the concept is actually harder than how DNA reproduces itself because DNA holds information in an essentially straight line (even though it is in 3d space), meanwhile my 3d system is not even a tree but a complicated mesh of orbs and links.
What is the purpose of all this? I am not trying to find a solution to any problem, so I have no criteria for a fitness-evaluator. This project is purely for entertainment and thought-stimulation. I am creating an open-ended evolutionary system. The organisms will reproduce through their own means, and there will not be any system that selects which ones are the most fit, like in real life. Instead the selection of the fittest will emerge itself in the system. There will be a large but limited number of orbs in the system, and as more orbs are being requested through reproduction, older orbs will be removed. What will the results of this? I predict that at first evolution will favour those algorithms that are simpler and reproduce the fastest. And then I predict that there will emerge an algorithm that is more complicated than the first but will be faster at reproducing because it will be construct several copies of itself at once. And then I predict there will be a very simple, parasitic algorithm that will invade other algorithms (in 3d space) "rewire" them in a way so that the host algorithm is tricked into reading and reproducing the parasite algorithm instead of itself. But that's just fantasizing, and the real results will probably be much different from what I've predicted.
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I just got an idea from DNA - maybe a part of the algorithm will break off from the rest of the organism, and then copy this chunk that it's broken off of?
Mmmm... but then how do I get it to reproduce this piece that's broken off...
I need ANY working idea for how to make a self-reproducing 3d-mesh/algorithm/organism. Once it's working and reproducing, I can introduce random mutations into the system, and eventually it would create other ways of reproducing.... But I need a complete and simple idea. I've been racking my brain trying to figure this out.
I could have just made a text-based evolutionary system based on Core Wars where the algorithms are stored as a sequence of commands, but I don't want to because I want to be able to point to a part of the screen and say "Oh, that's a parasite, and its rewiring that organism there, and that organism over there looks like a mutation of the XZ-Alpha-3 type because it has that unique triangle-shaped array of orbs"...
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Take a look at the JohnnyVon demo. You might find some inspiration there.
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Hmm... this is just an idea that popped into a tired mind, probably inspired by the X-like shape of chromosomes, but why not create a double function, each half of which copies the other half as part of its functionality?
For that matter, why nit start with an algorithm that does nothing but reproduce itself with mutation, and let that evolve - you may find that other functions arise as it evolves.
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# UGC-NET | UGC NET CS 2015 Dec – III | Question 73
Consider a language A defined over the alphabet $\sum$ = {0, 1} as The expression &space;=&space;0&space;\right&space;\}” title=”A = \left \{ 0^{\left \lfloor n / 2 \right \rfloor}1^{n}: n> = 0 \right \}” /> The expression $\left&space;\lfloor&space;n&space;/&space;2&space;\right&space;\rfloor$ means the floor of n/2, or what you get by rounding n/2 down to the nearest integer.
Which of the following is not an example of a string in A?
(A) 011
(B) 0111
(C) 0011
(D) 001111
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# Real world example of non-interactive zero knowledge proofs?
I am very interested in finding out more about the non-interactive zero knowledge proofs after reading the simplified and abstract examples about zero knowledge proofs themselves here.
For the NI-ZK, most answers over here are either very technical or on a very generic level that i have a hard time understanding it.
I do understand the concept for using NI-ZK with a common shared string (CRS): Instead of having a challenge from the Verifier, the Prover simulates it. But the CRS is completely known to the Prover and he can look at the consequence of possible commitments, refraining from them if he may be wrong, can't he? If i use the example on wikipedia for ZK proofs:
In this story, Peggy has uncovered the secret word used to open a magic door in a cave. The cave is shaped like a ring, with the entrance on one side and the magic door blocking the opposite side. Victor wants to know whether Peggy knows the secret word; but Peggy, being a very private person, does not want to reveal her knowledge (the secret word) to Victor or to reveal the fact of her knowledge to the world in general.
They label the left and right paths from the entrance A and B. First, Victor waits outside the cave as Peggy goes in. Peggy takes either path A or B; Victor is not allowed to see which path she takes. Then, Victor enters the cave and shouts the name of the path he wants her to use to return, either A or B, chosen at random. Providing she really does know the magic word, this is easy: she opens the door, if necessary, and returns along the desired path.
Now if we replace Victor with a CRS, Peggy would f.e. enter at entry A and look at a position that is the equivalent of her decision (or where would she look?) whether she should leave the cave at A or B, in this case B. But why doesnt she just look it up beforehand? She has the CRS and she knows where she would have to look at if she took either sides - she can just see that if she goes to A first, she is supposed to come out at B. So instead of really committing to A, she secretly sneaks into B and gets out there without having to pass the magic gate.
The argument about Peggy being able to start over is not about NI-ZK proofs, it is about zero knowledge in general.
Zero knowledge does not give you a definite certainty that Peggy actally has the claimed knowledge. In your example, the probability of Peggy being able to cheat is $1/2$ after one round. If we do this for two rounds, the probability that she cheated twice succssfully is $1/4$. This goes on, until there is a satisfying probability, e.g. that Peggy could only cheat with probability of $1/2^{80}$ for 80 rounds.
Now, if we require in a NI-ZK proof that Peggy created a transcript with 80 rounds, she would have to start over for $2^{80}$ times to find such a transcript, if she doesn't have the ability to go back just one round. This has to be enforced, which is done in the Fiat Shamir heuristic by hashing all the previous results. In the CRS model this is achieved by defining the reference string only with a certain distribution and granting no one authority over this choice. This assumption implies, that Peggy can not just go back one step. If she starts over from the beginning, the entire string would change.
From a practical point of view, this is a challenge. The most commonly suggested practical solution would be a trusted third part, like for certificate authorities. But the effort of creating and publishing common reference strings is quite a lot more complex than building a PKI.
For some interesting arguments about this, and a possibly different realization I would suggest reading Cryptography in the Multi-string Model by Groth and Ostrovsky. Especially the introduction is quite interesting and their proposed new setup assumption.
Considering most usages of a CRS in the sense of NI-ZK is that there is usually some property, which is then used in order to prove the ZK property. In regular ZK you give the simulator some kind of advantage, e.g. being able to rewind one step. In this model you use some kind of commitments, which are binding in the real protocol but the advantage of the simulator is that he can decomit his previous commitsments to anything.
The prover can "look at the ... be wrong".
If she doesn't "just look it up beforehand", then that's because
the challenge-function was chosen in some extremely clever way,
and I'm not aware of any reason why such a way should exist.
(Alternatively, they're just after arguments, rather than proofs.)
There are two known ways of constructing NIZK Proofs:
One plugs XB-enhanced trapdoor approximate-bijections
into this paper and these two lecture-note pdfs.
The other uses a commitment scheme that will be binding for
real setup strings and equivocable for simulation setup strings.
I'm not aware of any "real world example"s for understanding the concept.
• The prover can "look at the ... be wrong" Whats that supposed to mean? then that's because the challenge-function was chosen in some extremely clever way thats exactly the question. One plugs XB-enhanced trapdoor approximate-bijections into this [.. to end] is quite technical and not anywhere close to a real world example, for my understanding. – James Cameron Sep 8 '16 at 6:15
• See your question: That's supposed to mean The prover can "look at the consequence of possible commitments, refraining from them if he may be wrong".. I don't see anything like then that's because the challenge-function was chosen in some extremely clever way in your question. – user991 Sep 8 '16 at 6:22
• No but the core of the question was why Peggy cant cheat. You do say that you arent aware of such ways, but then how do CRS is used instead? There we get to the explanations below that are too techy. – James Cameron Sep 8 '16 at 6:35
• For one approach, the CRS+[almost-bijection] yield blocks so that, except with exponentially small probability, enough blocks encode directed cycle graphs, for Hamiltonian cycle. For the other approach, the CRS is the public-key for a commitment scheme that allows one to prove relations between committed values, such that for real CRSes the commitment is binding and for simulation CRSes the commitment is equivocable. – user991 Sep 8 '16 at 7:12
Any standard Digital Signature algorithm is basically a Zero-Knowledge Proof of Knowledge of the Private Key corresponding to the commonly known Public Key, where the Verifier challenge is derived in part from the message to be signed.
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My Math Forum Verifying 3 Hypotheses of Rolle's Theorem
Calculus Calculus Math Forum
January 7th, 2017, 07:59 AM #1 Newbie Joined: Jan 2017 From: Texas Posts: 1 Thanks: 0 Verifying 3 Hypotheses of Rolle's Theorem Need help on this problem- Verify that the function satisfies the three hypotheses of Rolle's Theorem on the given interval. Then find all numbers c that satisfy the conclusion of Rolle's Theorem. f(x) = cos2x, [pi/8, 7pi/8]
January 7th, 2017, 08:16 AM #2 Math Team Joined: Jul 2011 From: Texas Posts: 2,097 Thanks: 972 start by stating the three hypotheses of Rolle's Theorem (a special case of the MVT) ...
January 7th, 2017, 01:37 PM #3 Math Team Joined: Jan 2015 From: Alabama Posts: 2,210 Thanks: 555 Okay, what are those "three hypotheses"? What, precisely, does "Rolle's theorem say? Thanks from ProofOfALifetime
January 13th, 2017, 11:51 PM #4 Newbie Joined: Jan 2017 From: Nigeria Posts: 5 Thanks: 0 Math Focus: Real Analysis and Fluid Mechanics This is the Mean Value Theorem analytic proof, it will help you figure out 3 hypothesis of the Rolles Theorem, since the Mean Value Theorem is connected to the Rolles Theorem https://mymathware.blogspot.nl/2017/...proof.html?m=1
January 13th, 2017, 11:53 PM #5 Newbie Joined: Jan 2017 From: Nigeria Posts: 5 Thanks: 0 Math Focus: Real Analysis and Fluid Mechanics Rolle's theorem essentially states that any real-valued differentiable function that attains equal values at two distinct points must have a stationary or critical point somewhere between them—that is, a point where the first derivative (the slope of the tangent line) is equal to zero. https://mymathware.blogspot.nl/2017/...eorem.html?m=1
January 14th, 2017, 04:42 AM #6 Math Team Joined: Jan 2015 From: Alabama Posts: 2,210 Thanks: 555 So, again, what are the "three hypotheses"? Thanks from skeeter
Tags hypotheses, rolle, theorem, verifying
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[Template] Function dependencies?
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I have an octree template. It stores pointers to spatial objects in the tree. However once I go with templates I no longer need to specify the base interface class that enforces the types shape. For example I call Node->GetSphere(); in my code (Where Node is the type). Is this sloppy or the way it''s meant to be? I mean the compiler _does_ pick it up if I don''t define needed functions, not the the message is 100% intuitive, but I am getting use to it. Thoughts? Chris Brodie http:\\fourth.flipcode.com
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I think I know where you are coming from, ...ish.
1. If a templated function is never used then it will not be compiled into the code.
2. If you fail to impliment a function in the class that gets templated then it will cause an error, however there are a couple of ways to check at compile time that this is true... ie:
The microsoft way.... (MSVC specific)
template< class T > struct foo{ void Func() { __if_exists(Function) { Function(); } __if_not_exists(Function) { cout << "no func defined" << endl; } }};struct bar{ void Function();};
The above is quite funky but you still have no way to guarentee that the function matches the correct type required. In this case you can use some compile time assertions.....
// declare a template function (though it never gets called, the// params and return types are evaluated at compile time)template <class T> char ERROR_Types_Are_Not_The_Same( T );// a little macro to take the type, the variable itself and an// error message if the variable doesn't match the type...#define CTASSERT_TYPE( _TYPE_, _VAR_, _MSG_ ) { typedef _TYPE_ _MSG_##_##_VAR_; \ (void)sizeof( ERROR_Types_Are_Not_The_Same < _MSG_##_##_VAR_ >( _VAR_ ) ); }// our foo struct can now do something like this to ensure that the Function type matches the type we are expecting....template< class T > struct foo{ void Func() { __if_exists(Function) { // make a little typedef for the function pointer type typedef void (*Func)(void); // a compile time check that incurs zero overhead and compares the function pointers // to ensure they are compatible... CTASSERT_TYPE(Func,Function,Function_Is_Not_Of_Correct_Type); Function(); } }};
Basically this works on the principle that everything inside of the sizeof() operator is only checked at compile time. This essentually means that it all gets compiled out of the code and therefore incurs zero overhead. You can do this a bit more elegantly if you have access to partial template specialisation. I'd advise buying the book Modern C++ Design by Andrei Alexandrescu: ISBN 0-201-70431-5 it is by far the best book available on the subject and explains the ideas a lot better than I can.
Hope that's of some use....
[edited by - RobTheBloke on August 27, 2002 10:22:42 AM]
[edited by - RobTheBloke on August 27, 2002 10:23:36 AM]
[edited by - RobTheBloke on August 27, 2002 10:24:28 AM]
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check the back slashes on the macro the source tags just seem to remove them....
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Pokemon
Show off your games, demos and other (playable) creations.
Party member
Posts: 100
Joined: Mon Dec 12, 2011 8:44 pm
Pokemon
Ok so what I decided to make was a very simple version of the Pokemon battle system. There's no AI at the moment, and there is certainly a lot to be implemented, but I think it's a good start. Enjoy!
And excuse inappropriate names.
Last edited by jradich on Fri May 11, 2012 11:05 am, edited 1 time in total.
Losing a friend's trust is the fastest way to lose a friend, forever. FOREVER!
Roland_Yonaba
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Posts: 1563
Joined: Tue Jun 21, 2011 6:08 pm
Contact:
Re: Pokemon
This is definitely not a good start.
I looked at it, and found many code repetitions. That is not how classes work.
Besides, you are using a fairly simple class system.Then take full advantage of it.
Note that yours classes (Twattle and Vagikarp) have exactly the same methods, but you repeated them. I expect when you'll next have to add new species, you will rewrite down the same stuff, and it will result in a extremely long source code.
Why not move all methods (draw, fight, level up, faint) into the basePoke class ? And then instanciate Vagikarp and Twattle ?
Code: Select all
class "BasePoke" {
hp=100, name="Base",user=true,enemy=false,level=1,attacking=false,attackname='fap'
}
function BasePoke:__init(hp,name,user,enemy,level,attacking,attackname)--Sets how you would name a custom pokemon
--etc
end
function BasePoke:draw()
--etc
end
function BasePoke:fight()
--etc
end
function BasePoke:levelup()
--etc
end
function BasePoke:faint()
--etc
end
Vagikarp=BasePoke:new(100,"Vagikarp",true,false,1,false,'fap')
Twattle=BasePoke:new(100,"Twattle",false,true,1,false,'fap')
And that's it. Then you should be able to use Twattle:draw(), Vagikarp:levelup(), etc...the same way than before.
Don't forget to consider that inside methods, you can call other methods, just use self.
Example:
Code: Select all
function BasePoke:fight()
if fight==true then
--etc etc
end
self:levelup()
--etc etc
end
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Posts: 100
Joined: Mon Dec 12, 2011 8:44 pm
Re: Pokemon
Roland_Yonaba wrote:This is definitely not a good start.
I looked at it, and found many code repetitions. That is not how classes work.
Besides, you are using a fairly simple class system.Then take full advantage of it.
Note that yours classes (Twattle and Vagikarp) have exactly the same methods, but you repeated them. I expect when you'll next have to add new species, you will rewrite down the same stuff, and it will result in a extremely long source code.
Why not move all methods (draw, fight, level up, faint) into the basePoke class ? And then instanciate Vagikarp and Twattle ?
Code: Select all
class "BasePoke" {
hp=100, name="Base",user=true,enemy=false,level=1,attacking=false,attackname='fap'
}
function BasePoke:__init(hp,name,user,enemy,level,attacking,attackname)--Sets how you would name a custom pokemon
--etc
end
function BasePoke:draw()
--etc
end
function BasePoke:fight()
--etc
end
function BasePoke:levelup()
--etc
end
function BasePoke:faint()
--etc
end
Vagikarp=BasePoke:new(100,"Vagikarp",true,false,1,false,'fap')
Twattle=BasePoke:new(100,"Twattle",false,true,1,false,'fap')
And that's it. Then you should be able to use Twattle:draw(), Vagikarp:levelup(), etc...the same way than before.
Don't forget to consider that inside methods, you can call other methods, just use self.
Example:
Code: Select all
function BasePoke:fight()
if fight==true then
--etc etc
end
self:levelup()
--etc etc
end
Thanks
Changed it.
love pokemon.love
Losing a friend's trust is the fastest way to lose a friend, forever. FOREVER!
rokit boy
Party member
Posts: 198
Joined: Wed Jan 18, 2012 7:40 pm
Re: Pokemon
Roland_Yonaba wrote:This is definitely not a good start.
I looked at it, and found many code repetitions. That is not how classes work.
Besides, you are using a fairly simple class system.Then take full advantage of it.
Note that yours classes (Twattle and Vagikarp) have exactly the same methods, but you repeated them. I expect when you'll next have to add new species, you will rewrite down the same stuff, and it will result in a extremely long source code.
Why not move all methods (draw, fight, level up, faint) into the basePoke class ? And then instanciate Vagikarp and Twattle ?
Code: Select all
class "BasePoke" {
hp=100, name="Base",user=true,enemy=false,level=1,attacking=false,attackname='fap'
}
function BasePoke:__init(hp,name,user,enemy,level,attacking,attackname)--Sets how you would name a custom pokemon
--etc
end
function BasePoke:draw()
--etc
end
function BasePoke:fight()
--etc
end
function BasePoke:levelup()
--etc
end
function BasePoke:faint()
--etc
end
Vagikarp=BasePoke:new(100,"Vagikarp",true,false,1,false,'fap')
Twattle=BasePoke:new(100,"Twattle",false,true,1,false,'fap')
And that's it. Then you should be able to use Twattle:draw(), Vagikarp:levelup(), etc...the same way than before.
Don't forget to consider that inside methods, you can call other methods, just use self.
Example:
Code: Select all
function BasePoke:fight()
if fight==true then
--etc etc
end
self:levelup()
--etc etc
end
Thanks
Changed it.
love pokemon.love
u wot m8
Party member
Posts: 100
Joined: Mon Dec 12, 2011 8:44 pm
Re: Pokemon
rokit boy wrote: addthat to SYS post.
Losing a friend's trust is the fastest way to lose a friend, forever. FOREVER!
misstr87
Prole
Posts: 1
Joined: Mon May 14, 2012 11:46 am
Re: Pokemon
Some good basic advice there for beginners.
Thanks.
And hope the pokemon project continues
Party member
Posts: 100
Joined: Mon Dec 12, 2011 8:44 pm
Re: Pokemon
love pokemon.zip
New Update. Pictures are now possible, and there are multiple attacks.
Losing a friend's trust is the fastest way to lose a friend, forever. FOREVER!
Roland_Yonaba
Inner party member
Posts: 1563
Joined: Tue Jun 21, 2011 6:08 pm
Contact:
Re: Pokemon
I haven't run it yet, I just took a look at the code. Looks nicer than before.
Party member
Posts: 100
Joined: Mon Dec 12, 2011 8:44 pm
Re: Pokemon
Roland_Yonaba wrote:I haven't run it yet, I just took a look at the code. Looks nicer than before.
Thanks.
Oh and also, making pokemon is incredibly easy. Just write:
Code: Select all
YourPoke=BasePoke:new(hp,name,user,enemy,level,attackname1,attackname2,pic)
And call "YourPoke:fight()" and "YourPoke:draw()" in the proper functions. Remember, though, there are only two attacks, though new attacks are easy, too.
Losing a friend's trust is the fastest way to lose a friend, forever. FOREVER!
Party member
Posts: 100
Joined: Mon Dec 12, 2011 8:44 pm
Re: Pokemon
love pokemon.love
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# How accurate is pythagorean theorem?
Tags:
1. Jun 6, 2015
### christian0710
I'm just wondering: How accurate is the pythagaroen theorem? To what exact decimal point is the calculated length of the
hypotenus in a triangle 100% corect when using pythagarean theorem? It must not be 100% accurate to an infinite descimal point since it's called a "Theorem"
2. Jun 6, 2015
### Mentallic
That isn't what theorems in Mathematics mean. Pythagoras' theorem is perfectly accurate. In fact, the equality sign = is perfect. If Pythagoras' theorem was "nearly" perfect but not quite, then it wouldn't be using an equality, it would instead have the approximate sign $a^2+b^2\approx c^2$
The only thing that is imperfect are our decimal approximations to irrational numbers, hence why for a right triangle with short legs of length 1, the length of the hypotenuse is
$$\sqrt{1^2+1^2}$$
$$=\sqrt{2}$$
$$=1.41421356...$$
$$\approx 1.41421356$$
$$\approx 1.414$$
Note where the equality is used, and where the approximation is used.
3. Jun 6, 2015
### chingel
When the space is Euclidean (is not curved), the lines are perfectly straight and the angle is a perfect right angle, the theorem is exact. There is no 'rounding' in deriving the theorem, only the assumptions have to hold (Euclidean space, straight lines, right angle, maybe something more I'm missing).
In real life, you have all kinds of issues, space is curved, the angle is a little off, you can't measure with infinite accuracy (uncertainty principle etc).
But for usual applications, these effects can be ignored because they have a smaller effect than your measurement device can measure, i.e. you want to calculate the diagonal of your table after measuring the sides.
4. Jun 6, 2015
### christian0710
Thank you very much for the explanation! So it's the square root of numbers like 2,3,5 that's that gives us an infinite number of descimal points which is not defined,but roundede up on a calculator?
5. Jun 6, 2015
### Mentallic
Yes. $\sqrt{n}$ is precisely the number that when squared, gives us n. It just turns out that to express most of these numbers in decimal form, you need an infinite number of non-repeating digits.
Not defined is not the best word to be using here, but you seem to have the gist of it.
Also, calculators are a bit more interesting in their approximations than the simple rule of "rounding up" which isn't necessarily correct either. As reference, I'll post $\pi$ here to 15 decimal places and see what arises from it.
$$\pi = 3.141592653589793...$$
Now, my standard scientific calculator which can display 10 digits at a time shows $\pi$ as
$$\pi \approx 3.141592654$$
Which if we compare to earlier, we can see is rounding the value up. But we can delve further into this. If I subtract 3 from the value to get 0.141592654 and then multiply by 10, I'm essentially shifting the digits up and uncovering another digit that was previously hidden. Try it on your calculator, and keep repeating it to see how much the calculator actually holds and what happens at the end of its storage compared to the actual value of $\pi$.
6. Jun 6, 2015
### christian0710
My ti89 displayet 3.14159 however if i copy and paste this number it shows the total amount of decimals which is 3.1415926535898 if I subtract 3.141592 and multiply with 10^6 then it does not recover more decimal points, so I guess the copy paste displays the maximum amount of decimals.
7. Jun 6, 2015
### Mentallic
Alright so assuming your calculator actually calculates to 13 decimal places (dp) for all values and isn't just using a stored value of pi and a other hard-coded constants, I searched for an irrational that has 0-4 on the 14th dp such that a 13dp display would round down instead. Ironically, I actually had to search through all of the square root irrationals until $\sqrt{14}$ finally had the property I was looking for. Try that and see what your calculator displays for this one. For reference:
$$\sqrt{14} = 3.74165738677394...$$
The 14th dp is the 4 at the end.
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# Regime Switching / Change in Structural Parameter in time
Hi community,
in one model i am trying to put into dynare, i would like to simulate that a structural exogeneous Parameter is changing after 10 periods (=from period 11 on)
In other words, i would like to set the exogeneous interest rate during the first 10 periods at a Level of 1.01 (=1%) and from period 11 on until period 100 to Level 1.04 (=4%)
Could anyone support me with some hints on how to do that ?
Many thanks in Advance and have a nice week
Simon
Hi, you need to provide more context. Is the change permanent or temporary. Do people know/take into account that the parameter can switch or is it a once and for all change? Is the model stochastic or perfect forsight?
Thank you for your answer jpfeifer,
According to your Questions:
1. the “Change” Looks as follows: from period 0 till period 10, the structural exogeneous Parameter (here: interest rate) is supposed to be “1.01”, and from period 11 till 100 (end of the Simulation) the same structural exogeneous Parameter is supposed to take the value 1.04. In other words, the structural exogeneous Parameter is temporary low (for the first 10 periods)
2. yes, People know that the Parameter can switch
3. the model is perfect foresight
Many many thanks in advance and kind regards
Simon
Dear Community,
did someone have the Chance to think About my Question?
Many thanks in Advance & Kind regards
Simon
There is some tension in your answer. If you do perfect foresight, then agents will know exactly when the parameter will switch. This switch will be perfectly anticipated. Is that what you want?
In any case, it sounds like a straightforward transition exercise. Use `initval` and `endval` to set the initial and terminal steady state and a `shocks`-block to define the parameter change.
Many thanks for your answer jpfeifer,
At the Moment, i am on a Seminar that will take two more weeks.
after that, I will incorporate your Suggestion.
Once again many thanks
Simon
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# Open-sourcing My Dotfiles
2016-12-04*
programming
# Newsflash!
I just open-sourced my dotfiles! The repo is probably one of the older Git repos out there and has over 2000 commits! The first commit was made in March 10, 2009, over 7 years ago. Mind you, even by that time I had accumulated a whole suite of configs, and I even had an exotic setup with XMonad 0.8.1!
Here are some takeaways:
## Maintain clean, meaningful commit history
It’s nice to have clean commit history — every line of code can answer why it exists. You are doing yourself a disservice (in the long run) if you have overly terse commit messages. Generally speaking, be as verbose as you can be, but within reason. Keeping the first line of the commit message down to less than 80 characters goes a long way in setting the tone for any auxiliary paragraphs that follow.
I am quite proud of the richness of my commit messages. Pretty much everything makes sense and I don’t have to use Google to reason about my choices.
## Keep your build system simple
I’ve used a stupidly simple build system of creating symlinks to “install” my configuration — all with the help of a single Makefile. It’s not very flexible, and to be honest my Shell sklls are much better than what they used to be such that I could replace this setup with a script. But alas, the need for such a change has not been serious enough to warrant it.
Moreover, having a simple “build” system ensures robustness; the more I get older, the more I value systems that have a long “bit-rot halflife”. I admire Knuth’s TEX system for this very same reason. And this is the same reason why I will probably not use anything higher-level than a shell script for the job.
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# Graphing reciprocals of linear functions
#### Everything You Need in One Place
Homework problems? Exam preparation? Trying to grasp a concept or just brushing up the basics? Our extensive help & practice library have got you covered.
#### Learn and Practice With Ease
Our proven video lessons ease you through problems quickly, and you get tonnes of friendly practice on questions that trip students up on tests and finals.
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##### Intros
###### Lessons
1. Graph $f(x)= \frac{1}{x}$
##### Examples
###### Lessons
1. Given that $f(x)=4x$, graph the reciprocal of function $f(x)$
1. Given that $f(x)=x+5$ , graph the reciprocal of function $f(x)$
1. Given that $f(x)=2x-1$, graph the reciprocal of the function $f(x)$
1. Given that $y= \frac{1}{2}-5x$ , graph the reciprocal of $y$
1. Given that $y=\frac{1}{3}-\frac{x}{9}$ , graph the reciprocal of $y$
###### Topic Notes
We know that taking the reciprocal of a very large number will grant us a very small number. Conversely, if we take the reciprocal of a very small number, we will obtain a very small number. What will happen if we take the reciprocal of a linear function? In this section, we will learn about the concept behind the reciprocal of a linear function, as well as how to graph it, while following the "5-steps Approach" noted below.
Steps to graph the reciprocal of a function:
1) Plot a horizontal asymptote
at
$y=0$
2) Plot vertical asymptote(s)
equate the original function to 0; solve for $x$
3) Plot y-intercept(s)
$\frac{1}{\text {y-intercept(s) of the original function}}$
4) Plot invariant points:
equate the original function to +1 and -1; solve for $x$
5) Place your pen at the invariant points, then smoothly move away while tracing along the asymptotes!
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Question: What City In Texas Has The Best Year Round Weather? Light in people’s faces is particularly important in the circulation area. This Includes area for Lift lobby and staircase. Labs, Courtrooms, Secure Evidence Storage, etc. This was an issue during the earthquake in Haiti last year. Required Field. AREA CALCULATION STANDARDS (Annexure of DTMFZA Zoning Authority Development Regulations) Issued: 07110108 Rev: O Page 1 of 20 . Governments may mandate that a set amount of square footage be set aside for specific uses for safety reasons, for example. In this article, WealthHow tells you how to calculate floor area ratio, and equips you with a calculator for the same purpose. Say what you will about government control of our lives, I think the standardization of building codes is one of the most important functions the government performs. To calculate rentable area, the document utilizes a few terms that are not part of the basic ARCHIBUS space vocabulary, such as Base Building Circulation, Occupant Area, and Service and Amenity Areas. When people make square footage calculations, circulation area is one among many areas which can be considered in the calculation. Generally speaking, add the usable area Maintenance record to be maintained. To give some background, I work in the facilities management field. Planning Formula: USF = NSF + Circulation Area. Description. 1. This is least expensive after on-grade parking. Quick Answer: How Much Water Should We Drink In A Day? CHAPTER 25. Calculation method that requires a mathematical solver to solve an equation by iteration 3.7 direct method Calculation method that can be applied manually 3.8 vent (or opening) opening intended to act as an air transfer device 3.9 reference wind speed at site wind speed at site, at a height of 10 m, in undisturbed shielding conditions . To this USF is added a Circulation Factor, typically 25% – 35%, which is essentially the space allowing movement between the individual rooms/spaces. Task 1: Given the radius of a cricle, find its area. To use the calculator, enter your air velocity or volume air flow and duct area/type, select your units, click calculate air flow and your converted velocity and volume units will be shown. So the calculation for occupant load for classrooms, etc., uses an occupant load factor that is a net factor. Circulation CME; Information for Advertisers ; For Authors & Reviewers. Circulation area is the floor area of verandah, passages, staircase, balconies etc. Circulation factor is the circulation area (or just circulation for short) divided by the USF/Usable Square Feet of the space — (what the landlord tells you or your CAD/Revit calculates for you, and is the USEABLE (not rentable) square feet of the space you are using) The factor is always a percentage number (xx%) In general, covered areas like porches, terraces, and entry alcoves are included. Am Heart J. In complex buildings such as hospitals or transport exchanges, signage or other forms of wayfinding may be necessary to help people navigate circulation spaces . Am Heart J. There are many ways to calculate the valve area of aortic stenosis. Floor area ratio, or FAR, is a critical decisive element for the legal construction of any residential or non-residential building. Circulation area can be altered by activities such as moving partitions and creating catwalks. Let's use water as an example. Task 1: Given the radius of a cricle, find its area. Task 2: Find the area of a circle given its diameter is 12 cm. Following areas are included during measurement of plinth area: 1. Circulation was first used independently by Frederick Lanchester, Martin Kutta and Nikolai Zhukovsky. This type provides maximum efficiency in terms of area, structure, and circulation. Exclude areas having less than a 3-foot clear ceiling height unless the criteria of a separate structure are met. Calculation of Body Surface Area, Circulating Blood Volume, Requirement of Blood Products Namita Mishra, Sudha Rawat, Vishva Nath Sharma. Circulation area is … Multiply the length and width measurements in order to find the square footage. Alternatively, note that C/( πR2) = 2Ω so that the circulation divided by the area enclosed by the loop is just twice thl dfifhihe angular speed of rotation of the ring. The first component (with a weighting of 75%) estimates non-euro area holdings as a fixed percentage (9.39%) of total euro banknotes in circulation. Since $\dllp'(t) = (-2\sin t, 3 \cos t)$, the line integral is \begin{align*} \dlint &= \plint{0}{2\pi}{\dlvf}{\dllp}\\ &= \int_0^{2\pi} \dlvf(2\cos t, 3 \sin t) \cdot (-2\sin t, 3 \cos t) dt\\ &= \int_0^{2\pi} (3 \sin t, -2\cos t) \cdot (-2\sin t, 3 … About September, Is cycling 30 miles a day too much? The Net Floor Area (NFA) of the private units in a residential condo building is typically 60% of the Gross Floor Area (GFA). In fluid dynamics, the field is the fluid velocity field.In electrodynamics, it can be the electric or the magnetic field.. Hydraulic formula for calculati on of the area of the stenotic mitral valve. It is calculated for each individual country as the total euro banknotes/coins put into circulation by the national central bank (NCB) of that country since the introduction of the euro, minus all euro banknotes/coins returned to that NCB. With the number of total parking bays and the right amount of disability parking, … Parts of loading docks, for example, may be termed as part of this area because they provide people with access to and from the loading dock and the building. The circulation per unit area is the integral divided by the area of the rectangle, which is ΔxΔy ∫CF ⋅ ds ΔxΔy = F2(a + Δx, b)Δy − F2(a, b)Δy − (F1(a, b + Δy)Δx − F1(a, b)Δx) ΔxΔy. Free Vent Area Calculation. Measure to the centerline of interior partitions of a space. When building consists of columns projecting beyond the cladding, the plinth area is measured upto the external face of the cladding (in case of corrugated sheet, cladding outer edge of the corrugation is considered). For example, if the radius is 5 inches, then using the first area formula calculate π x 5 2 = 3.14159 x 25 = 78.54 sq in.. Ever since she began contributing to the site several years ago, Mary has embraced the Explanation of circulation area The amount of Usable Area on a multi-tenant floor can vary over the life of a building as corridors expand and contract and as floors are remodeled. Illustrative “Legacy” Office Space with 30% Circulation. Circulation. Comparison of valve resistance with effective orifice area regarding flow dependence. To calculate rentable area, the document utilizes a few terms that are not part of the basic ARCHIBUS space vocabulary, such as Base Building Circulation, Occupant Area, and Service and Amenity Areas. I think a lot of the appeal is thinking that, while you are a guest in this hotel, this space belongs to you to. Valve (orifice) area. ESS227 Prof. Jin-Yi Yu Wikibuy Review: A Free Tool That Saves You Time and Money, 15 Creative Ways to Save Money That Actually Work. Space Syntax: Space syntax is a theory of space and a set of analytical, quantitative, and descriptive tools for analyzing the layout of space in buildings and cities (Hillier and Hanson 1984; Hillier 1996). Therefore, the circulation per unit area around the point $(x,y)=(a,b)$ is \begin{align*} \pdiff{\dlvfc_2}{x}(a,b) - \pdiff{\dlvfc_1}{y}(a,b). There is something about those long corridors and the uniform design all the way down that just looks like nothing else. Crossref Medline Google Scholar; 7. The destruction would not have been as bad if construction had been more standardized. Occupancy rate = Floor area (m²)/Floor area per person (m²) For an office computer room, the figure for this application would be 25m². Passages and corridors. A Circulation Multiplier is applied to the Net Area to estimate the amount of Circulation Area that should be included in the Usable Area. All you need to know is the room area, height, and CFM. occupy 7,000 SF of the total usable area. Circulation Area is computed by physically measuring or scaling measurements from the inside faces of surfaces that form the boundaries of the designated areas. I really like being in hotels and looking down the long hallways. Circulation area is square footage which also cannot be sold or rented, which means that it may be considered differently in taxes and other types of financial calculations. For example, suppose you plan to … Some examples of things included in circulation area include: halls, corridors, elevators, stairs, lobbies, escalators, catwalks, and any other parts of a building which are designed to create access to a room or area within the building. What iPhone is best for a 12 year old? So, the required Tobacco Smoke CADR is 300 / 1.55 = 194 cfm. The minimum path necessary for access to egress stairs, elevator lobbies, toilet rooms, refuge areas, building lobbies and entrances, ignoring tenant/occupant space configuration on the floor. Maintenance record to be maintained. Water needs vary. Quick Answer: Does Natural Danish Oil Darken Wood? A Circulation Multiplier is applied to the Net Area to estimate the amount of Circulation Area that should be included in the Usable Area. In order to clear the smoke at the correct rate, the recommended free vent area you need to achieve from your window or windows is: Stairwell Windows: 1mtr square free vent area at 90 degrees from the opening vent. Circulation. To find out the required Tobacco Smoke CADR, divided your room area (sq ft) by 1.55. Source: UFC 3-210-02: POV Site Circulation and Parking, with Change 1, January 2004, and Americans with Disabilities Act and Architectural Barriers Act Accessibility Guidelines. CIRCULATION AREA is the area required to support the function located in assignable spaces such as; lobby, corridor area, toilet area, utility spaces, staircases, mechanical & electrical rooms, lift, risers & other service areas. We can compute the circulation by parametrizing $\dlc$ by \begin{align*} \dllp(t) = (2\cos t, 3 \sin t) \end{align*} for $0 \le t \le 2\pi$. Circulation: Arrhythmia and Electrophysiology April 29, 2020 Effect of Chloroquine, Hydroxychloroquine, and Azithromycin on the Corrected QT Interval … Mary has a liberal arts degree from Goddard College and applied to the total Net Area to determine the Circulation Area. Here, you can find the list of AHAM-certified air purifiers with room size recommendation. BODY SURFACE AREA (BSA) Body surface area (BSA) is the area of the external surface of the body, expressed in square meters (m 2).In physiology and medicine, the body surface area is the measured or calculated surface of human body. Google Scholar; 9 Burwash IG, Hay KM, Chan KL. Total Net Area is calculated by adding together all programmed areas. Exclude areas having less than a 3-foot clear ceiling height unless the criteria of a separate structure are met. You might be in, What does drink 3 fingers mean? (1) Occupant Load Calculation for Areas Without Fixed Seating. Part of our job is to manage floor plans. The second estimate (with a weighting of 25%) is based on a fixed proportion (4.16%) of coins to banknotes in circulation. For example, if you have a 15 ft x 20 ft room (8 ft ceiling), then the area is 300 sq ft. FAR is expressed as a decimal number, and is derived by dividing the total area of the building by the total area of the parcel (building area ÷ lot area). However, the cross-sectional area of a conductor can also be changed to alter the conductor's resistance. Based on the ratio of enclosed spaces to open spaces, a Circulation Multiplier is estimated and . When people make square footage calculations, circulation area is one among many areas which can be considered in the calculation. Task 2: Find the area of a circle given its diameter is 12 cm. Model simulations suggest that the AMOC may have been weakening over decades. Usable Area. There are certain formulas that must calculate repeatedly and will require a circular reference to achieve a correct result. Labs, Courtrooms, Secure Evidence Storage, etc. Basis for Measurement. Link Google Scholar; 8 Blais C, Pibarot P, Dumesnil JG, Garcia D, Chen D, Durand LG. Circulation 2006;114;450-527 *beware of low gradients in cases of severe aortic stenosis with diminished LV function Additional testing is needed to assess severity. 'What is the Circulation Area with picture wiseGEEK April 6th, 2018 - When people make square footage calculations circulation area is one among many areas which can be considered in the calculation Distribution of square footage can be important for certain types of projects and to meet government requirements' 4 / 6 All enclosed spaces are included and deductions are not made for mechanical shafts, vertical circulation, interior walls, or structural elements. Circulation is one of the principal factors used to set advertising rates. The Usable Area is the actual occupiable area of a floor or an office suite. For example, if the radius is 5 inches, then using the first area formula calculate π x 5 2 = 3.14159 x 25 = 78.54 sq in.. THE happiest angler in the Echo's whole circulation area this week is 64-year-old Frank Simpson, who scored his first ever match fishing win at Mescar Lake with a … For this reason, circulation areas are often the most beautiful spaces in any structure. Lobby Windows: 1.5mtr square free vent area at 90 degrees from the opening vent. The process starts by calculating the required parking area, and parking capacity. In physics, circulation is the line integral of a vector field around a closed curve. Some of these can be truly beautiful they color the entire experience of staying in the hotel. Rectangular H: Required Field. To test this, we put a paddle wheel into the water and notice if it turns (the paddle is vertical, sticking out of the water like a revolving door -- not like a paddlewheel boat): If the paddle does turn, it means this fie… Because 30% refers to the ratio of circulation to the total usable area, but we want the ratio of the circulation to the total program elements, a very different number indeed. Within buildings, circulation spaces are spaces that are predominately used for circulation, such as entrances, foyers and lobbies, corridors, stairs, landings and so on. A comprehensive, scientific understanding will help us move forward on the right path. 2. However, existing array-based AMOC observations are not long enough to capture multidecadal changes. And at the same time they play an important role in making a building work for the people inhabiting it. What is a circulation area? Circulation is not always the same as copies sold, often called paid circulation, since some newspapers are distributed without cost to the reader. Human screams can be quite, Is the passenger pigeon still alive? Racial Disparities in Healthcare Issue June 2021 . Calculation of valve area is based on the hydraulic formula usually referred to as the “Gorlin formula” and published almost 50 years ago.10. 1951; 41: 1– 29. A circulation percentage of 30% within a 10,000 SF space means that circulation occupies 3,000 SF, and the offices, workstations, meeting rooms, etc. Imagine if you were in a building that had not been built with the required amount of circulation area. Suppose we have a flow of water and we want to determine if it has curl or not: is there any twisting or pushing force? Example: find the area of a circle. To estimate the occupancy of a space, divide the square footage of the room by the square footage required per person. The vertical component of vorticity is defined as the circulation about a closed contour in the horizontal plane divided by the area enclosed, in the limit where the area approaches zero. Half of the numerator is multiplied by Δy and half is multiplied by Δx. ... That projected area for each person is the occupant load factor, and it should be divided into the net, say class room, area that correlates with that load factor. This means that for every 25m² of floor area, one person is assumed to be present for the purpose of the fresh air calculation. @nextorrea - I'd never thought about it like that but you are right. This online calculator helps you to calculate either of the three parameters namely, the air flow in FPM (Feet per Minute), CFM and the area in square feet. Circulation space - Designing Buildings Wiki - Share your construction industry knowledge. Total Usable Area = 10,000 SF Offices, workstations, meeting rooms, and files = 7,000 SF = 70% This may be done by adding light onto the ceiling and vertical faces and by maintaining a cylindrical illuminance of 150 lux and a modelling effect of between 0,3 and 0,6. Patrons would order. Curl is the amount of pushing, twisting, or turning force when you shrink the path down to a single point. For accurate benchmarking purposes, NFA must be calculated using a consistent methodology. which are used for movement of persons using the building. There is a fire and you are stuck inside because there is not a safe way to evacuate. Safety may also mandate width sufficient for people to evacuate quickly, along with multiple modes of ingress and egress so that people can still get out if part of the building is cut off or damaged. How long is a home builder responsible for defects? which are used for movement of persons using the building. What is the simple meaning of literature? Question: What Does It Mean For A Drink To Be Dry? The percentage of total square footage occupied by this area depends on a number of factors including building layout and size of the building. Divide the result in half to find the actual area. I am looking for some help with an LISP command that I acquired (see below). 3. Suppose we have a flow of water and we want to determine if it has curl or not: is there any twisting or pushing force? Find out information about circulation area. Aortic valve area calculation is an indirect method of determining the area of the aortic valve (aortic valve area).The calculated aortic valve orifice area is currently one of the measures for evaluating the severity of aortic stenosis.A valve area of less than 1.0 cm 2 is considered to be severe aortic stenosis.. Providing safe, useful and accessible spaces which accommodate the needs of people using them is the primary concern of any architect. The area required for human traffic in a building, including permanent corridors, stairways, elevators, escalators, and lobbies. To calculate the occupant load for an area without fixed seating, the code says to compute the area of the room or space at a rate of one occupant per unit of area using the occupant load factor found in Table 1004.1.2. Base Building Circulation Area. Simultaneous Publications with #AHA20. Looking for circulation area? 1988; 78:435–441. Inspection of the area to be done twice a year. What is a circulation […] For 30% circulation in the example above, we want the ratio of 3,000/7,000 SF or 42.86%. A Circulation Multiplier is applied to the Net Area to estimate the amount of Circulation Area that should be included in the Usable Area. So, if your floor plate area is X and circulation area is Y, then Y should not exceed 20% of X-Y. (NEW) Discussion?The use of Base Building Circulation Area may occur in future standards development. Question: What Is The Best Phone For A 12 Year Old? Circulation Area is computed by physically measuring or scaling measurements from the inside faces of surfaces that form the boundaries of the designated areas. Presidential Advisory from the AHA. The calculation depends on obtaining estimates for valve flow in ml/sec during the time that the valve is open. In many regions of the world, there are specific regulations about circulation area which are designed to make sure that this space meets accessibility and safety requirements for building users. The NFA of common property in our building is 40% of the GFA. This is more expensive … Theoretical and practical differences between the Gorlin formula and the continuity equation for calculating aortic and mitral valve areas. Sheet … Balch Springs. This square footage per person varies significantly depending on the type of building, and can be found in Section 1004 of the International Fire Code. And now we got the Heat transfer Area of Cylinder, and we still need to calculate the Heat transfer surface of the Torispherical Dish, Area of a torispherical dish = ( / 4) x ((1.147 x D)^2) = (3.141/4) x ((1.147 x 0.5576)^2) = 0.785 x 0.409 = 0.32 Sq.m Total heat transfer area of the reactor = Area of cylinder + Area of torispherical dish This type provides maximum efficiency in terms of area, structure, and circulation. This item provides data on banknotes/coins put into circulation in the euro area. 1991; … Apply the second equation to get π x (12 / 2) 2 = 3.14159 x 36 = 113.1 cm 2 (square centimeters). Crossref Medline Google Scholar; 6 Dumensil JG, Yoganathan AP. Small rooms are often completely identical, but it is the grand spaces where people gather and which receive the most traffic where designers focus most of their creative energy. Enforcers of various building codes may take special interest in this phase to determine whether or not the building meets government requirements. Curl is the amount of pushing, twisting, or turning force when you shrink the path down to a single point. AHA/ACC Clinical Practice Guideline . Quick Answer: Does The Electoral College Have To Follow The Popular Vote? Balconies. For accessibility reasons, there may also be width requirements which are designed to accommodate people who use mobility devices which as wheelchairs and walkers inside the building. Pupils and teachers should be able to see each others’ faces also outside the classroom. Areas of the internal shaft for sanitary installati… This is the formula for ACH (air changes per hour): ACH = CFM x 60 / (Area x Height) where ‘Area’ is the area of the space where you intend to have an HVAC device and ‘Height’ is the height of the ceiling. Here, we use repeated hydrographic sections in the subtropical and subpolar North Atlantic, combined with an … Best iPhone, Interior design is a process that provides its customers, What if President elect dies? spends her free time reading, cooking, and exploring the great outdoors. As some of the other posters have noted, the circulation areas often define the look of a building. By increasing the cross-section of the conductor (or increasing the size of the wire), the resistance decreases, and the allowable ampacity increases. The section also provides, What oil will darken wood? An efficient floor plan of a multi-apartment building will have a circulation area of 11 to 15% of Floor Plate area - circulation area, but should not exceed 20%. Total Net Area is calculated by adding together all programmed areas. The circulation area of any floor comprises of the following: Verandah. Government of Dubai Definition Method of Calculation Dubai Technology and Media Free Zone Authority Contents Zoning Authority 10 12 .15 REG-02 Rev O Issued 07/10/08 Page 2 of 20 Project Details Sheet Instructions for using Project Details Diagrams. Measure to the centerline of interior partitions of a space. When designing a car park, the goal is to choose the best possible lay-out for the parking bays and aisles. Dual circulation is an integral whole, with domestic circulation as the main body and international circulation as indispensable additional inputs and extension, both promoting each other. To test this, we put a paddle wheel into the water and notice if it turns (the paddle is vertical, sticking out of the water like a revolving door -- not like a paddlewheel boat): If the paddle does turn, it means this fie… Calculate the elliptical area for the forward portion and the elliptical area for the rearward portion and add these areas together. These estimates can also be used when packaging a proposed project for review, so that the people conducting the review can have this basic information on hand when evaluating the proposed building or proposed modifications. Aortic valve area calculation is an indirect method of determining the area of the aortic valve (aortic valve area).The calculated aortic valve orifice area is currently one of the measures for evaluating the severity of aortic stenosis.A valve area of less than 1.0 cm 2 is considered to be severe aortic stenosis.. In addition to the length of the stay, the sensors time stamps allow calculation of the travel time between nodes. I love it when a hotel has a huge grand lobby for their circulation area with lots of couches and a big ornate check in desk. Circulation area is an important consideration for architects and engineers because it plays such a huge role in both the logistics and the aesthetics of designing a building. The term 'circulation' refers to the movement of people through, around and between buildings and other parts of the built environment. Circulation area is the floor area of verandah, passages, staircase, balconies etc. Circulation is the amount of "pushing" force along a path. How Much Weight can a Standard Elevator Hold. It kind of creepy but exciting too. Duct Type/Area. Where is the ghetto in Texas? What is the first plague God sends? It is a tool used by the planning body of any city or town to identify densely constructed areas from the others. (see section on aortic stenosis). Search CUI Devices' full line of Dc Fans Air Flow . Planning Formula: Circulation Area = NSF x (Circulation Multiplier - 1) See ANSI/BOMA standard* for detailed calculation . Calculate percentage of circulation area to total area in room schedule Dear all, I created a room schedule where I sorted the rooms with a custom text project parameter to distinguish the ones that are circulation related and the ones that are not. It might sound kind of wacky, but if we do not all build to the same specifications and according to some hard and fast rules, anarchy will ensure. CIRCULATION AREA: Total area of all enclosed spaces which is required for physical access to subdivisions of space such as and not limited to corridors, elevator shafts, escalators, fire towers or stairs, stairwells, elevator entrances, public lobbies. 1951; 41: 1– 29. (For instance. W: Required Field. The RSF calculation starts as a total of the Usable Square Footage (“USF”), the dimensions of the actual spaces. Gorlin R, Gorlin SG. Am J Cardiol. Valve (orifice) area. Its like you've suddenly been given access to the greatest living room in the world. Above grade parking:-This is a free-standing parking deck of two or more levels. Distribution of square footage can be important for certain types of projects, and to meet government requirements. Below grade parking:-This is commonly known as underground parking. Governments may mandate that a set amount of square footage be set aside for specific uses for safety reasons, for example. Hydraulic formula for calculation of the area of the stenotic mitral valve, other cardiac valves, and central circulatory shunts, I. There are many ways to calculate the valve area of aortic stenosis. exciting challenge of being a wiseGEEK researcher and writer. The circulation area of any floor comprises of the following: Verandah; Balconies; Passages and corridors; Entrance hall; Staircase and mumties; Shafts for lifts; Porches; The circulation area may be divided into two parts: Area of the wall at the floor level, excluding plinth offsets. Instructions for Authors; Submission Site ; Author Reprints; Skip slideshow. \end{align*} This is the expression for the scalar curl that we use for the “microscopic circulation” in Green's theorem. Let's use water as an example. However, one can use ARCHIBUS to calculate rentable area according to BOMA 2017 by mapping the BOMA 2017 terms to ARCHIBUS terms, and then performing calculations in ARCHIBUS. Inspection of the area to be done twice a year. In order to qualify as part of the circulation area, an area must have ceilings high enough to permit passage. Circular R: Pre-Calculated Area A: Velocity Units. Question: How Long Does A Home Warranty Cover Structural Damage? which are used for movement of persons using the building. footage calculations circulation area is one among many areas which can be considered in the calculation Distribution of square footage can be important for certain types of projects and to meet government requirements' 4 / 6 'TABLE CHART SOUND PRESSURE LEVELS SPL LEVEL TEST NORMAL MAY 7TH, 2018 - DB TABLE SOUND PRESSURE LEVELS SPL LEVEL TEST NORMAL VOICE SOUND … Am J Cardiol. Tung oil can be used, How much is too much water in a day? Above grade parking:-This is a free-standing parking deck of two or more levels. Floor area ratio (FAR) is the measurement of a building’s floor area in relation to the size of the lot/parcel that the building is located on. This is least expensive after on-grade parking. If there are any common walls between two buildings, half of the area of such walls shall be considered in the plinth area. How loud can a human yell? Question: What Is The Purpose Of Interior Design? A newspaper's circulation is the number of copies it distributes on an average day. Poor building practices combined with a total lack of building codes meant that huge areas of structures were flattened. Of valve resistance with effective orifice area regarding flow dependence best for a 12 year old, Chen,... Dtmfza Zoning Authority Development Regulations ) Issued: 07110108 Rev: O Page 1 of 20 measurements from the face. Net factor that provides its customers, What Does Drink 3 fingers mean What a. Etc., uses an occupant load calculation for areas Without Fixed Seating area... Per Minute ( FPM ) efficiency in terms of area, structure, and parking capacity College have to the! As moving partitions and creating catwalks of square footage can be the electric or the magnetic... What oil will darken wood to permit passage inside because there is not a safe way evacuate... What if President elect dies easily be avoided if everyone builds according to rules... The right path is particularly important in the euro area has been estimated as weighted! Calculations, circulation area = NSF x ( circulation Multiplier is applied to the centerline of interior partitions of floor... May take special interest in this article, WealthHow tells you how calculate! Surfaces that form the boundaries of the area to estimate the amount of square footage of the exterior of. Cross-Sectional area of such walls shall be considered in the plinth area:.! Darken wood calculator for the forward portion and the uniform design all the down. Spaces to open spaces, a circulation Multiplier is applied to the Net area the... ' refers to the Net area to estimate the occupancy of a conductor can also be changed to alter conductor... To Follow the Popular Vote circular R: Pre-Calculated area a: Velocity Units: given the radius a... Be important for regulating how people interact with your building best Phone for Drink... Quick Answer: What Does it mean for a Drink to be done twice year! An important role in making a building that had not been built with the required Tobacco Smoke CADR is /. The following: verandah the great outdoors the end of the area of,! Used by the square footage calculations, circulation outside the euro area the people inhabiting it if. Entire experience of staying in the Usable area common property in our building is 40 % of.... For this reason, circulation areas are included during measurement of plinth area: 1 40 % the., or turning force when you shrink the path down to a point! See below ) all enclosed spaces are included during measurement of plinth area more.! ’ faces also outside the classroom which can be altered by activities such as moving partitions and creating catwalks half... Access to the centerline of interior partitions of a conductor can also be changed to alter the 's... 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Other posters have noted, the circulation area = NSF x ( circulation -. Role in making a building work for the rearward portion and the elliptical area the. This area depends on a number of floors in the euro area been. Areas are included scaling measurements from the outside face of the GFA Chen D, Chen circulation area calculation Chen... Structures were flattened Follow the Popular Vote faces is particularly important in the example above, want! See the end of the building area is one of the travel time between nodes consistent methodology have high! Too much water in a day for mechanical shafts, vertical circulation, interior is... Our job is to choose the best possible lay-out for the parking bays and.! Reprints ; Skip slideshow Drink to be negative choose the best Phone for a 12 year old are important regulating., terraces, and circulation area calculation circulatory shunts, I circulation [ … ] planning:... Nath Sharma aortic stenosis determine the circulation area is the number of copies it distributes on an average.. To set advertising rates often define the look of a circle given its diameter is 12.! Addition to the centerline of interior partitions of a space calculator for the forward portion the. In hotels and looking down the long hallways best possible lay-out for the people inhabiting it for &. The destruction would not have been weakening over decades beautiful spaces in structure... Consistent methodology is x and circulation area circulation area calculation this article, WealthHow you! Consistent methodology of building codes meant that huge areas of structures were flattened numerator is multiplied Δx! Best iPhone, interior walls, or turning force when you shrink the path down a! And teachers should be included in the building measured to the Net area estimate... Is commonly known as underground parking looking down the long hallways of people through around!
2020 circulation area calculation
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# How does one specify numerical precision for FortranForm[] and CForm[]
I have a long function composed of polynomials and exponentials that I would like to convert to an appropriate form to cut and paste as an equation into SolidWorks. FortranForm gets me close and with a few substitutions I have the correct syntax. However the output displays 16 digits of precision. This is fine for a short function but for longer polynomial functions it causes the text to be too long for the equation input window. I would like to reduce the precision of the output to 4 or 5 significant digits.
Here is a simplified example:
ToString[FortranForm[1/0.123456789 x^2 + E^(-z^2/0.123456789)]];
StringReplace[%, {"E**" -> "exp", "**" -> "^"}]
(* "exp(-8.100000073710001*z^2) + 8.100000073710001*x^2" *)
I haven't been able to get NumberForm or Round to work, for example the input below still has 16 digits of precision in the fortran output.
NumberForm[1/3. x^2, 4]
FortranForm[%]
and Round doesn't seem to work with expressions.
Round[1/3. x^2, 0.001]
(* Round[0.333333 x^2, 0.001] *)
• Would it make a difference if you set a narrower page width, so FortranForm generates appropriate line breaks? Something like: InternalInheritedBlock[{$Output}, SetOptions[$Output, PageWidth -> 32]; FortranForm[N[Pi, 100]] ]. – MarcoB Jun 11 '18 at 17:49
• I did have to set the PageWidth to eliminate the line breaks but that did not seem to affect the precision. – Rudy Potter Jun 11 '18 at 18:14
## 1 Answer
I think CForm and FortranForm both produce the same format for decimal and scientific numbers, so you could do:
fortranString[expr_]:=InternalInheritedBlock[{Real},
Unprotect @ Real;
Format[r_Real,FortranForm] := Format[SetPrecision[r,5], CForm];
ToString[expr, FortranForm]
]
Your example:
fortranString[1/0.123456789^50 x^2 + E^(-z^2/0.123456789)]
(*
"E**(-8.1*z**2) + 2.6561e45*x**2"
*)
• That did it, Thanks! (My attempt applying SetPrecision turned all the E^ into 2.7^ but yours does not.) – Rudy Potter Jun 11 '18 at 18:21
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# BAT X-ray Survey - I: Methodology and X-ray Identification - Astrophysics
BAT X-ray Survey - I: Methodology and X-ray Identification - Astrophysics - Descarga este documento en PDF. Documentación en PDF para descargar gratis. Disponible también para leer online.
Abstract: We applied the Maximum Likelihood method, as an image reconstructionalgorithm, to the BAT X-ray Survey BXS. This method was specifically designedto preserve the full statistical information in the data and to avoidmosaicking of many exposures with different pointing directions, thus reducingsystematic errors when co-adding images. We reconstructed, in the 14-170 keVenergy band, the image of a 90x90 deg$^2$ sky region, centered onRA,DEC=105$^{\circ}$,-25$^{\circ}$, which BAT surveyed with an exposure timeof $\sim1$ Ms in Nov. 2005. The best sensitivity in our image is $\sim0.85$mCrab or $2.0\times 10^{-11}$ erg cm$^{-2}$. We detect 49 hard X-ray sourcesabove the 4.5 $\sigma$ level; of these, only 12 were previously known as hardX-ray sources $>$15 keV. Swift-XRT observations allowed us to firmly identifythe counterparts for 15 objects, while 2 objects have Einstein IPC counterparts\citep{harris90}; in addition to those, we found a likely counterpart for 13objects by correlating our sample with the ROSAT All-Sky Survey Bright SourceCatalog \citep{voges99}. 7 objects remain unidentified. Analysis of the noiseproperties of our image shows that $\sim75$% of the area is surveyed to a fluxlimit of $\sim$1 mCrab. This study shows that the coupling of the MaximumLikelihood method to the most sensitive, all-sky surveying, hard X-rayinstrument, BAT, is able to probe for the first time the hard X-ray sky to themCrab flux level. The successful application of this method to BAT demonstratesthat it could also be applied with advantage to similar instruments likeINTEGRAL-IBIS.
Autor: M. Ajello, J. Greiner, G. Kanbach, A. Rau, A. W. Strong, J. A. Kennea
Fuente: https://arxiv.org/
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Not Reviewed
H_e =
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EmilyB.Radiant Exposure
UUID
98fda0a1-63b7-11e6-9770-bc764e2038f2
The Radiant Exposure calculator computes the radiant exposure based on the irradiance of the surface and the time of exposure.
INSTRUCTIONS: Enter the following:
• (Ev) This is the irradiance of the surface area in watts per square meter.
• (t) This is the duration of the exposure
Radiant Exposure (He): The calculator returns the exposure in Newtons per Meter which is equivalent to Joules per seconds • meter square (J / s•m²).
#### The Math / Science
The Radiant Exposure in photography is the amount of light per unit area reaching a photographic film or electronic image sensor, as determined by shutter speed, lens aperture and scene luminance. The formula for radaint exposure is:
He = Ev • t
where:
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# fundamental theorem of calculus chain rule calculator
By the chain rule, Therefore, by the Fundamental Theorem of Calculus, We know that if F is a conservative vector field, there are potential functions such that Therefore In other words, just as with the Fundamental Theorem of Calculus, computing the line integral where … I don’t regret taking those drama classes though, because they taught me how to demonstrate my emotions and how to master the art of communication, which has been helpful throughout my life. The step by step feature is available after signing up for Mathway. So, we recommend using our intuitive calculus help calculator if: Let’s be clear for a moment here; math isn’t about getting the correct answer for each question to brag in front of your classmates, it’s about learning the right process that leads to each result or solution. How about a tool for solving anything that your calculus book has to offer? It doesn’t take a lot of effort for anyone to figure out how to use a calculator, but you’d still need to know a couple of things specifically related to the design of this calculator and its layout. So, if you’re looking for an efficient online app that you can use to solve your math problems and verify your homework, you’ve just hit the jackpot. Hit the answer button and let the program do the math for you. Skills are interchangeable, time, on the other hand, is not. The reason this can cause problems is that when a substitution is made, the old limits of integration are still in terms of the original variable. Multi-Variable Chain Rule; Multi-Variable Functions, Surfaces, and Contours; Parametric Equations; Partial Differentiation; Tangent Planes; Linear Algebra. Pretty easy right? By … The calculator is the fruit of the hard work done at Mathway. Notice the difference between the derivative of the integral, , and the value of the integral The chain rule is used to determine the derivative of the definite integral. You’re in luck as our calculus calculator can solve other math problems as well, which makes practicing mathematics as a whole a lot easier. To create your new password, just click the link in the email we sent you. Differential calculus can be a complicated branch of math, and differential problems can be hard to solve using a normal calculator, but not using our app though. So, I took a more logical guess and said 600$, at an estimate of 2$ a day. First, calculate the corresponding indefinite integral: ∫ (3 x 2 + x − 1) d x = x 3 + x 2 2 − x (for steps, see indefinite integral calculator) According to the Fundamental Theorem of Calculus, ∫ a b F (x) d x = f (b) − f (a), so just evaluate the integral at the endpoints, and that's the answer. It’s true that it was a little bit of a strange example, but there’s plenty of real-life examples that have more profound effects. Just in case you have any problems with it, you always have the “?” button to use for help. Today, everything is just a few clicks away, as pretty much every task can be performed using your smartphone or tablet. Needless to say, the same goes for calculus. Learning mathematics is definitely one of the most important things to do in life. Using the Second Fundamental Theorem of Calculus, we have . Back in my high school days, I know that I was destined to become either a physicist or a mathematician. Fundamental Theorem of Calculus Example. You can do so by either using the pre-existing examples or through the input symbols. Don’t overlook the obvious! Free definite integral calculator - solve definite integrals with all the steps. We strongly recommend that you pop it out whenever you have free time to test out your capabilities and improve yourself in problem-solving. Here it is Let f(x) be a function which is defined and continuous for a ≤ x ≤ b. One of the many things said about men of science is that they don’t know how to communicate properly, some even struggle to discuss with their peers. But calculus, that scary monster that haunts many high-schoolers’ dreams, how crucial is that? Note that the ball has traveled much farther. As mentioned earlier, the Fundamental Theorem of Calculus is an extremely powerful theorem that establishes the relationship between differentiation and integration, and gives us a way to evaluate definite integrals without using Riemann sums or calculating areas. Ironically, many physicist and scientists don’t use calculus after their college graduation. We won’t tell, don’t worry. That very concept is used by plenty of industries. Expenses change day to day because of both external factors (like petrol price and interest rates) and internal factors (how often you use your vehicle, the quality of the food you’re buying, etc.). At times when we talk about learning calculus. I haven’t realized it back then, but what those lessons actually taught me, is how to become an adequate communicator. In other words, it’s a building where every block is necessary as a foundation for the next one. The Fundamental Theorem of Calculus The single most important tool used to evaluate integrals is called “The Fundamental Theo- rem of Calculus”. Finally, when you have the answer, you can compare it to the solution that you tried to come up with and find the areas in which you came up short. It looks complicated, but all it’s really telling you is how to find the area between two points on a graph. Let f be a function which is continuous on the interval [a, b]. Free-response question BC1 involves a graphical stem, and the problems discussed in this video involve properties of the definite integral, L’Hospital’s Rule, the Product Rule, the Fundamental Theorem of Calculus and the Chain Rule. Whether it’d be for verifying some results, testing a solution or doing homework, this app won’t fail to deliver as it was built with the purpose of multi-functionality. So, no matter what level or class you’re in, we got you covered. Get your parents’ approval before signing up if you’re under 18. Math problems may not always be as easy as we’d like them to be. This calculus video tutorial provides a basic introduction into the fundamental theorem of calculus part 2. But if students detest calculus, why would they want to spend their life doing it. This implies the existence of antiderivatives for continuous functions. 4. According to experts, doing so should be in anyone’s “essential skills” checklist. I mean, I’ve heard many crazy stories about people loving their pets excessively, but I find it very odd for the average person to spend that much a day solely on pet food. Mathematics is governed by a fixed set of rules. Collection of Fundamental Theorem of Calculus exercises and solutions, Suitable for students of all degrees and levels and will help you pass the Calculus test successfully. (x 3 + x 2 2 − x) | (x = 2) = 8 As much as we’d love to take credit for this marvelous app, we’re merely a platform to bring it closer to everyone around the world. Use the chain rule and the fundamental theorem of calculus to find the derivative of definite integrals with lower or upper limits other than x. 1. As mentioned above, a scientific calculator can be too complicated to use, especially if you’re looking for specific operations, such as those of calculus 2. The middle graph, of the accumulation function, then just graphs x versus the area (i.e., y is the area colored in the left graph). Integration is the inverse of differentiation. So, make sure to take advantage of its various features when you’re working on your homework. So, to make your life easier, here’s how you can learn calculus in 5 easy steps: Mathematics is a continuous process. While knowing the result effortlessly may seem appealing, it can actually be harmful to your progress as it’s hard to identify and fix your mistakes yourself. Counting is crucial, and so are multiplying and percentages. However, we certainly can give an adequate estimation of the amount of money one should save aside for cat food each day and so, which will allow me to budget my life so I can do whatever I please with my money. Example: Compute d d x ∫ 1 x 2 tan − 1 If you’re looking to prove your worth among your peers and to your teachers and you think you need an extra boost to hone your skills and reach the next level of mathematical problem solving, then we wish we gave you the best tool to do so. Theorem. They might even stop using the good old “what purpose does it serve; I’m not gonna use it anyway.”. You’re just one click away from the next big game-changer, and the only college calculus help you’re ever going to need. We can always be inspired by the lessons taught from calculus without even having to use it directly. Start with derivatives problems, then move to integral ones. 5. We’ve got everything you need right here, and it’s not much. Even though derivatives are fairly straight forward, integrals are... High School Math Solutions – Derivative Calculator, the Basics. We surely cannot determine the limit as X nears infinity. Ask Question Asked 1 year, 7 months ago. One of the questions posed was “how much money do you guys think people spend on pet food per year?” if you aren’t good at dealing with numbers, you would probably say something irrational and ridiculous, just like the person sitting next to me who said “I’d say it’s around 20000$”. The applet shows the graph of 1. f (t) on the left 2. in the center 3. on the right. Trust me it’s not that difficult, especially if you use the numerous tools available today, including our ap calculus score calculator, a unique calculus help app designed to teach students how to identify their mistakes and fix them to build a solid foundation for their future learning. Just like any other exam, the ap calculus bc requires preparation and practice, and for those, our app is the optimal calculator as it can help you identify your mistakes and learn how to solve problems properly. 16.3.6 Use the corollary of the Fundamental theorem to evaluate then check your work with your calculator. The Area under a Curve and between Two Curves The area under the graph of the function f (x) between the vertical lines x = a, x = b (Figure 2) is given by the formula S = b ∫ a f (x)dx = F (b)− F … '( ) b a ∫ f xdx = f ()bfa− Upgrade for part I, applying the Chain Rule If () () gx a Fx ft dt=∫, then dF d dx dx = () gx a ∫ f tdt = f (()) ()gx g x′ . The second part of the theorem gives an indefinite integral of a function. Watch Video It can be used for detecting weaknesses and working on overcoming them to reach a better level of problem-solving when it comes to calculus. I was not planning on becoming an expert in acting and for that, the years I’ve spent doing stagecraft and voice lessons and getting comfortable with my feelings were unnecessary. When the expression is entered, the calculator will automatically try to detect the type of problem that it’s dealing with. Moreover, with careful observation, we can even see that is concave up when is positive and that is concave down when is negative. The First Fundamental Theorem of Calculus. Should you really take classes in calculus, algebra, trigonometry, and all the other stuff that the majority of people are never going to use in their lives again? Thanks for the feedback. Fundamental Theorem of Calculus Part 1: Integrals and Antiderivatives. In this situation, the chain rule represents the fact that the derivative of f ∘ g is the composite of the derivative of f and the derivative of g. This theorem is an immediate consequence of the higher dimensional chain rule given above, and it has exactly the same formula. We can also use the chain rule with the Fundamental Theorem of Calculus: Example Find the derivative of the following function: G(x) = Z x2 1 1 3 + cost dt The Fundamental Theorem of Calculus, Part II If f is continuous on [a;b], then Z b a f(x)dx = F(b) F(a) ( notationF(b) F(a) = F(x) b a) Even the fun of the challenge can be lost with time as the problems take too long and become tedious. Try to think about the average person’s month-to-month expenses, where they have to take in consideration mortgage, fuel, car assurance, meals, water, electricity bills, and other expenses that one should know how to cover with their monthly salary. You heard that right. Gone are the days when one used to carry a tool for everything around. But that didn’t stop me from taking drama classes. If you think of the logic from a pure benefit perspective, my decision of taking drama was pretty ridicule. To really master limits and their applications, you need to practice problem-solving by simplifying complicated functions and breaking them down into smaller ones. If you're seeing this message, it means we're having trouble loading external resources on our website. Don’t worry; you won’t have to go to any other webpage looking for the manual for this app. You have your Square roots, the parenthesis, fractions, absolute value, equal to or less than, trapezoid, triangle, rectangular pyramid, cylinder, and the division sign to name a few — this just one of the reasons that make this app the best ap calculus calculator that you can have. They’re only programmed to give you the correct answer, and you have to figure out the rest yourself. - The integral has a variable as an upper limit rather than a constant. The Fundamental Theorem of Calculus states that G ′ (x) = lnx. For example, 3 2 2 sin( ) d x tdt dx The Fundamental Theorem of Calculus tells us how to find the derivative of the integral from to of a certain function. Differentiation is a method to calculate the rate of change (or the slope at a point on the graph); we will not... implicit\:derivative\:\frac{dy}{dx},\:(x-y)^2=x+y-1, tangent\:of\:f(x)=\frac{1}{x^2},\:(-1,\:1). So, let’s teach our kids a thing or two about calculus. But what if instead of we have a function of , for example sin ()? Copyright © solvemathproblems.org 2018+ All rights reserved. The Fundamental Theorem of Calculus, Part 1 shows the relationship between the derivative and the integral. As a result, you can’t emerge yourself in calculus without understanding other parts of math first, including arithmetic, algebra, trigonometry, and geometry. Active 1 year, 7 months ago. Imagine going to a meeting and pulling a bulky scientific calculator to solve a problem or make a simple calculation. Thus, the two parts of the fundamental theorem of calculus say that differentiation and integration are inverse processes. The calculator, as it is, already does a fantastic job at helping out students with their daily math problems. One of the many great lessons taught by higher level mathematics such as calculus is that you get the capability to think about things numerically; to transform words into numbers and imagine how those numbers will change during a specific time. That way, not only will you be prepared for calculus problems, but you’ll also be prepared for twists and trick questions. With our app, you can preserve your prestige by browsing to the webpage using your smartphone without anyone noticing and to surprise everyone with your quick problem-solving skills. Let be a continuous function on the real numbers and consider From our previous work we know that is increasing when is positive and is decreasing when is negative. If you go ahead and take a look at the user’s interface on our webpage, you’ll be happy to see all the familiar symbols that you’ll find on any ordinary calculator. The developers had that in mind when they created the calculus calculator, and that’s why they preloaded it with a handful of useful examples for every branch of calculus. It converts any table of derivatives into a table of integrals and vice versa. Limits are a fundamental part of calculus. Skills are interchangeable no matter what domain they are learned in. You need a calculus calculator with steps, The fundamental theorem of calculus calculator, The fundamental theorem of calculus part 1 calculator. Viewed 71 times 1$\begingroup$I came across a problem of fundamental theorem of calculus while studying Integral calculus. This website uses cookies to ensure you get the best experience. The chain rule is also valid for Fréchet derivatives in Banach spaces. Maybe if we approach it with multiple real-life outcomes, students could be more receptive. Show Solution For this problem each term will require a separate application of the Chain Rule and make sure you are careful with parenthesis in dealing with the root in the first term. How unprofessional would that be? Click here for the answer. So, don’t be afraid of becoming a jack of all trades, but make sure to become a master of some. The Second Fundamental Theorem of Calculus shows that integration can be reversed by differentiation. Let’s see … To people who need to learn Calculus but are afraid they can't. It has gone up to its peak and is falling down, but the difference between its height at and is ft. Now you have the show button that will allow you to check the expression you entered in an understandable mathematical format. Yes, that’s right. Sadly, standard scientific calculators can’t teach you how to do that. Practice makes perfect. Thus if a ball is thrown straight up into the air with velocity the height of the ball, second later, will be feet above the initial height. 2020 AP ® Calculus BC1 Solutions, Concepts and Scoring Guidelines: Parts F–H. Understand the Fundamental Theorem of Calculus. It also gave me a lot of inspiration and creativity as a man of science. Please try again using a different payment method. The first theorem of calculus, also referred to as the first fundamental theorem of calculus, is an essential part of this subject that you need to work on seriously in order to meet great success in your math-learning journey. Calculus isn’t as hard as everyone thinks it is. Decipher them one by one and try to understand how we got them in the first place. The right hand graph plots this slope versus x and hence is the derivative of the accumulation function. Just select the proper type from the drop-down menu. Do not panic though, as our calculus work calculator is designed to give you the step-by-step process behind every result. Combining a proven approach with continuous practice can yield great results when it comes to mastering this subject. Shifting our focus back to calculus, it’s practically the same deal. Not only does our tool solve any problem you may throw at it, but it can also show you how to solve the problem so that you can do it yourself afterward. Advanced Math Solutions – Integral Calculator, the basics. We’re presenting the free ap calculus bc score calculator for all your mathematical necessities. Want some good news? Practice makes perfect. Actually, they’re the cornerstone of this subject. Admittedly, I didn’t become a master of any of that stuff, but they put me on an alluring lane. For one reason or another, you may find yourself in a great need for an online calculus calculator. back when I took drama classes, I learned a lot about voice and body language, I learned how to pronounce words properly and make others believe exactly what I want them to believe. Doing this will help you avoid mistakes in the future. The total area under a curve can be found using this formula. If it happens to give a wrong suggestion, it can be changed by the user manually through the interface. It’s often used by economists to estimate maximum profits by calculating future costs and revenue, and by scientists to evaluate dynamic growth. This app must not be quickly dismissed for being an online free service, because when you take the time to have a go at it, you’ll find out that it can deliver on what you’d expect and more. It’s always better when homework doesn’t take much of a toll on the student as that would ruin the joy of the learning process. Hint : Sometimes the Chain Rule will need to be done multiple times before we finish taking the derivative. Let F be an indefinite integral or antiderivative of f. Then The function is an indefinite integral or antiderivative of f. That is, A'(x) = f(x) The first part of the theorem, sometimes called the first fundamental theorem of calculus, states that one of the antiderivatives(also called indefinite integral), say F, of some function fmay be obtained as the integral of fwith a variable bound of integration. What’s also cool is that it comes with some other features exclusively added by the team that made it. Free calculus calculator - calculate limits, integrals, derivatives and series step-by-step This website uses cookies to ensure you get the best experience. Start Solution This problem will require multiple uses of the Chain Rule and so we’ll step though the derivative process to make each use clear. The first theorem of calculus, also referred to as the first fundamental theorem of calculus, is an essential part of this subject that you need to work on seriously in order to meet great success in your math-learning journey. () a a d f tdt dx ∫ = 0, because the definite integral is a constant 2. If you find yourself incapable of surpassing a certain obstacle, remember that our calculator is here to help. Here are the few simple tips to know before you get started: First things first, you’ll have to enter the mathematical expression that you want to work on. Note that this graph looks just like the left hand graph, except that the variable is x instead of t. So you can find the derivativ… It can be used anywhere on your Smartphone, and it doesn’t require you to necessarily enter your own calculus problems as it comes with a library of pre-existing ones. You get many series of mathematical algorithms that come together to show you how things will change over a given period of time. Before moving to practice, you need to understand every formula first. Example problem: Evaluate the following integral using the fundamental theorem of calculus: How does fundamental theorem of calculus and chain rule work? The Fundamental Theorem of Calculus, Part 2 is a formula for evaluating a definite integral in terms of an antiderivative of its integrand. Combining the Chain Rule with the Fundamental Theorem of Calculus, we can generate some nice results. Not only is Mathway’s calculus calculator capable of handling simple operations and equations, but it can also solve series and other complicated calculus problems. Indeed, let f (x) be continuous on [a, b] and u(x) be differentiable on [a, b].Define the function There isn’t anything left or needed to be said about this app. Type in any integral to get the solution, free steps and graph This website uses cookies to ensure you get the best experience. To put it simply, calculus is about predicting change. That way, not only will you get the correct result, but you’ll also be able to know your flaws and focus on them while you’re practicing problem-solving. Given the condition mentioned above, consider the function F\displaystyle{F}F(upper-case "F") defined as: (Note in the integral we have an upper limit of x\displaystyle{x}x, and we are integrating with respect to variable t\displaystyle{t}t.) The first Fundamental Theorem states that: Proof If you’re stuck, do not hesitate to resort to our calculus calculator for help. Practice, Practice, and Practice! What makes our optimization calculus calculator unique is the fact that it covers every sub-subject of calculus, including differential. Then we need to also use the chain rule. There are several key things to notice in this integral. Let’s say it as it is; this is not a calculator for calculus, it is the best calculator for calculus. But just because they don’t use it in a direct way, that doesn’t imply that it’s not worth studying. Why bother using a scientific calculator to perform a simple operation such as measuring the surface area while you can simply do it following the clear instructions on our calculus calculator app? The abundance of the tools available at the user’s disposal is all anyone could ask for. Knowing how to handle numbers as they change during the time is indubitably a beneficial skill to acquire, and this is where the importance of learning calculus reveals itself. Some months ago, I had a silly board game with a couple of friends of mine. I thought about it for a brief moment and tried to analyze the situation saying that if you spend 20000$ a year on pet food that means that you’re paying around 60$a day. But they probably don't remember what it was like learning something like Calculus for the first time. Be it that you lost your scientific calculator, forgot it at home, can’t hire a tutor, etc. It’s free, it’s simple to use, and it has a lot to offer. Google Classroom Facebook Twitter The middle graph also includes a tangent line at xand displays the slope of this line. If it weren’t for my studies of drama, I wouldn’t have been able to develop the communication skills and have the level of courage that I’m on today. To give you a clearer idea, you should know that this app works as a: The variety of problems in which this calculator can be of assistance make it one of your best choices among all other calculus calculators out there. Message received. We often talk about the splendid job opportunities you can possibly get as a result. But if you truly want to have the ultimate experience using the app, you should sign up with Mathway. T. The correct answer I assume was around 300 to 500$ a year, but hey, I got very close to it. If you want to really learn calculus the right way, you need to practice problem-solving on a daily basis, as that’s the only way to improve and get better. 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At Mathway becoming a jack of all trades, but they put me on an alluring lane guess said. I came across a problem or make a simple calculation if students detest,. Afraid of becoming a jack of all trades, but what if instead of we have key things to in... Not a calculator for help doing so should be in anyone ’ also! A calculator for fundamental theorem of calculus chain rule calculator to our calculus work calculator is designed to give you the correct I. ) a a d f tdt dx the first place how crucial is that say, the Fundamental of... Has a lot to offer in anyone ’ s free, it ’ free... Solve a problem of Fundamental Theorem of calculus say that Differentiation and integration are inverse processes tedious. Mathematical format meeting and pulling a bulky scientific calculator, as our work! Clicks away, as pretty much every task can be lost with time as the problems too! Anyone ’ s say it as it is, already does a fantastic job at helping out students with daily! And their applications, you always have the “? ” button to use for help you have problems!
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# Bombing of Königsberg problem
A well-known problem in graph theory is the Seven Bridges of Königsberg. In Leonhard Euler's day, Königsberg had seven bridges which connected two islands in the Pregel River with the mainland, laid out like this:
And Euler proved that it was impossible to find a walk through the city that would cross each bridge once and only once. And more generally, that an Eulerian path does not exist for a graph with more than two nodes of odd degree.
World War 2 changed the topology of the city by destroying two of the bridges. (It also brought other changes such as the transfer of the territory from Germany to Russia, but this is not topologically relevant.) This lead to the similar but solvable problem of the Five Bridges of Kaliningrad.
I doubt that the British and Soviet air forces made creation of an Euler walk a priority when they were conducting their attacks. But if they had, one would have to criticize their inefficiency, because they could have achieved the same objective by bombing only one bridge:
Generalizing the problem:
What is the minimum number of edges that need to be removed from a graph so that the remaining edges form an Eulerian path?
My first conjecture was that it would be half the number of “extra” odd degree nodes. However, a simple X-shaped graph provides a counterexample: There are 4 odd nodes (and 1 even one), but two edges need to be removed, not just one.
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If you want a universal minimum (i.e. one that works for all graphs) then you'll be disappointed. Suppose there is some real number $0\le \alpha\le 1$ such that we can remove $\alpha$-percent of the edges of any graph to produce an Eulerian graph. Your example of the X-shaped graph generalized into a star-shaped graph with one central vertex and many leaves. For such a graph, all but two edges must be removed and so we must have $\frac{n-2}{n} \le \alpha$. Letting $n$ grow arbitrarily large, we must have $\alpha = 1$. There needs to be more restrictions on the problem I think. – EuYu Jun 13 '13 at 0:12
If there are $D$ vertices with odd degree, then you must remove at least $\frac{D-2}{2}$ vertices, though you will likely need to remove more than that. – Calvin Lin Jun 13 '13 at 0:14
@EuYu: Good point. I suppose what I meant is, "Given an arbitrary connected graph, is there a non-brute-force algorithm for finding the number of edges to remove?" – Dan Jun 13 '13 at 0:22
Sorry, I meant remove at least $\frac{D-2} { 2}$ edges (not vertices). – Calvin Lin Jun 13 '13 at 0:25
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Acknowledgements: Thanks to Ben S. for correcting some mistakes.
## 1.
### a)
See Question 8 of Chapter 3.
### b)
First, note that
$\begin{multline} S_{n}^{2}=\frac{1}{n-1}\sum_{i}\left(X_{i}-\bar{X}\right)^{2}=\frac{1}{n-1}\sum_{i}\left(X_{i}^{2}-2X_{i}\bar{X}+\bar{X}^{2}\right)\\ =\frac{1}{n-1}\sum_{i}X_{i}^{2}-\frac{n}{n-1}\bar{X}^{2}=c_{n}\frac{1}{n}\sum_{i}X_{i}^{2}-d_{n}\bar{X}^{2} \end{multline}$
where $$c_{n}\rightarrow1$$ and $$d_{n}\rightarrow1$$. By the WLLN, $$n^{-1}\sum_{i}X_{i}^{2}$$ and $$\bar{X}^{2}$$ converge, in probability, to $$\mathbb{E}[X_{1}^{2}]$$ and $$\mu^{2}$$. By Theorem 5.5 (d), $$c_{n}n^{-1} \sum_{i}X_{i}^{2}$$ and $$d_{n}\bar{X}^{2}$$ converge, in probability, to the same quantities. Lastly, by Theorem 5.5 (a), $$S_{n}^{2}$$ converges, in probability, to $$\mathbb{E}[X_{1}^{2}]-\mu^{2}=\sigma^{2}$$.
## 2.
Suppose $$X_{n}$$ converges to $$b$$ in quadratic mean. By Jensen’s inequality,
$\mathbb{E}\left[X_{n}-b\right]^{2} \leq\mathbb{E}\left[\left|X_{n}-b\right|\right]^{2} \leq\mathbb{E}[(X_{n}-b)^2] \rightarrow0$
Therefore, $$\mathbb{E}X_{n}\rightarrow b$$. Next, note that
$\mathbb{E}[(X_{n}-b)^2] =\mathbb{E}\left[X_{n}^{2}\right]-2b\mathbb{E}\left[X_{n}\right]+b^{2} =\mathbb{V}(X_{n})+\mathbb{E}[X_{n}]^{2}-2b\mathbb{E}\left[X_{n}\right]+b^{2}$
Taking limits of both sides reveals $$\lim_{n}\mathbb{V}(X_{n})=0$$. As for the converse, we can apply the limits $$\lim_{n}\mathbb{E}[X_{n}]=b$$ and $$\lim_{n}\mathbb{V}(X_{n})=0$$ directly to the equation above.
## 3.
Since the expectation of $$\bar{X}$$ is $$\mu$$ and the variance of $$\bar{X}$$ converges to zero, the desired result is obtained by an application of our findings in Problem 2.
Alternatively, taking a more direct approach, note that
$\begin{multline} \mathbb{E}\left[\left(\bar{X}-\mu\right)^{2}\right]=\mathbb{E}\left[\bar{X}^{2}-2\mu\bar{X}+\mu^{2}\right]=\mathbb{E}\left[\bar{X}^{2}\right]-\mu^{2}\\ =\frac{1}{n^{2}}\mathbb{E}\left[\sum_{i}X_{i}^{2}+\sum_{i,j\colon i\neq j}X_{i}X_{j}\right]-\mu^{2}=\frac{1}{n}\mathbb{E}\left[X_{1}^{2}\right]+\frac{n-1}{n}\mathbb{E}\left[X_{1}X_{2}\right]-\mu^{2}. \end{multline}$
Taking the limit,
$\mathbb{E}[(\bar{X}-\mu)^{2}] \rightarrow\mathbb{E}\left[X_{1}X_{2}\right]-\mu^{2} =\mathbb{E}\left[X_{1}\right]\mathbb{E}\left[X_{2}\right]-\mu^{2} =0.$
## 4.
Let $$\epsilon>0$$. For $$n$$ sufficiently large,
$\mathbb{P}(\left|X_{n}-0\right|>\epsilon)=\mathbb{P}(X_{n}>\epsilon)=\mathbb{P}(X_{n}=n)=1/n^{2}\rightarrow0$
and hence $$X_{n}$$ converges in probability. However,
$\mathbb{E}\left[\left(X_{n}-0\right)^{2}\right]=\mathbb{E}\left[X_{n}^{2}\right]\ge\mathbb{E}\left[X_{n}^{2}I_{\{X_{n}=n\}}\right]=n^{2}\mathbb{P}(X_{n}=n)=1$
and hence $$X_{n}$$ does not converge in quadratic mean.
## 5.
It is sufficient to prove the second claim since convergence in quadratic mean implies convergence in probability. Similarly to Problem 3, we can define $$Y_i = X_i^2$$ and apply our findings in Problem 2 to $$\bar{Y}$$.
Alternatively, taking a more direct approach, note that
$\left(\frac{1}{n}\sum_{i}X_{i}^{2}-p\right)^{2}=\frac{1}{n^{2}}\sum_{i}X_{i}^{4}+\frac{1}{n^{2}}\sum_{i,j\colon i\neq j}X_{i}^{2}X_{j}^{2}-\frac{2}{n}p\sum_{i}X_{i}^{2}+p^{2}.$
Taking expectations, and using the fact that $$X_i^k = X_i$$ and $$\mathbb{E} X_i = p$$,
$\begin{multline} \mathbb{E}\left[\left(\frac{1}{n}\sum_{i}X_{i}^{2}-p\right)^{2}\right]=\frac{1}{n}\mathbb{E}\left[X_{1}^{4}\right]+\frac{n-1}{n}\mathbb{E}\left[X_{1}^{2}\right]\mathbb{E}\left[X_{2}^{2}\right]-2p\mathbb{E}\left[X_{1}^{2}\right]+p^{2}\\ =\frac{1}{n}p+\frac{n-1}{n}p^{2}-p^{2}\rightarrow p^{2}-p^{2}=0. \end{multline}$
## 6.
Letting $$F$$ denote the CDF of a standard normal distribution, by the CLT,
$\begin{multline} \mathbb{P}\left(\frac{X_{1}+\cdots+X_{100}}{100}\geq68\right)\\ =\mathbb{P}\left(\frac{\sqrt{100}}{2.6}\left(\frac{X_{1}+\cdots+X_{100}}{100}-68\right)\geq0\right)\approx 1-F(0)=0.5. \end{multline}$
## 7.
Let $$f>0$$ be a function and $$\epsilon>0$$ be a constant. Then,
$\mathbb{P}(\left|f(n)X_{n}-0\right|>\epsilon)=\mathbb{P}(X_{n}>\epsilon/f(n))\leq\mathbb{P}(X_{n}\neq0)=1-\exp(-1/n)\rightarrow0.$
It follows that $$f(n) X_n$$ converges to zero in probability. Take $$f = 1$$ for Part (a) and $$f(n) = n$$ for (b).
## 8.
Letting $$F$$ denote the CDF of a standard normal distribution, by the CLT,
$\begin{multline} \mathbb{P}(Y<90)=\mathbb{P}(X_{1}+\cdots+X_{100}<90)\\ =\mathbb{P}\left(\frac{\sqrt{100}}{1}\left(\frac{X_{1}+\cdots+X_{100}}{100}-1\right)<-1\right)\approx F(-1) \end{multline}$
## 9.
Let $$\epsilon>0$$. Then,
$\mathbb{P}(\left|X_{n}-X\right|>\epsilon)\leq\mathbb{P}(X_{n}\neq X)=1/n\rightarrow0.$
Therefore, $$X_{n}$$ converges in probability (and hence in distribution) to $$X$$. On the other hand,
$\begin{multline} \mathbb{E}\left[\left(X-X_{n}\right)^{2}\right]=\mathbb{E}\left[\left(X-e^{n}\right)^{2}I_{\{X_{n}\neq X\}}\right]\\ =\mathbb{E}\left[1-2Xe^{n}+e^{2n}\right]\mathbb{P}(X_{n}\neq X)=\frac{1+e^{2n}}{n}\rightarrow\infty. \end{multline}$
## 10.
Since $$1\leq x^{k}/t^{k}$$ whenever $$x\geq t>0$$, it follows that
$\mathbb{P}(Z>t) =\mathbb{E}\left[I_{\{Z > t\}}\right] \leq\mathbb{E}\left[I_{\{Z > t\}} \left(\frac{Z}{t}\right)^k \right] \leq\frac{\mathbb{E}\left[I_{\{Z > t\}} \left|Z\right|^k \right]}{t^k}$
Therefore, since the distribution is symmetric,
$\mathbb{P}(|Z|>t) =2\mathbb{P}(Z>t) \leq\frac{\mathbb{E}\left[\left|Z\right|^{k}\left(I_{\{Z>t\}}+I_{\{Z<-t\}}\right)\right]}{t^{k}} \leq\frac{\mathbb{E}\left|Z\right|^{k}}{t^{k}}.$
Note that we only used symmetry in establishing the above and hence the result is more general than the problem description implies.
## 11.
First, note that $$X$$ is almost surely zero. Let $$\epsilon>0$$ and $$Z$$ be a standard normal random variable. Then,
$\mathbb{P}(\left|X_{n}-X\right|>\epsilon)=\mathbb{P}(\left|X_{n}\right|>\epsilon)=\mathbb{P}(\left|Z\right|>\epsilon\sqrt{n})\leq\frac{\mathbb{E}\left[Z^{2}\right]}{\epsilon^{2}n}=\frac{1}{\epsilon^{2}n}\rightarrow0.$
Therefore, $$X_{n}$$ converges in probability (and hence in distribution) to zero.
## 12.
Let $$F$$ be the CDF of an integer valued random variable $$K$$. Let $$k$$ be an integer. It follows that $$F(k)=F(k+c)$$ for all $$0\leq c<1$$. We use this observation multiple times below.
To prove the forward direction, suppose $$X_{n}\rightsquigarrow X$$. By definition, $$F_{X_{n}}\rightarrow F_{X}$$ at all points of continuity of $$F_{X}$$. Therefore,
$\begin{multline} \mathbb{P}(X_{n}=k)=F_{X_{n}}(k+1/2)-F_{X_{n}}(k-1/2)\rightarrow F_{X}(k+1/2)-F_{X}(k-1/2)\\=\mathbb{P}(X=k). \end{multline}$
To prove the reverse direction, suppose $$\mathbb{P}(X_{n}=k)\rightarrow\mathbb{P}(X=k)$$ for all integers $$k$$. Let $$j$$ be an integer and note that
$F_{X_{n}}(j)=\sum_{k\leq j}\mathbb{P}(X_{n}=k)\rightarrow\sum_{k\leq j}\mathbb{P}(X=k)=F_{X}(j)$
and hence $$X_{n}\rightsquigarrow X$$ as desired.
## 13.
First, note that
$F_{X_{n}}(x)=\mathbb{P}(\min\left\{ Z_{1},\ldots,Z_{n}\right\} \leq x/n)=1-\mathbb{P}(Z_{1}\geq x/n)^{n}.$
If $$x\leq0$$, it follows that $$F_{X_{n}}(x)=0$$. Otherwise,
$\begin{multline} \mathbb{P}(Z_{1}\geq x/n)^{n}=\left(1-\mathbb{P}(Z_{1}\leq x/n)\right)^{n}=\left(1-\int_{0}^{x/n}f(z)dz\right)^{n}\\ =\left(1-f(c_{n})\frac{x}{n}\right)^{n}=\left(e^{-f(c_{n})x/n}+O(n^{-2})\right)^{n}\rightarrow e^{-\lambda x}. \end{multline}$
Therefore, $$F_{X_{n}}(x)\rightarrow(1-e^{-\lambda x})I_{(0,\infty)}(x)$$ and hence $$X_{n}$$ converges in distribution to an $$\operatorname{Exp}(\lambda)$$ random variable.
## 14.
By the CLT
$\frac{\sqrt{n}}{\sigma}\left(\bar{X}-\mu\right)=\frac{\sqrt{n}}{1/\sqrt{12}}\left(\bar{X}-\frac{1}{2}\right)\rightsquigarrow N(0,1).$
Let $$g(x)=x^{2}$$ so that $$g^{\prime}(x)=2x$$. By the delta method,
$\frac{\sqrt{n}}{\left|g^{\prime}(\mu)\right|\sigma}\left(g(\bar{X})-g(\mu)\right)=\frac{\sqrt{n}}{1/\sqrt{12}}\left(Y_{n}-\frac{1}{4}\right)\rightsquigarrow N(0,1).$
## 15.
Define $$g:\mathbb{R}^{2}\rightarrow\mathbb{R}$$ by $$g(x)=x_{1}/x_{2}$$. Then, $$\nabla g(x)=(1/x_{2},-x_{1}/x_{2}^{2})^{\intercal}$$. Define $$\nabla_{\mu}=\nabla g(\mu)$$ for brevity. By the multivariate delta method,
$\sqrt{n}\left(Y_{n}-\frac{\mu_{1}}{\mu_{2}}\right)\rightsquigarrow N(0,\nabla_{\mu}^{\intercal}\Sigma\nabla_{\mu})=N(0,\Sigma_{11}/\mu_{2}^{2}-2\Sigma_{12}\mu_{1}/\mu_{2}^{3}+\Sigma_{22}\mu_{1}^{2}/\mu_{2}^{4}).$
## 16.
Let $$X_{n},X,Y\sim N(0,1)$$ be IID with $$X_{n}=Y_{n}$$. Trivially, $$X_{n}\rightsquigarrow X$$ and $$Y_{n}\rightsquigarrow Y$$. However, $$\mathbb{V}(X_{n}+Y_{n})=\mathbb{V}(2X_{n})=4$$ while $$\mathbb{V}(X+Y)=2$$ and hence $$X_{n}+Y_{n}$$ does not converge in distribution to $$X+Y$$.
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Timezone: »
Poster
Isometric 3D Adversarial Examples in the Physical World
yibo miao · Yinpeng Dong · Jun Zhu · Xiao-Shan Gao
@
Recently, several attempts have demonstrated that 3D deep learning models are as vulnerable to adversarial example attacks as 2D models. However, these methods are still far from stealthy and suffer from severe performance degradation in the physical world. Although 3D data is highly structured, it is difficult to bound the perturbations with simple metrics in the Euclidean space. In this paper, we propose a novel $\epsilon$-isometric ($\epsilon$-ISO) attack method to generate natural and robust 3D adversarial examples in the physical world by considering the geometric properties of 3D objects and the invariance to physical transformations. For naturalness, we constrain the adversarial example and the original one to be $\epsilon$-isometric by adopting the Gaussian curvature as the surrogate metric under a theoretical analysis. For robustness under physical transformations, we propose a maxima over transformation (MaxOT) method to actively search for the most difficult transformations rather than random ones to make the generated adversarial example more robust in the physical world. Extensive experiments on typical point cloud recognition models validate that our approach can improve the attack success rate and naturalness of the generated 3D adversarial examples than the state-of-the-art attack methods.
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# Processing in Detail with DUI¶
## Introduction¶
This tutorial follows the same steps as the command line tutorial Processing in Detail, but here processing will be driven graphically through the DIALS User Interface, DUI.
DUI is part of CCP4 and can be launched from ccp4i2 by selecting the relevant icon from within the “Integrate X-ray images” task folder.
## Tutorial data¶
The following example uses a Beta-Lactamase dataset collected using beamline I04 at Diamond Light Source, and reprocessed especially for these tutorials.
Hint
If you are physically at Diamond on the CCP4 Workshop, then this data is already available in your training data area. After typing module load ccp4-workshop you’ll be moved to a working folder, with the data already located in the tutorial-data/summed subdirectory.
The data is otherwise available for download from . We’ll only be using the first run of data in this tutorial, C2sum_1.tar, extracted to a tutorial-data/summed subdirectory.
### Import¶
When DUI starts up you will initially be presented by a window like the following.
At this stage you can resize various parts of the interface until you are comfortable with the layout, but you can’t do much else until the dataset is imported by DIALS. To do this, click on the “Select File(s)” button, navigate to the location of the images and select any one of them. DUI will automatically determine the filename template and will show that with a wildcard in the text box. If there are problems with this template it is possible to edit this before importing the images. Otherwise, just click the “Run” button to proceed.
What happens then is that that the metadata are read for all the images in the dataset. If these are consistent, then the dataset is imported and initial models for the “Beam”, “Scan” and “Detector” are created. The images are now also displayed within the “Image View” tab. You can adjust the contrast and colour scheme by controls under the “Palette Tuning” pull-down.
### Find Spots¶
The first “real” task in any processing using DIALS is the spot finding. To run this job, click on the “find” button at the upper-left of the window. When you do this you will see a new node will be created in the “History Tree”. This node is currently green, which indicates that it has not been run yet. By contrast, the import step is blue, which means this has been run. In general, it is always possible to navigate between each step of processing by clicking on the relevant position in the history. Advanced users will find this gives a great deal of control, allowing them to keep track of complex history, including parallel branches.
Spot-finding, like most of the other processing steps in DUI, presents user parameters at two levels of detail. The “Simple” tab contains the basic parameters that are the most commonly changed. The “Advanced” tab contains all of those again, plus other parameters that may be required for expert use with challenging data sets. In many cases, however, the default settings are fine.
Note that spot-finding is done on every image in the dataset. This means the job can take some time, but by default it will be run in parallel using multiple processors. To proceed, press the “Run” button below the input parameters.
Once the job is finished, the image viewer will display small green boxes around the pixels that have been marked as strong. It is also useful to click on the “Report View” tab and scroll down to the “Analysis of strong reflections”. This shows a graph of the number of strong spots found per image. In this case there is a pretty steady rate of around 150 spots found on each image. If instead we had seen the number of strong spots drop off over the dataset, or otherwise show large variability we would start to worry about issues such as radiation damage or a poorly-centred crystal.
The cyan button at the bottom left of the graph opens a help window with a description of how the appearance of this plot may be affected by various data collection issues. In the “Log Text” window you can see the text output from the dials.find_spots program, which also includes an ASCII-art version of this plot.
The default parameters for spot finding usually do a good job for Pilatus images, such as these. However they may not be optimal for data from other detector types, such as CCDs or image plates. If you have a case where spot-finding has gone badly, it may be helpful to debug using the dials.image_viewer and dials.reciprocal_lattice_viewer, which can be launched via buttons shown on the “External Tools” tab.
In particular, the effect of changing the spot-finding parameters can be explored interactively with the dials.image_viewer. The image mode buttons at the bottom of the “Settings” window allow a preview of how the parameters affect the spot finding algorithm. The final image, (“threshold”) is the one on which spots were found, so ensuring this produces peaks at real diffraction spot positions will give the best chance of success.
The second external viewer, the dials.reciprocal_lattice_viewer, displays the strong spots in 3D, after mapping them from their detector positions to reciprocal space. In a favourable case you should be able to see the crystal’s reciprocal lattice by eye in the strong spot positions. Some practice may be needed in rotating the lattice to an orientation that shows off the periodicity in reciprocal lattice positions.
Although the reciprocal spacing is visible, in this data, there are clearly some systematic distortions. These will be solved during indexing.
### Indexing¶
The next step will be indexing of the strong spots. Click on the “Index” button to move on to this step, and form a new node in the history tree. Here we see that the simple parameters allow only to select between different “Indexing Methods”, the default of which is the 3D FFT algorithm. The other options are the 1D FFT (DPS) algorithm and a special version of the 3D FFT called real_space_grid_search, which is particularly useful for narrow wedges containing multiple lattices, but requires a known cell and space group to be set under the “Advanced” parameters. If we do know the cell and space group, these can also be set as hints for either of the other two indexing algorithms. This can help in difficult cases and will be used to constrain the lattice during refinement. Otherwise indexing and refinement will be carried out in the primitive lattice using space group P1.
In this case, keep the method set to the default fft3d and click “Run” to start the indexing job. Once the job has finished running, you will see that the experimental models have now been completed with a “Crystal” model.
Now let’s click through the tabs of output. First, on the image viewer you will now see that indexed strong spots are assigned Miller indices. If you also click on the “Predictions” checkbox, under “Reflection Type” you will in addition see centroid positions and Miller indices for all predicted reflections, not just the strong spots.
Moving to the “Log Text” tab, it is worth reading through the output to understand what the indexing program has done. Inspecting the beginning of the log shows that the indexing step is done at a resolution lower than the full dataset; 1.84 Å:
Found max_cell: 94.4 Angstrom
Setting d_min: 1.84
FFT gridding: (256,256,256)
The resolution limit of data that can be used in indexing is determined by the size of the 3D FFT grid, and the likely maximum cell dimension. Here we used the default 256³ grid points. These are used to make an initial estimate for the unit cell parameters.
What then follows are ‘macro-cycles’ of refinement where the experimental model is first tuned to get the best possible fit from the data, and then the resolution limit is reduced to cover more data than the previous cycle. 16 parameters of the diffraction geometry are tuned: 6 for the detector, one for beam angle, 3 crystal orientation angles and the 6 triclinic cell parameters. At each stage only 36000 reflections are used in the refinement job. In order to save time, a subset of the input reflections are used - by default using 100 reflections for every degree of the 360° scan.
We see that the first macrocycle of refinement makes a big improvement in the positional RMSDs:
-------------------------------------------------
| Step | Nref | RMSD_X | RMSD_Y | RMSD_Phi |
| | | (mm) | (mm) | (deg) |
-------------------------------------------------
| 0 | 36000 | 0.54219 | 0.51246 | 0.1574 |
| 1 | 36000 | 0.22912 | 0.26037 | 0.13876 |
| 2 | 36000 | 0.10706 | 0.13611 | 0.12523 |
| 3 | 36000 | 0.057653 | 0.062184 | 0.10585 |
| 4 | 36000 | 0.051918 | 0.053799 | 0.10378 |
| 5 | 36000 | 0.051683 | 0.053837 | 0.10385 |
| 6 | 36000 | 0.051676 | 0.053843 | 0.10385 |
| 7 | 36000 | 0.051676 | 0.053843 | 0.10385 |
-------------------------------------------------
Second and subsequent macrocycles are refined using the same number of reflections, but after extending to higher resolution. The RMSDs at the start of each cycle start off worse than at the end of the previous cycle, because the best fit model for lower resolution data is being applied to higher resolution reflections. As long as each macrocyle shows a reduction in RMSDs then refinement is doing its job of extending the applicability of the model out to a new resolution limit, until eventually the highest resolution strong spots have been included. The final macrocycle includes data out to 1.30 Å and produces a final model with RMSDs of 0.050 mm in X, 0.049 mm in Y and 0.104° in φ, corresponding to 0.29 pixels in X, 0.28 pixels in Y and 0.21 image widths in φ.
Despite the high quality of this data, we notice from the log that at each macrocycle there were some outliers identified and removed from refinement as resolution increases. Large outliers can dominate refinement using a least squares target, so it is important to be able to remove these. More about this is discussed below in Refinement. It’s also worth checking the total number of reflections that were unable to be assigned an index:
------------------------------------
| Imageset | #indexed | #unindexed |
------------------------------------
| 0 | 102151 | 1629 |
------------------------------------
because this can be an indication of poor data quality or a sign that more care needs to be taken in selecting the indexing parameters.
Now the “Report View” contains more information than just after spot-finding. The “Spot count per image” plot also contains information about the number of indexed spots. In addition there are heat maps giving information about the positions of indexed and unindexed spots. Here we see that most of the unindexed spots are found in the region around the rotation axis. The “Analysis of reflection centroids” plots provide lots of detail regarding how well the predicted spot positions match the observed positions, both in image space and as a function of the position within the rotation scan.
After indexing it can be useful to inspect the reciprocal lattice again under the “External Tools”. Now indexed/unindexed spots are differentiated by colour, and it is possible to see which spots were marked by dials.refine as outliers. If you have a dataset with multiple lattices present, it may be possible to spot them in the unindexed reflections.
In this case, we can see that the refinement has clearly resolved whatever systematic error was causing distortions in the reciprocal space view, and the determined reciprocal unit cell fits the data well:
### Bravais Lattice Refinement¶
Since we didn’t know the Bravais lattice before indexing, we can now determine likely candidates - by taking the results of the P1 autoindexing, and running refinement with all of the possible Bravais settings applied. You can then choose your preferred solution. This step is accessed by the “Lattice” button on the left of the DUI window. As before, run this without altering any of the defaults, as they are suitable for the majority of data sets.
Once the job has run, a window will pop up containing scoring data and the unit cell for each Bravais setting.
The scores for each setting include max δ (a metric fit measured in degrees), RMSDs (in mm), and the best and worse correlation coefficients for data related by symmetry elements (the symmetry elements implied by the lowest symmetry space group from the Bravais setting). This uses the raw spot intensity measurement from the spot-finding procedure (uncorrected and unscaled) but provides a very useful check to see if the data does appear to adhere to the proposed symmetry operators.
DIALS uses an heuristic to determine which solutions are acceptable or not, indicated on this window by either a green highlighted “Y” or a red highlighted “N”. In addition, the single “best” solution (the highest symmetry of the acceptable results) is pre-selected (highlighted in blue). To pick this solution, simply click “OK” while the chosen solution is highlighted. This will automatically apply the symmetry constraints and will reindex the reflections ready for further refinement.
### Refinement¶
The model is already refined during indexing, but we can also add explicit refinement steps here. This is beneficial because it will use all reflections in refinement rather than a subset, uses a more sophisticated outlier rejection algorithm and will later allow us to fit a scan-varying model of the crystal.
We start by refining a static model including the monoclinic constraints from our chosen reindexed solution. For this we leave “Scan Varying Refinement” as “False”. There are various choices of outlier rejection algorithm allowed by refinement. The default selection of “auto” will choose the “mcd” algorithm for a rotation scan like this, which performs outlier rejection on the X, Y and φ residuals simultaneously, taking into account the multivariate nature of the data. This is appropriate for the ‘polishing up’ stage of refinement, whereas before during indexing the rougher, but less computationally expensive “tukey” option was used instead.
As before, click “Run” to start the job. The “Log Text” is familiar from the indexing stage. We see that all strong reflections were used in refinement, providing a small reduction in RMSDs. However, the refined model is still static over the whole dataset. We would like to do an additional refinement job at this point, to fit a more sophisticated model for the crystal, allowing small misset rotations to occur over the course of the scan. There are usually even small changes to the cell dimensions (typically resulting in a net increase in cell volume) caused by exposure to radiation during data collection. To account for both of these effects we can extend our parameterisation to obtain a smoothed scan-varying model for both the crystal orientation and unit cell.
This means running a further refinement job starting from the output of the previous job. To do that, note that the current “refine” node is a completed job and the parameters we entered are now greyed-out and cannot be edited. To do a second refinement starting from this point we simply click on the “refine” button again, opening a new green node in the history tree. Here we can select “Scan Varying Refinement” as “True” and click “Run” again to start the job.
The log output shows a decrease in each dimension, but especially in Y.
-------------------------------------------------
| Step | Nref | RMSD_X | RMSD_Y | RMSD_Phi |
| | | (mm) | (mm) | (deg) |
-------------------------------------------------
| 0 | 90631 | 0.046367 | 0.047677 | 0.10256 |
| 1 | 90631 | 0.043558 | 0.040853 | 0.10123 |
| 2 | 90631 | 0.041581 | 0.039963 | 0.10065 |
| 3 | 90631 | 0.041011 | 0.039795 | 0.10033 |
| 4 | 90631 | 0.040882 | 0.039801 | 0.10012 |
| 5 | 90631 | 0.040855 | 0.039803 | 0.10002 |
| 6 | 90631 | 0.04085 | 0.039803 | 0.1 |
| 7 | 90631 | 0.040849 | 0.039804 | 0.099999 |
-------------------------------------------------
The final RMSDs are less than a quarter of a pixel in both X and Y, and just under a fifth of a pixel in φ. This is about as good as we can expect from a high quality Pilatus data set such as this.
In the “Report View” we can now see plots of how the cell and orientation changes during the scan. The smoothness of these plots is guaranteed by the smoother model used by dials.refine. However, we are satisfied that this model is sufficient to match real changes present in the dataset because of the very low overall RMSDs.
Other useful plots in the report are:
• Difference between observed and calculated centroids vs phi, which shows how the average residuals in each of X, Y, and φ vary as a fuction of φ. If scan-varying refinement has been successful in capturing the real changes during the scan then we would expect these plots to be straight lines.
• Centroid residuals in X and Y, in which the X, Y residuals are shown directly. The key point here is to look for a globular shape centred at the origin.
• Difference between observed and calculated centroids in X and Y, which show the difference between predicted and observed reflection positions in either X or Y as functions of detector position. From these plots it is very easy to see whole tiles that are worse than their neighbours, and whether those tiles might be simply shifted or slightly rotated compared to the model detector.
In this tutorial, we see no overall increase in all three cell parameters. If significant cell volume increases had been observed that might be indicative of radiation damage. However we can’t yet conclude that there is no radiation damage from the lack of considerable change observed.
### Integration¶
After the refinement is done the next step is integration. Click on the “integrate” button to move to this job. Mostly, the default parameters are fine for Pilatus data, which will perform XDS-like 3D profile fitting while using a generalized linear model in order to fit a Poisson-distributed background model. As with spot-finding, the number of processes can be set >1 to speed the job up (but DUI will have selected a suitable default). Click “Run” to start integration. This is the most computationally-demanding stage of processing, so it will take a while to complete.
Checking the “Log Text” output, we see that after loading in the reference reflections, new predictions are made up to the highest resolution at the corner of the detector. This is fine, but if we wanted to we could have adjusted the resolution limits using parameters d_min and d_max under prediction in the “Advanced” parameters tab. The predictions are made using the scan-varying crystal model from the previous step. As this scan-varying model was determined in advance of integration, each of the integration jobs is independent and we can take advantage of true parallelism during processing.
The profile model is then calculated from the reflections file resulting from refinement. First reflections with a too small ‘zeta’ factor are filtered out. This essentially removes reflections that are too close to the spindle axis. In general these reflections require significant Lorentz corrections and as a result have less trustworthy intensities anyway. From the remaining reflection shoeboxes, the average beam divergence and reflecting range is calculated, providing the two Gaussian width parameters $$\sigma_D$$ and $$\sigma_M$$ used in the 3D profile model.
Following this, independent integration jobs are set up. These jobs overlap, so reflections are assigned to one or more jobs. What follows are blocks of information specific to each integration job.
After these jobs are finished, the reflections are ‘post-processed’, which includes the application of the LP correction to the intensities. Then summary tables are printed giving quality statistics first by frame, and then by resolution bin.
On the “Image View” tab we can now see integration “shoeboxes” around the spots, not just tight boxes around the strong pixels. If all stages up to this point have gone well, then the boxes should be centred on the strong pixels and should extend beyond the strong pixels to include pixels used for local background level determination.
The “Report View” now contains additional plots under the “Analysis of reflection intensities” and “Analysis of reference profiles” sections. It is worth checking through these, particularly paying attention to the following:
• Reflection and reference correlations binned in X/Y. These are useful companions to the plots of centroid residual as a function of detector position above. Whereas the above plots show systematic errors in the positions and orientations of tiles of a multi-panel detector, these plots indicate what effect that (and any other position-specific systematic error) has on the integrated data quality. The first of these plots shows the correlation between reflections and their reference profiles for all reflections in the dataset. The second shows only the correlations between the strong reference reflections and their profiles (thus these are expected to be higher and do not extend to such high resolution).
• Distribution of I/Sigma vs Z. This reproduces the $$\frac{I}{\sigma_I}$$ information versus frame number given in the log file in a graphical form. Here we see that $$\frac{I}{\sigma_I}$$ is fairly flat over the whole dataset, which we might use as an indication that there were no bad frames, not much radiation damage occurred and that scale factors are likely to be fairly uniform.
At this point we could export the integrated data set in MTZ format, however we will continue with this tutorial to demonstrate scaling within DIALS.
### Checking the symmetry¶
After integration we can return to our hypothesis of the space group of the crystal. Although we made an assessment of that when we chose a Bravais lattice after indexing, we now have better, background-subtracted, values for the intensities, and for all reflections, not just the strong spots. So, it is prudent to repeat the assessment to see if there is any indication that our initial assessment should be revised.
This job is run using the “symmetry” button. We will again run with default settings only. Once the job is finished, check the “Log Text” output. The most important part here is the table printed at the end:
Scoring all possible sub-groups
---------------------------------------------------------------------------------------------
Patterson group Likelihood NetZcc Zcc+ Zcc- CC CC- delta Reindex operator
---------------------------------------------------------------------------------------------
C 1 2/m 1 *** 0.909 9.72 9.72 0.00 0.97 0.00 0.0 -a,b,-c
P -1 0.091 0.11 9.77 9.66 0.98 0.97 0.0 -x-y,-x+y,-z
---------------------------------------------------------------------------------------------
Best solution: C 1 2/m 1
Here we see clearly that the best solution is given by C 1 2/m 1, with a high likelihood, in agreement with the result from dials.refine_bravais_settings. As we remain confident with this choice, we now continue to scaling.
### Scaling¶
Before the data can be reduced for structure solution, the intensity values must be corrected for experimental effects which occur prior to the reflection being measured on the detector. These primarily include sample illumination/absorption effects and radiation damage, which result in symmetry-equivalent reflections having unequal measured intensities (i.e. a systematic effect in addition to any variance due to counting statistics). Thus the purpose of scaling is to determine a scale factor to apply to each reflection, such that the scaled intensities are representative of the ‘true’ scattering intensity from the contents of the unit cell.
During scaling, a scaling model is created, from which we derive scale factors for each reflection. By default, three components are used to create a physical model for scaling (model=physical), in a similar manner to that used in the program aimless. This model consists of a smoothly varying scale factor as a function of rotation angle, a smoothly varying B-factor to account for radiation damage as a function of rotation angle and an absorption surface correction, dependent on the direction of the incoming and scattered beam vector relative to the crystal.
Let’s scale the Beta-lactamase dataset, after setting a resolution cutoff (d_min) of 1.4. This job is created by clicking the “scale” button. Enter 1.4 in the d_min field in the “simple” tab and click “Run” to start the job.
As can be seen from the “Log Text”, 70 parameters are used to parameterise the scaling model for this dataset. A subset of reflections are selected to be used in scaling model minimisation, which helps to speed up the algorithm (the model is used to calculate scales for all reflections at the end). Outlier rejection is performed at several stages, as outliers have a disproportionately large effect during scaling and can lead to poor scaling results. During scaling, the distribution of the intensity uncertainties are also analysed and an error model is optimised to transform the intensity errors to an expected normal distribution. An error estimate for each scale factor is also determined based on the covariances of the model parameters. At the end of the output, a table and summary of the merging statistics are presented, which give indications of the quality of the scaled dataset.
----------Overall merging statistics (non-anomalous)----------
Resolution: 69.19 - 1.40
Observations: 274799
Unique reflections: 41140
Redundancy: 6.7
Completeness: 94.11%
Mean intensity: 80.7
Mean I/sigma(I): 15.4
R-merge: 0.065
R-meas: 0.071
R-pim: 0.027
#### Inspecting the results¶
To see what the scaling is telling us about the dataset, plots of the scaling model should be viewed. These are visible within the “Report View” tab, at the bottom under “Analysis of scaling model”.
What is immediately apparent is the periodic nature of the scale term, with peaks and troughs 90° apart. This indicates that the illumated volume was changing significantly during the experiment: a reflection would be measured as twice as intense if it was measured at rotation angle of ~120° compared to at ~210°. The absorption surface also shows a similar periodicity, as may be expected. What is less clear is the form of the relative B-factor, which also has a periodic nature. As a B-factor can be understood to represent radiation damage, this would not be expected to be periodic, and it is likely that this model component is accounting for variation that could be described only by a scale and absorption term. To test this, we can repeat the scaling process but turn off the decay_term.
To do this in DUI, click “Retry” to set up a new scaling job continuing from the successful symmetry-determination step. This time, enter the “Advanced” tab and under parameterisation change the value of decay_term to False. Now click “Run” to start the job.
----------Overall merging statistics (non-anomalous)----------
Resolution: 69.19 - 1.40
Observations: 274578
Unique reflections: 41140
Redundancy: 6.7
Completeness: 94.11%
Mean intensity: 76.6
Mean I/sigma(I): 16.0
R-merge: 0.064
R-meas: 0.070
R-pim: 0.027
By inspecting the statistics in the output, we can see that removing the decay term has had the effect of causing around 200 more reflections to be marked as outliers (taking the outlier count from 0.72% to 0.80% of the data), while improving some of the R-factors and mean I/sigma(I). Therefore it is probably best to exclude the decay correction for this dataset. Other options which could be explored under the “Advanced” tab are the numbers of parameters used for the various components, for example by changing the scale_interval, or by adjusting the outlier rejection criterion with a different outlier_zmax.
### Exporting as MTZ¶
Once we are happy with the results from scaling, the data can be exported as an unmerged mtz file, for further symmetry analysis with pointless or to start structural solution.
To do this, click on the “export” button. This gives the option of an mtz output name and the option to output scaled intensities. Make sure that box is ticked otherwise the exported MTZ will only contain intensities from integration.
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Albion College Mathematics and Computer Science Colloquium
Title: Math in the Madness Speaker: Luke Walker, 12 Mathematics Major Abstract: Learning math can be unexciting for the average teenager. To spark an interest in this amazing subject, teachers must find ways to connect math to the interests of the students. My passion in math and basketball, so I combined these two into a lesson on the statistics behind the NCAA Basketball Tournament in March. Students will explore the field of the tournament to determine their winner based on the calculation they perform each round. Through all the fun, students practice calculating percentage, average, mean, median, and mode while also learning to think outside of the norm. Location: Palenske 227 Date: 4/26/2012 Time: 3:50 PM
@abstract{MCS:Colloquium:LukeWalker12:2012:4:26,
author = "{Luke Walker, 12}",
title = "{Math in the Madness}",
address = "{Albion College Mathematics and Computer Science Colloquium}",
month = "{26 April}",
year = "{2012}"
}
`
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# How does Geothermal Energy make Electricity
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495 views
Most of the electricity power plants use geothermal energy to generate electricity of hydroelectric energy and are reducing the use of fossil fuels to generate electricity. One of the widely used energy to make electricity is known as the geothermal energy.
We would be discussing about the process of converting geothermal energy into electricity. So how does geothermal energy make electricity?
When you look into geothermal plants, you would find that their are different techniques to convert geothermal energy into electricity:
-Dry Steam
-Flash Steam
-Binary cycle
The common factor among all these three techniques is that they use steam turbines for the generation of electricity.
## Conversion of geothermal energy into electricity:
The water which is used in the geothermal power plants is heated and then sent through pipes with force to rotate the steam turbines. The heat of the water is used to rotate the turbines and also the circular motion of the water around the turbines also rotates the turbines. The steam which is generated from the water due to heat and rapid movement helps in the motion of the turbines as well.
The turbines were placed around the magnet which results in electromagnetic induction. This electromagnetic induction coarses the flow of electrons from one place to another which causes the current to flow.
Under normal circumstances, the heat of the earth is used to heat the water instead of using any fuels to heat the water.
That is why, the process of creating electricity using geothermal energy is completely natural. It does not make use of any fossil fuels.
Before the water is passed through the turbines, it is passed through the rock catchers because if there is any rock in the water which is sent through the turbines, there can be a great damage to the turbines which are used to generate electricity.
Depending on the type of the geothermal plant out of three which we have discussed above, the procedure could be modified slightly according to the type of the techniques used.
So, as you can see geothermal energy is one of the cleanest forms of electricity generation which can be used. It helps in ensuring that there is no pollution of any kind during the generation of electricity using geothermal energy.
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#### Problem 18E
The velocity graph of a car accelerating from rest to a speed of 120 km/h over a period of 30 seconds is shown. Estimate the distance traveled during this period.
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# The probability that the team A will win by scoring in the last ball.
Two teams A and B play an one-day cricket match.The match is in the last-over and the situation of the match is as follows:
Four balls are to be bowled by the team B;no wides or no-balls are bowled;the team A has to score 16 runs to win.Runs that can be scored off a ball are 0,1,2,3,4,5 and 6.Find the probability that the team A will win by scoring in the last ball.
Since the 4 balls are left,so the probability of winning in the last ball has 6 cases.
(1)15 runs in first three balls and 1 run in the last ball.
(2)14 runs in first three balls and 2 run in the last ball.
(3)13 runs in first three balls and 3 run in the last ball.
(4)12 runs in first three balls and 4 run in the last ball.
(5)11 runs in first three balls and 5 run in the last ball.
(6)10 runs in first three balls and 6 run in the last ball.
I let $x_1,x_2,x_3$ as the run in the fourth last,third last,second last balls respectively such that $0\leq x_1,x_2,x_3\leq 6$.
$(1)x_1+x_2+x_3=15\Rightarrow y_1+y_2+y_3=3$,here $y_1,y_2,y_3\geq 0$ Probability in this case is $\binom{5}{2}\times \frac{1}{6}$
In the same way,taking all six cases,$\binom{5}{2}\times \frac{1}{6}+\binom{6}{5}\times \frac{1}{6}+\binom{7}{6}\times \frac{1}{6}+\binom{8}{7}\times \frac{1}{6}+\binom{9}{8}\times \frac{1}{6}+\binom{10}{9}\times \frac{1}{6}$
But my answer is coming wrong.Correct answer given is $\frac{405}{7^4}$.Is my logic totally wrong.Where have i gone wrong.Please help me.
Assuming the second to sixth terms in your final equation are just typos, there are two mistakes in your solution.
(1) Your understanding of the question is wrong. This is not really a mathematical error though. $A$ wins for any score greater than or equal to $16$ instead of only $16$ due to the nature of a scoring based game.
For example, your first case "15 runs in first three balls and $1$ run in the last ball." should actually be "15 runs in first three balls and $1$ ~ $6$ run in the last ball." And so on.
(2) The way you let $y = 6-x$ is a great way of simplifying calculations, however you forgot to restrict $y \leq6$ as well for the last two cases.
When $y_1+y_2+y_3=7$ or larger $y$ is actually possible to be 7 and generate negative $x$ so you need to substract these cases.
For sum of $y$ = $7$, there are {$7,0,0$} {$0,7,0$} {$0,0,7$}, three cases.
For sum of $y$ = $8$, there are permutations of {$8,0,0$} and permutations of {$7,1,0$}, totally nine cases.
Overall there are
${5\choose2}\times6 +{6\choose2}\times5 + {7\choose2}\times4 + {8\choose2}\times3 + ({9\choose2}-3)\times2 + ({10\choose2}-9)\times1$ = 405 cases.
Hence the possibility is $\frac{405}{7^4}$.
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# Speed with which m must move for M to stay at rest
1. Oct 6, 2008
### shimizua
1. The problem statement, all variables and given/known data
A 2.80 kg mass, m, on a frictionless table is moving in a circle with radius 0.440 m at a constant speed. m is attached to a 5.40 kg mass, M, by a cord through a hole in the table. Find the speed with which m must move for M to stay at rest.
for this one i am completely lost.
2. Oct 6, 2008
### Rake-MC
F=ma
You can apply this for the force pulling down on the mass at the end of the cord under the table
The other force is centripetal force caused by the rotational motion of the mass on the table.
This is governed by $$F = \frac{mv^2}{r}$$
Let those two formulae equal each other and you have your answer.
3. Oct 6, 2008
### shimizua
so it would be ma=mv^2/r? and then the m's would cancel out and you be be left with a=v^2/r. but then i only know r and not v or a so how do i find that?
4. Oct 7, 2008
### Rake-MC
no... because the masses in the equations are not the same..
Mass on the table = m
Mass under table = M
so you have Ma = mv^2/r
where M does not equal m
5. Oct 7, 2008
### shimizua
but what about having a and v since i dont have either of those, how do i find that
6. Oct 7, 2008
### Rake-MC
Don't be so sure. The Ma side of the equation applies to the block hanging under the table.
What force do you think is pulling it towards the ground? This gives you a.
And I know you don't have v, this is what you're solving for, you want to re-arrange your equation such that it is in the form of v = [...]
7. Oct 7, 2008
### shimizua
so would u use gravity for a? or like gravity times mass?
8. Oct 7, 2008
### Rake-MC
gravity is a measure of acceleration (ms-2). If you multiply it by mass, you have ma = F.
9. Oct 7, 2008
### shimizua
lol, sorry for all of the questions but i am still a little lost.
so i could just do a/g=m2/(m1+m2) to get a right?
10. Oct 7, 2008
### Rake-MC
Okay yes you are still a little lost.
we have F = Ma and F = mv2/r
We are given M, m and r and we are solving for v.
we will let the two equations equal each other. The only thing that remains in a.
Now notice that the left hand side of the equation represents the mass under the table. The acceleration affecting this mass IS gravity. Therefore a = g.
11. Oct 7, 2008
### shimizua
thank you so much. hahaha i am def going to write this down so i will not have to ask about it again.
12. Oct 7, 2008
### Rake-MC
Haha don't worry about it, it's what we're here for.
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Article Contents
Article Contents
# Existence and blow up of small amplitude nonlinear waves with a negative potential
• Consider a nonlinear wave equation in three space dimensions with zero mass together with a negative potential. If the potential is sufficiently short-range, then it does not alter the global existence of small-amplitude solutions. On the other hand, if the potential is sufficiently large, it will force some solutions to blow up in a finite time.
Mathematics Subject Classification: 35L70, 35B35.
Citation:
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# What is the molar mass of CdI_2?
Well, the molar mass of $H g {I}_{2} = 545.40 \cdot g \cdot m o {l}^{-} 1$.
$\left(2 \times 126.9 + 200.6 \cdot g \cdot m o {l}^{-} 1\right) \cdot g \cdot m o {l}^{-} 1$ $=$ $454.40 \cdot g \cdot m o {l}^{-} 1$.
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1. Apr 15, 2010
### triac
1. The problem statement, all variables and given/known data
Hi!
I'm stuck on this one too.
"A rock contains the radioactive isotope Rb-87. A piece of this rock contained 2,05*10^20 Rb-87 atoms and 8,25*10^20 decay products per kg. The half-life of Rb-87 is 47*10^9 years. How old is the rock?"
2. Relevant equations
$$N(t)=N_0(\frac{1}{2})^{t/T_{1/2}}$$
3. The attempt at a solution
No nuclei (or atoms) disappear, they only decay. Therefore, the initial number of Rb-87 atoms must have been 2,05*10^20+8,25*10^20 per kg. Let's assume that we have a sample of 1 kg (the mass will change with decay, but not the number of atoms, so it doesn't matter).
Now, let x be its age. Then we can write
2,05*10^20=(2,05*10^20+8,25*10^20)(1/2)^(t/T_{1/2}) which gives us that x is approximately 22,9*10^9 years. However, in the key it says 1,7*10^9 years. What's wrong here?
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
2. Apr 16, 2010
### Rajini
Is your question is fully written?
Please don't use comma or dot in numbers..? it is confusing..(Germany it is common, using comma for dot)
in your case 't' should be age..but why you introduced 'x'..
is this problem is solved?
$$47\times10^9 {\rm or} 4.7\times10^9$$?
Last edited: Apr 16, 2010
3. Apr 16, 2010
### py1sl
I dont think you have all the data written down correctly, if 1.7*10^9 is the answer. If you start with 10.3 atoms after one half life there would be 5.15 atoms left and 47*10^9 years would have past but you have fewer atoms left and an answer of less time which is clearly wrong. Unless i haven't understood the question
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## Lesson 2 Problem Solving Practice Slope Answer Key
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Lesson 22 Solve Word Problems Involving Conversions Lesson 22 Solve Word Problems Involving Conversions Students read a word problem and answer a series of questions designed to guide them through converting units in the same measurement system. 2 Extra Practice Lesson 7. (ISBN 978-1-939246-02-8) This is a HARD COPY that will be mailed to your shipping address. Problem 7: A golden-colored cube is handed to you. Course 3 • Chapter 2 Equations in One Variable 31 Lesson 4 Homework Practice Solve Equations with Variables on Each Side Solve each equation. In this part of Lesson 2, we will focus on the first type of problem - sometimes referred to as horizontally launched projectile problems. LESSON 19-1 Practice and Problem Solving: A/B 1. x 2 y 4 Word Problem Practice Solving Systems of Equations By Graphing 1. The exercises are designed to aid your study of mathematics by reinforcing important mathematical skills needed to succeed in the everyday world. On this page you can read or download lesson 7 homework practice ratio and rate problems answer key in PDF format. From 2 to 13 years, the growth rate for children is generally Practice A 12-5 Direct. Lesson 8 skills practice slope Lesson 8 skills practice slope. 18 months at an interest rate of 6. Brittney’s Lesson 3-1 The. In basic geometry, we examine. LESSON 10-2 Practice and Problem Solving: A/B 1. (−1, −3) 2. y w xM 6a5d el 4wPiztDhV eIXnCfliDnXiztde o tA5l BgWedb4rMa0 U1D. Given two points on a line, (x 1, y 1) and (x 2, y 2), we can find the slope (m) of the line using the following formula: m = (y 2 - y 1)/(x 2. Students, teachers, parents, and everyone can find solutions to their math problems instantly. Problem Solving Slope-Intercept Form The cost of food for an honor roll dinner is $300 plus$10 per student. ) = 2 • 4 • 4 (Rewrite 16 as the product of 2 numbers. How many are there? 2. 6E Lesson 6 – 3. 9D Lesson 7 – 3. We will consider problems involving lines and then problems involving triangles. Lesson 3: Video Multiplying and Dividing Rational Numbers Lesson Quiz 2-3. Describe the solution set of the compound inequality 7. Problem 7: A golden-colored cube is handed to you. Problem 5: What percent of 56 is 14? Identify: 56 is the whole and will replace OF in our proportion. LESSON 4-2 Practice and Problem Solving: A/B 1. Long drives, she travels 3 miles. The graph of f(x) = 7x + 2 is translated up 2 units and the line is steeper by a factor of 7. We are dedicated in building the best dynamic Math Worksheets for our users. Key Curriculum Press Lesson 0. 4 Fitting a Line to Data 5. Algebra 1 answer key, scott foresman math, online rational equations calculator. Make sense of problems and persevere in solving them. Course 3 • Chapter 2 Equations in One Variable 31 Lesson 4 Homework Practice Solve Equations with Variables on Each Side Solve each equation. Simplify the problem by raising 2 to the fifth power. Home Link 2 1 Math Masters, p. A plumber charges $45 plus$39. If you don't see any interesting for you, use our search form on bottom ↓. 13x = 32 + 5x 3. How many inches long is the. Slope 2 3; 6, 7 c. Practice and Problem Solving: Tell whether each function is written in standard form. Basic Lesson Demonstrates the concept of variations and the use of cross multiplication. Melissa Cueto M. Make sure to show your work. That’s my final answer, and it makes sense. One quarter is 191 200 inch in diameter. 7) (3, 0), (−11 , −15) 8) (19 , −2), (−11 , 10) -1- ©l q2Z0 u1u2 m YK4uet LaH XSSoVfCttw7aRrQed bLPLpCH. com is the best site to check-out!. y ˚ 8 9 ˛ x b. Because it is so important, robotic security cameras are placed at the locations of the dots in the figure.
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## Design IIR Band-Reject Filters
January 17, 2018
In this post, I show how to design IIR Butterworth band-reject filters, and provide two Matlab functions for band-reject filter synthesis. Earlier posts covered IIR Butterworth lowpass [1] and bandpass [2] filters. Here, the function br_synth1.m designs band-reject filters based on null frequency and upper -3 dB frequency, while br_synth2.m designs them based on lower and upper -3 dB frequencies. I’ll discuss the differences between the two approaches later in this...
## Design IIR Bandpass Filters
In this post, I present a method to design Butterworth IIR bandpass filters. My previous post [1] covered lowpass IIR filter design, and provided a Matlab function to design them. Here, we’ll do the same thing for IIR bandpass filters, with a Matlab function bp_synth.m. Here is an example function call for a bandpass filter based on a 3rd order lowpass prototype:
N= 3; % order of prototype LPF fcenter= 22.5; % Hz center frequency, Hz bw= 5; ...
## Design IIR Butterworth Filters Using 12 Lines of Code
While there are plenty of canned functions to design Butterworth IIR filters [1], it’s instructive and not that complicated to design them from scratch. You can do it in 12 lines of Matlab code. In this article, we’ll create a Matlab function butter_synth.m to design lowpass Butterworth filters of any order. Here is an example function call for a 5th order filter:
## Simplest Calculation of Half-band Filter Coefficients
Half-band filters are lowpass FIR filters with cut-off frequency of one-quarter of sampling frequency fs and odd symmetry about fs/4 [1]*. And it so happens that almost half of the coefficients are zero. The passband and stopband bandwiths are equal, making these filters useful for decimation-by-2 and interpolation-by-2. Since the zero coefficients make them computationally efficient, these filters are ubiquitous in DSP systems.
Here we will compute half-band...
## There's No End to It -- Matlab Code Plots Frequency Response above the Unit Circle
Reference [1] has some 3D plots of frequency response magnitude above the unit circle in the Z-plane. I liked them enough that I wrote a Matlab function to plot the response of any digital filter this way. I’m not sure how useful these plots are, but they’re fun to look at. The Matlab code is listed in the Appendix.
This post is available in PDF format for easy...
## Modeling a Continuous-Time System with Matlab
Many of us are familiar with modeling a continuous-time system in the frequency domain using its transfer function H(s) or H(jω). However, finding the time response can be challenging, and traditionally involves finding the inverse Laplace transform of H(s). An alternative way to get both time and frequency responses is to transform H(s) to a discrete-time system H(z) using the impulse-invariant transform [1,2]. This method provides an exact match to the continuous-time...
## Canonic Signed Digit (CSD) Representation of Integers
February 18, 2017
In my last post I presented Matlab code to synthesize multiplierless FIR filters using Canonic Signed Digit (CSD) coefficients. I included a function dec2csd1.m (repeated here in Appendix A) to convert decimal integers to binary CSD values. Here I want to use that function to illustrate a few properties of CSD numbers.
In a binary signed-digit number system, we allow each binary digit to have one of the three values {0, 1, -1}. Thus, for example, the binary value 1 1...
## Matlab Code to Synthesize Multiplierless FIR Filters
This article presents Matlab code to synthesize multiplierless Finite Impulse Response (FIR) lowpass filters.
A filter coefficient can be represented as a sum of powers of 2. For example, if a coefficient = decimal 5 multiplies input x, the output is $y= 2^2*x + 2^0*x$. The factor of $2^2$ is then implemented with a shift of 2 bits. This method is not efficient for coefficients having a lot of 1’s, e.g. decimal 31 = 11111. To reduce the number of non-zero...
## The Power Spectrum
October 8, 2016
Often, when calculating the spectrum of a sampled signal, we are interested in relative powers, and we don’t care about the absolute accuracy of the y axis. However, when the sampled signal represents an analog signal, we sometimes need an accurate picture of the analog signal’s power in the frequency domain. This post shows how to calculate an accurate power spectrum.
Parseval’s theorem [1,2] is a property of the Discrete Fourier Transform (DFT) that...
## Digital PLL's -- Part 2
June 15, 2016
In Part 1, we found the time response of a 2nd order PLL with a proportional + integral (lead-lag) loop filter. Now let’s look at this PLL in the Z-domain [1, 2]. We will find that the response is characterized by a loop natural frequency ωn and damping coefficient ζ.
Having a Z-domain model of the DPLL will allow us to do three things:
Compute the values of loop filter proportional gain KL and integrator gain KI that give the desired loop natural frequency and...
## Design IIR Highpass Filters
February 3, 2018
This post is the fourth in a series of tutorials on IIR Butterworth filter design. So far we covered lowpass [1], bandpass [2], and band-reject [3] filters; now we’ll design highpass filters. The general approach, as before, has six steps:
Find the poles of a lowpass analog prototype filter with Ωc = 1 rad/s. Given the -3 dB frequency of the digital highpass filter, find the corresponding frequency of the analog highpass filter (pre-warping). Transform the...
## Canonic Signed Digit (CSD) Representation of Integers
February 18, 2017
In my last post I presented Matlab code to synthesize multiplierless FIR filters using Canonic Signed Digit (CSD) coefficients. I included a function dec2csd1.m (repeated here in Appendix A) to convert decimal integers to binary CSD values. Here I want to use that function to illustrate a few properties of CSD numbers.
In a binary signed-digit number system, we allow each binary digit to have one of the three values {0, 1, -1}. Thus, for example, the binary value 1 1...
## Modeling a Continuous-Time System with Matlab
Many of us are familiar with modeling a continuous-time system in the frequency domain using its transfer function H(s) or H(jω). However, finding the time response can be challenging, and traditionally involves finding the inverse Laplace transform of H(s). An alternative way to get both time and frequency responses is to transform H(s) to a discrete-time system H(z) using the impulse-invariant transform [1,2]. This method provides an exact match to the continuous-time...
## Design a DAC sinx/x Corrector
This post provides a Matlab function that designs linear-phase FIR sinx/x correctors. It includes a table of fixed-point sinx/x corrector coefficients for different DAC frequency ranges.
A sinx/x corrector is a digital (or analog) filter used to compensate for the sinx/x roll-off inherent in the digital to analog conversion process. In DSP math, we treat the digital signal applied to the DAC is a sequence of impulses. These are converted by the DAC into contiguous pulses...
## Filter a Rectangular Pulse with no Ringing
To filter a rectangular pulse without any ringing, there is only one requirement on the filter coefficients: they must all be positive. However, if we want the leading and trailing edge of the pulse to be symmetrical, then the coefficients must be symmetrical. What we are describing is basically a window function.
Consider a rectangular pulse 32 samples long with fs = 1 kHz. Here is the Matlab code to generate the pulse:
N= 64; fs= 1000; % Hz sample...
## Design IIR Band-Reject Filters
January 17, 2018
In this post, I show how to design IIR Butterworth band-reject filters, and provide two Matlab functions for band-reject filter synthesis. Earlier posts covered IIR Butterworth lowpass [1] and bandpass [2] filters. Here, the function br_synth1.m designs band-reject filters based on null frequency and upper -3 dB frequency, while br_synth2.m designs them based on lower and upper -3 dB frequencies. I’ll discuss the differences between the two approaches later in this...
## Digital PLL’s, Part 3 – Phase Lock an NCO to an External Clock
Sometimes you may need to phase-lock a numerically controlled oscillator (NCO) to an external clock that is not related to the system clocks of your ASIC or FPGA. This situation is shown in Figure 1. Assuming your system has an analog-to-digital converter (ADC) available, you can sync to the external clock using the scheme shown in Figure 2. This time-domain PLL model is similar to the one presented in Part 1 of this series on digital PLL’s [1]. In that PLL, we...
## Decimator Image Response
Note: this is an improved version of a post I made to the dsp forum a few weeks ago.
This article presents a way to compute and plot the image response of a decimator. I’m defining the image response as the unwanted spectrum of the impulse response after downsampling, relative to the desired passband response.
Consider a decimate-by-4 filter with fs= 4 Hz, to which we apply the signal spectrum shown in Figure 1. The desired signal is the CW component at 0.22 Hz,...
## ADC Clock Jitter Model, Part 1 – Deterministic Jitter
April 16, 2018
Analog to digital converters (ADC’s) have several imperfections that affect communications signals, including thermal noise, differential nonlinearity, and sample clock jitter [1, 2]. As shown in Figure 1, the ADC has a sample/hold function that is clocked by a sample clock. Jitter on the sample clock causes the sampling instants to vary from the ideal sample time. This transfers the jitter from the sample clock to the input signal.
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# Weight and acceleration due to gravity (Page 3/3)
Page 3 / 3
1. Divide into pairs and explain Galileo's experiment to your friend.
2. Write down an aim and a hypothesis for Galileo's experiment.
3. Write down the result and conclusion for Galileo's experiment.
## Research project : experimental design
Design an experiment similar to the one done by Galileo to prove that the acceleration due to gravity of an object is independent of the object's mass. The investigation must be such that you can perform it at home or at school. Bring your apparatus to school and perform the experiment. Write it up and hand it in for assessment.
## Case study : determining the acceleration due to gravity 1
Study the set of photographs alongside showing the position of a ball being dropped from a height at constant time intervals. The distance of the ball from the starting point in each consecutive image is observed to be: ${x}_{1}=0$ cm, ${x}_{2}=4,9$ cm, ${x}_{3}=19,6$ cm, ${x}_{4}=44,1$ cm, ${x}_{5}=78,4$ cm and ${x}_{6}=122,5$ cm. Answer the following questions:
1. Determine the time between each picture if the frequency of the exposures were 10 Hz.
2. Calculate the velocity, ${v}_{2}$ , of the ball between positions 1 and 3.
${v}_{2}=\frac{{x}_{3}-{x}_{1}}{{t}_{3}-{t}_{1}}$
3. Calculate the velocity, ${v}_{5}$ , of the ball between positions 4 and 6.
${v}_{5}=\frac{{x}_{6}-{x}_{4}}{{t}_{6}-{t}_{4}}$
4. Calculate the acceleration the ball between positions 2 and 5.
$a=\frac{{v}_{5}-{v}_{2}}{{t}_{5}-{t}_{2}}$
5. Compare your answer to the value for the acceleration due to gravity ( $9,8\phantom{\rule{2pt}{0ex}}m·$ s ${}^{-2}$ ).
The acceleration due to gravity is constant. This means we can use the equations of motion under constant acceleration that we derived in motion in one dimension to describe the motion of an object in free fall. The equations are repeated here for ease of use.
$\begin{array}{ccc}\hfill {v}_{i}& =& \mathrm{initial velocity}\left(\mathrm{m}·{\mathrm{s}}^{-1}\right)\mathrm{at}\phantom{\rule{2pt}{0ex}}\mathrm{t}=0\mathrm{s}\hfill \\ \hfill {v}_{f}& =& \mathrm{final velocity}\left(\mathrm{m}·{\mathrm{s}}^{-1}\right)\mathrm{at time}\phantom{\rule{2pt}{0ex}}\mathrm{t}\hfill \\ \hfill \Delta x& =& \mathrm{displacement}\left(\mathrm{m}\right)\hfill \\ \hfill t& =& \mathrm{time}\left(\mathrm{s}\right)\hfill \\ \hfill \Delta t& =& \mathrm{time interval}\left(\mathrm{s}\right)\hfill \\ \hfill g& =& \mathrm{acceleration}\left(\mathrm{m}·{\mathrm{s}}^{-2}\right)\hfill \end{array}$
${v}_{f}={v}_{i}+gt$
$\Delta x=\frac{\left({v}_{i}+{v}_{f}\right)}{2}t$
$\Delta x={v}_{i}t+\frac{1}{2}g{t}^{2}$
${v}_{f}^{2}={v}_{i}^{2}+2g\Delta x$
## Experiment : determining the acceleration due to gravity 2
Work in groups of at least two people.
Aim: To determine the acceleration of an object in freefall.
Apparatus: Large marble, two stopwatches, measuring tape.
Method:
1. Measure the height of a door, from the top of the door to the floor, exactly. Write down the measurement.
2. One person must hold the marble at the top of the door. Drop the marble to the floor at the same time as he/she starts the first stopwatch.
3. The second person watches the floor and starts his stopwatch when the marble hits the floor.
4. The two stopwatches are stopped together and the two times substracted. The difference in time will give the time taken for the marble to fall from the top of the door to the floor.
5. Design a table to show the results of your experiment. Choose appropriate headings and units.
6. Choose an appropriate equation of motion to calculate the acceleration of the marble. Remember that the marble starts from rest and that it's displacement was determined in the first step.
7. Write a conclusion for your investigation.
1. Why do you think two stopwatches were used in this investigation?
2. Compare the value for acceleration obtained in your investigation with the value of acceleration due to gravity ( $9,8\phantom{\rule{2pt}{0ex}}m·s{}^{-2}$ ). Explain your answer.
A ball is dropped from the balcony of a tall building. The balcony is $15\phantom{\rule{2pt}{0ex}}m$ above the ground. Assuming gravitational acceleration is $9,8\phantom{\rule{2pt}{0ex}}m·s{}^{-2}$ , find:
1. the time required for the ball to hit the ground, and
2. the velocity with which it hits the ground.
1. It always helps to understand the problem if we draw a picture like the one below:
2. We have these quantities:
$\begin{array}{ccc}\hfill \Delta x& =& 15\phantom{\rule{4pt}{0ex}}\phantom{\rule{0.166667em}{0ex}}\mathrm{m}\hfill \\ \hfill {v}_{i}& =& 0\phantom{\rule{4pt}{0ex}}\phantom{\rule{0.166667em}{0ex}}\mathrm{m}·{\mathrm{s}}^{-1}\hfill \\ \hfill g& =& 9,8\phantom{\rule{4pt}{0ex}}\phantom{\rule{0.166667em}{0ex}}\mathrm{m}·{\mathrm{s}}^{-2}\hfill \end{array}$
3. Since the ball is falling, we choose down as positive. This means that the values for ${v}_{i}$ , $\Delta x$ and $a$ will be positive.
4. We can use [link] to find the time: $\Delta x={v}_{i}t+\frac{1}{2}g{t}^{2}$
5. $\begin{array}{ccc}\hfill \Delta x& =& {v}_{i}t+\frac{1}{2}g{t}^{2}\hfill \\ \hfill 15& =& \left(0\right)t+\frac{1}{2}\left(9,8\right){\left(t\right)}^{2}\hfill \\ \hfill 15& =& 4,9\phantom{\rule{3.33333pt}{0ex}}{t}^{2}\hfill \\ \hfill {t}^{2}& =& 3,0612...\hfill \\ \hfill t& =& 1,7496...\hfill \\ \hfill t& =& 1,75\phantom{\rule{3.33333pt}{0ex}}s\hfill \end{array}$
6. Using [link] to find ${v}_{f}$ :
$\begin{array}{ccc}\hfill {v}_{f}& =& {v}_{i}+gt\hfill \\ \hfill {v}_{f}& =& 0+\left(9,8\right)\left(1,7496...\right)\hfill \\ \hfill {v}_{f}& =& 17,1464...\hfill \end{array}$
Remember to add the direction: ${v}_{f}=17,15\phantom{\rule{2pt}{0ex}}m·s{}^{-1}$ downwards.
By now you should have seen that free fall motion is just a special case of motion with constant acceleration, and we use the same equations as before. The only difference is that the value for the acceleration, $a$ , is always equal to the value of gravitational acceleration, $g$ . In the equations of motion we can replace $a$ with $g$ .
## Gravitational acceleration
1. A brick falls from the top of a $5\phantom{\rule{2pt}{0ex}}m$ high building. Calculate the velocity with which the brick reaches the ground. How long does it take the brick to reach the ground?
2. A stone is dropped from a window. It takes the stone $1,5\phantom{\rule{2pt}{0ex}}s$ to reach the ground. How high above the ground is the window?
3. An apple falls from a tree from a height of $1,8\phantom{\rule{2pt}{0ex}}m$ . What is the velocity of the apple when it reaches the ground?
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
Preparation and Applications of Nanomaterial for Drug Delivery
Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
how did you get the value of 2000N.What calculations are needed to arrive at it
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# help complete this attempt (real analysis)
Given 2 quadratic functions $f$ and $g$, such that $f(x)=ax²+bx+c$ and $g(x)=cx²+bx+a$ and the absolute value of $f(x)$ is less than $1$, for each $x\in[-1,1]$, prove that the absolute value of $g(x)$ is less than $2$, for each $x\in[-1,1]$.
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And where is the attempt? – t.b. Jun 28 '11 at 14:04
@Theo: $\varnothing$ :-P – Asaf Karagila Jun 28 '11 at 14:29
Are you sure the question states that the absolute value of $f(x)$ is less than 1 or less than or equal to 1? – user38268 Jun 28 '11 at 14:58
Here is a complete solution; but note that a lot of thinking went into reducing the number of cases to consider.
Multiplying through by $-1$, if necessary, we may assume $a\geq c$, and replacing $x$ by $-x$, if necessary, we may assume $b\geq0$.
We note first that $c=f(0)\geq -1$ and $a+b+c=f(1)\leq 1$, therefore $$a\leq 1 -c-b\leq 2-b\ .\qquad(1)$$
Now $g(x)-f(x)=(a-c)(1-x^2)\geq0$; whence we already now that $g(x)\geq f(x)\geq-1$. If $\arg\max_x g(x)\in\{\pm 1\}$ then for all $x$ we have $g(x)\leq\max g(\pm1)=\max f(\pm1)\leq 1$.
So it remains the case that $\arg\max_x g(x)=:\tau \in\ ]{-1},1[\$. In this case we have a local maximum at $\tau$, whence necessarily $g'(\tau)=0$ and $c<0$. It follows that $${b\over 2|c|}=\tau<1\ .\qquad (2)$$ Using (1) and (2) we now obtain $$\max_x\ g(x)=g(\tau)= a-{b^2\over 4c}=a+{b^2\over 4|c|} \leq 2 -b +{b\over2}\leq 2\ ,$$ as stated.
The pair $f(x):=2x^2-1$, $\ g(x):=2-x^2$ shows that the proven inequality is sharp.
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Really I did not understand why c is ncessarily negative in the second case, can you explain more – Mopzer Moreena Jun 29 '11 at 15:16
@Mopzer: $g$ has a local maximum, so its leading coefficient $c$ is negative. – TonyK Jun 29 '11 at 17:01
thanks Tony, yeah it does by using the theorem of parabola extremum – Mopzer Moreena Jun 30 '11 at 10:54
Your two functions are parabolas. The only way the extreme values are inside of $[-1,1]$ is if the vertex $x_0$ of the parabola satisfies $|x_0|<1$ . Then the maximum is reached at the vertex of the parabola if $a<0$ —parabola opens downwards —and the minimum is reached inside when $a>0$. Otherwise, if the vertex of the parabola is not in $[-1,1]$, i.e., if the x-coordinate satisfies $|x|>1$ ,the maximum values will happen at the boundaries, i.e., at either $x=1$ or at $x=-1$. The case of the parabola opening downwards (when $a<0$) comes down to finding the "standard form" $y-y_0=c(x- x_0)^2$, and determining the values of $x_0$ and $y_0$. Check what happens to $f'(x)$ left- and right- of the vertex of the parabola.
Otherwise, the extreme values happen at the endpoints $\{-1,1\}$ , so that the extreme values will be $a+b+c$ (at $x=1$), and/or $a-b+c$ (at $x=-1$) unless $a<0$ and, when you express the parabola in standard form, $|x_0|<1$.
So if $a<0$ and $|x_0|<1$ , then $y_0$ is the maximum value; otherwise, the max will be reached at either of the endpoints, and will be either $a+b+c$ , or $a-b+c$
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Hey, you're making progress :) Only one fix this time! To typeset $\{-1,1\}$ use $\{-1,1\}$ – t.b. Jun 28 '11 at 16:03
Thanks, Theo, even I did not want to read my old messages :). Just hope the content is also good. – gary Jun 28 '11 at 16:07
No complaints contentwise. – t.b. Jun 28 '11 at 16:09
Thanks again for your help and patience, Theo. – gary Jun 28 '11 at 16:11
@: D Lim; you are right, less or equal to 1 for f(x), and less or equal to 2 for g(x) – Mopzer Moreena Jun 28 '11 at 19:51
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# Expected value of repeated bets
Suppose a game where a fair coin is flipped. Tails pays out $$0.5$$x the bet, and heads pays out $$1.75$$x. With a bet of \$1, the expected value of a single game is $$0.5\cdot 0.5 + 0.5\cdot 1.75 = 1.125$$. So if $$1000$$ games of this is played, \$125 is expected to be gained.
But what if instead of betting \$1, the entire bankroll is bet? So starting with \$100, after a single round there is a $$50$$% chance of \$50, and a $$50$$% chance of \$175. Regardless of outcome, the entire amount is bet again, so if it was heads, the next round \$175 would be bet and so on. What is the expected amount after $$1000$$ rounds? My intuition is that you'd end up with \$0, since it would take more than one win to make up for a single loss as $$0.5\cdot 1.75 = 0.875$$, and when I run a few trials programmatically it would come out to $$0$$ (as in, $$<\epsilon$$) each time, but I don't know how to prove this
• How can you ever reach zero if you always receive some positive payout? – Math1000 Feb 26 at 18:54
• well, not zero but a negligible amount, once you start splitting pennies – Ally Feb 26 at 19:59
Computer simulations suggest that the bankroll is expected to get vanishingly small after the 1000 rounds. This is false, as will be shown below.
Without loss of generality let the initial bankroll be 1. If the number of heads obtained is $$k$$, the bankroll after playing 1000 rounds is $$(1.75)^k(0.5)^{1000-k}$$ Thus the expected value of the bankroll after 1000 rounds is $$\sum_{k=0}^{1000}\frac{\binom{1000}k}{2^{1000}}(1.75)^k(0.5)^{1000-k}$$ $$=\sum_{k=0}^{1000}\binom{1000}k(0.875)^k(0.25)^{1000-k}$$ $$=(1.125)^{1000}\sum_{k=0}^{1000}\binom{1000}k(0.875/1.125)^k(0.25/1.125)^{1000-k}$$ The sum in the last expression is over the probabilities of every possible result in a binomial $$n=1000,p=0.875/1.125$$ random variable, so is 1. The whole thing reduces to $$1.125^{1000}=1.421\cdot10^{51}$$.
This is counterintuitive, and the reason for this is that the distribution of ending values has a very long tail (positive-skewed) which is not very apparent from the simulation.
• what is this method of using the base-2 logarithm called? I was trying to find material on this to read about it but not sure what to search for – Ally Feb 26 at 20:04
• @TannerSwett Does this please you now? – Parcly Taxel Feb 27 at 4:21
• Looks good now, thanks for making the edits! – Tanner Swett Feb 27 at 4:23
Let $$\ X_n\ = 1\$$ if the $$\ n^\mathrm{th}\$$ toss is heads, $$\ 0\$$ otherwise, $$\ Y_n = 0.5 + 1.25\,X_n\$$, and $$\ B_{n-1}\$$ the player's bankroll just before the $$\ n^\mathrm{th}\$$ toss. His bankroll $$\ B_n\$$ after that toss will then be $$\ B_n = Y_n\,B_{n-1}\$$. After $$\ 1000\$$ tosses, the bankroll will therefore be $$\ B_{1000}=Y_{1000}Y_{999}\dots Y_1B_0\$$. Since $$\ Y_1, Y_2, \dots, Y_{1000}\ \$$ are independent, all with mean $$\ \frac{1}{2}\times 0.5 + \frac{1}{2}\times 1.75 = 1.125\$$, then $$\ E\left(B_{1000}\right) = 1.125^{1000}\,B_0\approx 1.4\times 10^{51}\,B_0\$$.
Thus, far from becoming vanishingly small, the player's expected bankroll will instead grow enormously large. The key here is that the "average" amount by which the bankroll gets multiplied on each toss is the arithmetic mean of $$\ 0.5\$$ and $$\ 1.75\$$—which is greater than $$\ 1\$$—, rather than their geometric mean of $$\ \sqrt{0.5\times1.75}\approx 0.94 \$$.
Additional observation: Despite the player's huge expected bankroll after $$\ 1000\$$ tosses, he or she only has a very small chance of coming out ahead. If there are $$\ n\$$ heads and $$\ 1000-n\$$ tails in the the $$\ 1000\$$ tosses, the player's bankroll at the end will be $$\ 0.5^{1000-n}\,1.75^n\,B_0\$$, and this is greater than $$\ B_0\$$ if and only if $$\ n\ge 554\$$. The probability that the player will come ot ahead is thus $$\ \sum_{n=554}^{1000} {1000 \choose n}2^{-1000}\approx\ 3.5\times 10^{-4}\$$, or about $$\ 1\$$ in $$\ 2800\$$.
After 1000 rounds, we cannot say, what the expected emount is, as that is down to chance.
However, taking the worst possible outcome, every time (that is, losing), you will be left, with an extremely small number, which, however, is unequal to zero.
The amount will only me equal to zero, is if someone loses the game, an infinite amount of times
• "we cannot say as that is down to chance"... You absolutely can figure out what the expected amount will be. Expected Value is quite rigorously defined and is a rather important and useful thing which is commonly studied for questions related to random process and random variables. – JMoravitz Feb 26 at 20:05
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# Prove the soundness of propositional logic without using induction?
I want to prove the soundness of propositional logic without using induction. I think I can do that via a process that's basically universal introduction (i.e., demonstrate something about an arbitrary member and infer that it applies universally). As an example of this approach I've picked a new inference rule to prove.
#### My questions:
• Is my approach valid?
• Is my proof correct?
• Is it easy to understand and follow?
• Do I need to add more/less detail?
• How else can I improve it?
#### Proof
We want to show that $$\boxed{\dfrac{\Gamma_1 ,\,\phi\vdash\psi\quad\Gamma_2,\,\lnot\phi\vdash\psi}{\Gamma_1 ,\,\Gamma_2\vdash\psi}}\def\pa {((\Gamma_1\land\phi)\longrightarrow\psi)} \def\pb {((\Gamma_2\land\lnot\phi)\longrightarrow\psi)} \def\ca {((\Gamma_1\land\Gamma_2)\longrightarrow\psi)} \def\lto {\longrightarrow} \def\val#1{V_\mathscr{I}( #1 )} \def\pli {\text{PL-interpretation, \mathscr{I}},} \def\inp{\mathscr{I}}$$ is a truth preserving inference rule without using induction. To do so we'll convert the rule to an axiom schema and show that it's valid.
#### Conversion Rules
##### Symbols
• "," and the "$$\quad$$" convert to conjunction
• "$$\vdash$$" and the vinculum convert to implication
• $$\Gamma$$, with or without subscript, is a finite set of conjoined wffs, so it's simply a wff with it's own valuation rules
##### Valuation of $$\Gamma$$
• $$\val{\Gamma}=1$$ iff, for all $$\gamma\in\Gamma$$, it's the case that $$\val{\gamma}=1$$ and $$\Gamma\neq\emptyset$$
#### Conversion
• $$\Gamma_1 ,\,\phi\vdash\psi:= \pa$$
• $$\Gamma_2,\,\lnot\phi\vdash\psi:= \pb$$
• $$\Gamma_1 ,\,\Gamma_2\vdash\psi:= \ca$$
• Putting it all together, the rule converts to - $$\boxed{((\pa\land\pb)\lto\ca)}$$
#### Proof that the axiom schema is valid
1. Assume for reductio that $$\val {((\pa\land\pb)\lto\ca)}=0$$
2. It follows from (1) that $$\val {\ca}=0$$
3. It follows from (2) that $$\val{\Gamma_1}=1$$, $$\val{\Gamma_2}=1$$, and $$\val{\psi}=0$$
4. It follows from (1) that $$\val{\pa}=1$$
5. It follows from (3) and (4) that $$\val{\phi}=0$$
6. It follows from (5) that $$\val{\lnot\phi}=1$$
7. It follows from (1) that $$\val{\pb}=1$$
8. It follows from (3) and (6) that $$\val{\pb}=0$$, which contradicts (7) $$\boxed{}$$
##### Example Proof
$$\begin{array}{lrcll} 1.&\phi&\vdash &\phi &\text{RA}\\ 2.&\phi&\vdash &\lnot\psi\lor\phi &\text{1, \lorI}\\ 3.&\phi&\vdash &\psi\to\phi &\text{2, Abbrv}\\ 4.&\phi&\vdash & (\phi\to\psi)\lor(\psi\to\phi) &\text{3, \lor I}\\ 5.&\lnot\phi&\vdash &\lnot\phi &\text{RA}\\ 6.&\lnot\phi&\vdash &\lnot\phi\lor\psi &\text{5, \lorI}\\ 7.&\lnot\phi&\vdash &\phi\to\psi &\text{6, Abbrv}\\ 8.&\lnot\phi&\vdash & (\phi\to\psi)\lor(\psi\to\phi) &\text{7, \lor I}\\ 9.&\emptyset &\vdash & (\phi\to\psi)\lor(\psi\to\phi) &\text{4, 8 New Rule}\\ \end{array}$$
• ... provided that $\Gamma_i$ are finite sets of formulas, otherwise $\Gamma_i \land \phi$ is not a formula. Sep 10 '21 at 10:00
• @Mauro ALLEGRANZA, I've explicitly mentioned that $\Gamma$ is a finite set, but I could draw further attention to it if it makes things clearer? Sep 10 '21 at 10:07
• But this is the key-point: if the number of cases is finite, we do not need induction. It is enough to check them one-by-one. Sep 10 '21 at 10:09
• @Mauro ALLEGRANZA, while each substitution instance is composed of a finite number of wffs, there are an infinite number of substitution instances, though Sep 10 '21 at 10:18
• To prove your concerned sentential soundness theorem, normally we need to use RAA to prove each logical connective's soundness case by case. Since each connective has different meaning, what's the most important principle upon which you can claim your single universal inference rule can ensure each row of the truth table of every connective to acting exactly according to their respective definition? The famous XOR connective, for example, has a very peculiar elimination rule, how your proposed theory can ensure it's not acting like the usual wrong elimination rule mimic disjunction syllogism? Sep 10 '21 at 23:35
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## Woodin conference
March 11, 2015
The conference in honor of Hugh Woodin’s 60th birthday will take place at Harvard University, on March 27-29, 2015. The meeting is partially supported by the Mid-Atlantic Mathematical Logic Seminar and the National Science Foundation. Funding is available to support participant travel. Please write to woodinbirthdayconference@gmail.com to apply for support, and to notify the organizers if you are planning to attend.
The list of speakers is as follows:
• H. Garth Dales
• Qi Feng
• Matthew D. Foreman
• Ronald Jensen
• Alexander S. Kechris
• Menachem Magidor
• Donald A. Martin
• Grigor Sargsyan
• Theodore A. Slaman
• John R. Steel.
We expect to publish proceedings of the conference, together with select additional research and survey papers, through the series Contemporary Mathematics, of the AMS. The editors of the proceedings are myself, James Cummings, Peter Koellner, and Paul Larson. Please contact me for information regarding the proceedings.
Additional information can be found at the conference website.
## On proofs and more
October 17, 2013
This is a transcript of an exchange on Twitter on what mathematicians and others expect from proofs. (A previous exchange on a different topic is here. Twitter produces surprisingly nice results sometimes. What follows is a bit meandering, but interesting points are made.)
It began at 7:56 am – 27 Jun 13, with the twitter account of Republic of Mathematics (a website started by Gary Davis) quoting from Bill Thurston‘s great essay On proof and progress in mathematics. The quoted sentence was a short excerpt from the following:
The question is not even “How do mathematicians make progress in mathematics?”
Rather, as a more explicit (and leading) form of the question, I prefer “How do mathematicians advance human understanding of mathematics?”
This question brings to the fore something that is fundamental and pervasive: that what we are doing is finding ways for people to understand and think about mathematics.
To this, the account of The True Beauty of Math replied with “[Which is] why computer proofs [are of] little use.” Republic of Mathematics objected to this position, and quoted from an article by Sara Billey titled Computer Proofs. What is the value of computer assisted proofs?. The quoted sentence is an excerpt from:
Some mathematicians have tried to protect their egos by proposing that human proofs are superior to computer assisted proofs. They claim that we don’t learn as much from computer assisted proofs as we do from a human proof. They claim computer proofs can be difficult to verify. They claim computer proofs are less elegant. I find these complaints to be naive.
This was retweeted by Guy Longsworth, which is how I saw it.
## Woodin’s proof of the second incompleteness theorem for set theory
November 4, 2010
[Edit, Oct. 1, 2013: Robert Solovay has pointed out an inaccuracy in my presentation of Woodin’s argument: Rather than simply requiring that $P$ is a hereditary property of models, we must require that $\mathsf{ZFC}$ proves this. A corrected presentation of the argument will be posted shortly.]
As part of the University of Florida Special Year in Logic, I attended a conference at Gainesville on March 5–9, 2007, on Singular Cardinal Combinatorics and Inner Model Theory. Over lunch, Hugh Woodin mentioned a nice argument that quickly gives a proof of the second incompleteness theorem for set theory, and somewhat more. I present this argument here.
The proof is similar to that in Thomas Jech, On Gödel’s second incompleteness theorem, Proceedings of the American Mathematical Society 121 (1) (1994), 311-313. However, it is semantic in nature: Consistency is expressed in terms of the existence of models. In particular, we do not need to present a proof system to make sense of the result. Of course, thanks to the completeness theorem, if consistency is first introduced syntactically, we can still make use of the semantic approach.
Woodin’s proof follows.
## Set theory seminar -Richard Ketchersid: Quasiiterations I. Iteration trees
January 19, 2009
In October 24-November 14 of 2008, Richard Ketchersid gave a nice series of talks on quasiiterations at the Set Theory Seminar. The theme is to correctly identify `nice’ branches through iteration trees, and to see how difficult it is for a model to compute these branches. Richard presented a prototypical result in this area (due to Woodin) and a nice application (due to Jackson and Ketchersid). This post will be far from self-contained, and only present some of the definitions.
[Edit Sep. 25, 2010: My original intention was to follow this post with two more notes, on Woodin’s result and on the Jackson-Ketchersid theorem, but I never found the time to polish the presentation to a satisfactory level, so instead I will let the interested reader find my drafts at Lucien’s library.]
I’ll assume known the notions of extender and Woodin cardinal, and associated notions like the length or strength of an extender. A good reference for this post is Donald Martin, John Steel, Iteration trees, Journal of the American Mathematical Society 7 (1) 1994, 1-73. As usual, all inaccuracies below are mine. Some of the notions below are slightly simpler than the official definitions. These notions are all due to Donald Martin, John Steel, and Hugh Woodin.
In this post I present the main notions (iteration trees and iterability) and close with a quick result about the height of tree orders. The order I follow is close to Richard’s but it differs from his presentation at a few places.
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# Stability of a balanced plank
## Homework Statement
A uniform plank of thickness 2d and weight W is balanced horizontally across the top of a circular cylinder of radius r, whose axis is horizontal and perpendicular to the length of the plank. Prove that the gain of potential energy when the plank is turned without slipping through an angle theta in a vertical plane is:
$W(r \sin{\theta} -(r+d)(1-\cos{\theta}))$
## Homework Equations
Clearly this is a problem of gravitational potential energy, U = mgh.
## The Attempt at a Solution
I drew the diagrams, which are attached.
So from diagram 2, I get
$\inc U=W(\frac{r+d}{\sin{\theta}}-(r+d)-((r+d)\tan{\theta}-r\theta)\sin{\theta}))$
which is completely unlike what is to be proved.
I'm at a dead end and not sure how to proceed, thanks in advance for the help!
#### Attachments
• 3.8 KB Views: 426
• 14.9 KB Views: 476
## Answers and Replies
Related Introductory Physics Homework Help News on Phys.org
verty
Homework Helper
I think the length $$\frac{r+d}{sin\theta}$$ is incorrect. If you fix that, you should be able to simplify the formula considerably.
I would also suggest trying to find an easier formula by adding two lengths.
Hmm ok I see that it should have been $\frac{r+d}{\cos{\theta}}$ instead.
But I'm not sure how to find the two lengths to add to get the correct answer still.
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Related Searches
Definitions
# Multiplicative group of integers modulo n
In modular arithmetic the set of congruence classes relatively prime to the modulus n form a group under multiplication called the multiplicative group of integers modulo n. It is also called the group of primitive residue classes modulo n. In the theory of rings, a branch of abstract algebra, it is described as the group of units of the ring of integers modulo n. (Units refers to elements with a multiplicative inverse.)
This group is fundamental in number theory. It has found applications in cryptography, integer factorization, and primality testing. For example, by finding the order (ie. the size) of the group, one can determine if n is prime: n is prime if and only if the order is n – 1.
## Group Axioms
It is easy to establish that under multiplication the congruence classes (mod n) which are relatively prime to n satisfy the axioms for an Abelian group.
Closure: if a and b are both relatively prime to n, so is ab.
Inverse: Finding x satisfying ax ≡ 1 (mod n) is the same as solving ax + ny = 1, which can be done by Euclid's algorithm.
Identity: 1 is the identity.
Associativity and commutivity: These follow from the corresponding facts for the integers, plus the fact that the map that takes an integer to its congruence class (mod n) is a ring homomorphism.
## Notation
The ring of integers (mod n) is denoted $mathbb\left\{Z\right\}/nmathbb\left\{Z\right\}$ or $mathbb\left\{Z\right\}/\left(n\right)$ (ie. the ring of integers modulo the ideal nZ = (n) consisting of the multiples of n) or by $mathbb\left\{Z\right\}_n.$ Depending on the author its group of units may be written $\left(mathbb\left\{Z\right\}/nmathbb\left\{Z\right\}\right)^*,$ $\left(mathbb\left\{Z\right\}/nmathbb\left\{Z\right\}\right)^times,$ $U\left(mathbb\left\{Z\right\}/nmathbb\left\{Z\right\}\right),$ or similar notations. This article uses $\left(mathbb\left\{Z\right\}/nmathbb\left\{Z\right\}\right)^times.$
## Structure
### Powers of 2
Modulo 2 there is only one relatively prime congruence class, 1, so $\left(mathbb\left\{Z\right\}/2mathbb\left\{Z\right\}\right)^times cong \left\{1\right\}$ is trivial.
Modulo 4 there are two relatively prime congruence classes, 1 and 3, so $\left(mathbb\left\{Z\right\}/4mathbb\left\{Z\right\}\right)^times cong C_2,$ the cyclic group with two elements.
Modulo 8 there are four relatively prime classes, 1, 3, 5 and 7. The square of each of these is 1, so $\left(mathbb\left\{Z\right\}/8mathbb\left\{Z\right\}\right)^times cong C_2 times C_2,$ the Klein four-group.
Modulo 16 there are eight relatively prime classes 1, 3, 5, 7, 9, 11, 13 and 15. $\left\{pm 1, pm 7\right\}cong C_2 times C_2,$ is the 2-torsion subgroup (ie. the square of each element is 1), so $\left(mathbb\left\{Z\right\}/16mathbb\left\{Z\right\}\right)^times$ is not cyclic. The powers of 3, $\left\{1, 3, 9, 11\right\}$ are a subgroup of order 4, as are the powers of 5, $\left\{1, 5, 9, 13\right\}.;;$ Thus $\left(mathbb\left\{Z\right\}/16mathbb\left\{Z\right\}\right)^times cong C_2 times C_4.$
The pattern shown by 8 and 16 holds for higher powers 2k, k > 2: $\left\{pm 1, 2^\left\{ k-1\right\} pm 1\right\}cong C_2 times C_2,$ is the 2-torsion subgroup (so $\left(mathbb\left\{Z\right\}/2^kmathbb\left\{Z\right\}\right)^times$ is not cyclic) and the powers of 3 are a subgroup of order 2k – 2, so $\left(mathbb\left\{Z\right\}/2^kmathbb\left\{Z\right\}\right)^times cong C_2 times C_\left\{2^\left\{k-2\right\}\right\}.$
### Powers of odd primes
For powers of odd primes pk the group is cyclic: $;;\left(mathbb\left\{Z\right\}/p^kmathbb\left\{Z\right\}\right)^times cong C_\left\{p^\left\{k-1\right\}\left(p-1\right)\right\} cong C_\left\{varphi\left(p^k\right)\right\}.$
### General composite numbers
The Chinese remainder theorem says that if $;;n=p_1^\left\{k_1\right\}p_2^\left\{k_2\right\}p_3^\left\{k_3\right\}dots, ;$ then the ring $mathbb\left\{Z\right\}/nmathbb\left\{Z\right\}$ is the direct product of the rings corresponding to each of its prime power factors:
$mathbb\left\{Z\right\}/nmathbb\left\{Z\right\} cong mathbb\left\{Z\right\}/\left\{p_1^\left\{k_1\right\}\right\}mathbb\left\{Z\right\}; times ;mathbb\left\{Z\right\}/\left\{p_2^\left\{k_2\right\}\right\}mathbb\left\{Z\right\} ;times; mathbb\left\{Z\right\}/\left\{p_3^\left\{k_3\right\}\right\}mathbb\left\{Z\right\}dots;;$
Similarly, the group of units $\left(mathbb\left\{Z\right\}/nmathbb\left\{Z\right\}\right)^times$ is the direct product of the groups corresponding to each of the prime power factors:
$\left(mathbb\left\{Z\right\}/nmathbb\left\{Z\right\}\right)^timescong \left(mathbb\left\{Z\right\}/\left\{p_1^\left\{k_1\right\}\right\}mathbb\left\{Z\right\}\right)^times times \left(mathbb\left\{Z\right\}/\left\{p_2^\left\{k_2\right\}\right\}mathbb\left\{Z\right\}\right)^times times \left(mathbb\left\{Z\right\}/\left\{p_3^\left\{k_3\right\}\right\}mathbb\left\{Z\right\}\right)^times dots;.$
## Order
The order of the group is given by Euler's totient function: $| \left(mathbb\left\{Z\right\}/nmathbb\left\{Z\right\}\right)^times|=varphi\left(n\right).$ This is the product of the orders of the cyclic groups in the direct product.
## Exponent
The exponent is given by the Carmichael function $lambda\left(n\right),$ the least common multiple of the orders of the cyclic groups. This means that if a and n are relatively prime, $a^\left\{lambda\left(n\right)\right\} equiv 1 pmod n.$
## Generators
$\left(mathbb\left\{Z\right\}/nmathbb\left\{Z\right\}\right)^times$ is cyclic if and only if $varphi\left(n\right)=lambda\left(n\right).$ This is the case for n a power of an odd prime, twice a power of an odd prime, 2, or 4. In this case a generator is called a
Since all the $\left(mathbb\left\{Z\right\}/nmathbb\left\{Z\right\}\right)^times,$ n = 1, 2, ..., 7 are cyclic, another way to state this is: If n < 8 then $;\left(mathbb\left\{Z\right\}/nmathbb\left\{Z\right\}\right)^times$ has a primitive root. If n ≥ 8 $;\left(mathbb\left\{Z\right\}/nmathbb\left\{Z\right\}\right)^times$ has a primitive root unless n is divisible by 4 or by two distinct odd primes.
In the general case there is one generator for each cyclic direct factor.
## Table
This table shows the structure and generators of $\left(mathbb\left\{Z\right\}/nmathbb\left\{Z\right\}\right)^times$ for small values of n. The generators are not unique (mod n); e.g. (mod 16) both {–1, 3} and {–1, 5} will work. The generators are listed in the same order as the direct factors.
For example take n = 20. $varphi\left(20\right)=8$ means that the order of $\left(mathbb\left\{Z\right\}/20mathbb\left\{Z\right\}\right)^times$ is 8 (i.e. there are 8 numbers less than 20 and coprime to it); $lambda\left(20\right)=4$ that the fourth power of any number relatively prime to 20 is ≡ 1 (mod 20); and as for the generators, 19 has order 2, 3 has order 4, and every member of $\left(mathbb\left\{Z\right\}/20mathbb\left\{Z\right\}\right)^times$ is of the form 19a × 3b, where a is 0 or 1 and b is 0, 1, 2, or 3.
The powers of 19 are {±1} and the powers of 3 are {3, 9, 7, 1}. The latter and their negatives (mod 20), {17, 11, 13, 19} are all the numbers less than 20 and prime to it. The fact that the order of 19 is 2 and the order of 3 is 4 implies that the fourth power of every member of $mathbb\left\{Z\right\}_\left\{20\right\}^times$ is ≡ 1 (mod 20).
Group Structure of (Z/nZ)×
$n$ $\left(mathbb\left\{Z\right\}/nmathbb\left\{Z\right\}\right)^times$ $varphi\left(n\right)$ $lambda\left(n\right)$ generators $n$ $\left(mathbb\left\{Z\right\}/nmathbb\left\{Z\right\}\right)^times$ $varphi\left(n\right)$ $lambda\left(n\right)$ generators
2 {1} 1 1 1 33 C2×C10 20 10 10, 2
3 C2 2 2 2 34 C16 16 16 3
4 C2 2 2 3 35 C2×C12 24 12 6, 2
5 C4 4 4 2 36 C2×C6 12 6 19, 5
6 C2 2 2 5 37 C36 36 36 2
7 C6 6 6 3 38 C18 18 18 3
8 C2×C2 4 2 7, 3 39 C2×C12 24 12 38, 2
9 C6 6 6 2 40 C2×C2×C4 16 4 39, 11, 3
10 C4 4 4 3 41 C40 40 40 6
11 C10 10 10 2 42 C2×C6 12 6 13, 5
12 C2×C2 4 2 5, 7 43 C42 42 42 3
13 C12 12 12 2 44 C2×C10 20 10 43, 3
14 C6 6 6 3 45 C2×C12 24 12 44, 2
15 C2×C4 8 4 14, 2 46 C22 22 22 5
16 C2×C4 8 4 15, 3 47 C46 46 46 5
17 C16 16 16 3 48 C2×C2×C4 16 4 47, 7, 5
18 C6 6 6 5 49 C42 42 42 3
19 C18 18 18 2 50 C20 20 20 3
20 C2×C4 8 4 19, 3 51 C2×C16 32 16 50, 5
21 C2×C6 12 6 20, 2 52 C2×C12 24 12 51, 7
22 C10 10 10 7 53 C52 52 52 2
23 C22 22 22 5 54 C18 18 18 5
24 C2×C2×C2 8 2 5, 7, 13 55 C2×C20 40 20 21, 2
25 C20 20 20 2 56 C2×C2×C6 24 6 13, 29, 3
26 C12 12 12 7 57 C2×C18 36 18 20, 2
27 C18 18 18 2 58 C28 28 28 3
28 C2×C6 12 6 13, 3 59 C58 58 58 2
29 C28 28 28 2 60 C2×C2×C4 16 4 11, 19, 7
30 C2×C4 8 4 11, 7 61 C60 60 60 2
31 C30 30 30 3 62 C30 30 30 3
32 C2×C8 16 8 31, 3 63 C6×C6 36 6 2, 5
## References
The Disquisitiones Arithmeticae has been translated from Gauss's Ciceronian Latin into English and German. The German edition includes all of his papers on number theory: all the proofs of quadratic reciprocity, the determination of the sign of the Gauss sum, the investigations into biquadratic reciprocity, and unpublished notes.
## External links
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Given complete linkage and selective neutrality, mtDNA gene trees are expected to correspond to intraspecific phylogenies, and mtDNA diversity will reflect population size. In Chapter 4 we considered a number of exceptions and extensions to Mendel’s principles of heredity. LSC promotes equal access to justice by providing funding to 133 independent non-profit legal aid programs in every state, the District of Columbia, and U. Order your agriculture equipment parts online - it's easy on MG Equipment Company!. In this paper we focus on two objectives, introduced recently to give insight into the performance of average-linkage clustering: a similarity based HC objective proposed by [Moseley and Wang, 2017. a) Single-linkage or nearest-neighbour clustering merges clusters based on the (dis)similarities between their nearest members. Multiplication of data sources within heterogeneous healthcare information systems always results in redundant information, split among multiple databases. 2 Ward (1963) proposed an agglomerative hierarchical clustering procedure in. 2015 JF-STOLL GX3205SM. Hierarchical clustering. In the event that a distance function that depends on the entire data is desirable, one can simply use an extension operator that is not local. 5% of regions covered by SNP genotypes) for our analyses in the largest available AGRE sample. This method is much like the single linkage, but instead of using the minimum of the distances, we use the maximum. 5 0 0 Observed partial linkage 0. and Liu, Y and Low, P S}, abstractNote = {A length polymorphism at the 5{prime} untranslated region of exon 1 and an RFLP (Dde I) in intron 5 (nt 160) of the ATIII gene were amplified by. Genes are further apart, such that, at Prophase I during Pachytene, cross-over events can separate original combinations. Our objective is to detect exact and approximate duplicates within identity records, in order to attain a better quality of information and to permit cross-linkage among stand-alone and clustered databases. Now 29 years later we have grown, but still have the same philosophy, to ensure that our products will stand up to the high demands of the top level racers around the world and the average riders alike. Columns 1 and 2 of Z contain cluster indices linked in pairs to form a binary tree. (1) Single Linkage – The distance between two clusters is defined as the shortest distance between two points in each cluster. Imm also developed a bar-linkage computer for aircraft power vs. In the general case, the complexity is. d (13)2 = min {d 12 , d 32. ADVERTISEMENTS: Linkage of Genes : Complete Linkage, Incomplete Linkage and Significances! Complete Linkage: If linkage is complete, there should be parental combinations only and no recombination. The linkage criterion determines which distance to use between sets of observation. The following common linkage methods are available: single, complete, average, Ward's method, centroid, median, and weighted average. 2 Complete Linkage The complete linkage measure d cl(C 1;C 2) = max x 12C 1;x 22C 2 d(x 1;x 2) judges cluster similarity by. Suppose that for a particular data set, we perform hierarchical clustering using single linkage and using complete linkage. 38 so 38% of chromatids are recombinant for m and w. Complete Linkage. 2 showed suggestive linkage in our extended. There are several criteria in use, such as single linkage, complete linkage, average linkage, and Ward's method (or centroid linkage). dendrogram, and I found that scipy. Truck Restoration, Custom & Rebuilds. We selected the complete linkage criteria to ensure that the maximum distance between elements of each cluster was under the threshold t. Average Linkage. Details of results for linkage, association, and conditional logistic regression analyses are described by Charoen et al. This single-link merge criterion is local. Squared Euclidean distances are commonly used with this. Alternative linkage schemes include complete linkage clustering, average linkage clustering (UPGMA and WPGMA), and Ward's method. On Tuesday, the Louisiana Department of Health announced 1,356 new infections, the largest single-day increase in cases since April 9. to view this website, please verify your age: |. bam file? thank you! clustering next-gen sam bam • 1. Sass home; About Sass; Membership; Shooting Events; Sass Mercantile; Sass Wire forums. Average linkage within groups. It is the basis for the synthesis of all the important biological macromolecules (carbohydrates, proteins, lipids, nucleic acids) from their simpler sub-units. The original matrix has 56,000 rows (genes) and 7 columns (treatments). There is no Krishna-Christ linkage; the topic is ridiculous; it is not even worth investigating. Linkage and crossing over can be seen as processes that have opposite effects: linkage keeps particular genes together, and crossing over mixes them up. Features: • Compatible with Invensys (Barber-Colman) 2-1/2 to 6 in. After quadrats 1 and 2 are joined, centroid(1,2) = [(15+12)/2, (9+8)/2] = (13. Complete Linkage Kit Comes with 3” all thread, 2 ball joint linkages, nuts and spacers. Single linkage method controls only nearest neighbours similarity. New “low-friction” linkage bearing seals made by SKF provide noticeably freer linkage action, offering better suspension feel and performance throughout the shock stroke. We ask the question “does a single locus segregate as. single linkage is fast, and can perform well on non-globular data, but it performs poorly in the presence of noise. Some linkage methods are single linkage (a), complete linkage (b), and average linkage(c): In the Minitab dialog window for Cluster Observations, you must specify the final partition via a final number of clusters or by a level of similarity. T = clusterdata(X,cutoff) returns cluster indices for each observation (row) of an input data matrix X, given a threshold cutoff for cutting an agglomerative hierarchical tree that the linkage function generates from X. This includes all synthetic, recombinant and purified SubstanceProteins of defined sequence, whether the use is therapeutic or prophylactic. Single, complete and average and centroid linkage Figure 4. Between-cluster distance has been formalized in a variety of ways. Complete linkage is strongly. ALl genes that do not determine sex are said to be autosomal chromosomes. You may do so in any reasonable manner, but. The closely located genes show strong linkage while the widely located ones show weak linkage. The new object is then assigned to the nearest cluster. During the past few decades, several molecular linkage maps of tomato have been developed mainly based on interspecific crosses between the cultivated and related wild species of tomato (for a complete list see Foolad 2007). Both of these could be improved upon at some point in the future if required. So, with single linkage, they would fuse at a height of 1, and with complete linkage, they would fuse at a height of 4. Expected if complete linkage 0. Complete Linkage. 上一篇博客介绍了single linkage是让所有簇的距离为簇间结点最短的距离，同时每一次合并所有簇间距中最短的那一个。而complete则是让簇间结点距离最长的作为簇间距，并且每一次让所有簇间距最短两簇进行合并，因此实际上不是都找最长，而是最长最短。. For example, the distance between clusters “r” and “s” to the left is equal to the length of the arrow between their two closest points. Pseudocode for Hierarchical Clustering. (vi) Closely linked genes more often tend to remain in original combination in the course of inheritance (complete linkage) but some-times genes change linkage groups in some meiotic cells because of exchange of chromosome segments). • Required for rated close-off pressure on 5 and 6 in. The linkage criterion determines the distance between sets of observations as a function of the pairwise distances between observations. Linkage In addition to a distance measure, the type of linkage needs to be applied. which was focused on single-linkage hierarchical clustering, to the case of com-plete linkage by devising an ad hoc poisoning attack. Under a parametric model (with penetrances of 0. Any genes in this non-recombining region, the "male specific Y" (which is 95% of the Y), will be "completely linked", in other words, they cannot be broken apart by recombination and are inherited as a single unit. Complete Linkage : In complete link hierarchical clustering, we merge in the members of the clusters in each step, which provide the smallest maximum pairwise distance. ADVERTISEMENTS: Read this article to learn about the Genetics linkage: Types, Groups and Characteristics of Genetics Linkage ! Linkage is the phenomenon of certain genes staying together during inheritance through generations without any change or separation due to their being present on the same chromosome. Complete-link clustering can also be described using the concept of clique. d (r, s) = min (d i s t (x r i, x s j)), i ∈ (i,, n r), j ∈ (1,, n s) Complete linkage, also called farthest neighbor, uses the largest distance between objects in the two clusters. Linkage ratio If complete linkage, a unique F2 phenotypic ratio results. Most of the time this is not the problem of the governor, but a partially plugged carburetor, usually a plugged idle circuit or worn linkage. Today, we will discuss the linkage of qualitative traits. In the case of male Drosophila there is a complete absence of recombinant types due to the absence of crossing over. dictated by the distance and linkage measures chosen by the researcher. This type of linkage however is rare in nature and occurs due to absence of cross over male individuals of Drosophila and certain other insects show complete linkage. Repeat : Merge two closest clusters. Complete linkage: If the parental combination of characters appear together for two or more generation in a continuous manner and regular manner. Introduction to Linkage of Gene 2. So, with single linkage, they would fuse at a height of 1, and with complete linkage, they would fuse at a height of 4. 2 are often not ideal for expression data, and overriding the defaults requires explicit calls to hclust and as. Complete Linkage - distance is computed between the two LEAST similar parts of clusters (two most distant points). Thus, this warrants pruning of genotyping data for high LD. ADVERTISEMENTS: Linkage of Genes : Complete Linkage, Incomplete Linkage and Significances! Complete Linkage: If linkage is complete, there should be parental combinations only and no recombination. This rule will, in a sense, string objects together to form clusters, and the resulting clusters tend to represent long "chains. R # # Enter the data as a matrix and assign variable names. Starting from a matrix of dissimilarities, linkage() calculates its dendrogram with the most commonly used agglomerative hierarchical clustering methods, e. 222 is obtained by complete linkage, which shows that complete linkage is better than single linkage and average linkage in the case of subgrouping based on Sidoarjo regency livestock types. The parallelization strategy naturally becomes. STT592-002: Intro. 0 International license. PERBANDINGAN KINERJA METODE SINGLE LINKAGE, METODE COMPLETE LINKAGE DAN METODE K-MEANS DALAM ANALISIS CLUSTER skripsi disajikan sebagai salah satu syarat untuk memperoleh gelar Sarjana Sain Program Studi Matematika oleh Yani Soraya 4150407024 JURUSAN MATEMATIKA FAKULTAS MATEMATIKA DAN ILMU PENGETAHUAN ALAM UNIVERSITAS NEGERI SEMARANG 2011 PENGESAHAN Skripsi yang berjudul Perbandingan Kinerja. Full Emoji List, v13. Multiplication of data sources within heterogeneous healthcare information systems always results in redundant information, split among multiple databases. As usual the brass slider does not need any grease or oil. The single linkage dendogram of the data generated by "Example Assignment" would appear as follows. Ch7: Linkage, recombination, and eukaryotic gene mapping • If humans have 23 pairs of chromosomes,and if chromosomes carry genes, is it possible for all genes to independentlyassort? NO some genes are linked. These particular genes do not generate combinations that are non-parental. So, with single linkage, they would fuse at a height of 1, and with complete linkage, they would fuse at a height of 4. Terpilih kelompok 1 dan 3, sehingga kedua kelompok ini digabungkan. Coupling and repulsion phenomenon was concerned with a) Crossing over b) Mutation c) Linkage d) All of these 10. Consequently, they use probabilistic linkage in their study. I Thus, the magnitude of the coe cient is important, not the sign. JF STOLL GXF 3205 P, front disc mower with conditioner, year: 2015, width 3. between two clusters). Clustered Iris data set (the labels give the true flower species) virgnca versicolor. Anno: 2015. Treating the sample correlations as similarity measures, cluster the stocks using the single linkage and complete linkage hierarchical procedures. Single, complete and average and centroid linkage Figure 4. (a) noncrossovers, single crossovers, double crossovers (b) 8. dictated by the distance and linkage measures chosen by the researcher. hierarchical-clustering: Fast algorithms for single, average/UPGMA and complete linkage clustering. 'single' Nearest distance (single linkage method) 'complete' Furthest distance (complete linkage method) seqlinkage incorrectly doubled the input pairwise. Supervised Hierarchical Clustering with Exponential Linkage Nishant Yadav 1Ari Kobren Nicholas Monath Andrew McCallum1 Abstract In supervised clustering, standard techniques for learning a pairwise dissimilarity function often suffer from a discrepancy between the training and clustering objectives, leading to poor cluster quality. With the "Upload Multiple Files" option, you can flip through heatmaps from several data files for time series analysis or other comparisons. AgglomerativeClustering doesn't return the distance between clusters and the number of original observations, which scipy. Hierarchical clustering (1,983 words) no match in snippet view article find links to article {\mathcal {O}}(n^{2})} ) are known: SLINK for single-linkage and CLINK for complete-linkage clustering. To install the pop-up linkage, first place the pop-up in the drain hole making sure that the slot in the bottom of the pop-up is facing the back of the sink. I Thus, the magnitude of the coe cient is important, not the sign. This is the first complete linkage genome scan for agoraphobia and the third for PD. There are others mentioned in the literature, but these are the best-known methods. The distance metric to use in the case that y is a collection of observation vectors; ignored otherwise. with appropriate data structures. Complete Edelbrock. 2015 JF-STOLL GX3205SM. A standalone bar-linkage mechanism to solve linear equations was patented by Arnold Spilker. A total of 407 markers (389 autosomal, 18 X chromosome) were genotyped. bolens parts: new & used parts for all types of bolens and troy-bilt equipment: bolens iseki diesel parts now in stock ! walk behind mowers, tillers, snowblowers. Each point in the linkage disequilibrium matrix Linkage disequilibrium analyses. 1 under a dominant model of inheritance. With a heap the. It has been called a “linkage function,” (see, for example, [3] and. The key idea is to reduce the single-linkage hier-archical clustering problem to the minimum spanning tree (MST) problem in a complete graph constructed by the in-put dataset. (ii) Incomplete Linkage: Incomplete linkage is exhibited by those genes which produce some percentage of non-parental combinations. Florida Single Audit Act The Florida Single Audit Act, Section 215. On the other hand, complete and average linkage tend to yield more balanced, attractive clusters. w+ m+ w m 2. The new object is then assigned to the nearest cluster. Thus we introduce an approach to supervised hierarchical clustering that smoothly interpolates between single, average, and complete linkage, and we give a training procedure that simultaneously learns a linkage function and a dissimilarity function. Metode Single Linkage. Textbook solution for Essentials Of Business Analytics 1st Edition Camm Chapter 6 Problem 3P. The notorious chaining tendency of single linkage can be alleviated by specifying the TRIM= option (Wishart 1969, pp. Complete linkage is when two genes are on a single pair of homologs and no exchange occurs, producing non crossover/parental gametes. In this study, we used a pruned set of single nucleotide polymorphism (SNP) markers providing nearly complete linkage information on all autosomes and the X chromosome (information content greater than 0. They are also handy to transport as they slot neatly into a box that can be carried on the tractor’s front linkage. Second, when using a connectivity matrix, single, average and complete linkage are unstable and tend to create a few clusters that grow very quickly. The closer the physical location of two genes on the DNA, the less likely they are to be separated by a crossing-over event. 2015 JF-STOLL GX3205SM. 0 054 051 0. The first molecular linkage map of tomato was published in 1986, which included 18 isozyme and 94 DNA markers. linkage : {"average", "complete"}, optional, default: "complete" Which linkage critera to use. Quick Features 130mm of rear LTS (Linkage Tuned Suspension), 130mm front travel Single Flip-Chip takes you from low and fast to lower and faster Threaded bottom bracket Carbon main frame / alloy rear swingarm Room for a water. The strength of linkage is determined by distance between two genes. linkage synonyms, linkage pronunciation, linkage translation, English dictionary definition of linkage. 66 D ij = 1 if d( x i. between single, average, and complete linkage, and we give a training procedure that simulta-neously learns a linkage function and a dissimi-larity function. Agglomerative hierarchical cluster tree, returned as a numeric matrix. 4 Properties. Quick start Cluster analysis of data Single-linkage cluster analysis using Euclidean distance of v1, v2, v3, and v4 cluster singlelinkage v1 v2 v3 v4 Same as above cluster single v1 v2 v3 v4. noun a linkage which has one, two, etc. As the clustering proceeds, each observation in turn is added to the starting cluster (4,5). Incomplete. We obtain two dendrograms. Both the single linkage method and the complete linkage method can be described on the basis of the threshold graph: Single linkage clustering is equivalent to seeking maximally connected sub-graphs (components), while complete linkage clustering corresponds to finding maximally complete sub-graphs (cliques). Genes showing complete linkage are closely located in the same chromosome. The default settings for heatmap. Specific distance and linkage measures are reviewed, including a discussion of how these choices can influence the clustering process by comparing three common linkage measures (single linkage, complete linkage, average linkage). Single linkage and complete linkage are two algorithms of agglomerative Hierarchical clustering. Dendrograms. The association analysis indicated that the AA genotype ewes had larger litter size at average parity than those with the BB genotype. It is rather obvious that that they. This is the minimal changes required to get single linkage working. Pathobiology Genetics. A greater number of nucleotide-binding leucine-rich repeat disease resistance genes were present in genomes of taro than the duckweed, ∼391 versus ∼70 (∼182 and ∼46 complete). 4 GENETIC MAPPING AND LINKAGE - Single Nucleotide Polymorphisms are single nucleotide changes. complete linkage single linkage Shortest complete linkage distance. Importantly, it contains a new parameterized method named versatile linkage, which includes single linkage, complete linkage and average linkage as particular cases, and which naturally defines two new methods, geometric linkage and harmonic linkage (hence the convenience to rename average linkage as arithmetic linkage, to emphasize the. Genetic linkage, Complete linkage, on the same chromosome so that they are unlikely to be separated during crossing over and tend to be inherited as a single unit. 5% of regions covered by SNP genotypes) for our analyses in the largest available AGRE sample. Results: In association analysis, the lowest GEE p-value for hemostatic factors was p = 4. These clusters can appear spread-out. Complete Linkage. Complete Linkage The following method is obtained by specifying METHOD=COMPLETE. BAVERAGE is the default and can also be requested with keyword DEFAULT. Each point in the linkage disequilibrium matrix Linkage disequilibrium analyses. Exercise 2:. This class includes commonly-used algorithms such as single-linkage, average-linkage, complete-linkage, and Ward's method. Linkage is greatly used by geneticists to map the entire genome. It is rather obvious that that they. bolens parts: new & used parts for all types of bolens and troy-bilt equipment: bolens iseki diesel parts now in stock ! walk behind mowers, tillers, snowblowers. In "complete" linkage: distance between clusters is max. Indeed, experience has shown that a given step may increase the number of true positives and reduce the number of false positives in one year, and that the same step in. Usually linkage group as a whole passes into gamete during gametogenesis and are inherited together. example, single linkage defines cluster distance by the shortest edge between members of the clusters, while complete linkage uses the longest between cluster edge to define the distance between clusters. As agencies announce they are examining whether vitamin D supplements could reduce the risk of covid-19, Ingrid Torjesen finds out what the existing evidence shows Public health agencies in England and Scotland are conducting urgent reviews into the potential for vitamin D to reduce the risk of covid-19. Single linkage clustering, on the other hand, computes the similarity between two groups as the similarity of the closest pair of observations between the two groups. The single linkage dendogram of the data generated by "Example Assignment" would appear as follows. Balancing Chemical Equations - PhET Interactive Simulations. Single Linkage Distance Between Clusters Complete Linkage (Farthest Neighbor) The Complete Linkage distance between two clusters is defined as the distance between the most distant (farthest) pair of objects in the two clusters (one object in each cluster). Commas are placed between genes that are linked b. Mahfouz PhD, Department of Computer and Systems Engineering Faculty of Engineering, Alexandria University Alexandria, Egypt [email protected] the number of clusters to be known in advance as partitioning algorithm do. Complete Linkage. Which of the dendrograms below is the result of single-linkage, which is the result of average-linkage, and which is the result of complete-linkage? Explain your reasoning. The mapping population data revealed 16 major linkage groups with 520 markers, and 10 quantitative trait loci (QTLs) significantly associated with Taro Leaf. PERBANDINGAN KINERJA METODE SINGLE LINKAGE, METODE COMPLETE LINKAGE DAN METODE K-MEANS DALAM ANALISIS CLUSTER skripsi disajikan sebagai salah satu syarat untuk memperoleh gelar Sarjana Sain Program Studi Matematika oleh Yani Soraya 4150407024 JURUSAN MATEMATIKA FAKULTAS MATEMATIKA DAN ILMU PENGETAHUAN ALAM UNIVERSITAS NEGERI SEMARANG 2011 PENGESAHAN Skripsi yang berjudul Perbandingan Kinerja. Complete Linkage. Linkage disequilibrium (LD) is the nonrandom association of alleles of different loci. Second, when using a connectivity matrix, single, average and complete linkage are unstable and tend to create a few clusters that grow very quickly. Several linkage criteria can be used in building the hierarchy (dendrogram) such as single linkage, complete linkage or average linkage Abstract— Ease of interpretation of results makes hierarchical clustering algorithms suitable for many applications. Exercise 2:. We sell used Scout parts as well as common upgrades such as disc brake conversion kits, lift kits, fuel injection kits, offroad steering upgrades and roll bars. The computer POST (power-on self-test) checks a computer's internal hardware for compatibility and connection before starting the remainder of the boot process. Record linkage is the process of identifying records that refer to the same real-world entities in situations where entity identifiers are unavailable. Where you'll always get a warm welcome. Package shipments are currently taking 3-5 days to process, and UPS/USPS are also experiencing extended delivery times. Hierarchical clustering¶ Complete linkage It is usually somewhat intermediate between complete and single linkage. Linkage Rule There are 8 Linkage Rules listed here, single linkage, complete linkage, median linkage, centroid linkage, unweighted pair-group average, weighted pair-group average, flexable-beta method, and ward's method. A complete binary tree is a binary tree in which every level, except possibly the last, is completely filled, and all nodes are as far left as possible. Slider Crank Mechanism. Importantly, it contains a new parameterized method named versatile linkage, which includes single linkage, complete linkage and average linkage as particular cases, and which naturally defines two new methods, geometric linkage and harmonic linkage (hence the convenience to rename average linkage as arithmetic linkage, to emphasize the. If there is no recombination (complete linkage) then the two genes are essentially inherited as if they were a single gene. Single linkage and complete linkage clustering are popular in use. Single linkage and complete linkage are two popular examples of agglomerative clustering. Complete Linkage: For two clusters R and S, the single linkage returns the maximum distance between two points i and j such that i belongs to R and j belongs to S. Algorithms for hierarchical clustering normally differ by the linkage function used. The average-linkage algorithm is a compromise between the single-linkage and complete-linkage algorithm. Define linkage. A linkage is said to be complete when two or multiple characteristics are inherited and normally surface in two or further generations in their parental or original combinations, they are known as complete linkage. 1/1) Lower Link Lift Arm - Complete (Cat. 728 for the variant (112Ala/233Ile) allele. Crank and slotted link mechanism. Thus, this warrants pruning of genotyping data for high LD. This linkage may be used to detect high values in your dataset which may be outliers as they will be merged at the end. You may do so in any reasonable manner, but. The clustering can be obtained based on educational participation rank, that are low educational participation cluster that consist of Papua Province, and high educational participation cluster. Single linkage clustering: Find the minimum possible distance between points belonging to two different clusters. There is now a single linkage peak around 75cM (LOD of 2. (1,2,3) and (4,5) (whichever way the ties are resolved). One popular class of hierarchical algorithms is linkage-based algorithms. Incomplete. Complete Linkage: Genes are so close together as part of the same structure that they are inherited as a single unit. They are also handy to transport as they slot neatly into a box that can be carried on the tractor’s front linkage. They used two indices of fit (roughly Euclidean distances between the true membership vector and the clustering vector produced by the algorithm) as criteria to compare the. For example, they both found similar groups that were classified by PDB for oxidoreductase, nitrite reductase and ligase. , single, complete, average linkage). Example: Drosophila melanogaster. Hotels have been told that they must not offer free single-use plastic items by 2025. The algorithm begins with a forest of clusters that have yet to be used in the hierarchy being formed. The single-actuator linkage is well suited for this application, due to the cost-sensitive nature of wheelchair products. GAO provides fact-based, nonpartisan information to Congress. (a) Single linkage looks at minimum distance between all inter-group pairs. We will evaluate efficiency of record linkage algorithm using complete linkage hierarchical clustering over single linkage clustering. Single linkage and complete linkage are two algorithms of agglomerative Hierarchical clustering. 00 This is a Brand new boot for use with the overdrive unit (Warn, Husky, Saturn, ATV) and a dual stick D-18. FIND A DEALER Retailers Where to Buy. Specific distance and linkage measures are reviewed, including a discussion of how these choices can influence the clustering process by comparing three common linkage measures (single linkage, complete linkage, average linkage). Differences in the resulting dendrogram structure are discussed. The recent completion of a genome-wide scan using the Affymetrix GeneChip Human Mapping 100K single nucleotide polymorphism (SNP) set on participants in the Framingham Heart Study offered the opportunity to conduct a genome-wide association study (GWAS) and linkage scan for variants that influence hemostatic factors and hematological phenotypes. Any genes in this non-recombining region, the "male specific Y" (which is 95% of the Y), will be "completely linked", in other words, they cannot be broken apart by recombination and are inherited as a single unit. ManageEngine ADSelfService Plus is an integrated self-service password management and single sign on solution. Briggs & Stratton is open for business. The resulting diagram from a Hierarchical Cluster Analysis is called a dendrogram , in which data are nested into brackets of increasing dissimilarity. Complete Linkage. English Turkish online dictionary Tureng, translate words and terms with different pronunciation options. The Central Board of Direct Taxes (CBDT) through a notification also extended the time limit by. Single linkage method controls only nearest neighbours similarity. Three complete linkage SNPs at −534A/G, −407T/G, and −332C/T were detected, resulting in three genotypes (namely, AA, AB, and BB). The linkage criterion determines the distance between sets of observations as a function of the pairwise distances between observations. Single-linkage: calculates the minimum distance between the clusters before merging. for divisive clustering, which makes them too slow for large data sets. Complete linkage: similar to single linkage except the cluster formation criterion is maximum case distance (furthest rather than nearest neighbor). • Lloyd’s method • Single linkage, Complete linkage • Initialization techniques (random, furthest traversal, k-means++). Complete linkage is rare, usually some crossing-over. If cluster A is the set of objects A 1,A 2, ,A m and cluster B is B 1,B 2, ,B n, the. Maximum linkage = complete linkage Minimum linkage = single linkage. Florida Single Audit Act The Florida Single Audit Act, Section 215. (2) Single-linkage (or edge-linkage) almost classifies all the points into a same cluster. Hi, is there any existing software to cluster aligned reads form a. which was focused on single-linkage hierarchical clustering, to the case of com-plete linkage by devising an ad hoc poisoning attack. The Complete Guide to Creating Symbolic Links (aka Symlinks) on Windows Chris Hoffman @chrisbhoffman Updated April 20, 2020, 3:51pm EDT Windows 10, 8, 7, and Vista all support symbolic links—also known as symlinks—that point to a file or folder on your system. complete linkage) within PDS1 or PDS2, and adjacent SNPs were within 1 kb on a contig, a single SNP was chosen to represent the group for linkage analysis. A linkage is said to be complete when two or multiple characteristics are inherited and normally surface in two or further generations in their parental or original combinations, they are known as complete linkage. The group (series-series) similarity scores are based on a linkage method from hierarchical cluster analysis (e. Linkage Rule There are 8 Linkage Rules listed here, single linkage, complete linkage, median linkage, centroid linkage, unweighted pair-group average, weighted pair-group average, flexable-beta method, and ward's method. complete linkage: Genetics An inheritance pattern for 2 gene loci on the same chromosome, in which the observed crossover frequency between the loci is zero. COM stocks 1961-1980 International Harvester Scout 80, 800, 800A, 800B, Scout II, Terra and Traveler parts. Key words: spinal cord injury, steering linkage, wheelchair. T = clusterdata(X,cutoff) returns cluster indices for each observation (row) of an input data matrix X, given a threshold cutoff for cutting an agglomerative hierarchical tree that the linkage function generates from X. Linkage "a" and "b" are too close to each other. Single linkage, also called nearest neighbor, uses the smallest distance between objects in the two clusters. The resulting diagram from a Hierarchical Cluster Analysis is called a dendrogram , in which data are nested into brackets of increasing dissimilarity. 1959 – 1961 3 x 2 3 x 2 Factory cast iron intake, glass beaded and painted correct Cadillac blue, correct Rochester carburetors completely cosmetically and mechanically restored, linkage and fuel lines, NO air cleaner. This is the first complete linkage genome scan for agoraphobia and the third for PD. which was focused on single-linkage hierarchical clustering, to the case of com-plete linkage by devising an ad hoc poisoning attack. ! "# \$ % Comparison of single and complete linkage clustering with the hierarchical factor classification of variables S. Modello: GX3205SM. Proximity between two clusters is the proximity between their two most distant objects. On the basis of this definition of distance between clusters, at each stage of the process we combine the two clusters that have the. metric str or function, optional. Single linkage and complete linkage clustering are popular in use. Observe that single-linkage, average-linkage, and complete-linkage are linkage-based clustering functions. Hierarchical Clustering better than Average-Linkage Moses Charikar Vaggos Chatziafratis Rad Niazadeh Abstract Hierarchical Clustering (HC) is a widely studied problem in exploratory data analysis, usually tackled by simple ag-glomerative procedures like average-linkage, single-linkage or complete-linkage. With single linkage, the distance between two clusters is defined by the distance between the closest (in attribute space) observations from each cluster. The leaf nodes are numbered from 1 to m. We will concentrate on the first steps, establishing linkage and measuring linkage distance. cpp ( File view ) From： Hierarchical clustering implementation (complete linkage, single linkage) Description： Complete-linkage clustering is one of several methods of agglomerative hierarchical clustering. It can sometimes produce clusters where observations in different clusters are closer together than to observations within their own clusters. * Add single linkage to digits agglomerative clustering example. For example, the distance between clusters "r" and "s" to the left is equal to the length of the arrow between their two closest points. it Abstract. new organism Loci % recombination a-b 50 Data allow genes to be put in two linkage groups: a-c 10 a-d 50 a 10 c b 20 d b-c 50 b-d 20 c-d 50. In this paper, we present DiSC, a Distributed Single-linkage hierarchical Clustering algorithm using MapReduce frame-work. Ward is the most effective method for noisy data. It has been called a "linkage function," (see, for example, [3] and. d (13)2 = min {d 12 , d 32. Between-cluster distance has been formalized in a variety of ways. In addition, the complete linkage method does not have the tendency to chain, as does the single linkage method. Single, complete and average and centroid linkage Figure 4. The results showed the single linkage and complete linkage method, the same optimal cluster was obtained, that is 2 clusters with Davies Bouldin index about 0. Methodology: Data linkage 1 Data linkage Data linkage is a part of the process of data integration – linking combines the input sources (census, sample surveys and administrative data) into a single population, but integration also processes this population to remove duplicates/mis-matches. This chart provides a list of the Unicode emoji characters and sequences, with images from different vendors, CLDR name, date, source, and keywords. Three complete linkage SNPs at −534A/G, −407T/G, and −332C/T were detected, resulting in three genotypes (namely, AA, AB, and BB). min(D man) = min(d 13) = 1. This will give you a new perspective on the effect the decision of the linkage method has on your resulting cluster analysis. Complete linkage is strongly biased toward producing clusters with roughly equal diameters, and it can be severely distorted by moderate outliers (Milligan 1980). bam file with a single-linkage clustering, or a script which generates a distance matrix from a. Robust Multiple Structures Estimation with J-linkage Roberto Toldo and Andrea Fusiello Dipartimento di Informatica, Universita di Verona Strada Le Grazie 15, 37134 Verona, Italy andrea. Types of linkage. w+ m+ w m 2. We take advantage of scipy. Complete linkage clustering. Properties: Single and complete linkage can have problems with chaining and crowding, respectively, but average linkage doesn't Cutting an average linkage tree provides no interpretation, but there is a nice interpretation for single, complete linkage trees Average linkage is sensitive to a monotone transformation. Single linkage, also called nearest neighbor, uses the smallest distance between objects in the two clusters. Differences in the resulting dendrogram structure are discussed. Mahfouz PhD, Department of Computer and Systems Engineering Faculty of Engineering, Alexandria University Alexandria, Egypt [email protected] the number of clusters to be known in advance as partitioning algorithm do. BAVERAGE is the default and can also be requested with keyword DEFAULT. The distance of this linkage would be 16. I am guessing you are asking in terms of genes,so in pleiotropic condition a single gene will control more than one phenotypic character or in simple terms a single gene will control more than one trait of the organism. Quick start Cluster analysis of data Single-linkage cluster analysis using Euclidean distance of v1, v2, v3, and v4 cluster singlelinkage v1 v2 v3 v4 Same as above cluster single v1 v2 v3 v4. Complete Linkage: In complete linkage, we define the distance between two clusters to be the maximum distance between any single data point in the first cluster and any single data point in the second cluster. Other than that Average linkage and Centroid linkage. 'single' Nearest distance (single linkage method) 'complete' Furthest distance (complete linkage method) seqlinkage incorrectly doubled the input pairwise. If both objects are clusters, the distance between the two closest members are used. llows anisotropic and non-convex shapes. The results showed the single linkage and complete linkage method, the same optimal cluster was obtained, that is 2 clusters with Davies Bouldin index about 0. The group (series-series) similarity scores are based on a linkage method from hierarchical cluster analysis (e. I Thus, the magnitude of the coe cient is important, not the sign. Indeed, experience has shown that a given step may increase the number of true positives and reduce the number of false positives in one year, and that the same step in. We accomplish this with a novel Exponential Linkage function that has a learnable parameter that controls the interpolation. White Latin Americans (19,383 words) exact match in snippet view article find links to article 1177/0091270006287586. (a) noncrossovers, single crossovers, double crossovers (b) 8. Therefore, you could conclude that the single linkage method did a fine job identifying the outliers. w+ m+ w m 2. org is recommended. Now 29 years later we have grown, but still have the same philosophy, to ensure that our products will stand up to the high demands of the top level racers around the world and the average riders alike. These particular genes do not generate combinations that are non-parental. Section: 7. If there is no recombination (complete linkage) then the two genes are essentially inherited as if they were a single gene. Single, complete, and average linkage method The distance between two objects is defined to be the smallest distance possible between them. (2) Complete Linkage - The distance between two clusters is defined as the longest distance between two points in each cluster. Shape of clusters Single-linkage a. This helps - Use Distance Matrix in scipy. DWSP211 Drive Description Dw Linkage Dwsp211 Brand Drive. R # # Enter the data as a matrix and assign variable names. 2 metre cut, A frame linkage linkage mounted, 540 PTO, hydraulic lift up / down with single spool ram & springs with contour float, manual quick release fold down side sheets for work / transport, front transport. Generally speaking, the AC describes the strength of the clustering structure. We understand that your time is valuable. Access to this program and data are tracked and recorded. In genetics complete linkage is defined as the state in which two loci are so close together that alleles of these loci are virtually never separated by crossing over. 38 * 10−85) and β57A (OR 3. 5 00 (a) (b) (c). It is blasphemy to suggest such a connection. Draw the dendrogram and compare the. Another thing to consider is the distance between instances when we have already two or more instances in a cluster. FIND A DEALER Retailers Where to Buy. There are others mentioned in the literature, but these are the best-known methods. Shop online for thousands of great items like the Off Road. Single nucleotide polymorphisms, frequently called SNPs (pronounced “snips”), are the most common type of genetic variation among people. Complete linkage tends to find compact clusters of approximately equal diameters. the dendrograms below is the result of single-linkage, which is the result of average-linkage, and which is the result of complete-linkage? Explain your reasoning. View this article on JSTOR. The notorious chaining tendency of single linkage can be alleviated by specifying the TRIM= option (Wishart 1969, pp. During reproduction, chromosomes on the same chromosome pair, exchange sections of DNA. However they suffer from high computational cost and their inability to recover early bad. This process is experimental and the keywords may be updated as the learning algorithm improves. Linkage Groups 3. , joins the clustersAandBwith the smallest. But, if all inter-observations distance are equal to 2, we would have that the single and complete linkage dissimilarities between $$\{1,2,3\}$$ and $$\{4,5\}$$ are equal to 2. What percent recombination would be expected in the offspring of this type of cross?. With a the runtime of the general case can be reduced to at the cost of further increasing the memory requirements. The notorious chaining tendency of single linkage can be alleviated by specifying the TRIM= option (Wishart 1969, pp. Repeat step third until one single cluster remains. A total of 407 markers (389 autosomal, 18 X chromosome) were genotyped. In contrast, top-down methods directly define similarity between clusters. For example, a SNP may replace the nucleotide cytosine (C) with the nucleotide thymine (T) in a certain stretch of DNA. 97, Florida Statutes, establishes state audit and accountability requirements for state financial assistance provided to nonstate entities. b) Complete or farthest-neighbour clustering takes into account the dissimilarities between the most distant members of a cluster. In the complete linkage, also called farthest neighbor, the clustering method is the opposite of single linkage. SINGLE LINKAGE: In single linkage, the distance between two clusters is defined as the minimum distance between a point in one cluster and a point in the other cluster. The test cross shows that these autosomal genes exhibit linkage; they do not assort independently. We sell used Scout parts as well as common upgrades such as disc brake conversion kits, lift kits, fuel injection kits, offroad steering upgrades and roll bars. Purchase the complete fuel injection conversion or get the individual sub-system packages to create your own custom conversion kit! Fine tune your creation with free REDLINE fuel injection software and documentation!. One drawback is that groups with close pairs can merge sooner than is optimal, even if those groups have overall dissimilarity. Results: We observed an independent replication of previously observed linkage at chromosome 20p13 (P<0. New “low-friction” linkage bearing seals made by SKF provide noticeably freer linkage action, offering better suspension feel and performance throughout the shock stroke. Estimating linkage disequilibrium from genotypes under Hardy-Weinberg equilibrium Tin-Yu J. Not surprisingly, complete linkage tends to produce very compact clusters. Complete linkage tends to be less desirable when there is a considerable amount of noise present in the data. A standalone bar-linkage mechanism to solve linear equations was patented by Arnold Spilker. Complete Linkage in Drosophila melanogaster is an article from The American Naturalist, Volume 56. Complete linkage In this method, the distance between one cluster and another cluster should be equal to the greatest distance from any member of one cluster to any member of the other cluster. (single linkage), (complete linkage), (group average). Draw the dendrogram and compare the results. Key words: spinal cord injury, steering linkage, wheelchair. However they suffer from high computational cost and their inability to recover early bad. The Central Board of Direct Taxes (CBDT) through a notification also extended the time limit by. Complete Linkage: Genes are so close together as part of the same structure that they are inherited as a single unit. (d) Centroid linkage first computes the centroid of each group and then looks at the distance between them. The SNP assays are based on KASPar genotyping chemistry and were detected using the Fluidigm dynamic array platform. Complete linkage is when two genes are on a single pair of homologs and no exchange occurs, producing non crossover/parental gametes. Complete Linkage. * Update documentation to reflect the addition of single linkage. This helps - Use Distance Matrix in scipy. single_linkage_cluster. omplete-linkage: consider the distance between one cluster and another cluster to be equal to thegreatest distance from any member of one. Morgan further suggested that coupling and repulsion are two aspects of single phenomenon called linkage. In the case of male Drosophila there is a complete absence of recombinant types due to the absence of crossing over. In complete linkage, there is virtually no recombination. Condizione: Usati. The mapping population data revealed 16 major linkage groups with 520 markers, and 10 quantitative trait loci (QTLs) significantly associated with Taro Leaf. If single cross over occurs 100% of time,. 1:2:1 phenotypic and genotypic ratio Test cross produces 1:1 ratio Genetic Symbols P1 : bw hv+ bw+ hv X bw hv+ bw+ hv Genes located on same chromosome will show evidence of linkage. min(D man) = min(d 13) = 1. Genes that are on the same chromosome as those that determine sex are said to be sex-linked. To improve mapping efficiency, when two or more SNPs were identical (i. Introduction to Linkage of Gene: Morgan (1910) noted similar behaviour in Drosophila melanogaster (Fig. See the Linkage Methods section below for full descriptions. (a) Single linkage looks at minimum distance between all inter-group pairs. Hierarchical clustering. In linkage with there are still two genes on a single homolog but they are usually far apart and cross over occurs between two nonsister chromatids, producing a mixture of recombinant/crossover gametes and. Quick Features 130mm of rear LTS (Linkage Tuned Suspension), 130mm front travel Single Flip-Chip takes you from low and fast to lower and faster Threaded bottom bracket Carbon main frame / alloy rear swingarm Room for a water. Two-phase solution for clustering large datasets When working with a large number of observations, the computations for a hierarchical cluster solution may take hours to complete, making this solution less feasible. Complete linkage avoids chaining but suffers from crowding meaning that Clusters are compact, but not far enough apart. Using complete linkage two. It is the basis for the synthesis of all the important biological macromolecules (carbohydrates, proteins, lipids, nucleic acids) from their simpler sub-units. “The community is the driver for this,” said O’Neal. If single cross over occurs 100% of time,. The Central Board of Direct Taxes (CBDT) through a notification also extended the time limit by. , Washington, DC 20410 Telephone: (202) 708-1112 TTY: (202) 708-1455. 9 Construction of an SSR Map of Cassava Based upon Linkage Analysis in a F2 Cross Derived from Non-Inbred Parents and QTL Mapping of Early Bulking. Average and complete linkage are generally preferred over single linkage, as they tend to yield more balanced dendrograms. 1/2) Lower Link Lift Arm - Complete (Cat. The range of values the linkage disequilibrium coe cient can take on varies with. complete linkage, reliability is compromised, as it is difficult to apply the algorithm to other years, databases, or geographic settings. 80 * 10−84) to be associated with T1D in the DQ8/9 cluster representing all ten residues (α22, α23, α44, α49, α51, α53, α54, α73, α184, β57) due to complete linkage disequilibrium (LD) of α44 with eight such residues. The Sensor offers astounding performance and ride quality at a fraction of the price compared to the competition. Now 29 years later we have grown, but still have the same philosophy, to ensure that our products will stand up to the high demands of the top level racers around the world and the average riders alike. Start studying Linkage, recombination and eukaryotic gene mapping. single, complete, average linkage. This solution helps domain users perform self-service password reset, self-service account unlock, employee self-update of personal details (e. 1/2) Lower Link Lift Arm - Complete (Cat. This feature is not available right now. See Chromosome , Crossing over , Gene , Inheritance, Linkage, Locus, Nonlinkage, Partial linkage. There is always parental combination's and no any recombination. Single-piece cast. Robust Multiple Structures Estimation with J-linkage Roberto Toldo and Andrea Fusiello Dipartimento di Informatica, Universita di Verona Strada Le Grazie 15, 37134 Verona, Italy andrea. In genetics, complete linkage is defined as the state in which two loci are so close together that alleles of these loci are virtually never separated by crossing over. In a letter dated June 19, the financial commissioner of the railways told the general managers of all the zones that the traffic earnings of the national transporter dropped by 58 per cent at the end of May, as compared to the corresponding period of the previous year. Conclusions—Multiple linkage regions support the notion that risk for hypertension is due to multiple (ie, oligogenic) susceptibility loci. The distance between two clusters is equal to the distance of the closest elements from the two clusters The single linkage has practical disadvantages and is therefore not selectable in ClinProTools 14. Types of Linkage: Depending upon the presence or absence of new combinations or non- parental combinations, linkage can be of two types: (i) Complete Linkage: • If two or more characters are inherited together and consistently appear in two or more generations in their original or parental combinations, it is called complete linkage. See the Linkage Methods section below for full descriptions. (dij) •Single linkage clustering is sometimes called “nearest neighbor” clustering. Linkage and crossing over can be seen as processes that have opposite effects: linkage keeps particular genes together, and crossing over mixes them up. Complete Linkage. There is no single best statistic that quantifies the extent of LD. for divisive clustering, which makes them too slow for large data sets. Introduction to Linkage of Gene: Morgan (1910) noted similar behaviour in Drosophila melanogaster (Fig. Alternative linkage schemes include complete linkage clustering, average linkage clustering (UPGMA and WPGMA), and Ward's method. The restaurant industry must reduce the use of single-use plastic items by 30%. Single- Average- and Complete-Linkage. White Latin Americans (19,383 words) exact match in snippet view article find links to article 1177/0091270006287586. " Wishart (1969), however, reports that single linkage failed to de-. Income tax return filed by certain citizens or residents of the United States. 1:2:1 phenotypic and genotypic ratio Test cross produces 1:1 ratio Genetic Symbols P1 : bw hv+ bw+ hv X bw hv+ bw+ hv Genes located on same chromosome will show evidence of linkage. Menghitung jarak antar kelompok (1 dan 3) dengan kelompok lain yang tersisa, yaitu 2, 4 dan 5. The frequency of crossing over has been used to construct linkage maps. (a) Single linkage looks at minimum distance between all inter-group pairs. summary, linkage adds a further complication to interpreta-tions of the results of genetic crosses. During reproduction, chromosomes on the same chromosome pair, exchange sections of DNA. 01), while loci at 6q27 and 8q13. There is no Krishna-Christ linkage; the topic is ridiculous; it is not even worth investigating. This method is much like the single linkage, but instead of using the minimum of the distances, we use the maximum. A greater number of nucleotide-binding leucine-rich repeat disease resistance genes were present in genomes of taro than the duckweed, ∼391 versus ∼70 (∼182 and ∼46 complete). Types of linkage. Other than that Average linkage and Centroid linkage. ADD COMMENT • link written 5 months ago by chrchang523 ♦ 6. valve bodies. All thread is 5/16"-24. The opposite method is complete-linkage (complete) which measures distance based on the furthest away elements in each cluster. Single linkage and complete linkage are two popular examples of agglomerative clustering. 上一篇博客介绍了single linkage是让所有簇的距离为簇间结点最短的距离，同时每一次合并所有簇间距中最短的那一个。而complete则是让簇间结点距离最长的作为簇间距，并且每一次让所有簇间距最短两簇进行合并，因此实际上不是都找最长，而是最长最短。. Costruttore: JF. Linkage "a" and "b" are too close to each other. Determining clusters. cluster analysis. These approaches include: • Using map locations to distinguish different genes with similar sequences, mutant phenotypes, or functions. In single linkage, the distance between two clusters A and B is computed as the minimum distance between a point (i. Complete‐Linkage and general algorithms defined by d(A i, B) = = α,d(A i, A r)±α s d(A i, A s)±βd(A r, A s) will be discussed in two subsequent papers. Jim Carter's OldChevyTrucks. Likewise, (b) is for complete-linkage, and (c) must be for average-linkage. 117-120); Read about this test and do the problems! What if linkage is not very tight, and the percentage of recombinant classes approaches 50%?. The shortest of these links that remains at any step causes the fusion of the two clusters whose elements are involved. A drawback of this choice is that single Linkage may yield long, extended clusters or chains of points (\chaining") so points in the same cluster could be quite dissimilar. (a) We don't have enough information to tell. The closer the genes are located, stronger is the linkage. This class includes commonly-used algorithms such as single-linkage, average-linkage, complete-linkage, and Ward's method. A slider crank mechanism, sometimes referred to as a three-bar-linkage, can be thought of as a four bar linkage where one of the links is made infinite in length. which was focused on single-linkage hierarchical clustering, to the case of com-plete linkage by devising an ad hoc poisoning attack. Für andere Fusionierungsmethoden wie Average-Linkage sind keine effizienten Algorithmen bekannt. In genetics, complete linkage is defined as the state in which two loci are so close together that alleles of these loci are virtually never separated by crossing over. In the complete linkage method, D(r,s) is computed as. This will give you a new perspective on the effect the decision of the linkage method has on your resulting cluster analysis. Types of Linkage: Depending upon the presence or absence of new combinations or non- parental combinations, linkage can be of two types: (i) Complete Linkage: • If two or more characters are inherited together and consistently appear in two or more generations in their original or parental combinations, it is called complete linkage. The distance between the two clusters is the same as the distance between the two rows (or. Significance of Linkage 1. Costruttore: JF. dictated by the distance and linkage measures chosen by the researcher. Coupling and repulsion phenomenon was concerned with a) Crossing over b) Mutation c) Linkage d) All of these 10. New “low-friction” linkage bearing seals made by SKF provide noticeably freer linkage action, offering better suspension feel and performance throughout the shock stroke. Observe that single-linkage, average-linkage, and complete-linkage are linkage-based clustering functions. Pathobiology Genetics. Z = linkage(Y,'single') If 0 < c < 2, use cluster to define clusters from Z when inconsistent values are less than c. I Thus, the magnitude of the coe cient is important, not the sign. Complete Linkage: When the genes are very close to each other they have no chance of separation from each other and are always transmitted together, generation to generation. The original matrix has 56,000 rows (genes) and 7 columns (treatments). Sequen-tial RANSAC - sequentially apply RANSAC and remove the inliers from the complete linkage and average linkage [10]. To calculate the distance between two clusters, each possible combination of two rows (or columns) between the two clusters is compared. We pay attention solely to the area where the two clusters come closest to each other. - check the tail slider for smooth running. Candidate SNP selection. Hierarchical clustering Similarity between clusters 29 complete linkage single linkage. two single data points in each cluster, and complete linkage, where the distance is the minimum similarity between any two single data points in each cluster. Access to this program and data are tracked and recorded. Für andere Fusionierungsmethoden wie Average-Linkage sind keine effizienten Algorithmen bekannt. Draw the dendrogram and compare the. def HC(data, meth, metr, num_clust): # Mahalanobis Hierarchycal Clustering # data: the set of variables used to perform the clustering analysis # method: method to perform the HCA [single(default), complete, average, weighted, average, centroid, median, ward] # metric: the metric to perform the HCA [euclidean(default), mahalanobis] # num_clust: predefined number of clusters, if not present. Where you'll always get a warm welcome. Single Cell RNA-seq Clustering Ahmed Mahfouz Leiden Computational Biology Center, LUMC Delft Bioinformtaics Lab, TU Delft. In the complete linkage method, D(r,s) is computed as. Values closer to 0 suggest less well-formed clusters such as the single linkage dendrogram in Figure 21. Not surprisingly, complete linkage tends to produce very compact clusters. Proximity between two clusters is the proximity between their two most distant objects. The gene located in different chromosome are unlinked genes. Join Target Circle to start receiving the benefits of voting and services. Start studying Linkage, recombination and eukaryotic gene mapping. There seems to be some discrepancy in whether single-link or complete-link is sensitive to outliers. There are various methods available: Ward method (compact spherical clusters, minimizes variance) Complete linkage (similar clusters) Single linkage (related to minimal spanning tree) Median linkage (does not yield monotone distance measures) Centroid linkage (does.
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July 14, 2020
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# How would you calculate a contour integral? [duplicate]
How would you be able to calculate a contour integral of 1/(((z-1)^2)*(z-i)) over the contour |z-1| = 1? Not sure how to type that in :/
With
f[z_] := 1/((z - 1)^2*(z - I))
the contour integral around a circle centered at z == 1 can be parameterized by arc length around the circle.
Integrate[(f[z] /. z -> 1 + Cos[t] + I Sin[t]) D[1 + Cos[t] + I Sin[t], t],
{t, 0, 2 Pi}]
(* π *)
which, of course, is equal to the residue at z == 1, multiplied by 2 π I. (The pole at z == I is outside the contour and so does not contribute to the integral.)
Alternatively, the integral can be performed over a region.
reg = ParametricRegion[{1 + Cos[t], Sin[t]}, {{t, 0, 2 Pi}}];
Integrate[(f[z] /. z -> x + I y) (-y + I (x - 1)), {x, y} ∈ reg]
which gives the same result. (-y + I (x - 1)) is the equivalent of D[1 + Cos[t] + I Sin[t], t].
f[z_] := 1/((z - 1)^2*(z - I))
So for a=1, you get Pi
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# Collision detection with a tilemap- raping the walls
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13 replies to this topic
### #1keelx Members
Posted 22 May 2011 - 03:39 PM
I am using Java to make a 2d platformer, and I am having trouble with the enemies. I am using rectangles and the intersects method to detect collisions. I just want the enemy to run back and forth between two bricks.
Here is my code:
public void checkCollision() {
for (int i = 0; i < map.tiles.size(); i++) { //map.tiles is an ArrayList of the tiles on the map and size() is how many tiles are on the map, we are looping through each tile
Tile temp = map.tiles.get(i); //temp is the handle to the tile we are on (not physically on, but what tile we are on in the loop)
if (map.tiles.get(i).index != 0) { //checking if the tile is not blank, if there is actually a brick or whatever there
if (bounds.intersects(temp.bounds)) { //bounds is the rectangle that encapsulates the object. It is used for detecting intersections.
if (getX()+32>temp.x) //getX(), the x coordinate of the object, temp.x, the x coordinate of the current tile
setX(getX()-1); //push the enemy outside of the block, so he doesn't collide with it again.
if (getX()<temp.x+32)
setX(getX()+1); //same, but in opposing direction if he is going the other way
hspeed = hspeed * -1; //reverse the horizontal speed
}
}
}
}
The enemy is suffering from what I call 'rape the wall syndrome', meaning he is just bouncing back and forth, on a single brick like this when he runs into one:
[~] = enemy
[_] = brick
vertical order of frames:
[~] [_]
[~][_]
[~]_]
[~][_]
[~] [_]
[~][_]
[~]_]
[~][_]
[~] [_]
and repeat.
More about the enemy- he is a 32x32 square, and he is raised one pixel above the tiles on the ground so as not to collide with them, and have him reverse his direction in the same spot. In his 'main loop' his X coordinate is updated by his vertical speed like so:
setX(getX()+hspeed)
I don't know why he is doing this, and I don't know how to solve this. Thanks in advance to whomever can help me find the solution.
### #2coderWalker Members
Posted 22 May 2011 - 06:27 PM
Man not trying to be harsh but drop the list totally first.
This will cause LOTS of problems with Logic, Performance and Memory!
Lets say you want to have a level the size of Super Mario Bros 1-1:
Memory:
This level contains 5936 blocks!
If thier represented by ID's that would only be 5936 bytes
Performance:
Meaning for everything on screen that's moving you are going to have loop running that loops with all those statements 5936 times!
Again in the Mario level you would be lucky to get 60fps even with something this simple.
Implementation:
The best way to do this is to have a unique id for each tile.
A great practive for collision would be to have 0 for an empty block.
Then have a 2d Array and store all the id's in the array.
Drawing:
Basically when drawing have 2 for loops to draw the level, very simple.
Very Fast! Collision Routine:
(This routine will always take the same amount of time,
while the above routine gets Exponencially slower when the number of blocks increases)
Then when checking for collision just:
Take the coordinate of the enemy and say for example
bool collide()
{
if (tile[enemy.x-1,enemy.y] != 0)
{
return true;
}
else if (tile[enemy.x+1,enemy.y] != 0)
{
return true;
}
else
{
return false;
}
}
CoderWalker
Webstrand
### #3keelx Members
Posted 22 May 2011 - 06:53 PM
Well, what if he is not moving an entire block over every frame? He is only moving two pixels, and your method suggests that there is a block at each pixel, but there is a block at every 32 pixels. Also, the enemy (and the player) are rectangles, not pixels.
### #4coderWalker Members
Posted 22 May 2011 - 07:00 PM
Most games use the tiles for static objects (ones that dont move).
Sorry I didn't mention this.
Meaning the enemy would not be drawn with the map.
if each tile is 32x32.
The solution is:
tile.x = enemy.x/32;
tile.y = enemy.y/32;
Since this is integer division they will get "Snapped" to 32x32 positions
ex:
1 = 36/32;
0 = 14/32;
then check tile[1][0]
Anything that is Dynamic (not snapped to the map) should:
Draw itself
Have Actual Pixel coordinates instead of tile coordinates
What I talked about above was implemented in many games Mario, Zelda, Donkey Kong.
Evidence:
Attached to map?
--------------------------
Soldier - No
Weapon - No
Grass - Yes
Dirt - Yes
Fence - Yes
Plant - Yes
Webstrand
### #5/ owl Banned
Posted 22 May 2011 - 07:06 PM
Codewalker's advice is good. I'd take it.
@codewalker: Asking for members to rate you up is like asking a girl to get in your pants. The chances of it working (for free) are negligible. Removing any suggestion to it from your post will increase your chances.
I like the Walrus best.
### #6Trienco Members
Posted 23 May 2011 - 10:52 PM
if (getX()+32>temp.x) //getX(), the x coordinate of the object, temp.x, the x coordinate of the current tile
setX(getX()-1); //push the enemy outside of the block, so he doesn't collide with it again.
if (getX()<temp.x+32)
setX(getX()+1); //same, but in opposing direction if he is going the other way
Are you sure +/- 1 is going to be enough to not still be stuck and just causing collisions (or rather intersections) every single frame? You could also consider the amount of movement and how much of that was used to get to the block and how much is left and should be used to move away from the block (at least pretend the world doesn't stand still between frames).
Since I have no idea if your getX returns the left side or the center, I'll take a guess that X is left and 32 is the size of the object AND block (consider using a named const instead of magic numbers).
So,
insideX = getX() + 32 - temp.x
is how far inside the block you ended up. Now either your new position should be
setX(temp.x - insideX)
(since you should bounce off when you HIT the block, not when you've already moved halfway into it)
or at the very least
setX(temp.x) (-1 if you want to be on the safe side).
f@dzhttp://festini.device-zero.de
### #7keelx Members
Posted 23 May 2011 - 11:18 PM
Very nice! Now I don't have to rewrite my entire game code. But now he bounces one direction but goes straight through blocks in the other direction.
### #8Trienco Members
Posted 24 May 2011 - 10:53 AM
Very nice! Now I don't have to rewrite my entire game code. But now he bounces one direction but goes straight through blocks in the other direction.
How did you adjust the lines for the other direction? The code above only handles collision to the right.
f@dzhttp://festini.device-zero.de
### #9Sirisian Members
Posted 24 May 2011 - 12:49 PM
Oddly enough I was bored yesterday and made this. Right click view source and look at the update code to see how to handle tile collision and response between an entity. It might help you to see what your problem is. In regards to an enemy moving left until he hits the edge and reversing you just do Math.floor(entity.Position / tileSize) or in C++:
int tileX = static_cast<int>(entity.x / tileSize);
int tileY = static_cast<int>(entity.y / tileSize);
if (!map[tileX - 1][tileY + 1].Collidable)
{
// Set internal minimum move left to tileX * tilesize which is the far left of the current tile the entity is on.
}
For moving right:
if (!map[tileX + 1][tileY + 1].Collidable)
{
// Set internal minimum move left to tileX * tilesize + tileSize which is the far left of the current tile the entity is on.
}
Also if your enemy collides with a tile and the minimum translation distance for x is > 0 then just set direction to right and if it's < 0 set it to left.
### #10keelx Members
Posted 24 May 2011 - 06:26 PM
Okay. Solved. Although I think the best way would have been the 2d array of tiles, and the separation of object-space and tile-space, however as to avoid a complete rewrite, my new working code is as follows:
public void checkCollision() {
for (int i = 0; i < map.tiles.size(); i++) {
Tile temp = map.tiles.get(i);
float insideX;
if (map.tiles.get(i).index != 0) {
if (bounds.intersects(temp.bounds)) {
if (hspeed > 0 && !collided) {
insideX = (getX() + 32) - temp.x;
setX(temp.x - insideX - 32);
hspeed = -1;
collided = true;
break;
}
if (hspeed < 0 && !collided) {
insideX = getX() - temp.x + 32;
setX(temp.x + insideX - 24);
hspeed = 1;
collided = true;
break;
}
collided = false;
}
}
}
}
### #11Trienco Members
Posted 24 May 2011 - 11:09 PM
int tileX = static_cast<int>(entity.x / tileSize);
int tileY = static_cast<int>(entity.y / tileSize);
I see trouble unless you limit it to positive numbers. -1 should always be in tile -1 but ends up 0 unless you add special handling for negative coordinates.
Shouldn't the left side be:
insideX = (temp.x + 32) - getX(); (the right block edge minus the objects left edge
setX(temp.x + insideX); (the right block edge plus the amount you have to bounce back)
The numbers in your code confuse me. Where did 24 come from? Why the additional -32 after colliding to the right?
Wait, I noticed one bug in my code. You need to
setX(temp.x - 2*insideX)
(once to move "out" of the block, twice to where you should be after bouncing back)
So the fixed collision to the left would be:
setX(temp.x + 2*insideX)
And to exploit the neatness of math
if (hspeed > 0)
insideX = (getX() + 32) - temp.x;
else
insideX = getX() - (temp.x + 32);
//or more compact and confusing: insideX = getX() - temp.x + (hSpeed * 32);
setX(getX() - 2*insideX);
edit: damn non-descriptive names. getX, not temp.x
f@dzhttp://festini.device-zero.de
### #12keelx Members
Posted 24 May 2011 - 11:38 PM
Yes, that was what I had done originally (with the exception of multiplying the insideX by 2), but for some reason the enemy would then start flickering left and right, without moving, when he touched a wall. After much trial and error, this is what I got. Not entirely sure why it works, but it does. And as for subtracting 32, it made for smoother looking animation. Same for adding 24, but if I added 32 on that side it would end up on the other opposite of the block.
### #13Sirisian Members
Posted 25 May 2011 - 10:27 AM
int tileX = static_cast<int>(entity.x / tileSize);
int tileY = static_cast<int>(entity.y / tileSize);
I see trouble unless you limit it to positive numbers. -1 should always be in tile -1 but ends up 0 unless you add special handling for negative coordinates.
Indexes into an array are normally restricted to positive numbers. You see in a tile based game the tile map is normally represented by a multidimensional array either indexed with y * width + x or using jagged arrays so the restriction of only positive numbers is there.
### #14Trienco Members
Posted 25 May 2011 - 11:20 PM
Indexes into an array are normally restricted to positive numbers.
Yes, but entity positions aren't tile indices and you add implicit limitations to one level of abstraction (object positions) based on a much lower level (layout in memory). I'll not bring up languages where negative indices are perfectly normal and common, since we're talking Java here. Still, you should point out that your code requires positions to be limited to values >tileSize and <worldsize - tileSize not only to prevent "falling into the void", but because any position on or beyond the first or last tile will cause access violations.
Yes, that was what I had done originally (with the exception of multiplying the insideX by 2), but for some reason the enemy would then start flickering left and right, without moving, when he touched a wall. After much trial and error, this is what I got.
You should avoid the "trial and error shortcut", it will only result in stuff that looks like a working solution, but blows up every 100th time (which is the most annoying kind of bug to find). What if your randomly chosen value of 24 only works for 75% of cases, because for the remaining situations it should have been 25 or 29? What if in two weeks you suddenly see objects flickering twice and then pop out on the wrong side of a wall?
Programming means analyse and understand the problem, create an informed plan to solve it and then implement that plan. Seriously, you are currently dealing with the most trivial kind of collision you will ever get. It's only 1 dimensions and you're not even checking for actual collision, but for intersection (which can easily fail with fast moving objects and/or variable time steps). If you decide you can't be bothered to wrap your head around the problem while it's simple, how do you expect to get anything done when you add a second dimensions or want objects that move more than 32 pixels per frame or add thinner blocks as walls?
Look at how you calculate insideX for collisions to the left and right: Notice something? It's the exact same thing. You add 32 and subtract temp.x. The whole point is to determine how much you overlap the block and how much to bounce back. While you _could_ still exploit the fact that apparently everything is 32 wide, but that's not going to make the code any easier to understand.
You can of course ignore the idea of "how close was the object to the block before colliding and how much of it's movement should have already happened in the other direction", in that case you don't need any calculation except for placing the object to the left (temp.x - 32) or to the right (temp.x + 32) and change the direction (hspeed *= -1). Just make sure to use a method you actually understand and could explain (nothing more embarassing than answering "that's cool, how does it work" with "uhm... no idea, I just randomly put some stuff there until it kind of worked most of the time").
f@dzhttp://festini.device-zero.de
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## Tuesday, September 24, 2013
### Bye
I've got a new job with Pine River, and I want my new colleagues to know I'm not going to blab about anything that comes up, so blogging is now really over. Of course, if you bump into me you can always buy me drinks and try to get me spill the beans (about non-proprietary matters) but I should warn you, I can drink a lot of beer. Best.
## Tuesday, September 17, 2013
### Historical CBO Budget Projection Highlights Bias
Recently the CBO issued its annual budget projection, and it's pretty benign for the next decade, then climbs at a pretty measured pace.
Yet, note that in the last recession our debt relative to GDP doubled. Given that economists still don't have a good model for predicting business cycles let alone avoiding recessions, we can expect more of them. I think the odds that we elect a modern-day Calvin Coolidge next term are much smaller than the odds the deficit will increase dramatically when the next recession hits.
Consider that in 2007, before anyone saw any hint of the 2008 crisis, the debt was actually projected to fall, but we know how that turned out (black are actual historicals). That is, they never anticipate recessions, though we all know they haven't been abolished. I pulled these numbers out of the 2007 'wayback machine' which is a great way to hold large institutions accountable, because for some reason the CBO doesn't keep their historical forecasts on their current site (maybe the NSA can get Google to scrape them away?). Liberals who happen to be economists (eg, Brad DeLong) think the latest objective projections prove we have no budget worries. I guess some people really do think This Time It's Different.
## Sunday, September 08, 2013
### MSCI Quality Index
I was unaware MSCI had beaten AQR to the punch by producing a boatload of quality indices last spring. These are applied worldwide, so they are necessarily more parsimonious than AQRs...but jeez, these are really barebones:
1) Net Income/Book Equity
2) Debt/Book Equity
3) Earnings volatility over 5 years
Instructively, they Winsorize the data, which everyone should do to financial ratios (ie, truncate extremums). But, book equity in the denominator? Earnings volatility over 5 years? Those seem like bad choices, and AQR's quality index will be superior.
I have a feeling MSCI is a bit confused, as they have another tab noting their 'Risk Premia Indexing', which they note
An accumulating body of empirical research has found positive gross excess returns from exposure to factors (or risk premia) such as Value, Momentum, Low Size (small firms), and Low Volatility stocks. The studies show that these factors historically have improved return-to-risk ratios. Today, interest in risk premia (also known as smart beta or alternative beta) has been widespread across the institutional investor community.
In other words, risk premia are really return premiums, because predictable returns only come from risk (in theory). But then, they also 'improve return-to-risk ratios', because, as we all know, these factors aren't risk in any obvious way, so strangely they all have 'excess return' premia. Indeed, 'value and size were initially thought to be due to distress risk, which would show up only episodically. Alas, 'quality' is basically an metric of anti-distress, and this generates a return premium, which MSCI occasionally calls a 'risk premium'...so basically whatever asset outperforms over the next 20 year period will ex post be declared risky.
The risk-begets-return model of economics is clearly nonfalsifiable amongst current financial academics and their coterie. They still say interesting things on occasion, so it doesn't render them useless, but it definitely impairs their ability to see and interpret reality.
## Wednesday, September 04, 2013
### de Botton on Status Anxiety
I find Alain de Botton's approach to philosophy rather refreshing, because one senses his genuine lack of certainty, and appreciation of discovering, in his works. He's interested in applying virtue for daily betterment, and the search for meaning, two very important goals in my life. Interestingly he was insightfully quoted in a NYT review of Sophie Fontanel's self-indulgent book on her self-induced celibacy, which highlighted his breadth and profundity (de Botton's quip was basically that 'sex is messy, get over it').
Anyway, here's de Botton on status anxiety. He argues that status anxiety is worse than ever because now we believe we are less constrained by our birth, more responsible for our fate. Paul Krugman agrees with this view of life, but like most economists, can't take this to it's ultimate implication, that this this leads to a zero risk premium, which when combined with the various attractions of sexy stocks, leads to high risk assets have lower-than-average returns (see my book The Missing Risk Premium).
## Sunday, September 01, 2013
### How to Maximize Lottery Revenue
As a proponent of the idea that people are oriented towards their relative success, not absolute wealth, I think this lottery idea is fiendishly clever. Here's a description from TheWeek of a clever way to capitalize on this instinct:
A salient example is the "Postcode Lottery" in the Netherlands. Weekly it awards a "Street Prize" to one postal code, the Dutch equivalent of a zip code, chosen at random. When a postal code (usually about 25 houses on a street) is drawn, everybody who played the lottery in that code wins about \$12,500 or more. Those living there who neglected to buy a ticket win nothing — except the chance to watch their neighbors celebrate.
In a 2003 study, researchers in the Netherlands noted that fear of regret played a significantly larger role in the Postcode Lottery than in a regular lottery. It was not the chance of winning that drove the players to buy tickets, the researchers found, it was the idea that they might be forced to sit on the sidelines contemplating missed opportunity.
Todd Mitton and Keith Vorkink from (boring) BYU published Why Do Firms With Diversification Discounts Have Higher Expected Returns? Their answer: no skew. People will pay up for lottery tickets, but if you take those dreams away, it becomes an asset that is neglected. They find diversified firms offers less skew, and diversification discounts are significantly greater when the diversified firm offers less skewness than typical focused firms in similar business segments. They suggest a substantial proportion of the excess returns received on discount firms relative to premium firms can be explained by differences in exposure to skewness.
The implication is clear: people pay a premium for volatile stocks that have stories and potential. Conditional upon playing in a risky game, such as equities, there's not a return premium for risk, there's a premium for boring.
## Sunday, August 25, 2013
### AQR's Quality at a Reasonable Price
Our intrepid equity researchers at AQR have come out with a new paper adding to the color on how to pick a strategy given value considerations. In Asness, Frazzini and Pedersen's latest paper, Quality Minus Junk, they first try to create a 'quality' metric, and then try to meld it with value.
Quality is defined very clearly as the composite of 4 factors (each of which is made up of 3-5 ratios):
• Profitability (eg, Net Income/Assets)
• Growth (eg, change in Profitability)
• Safety (eg, volatility, leverage)
• Payout (eg, equity issuance, dividend payout)
They find that
1) Stocks with higher 'quality' have higher market/book ratios (higher price ceteris paribus)
2) A long-short portfolio, where one goes long high quality, short low quality, generates significant, positive excess and total returns
They assert that a value-quality portfolio that tries to balance quality with value has nice properties, and the Sharpe maximizing combination is about 70% quality, 30% value. This is coming from Asness, who is a pretty big value proponent, so I think this is rather telling (value losing it's pre-eminence!).
Their quality metric has a kitchen-sink aspect to it, with about 20 ratios that go into those 4 different groupings. I could imagine many people would find this an attractive framework to develop and tweak their own quality metric, substituting for various ratios, or subtle changes to the functional form. Haugen and Baker's (2008) Case Closed, and Zack's Handbook of Investment Anomalies are good places to look for alternative ratios.
I would like to see how this QMJ factor compares to Analytic Investor's Volatile Minus Stable (VMS) factor...they seem similar, though obviously 1) they are negatively correlated and 2) the VMS factor is simply a vol factor, which is just one part of the 'quality' metric.
Lastly, I love the little note at the end:
Our results present an important puzzle for asset pricing: We cannot tie the returns of quality to risk
By construction their return-generating metric seems patently 'anti-risk', as quality implies 'low risk'. The risk-begets-return theory obviously has a lot of intuition, because empirically it's counterfactual when not irrelevant. I think if you divided the data described by asset pricing theory into 'puzzles' and 'consistent', it's mainly puzzles.
### Economath and the Drake Equation
There were several posts last week on the hypothesis that there's too much emphasis on mathematical modeling in modern economics. Most said yes (Dave Hendersen, Bryan Caplan, Noahpundit, Robin Hanson, The NewYorkTimes), though Krugman said no.
Krugman's experience is very pertinent as his Nobel Prize winning model on increasing returns to scale is a good example of obtuse economodeling: its thesis was known before being the basis of the centuries-old infant industry argument, and after Krugman it was no easier to apply. Consider Detroit, a popular application for regional increasing returns when applied to autos in the early 20th century: what were the key conditions that allowed it enjoy increasing returns to scale in the early 20th century, but then decreasing returns to scale later in the century? He doesn't say.
Krugman responded that his theory changed the debate, because it showed--under certain parameterizations--that increasing returns to scale can be an argument for lower trade barriers! While true, this is a possibility, not a probability, and those who believe in increasing returns to scale invariably are more inclined to believe in selective tariffs, that is, they don't use Krugman's model to support free trade but rather increased protection. So, it hasn't changed the debate and is counter to his assertion that his New Trade Theory is "probably the main story" in import-export arguments for decreasing trade restrictions; his new model has not changed the debate at all, merely added another obscure reference to the confabulators. Increasing returns to scale remains 1) a fringe argument and 2) used primarily to support trade restrictions, as it was in the 1900s before Krugman's New Trade Theory model.
Krugman is a very smart person, but the fact he can't see this highlights that the greatest lies we tell are the ones we tell ourselves, because he clearly has the capacity to see slight inconsistencies and flaws in others (he's a meticulous advocate against his opponents).
I think a lot of math in econ is like the cargo cult phenomenon, where people see correlations (planes and cargo) and suppose the essence of something is one of those correlations (eg, build models of planes, and cargo will show up). Thus, just as naive people think the essence of a good poem is rhyming, naive economists think that setting up a hypothesis as if one were deriving the Dirac equation or special relativity seems like the essence of a science. Unfortunately, economic equations rarely work out that way.
Consider the Drake equation.
$N = R_{\ast} \cdot f_p \cdot n_e \cdot f_{\ell} \cdot f_i \cdot f_c \cdot L$
Where
N = the number of civilizations in our galaxy with which communication might be possible
R* = the average number of star formation per year in our galaxy
fp = the fraction of those stars that have planets
np= The number of planets, per solar system, with an environment suitable for life
etc.
None of the terms can be known, and most cannot even be estimated. As a result, the Drake equation can have any value from a hundred billion to zero. An expression that can imply anything implies nothing. I mean, this formulation is worthy of writing down, but it's very different than the Dirac equation or Newton's laws, even though at some level there's a similarity.
I remember teaching a money and banking course, and a fun way to get the kids introduced to economic models is to show them the Baumol-Tobin money demand model. This can be derived from some simple assumptions, and applies calculus to the maximization function individuals would apply, generating the equation:
$M= \left ( \frac {CY} {2i} \right )^{\frac {1} {2}}$
Where
M=money demand
C=cost of withdrawing money
Y=Total income
i=interest rate
All very rigorous and tidy. Yet, it doesn't help predict interest rates, or the size of money aggregates. It's empirically vacuous, because it simply doesn't fit the data.
That's one of the more concrete equations. Most equations are like this one for money demand:
Basically one merely argues what arguments should be in the function and then the derivatives on those arguments. Thus, the first argument is 'permanent income' Yp, and the first derivative here is positive. Yet, the parameters can vary wildly, and may even be endogenous themselves. At the end of the day, atheoretical vector-autoregressions do a better job predicting any of these variables.
Yet, for all the insufficiency of mathematics in creating a good science, sociologists show that an absence of rigor doesn't seem to be any better. I think this highlights there's no delusion greater than the notion that method can make up for lack of common sense. Ultimately, there is no method but to be very intelligent.
## Tuesday, August 13, 2013
### Is The Low Vol Anomaly Really a Skew Effect?
The idea that low volatility stocks have higher returns than high volatility stocks is difficult for economists to digest, because it's so hard to square with standard theory. It brings to mind Dostoyevsky's line "If God is dead, then everything is permitted." Similarly, when one sees their favored theory as being abandoned, it seems like all explanation is lost and chaos reigns. Yet, when a wrong theory is adopted, well, as the ever-logical Bertrand Russel used to note, if 1+1=1, everything is both true and untrue. We need a framework to evaluate reality, and it has to be consistent.
Alas, many frameworks are largely untrue, leading to inconsistencies and explanations that are transparently tendentious. The sign of a bad Weltanschauung is that explanations for reality become more and more convoluted, like epicycles in Ptolemaic astronomy. I'll gladly enjoy the hypocrisy of those who don't share my worldview because, as the Detroit bankruptcy has reminded us (eg, its bankruptcy blamed on too much or too little gov't), people might admit tactical errors, but they'll go to their grave with their worldview (see Max Planck).
Consider the recent papers arguing that low volatility is really just a skew effect, in which case their worldview is safe. In the recent Journal of Economic Perspectives, longtime behavioral finance academic Nicholas Barberis wrote a paper on Kahneman and Tversky's prospect theory (that's Nobel prize winning Danny Kahneman, who's unimpeachability seems somewhere around that of Nelson Mandela) It's helpful to note that this insight is 34 years old, because many seem to all think these newfangled behavioural insights are going to revolutionize economics as if they haven't been applied continuously over the past generation.
Barberis goes over his Barberis and Huang (2008) model where prospect theory is used to motivate the hypothesis that a security’s skewness in the distribution of its returns will be priced. A positively skewed security— a security whose return distribution has a right, upper, tail is longer than its left tail—will be overpriced relative to the price it would command in an economy with standard investors. As a result, investors are willing to pay a high price for lottery-ticket type stocks.
Barberis references several papers, including Bali, Cakici, and Whitelaw (2011), and Conrad, Dittmar, and Ghysels (here's the 2009 version, though a more recent version was just published in the Journal of Finance). He also finds it relevant to the underperformance of IPOs, the low average return of distressed stocks, of bankrupt stocks, of stocks traded over the counter, and of out-of-the-money options (all of these assets have positively skewed returns); the low relative valuations of conglomerates as compared to single-segment firms (single-segment firms have more skewed returns); and the lack of diversification in many household portfolios (households may choose to be undiversified in positively skewed stocks so as to give themselves at least a small chance of becoming wealthy).
It seems like an orthogonal way to address these puzzles compared to the constrained rational approach offered by Betting Against Beta, but there's a problem, and it's that the well-know equity risk premium has a negative skew relative to what's considered less premium-worthy, long-term bonds. That is, equities in general have a lower (ie, more negative) skew than bonds, and this is the most prominent 'risk premium', so it must not be an exception to a rule.
US Monthly Data 1962-2013
10-year US T-Bond SP500 Index AnnRet 7.05% 7.28% AnnStdev 6.86% 15.05% Skew 61.09% -42.16%
Note that indices have negative skew while individual stocks have positive skew. This is because correlations go up in down markets, and this predictable tendency creates a problem for idiosyncratic skew pricing models. That is, in the CAPM and other asset pricing models, risk factors have prices that are linear in the covariances, otherwise there is arbitrage, the essence of the Arbitrage Pricing Theory: whatever risks are priced, they are based on additive moments, so risk and returns are linear functions. Now we have priced risks that are not just diversifiable, but change sign depending on what else is in the portfolio. If true, there is an implausible level of profit to be had from buying portfolios and selling the constituents.
As an ivy league confabulator Barberis deftly ignores this inconsistency and instead notes that the equity risk premium makes perfect sense given Benartzi and Thaler’s (1995) idea that if you focus only on the net changes in wealth (technically, U(x) vs. U(w+x)), you can get this to work in cumulative prospect theory, because losses hurt more than gains, so one gets paid to take risk in this case.
Alas, there's a limit to how much skew and variance can both be priced in the same universe, where people love positive skew and hate variance. If skew explains most of the volatility anomaly, that implies people can't be globally risk averse because they would like extremum up-moves too much, and these happen proportionally more for volatile stocks. Yet if that's true there's no risk premium of any sort, because people would simply buy single assets or derivatives and have no incentive to mitigate risk via bundling and arbitrage. This has been shown formally by Levy, Post, and van Vliet (2003), but it should be intuitive: skew is positively correlated with volatility for stocks with lognormal returns, so there's a point at which one's love of skew dominates one's fear of volatility. If that point is reached, volatility is always less costly than skew is beneficial. This constrains the size of the skew-loving effect to be an order of magnitude less than the risk premium if global risk aversion exists. If global risk aversion does not exist, then the rest of the general framework presented in simply meaningless.
So we have prospect theory explaining the overpricing of high volatility stocks due to skew, the underpricing of equity indices due to 'narrow framing.' One could add that prospect theory is used to explain why people overpay for longshots at the horse track, in that 'decisions weights' applied to payoffs prospect theory are observationally equivalent to overoptimistic probability assessments (see Snowberg and Wolfers (2010)), and that Danny Kahneman is an admirer of Nassim Taleb's Black Swan theory, which argues that small probability events are generally underappreciated. In other words, whatever the probability density function and expected return, it's explained by prospect theory.
Skew also shows up also in the recent publication of Conrad, Dittmar, and Ghysels (2013), who are incredibly meticulous in their analysis of how skew relates to future returns, highlighting what three top researchers over several years can do to data. Yet, they then ignore the elephant in the room, that is, if volatility is negatively priced and skew is positively priced, how do these both exist in equilibrium? It should be hard for these authors to say they don't care, because they are very exhaustive in their analysis, noting at one point:
We use several methods to estimate [the stochastic discount function] Mt(τ) that allow for higher co-moments to influence required returns. These methods differ in the details of specific factor proxies, the number of higher co-moments allowed, and the construction of the SDF.
Alas, as usual in analysis of SDFs, there is no take-away input one can use to measure risk, no soon-to-be-indispensable tool, just a promise that this has all been vouchsafed against high-falutin theory and so 'it's all good.' Consistency is a good thing, but only in certain dimensions. One of the authors, Dittmar (2002), wrote a very nice paper for the Journal of Finance in 2002 noting that if you restricts a non-linear pricing kernel to obey the risk-aversion needed to ensure that the market portfolio is the optimal portfolio, the explanatory power goes away of higher moments. With all the abstruse checks in this paper, one would think he might want to address that issue, but instead he ignores it.
I'm sure former JoF editor Cam Harvey read this while nodding approvingly throughout (he's referenced every other page, and a big believer that risk explains most everything in finance). While understanding SDFs and their risk premiums won't help you get a job at a hedge fund, it will help you get published and be popular among publishing academics.
I agree that skew is important, as it measures the upside potential that delusional lottery-ticket buying investors love, and because of relative wealth preferences, arbitrage is costly and their footprint remains. That's a mathematically consistent story. Skew loving effects can't exist on the same par with variance hating effects in any consistent story about asset returns. Is this important? Consistency can be overdone, but I don't think this is foolish because one tends to see what one believes rather than vice versa, and I think there's more power and predictability in viewing volatility as merely a desirable attribute for delusional investors, as opposed to something that pays you a premium.
Paradoxically, behavioral refinements such as prospect theory are preventing needed outside-the-box adjustments and are used to maintain a defective status quo, one that has been wrong on a profound empirical issue for 50 years (ie, the risk premium). These putative revolutionary insights allow academics to wax eloquent on how their complex paradigm handles subtleties such as any of those 50 behavioral quirks, and outside commentators are pleased to be part of a new vanguard, obliviously marching in basically the same, pointless, confabulating path.
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# Problem: An important step in the industrial production of hydrogen is the reaction of carbon monoxide with water (i.e. the water gas shift reaction):CO (g) + H2O (g) ⇋ CO2 (g) + H2 (g)Use the law of mass action to write the equilibrium expression for this reaction.
###### FREE Expert Solution
We are asked to give the equilibrium expression of the given reaction.
The general form of an equilibrium expression is:
$\overline{){{\mathbf{K}}}_{{\mathbf{c}}}{\mathbf{=}}\frac{\mathbf{\left[}\mathbf{products}\mathbf{\right]}}{\mathbf{\left[}\mathbf{reactants}\mathbf{\right]}}}$
Given reaction:
CO (g) + H2O (g) ⇋ CO2 (g) + H2 (g)
(reactants) (products)
When writing equilibrium expressions:
99% (480 ratings)
###### Problem Details
An important step in the industrial production of hydrogen is the reaction of carbon monoxide with water (i.e. the water gas shift reaction):
CO (g) + H2O (g) ⇋ CO2 (g) + H2 (g)
Use the law of mass action to write the equilibrium expression for this reaction.
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# Can I create these sequences with Sage?
Let me start by confessing that I know next to nothing about math and really don't know exactly how to describe what I'm trying here; I hope the below description is adequate.
I am attempting to create a list of all possible 7-digit sequences of the numbers 1, 2, and 3 that add up to 12 (e.g., 1-2-2-3-1-1-2, 2-1-2-3-1-1-2, 2-1-2-3-1-2-1, etc.). I'm not looking merely for the number of sequences, but the actual sequences themselves. Is this something that Sage is capable of and, if so, how would I go about accomplishing this? If not
any suggestions?
Thank you in advance.
Greg
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Sort by » oldest newest most voted
There is a marvelous combinatorial class called IntegerListsLex that returns an iterator (a type of python object that you can iterate over) for lists of integers subject to certain constraints. Try entering IntegerListsLex? at the Sage command line to see the documentation.
Below is an iterator that does the job. The length parameter limits to sequences of length 7, the floor parameter limits the integers to be larger or equal to 1, the ceiling parameter limits the integers to be less or equal to 3.
sage: L = IntegerListsLex(12, length=7, floor=[1]*7, ceiling=[3]*7)
sage: for l in L:
....: print l
....:
[3, 3, 2, 1, 1, 1, 1]
[3, 3, 1, 2, 1, 1, 1]
[3, 3, 1, 1, 2, 1, 1]
[3, 3, 1, 1, 1, 2, 1]
etc...
The Lex part of the name indicates that the lists are returned in lexicographic order.
Note that if you sort the sequences, convert to tuples (so that they are hashable), and then uniquify them you get the 3 basic sequences that generate the whole collection by reordering:
sage: L = IntegerListsLex(12, length=7, floor=[1]*7, ceiling=[3]*7)
sage: LS = [ tuple(sorted(i)) for i in L ]
sage: uniq(LS)
[(1, 1, 1, 1, 2, 3, 3), (1, 1, 1, 2, 2, 2, 3), (1, 1, 2, 2, 2, 2, 2)]
Update changed ceiling option to limit the sequence to 1's, 2's, and 3's
more
Thanks so much for the info, Benjamin. I have Sage installed and was able to generate the lists. Looking at the result, starting with [6, 1, 1, 1, 1, 1, 1], I think I should have been clearer in my description. I want to limit the numbers used to 1, 2, and 3 (i.e., no 4, 5, 6, etc.), THEN create all possible orderings of those numbers. The results I generated include 4, 5, and 6, and are not limited to 1, 2, and 3. I did a little background work on this, and it looks as though there are basically three series that will fulfill my criteria: [1+1+1+1+2+3+3], [1+1+1+2+2+2+3], and [1+1+2+2+2+2+2] (there could be another I've missed, perhaps?). Having those basic series (I apologize if my terminology is confusing ...(more)
( 2011-09-18 19:57:45 +0200 )edit
Sure, it's an easy change. I somehow missed that part of your question. I've updated the answer now.
( 2011-09-18 23:53:45 +0200 )edit
That seems to have done the trick; thanks so much, Benjamin!
( 2011-09-19 01:15:08 +0200 )edit
A composition of a nonnegative integer n is an ordered list of positive integers with total sum n, so you can call Compositions, specifying the maximum term:
sage: P = Compositions(4, max_part=2)
sage: P.list()
[[2, 2], [2, 1, 1], [1, 2, 1], [1, 1, 2], [1, 1, 1, 1]]
sage: list(P)
[[2, 2], [2, 1, 1], [1, 2, 1], [1, 1, 2], [1, 1, 1, 1]]
sage: list(Compositions(12, length=7, max_part=3))
[[3, 3, 2, 1, 1, 1, 1], [3, 3, 1, 2, 1, 1, 1], [3, 3, 1, 1, 2, 1, 1], ...
more
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Outlook: BlackRock Taxable Municipal Bond Trust Common Shares of Beneficial Interest is assigned short-term Ba1 & long-term Ba1 estimated rating.
Dominant Strategy : Sell
Time series to forecast n: 17 Jan 2023 for (n+1 year)
Methodology : Transductive Learning (ML)
## Abstract
BlackRock Taxable Municipal Bond Trust Common Shares of Beneficial Interest prediction model is evaluated with Transductive Learning (ML) and Logistic Regression1,2,3,4 and it is concluded that the BBN stock is predictable in the short/long term. According to price forecasts for (n+1 year) period, the dominant strategy among neural network is: Sell
## Key Points
1. Why do we need predictive models?
2. How do you pick a stock?
3. Technical Analysis with Algorithmic Trading
## BBN Target Price Prediction Modeling Methodology
We consider BlackRock Taxable Municipal Bond Trust Common Shares of Beneficial Interest Decision Process with Transductive Learning (ML) where A is the set of discrete actions of BBN stock holders, F is the set of discrete states, P : S × F × S → R is the transition probability distribution, R : S × F → R is the reaction function, and γ ∈ [0, 1] is a move factor for expectation.1,2,3,4
F(Logistic Regression)5,6,7= $\begin{array}{cccc}{p}_{a1}& {p}_{a2}& \dots & {p}_{1n}\\ & ⋮\\ {p}_{j1}& {p}_{j2}& \dots & {p}_{jn}\\ & ⋮\\ {p}_{k1}& {p}_{k2}& \dots & {p}_{kn}\\ & ⋮\\ {p}_{n1}& {p}_{n2}& \dots & {p}_{nn}\end{array}$ X R(Transductive Learning (ML)) X S(n):→ (n+1 year) $∑ i = 1 n a i$
n:Time series to forecast
p:Price signals of BBN stock
j:Nash equilibria (Neural Network)
k:Dominated move
a:Best response for target price
For further technical information as per how our model work we invite you to visit the article below:
How do AC Investment Research machine learning (predictive) algorithms actually work?
## BBN Stock Forecast (Buy or Sell) for (n+1 year)
Sample Set: Neural Network
Stock/Index: BBN BlackRock Taxable Municipal Bond Trust Common Shares of Beneficial Interest
Time series to forecast n: 17 Jan 2023 for (n+1 year)
According to price forecasts for (n+1 year) period, the dominant strategy among neural network is: Sell
X axis: *Likelihood% (The higher the percentage value, the more likely the event will occur.)
Y axis: *Potential Impact% (The higher the percentage value, the more likely the price will deviate.)
Z axis (Grey to Black): *Technical Analysis%
## IFRS Reconciliation Adjustments for BlackRock Taxable Municipal Bond Trust Common Shares of Beneficial Interest
1. The expected credit losses on a loan commitment shall be discounted using the effective interest rate, or an approximation thereof, that will be applied when recognising the financial asset resulting from the loan commitment. This is because for the purpose of applying the impairment requirements, a financial asset that is recognised following a draw down on a loan commitment shall be treated as a continuation of that commitment instead of as a new financial instrument. The expected credit losses on the financial asset shall therefore be measured considering the initial credit risk of the loan commitment from the date that the entity became a party to the irrevocable commitment.
2. An entity can also designate only changes in the cash flows or fair value of a hedged item above or below a specified price or other variable (a 'one-sided risk'). The intrinsic value of a purchased option hedging instrument (assuming that it has the same principal terms as the designated risk), but not its time value, reflects a one-sided risk in a hedged item. For example, an entity can designate the variability of future cash flow outcomes resulting from a price increase of a forecast commodity purchase. In such a situation, the entity designates only cash flow losses that result from an increase in the price above the specified level. The hedged risk does not include the time value of a purchased option, because the time value is not a component of the forecast transaction that affects profit or loss.
3. In almost every lending transaction the creditor's instrument is ranked relative to the instruments of the debtor's other creditors. An instrument that is subordinated to other instruments may have contractual cash flows that are payments of principal and interest on the principal amount outstanding if the debtor's non-payment is a breach of contract and the holder has a contractual right to unpaid amounts of principal and interest on the principal amount outstanding even in the event of the debtor's bankruptcy. For example, a trade receivable that ranks its creditor as a general creditor would qualify as having payments of principal and interest on the principal amount outstanding. This is the case even if the debtor issued loans that are collateralised, which in the event of bankruptcy would give that loan holder priority over the claims of the general creditor in respect of the collateral but does not affect the contractual right of the general creditor to unpaid principal and other amounts due.
4. The risk of a default occurring on financial instruments that have comparable credit risk is higher the longer the expected life of the instrument; for example, the risk of a default occurring on an AAA-rated bond with an expected life of 10 years is higher than that on an AAA-rated bond with an expected life of five years.
*International Financial Reporting Standards (IFRS) adjustment process involves reviewing the company's financial statements and identifying any differences between the company's current accounting practices and the requirements of the IFRS. If there are any such differences, neural network makes adjustments to financial statements to bring them into compliance with the IFRS.
## Conclusions
BlackRock Taxable Municipal Bond Trust Common Shares of Beneficial Interest is assigned short-term Ba1 & long-term Ba1 estimated rating. BlackRock Taxable Municipal Bond Trust Common Shares of Beneficial Interest prediction model is evaluated with Transductive Learning (ML) and Logistic Regression1,2,3,4 and it is concluded that the BBN stock is predictable in the short/long term. According to price forecasts for (n+1 year) period, the dominant strategy among neural network is: Sell
### BBN BlackRock Taxable Municipal Bond Trust Common Shares of Beneficial Interest Financial Analysis*
Rating Short-Term Long-Term Senior
Outlook*Ba1Ba1
Income StatementCCaa2
Balance SheetCaa2C
Leverage RatiosBa1C
Cash FlowCCaa2
Rates of Return and ProfitabilityBaa2Caa2
*Financial analysis is the process of evaluating a company's financial performance and position by neural network. It involves reviewing the company's financial statements, including the balance sheet, income statement, and cash flow statement, as well as other financial reports and documents.
How does neural network examine financial reports and understand financial state of the company?
### Prediction Confidence Score
Trust metric by Neural Network: 77 out of 100 with 516 signals.
## References
1. Imbens GW, Lemieux T. 2008. Regression discontinuity designs: a guide to practice. J. Econom. 142:615–35
2. Abadie A, Diamond A, Hainmueller J. 2010. Synthetic control methods for comparative case studies: estimat- ing the effect of California's tobacco control program. J. Am. Stat. Assoc. 105:493–505
3. S. Bhatnagar. An actor-critic algorithm with function approximation for discounted cost constrained Markov decision processes. Systems & Control Letters, 59(12):760–766, 2010
4. Abadie A, Diamond A, Hainmueller J. 2015. Comparative politics and the synthetic control method. Am. J. Political Sci. 59:495–510
5. Athey S, Blei D, Donnelly R, Ruiz F. 2017b. Counterfactual inference for consumer choice across many prod- uct categories. AEA Pap. Proc. 108:64–67
6. C. Szepesvári. Algorithms for Reinforcement Learning. Synthesis Lectures on Artificial Intelligence and Machine Learning. Morgan & Claypool Publishers, 2010
7. D. Bertsekas. Dynamic programming and optimal control. Athena Scientific, 1995.
Frequently Asked QuestionsQ: What is the prediction methodology for BBN stock?
A: BBN stock prediction methodology: We evaluate the prediction models Transductive Learning (ML) and Logistic Regression
Q: Is BBN stock a buy or sell?
A: The dominant strategy among neural network is to Sell BBN Stock.
Q: Is BlackRock Taxable Municipal Bond Trust Common Shares of Beneficial Interest stock a good investment?
A: The consensus rating for BlackRock Taxable Municipal Bond Trust Common Shares of Beneficial Interest is Sell and is assigned short-term Ba1 & long-term Ba1 estimated rating.
Q: What is the consensus rating of BBN stock?
A: The consensus rating for BBN is Sell.
Q: What is the prediction period for BBN stock?
A: The prediction period for BBN is (n+1 year)
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Joined: 25 Feb 2013
Posts: 1217
Location: India
GPA: 3.82
### Show Tags
16 Dec 2017, 11:22
Erjan_S wrote:
Bunuel wrote:
Braving the Binomial Probability
BY Karishma, VERITAS PREP
Question 2: For one toss of a certain coin, the probability that the outcome is heads is 0.7. If the coin is tossed 6 times, what is the probability that the outcome will be tails at least 5 times?
Solution: This question is very similar to the questions we saw in the Probability book. The only difference is that we are not tossing a fair coin. The probability of getting heads is 0.7 not 0.5. So the probability of getting tails must be 0.3 since the total probability has to add up to 1.
The only acceptable cases are those in which we get ‘tails’ on all 6 tosses or we get tails on exactly 5 of the 6 tosses.
P(Tails on all 6 tosses) = $$(0.3)*(0.3)*(0.3)*(0.3)*(0.3)*(0.3) = (0.3)^6$$
P(Tails on exactly 5 tosses and Heads on one toss) = $$(0.3)^5*(0.7)*6$$
We multiply by 6 because 5 tails and 1 heads can be obtained in 6 different ways: HTTTTT, THTTTT, TTHTTT, TTTHTT, TTTTHT, TTTTTH
Probability that the outcome will be tails at least 5 times = Probability that the outcome will be tails 5 times + Probability that the outcome will be tails 6 times
Probability that the outcome will be tails at least 5 times = $$(0.3)^6 + (0.3)^5*(0.7)*6$$
I am confused by this multiplication by 6 - why when probability that the outcome will be tails 6 times we simply (0.3)^6 and when probability that the outcome will be tails 5 times - we do multiply (0.3)^5*(0.7) by 6? Why tails on all 6 tosses is (0.3)^6 without any multiplication by 6! and Tails on exactly 5 tosses and heads on one toss needs to be multliplied by 6?
Hi Erjan_S
when you are sure that only tails will occur, then the situation will look like TTTTTT i.e only 1 possibility. Hence probability of tails on all 6 counts will be (0.3)^6
but when 5 tails and 1 heads has to occur, then this situation throws up additional possibilities -
HTTTTT, the probability of this possibility is 0.7*(0.3)^5
THTTTT, the probability of this possibility is 0.7*(0.3)^5
TTHTTT, the probability of this possibility is 0.7*(0.3)^5
TTTHTT, the probability of this possibility is 0.7*(0.3)^5
TTTTHT, the probability of this possibility is 0.7*(0.3)^5
TTTTTH, the probability of this possibility is 0.7*(0.3)^5
Hence total possibility in this case will be sum of all the above possibilities, which simply means you multiply 0.7*(0.3)^5 by 6
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### Show Tags
05 Feb 2018, 06:40
Bunuel wrote:
When Does Order Matter?
BY Karishma, VERITAS PREP
I have to admit that probability is confusing. The problem is not so much that students find it hard to understand as that teachers find it hard to explain. There are subtle points in a probability question that make all the difference in the world and it takes a ton of ingenuity to explain them in a manner that others understand. You either get the point right away, or you don’t.
Here, I will try to explain a probability concept I have always found very difficult to explain in person so the fact that I am attempting to explain it in a post is making me queasy. Nevertheless, the concept is important and I think it deserves a post.
Let me give you two questions:
Question 1: First 15 positive integers are written on a board. If two numbers are selected one by one from the board at random (the numbers are not necessarily different), what is the probability that the sum of these numbers is odd?
Question 2: There is a group of people consisting of 10 men and 6 women. Among these 16 people, there are 4 married couples (man-woman couples). If one man and one woman are selected at random, what is the probability that a married couple gets selected?
Now, let me give you the solutions to these questions. Note that the two solutions are different. We will discuss the reasons behind the difference today.
Solution 1:
Numbers: 1, 2, 3, 4, …, 13, 14, 15
When will the sum of two of these numbers be odd? When one number is odd and the other is even.
P(Sum is Odd) = P(First number is Odd)*P(Second number is even) + P(First number is Even)*P(Second number is Odd)
P(Sum is Odd) = (8/15)*(7/15) + (7/15)*(8/15) = 112/225
Solution 2:
P(Selecting a Married Man) = 4/10
P(Selecting the Wife of that Man) = 1/6
P(Married Couple is Selected) = (4/10)*(1/6) = 4/60
The question I come across here is this: Why is the second question not solved the way we solved the first question? After all, selecting two things together is the same as selecting them one after another (explained in your Combinatorics book) i.e. why don’t we solve the second question in this way:
P(Selecting a Married Couple) = P(Selecting a Married Man)*P(Selecting the Wife of that Man) + P(Selecting a Married Woman)*P(Selecting the Husband of the Woman)
= (4/10)*(1/6) + (4/6)*(1/10)
Other than the fact that it gives the wrong answer, why can’t we solve it like this? Because the order doesn’t matter here. It doesn’t matter whether we pick the husband first or the wife first. The end result is the same. After we pick either one, the probability of picking the other one stays the same. The two selections have to be made from two different groups. They cannot be made from the same group (contrary to the first question). It doesn’t matter whether you catch hold of the man first or the woman first.
In the first question, the probability of picking the correct second number depends on what you picked in the first selection. Hence we consider the order. I will explain this by trying to solve the first question the way we solved the second question:
On first selection, we can pick any number so the probability is 1. The second selection depends on what you selected in the first pick. If you selected an odd number in the first pick, the probability of selecting an even number is 7/15. If you selected an even number in the first pick, the probability of selecting an odd number is 8/15. So what do you do? Do you use 7/15 or 8/15 with 1? You cannot say so you must take individual cases.
Case 1: Select an odd number and then an even number: (8/15) * (7/15)
Case 2: Select an even number and then an odd number: (7/15) * (8/15)
The total cases considered here are 15*15 (select first number in 15 ways and select the second number in 15 ways since the second number can be the same as the first number). In 8*7 ways, you will select an odd number and then an even number. In 7*8 ways, you will select an even number and then an odd number. In both the cases, the sum will be odd. This gives us a probability of (56+56)/225 = 112/225
The total probability of 1 is obtained as follows:
1 = P(first number odd, second number even) + P(first number even, second number odd) + P(first number odd, second number odd) + P(first number even, second number even)
= 56/225 + 56/225 + 64/225 + 49/225 = 1
We are only interested in the 56/225 + 56/225 part.
In the second question, we need to select a couple. Here, it doesn’t matter whether you select the man first or the woman; the two member types are different and there is only one way in which you can select the corresponding partner. You cannot select two members of the same type e.g. two men or two women. Hence we don’t need to bother with calculating the different cases of selecting the man first or the woman first.
Of course, even if we do it, we will get the correct answer. Let me show you the calculation.
The total number of ways of selecting a man and a woman are ‘select a man in 10 ways’ and ‘a woman in 6 ways’. Then ‘select a woman in 6 ways’ and ‘then a man in 10 ways’ i.e. total 120 ways. To select a couple, you can select a man in 4 ways and the woman in 1 way. You can select a woman in 4 ways and the man in 1 way. So total 4 + 4 = 8 ways.
Probability of selecting a couple = 8/120 = 4/60 (same as before).
To sum it, the two questions are quite different.
In the first question, you have two groups of numbers: Even Numbers and Odd Numbers
You can select the two numbers from different groups or from the same group. Hence the total number of cases is 15*15. Also, you can select the same number again.
In the second question, you have two groups of members: Men and Women
You must select the two members from different groups. You cannot select two men or two women. Hence the total number of cases is only 10*6 (and not 16*15). You cannot select the same member again.
In case of confusion, just use the combinations approach rather than probability. You will invariably get the correct answer.
In question 1, why is it said 'the numbers are not necessarily different'? Isn't it ambiguous to say it? Wouldn't the probabilty change if some of the numbers repeat?
_________________
Best,
Spiritual Yoda
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# What is the minimal codes to create my own \documentclass derived from LaTeX's book?
I want to invoke my document class as follows:
\documentclass[cymk,dvips,dvipsnames]{mybook}
Internally mybook.cls includes my own packages such as mytables.sty, mypstricks.sty, etc.
What is the minimal code to create this document class?
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\ProvidesClass{mybook}[2011/01/20 v 0.01 my own class (hv)]
\DeclareOption*{\PassOptionsToClass{\CurrentOption}{book}}% or whatever
\ProcessOptions\relax
%
\LoadClass[% the default options
]{book}
\RequirePackage{xcolors}
\RequirePackage{...}
...
\endinput
And, of course, a comment statement of the license (LPPL) at the beginning, and preferrably some basic info about the file if you plan to ever publish it anywhere.
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thanks for this solution. I am trying to use it now. – xport Jan 20 '11 at 9:14
I was under the impression that classes need to have: \NeedsTeXFormat{LaTeX2e}[<date of some version>] as the first statement. Did you leave it out? Or am I in error? Thanks. – bev Mar 9 '11 at 0:45
@bev: that is done by the loaded class and already nowadays no more a problem. – Herbert Mar 9 '11 at 6:32
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Computes various measures of trend and seasonality of a time series based on an STL decomposition. The number of seasonal periods, and the length of the seasonal periods are returned. Also, the strength of seasonality corresponding to each period is estimated. The mstl function is used to do the decomposition.
stl_features(x, ...)
## Arguments
x
a univariate time series.
...
Other arguments are passed to mstl.
## Value
A vector of numeric values.
Rob J Hyndman
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# How to compare yearly categorical data?
I have location data from 5 years (>30,000 points). Each location is given a classification name (in my case vegetation classes). These vegetation classes were assigned by intersecting the locations with vegetation maps representing 5 different years. The vegetation categories changed due to clear cut practices. Basically, some old forest areas in 2007 became clear cut in 2008, and others in 2009 became clear cuts in 2010 etc. Also, some old clear cuts became young forests because they grew back. What I want to know is, if I used just 1 map (say 2007) - how many wrong classifications will I have per year after this? Because what was once forest in 2007 could be clear cut in 2009. Knowing that the location from year 2009 is a clear cut, using the map from 2007 - was it misclassified? I will want to test this for each year. Basically, is there a map that minimizes miss-classification of points.
In the data set for each XY location, I have Year location was collected, clearcut07, clearcut08, clearcut09..., Names07, Names08, Names09....
The clearcut07 is binary indicator of 1 or 0, if was really a clear cut that year. the Names07 are the categorical names for all locations (all 5 years of points - if they were to be classified from that year). So, all locations obtained are given the attributes from the 2007 map, then the 2008 map then 2009, etc.
I need a summary of counts of locations within each category. A count of locations that were properly classified as clear cuts from each year of the data. A count of locations from data collected, for example, in 2008 that were classified as clear cuts from each map? And a count of locations that were incorrectly classified as clear cuts from each map when I know the correct classification.
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## migrated from stackoverflow.comSep 14 '12 at 15:37
This question came from our site for professional and enthusiast programmers.
it sounds like you want to display the map that corresponds to the year that is most reflective of the sample as a whole. To me, this sounds like you want to be able to show how forest vegetation changes over time in addition to providing some numerical summary of that. However if I interpret your question directly then it seems like you want to do a multiple regression that uses longitudinal data.
Assuming visualization and summary is the goal using R. I'm recommending changing your data around with sqldf and visualizing it with ggplot2.
If I've understood what you're trying to do then this is a two step process. Step 1 is data set revision. Step 2 is visualization.
The first step seems like you need to do a 'one-to-many' command to add information to your dataset that answers: "summary of counts of locations within each category", "count of locations that were classified as clear cuts from each year of the data, "locations classified as clear cuts from each map".
You want to do this in R so here's the script: [ggplot2 package has been acting up for the record]
# library
library(sqldf)
library(munsell)
library(ggplot2,require(munsell))
library(plyr)
# Replicating location dataset
Roll= c("A","B","C","D","E","F", "G","H","I","J")
RollTide= function(n) sample(Roll[1:10],n,replace=T)
c<-RollTide(100)
data <- data.frame(Year = sample(1:5, 100,replace=T), Location = c,
Clearcut = sample(0:1, 100, replace=T))
#per category
sum_count_loc<-sqldf("select *, sum(Clearcut)as TotLocal from data group by Location,Year")
#per year
clear_cut_loc<-sqldf("select *, count(Clearcut)as CountLoc from data group by Location,Year")
#per map
clear_cut_map<-sqldf("select * from data group by Year, Clearcut,Location")
# visualize data
df.m <- melt(data)
ggplot(df.m) + geom_freqpoly(aes(x = value, y = ..density.., colour = variable))
ggplot(df.m) + geom_density(aes(x = value, colour = variable)) + labs(x = NULL) + opts(legend.position = "none") + opts(title = "Forest Vegetation Over Time")
# check
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thank you for your answer, I will go try this. I appreciate your attempt to filter through what I needed. Its very hard for me to explain. – Kerry Sep 16 '12 at 8:51
@Kerry The code I gave you for the step2 visualization part is buggy. The libraries you need to use are: "sqldf", "ggplot2", "reshape2". My idea was to visualize it using a heatmap. The problem is the way the data set "data" is constructed. I think if you use your original spreadsheet, idk what that is exactly, it should work. B/C there are separate columns for clearcut. ex. Clearcut08. If you give me more information I can help you more. – Tyler Sep 16 '12 at 18:03
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# A Four Product
The product of any 4 consecutive integers must be divisible by $$\text{_________}.$$
×
Problem Loading...
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Set Loading...
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# What notions are used but not clearly defined in modern mathematics?
"Everyone knows what a curve is, until he has studied enough mathematics to become confused through the countless number of possible exceptions."
Felix Klein
What notions are used but not clearly defined in modern mathematics?
To clarify further what is the purpose of the question following is another quote by M. Emerton:
"It is worth drawing out the idea that even in contemporary mathematics there are notions which (so far) escape rigorous definition, but which nevertheless have substantial mathematical content, and allow people to make computations and draw conclusions that are otherwise out of reach."
The question is about examples for such notions.
The question was asked by Kakaz
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In mathematics, by mathematicians. Everything is clear? I suppose mathematics is still live nowadays... – kakaz Feb 25 '11 at 21:08
"Everything is well defined in modern mathematics" - We really don´t know that for sure yet (i.e. consistency of ZFC)... "Mathematics is more about correctness than about truth." -I would argue that it is more about the relative truth. Being about correctness is contains too much of a self-purpose... – efq Feb 26 '11 at 1:55
This question has a meta thread: tea.mathoverflow.net/discussion/968/… – JBL Feb 26 '11 at 1:57
@Qiaochu - I understand Your position, but I think nobody have to fear vagueness in such situation. It is just another soft-question on math-overflow. It is some fun. I know that You are professionals but I am an amateur. I would like to play with mathematics. Usually it is worth of mention what do we use without proper definition, fighting between intuition and complicated formalism, and possibly why. I do not understand why question which is obviously interesting and have potential to broaden horizons for many people is so controversial. – kakaz Feb 27 '11 at 19:05
Closing this thread seems more like punishing someone for being an amateur rather than enhancing the quality of the site. Now and throughout history, I believe, a large percentage of the most interesting mathematics revolves precisely around those notions that are used but not (yet) clearly defined. A big list of such subjects seems extremely valuable to me. Vote to reopen. – Louigi Addario-Berry Mar 1 '11 at 16:01
In analysis, the concept of a limit at infinity vs. a limit at a real number $r$.
Typically, there is a whole list of definitions of various limits $\lim_{x \rightarrow a} f(x) = b$, depending on whether $a$ and $b$ are ordinary reals or $\pm \infty$. You may have 9 separate definitions of the limit, one for each case. This situation repeats itself any time a limit is used implicitly, for example if an integral converges to a real or to $\pm \infty$, a series converges, and so on.
Everyone knows that these definitions are really the same, but it seems more cumbersome to have a single unified definition than to have separate definitions that are, informally, the same concept. It is this covert "intuitive sense" in which all the definitions are the same that is not clearly defined.
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I don't think this is in the spirit of the question. If one really insists, it is possible to provide a compact definition of those "nine different limits" (consider extended reals, and so on...), and there is absolutely nothing unclear about it. – Mariano Suárez-Alvarez Sep 30 '11 at 2:18
The definition of a number is kinda fuzzy. Is the alph null a number?
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Perhaps this answer is being voted down because $\aleph_0$ is a cardinal number, which is a well-defined notion. But the issue of which algebraic objects constitute "number systems" is kinda fuzzy, yes. – Mark Grant Feb 26 '11 at 7:40
Nobody really uses the notion "number" in modern mathematics without specifying what kind of number is meant. – darij grinberg Feb 26 '11 at 10:08
@darij: doesn't that support what Frank is saying? – Carl Mummert Feb 27 '11 at 1:32
The question asks for things that are both poorly-defined and used; darij is pointing out that it is not used. – JBL Feb 28 '11 at 3:36
How about the natural numbers $\mathbf N$ and the real numbers $\mathbf R$? If they were both clearly defined, then (for example) mathematicians would agree that the continuum hypothesis has a truth value, even if that value is not known. But there's not such agreement, and some will dispute that CH even has a meaning, much less a truth value. Even keeping it just to $\mathbf N$, all "definitions" that I know of are circular (e.g. the Peano axioms in second order logic just kick the unclarity up to the level of the predicates that the induction axiom quantifies over). The Hilbert $\omega$-rule is similarly self-referential. Yet we (mostly) agree that all arithmetic formulas (even, say, $\Pi^0_{100}$ formulas) do have truth values, that there's a (not effectively describable) first-order theory of "true arithmetic", etc. It just comes back to "the naturals, I mean the ordinary naturals, you know, 0, 1, 2, 3..." which comes across as a little bit faith-based ;-).
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# Collapsing Wavefunctions
One of the postulates is that all measurable quantities in a quantum system are represented mathematically by so called observables. An observable is thus a mathematical object, more specifically a real linear operator whose 'eigenstates' form a complete set. This essentially means that any quantum state can be expressed as a linear combination of these eigenstates of the observable.
A simple example of an observable is the spin operator. If we apply the postulate to this case it simply means that any spin state can be expressed as a combination of the eigenstates of the spin operator. If we are talking about the spin of an electron, for example, the eigenstates are 'spin up' and 'spin down' (naively one could think of an electron spinning counterclockwise or clockwise, respectively). So any spin state can be seen as a linear combination of these spin up and spin down states.
Now, when we do a measurement of the spin of a particular electron, we find out what the spin of the electron is at that moment. Another postulate states that the only possible outcomes of such a measurement is an eigenstate. So the only possible results of measuring the spin of an electron is either spin up, or spin down. After this measurement we thus know that the electron has one of these spins, it's previous spin state has 'collapsed' onto one of these states.
Now there are other postulates which explicitly tell us exactly how the state of a quantum system evolves with time. So if we wait a while after we measured the spin state of the electron, it's spin state might have changed if for example it interacts with some other particle. Using the laws of quantum mechanics, we can thus calculate the probabilities of measuring spin up or spin down at a later time. So quantum mechanics really does not state anything about quantum states being constantly observed, or about observation apart from measurement at all for that matter. It is only concerned with measurements of states and evolution of states over time.
Say that we have a complex quantum system consisting of many parts (particles, fields, etc). We can measure some properties of this system at the outset, providing us with a specific initial state of the system. These different parts of the system then might go on to interact with each other and evolve by the laws of quantum mechanics into some new state (i.e. by the Schrödinger equation). After this, we can do new measurements, and we can in principle calculate, exactly, the probabilities of the different possible outcomes of each of these measurements. When we do these new measurements, the probabilities stop being probabilities however and we get a new definite state, the previous 'probabilistic state' has 'collapsed' (the probabilistic state being a linear combination of eigenstates, and the collapsed state a specific eigenstate).
Quantum mechanics really does not state anything about quantum states being constantly observed, or about observation apart from measurement at all for that matter. It is only concerned with measurements of states and evolution of states over time.
## How to Collapse the Wavefunction
From the wavefunction, one can determine the probability that a measurement performed on the system will yield a particular result. For instance, if we know the wavefunction of an electron in a hydrogen atom, we can find the probability that a measurement of the electron's position will find it at 1 angstrom away from the nucleus. We can also find the probability that the electron will be found 1 meter away from the nucleus (it's very low), or half an angstrom away from the nucleus (probably higher).
A counter-intuitive property of quantum systems is that each state can be expressed as a linear combination of other eigenstates (this is known as a "superposition"). Given a sufficiently large number of states, every other state can be expressed as a superposition of the original states. When a measurement of a property is carried out, the wavefunction "collapses" to one of the state with a defined value for that property, and the measurement corresponding to that particular state is observed. A system in a superposition of states 1, 3, 5, and 6 might collapse to state 3. The probability of collapsing to a given state is determined by the wavefunction of the system before the collapse.
Example $$\PageIndex{1}$$: A Three-state Wavefunction
Suppose that we have a hypothetical quantum system with several eigenstates with well-defined momentum represented by $$|1\rangle$$, $$|2\rangle$$, $$|3\rangle$$, $$|4\rangle$$, .... Now suppose that the wavefunction of the system $$| \Psi \rangle$$ is in a superposition of states $$|1\rangle$$, $$|2\rangle$$, and $$|6\rangle$$, i.e.,
$| \Psi \rangle = c_1 | 1 \rangle + c_2 | 2 \rangle + c_6 | 6 \rangle$
Now, say that we want to measure the momentum of the system, which involves applying the momentum operator on the wavefunction. The instant we make the measurement, the wavefunction collapses to one of the three basis eigenstates. If it collapses to $$| 1 \rangle$$, then the value for momentum associated with the state is measured. If it collapses to $$|2\rangle$$ instead, we measure a value for momentum associated with that state and the same is true if the system collapses to state $$|(6\rangle$$.
Note that a superposition of states is never actually observed, since the system collapses to a single state at the instant that a measurement takes place. The superposition can be interpreted as the description of potential measurement outcomes, while the state of the system after the collapse takes place is the actual realized outcome. The collapse can thus be defined as the transition between the potential and the actual. However, the situation is a bit more complicated than that, since whether something is a "superposition" of states, or a "pure" state, depends on the property being measured. A state with a well-defined position will be a superposition of states with well-defined momentum, and a state with well-defined momentum will be a superposition of states with well-defined position (the fact that no state is both well-defined is related to the Heisenberg uncertainty principle).
Example $$\PageIndex{2}$$: Schrödinger's Cat
A cat, a flask of poison, and a radioactive source are placed in a sealed box. If an internal monitor detects radioactivity (i.e., a single atom decaying), the flask is shattered, releasing the poison that kills the cat. The Copenhagen interpretation of quantum mechanics implies that after a while, the cat is simultaneously alive and dead. Yet, when one looks in the box, one sees the cat either alive or dead, not both alive and dead. This poses the question of when exactly quantum superposition ends and reality collapses into one possibility or the other.
Figure $$\PageIndex{1}$$: Schrödinger's cat. Image used with permission from Wikipedia.
Schrödinger's cat poses the question, "when does a quantum system stop existing as a superposition of states and become one or the other?" (More technically, when does the actual quantum state stop being a linear combination of states, each of which resembles different classical states, and instead begin to have a unique classical description?)
Collapse is fancy terminology for measurement
## Five Interpretations of Collapsing Wavefunctions
So far, the discussion of quantm mechanics has focused on the numerics and mathematics, but there exists a rich effort in discussing the nature of the quantum mechanical world which often more philosophical than mathematical. Below are the five most popular interpretations of quantum mechanics, which the definition of "measurement" differ and hence the interpretation of the wavefunction collapse also differ.
1. The Copenhagen/von Neuman interpretations argues the collapse of the wave function is triggered by the observer. This person has the special property which no other object in universe is capable of. In the Copenhagen interpretation, the collapse can be triggered by any system which is connected to the observer, including the measurement apparatus and external medium (if the observer is not isolated from it). the Copenhagen interpretation his is the most popular interpretation of quantum mechanics.
2. The the von Neuman interpretation argues the collapse of the wave function happens when the observer feels any feeling depended on the measured value.
3. The Bohm interpretation argues collapse of the wavefunction happens when the observer introduces into the measured system some perturbation, which is inevitable when performing the measurement. The difference between the measurement and any other interaction is in that the perturbation introduced by measurement is unknown beforehand.
4. The Relational interpretation argue the collapse happens when the interaction affects the ultimate measurement performed by ultimate observer on the universal wavefunction at infinite future. As such, for the collapse to happen, the result of interaction should somehow affect the external medium, the stars, etc, either now or in the future, rather than being lost.
5. The Many-worlds interpretation argues the wavefunction collapse never happens. Instead what the observer perceives as the collapse is just the event of entanglement of the observer with the observed system.
Figure $$\PageIndex{2}$$: The quantum-mechanical "Schrödinger's cat" paradox according to the many-worlds interpretation. In this interpretation, every event is a branch point. The cat is both alive and dead—regardless of whether the box is opened—but the "alive" and "dead" cats are in different branches of the universe that are equally real but cannot interact with each other. Imge used with permission from Wikipedia.
## Contributors
• Anixx (Stack Exchange)
• dape (Stack Exchange)
• Wikipedia
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# If f(x)= 2 x^2 - 3 x and g(x) = 2e^x + 1 , how do you differentiate f(g(x)) using the chain rule?
Mar 29, 2016
Hi there!
It looks like you're being asked to differentiate a composite function! You'll want to start off by subbing g(x) into f, then you can differentiate.
#### Explanation:
Starting off with composing the function, or subbing g(x) into f. Take g(x) and sub it into wherever there is an x:
$f \left(g \left(x\right)\right) = 2 {\left(2 {e}^{x} + 1\right)}^{2} - 3 \left(2 {e}^{x} + 1\right)$
Now you can differentiate using the chain rule. Let's differentiate:
Remember that when differentiating something of the form:
$f \left(x\right) = a {\left(g \left(x\right)\right)}^{n}$, by the chain rule you get:
$f ' \left(x\right) = a n {\left(g \left(x\right)\right)}^{n - 1} \cdot g ' \left(x\right)$
Now, using the same principle you can differentiate:
$f \left(g \left(x\right)\right) ' = 4 {\left(2 {e}^{x} + 1\right)}^{1} \cdot 2 {e}^{x} - 3 {\left(2 {e}^{x} + 1\right)}^{0} \cdot 2 {e}^{x}$
Simplifying we get:
$f \left(g \left(x\right)\right) ' = 4 \left(2 {e}^{x} + 1\right) \cdot 2 {e}^{x} - 3 \left(1\right) \cdot 2 {e}^{x}$
And that's pretty much it! Most teachers/professors are okay with leaving it unsimplified, as cleaning it up doesn't change the answer.
Hopefully you've understood how the chain rule was applied! If you have any questions, let me know!
Mar 29, 2016
Using a less formal notation on part only as a demonstration
#### Explanation:
Consider : $2 {\left(2 {e}^{x} + 1\right)}^{2}$
Let $u = 2 {e}^{x} + 1$ then $\frac{\mathrm{du}}{\mathrm{dx}} = 2 {e}^{x}$
Let $y = 2 {u}^{2}$ then $\frac{\mathrm{dy}}{\mathrm{du}} = 4 u$
But $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{dx}}$
so $\frac{\mathrm{dy}}{\mathrm{dx}} = 4 u \times 2 {e}^{x} \text{ "=" } 4 \left(2 {e}^{x} + 1\right) \times 2 {e}^{x}$
so $\frac{\mathrm{dy}}{\mathrm{dx}} = 8 {e}^{x} \left(2 {e}^{x} + 1\right)$
Which can be worked on further
This is only a demonstration of method
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Electric Charge from a ring?
1. Sep 2, 2005
In a uniformly charged ring of radius r, if there is an imaginary line from the center of the ring extending outward
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At the center of the ring Im told the charges cancel so the net charge is zero. And at infinity, the charge is so small it approaches zero.
Going out from the ring, there is some point X where the charge is greatest.
Why does this effect take place? How to figure out the field energy lines that have an effect on point X?
tia
2. Sep 2, 2005
mukundpa
I think you have to determine maximum Elecric field on the axis of the ring
3. Sep 2, 2005
Kazza_765
You can work this out just using coulombs law. Work out the force on a particle at a distance s due to an infinitismal point of the ring and then integrate around the ring. If you do it right you'll get.
$$E=K\frac{qz}{(z^2+R^2)^\frac{3}{2}}$$
where q is the total charge on the ring and z is the distance along the perpendicular axis.
I don't think this really answers your question but perhaps if you plot it, it might be of some help. It at least explains why there is a point where the electric field is at a maximum.
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# If a group of wizards is stranded on a small jungle island with deadly tigers, how can they survive a month? [closed]
Tragedy! 10 Novice wizards are stranded on a jungle island, and they need to stay alive until the next tour comes in a month. They know only the most elementary spells, and the waters around them have been contaminated with scorcium, an element that hurts magical creatures on contact, so they can't swim back.
The island is home to a fierce species of tiger, whose teeth are made of scorcium. The tigers are diverse: About 940 in an island that is about 9 square miles large. The only food source is a type of berry that will give a small amount of magic to the consumer. The tigers don't eat them. There are several water sources that the tigers don't use, as they have no scorcium in the waters. The tigers like to prowl around the center of the island, where all the berries and water sources are.
The wizards can:
• Make fireballs about the size of tennis balls, which use a small amount of magic and take away a tenth of the user's magic source.
• They can also make a small rainstorm that drains all of their magic.
The berries replenish one twentieth of the wizard's magic. How can they survive for 1 month, if they even can, and what is the best way to fight the tigers? (For the story, it is best if at least 3 survive.)
## closed as off-topic by Mołot, SRM, dot_Sp0T, Frostfyre, James♦Jan 19 '17 at 19:18
This question appears to be off-topic. The users who voted to close gave this specific reason:
If this question can be reworded to fit the rules in the help center, please edit the question.
• "The only food scource is a type of berry that will give a small amount of magic to the consumer. The tigers don't eat them" then what do the tigers eat? Surely the wizards can eat that. – Zxyrra Jan 19 '17 at 6:32
• How smart are the novices? Are they trained in survival practices? Could they make traps, spears, bows and simple constructs? Any conflict between themselves? Do they have enough common sense? In this setting, magic is only an convienience. – Ngoc Jan 19 '17 at 6:44
• 9 sq mi is about 23 sq km. 940 tigers in 23 sq km is 41 per sq km. This is pretty much the same population density as real-world American Great Planes bison, at about 46 per sq km, whereas a reasonable rule of thumb is that predators need ten times their own biomass in prey. Ignoring territory size, that would lead to a tiger population closer to 100 than 1,000 animals, or an island more like 100 sq mi than 10 sq mi. – a CVn Jan 19 '17 at 8:58
• 940 tigers on 9 sq mi island. This island must be have plenty of 'prey' animals around. Just eat that small critters and make a bonfire – Hariz Rizki Jan 19 '17 at 12:22
The boring answer is tigers won't attack humans if they're smart, so it's not a danger.
Like most animals tigers have an innate fear of the unknown, which humans represent. Their instinct will be to avoid humans because they don't know what the humans are or whether they are prey. The fact that humans are tall (due to standing on two legs) will likely cause most animals to misjudge the threat level of humans as more dangerous then we are physically. Only very hungry tigers will attack the humans at all.
Then you throw in the ability for these humans to create more unknown scary things and you're pretty safe. A single fireball is all you need to 'defeat' the tigers. It doesn't have to even hit them, it's Fire! All animals are afraid of Fire, Fire is very dangerous in the wild! When these unknown huge possible-predators respond to your presence by flinging something as dangerous as FIRE at you, and you have no way of knowing that this comes in limited supply, you do not get closer, you run the other direction fast!
Put simply the humans are too scary an unknown for tigers to hunt them if they are smart. Part of being smart would of course mean making sure they don't look like attractive targets to a tiger hungry enough to be willing to take a risk (which as other pointed out your tigers would be pretty hungry). As such there are a few simple steps they need to take, but if taken they should be fine.
I'll also point out that tigers are a solitary species. As such humans would only have to worry about 1-3 tigers at once, depending on whether a female & male share a territory and if they foolishly set up camp right at the border of two males territories. Kill or, more likely, scare those few tigers into not dealing with you and you're safe.
1) don't go anywhere alone. Stick in groups of two or three. Most prey species institutionally are drawn to single animals, especially animals they identify as a herd species (which your novices are and the tigers may have realized by seeing their 'herd' working together). By having a few together that instinct to attack the isolated straggler isn't started. Furthermore if the tiger does attack one the other can fireball to scare it away before it has time to land a killing blow.
2) Keep fire at the base camp! A few simple torches around the base camp, or even a large camp fire with the novices intentionally sleeping with their backs to a wall/cliff/ocean and fire between them and the rest of the jungle, will keep any predator from coming close during night, or even day. If we presume their novice abilities give them some extra talent with controlling and feeding a fire keeping one up and running all times may make sense, or using small torches when hunting for supplies.
3) Teach the tigers to be afraid! Go out of your way to scare tigers whenever they come close. Throw a little fire at any nearby (maybe not enough for a 'full' fireball, just a small fire is enough). Maybe just light up a few touches and walk towards the tigers in a group with touches slowly, enough to scare them away but not enough to trigger them to feel they need to fight to defend themselves. Even simpler, just throw rocks at them, they won't fully understand how the rocks work and it can still be scary to them, particularly if you're somewhere 'defended' by a large fire burning you can throw projectiles at them without risk of their attacking. The net effect is the tigers will learn to stay away from you fast.
4) Find or build fortification. Pick a safe water source and create a camp there. Put logs or rocks down to make it harder to come up towards your base camp. If they have a means to cut down trees cut them down around your camp, to fuel the fire you keep running and to create a clearing so it's harder for things to get close to you. Failing that find a spot that is easily secured, a cave, a cliff face you can put your back to, anything that lowers the areas that these animals can come close to you at.
In all honesty it should be relatively easy to keep normal tigers away from the novices, with food and danger of death from exposure being the real threats. If you want the tigers to be any kind of threat the best way is to ensure they are hungry (though you still need to widen your island, as pointed out that many tigers in one area is just unbelievable). If you suggest some recent drought or other effect killed off some of the prey so that the tigers are hungry you could potentially have tigers hungry enough to be willing to take a risk by attacking humans. But frankly I think realistically your going to have to put more creativity in justifying why the tigers would be a threat to humans given their instincts to avoid us then why humans can't defend themselves against tigers.
Considering that 940 tigers in 9 square miles will hardly find enough food to sustain themselves
A tigress may have a territory of 20 km2 (7.7 sq mi), while the territories of males are much larger, covering 60 to 100 km2 (23 to 39 sq mi). Source
their best bet would be to hide until the tigers starve.
• I WAS thinking that the island is like a wildlife preserve created by someone without the needed knowledge. And that tigers are edible. And that desperate predators are not something 10 Rincewinds can hide from. Sadly, this wouldn't work. – kaay Jan 19 '17 at 8:17
• This; OP needs to adjust some parameter here. 9 sq mi is about 23 sq km. 940 tigers in 23 sq km is 41 per sq km. This is pretty much the same population density as real-world American Great Planes bison, at about 46 per sq km, whereas a reasonable rule of thumb is that predators need ten times their own biomass in prey. Ignoring territory size, that would lead to a tiger population closer to 100 than 1,000 animals, or an island more like 100 sq mi than 10 sq mi. – a CVn Jan 19 '17 at 9:00
Burn the jungle.
Enjoy the tiger meat BBQ. Everyone knows scorcium only accumulates in bones.
Collect drinkable water from created rainstorms.
Get picked up and sold into slavery by the angry tour organizers who notice the large bonfire and come early to investigate.
Yay, survival!
Side ideas:
1. Use wind direction and rainstorms to leave a small part of the jungle intact, increasing their risk of being eaten on day one to gain a piece of land with its timber, roaming tigers, and no berries, for later use.
2. Whatever comes to eat the rest of the burnt tiger corpses, can probably also be eaten. Ah, who am I kidding - we're not talking about hardened survivalists here. Once the meat spoils, they're doomed.
3. They could, instead, try claiming a piece of the island center by just strolling in, parting the shopping-mall-density tiger crowd with a ring of burning whatever-they-can-carry, hoping the tigers come to tolerate them in time, but that is likely to accidentally turn into the "Burn the jungle" scenario anyway, with them in the center, rather than on the beach.
## Mundane (normal) means
Clearly magic is a hassle. Fireballs are energy-consuming for the amount of work required to get to the berries needed to fuel magic - and fire generally dissipates when not sustained by fuel. Without the proper means to sustain their magic, it can't be used as an efficient weapon until the tigers are disposed of - at which point they would no longer need a weapon. Rainstorms won't really help, unless water is incredibly scarce, because healthy rainwater should be abundant on the island (it's tropical). Therefore, compared to the alternatives, magic is kind of useless here.
Your mages are better off fashioning blades out of rock (obsidian, perhaps), making spears, and trapping game. They can eat whatever the tigers eat regularly, seeing as the tigers are ... happy stalking berry bushes... and each person only has to kill three or four tigers daily to eliminate them. They can knock out most of the tigers quickly by contaminating the water they drink (if they're in Central or South America, that may not be too hard), and perhaps the few immunies or lucky ones that survive will help you with your desired plot device.
I know this isn't ideal - a clever, magical solution seems like a better story option - but there's a common fallacy that magic should be applied to everything. Magic's so refined - so limited to a set criteria. Moral of the story: magic is great for accomplishing one task, but for the rest of what you want to do, mundane means work best.
• I was going to go with "pointy stick" but you've beaten me to it. People forget about such things as soon as magic comes along. – Separatrix Jan 19 '17 at 11:56
• The tigers don't eat the berries. Also they have scorcium in their saliva, which contaminates food. – KingraHoundoomJazz Jan 19 '17 at 23:31
• @Bubbles_as_Jazz perhaps that would be more appropriate for the question as opposed to a comment so others can use that information. – Zxyrra Jan 20 '17 at 1:09
9 square miles. So less than half of Manhattan. And 940 tigers. That gather in the centre of the island.
What are they? Stack-em tigers? Lego Tigers? Even If you take the smallest one, sumatran tiger, their length is 2 metres (I will go with rebel measurements from now on you imperial scum). Let's say the width is 0,5 metres. So that give us 940 square metres of tigers. Not to mentions their tails. So almost one square kilometre of 23 sq km island is taken by tigers. But put a little berry bush beetwen each one of them and some water source and you could easily take 1/4 of the island surface. Now give them some space to move. While tigers are graceful, they are also solitary animals. So they would not like to cross one another path.
And voila! Problem solved: most of the tigers falls from cliff into to water and drowns.
• 1 sq km = 1,000 x 1,000 sq m = 1,000,000 sq m, as opposed to 1,000 sq m that you seem to be using in your answer. So the island is about 23,000,000 sq m. Each tiger thus has $\frac{23\,000\,000}{940} \approx 24\,468~\text{m}^2$ to itself, if they divide the island perfectly fairly amongst themselves (which they do because magic). – a CVn Jan 19 '17 at 16:13
• SO FUNNY! LOLOLOLOL! – KingraHoundoomJazz Jan 19 '17 at 23:35
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# Addition of Signed Numbers
Problems like $(-2) + (-3) = -5$ and $\,(-3) + 5 = 2\,$ are easy for some people and hard for others. If they're easy for you, then try a few problems below, and then jump right to the exercises.
Otherwise, read on—and keep in mind that explaining something simple, in words, often ends up sounding very complicated!
Click the ‘New Problem’ button to practice adding signed numbers. Type your answer, then press ‘enter’.
This exercise was conceptualized and contributed to by Robert Fant, Technology Integration Enthusiast. Thanks, Robert!
## Signed Numbers
The phrase signed numbers refers to numbers that can be either positive (like $\,5\,$) or negative (like $\,-5\,$). That is, signed numbers are allowed to have a minus sign.
Every real number can be interpreted in two ways:
• as a position on a number line;
• as a movement.
Both of these interpretations—position and movement—are used when learning to add and subtract signed numbers. (Once you know how to do it, the process will become automatic and you probably won't think about the position and movement stuff.)
## Numbers as Position
The number $\,3\,$ can mean: go to position $\,3\,$ on the number line.
The number $\,-3\,$ can mean: go to position $\,-3\,$ on the number line.
## Numbers as Movement
Positive numbers can indicate movement to the right. For example, $\,3\,$ can mean: move $\,3\,$ units to the right.
Negative numbers can indicate movement to the left. For example, $\,-3\,$ can mean: move $\,3\,$ units to the left.
## Adding a Negative Number
When you add a negative number, you should put it in parentheses, unless it comes first. For example, the sum of $\,-3\,$ and $\,-1\,$ should be written as $\,-3 + (-1)\,$.
(Recall that the word sum refers to an addition problem.) If you want, you can optionally put that first negative number in parentheses, too: $\,(-3) + (-1)$.
## Size versus Sign
Every number has a size (its distance from zero). Every nonzero number has a sign (positive or negative).
For example:
The number $\,3\,$: its size is $\,3\,$, and its sign is positive.
The number $\,-3\,$: its size is $\,3\,$, and its sign is negative.
In the movement interpretation of a real number, the size tells us how far to move, and the sign tells us which direction to move.
## Using Position/Movement Ideas in an Addition Problem
Now we're ready to combine the position and movement ideas in an addition problem. The process is illustrated first with an example:
Consider the problem: $\,2 + (-3) + 5\,$
• The first number, $\,2\,$, indicates a position. Go to $\,2\,$ on the number line.
• Adding a negative number indicates movement to the left. Thus, adding $\,-3\,$ says to move $\,3\,$ units to the left.
• Adding a positive number indicates movement to the right. Thus, adding $\,5\,$ says to move $\,5\,$ units to the right.
• You end up at position $\,4\,$.
Thus, $\,2 + (-3) + 5 = 4\,$.
## The ‘Start at Zero’ Interpretation
Or, you can always start at zero! That is, write $2 + (-3) + 5$ as $0 + 2 + (-3) + 5$ . The first number indicates position, and the remaining numbers indicate movement. Start at $\,0\,$, move $\,2\,$ to the right, $\,3\,$ to the left, and $\,5\,$ to the right, ending up at $\,4\,$.
You should understand both interpretations, but in practice you can use whichever is more natural to you. The start at zero interpretation is used in the following discussion.
You probably don't want to be drawing number lines every time you need to do an addition of signed numbers problem. The good news is that every problem—no matter how many numbers are involved—boils down to either a two-number addition problem, or a two-number subtraction problem, which can then be done efficiently in your head. Keep reading!
## Adding Numbers with the Same Signs
When you add numbers with the same signs (both positive or both negative), then in your head you do an addition problem.
Here are two examples:
• $2 + 3$
(Both numbers are positive.)
Start at zero. Move to the right $\,2\,$, then to the right $\,3\,$. End up at $\,5\,$. Thus, $2 + 3 = 5\,$.
• $(-2) + (-3)$
(Both numbers are negative.)
Start at zero. Move to the left $\,2\,$, then to the left $\,3\,$. The total distance moved is $\,2 + 3 = 5\,$. You moved to the left, so you end up at $\,-5\,$. Thus, $\, (-2) + (-3) = -5\,$.
Notice that in both of these problems, you do an addition problem in your head, which gives the total distance moved. If you always move to the right, then the final answer is positive. If you always move to the left, then the final answer is negative.
## Adding Numbers with Different Signs
When you add numbers with different signs (one positive, one negative), then in your head you do a subtraction problem.
Here are two examples:
• $2 + (-3)$
(first number is positive, second number is negative)
Start at zero. Move $\,2\,$ to the right and $\,3\,$ to the left. You moved more to the left—how much more?
Answer: $\,3 - 2 = 1$
So you end up at $\,-1\,$. Thus, $\,2 + (-3) = -1\,$.
• $3 + (-2)$
(first number is positive, second number is negative)
Start at zero. Move $\,3\,$ to the right and $\,2\,$ to the left. You moved more to the right—how much more?
Answer: $\,3 - 2 = 1\,$.
So you end up at $\,1\,$.
Thus, $\,3 + (-2) = 1\,$.
The mental process is this:
Once you recognize that you're adding numbers with different signs, throw away (for the moment) all the signs, take the bigger number, and subtract the smaller number. This gives you the net distance traveled.
If you moved farther to the right, then your answer is positive. If you moved farther to the left, then your answer is negative.
Notice that when you add numbers with different signs, then in your head you do a subtraction problem.
## Five-Step Process for Adding Two Signed Numbers
When you add two signed numbers, you can follow this five step process. As you read through these steps, think of applying these questions to the problem $\,2 + (-3)\,$:
1. Step 1: What numbers are being added? (Answer: $\,2\,$ and $\,-3\,$)
2. Step 2: Do these numbers have the same sign or different signs? (Answer: different signs)
3. Step 3: In your head, will you be doing an addition or subtraction problem? (Answer: subtraction problem)
4. Step 4: Do the appropriate addition or subtraction problem. Answer: Throw away the signs, leaving you with $\,2\,$ and $\,3\,$. Subtract the smaller from the larger: $\,3 - 2 = 1\,$
5. Step 5: Is your answer positive or negative? Answer: The bigger number is negative, so the answer will be negative. So, $\,2 + (-3) = -1\,$.
## More Than Two Numbers Being Added
If there are more than two numbers being added, just turn it into a two-number problem in the first step, by combining the positive and negative numbers separately, like this:
$(-3) + 5 + (-2) + 1 + 4 + (-6)$
$= (\,(-3) + (-2) + (-6)\,) \,+\, (5 + 1 + 4)$ (re-group, re-order, to combine negative and positive separately)
$= -11 + 10$
$= -1$
## Practice
Here, you will practice addition problems of the form ‘$\,x + y\,$’ where $\,x\,$ and $\,y\,$ can be any of these numbers: $\,-10, -9, -8, \ldots, -1, 0, 1, \ldots, 8, 9, 10\,$. About half of the problems will involve variables!
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# Tutor profile: Youssef A.
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Youssef A.
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## Questions
### Subject:Machine Learning
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Question:
It is common to specify a multiple regression model when, in fact, interest centers in only one or a subset of the full set of variables- the remaining variables are often viewed as "controls". Suppose that a regression involves two sets of variables, $\textbf{X}_1$ and $\textbf{X}_2$. Prove the theorem: \textbf{Firsch-Waugh-Lovell} In the linear least squares regression of vector \textbf{y} on two subsets of variables $\textbf{X}_1$ and $\textbf{X}_2$, the subvector $\textbf{b}_2$ is the set of coefficients obtained when the residuals from a regression of \textbf{y} on $\textbf{X}_1$ alone are regressed on the set of residuals obtained when each column of $\textbf{X}_2$ is regressed on $\textbf{X}_1$.
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To prove this theorem, start with the normal equations \begin{gather} \begin{bmatrix} \textbf{X}^T_1\textbf{X}_1 & \textbf{X}^T_1\textbf{X}_2 \\ \textbf{X}^T_2\textbf{X}_1 & \textbf{X}^T_2\textbf{X}_2 \end{bmatrix} \begin{bmatrix} \textbf{b}_1 \\ \textbf{b}_2 \end{bmatrix} = \begin{bmatrix} \textbf{X}^T_1\textbf{y} \\ \textbf{X}^T_2\textbf{y} \end{bmatrix} \end{gather} Insert the formula for $\textbf{b}_1$ into the second normal equation and collect terms to arrive at: \begin{equation*} \textbf{b}_2=(\textbf{X}_2^T\textbf{M}_1\textbf{X}_2)^{-1}(\textbf{X}_2^T\textbf{M}_1\textbf{y}) \end{equation*} By exploiting the fact that $\textbf{M}_1$ is symmetric and idempotent, we have \begin{equation*} \textbf{b}_2=((\textbf{M}_1\textbf{X}_2)^T\textbf{M}_1\textbf{X}_2)^{-1}((\textbf{M}_1\textbf{X}_2)^T\textbf{M}_1\textbf{y}) \end{equation*} This process is commonly called partialling out or netting out the effect of $\textbf{X}_1$.
### Subject:Applied Mathematics
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Question:
Consider a continuous function $f: \mathbb{R}^n \rightarrow \mathbb{R}$. We are interested in the unconstrained optimization problem $\min_{\mathbf{x}\in \mathbb{R}^n}f(\mathbf{x})$ (a) Prove the Necessary condition" Let $f(\mathbf{x})$ be continuously differentiable. If $\mathbf{x}^* \in \mathbb{R}^n$ is a local minimum of $f(\mathbf{x})$, then $\nabla f(\mathbf{x}^*)=\mathbf{0}$ and $\nabla^2 f(\mathbf{x}^*)\geq \mathbf{0}$ (b) Prove the sufficient condition: Let $f(\mathbf{x})$ be twice continuously differentiable. If $\nabla f(\mathbf{x}^*)=\mathbf{0}$ and $\nabla^2 f(\mathbf{x})\geq \mathbf{0} \; \forall \mathbf{x} \in B(\mathbf{x}^*, \epsilon)$ then $\mathbf{x}^*$ is a local minimum.
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Youssef A.
(a) Since $\mathbf{x}^*$ is a local minimum, for $\forall \; \mathbf{d} \in \mathbb{R}^n, \lambda \geq 0$ sufficiently small, we have \begin{align*} f(\mathbf{x}^*) \leq f(\mathbf{x}^*+ \lambda \mathbf{d}) \end{align*} Pick $\lambda >0$ \begin{align*} 0 \leq \frac{f(\mathbf{x}^*+ \lambda \mathbf{d})-f(\mathbf{x}^*)}{\lambda} \end{align*} Take limits as $\lambda \rightarrow 0$ \begin{align*} 0 \leq \nabla f(\mathbf{x}^*)^T \mathbf{d} \; \forall \mathbf{d}\in \mathbb{R}^n \end{align*} Since \textbf{d} is arbitrary, we must have $\nabla f(\mathbf{x}^*)=0$. Since we have $\nabla f(\mathbf{x}^*)=0$, consider the Taylor series expansion \begin{align*} f(\mathbf{x}^*+ \lambda \mathbf{d})-f(\mathbf{x}^*) = \frac{1}{2}\lambda^2\mathbf{d}^T\nabla^2 f(\mathbf{x}^*)d + \lambda^2||\mathbf{d}||^2R(\mathbf{x}^*;\lambda\mathbf{d}) \end{align*} Hence \begin{align*} \frac{f(\mathbf{x}^*+ \lambda \mathbf{d})-f(\mathbf{x}^*)}{\lambda^2} = \frac{1}{2}\mathbf{d}^T\nabla^2 f(\mathbf{x}^*)\mathbf{d} + ||\mathbf{d}||^2R(\mathbf{x}^*;\lambda\mathbf{d}) \end{align*} If $\nabla^2 f(\mathbf{x}^*)< \mathbf{0}$, $\exists\bar{\mathbf{d}}:\bar{\mathbf{d}}^T\nabla^2 f(\mathbf{x}^*)\bar{\mathbf{d}} <0 \implies f(\mathbf{x}^*+ \lambda \mathbf{d})-f(\mathbf{x}^*) <0$ for $\lambda$ sufficiently small. (b) Consider a Taylor series expansion for all $\mathbf{x} \in B(\mathbf{x}^*, \epsilon)$. For some $\lambda \in [0,1]$ we have \begin{align*} f(\mathbf{x})=f(\mathbf{x}^*)^T(\mathbf{x-x^*})+\frac{1}{2}(\mathbf{x-x^*})^T\nabla^2 f(\mathbf{x}^*+\lambda (\mathbf{x-x^*}))(\mathbf{x-x^*}) \end{align*} Hence $f(\mathbf{x})\geq f(\mathbf{x}^*)$.
### Subject:Calculus
TutorMe
Question:
(a) Find the equation for the tangent line to $y = \ln{x}$ at the point $x = 1$. (b) Find the quadratic approximation $Q(x)$ to the function $f(x) = \ln{x}$ at $x=1$.
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Youssef A.
(a) The equation for the tangent line is $y = L(x)$ where $L(x)$ is the linear approximation of $y = \ln{x}$ at $x = 1$. Using the formula $L(x)=f(a)+f^{'}(a)(x-a)$, with $f(x) = \ln{x}$and $a=1$ and $f^{'}(x)=1/x$ we get $f(1) = ln(1) = 0$ and $f^{'}(1)=1$ $L(x)=0+1(x-1)=(x-1)$. (b) Using the formula $#Q(x)=f(a)+f^{'}(a)(x-1)+\frac{f^{''}(x-1)^2}{2} with$f(x)=\ln(x)$,. we get$f^{'}(x)=1/x$,$f^{''}(x)=-1/x^2$so$f^{''}=-1$and hence$Q(x)=(x-1)-\frac{(x-1)^2}{2}\$
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Constructing Volatility Smile from American Options
My question is about best practices for reconstructing volatility smiles for a fixed tenor from American option data. For simplicity/liquidity, I am currently considering options on SPY. I am ...
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I am trying to calculate the jump parameters for the Bates volatility jumps, specifically, the mean of the jump percentages, $\mu_j$. For the value of $J$, I am using jumps \$|\frac{s_{i}-s_{i-1}}{s_{i-...
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# Make header files in c++ when you come up with simple ideas!!
This topic is 4205 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
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It is unwise to put "using namespace std;" in a header file since you then force any other files that include your header file to have the contents of std dumped into their namespace.
Really, you would be better to split the above into a .h and a .cpp file, with the function prototype in the .h and the actual implementation into the .cpp file. You could then put "using namespace std;" at the top of the .cpp file without having to worry about affecting other units that use your code.
I appreciate that you are only just beginning, but since you seem to be here offering advice, you need to ensure that you are not suggesting bad practices to people who may have less experience than you.
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The most fun part is, if you created a function in a separate source file, you would have to make a function prototype in all the source files that used the function.
Not really. The proper way is to declare the function prototype in MyUnit.h, and define the function in MyUnit.cpp. To use a custom function, for example 'void Foo()', in another source file, you just include MyUnit.h and link to MyUnit.o(if you add MyUnit.cpp to your project, that happens automatically). Your way is mostly incorrect. Just try to include your .h file that contain function definitions in more than 1 source file in the same project. You will get a multiple definition error from the linker, since the function Foo() is defined in all object files that are linked together.
I appreciate your enthusiasm, but it would be better if you wait until you have sorted things out yourself before deciding to tutor beginners. I'm sure you don't want anyone to learn bad practices from you. It's bad karma :)
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#ifndef REVERSESTR_H#define REVERSESTR_H#include <string>inline std::string RevStr(const std::string& in){ std::string reversed; std::string::size_t len = in.length(); for(int i = 0; i < len; ++i) { reversed += in; } return reversed;}#endif//REVERSESTR_H
However, I'd actually just use the following line to achieve what you're trying to do (note: haven't tested this so it might not be 100% accurate):
std::string in = "Hello, World!";std::string rev(in.rbegin(),in.rend());
You could do one step better and look into Boost.StringAlgorithms
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Thank you guys!!!
See I wasn't a looser after all. An I really hope that beginners who go through my
entire comments(filed by you guys) so that even they learn the rigth way.
Again guys than you vey much!! I really appreciate it.
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Original post by daeridJust to adopt the helpful viewpoint here, try this refactoring:...You could do one step better and look into Boost.StringAlgorithms
Either way, you're making it much too complicated [smile]
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Original post by AbhinashSee I wasn't a looser after all.
Oh, you definitely got the right idea: as you start solving common problems, gathering your solutions into a library rather than cutting and pasting is definitely the way to go. However, you need to make sure that there isn't a better way to do it (e.g. using the standard C++ library) and that your code is of good quality. This is the kind of things that come with experience.
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Aww, aren't new programmers cute?
Also, don't abreviate names. Never sacrifice readability to save keystrokes.
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Original post by Zahlman
Quote:
Original post by daeridJust to adopt the helpful viewpoint here, try this refactoring:...You could do one step better and look into Boost.StringAlgorithms
Either way, you're making it much too complicated [smile]
Not if you need a reversed copy of the string. I was going to post reverse(), but then I realized that his function was returning a new string.
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Oh, well that's easy too:
std::string reversed_string(const std::string& input) { return std::string(input.rbegin(), input.rend());}
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# 3. (20 points) Computational Consider the following project and assume that you require a 10% return on your...
###### Question:
3. (20 points) Computational
Consider the following project and assume that you require a 10% return on your investment.
Year Cash Flow
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# An example needed of Schur's lemma
Reading Linear representations of finite groups by Serre, I need an example of the following:
Schur's lemma:
Let $$\rho^1\colon G \rightarrow GL(V_1)$$ and $$\rho^2\colon G \rightarrow GL(V_2)$$ be two irreducible representations of a group $$G$$, and let $$f$$ be a linear mapping of $$V_1$$ into $$V_2$$ such that $$\rho_s^2 \circ f = f \circ \rho_s^1$$ for all $$s \in G$$. Then:
(i) If $$\rho^1$$ and $$\rho^2$$ are not isomorphic, we have $$f=0$$
(ii) If $$V_1 = V_2$$ and $$\rho^1 = \rho^2$$, $$f$$ is scalar multiple of the identity.
Can someone form a concrete example so I can see what's going on?
• I've answered your questions before, and my sense is you are not ready for the answer. The map $f$ is a change of basis. Spend some time thinking about why the statement is true. – David Hill Apr 25 at 4:26
• @DavidHill I learn from examples as I see what is going on hence my questions on examples. it is rather obvious $f$ is a change of basis. – john Apr 25 at 10:55
• and actually I did understand the other solution which was presented, it was clearer. – john Apr 25 at 14:56
• I'm also interested in this question, can someone provide answer please! – Math Apr 29 at 12:31
If you translate Schur's Lemma into the language of representations of finite groups, you get the following.
Let $$G$$ be a finite group, $$\mathsf{k}$$ some field, and $$\rho_i : G \to \operatorname{GL}\left(V_i\right)$$ some irreducible representations of $$G$$ over $$\mathsf{k}$$. Then
1) If $$\rho_1$$ is not equivalent to $$\rho_2$$, then there are no non-trivial $$\mathsf{k}$$-linear maps $$T : V_1 \to V_2$$ intertwining $$\rho_1, \rho_2$$.
2) If $$\mathsf{k}$$ is algebraically closed, and $$\rho_1, \rho_2$$ are equivalent, then every non-trivial $$\mathsf{k}$$-linear map $$T : V_1 \to V_2$$ intertwining $$\rho_1, \rho_2$$ is an isomorphism.
By a linear map $$T : V_1 \to V_2$$ intertwining $$\rho_1, \rho_2$$, what I mean is that $$T\left( \rho_1(g) v \right) = \rho_2(g) T(v)$$ for all $$g \in G$$, $$v \in V$$, and two representations $$\rho_1, \rho_2$$ are called equivalent if there is a linear isomorphism intertwining them.
Schur's lemma is a direct result of the following observation
Let $$G$$ be a finite group, $$\mathsf{k}$$ some field, and $$\rho_i : G \to \operatorname{GL}\left(V_i\right)$$ some representations of $$G$$ over $$\mathsf{k}$$, and suppose that $$T : V_1 \to V_2$$ is a $$\mathsf{k}$$-linear map intertwining $$\rho_1$$, $$\rho_2$$. Then
1) $$(\rho_1, \operatorname{Ker}{T})$$ is a subrepresentation of $$(\rho_1, V_1)$$.
2) $$(\rho_2, \operatorname{Img}{T})$$ is a subrepresentation of $$(\rho_2, V_2)$$.
Now, if $$(\rho_i, V_i)$$ are irreducible, and $$T : V_1 \to V_2$$ is a $$\mathsf{k}$$-linear map intertwining $$\rho_1$$, $$\rho_2$$, then $$\ker{T}$$ is either $$\left\{0\right\}$$, or $$V_1$$. In the first case, $$T$$ is injective, whilst in the second case $$T$$ is the trivial map. Now suppose we have the first case, then since $$V_2$$ is irreducible, we have $$\operatorname{Img}T$$ is either $$\left\{0\right\}$$ or $$V_2$$. So our options are
1. $$V_1 = \left\{0\right\}$$, in which case $$T = 0$$,
2. $$V_1 \neq \left\{0\right\}$$ and $$T=0$$,
3. $$V_1 \neq \left\{0\right\}$$ and $$T \neq 0$$, in which case $$T$$ is injective and surjective, and so $$T$$ is a isomorphism intertwining $$V_1, V_2$$
Now suppose that we are in the third case, and $$V_1 = V_2$$, and that $$\mathsf{k}$$ is algebraically closed, and suppose that $$T : V_1 \to V_1$$ is a non-trivial intertwining map. Then by the above $$T$$ is an isomorphism, and since $$\mathsf{k}$$ is algebraically closed, $$T$$ has an eigenvector $$0 \neq v_1 \in V_1$$, with eigenvalue $$\lambda$$ say. Then it is easy to see that the $$\lambda$$-eigenspace of $$T$$ is a non-zero subrepresentation of $$V_1$$, and since $$V_1$$ is irreducible we have that $$V_1$$ is the $$\lambda$$-eigenspace of $$T$$, and so $$T(v) = \lambda v$$ for every $$v \in V_1$$, and so $$T = \lambda \operatorname{Id}_{V_1}$$, and moreover $$\lambda \neq 0$$. It follows that every non-trivial intertwining map $$T : V_1 \to V_1$$ is a scalar multiple of the identity.
Now, for a concrete example. Consider the representation of $$S_3$$ defined by $$\rho_V : S_3 \to \operatorname{GL}(\mathbb{C}^{3}) \ : \ \rho_V(\sigma)(ae_1 + be_2 + ce_3) = ae_{\sigma(1)} + be_{\sigma(2)} + ce_{\sigma(3)}.$$
Then $$\rho_V$$ is not irreducible, because the two subspaces
$$U = \left\{ae_1 + be_2 + ce_3 \mid a,b,c \in \mathbb{C} : a + b + c = 0 \right\}, \\ W= \left\{a(e_1 + e_2 + ce_3) \mid a \in \mathbb{C} \right\}$$
are both subrepresentations. Moreover, it is clear to see that $$U \cap W = \left\{ 0 \right\}$$, and so for dimension reasons we have $$V = \mathbb{C}^{3} = U \oplus W$$. Now, $$W$$ is one-dimensional and so must be irreducible, and take a moment to convince yourself that $$U$$ is irreducible too.
Now consider the representation $$\rho_t : S_3 \to \mathbb{C}^{\times}$$ such that $$\rho_t(\sigma) = t$$ for every $$\sigma$$, where $$t \neq 0$$. Suppose that $$T : \mathbb{C}^{\times} \to U$$ is an intertwining map, and suppose $$T(1) = ae_1 + be_2 + ce_3$$. Then since $$T$$ intertwines, we have $$t(ae_1 + be_2 + ce_3) = tT(1) = T(t) = T(\rho_t(\sigma)1) = \rho_V(\sigma)(T(1)) = \rho_V(\sigma)(ae_1 + be_2 + ce_3),$$
and so
$$tae_1 + tbe_2 + tce_3 = ae_{\sigma(1)} + be_{\sigma(2)} + ce_{\sigma(3)},$$
for every $$\sigma \in S_3$$. Taking $$\sigma = \operatorname{Id}$$ and $$\sigma = (12)$$ gives $$ta = a = b$$, and so $$a = b$$. Similarly you can show $$a = c$$ and $$c=b$$, but $$a + b + c =0$$, and so we get $$a = b = c =0$$. So every linear map intertwining $$\rho_t$$, and $$\rho_V$$ is zero, for whatever choice of $$\rho_t$$ we choose.
Now, suppose instead that $$T: \mathbb{C}^{\times} \to W$$ is an intertwining map, suppose that $$T(1) = a(e_1 + e_2 + e_3)$$. Then it is clear that $$T(x) = xa(e_1 + e_2 + e_3)$$, and so $$T$$ is a scalar multiple of the map $$I : \mathbb{C}^{\times} \to W$$ sending $$1$$ to $$e_1 + e_2 + e_3$$. This works regardless of the choice of $$t$$ (so long as $$t \neq 0$$ because $$(\rho_t, \mathbb{C}^{\times})$$ and $$(\rho_V, W)$$ are equivalent representations.
• For similar examples (in slightly different words, but only slightly), see also pure.au.dk/portal/files/120581284/intro_to_character_theory.pdf Example 4.3 and 4.4 (unfortunately, those are the old version of the notes as I don't have a good place to upload the newest version). – Tobias Kildetoft Apr 29 at 15:36
• This is good but can you translate it into my notations? – john Apr 30 at 11:46
• also I defined the intertwining map $T$ in my document the same way you defined it. I believe $f$ is also an intertwining map which has the same properties – john Apr 30 at 11:53
• also what is $\rho^1$ and $\rho^2$ in this example? – john Apr 30 at 11:58
• What you write as $\rho^1_s$, I write as $\rho_1(s)$. I have two examples. In the first case $\rho_1$ is the map $\rho_V : S_3 \to \operatorname{GL}(U)$ and $\rho_2$ is the map $\rho_t : S_3 \to \operatorname{GL}(\mathbb{C}) = \mathbb{C}^{\times}$. In the second case $\rho_2$ is the same, but now $\rho_1$ is the map $\rho_V : S_3 \to \operatorname{GL}(W)$ – Adam Higgins Apr 30 at 12:02
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# The Unapologetic Mathematician
## More on tensor products and direct sums
We’ve defined the tensor product and the direct sum of two abelian groups. It turns out they interact very nicely.
One thing we need is another fact about the tensor product of abelian groups. If we take three abelian groups $A$, $B$, and $C$, we can form the tensor product $A\otimes B$, and then use that to make $(A\otimes B)\otimes C$. On the other hand, we could have started with $B\otimes C$ and then built $A\otimes(B\otimes C)$. If we look at the construction we used to show that tensor products actually exist we see that these two groups are not the same. However, they are isomorphic.
To see this, let’s make a bilinear function from $(A\otimes B)\times C$ to $A\otimes(B\otimes C)$. By our construction, any element of $A\otimes B$ can be represented as a sum $\sum\limits_i a_i\otimes b_i$, so linearity says we just need to consider elements of the form $a\otimes b$. Define $f(a\otimes b,c)=a\otimes(b\otimes c)$. This induces a unique linear function given by $\bar{f}((a\otimes b)\otimes c)=a\otimes(b\otimes c)$ and extending to sums of such elements. Similarly we get a linear function $\bar{f}^{-1}(a\otimes(b\otimes c))=(a\otimes b)\otimes c)$, so we have an isomorphism of abelian groups. We can thus (somewhat) unambiguously talk about “the” tensor product $A\otimes B\otimes C$.
Now let’s take a collection of abelian groups $A_i$ with $i$ running over an index set $\mathcal{I}$, and let $B$ be any other abelian group. We want to consider the tensor product
$\left(\bigoplus\limits_{i\in\mathcal{I}}A_i\right)\otimes B$
Since the direct sum is a direct product of groups, it comes with projections $\pi_k:\bigoplus_i A_i\rightarrow A_k$. Since the free product is in general a subgroup of the direct sum (a proper subgroup when the index set is infinite), we also have injections $\iota_k:A_k\rightarrow\bigoplus_i A_i$ coming from the free product. We can use these to build homomorphisms
$\iota_i\otimes1_B:A_i\otimes B\rightarrow\left(\bigoplus\limits_{i\in\mathcal{I}}A_i\right)\otimes B$
applying $\iota_i$ to $A_i$ and the identity to $B$. By the universal property of direct sums (the one it gets from free products of groups) this gives us a homomorphism
$\alpha:\bigoplus\limits_{i\in\mathcal{I}}(A_i\otimes B)\rightarrow\left(\bigoplus\limits_{i\in\mathcal{I}}A_i\right)\otimes B$
On the other hand, for each $k$ we have a bilinear function sending $(a,b)$ in $\left(\bigoplus_i A_i\right)\times B$ to $\pi_k(a)\otimes b$ in $A_k\otimes B$. By the universal properties of tensor products this gives a linear function $\left(\bigoplus_i A_i\right)\otimes B\rightarrow A_k\otimes B$. The universal property of direct sums (the one it gets from direct products of groups) gives us a linear function
$\beta:\left(\bigoplus\limits_{i\in\mathcal{I}}A_i\right)\otimes B\rightarrow\bigoplus\limits_{i\in\mathcal{I}}(A_i\otimes B)$
Now there’s a lot of juggling of functions and injections and projections here that I really don’t think is very illuminating. The upshot is that $\alpha$ and $\beta$ are inverses of each other, giving us an isomorphism of the two abelian groups. There’s nothing really special about the left side of the tensor product either. A similar result holds if the direct sum is the right tensorand. We can even put them together to get the really nice isomorphism:
$\left(\bigoplus\limits_{i\in\mathcal{I}}A_i\right)\otimes\left(\bigoplus\limits_{j\in\mathcal{J}}B_j\right)\cong\bigoplus\limits_{(i,j)\in\mathcal{I}\times\mathcal{J}}(A_i\otimes B_j)$
Neat!
April 17, 2007 - Posted by | Abelian Groups, Algebra, Group theory
1. “More on tensor products and direct sums”:
The direct sum is a co-product, not a direct product. It therefore comes with injections, not with projections. It seems to me that the proof given here might be incorrect.
Shlomi
Comment by Shlomi | July 31, 2010 | Reply
2. The direct sum is both a product and a coproduct, and it comes with both injections and projections. As I say in the second link in the post, “When we restrict our attention to abelian groups, direct products and free products are the same thing”.
See also here, keeping in mind that an Abelian group is a $\mathbb{Z}$-module.
Comment by John Armstrong | July 31, 2010 | Reply
3. I also find this proof troublesome. I believe you are confusing the terms “direct sum” and “direct product”, and unnecessarily introducing the term “free product” (I assume you mean “free product of abelian groups”, in which case this would by definition be identical to the direct sum). As you are aware, the direct sum is a subgroup of the direct product, and only equal when $\mathcal{I}$ is finite. Thus, your construction of the map $\beta$ by arguing that a collection of maps to each $A_i\otimes B$ induces a map to $\bigoplus_{i\in\mathcal{I}}(A_i\otimes B)$ is incorrect when $\mathcal{I}$ is infinite.
I’m not disputing the theorem itself – the tensor product *does* distribute over arbitrary direct sums – but I believe your proof is flawed. For an example of a correct proof, see Theorem 5.4 on p.22 of http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/tensorprod.pdf
I also wanted to say I have used your site many times to understand things better, and you have an impressive command of an extremely broad collection of areas in math. I don’t intend to sound like I’m claiming you don’t know what you’re doing; I just think you made an honest mistake here.
Many thanks for your continued blathing,
Zev
Comment by Zev Chonoles | February 13, 2011 | Reply
• The direct sum of Abelian groups plays two roles that come from group theory, each of which gives different properties. I refer to those names (which were defined earlier and are available through tracing backlinks) to emphasize and use those properties.
Comment by John Armstrong | February 13, 2011 | Reply
• But the direct sum only plays those two roles, both product and coproduct, in the case of finitely many factors. Shlomi’s comment is not quite right; a direct sum $A=\bigoplus A_i$ (even one with infinitely many factors) does, of course, come with projections $\pi_i$ to each factor $A_i$. However, these projections do not serve to make $A$ the product of the $A_i$ if there are infinitely many $A_i$. In other words, given an infinite collection of (non-trivial) abelian groups $A_i$, their direct sum $A$ *does not* have the property that any collection of maps of abelian groups $f_i:X\rightarrow A_i$ factors uniquely through a map $f:X\rightarrow A$.
For example, let $I$ be any infinite set, let each $A_i=\mathbb{Z}$, let $X=\mathbb{Z}$, and let all the maps $f_i:X\rightarrow A_i$ be the identity map of $\mathbb{Z}$. But there is no map $f:X\rightarrow A$ such that $f_i=\pi_i \circ f$, because where could $f$ send $1\in X$ to? Not $(1,1,\ldots)$, because that is not an element of the direct sum.
Comment by Zev Chonoles | February 13, 2011 | Reply
• I would point to the Wikipedia page on biproducts, http://en.wikipedia.org/wiki/Biproduct, but it is, unhelpfully, poorly worded about this very issue. It is not made explicitly clear that when they say “In the category of abelian groups, biproducts always exist and are given by the direct sum.”, they are only referring to biproducts of finitely many factors. However, this is alluded to in other parts of the page.
Comment by Zev Chonoles | February 13, 2011 | Reply
• Let me clarify the first sentence of my antepenultimate post. I should say, “The direct sum always plays the role of the coproduct; it only additionally plays the role of the product in the case of finitely many factors.”
Comment by Zev Chonoles | February 13, 2011 | Reply
4. Zev has decided to passive-aggressively demand that I pay attention to him by complaining to me on Formspring.
So yes, Zev, go through and mentally add a finiteness condition. If you want to fill in the gap in the infinite case, start a weblog of your own and write up the proof.
I’m sorry that I don’t instantly bend to your whims, and that I have other projects and a day job that keep me from doing so. If you want more responsiveness, pay for my time.
Comment by John Armstrong | February 27, 2011 | Reply
• I truly do apologize if I seemed passive-aggressive; I don’t remember my phrasing, but I’m sure it could have been much better if I made you angry. However, your response to my (quite kindly worded) initial post indicated that you thought I was in error. Indeed, you were rather dismissive of both me and Shlomi. Perhaps I am suffering from a bad case of http://xkcd.com/386/, but all I wanted to do by contacting you on Formspring (since you have no email listed) was to follow up on this post – privately(!) – and see if you still felt there was no problem. I don’t feel that was too unreasonable, although I again apologize for my phrasing.
Respectfully,
Zev
Comment by Zev Chonoles | February 27, 2011 | Reply
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# Using Least Squares for curve fitting in Matlab
I have a problem right now to solve. I have a Gaussian peak, which I need to divide into three components (three Gaussian distribution). For those three component Gaussian distribution, I already know the mean and standard deviation, if I want to do this by using least squares, where should I start with in Matlab?
Thanks
1) Form a matrix, A of order $m$ x $3$, such that:
$$A=\left[f(x) \ \ g(x) \ \ h(x)\right]$$
where $f(x), g(x)$ and $h(x)$ are three column vectors such that each column is a sample of component guassian at $x$.
2) Form a column vector, $b$ of your original Gaussian pulse of order m.
3) Solve the system $x=((A'*A)^{-1}*A')*b$, where $x$ is your desired set of coefficients of order $3$ x $1$, that would minimize the error in the least square sense.
For fitting multiple Gaussians to data you could take a look at EM algorithm:
http://www.mathworks.com/matlabcentral/fileexchange/26184-em-algorithm-for-gaussian-mixture-model
http://en.wikipedia.org/wiki/Expectation%E2%80%93maximization_algorithm
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The coherent dynamics of many body quantum system is nowadays an experimental reality: by means of the cold atoms in optical lattices, many Hamiltonians and time-dependent perturbations can be engineered. In this Thesis we discuss what happens in these systems when a periodic perturbation is applied. Thanks to Floquet theory, we can see that -- if the Floquet spectrum obeys certain continuity conditions possible in the thermodynamic limit-- dephasing among Floquet quasi-energies makes local observables relax to a periodic steady regime described by an effective density matrix: the Floquet diagonal ensemble (FDE). By means of numerical examples on the Quantum Ising Chain and the Lipkin model, we discuss the properties of the FDE focusing on the difference among ergodic and regular quantum dynamics and on how this reflects on the thermal properties ($T=\infty$) of the asymptotic condition. We verify thermalization in the classically ergodic Lipkin model and we demonstrate that this effect is induced by the Floquet states being delocalized and obeying Eigenstate Thermalization Hypothesis.We discuss also, in the Ising chain case, the work probability distribution, whose asymptotic condition is not described by the form (Generalized Gibbs Ensemble) that FDE acquires for local obserbvables because of integrability. Dephasing makes some correlations invisible in the local observables, but they are still present in the system. We consider also the linear response limit: when the amplitude of the perturbation is vanishingly small, the Floquet diagonal ensemble is not sufficient to describe the asymptotic condition given by LRT. For every small but finite amplitude, there are quasi-degeneracies in the Floquet spectrum giving rise to pre-relaxation to the condition predicted by Linear Response; these phenomena are strictly related to energy absorption and boundedness of the spectrum.
Periodic driving of a coherent quantum many body system and relaxation to the Floquet diagonal ensemble / Russomanno, Angelo. - (2014 Oct 31).
Periodic driving of a coherent quantum many body system and relaxation to the Floquet diagonal ensemble
Abstract
The coherent dynamics of many body quantum system is nowadays an experimental reality: by means of the cold atoms in optical lattices, many Hamiltonians and time-dependent perturbations can be engineered. In this Thesis we discuss what happens in these systems when a periodic perturbation is applied. Thanks to Floquet theory, we can see that -- if the Floquet spectrum obeys certain continuity conditions possible in the thermodynamic limit-- dephasing among Floquet quasi-energies makes local observables relax to a periodic steady regime described by an effective density matrix: the Floquet diagonal ensemble (FDE). By means of numerical examples on the Quantum Ising Chain and the Lipkin model, we discuss the properties of the FDE focusing on the difference among ergodic and regular quantum dynamics and on how this reflects on the thermal properties ($T=\infty$) of the asymptotic condition. We verify thermalization in the classically ergodic Lipkin model and we demonstrate that this effect is induced by the Floquet states being delocalized and obeying Eigenstate Thermalization Hypothesis.We discuss also, in the Ising chain case, the work probability distribution, whose asymptotic condition is not described by the form (Generalized Gibbs Ensemble) that FDE acquires for local obserbvables because of integrability. Dephasing makes some correlations invisible in the local observables, but they are still present in the system. We consider also the linear response limit: when the amplitude of the perturbation is vanishingly small, the Floquet diagonal ensemble is not sufficient to describe the asymptotic condition given by LRT. For every small but finite amplitude, there are quasi-degeneracies in the Floquet spectrum giving rise to pre-relaxation to the condition predicted by Linear Response; these phenomena are strictly related to energy absorption and boundedness of the spectrum.
Scheda breve Scheda completa Scheda completa (DC)
Santoro, Giuseppe Ernesto
Russomanno, Angelo
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1963_7477_Thesis_SISSA_Russomanno.pdf
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Tipologia: Tesi
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Utilizza questo identificativo per citare o creare un link a questo documento: http://hdl.handle.net/20.500.11767/3904
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# Tutor profile: Gray O.
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## Questions
### Subject:Javascript Programming
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Question:
Consider the following code: $(\texttt{ (function() { var a = b = 10; })(); console.log(a); console.log(b); }$) What will be output to the console?
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The first thing we notice in this question is that we have a function literal that is being called immediately and then two $${\tt console.log}$$ statements that output the value of variables $$a$$ and $$b$$. The important thing to understand here is that the IIFE will interpret the line $${\tt var a = b = 10;}$$ as two separate assignments. The first will assign a local variable $$a$$, to the value of $$b$$. However, in order to get the value of $$b$$, it must first assign $$b$$ to be a global variable with the value 10. $$b$$ is a global variable because it does not have the keyword $${\tt var}$$ in front of it. Therefore, the console will output $(\texttt{ undefined 10 })$ $$a$$ is a local variable to the function literal, so it will not be defined in the global scope. $$b$$, on the other hand, is declared as a global variable, and so its value will still be 10 outside of the function.
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Two airplanes take off at the same time from the same airport. One is traveling at 350mph and the other is traveling at 575mph. How far apart are they from each other after 1 hour and 40 minutes if there is an angle of 30 degrees between their directions of flight?
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# From 4 Officers and 8 Jawans in How Many Ways Can 6 Be Chosen (I) to Include Exactly One Officer - Mathematics
From 4 officers and 8 jawans in how many ways can 6 be chosen (i) to include exactly one officer
#### Solution
From 4 officers and 8 jawans, 6 need to be chosen. Out of them, 1 is an officer.
Required number of ways =${}^4 C_1 \times {}^8 C_5 = 4 \times \frac{8!}{5! 3!} = 4 \times \frac{8 \times 7 \times 6 \times 5!}{5! \times 6} = 224$
Concept: Combination
Is there an error in this question or solution?
#### APPEARS IN
RD Sharma Class 11 Mathematics Textbook
Chapter 17 Combinations
Exercise 17.2 | Q 9.1 | Page 16
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This is “Gas Mixtures”, section 6.6 from the book Beginning Chemistry (v. 1.0). For details on it (including licensing), click here.
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## 6.6 Gas Mixtures
### Learning Objective
1. Learn Dalton’s law of partial pressures.
One of the properties of gases is that they mix with each other. When they do so, they become a solution—a homogeneous mixture. Some of the properties of gas mixtures are easy to determine if we know the composition of the gases in the mix.
In gas mixtures, each component in the gas phase can be treated separately. Each component of the mixture shares the same temperature and volume. (Remember that gases expand to fill the volume of their container; gases in a mixture continue to do that as well.) However, each gas has its own pressure. The partial pressureThe pressure that an individual gas in a mixture has. of a gas, Pi, is the pressure that an individual gas in a mixture has. Partial pressures are expressed in torr, millimeters of mercury, or atmospheres like any other gas pressure; however, we use the term pressure when talking about pure gases and the term partial pressure when we are talking about the individual gas components in a mixture.
Dalton’s law of partial pressuresThe total pressure of a gas mixture, $Ptot$, is equal to the sum of the partial pressures of the components, $Pi$. states that the total pressure of a gas mixture, Ptot, is equal to the sum of the partial pressures of the components, Pi:
Although this may seem to be a trivial law, it reinforces the idea that gases behave independently of each other.
### Example 15
A mixture of H2 at 2.33 atm and N2 at 0.77 atm is in a container. What is the total pressure in the container?
Solution
Dalton’s law of partial pressures states that the total pressure is equal to the sum of the partial pressures. We simply add the two pressures together:
Ptot = 2.33 atm + 0.77 atm = 3.10 atm
Test Yourself
Air can be thought of as a mixture of N2 and O2. In 760 torr of air, the partial pressure of N2 is 608 torr. What is the partial pressure of O2?
152 torr
### Example 16
A 2.00 L container with 2.50 atm of H2 is connected to a 5.00 L container with 1.90 atm of O2 inside. The containers are opened, and the gases mix. What is the final pressure inside the containers?
Solution
Because gases act independently of each other, we can determine the resulting final pressures using Boyle’s law and then add the two resulting pressures together to get the final pressure. The total final volume is 2.00 L + 5.00 L = 7.00 L. First, we use Boyle’s law to determine the final pressure of H2:
(2.50 atm)(2.00 L) = P2(7.00 L)
Solving for P2, we get
P2 = 0.714 atm = partial pressure of H2
Now we do that same thing for the O2:
(1.90 atm)(5.00 L) = P2(7.00 L) P2 = 1.36 atm = partial pressure of O2
The total pressure is the sum of the two resulting partial pressures:
Ptot = 0.714 atm + 1.36 atm = 2.07 atm
Test Yourself
If 0.75 atm of He in a 2.00 L container is connected to a 3.00 L container with 0.35 atm of Ne and the containers are opened, what is the resulting total pressure?
0.51 atm
One of the reasons we have to deal with Dalton’s law of partial pressures is because gases are frequently collected by bubbling through water. As we will see in Chapter 10 "Solids and Liquids", liquids are constantly evaporating into a vapor until the vapor achieves a partial pressure characteristic of the substance and the temperature. This partial pressure is called a vapor pressureThe partial pressure exerted by evaporation of a liquid.. Table 6.2 "Vapor Pressure of Water versus Temperature" lists the vapor pressures of H2O versus temperature. Note that if a substance is normally a gas under a given set of conditions, the term partial pressure is used; the term vapor pressure is reserved for the partial pressure of a vapor when the liquid is the normal phase under a given set of conditions.
Table 6.2 Vapor Pressure of Water versus Temperature
Temperature (°C) Vapor Pressure (torr) Temperature (°C) Vapor Pressure (torr)
5 6.54 30 31.84
10 9.21 35 42.20
15 12.79 40 55.36
20 17.54 50 92.59
21 18.66 60 149.5
22 19.84 70 233.8
23 21.08 80 355.3
24 22.39 90 525.9
25 23.77 100 760.0
Any time a gas is collected over water, the total pressure is equal to the partial pressure of the gas plus the vapor pressure of water. This means that the amount of gas collected will be less than the total pressure suggests.
### Example 17
Hydrogen gas is generated by the reaction of nitric acid and elemental iron. The gas is collected in an inverted 2.00 L container immersed in a pool of water at 22°C. At the end of the collection, the partial pressure inside the container is 733 torr. How many moles of H2 gas were generated?
Solution
We need to take into account that the total pressure includes the vapor pressure of water. According to Table 6.2 "Vapor Pressure of Water versus Temperature", the vapor pressure of water at 22°C is 19.84 torr. According to Dalton’s law of partial pressures, the total pressure equals the sum of the pressures of the individual gases, so
We solve by subtracting:
Now we can use the ideal gas law to determine the number of moles (remembering to convert temperature to kelvins, making it 295 K):
All the units cancel except for mol, which is what we are looking for. So
n = 0.0775 mol H2 collected
Test Yourself
CO2, generated by the decomposition of CaCO3, is collected in a 3.50 L container over water. If the temperature is 50°C and the total pressure inside the container is 833 torr, how many moles of CO2 were generated?
0.129 mol
Finally, we introduce a new unit that can be useful, especially for gases. The mole fractionThe ratio of the number of moles of a component in a mixture divided by the total number of moles in the sample., χi, is the ratio of the number of moles of component i in a mixture divided by the total number of moles in the sample:
(χ is the lowercase Greek letter chi.) Note that mole fraction is not a percentage; its values range from 0 to 1. For example, consider the combination of 4.00 g of He and 5.0 g of Ne. Converting both to moles, we get
The total number of moles is the sum of the two mole amounts:
total moles = 1.00 mol + 0.025 mol = 1.25 mol
The mole fractions are simply the ratio of each mole amount and the total number of moles, 1.25 mol:
The sum of the mole fractions equals exactly 1.
For gases, there is another way to determine the mole fraction. When gases have the same volume and temperature (as they would in a mixture of gases), the number of moles is proportional to partial pressure, so the mole fractions for a gas mixture can be determined by taking the ratio of partial pressure to total pressure:
$χi=PiPtot$
This expression allows us to determine mole fractions without calculating the moles of each component directly.
### Example 18
A container has a mixture of He at 0.80 atm and Ne at 0.60 atm. What are the mole fractions of each component?
Solution
According to Dalton’s law, the total pressure is the sum of the partial pressures:
Ptot = 0.80 atm + 0.60 atm = 1.40 atm
The mole fractions are the ratios of the partial pressure of each component and the total pressure:
Again, the sum of the mole fractions is exactly 1.
Test Yourself
What are the mole fractions when 0.65 atm of O2 and 1.30 atm of N2 are mixed in a container?
### Food and Drink App: Carbonated Beverages
Carbonated beverages—sodas, beer, sparkling wines—have one thing in common: they have CO2 gas dissolved in them in such sufficient quantities that it affects the drinking experience. Most people find the drinking experience pleasant—indeed, in the United States alone, over 1.5 × 109 gal of soda are consumed each year, which is almost 50 gal per person! This figure does not include other types of carbonated beverages, so the total consumption is probably significantly higher.
All carbonated beverages are made in one of two ways. First, the flat beverage is subjected to a high pressure of CO2 gas, which forces the gas into solution. The carbonated beverage is then packaged in a tightly-sealed package (usually a bottle or a can) and sold. When the container is opened, the CO2 pressure is released, resulting in the well-known hiss of an opening container, and CO2 bubbles come out of solution. This must be done with care: if the CO2 comes out too violently, a mess can occur!
If you are not careful opening a container of a carbonated beverage, you can make a mess as the CO2 comes out of solution suddenly.
The second way a beverage can become carbonated is by the ingestion of sugar by yeast, which then generates CO2 as a digestion product. This process is called fermentation. The overall reaction is
C6H12O6(aq) → 2C2H5OH(aq) + 2CO2(aq)
When this process occurs in a closed container, the CO2 produced dissolves in the liquid, only to be released from solution when the container is opened. Most fine sparkling wines and champagnes are turned into carbonated beverages this way. Less-expensive sparkling wines are made like sodas and beer, with exposure to high pressures of CO2 gas.
### Key Takeaways
• The pressure of a gas in a gas mixture is termed the partial pressure.
• Dalton’s law of partial pressure says that the total pressure in a gas mixture is the sum of the individual partial pressures.
• Collecting gases over water requires that we take the vapor pressure of water into account.
• Mole fraction is another way to express the amounts of components in a mixture.
### Exercises
1. What is the total pressure of a gas mixture containing these partial pressures: , , and ?
2. What is the total pressure of a gas mixture containing these partial pressures: PNe = 312 torr, PHe = 799 torr, and PAr = 831 torr?
3. In a gas mixture of He and Ne, the total pressure is 335 torr and the partial pressure of He is 0.228 atm. What is the partial pressure of Ne?
4. In a gas mixture of O2 and N2, the total pressure is 2.66 atm and the partial pressure of O2 is 888 torr. What is the partial pressure of N2?
5. A 3.55 L container has a mixture of 56.7 g of Ar and 33.9 g of He at 33°C. What are the partial pressures of the gases and the total pressure inside the container?
6. A 772 mL container has a mixture of 2.99 g of H2 and 44.2 g of Xe at 388 K. What are the partial pressures of the gases and the total pressure inside the container?
7. A sample of O2 is collected over water in a 5.00 L container at 20°C. If the total pressure is 688 torr, how many moles of O2 are collected?
8. A sample of H2 is collected over water in a 3.55 L container at 50°C. If the total pressure is 445 torr, how many moles of H2 are collected?
9. A sample of CO is collected over water in a 25.00 L container at 5°C. If the total pressure is 0.112 atm, how many moles of CO are collected?
10. A sample of NO2 is collected over water in a 775 mL container at 25°C. If the total pressure is 0.990 atm, how many moles of NO2 are collected?
11. A sample of NO is collected over water in a 75.0 mL container at 25°C. If the total pressure is 0.495 atm, how many grams of NO are collected?
12. A sample of ClO2 is collected over water in a 0.800 L container at 15°C. If the total pressure is 1.002 atm, how many grams of ClO2 are collected?
13. Determine the mole fractions of each component when 44.5 g of He is mixed with 8.83 g of H2.
14. Determine the mole fractions of each component when 9.33 g of SO2 is mixed with 13.29 g of SO3.
15. In a container, 4.56 atm of F2 is combined with 2.66 atm of Cl2. What are the mole fractions of each component?
16. In a container, 77.3 atm of SiF4 are mixed with 33.9 atm of O2. What are the mole fractions of each component?
1. 2.70 atm
2. 162 torr, or 0.213 atm
3. PAr = 10.0 atm; PHe = 59.9 atm; Ptot = 69.9 atm
4. 0.183 mol
5. 0.113 mol
6. 0.0440 g
7. $χHe=0.718$; $χH2=0.282$
8. $χF2=0.632$; $χCl3=0.368$
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# Py{} adds a none in the code
I have the following MWE using pythontex:
%& -shell-escape
\documentclass{article}
\usepackage{pythontex}
\newcommand{\printPython}[1]{
\py{print(r'Hello, #1')}%
\newline
}
\begin{document}
Say hello Python:
%\begin{python}%
%print r"Hello \LaTeX!"
%\end{python}%
\printPython{4}
\begin{pycode}
print(r"\centering")
print(r"\textit{A message from Python!}")
\end{pycode}
\end{document}
When compiling, the result is
Say hello Python:
Hello, 4 None
A message from Python!
Where does the None come from, and how do I remove it?
• I solved it by replacing \py with \pyc, but nevertheless I am interested in the reason here. – arc_lupus Mar 12 '16 at 11:15
\py inserts a string representation of whatever it is given. In this case, it was given the print() function, which returns None. So you get what was printed, plus the return value. Basically, \py is a shortcut so that using print() isn't necessary. You could just use \py{r'Hello, #1'}%.
\pyc is the inline equivalent of the pycode environment. It just executes code. So if you want it to insert anything in the document, you need to use print().
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# Revision history [back]
Hi,
This is because the two numbers do not live in the same ground field as you can see
sage: 1.414.parent()
Real Field with 53 bits of precision
sage: sqrt(2).n(digits=4).parent()
Real Field with 17 bits of precision
1.414 is not exactly represented in memory because it is not a diadic number. Two different diadic approximations with different precisions are different.
Vincent
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## Thursday, November 22, 2012
### Flexible language
I've been learning Lisp for few years now, and every Lisp book I read keeps saying that Lisp is a flexible language that you can extend to the degree when it fits naturally to your domain. It's easy to say, but what exactly does this phrase mean? After all, when you program in your non-Lisp language, don't you modify it for your domain problem? I've been thinking about it for a long time, and only recently I started to understand what flexibility really means. There is a difference between using the language and changing the language to solve a problem. In this post I will try to show the difference based on a simple example.
## Problem
Suppose you have a process that listens to a message queue. The messages are just ordinary maps. If the map contains certain keys, one or more handlers must be invoked. Here is a matrix that shows which handler is invoked for which key
For example, if the map has key a, then DocHandler and AlertHandler need to be called. If it has key b, then NoteHandler and AlertHandler are called. In reality there might be more keys and more handlers, but for simplicity we limit our example to three keys and three handlers.
## Java
Let's see how this can be implemented in Java. I chose Java just as an example of non-Lisp language. You can pick any other non-Lisp language instead.
public class SimpleMessageListener {
public List onMessage(Map message) {
if (isDoc(message)) {
}
if (isNote(message)) {
}
}
return result;
}
// Decision makers
private boolean isDoc(Map message) {
return message.containsKey("a") || message.containsKey("c");
}
private boolean isNote(Map message) {
return message.containsKey("b") || message.containsKey("c");
}
return message.containsKey("a") || message.containsKey("b");
}
// Handlers
private String handleDoc(Map message) {
return String.format("Document:%s:%s", message.get("a"), message.get("c"));
}
private String handleNote(Map message) {
return String.format("Note:%s:%s", message.get("b"), message.get("c"));
}
}
}
The internals of handle- methods might be very different in reality. Consider the fact they have the same structure as a coincidence. What is not coincidence though is the structure of is- methods. Those methods are identical indeed.
Is this code clean? I would say, no. The main issue is that it's split in three separate but closely related parts. If tomorrow I introduce another message key and a new handler, I have to change three places in the code. Another problem is the code duplication in two spots: a series of if-statements and a group of is- methods.
The last thing to notice about this code is that it's hard to see what kind of problem it's trying to solve. If I didn't provide a matrix which maps message keys to handlers, it would take even more time to figure out what the code is doing. Can we make this code better?
Let's rewrite it as follows:
public class FunctionalMessageListener {
private interface Handler {
void handle(Map message, List acc);
}
private class DocHandler implements Handler {
private boolean isDoc(Map message) {
return message.containsKey("a") || message.containsKey("c");
}
public void handle(Map message, List acc) {
}
}
private class NoteHandler implements Handler {
private boolean isNote(Map message) {
return message.containsKey("b") || message.containsKey("c");
}
public void handle(Map message, List acc) {
}
}
private class AlertHandler implements Handler {
return message.containsKey("a") || message.containsKey("b");
}
public void handle(Map message, List acc) {
}
}
private List<Handler> handlers() {
return Arrays.asList(new DocHandler(), new NoteHandler(), new AlertHandler());
}
public List onMessage(Map message) {
for (Handler handler : handlers()) {
handler.handle(message, result);
}
return result;
}
}
In this version we eliminated ugly if-series, and group together decision making logic and message handling. From that perspective the code became cleaner, but not necessarily clearer. Now it actually takes more effort to understand what the code is doing. Also, the duplication inside the is- methods is still there. We can fix it by extracting it to some abstract class or utility method. We can also use Java reflection within handlers() method to build a collection of handlers without explicitly specifying them. All these manipulations arguably make the code cleaner, but... one thing we'll never be able to fix is the separation between decision making logic and message handling. Whatever you do, there will always be the if-statement that checks if you need to process the message, and the message processing logic itself. Those two things will always be separate. Here is the point where we hit the language constraints.
## Clojure
Now let's try to solve the same problem in Lisp and see if we can fix the language to eliminate the last issue from the paragraph above. Here is the direct translation of the previous Java snippet to Clojure dialect of Lisp
(defn- doc-handler [msg]
(let [a (msg :a), c (msg :c)]
(when (or a c)
(format "Document:%s:%s" a c))))
(defn- note-handler [msg]
(let [b (msg :b), c (msg :c)]
(when (or b c)
(format "Note:%s:%s" b c))))
(let [a (msg :a), b (msg :b)]
(when (or a b)
(defn- handlers []
(defn on-message [msg]
(letfn [(handle [acc h]
(if-let [res (h msg)]
(conj acc res)
acc))]
(reduce handle [] (handlers))))
This code is already easier to read, but we can do even better. The separation between decision making logic and message handling is still there. At this point we should ask the question: what kind of code do we want to see there? And my answer is: I want to replace the -handler methods above with the following code
(handler doc [a c]
(format "Document:%s:%s" a c))
(handler note [b c]
(format "Note:%s:%s" b c))
(format "Alert:%s:%s" a b))
You see: no conditionals. Handlers are self-sufficient entities which know when they have to be applied and how. In their signatures they explicitly declare what kind of parameters they expect, and in the body they just use those parameters. No boilerplate, clean and simple. The beauty of Lisp is that you can actually implement that code. The way you do it is by creating a macro which will generate the appropriate functions. Creating a macro is not a simple task, I spent quite some time to get this one working, but it's worth of doing, because now the code is clean and clear.
We can make one additional step further by moving the handler declarations inside the handlers() method. (We need one small macro for that.) And here is the final solution
(defmacro handler [name args & body]
(fn [~'msg]
(let [~@(interleave args (map (fn [x] (get ~'msg ~(keyword x))) args))]
(when (or ~@args)
~@body))))
(defmacro build-handlers [& body]
(defn- handlers []
[~@body]))
(build-handlers
(handler doc [a c]
(format "Document:%s:%s" a c))
(handler note [b c]
(format "Note:%s:%s" b c))
(defn on-message [msg]
(letfn [(handle [acc h]
(if-let [res (h msg)]
(conj acc res)
acc))]
(reduce handle [] (handlers))))
As I said, the first macro might be cryptic, but look at the highlighted part. This is the essence of our problem, and it cannot be done any simpler. Suppose, we need to implement a new handler which should be called if key c is present in the message. Here what we would need to add to build-handler's body to implement this new requirement
(handler new [c]
...)
Simple, right? And what if a new key is added to the message which should be processed by document handler? Here is what we need to change
(handler doc [a c d]
...)`
We just add a new key to the function's parameter list. That's it — one-word change.
## Summary
Lisp is the most powerful programming language. By that I mean you can change the language in such a way that the solution to any particular problem can be expressed in the simplest possible way. By changing the language, you can remove all the barriers between the language and the problem domain. I hope I demonstrated this in my simple example.
## Resources
Clojure source code for this blog, along with the unit tests.
Eugene Kuleshov said...
You can do the same thing in Java. Just need containsAny() helper method and then make all your handlers declarative. You can write your own or use one from commons collections
Andrey Paramonov said...
Nice. Can you post the entire solution? I couldn't come up with anything simple.
Eugene Kuleshov said...
Something like this https://gist.github.com/4149886
Andrey Paramonov said...
Thanks Eugene for your example. Indeed you separated if- logic from handling logic, which makes your example better than my Java code above, but I think my Summary about Lisp expressiveness is still valid. Look at the signature of onHandle() methods. First, it's enforced by handle() method which is responsible for if- logic, so in a way if- logic leaks into handling logic. Second, all Handlers are coupled with message representation (i.e. map), which creates a small code duplication in every onHandle() implementation where you have to extract the data from the map.
Lisp code, on the other hand, does not have these limitations. (handler …) calls are completely unaware of message representation. Theoretically, I can copy some existing function definition, rename parameters accordingly, and paste it inside (build-handlers …) body. This will automatically transform the function into a message handler. I don't think we can achieve this level of flexibility in Java.
Eugene Kuleshov said...
The handle() method is just an external contract, which is carried trough... Anyways, you can't really compare strongly-typed language with dynamically typed one.
Baljeet said...
Very nice.
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# Proof for variation of Prim's and Kruskal's to find maximum-weight acyclic subgraph
I have been scratching my head to find good counter examples to the following problem:
Suppose we are given a directed graph G=(V,E) in which every edge has a distinct positive edge weight. A directed graph is acyclic if it has no directed cycle. Suppose that we want to compute the maximum-weight acyclic subgraph of G (where the weight of a subgraph is the sum of its edges' weights). Assume that G is weakly connected, meaning that there is no cut with no edges crossing it in either direction.
Here is an analog of Prim's algorithm for directed graphs:
Start from an arbitrary vertex s, initialize S={s} and F=∅.
While S≠V, find the maximum-weight edge (u,v) with one endpoint in S and one endpoint in V−S. Add this edge to F, and add the appropriate endpoint to S.
Here is an analog of Kruskal's algorithm. Sort the edges from highest to lowest weight. Initialize F=∅. Scan through the edges; at each iteration, add the current edge i to F if and only if it does not create a directed cycle.
Both algorithm fail.
There should be a 4 vertices graph with 2 cycles that demonstrate this.
So far I played with different weights for this:
A<----B
^⟍ ^
| ⟍ |
| ➘|
C<--- D
However I am not yet able to prove that both algorithms fail. I would love any suggestioin
This is a nice exercise. You should probably do it yourself, to get the learning benefit. I suggest that you enumerate all non-isomorphic graphs with 3 vertices or 4 vertices, and for each such graph, try playing with the weights to try to find a counter-example. Basically, keep trying what you've been doing, but be more systematic and exhaustive about it. You're on the right direction -- keep at it.
• so I had a good go at this and in what I was able to demonstrate is that both algorithms fail at one main point: they stop when they have a tree, yet they do not continue to add edges although they do not create a cycle. Is this your conclusion too? is there anything else to be seen? – superuseroi Apr 10 '15 at 16:30
• @superuseroi, Hmm, that doesn't sound entirely right. I believe you are right that the Prim's-variant will always stop with a tree, but that's not correct of the Kruskal's-variant: the Kruskal's-variant can output a non-tree in some cases (consider running it on a DAG, for instance). By the way, all it takes is one counterexample to prove that the algorithm doesn't work, so you don't have to characterize all failure cases to prove that both algorithms don't work. – D.W. Apr 10 '15 at 16:44
• @WD thank you so much for your response. As far as I can see because Kruskal's algo uses a disjoint set (unionfind) and will stop when there is 1 single set (meaning every set was U with every other) this happens only when we have collected V-1 edges. There can be a graph where there is a max acyclic subgraph which has more than V-1 edges. As for running it on a DAG I see that the tree could be weakly connected that's all. What do you see? thank you! – superuseroi Apr 11 '15 at 11:37
• Take another look at the definition of the "analog of Kruskal's algorithm". Note that it is not specified to use union-find. Also note that the real Kruskal's algorithm can be implemented using union-find, but the variant doesn't seem like it can. Anyway, CS.SE isn't a discussion forum and isn't a good place if you want a series of follow-up questions answered, so I'm going to sign off here -- this site is best for posting a single question and getting an answer to it (if you have other questions they should be posted separately). – D.W. Apr 11 '15 at 16:30
• Yes perfect I understand, thank you for your input it was great! – superuseroi Apr 11 '15 at 17:53
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## Potential Energy Operator given Hamiltonian Solution
STEP 0: Pre-Calculation Summary
Formula Used
Potential Energy Operator = Hamiltonian of System-Kinetic Energy Operator
V^ = Ĥ-T^
This formula uses 3 Variables
Variables Used
Potential Energy Operator - Potential Energy Operator corresponds to the classical interaction energies between particles in the system.
Hamiltonian of System - Hamiltonian of System specifies its total energy. The sum of its kinetic energy (that of motion) and its potential energy (that of position).
Kinetic Energy Operator - Kinetic Energy Operator is defined as the momentum operator divided by the mass of the particle.
STEP 1: Convert Input(s) to Base Unit
Hamiltonian of System: 16 --> No Conversion Required
Kinetic Energy Operator: 3 --> No Conversion Required
STEP 2: Evaluate Formula
Substituting Input Values in Formula
V^ = Ĥ-T^ --> 16-3
Evaluating ... ...
V^ = 13
STEP 3: Convert Result to Output's Unit
13 --> No Conversion Required
13 <-- Potential Energy Operator
(Calculation completed in 00.016 seconds)
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## < 5 Hamiltonian System Calculators
Molecular Potential Energy of Molecules
## Molecular Potential Energy of Molecules
Formula
"E" = "E"_{"bonds"}+"E"_{"dihedral"}+"E"_{"angle"}+"E"_{"non-bonded"}
Example
"30.5KJ/mol"="12KJ/mol"+"5.5KJ/mol"+"10.5KJ/mol"+"2.5KJ/mol"
Calculator
LaTeX
Molecular Potential Energy = Energy of Bond Lengths+Energy of Torsion Angle+Energy of Bond Angles+Energy of Non Bonded Atom
Molecular Potential Energy of Non-bonded pairs of Atoms
## Molecular Potential Energy of Non-bonded pairs of Atoms
Formula
"E" = "E"_{"electrostatic"}+"E"_{"van "der" waals"}
Example
"21KJ/mol"="10.55KJ/mol"+"10.45KJ/mol"
Calculator
LaTeX
Molecular Potential Energy = Energy of Electrostatic Force+Energy of Van Der Waals Force
Potential Energy Operator given Hamiltonian
## Potential Energy Operator given Hamiltonian
Formula
"V"^{"^"} = "Ĥ"-"T"^{"^"}
Example
"13"="16"-"3"
Calculator
LaTeX
Potential Energy Operator = Hamiltonian of System-Kinetic Energy Operator
Kinetic Operator given Hamiltonian
## Kinetic Operator given Hamiltonian
Formula
"T"^{"^"} = "Ĥ"-"V"^{"^"}
Example
"6"="16"-"10"
Calculator
LaTeX
Kinetic Energy Operator = Hamiltonian of System-Potential Energy Operator
Hamiltonian of System
## Hamiltonian of System
Formula
"Ĥ" = "T"^{"^"}+"V"^{"^"}
Example
"13"="3"+"10"
Calculator
LaTeX
Hamiltonian of System = Kinetic Energy Operator+Potential Energy Operator
## Potential Energy Operator given Hamiltonian Formula
Potential Energy Operator = Hamiltonian of System-Kinetic Energy Operator
V^ = Ĥ-T^
## What is Quantum Chemistry?
Quantum chemistry, also called molecular quantum mechanics, is a branch of chemistry focused on the application of quantum mechanics to chemical systems. Understanding electronic structure and molecular dynamics using the Schrödinger equations are central topics in quantum chemistry.
## How to Calculate Potential Energy Operator given Hamiltonian?
Potential Energy Operator given Hamiltonian calculator uses Potential Energy Operator = Hamiltonian of System-Kinetic Energy Operator to calculate the Potential Energy Operator, The Potential Energy Operator given Hamiltonian formula corresponds to the classical interaction energies between particles in the system. It is calculated by subtracting kinetic energy operator from Hamiltonian system. Potential Energy Operator is denoted by V^ symbol.
How to calculate Potential Energy Operator given Hamiltonian using this online calculator? To use this online calculator for Potential Energy Operator given Hamiltonian, enter Hamiltonian of System (Ĥ) & Kinetic Energy Operator (T^) and hit the calculate button. Here is how the Potential Energy Operator given Hamiltonian calculation can be explained with given input values -> 13 = 16-3.
### FAQ
What is Potential Energy Operator given Hamiltonian?
The Potential Energy Operator given Hamiltonian formula corresponds to the classical interaction energies between particles in the system. It is calculated by subtracting kinetic energy operator from Hamiltonian system and is represented as V^ = Ĥ-T^ or Potential Energy Operator = Hamiltonian of System-Kinetic Energy Operator. Hamiltonian of System specifies its total energy. The sum of its kinetic energy (that of motion) and its potential energy (that of position) & Kinetic Energy Operator is defined as the momentum operator divided by the mass of the particle.
How to calculate Potential Energy Operator given Hamiltonian?
The Potential Energy Operator given Hamiltonian formula corresponds to the classical interaction energies between particles in the system. It is calculated by subtracting kinetic energy operator from Hamiltonian system is calculated using Potential Energy Operator = Hamiltonian of System-Kinetic Energy Operator. To calculate Potential Energy Operator given Hamiltonian, you need Hamiltonian of System (Ĥ) & Kinetic Energy Operator (T^). With our tool, you need to enter the respective value for Hamiltonian of System & Kinetic Energy Operator and hit the calculate button. You can also select the units (if any) for Input(s) and the Output as well.
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Volume 336 - XIII Quark Confinement and the Hadron Spectrum (Confinement2018) - A: Vacuum structure and confinement
Conformal Perturbation description of Deconfinement.
M. Caselle,* N. Magnoli, A. Nada, M. Panero, M. Scanavino
*corresponding author
Full text: pdf
Pre-published on: 2019 September 24
Published on: 2019 September 26
Abstract
Conformal perturbation theory is a powerful tool to describe the behavior of statistical-mechanics models and quantum field theories in the vicinity of a critical point. In the past few years, it has been extensively used to describe two-dimensional models and recently has also been extended to three-dimensional models. We show here that it can also be used to describe the behavior of four-dimensional lattice gauge theories in the vicinity of a critical point. As an example, we discuss the two-point correlator of Polyakov loops close to the thermal deconfinement transition of $\mathrm{SU}(2)$ Yang-Mills theory. We show that the short-distance behavior of this correlation function (and, thus, of the interquark potential) is described very well by conformal perturbation theory. This method is expected to work with a similarly high accuracy for all critical points in the same universality class, including, in particular, the critical endpoint in the QCD phase diagram.
DOI: https://doi.org/10.22323/1.336.0042
Open Access
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