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# 20130327 (Wednesday, 27 March 2013)¶
After having seen the demo sites at lino-framework.org work even without the released version of Babel, I understood that the warning “Lino currently requires the development version of python-babel which requires some manual work as explained on their site in Setting up Babel from a Subversion Checkout, see 2013-03-23” in Installing a Lino development environment was wrong and removed it. Because the development version is needed only for extracting messages with pgettext and writing catalog files, but not for using them.
Today I still continued to work on the API for writing tested docs, used in Debts mediation and Miscellaneous.
Redesigned the signatures of lino.core.actors.Actor.request() and lino.ui.Site.login().
I had an idea which would enable us to replace calls like:
ses.show(debts.ExpensesByBudget.request(obj))
by:
ses.show(obj.expenses_by_budget())
Lino would have automagically converted a string like “ExpensesByBudget” to “expenses_by_budget” (using lino.util.uncamel()), and even use a new optimal class attribute Actor.related_name to give custom names. But I then dropped this feature because it creates an additional copy of the actor’s name. This would cause additional work when an actor needs to be renamed. So we refuse this feature deliberately.
Adapted appy_pod.Renderer to the new interface. Cool: added another new style of test in Debts mediation: print a document using appy_pod! It is a pity though that the appy_pod renderer has appearently no option to not catch any exception that occurs during the rendering. So the test isn’t yet really perfect.
• Wrote and tested lino_welfare.migrate.migrate_from_1_1_0()
• There was a funny little bug in north.dpy : it wrote the Python version (instead of settings.SITE.version) to SOURCE_VERSION.
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### Two-query PCP with subconstant errorTwo-query PCP with subconstant error
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Author Moshkovitz, Dana ♦ Raz, Ran Source ACM Digital Library Content type Text Publisher Association for Computing Machinery (ACM) File Format PDF Copyright Year ©2010 Language English
Subject Domain (in DDC) Computer science, information & general works ♦ Data processing & computer science Subject Keyword Probabilistically checkable proofs (PCP) ♦ Composition ♦ Label cover ♦ Locally decode/reject code (LDRC) Abstract We show that the NP-Complete language 3Sat has a PCP verifier that makes two queries to a proof of almost-linear size and achieves subconstant probability of error $ϵ=\textit{o}(1).$ The verifier performs only projection tests, meaning that the answer to the first query determines at most one accepting answer to the second query. The number of bits representing a symbol in the proof depends only on the error ϵ. Previously, by the parallel repetition theorem, there were PCP Theorems with two-query projection tests, but only (arbitrarily small) $\textit{constant}$ error and $\textit{polynomial}$ size. There were also PCP Theorems with $\textit{subconstant}$ error and $\textit{almost-linear}$ size, but a constant number of queries that is $\textit{larger}$ than 2. As a corollary, we obtain a host of new results. In particular, our theorem improves many of the hardness of approximation results that are proved using the parallel repetition theorem. A partial list includes the following: (1) 3Sat cannot be efficiently approximated to within a factor of $7/8+\textit{o}(1),$ unless P=NP. This holds even under almost-linear reductions. Previously, the best known NP-hardness factor was 7/8+ϵ for any constant ϵ>0, under polynomial reductions (Håstad). (2) 3Lin cannot be efficiently approximated to within a factor of $1/2+\textit{o}(1),$ unless P=NP. This holds even under almost-linear reductions. Previously, the best known NP-hardness factor was 1/2+ϵ for any constant ϵ>0, under polynomial reductions (Håstad). (3) A PCP Theorem with amortized query complexity 1 + $\textit{o}(1)$ and amortized free bit complexity $\textit{o}(1).$ Previously, the best-known amortized query complexity and free bit complexity were 1+ϵ and ϵ, respectively, for any constant ϵ>0 (Samorodnitsky and Trevisan). One of the new ideas that we use is a new technique for doing the $\textit{composition}$ step in the (classical) proof of the PCP Theorem, without increasing the number of queries to the proof. We formalize this as a composition of new objects that we call Locally Decode/Reject Codes (LDRC). The notion of LDRC was implicit in several previous works, and we make it explicit in this work. We believe that the formulation of LDRCs and their construction are of independent interest. ISSN 00045411 Age Range 18 to 22 years ♦ above 22 year Educational Use Research Education Level UG and PG Learning Resource Type Article Publisher Date 2008-06-01 Publisher Place New York e-ISSN 1557735X Journal Journal of the ACM (JACM) Volume Number 57 Issue Number 5 Page Count 29 Starting Page 1 Ending Page 29
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Source: ACM Digital Library
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## Algebra 1: Common Core (15th Edition)
$8.95 \times 10^{-12}$
Rearrange Factors: $= (1.78 \times 5.03)({10^{-7}}\times {10^{-5}})$ Simplify: $=8.9534 \times 10^{-12}$ Round to the appropriate number of significant digits: $\approx \boxed{8.95 \times 10^{-12}}$
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# What is the slope and y-intercept for the equation y=2x+6?
Mar 31, 2017
$y = 2 x + 6$ is the slope-intercept form of a linear equation, where $\frac{2}{1}$ is the slope and $6$ is the y-intercept.
#### Explanation:
Graph of $y = 2 x + 6$.
graph{y=2x+6 [-16.35, 15.67, -6.54, 9.48]}
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Permanent URL to this publication: http://dx.doi.org/10.5167/uzh-16770
# Straumann, N (2008). Energy extraction from black holes. In: Macias, A; Lämmerzahl, C; Camacho, A. Recent developments in gravitation and cosmology. 3rd Mexican Meeting on Mathematical and Experimental Physics. Melville NY, 75-93. ISBN 978-0-7354-0496-0 .
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## Abstract
In this lecture I give an introduction to the rotational energy extraction of black holes by the electromagnetic Blandford-Znajek process and the generation of relativistic jets. After some basic material on the electrodynamics of black hole magnetospheres, we derive the most important results of Blandford and Znajek by making use of Kerr-Schild coordinates, which are regular on the horizon. In a final part we briefly describe results of recent numerical simulations of accretion flows on rotating black holes, the resulting large-scale outflows, and the formation of collimated relativistic jets with high Lorentz factors
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Cantera 2.3.0
MultiPhase Class Reference
A class for multiphase mixtures. More...
#include <MultiPhase.h>
Collaboration diagram for MultiPhase:
[legend]
## Public Member Functions
MultiPhase ()
Constructor. More...
virtual ~MultiPhase ()
Destructor. More...
void addPhases (std::vector< ThermoPhase *> &phases, const vector_fp &phaseMoles)
Add a vector of phases to the mixture. More...
Add all phases present in 'mix' to this mixture. More...
void addPhase (ThermoPhase *p, doublereal moles)
Add a phase to the mixture. More...
size_t nElements () const
Number of elements. More...
void checkElementIndex (size_t m) const
Check that the specified element index is in range. More...
void checkElementArraySize (size_t mm) const
Check that an array size is at least nElements(). More...
std::string elementName (size_t m) const
Returns the name of the global element m. More...
size_t elementIndex (const std::string &name) const
Returns the index of the element with name name. More...
size_t nSpecies () const
Number of species, summed over all phases. More...
void checkSpeciesIndex (size_t k) const
Check that the specified species index is in range. More...
void checkSpeciesArraySize (size_t kk) const
Check that an array size is at least nSpecies(). More...
std::string speciesName (const size_t kGlob) const
Name of species with global index kGlob. More...
doublereal nAtoms (const size_t kGlob, const size_t mGlob) const
Returns the Number of atoms of global element mGlob in global species kGlob. More...
void getMoleFractions (doublereal *const x) const
Returns the global Species mole fractions. More...
void init ()
Process phases and build atomic composition array. More...
std::string phaseName (const size_t iph) const
Returns the name of the n'th phase. More...
int phaseIndex (const std::string &pName) const
Returns the index, given the phase name. More...
doublereal phaseMoles (const size_t n) const
Return the number of moles in phase n. More...
void setPhaseMoles (const size_t n, const doublereal moles)
Set the number of moles of phase with index n. More...
thermo_tphase (size_t n)
Return a reference to phase n. More...
void checkPhaseIndex (size_t m) const
Check that the specified phase index is in range Throws an exception if m is greater than nPhases() More...
void checkPhaseArraySize (size_t mm) const
Check that an array size is at least nPhases() Throws an exception if mm is less than nPhases(). More...
doublereal speciesMoles (size_t kGlob) const
Returns the moles of global species k. units = kmol. More...
size_t speciesIndex (size_t k, size_t p) const
Return the global index of the species belonging to phase number p with local index k within the phase. More...
size_t speciesIndex (const std::string &speciesName, const std::string &phaseName)
Return the global index of the species belonging to phase name phaseName with species name speciesName. More...
doublereal minTemp () const
Minimum temperature for which all solution phases have valid thermo data. More...
doublereal maxTemp () const
Maximum temperature for which all solution phases have valid thermo data. More...
doublereal charge () const
Total charge summed over all phases (Coulombs). More...
doublereal phaseCharge (size_t p) const
Charge (Coulombs) of phase with index p. More...
doublereal elementMoles (size_t m) const
Total moles of global element m, summed over all phases. More...
void getChemPotentials (doublereal *mu) const
Returns a vector of Chemical potentials. More...
void getValidChemPotentials (doublereal not_mu, doublereal *mu, bool standard=false) const
Returns a vector of Valid chemical potentials. More...
doublereal temperature () const
Temperature [K]. More...
void equilibrate (const std::string &XY, const std::string &solver="auto", double rtol=1e-9, int max_steps=50000, int max_iter=100, int estimate_equil=0, int log_level=0)
Equilibrate a MultiPhase object. More...
void setTemperature (const doublereal T)
Set the temperature [K]. More...
void setState_TP (const doublereal T, const doublereal Pres)
Set the state of the underlying ThermoPhase objects in one call. More...
void setState_TPMoles (const doublereal T, const doublereal Pres, const doublereal *Moles)
Set the state of the underlying ThermoPhase objects in one call. More...
doublereal pressure () const
Pressure [Pa]. More...
doublereal volume () const
The total mixture volume [m^3]. More...
void setPressure (doublereal P)
Set the pressure [Pa]. More...
doublereal enthalpy () const
The enthalpy of the mixture [J]. More...
doublereal IntEnergy () const
The internal energy of the mixture [J]. More...
doublereal entropy () const
The entropy of the mixture [J/K]. More...
doublereal gibbs () const
The Gibbs function of the mixture [J]. More...
doublereal cp () const
Heat capacity at constant pressure [J/K]. More...
size_t nPhases () const
Number of phases. More...
bool solutionSpecies (size_t kGlob) const
Return true is species kGlob is a species in a multicomponent solution phase. More...
size_t speciesPhaseIndex (const size_t kGlob) const
Returns the phase index of the Kth "global" species. More...
doublereal moleFraction (const size_t kGlob) const
Returns the mole fraction of global species k. More...
void setPhaseMoleFractions (const size_t n, const doublereal *const x)
Set the Mole fractions of the nth phase. More...
void setMolesByName (const compositionMap &xMap)
Set the number of moles of species in the mixture. More...
void setMolesByName (const std::string &x)
Set the moles via a string containing their names. More...
void getMoles (doublereal *molNum) const
Get the mole numbers of all species in the multiphase object. More...
void setMoles (const doublereal *n)
Sets all of the global species mole numbers. More...
Adds moles of a certain species to the mixture. More...
void getElemAbundances (doublereal *elemAbundances) const
Retrieves a vector of element abundances. More...
bool tempOK (size_t p) const
Return true if the phase p has valid thermo data for the current temperature. More...
Update the locally-stored composition within this object to match the current compositions of the phase objects. More...
void updatePhases () const
Set the states of the phase objects to the locally-stored state within this MultiPhase object. More...
## Private Member Functions
void calcElemAbundances () const
Calculate the element abundance vector. More...
double equilibrate_MultiPhaseEquil (int XY, doublereal err, int maxsteps, int maxiter, int loglevel)
Set the mixture to a state of chemical equilibrium using the MultiPhaseEquil solver. More...
## Private Attributes
vector_fp m_moles
Vector of the number of moles in each phase. More...
std::vector< ThermoPhase * > m_phase
Vector of the ThermoPhase pointers. More...
DenseMatrix m_atoms
Global Stoichiometric Coefficient array. More...
vector_fp m_moleFractions
Locally stored vector of mole fractions of all species comprising the MultiPhase object. More...
std::vector< size_t > m_spphase
Mapping between the global species number and the phase ID. More...
std::vector< size_t > m_spstart
Vector of ints containing of first species index in the global list of species for each phase. More...
std::vector< std::string > m_enames
String names of the global elements. More...
vector_int m_atomicNumber
Atomic number of each global element. More...
std::vector< std::string > m_snames
Vector of species names in the problem. More...
std::map< std::string, size_t > m_enamemap
Returns the global element index, given the element string name. More...
doublereal m_temp
Current value of the temperature (kelvin) More...
doublereal m_press
Current value of the pressure (Pa) More...
size_t m_nel
Number of distinct elements in all of the phases. More...
size_t m_nsp
Number of distinct species in all of the phases. More...
bool m_init
True if the init() routine has been called, and the MultiPhase frozen. More...
size_t m_eloc
Global ID of the element corresponding to the electronic charge. More...
std::vector< bool > m_temp_OK
Vector of bools indicating whether temperatures are ok for phases. More...
doublereal m_Tmin
Minimum temperature for which thermo parameterizations are valid. More...
doublereal m_Tmax
Minimum temperature for which thermo parameterizations are valid. More...
vector_fp m_elemAbundances
Vector of element abundances. More...
## Detailed Description
A class for multiphase mixtures.
The mixture can contain any number of phases of any type.
This object is the basic tool used by Cantera for use in Multiphase equilibrium calculations.
It is a container for a set of phases. Each phase has a given number of kmoles. Therefore, MultiPhase may be considered an "extrinsic" thermodynamic object, in contrast to the ThermoPhase object, which is an "intrinsic" thermodynamic object.
MultiPhase may be considered to be "upstream" of the ThermoPhase objects in the sense that setting a property within MultiPhase, such as temperature, pressure, or species mole number, affects the underlying ThermoPhase object, but not the other way around.
All phases have the same temperature and pressure, and a specified number of moles for each phase. The phases do not need to have the same elements. For example, a mixture might consist of a gaseous phase with elements (H, C, O, N), a solid carbon phase containing only element C, etc. A master element set will be constructed for the mixture that is the intersection of the elements of each phase.
Below, reference is made to global species and global elements. These refer to the collective species and elements encompassing all of the phases tracked by the object.
The global element list kept by this object is an intersection of the element lists of all the phases that comprise the MultiPhase.
The global species list kept by this object is a concatenated list of all of the species in all the phases that comprise the MultiPhase. The ordering of species is contiguous with respect to the phase id.
Definition at line 57 of file MultiPhase.h.
## ◆ MultiPhase()
MultiPhase ( )
Constructor.
The constructor takes no arguments, since phases are added using method addPhase().
Definition at line 21 of file MultiPhase.cpp.
## ◆ ~MultiPhase()
virtual ~MultiPhase ( )
inlinevirtual
Destructor.
Does nothing. Class MultiPhase does not take "ownership" (i.e. responsibility for destroying) the phase objects.
Definition at line 69 of file MultiPhase.h.
## Member Function Documentation
void addPhases ( std::vector< ThermoPhase *> & phases, const vector_fp & phaseMoles )
Add a vector of phases to the mixture.
See the single addPhases command. This just does a bunch of phases at one time
Parameters
phases Vector of pointers to phases phaseMoles Vector of mole numbers in each phase (kmol)
Definition at line 40 of file MultiPhase.cpp.
void addPhases ( MultiPhase & mix )
Add all phases present in 'mix' to this mixture.
Parameters
mix Add all of the phases in another MultiPhase object to the current object.
Definition at line 33 of file MultiPhase.cpp.
void addPhase ( ThermoPhase * p, doublereal moles )
Add a phase to the mixture.
This function must be called before the init() function is called, which serves to freeze the MultiPhase.
Parameters
p pointer to the phase object moles total number of moles of all species in this phase
Definition at line 49 of file MultiPhase.cpp.
## ◆ nElements()
size_t nElements ( ) const
inline
Number of elements.
Definition at line 98 of file MultiPhase.h.
References MultiPhase::m_nel.
## ◆ checkElementIndex()
void checkElementIndex ( size_t m ) const
Check that the specified element index is in range.
Throws an exception if m is greater than nElements()-1
Definition at line 716 of file MultiPhase.cpp.
References MultiPhase::m_nel.
## ◆ checkElementArraySize()
void checkElementArraySize ( size_t mm ) const
Check that an array size is at least nElements().
Throws an exception if mm is less than nElements(). Used before calls which take an array pointer.
Definition at line 723 of file MultiPhase.cpp.
References MultiPhase::m_nel.
## ◆ elementName()
std::string elementName ( size_t m ) const
Returns the name of the global element m.
Parameters
m index of the global element
Definition at line 730 of file MultiPhase.cpp.
References MultiPhase::m_enames.
Referenced by MultiPhaseEquil::MultiPhaseEquil().
## ◆ elementIndex()
size_t elementIndex ( const std::string & name ) const
Returns the index of the element with name name.
Parameters
name String name of the global element
Definition at line 735 of file MultiPhase.cpp.
References MultiPhase::m_enames, MultiPhase::m_nel, and Cantera::npos.
## ◆ nSpecies()
size_t nSpecies ( ) const
inline
Number of species, summed over all phases.
Definition at line 124 of file MultiPhase.h.
References MultiPhase::m_nsp.
## ◆ checkSpeciesIndex()
void checkSpeciesIndex ( size_t k ) const
Check that the specified species index is in range.
Throws an exception if k is greater than nSpecies()-1
Definition at line 745 of file MultiPhase.cpp.
References MultiPhase::m_nsp.
## ◆ checkSpeciesArraySize()
void checkSpeciesArraySize ( size_t kk ) const
Check that an array size is at least nSpecies().
Throws an exception if kk is less than nSpecies(). Used before calls which take an array pointer.
Definition at line 752 of file MultiPhase.cpp.
References MultiPhase::m_nsp.
## ◆ speciesName()
std::string speciesName ( const size_t kGlob ) const
Name of species with global index kGlob.
Parameters
kGlob global species index
Definition at line 759 of file MultiPhase.cpp.
References MultiPhase::m_snames.
## ◆ nAtoms()
doublereal nAtoms ( const size_t kGlob, const size_t mGlob ) const
Returns the Number of atoms of global element mGlob in global species kGlob.
Parameters
kGlob global species index mGlob global element index
Returns
the number of atoms.
Definition at line 764 of file MultiPhase.cpp.
References MultiPhase::m_atoms.
Referenced by MultiPhaseEquil::getComponents(), and MultiPhaseEquil::MultiPhaseEquil().
## ◆ getMoleFractions()
void getMoleFractions ( doublereal *const x ) const
Returns the global Species mole fractions.
Write the array of species mole fractions into array x. The mole fractions are normalized to sum to one in each phase.
Parameters
x vector of mole fractions. Length = number of global species.
Definition at line 769 of file MultiPhase.cpp.
References MultiPhase::m_moleFractions.
## ◆ init()
void init ( )
Process phases and build atomic composition array.
This method must be called after all phases are added, before doing anything else with the mixture. After init() has been called, no more phases may be added.
Definition at line 116 of file MultiPhase.cpp.
## ◆ phaseName()
std::string phaseName ( const size_t iph ) const
Returns the name of the n'th phase.
Parameters
iph phase Index
Definition at line 774 of file MultiPhase.cpp.
References Phase::id(), and MultiPhase::m_phase.
Referenced by MultiPhase::speciesIndex().
## ◆ phaseIndex()
int phaseIndex ( const std::string & pName ) const
Returns the index, given the phase name.
Parameters
pName Name of the phase
Returns
the index. A value of -1 means the phase isn't in the object.
Definition at line 780 of file MultiPhase.cpp.
References MultiPhase::m_phase, and MultiPhase::nPhases().
Referenced by MultiPhase::speciesIndex().
## ◆ phaseMoles()
doublereal phaseMoles ( const size_t n ) const
Return the number of moles in phase n.
Parameters
n Index of the phase.
Definition at line 790 of file MultiPhase.cpp.
References MultiPhase::m_moles.
## ◆ setPhaseMoles()
void setPhaseMoles ( const size_t n, const doublereal moles )
Set the number of moles of phase with index n.
Parameters
n Index of the phase moles Number of moles in the phase (kmol)
Definition at line 795 of file MultiPhase.cpp.
References MultiPhase::m_moles.
Referenced by vcs_MultiPhaseEquil::equilibrate_TP().
## ◆ phase()
ThermoPhase & phase ( size_t n )
Return a reference to phase n.
The state of phase n is also updated to match the state stored locally in the mixture object.
Parameters
n Phase Index
Returns
Reference to the ThermoPhase object for the phase
Definition at line 159 of file MultiPhase.cpp.
## ◆ checkPhaseIndex()
void checkPhaseIndex ( size_t m ) const
Check that the specified phase index is in range Throws an exception if m is greater than nPhases()
Definition at line 170 of file MultiPhase.cpp.
References MultiPhase::nPhases().
## ◆ checkPhaseArraySize()
void checkPhaseArraySize ( size_t mm ) const
Check that an array size is at least nPhases() Throws an exception if mm is less than nPhases().
Used before calls which take an array pointer.
Definition at line 177 of file MultiPhase.cpp.
References MultiPhase::nPhases().
## ◆ speciesMoles()
doublereal speciesMoles ( size_t kGlob ) const
Returns the moles of global species k. units = kmol.
Parameters
kGlob Global species index k
Definition at line 184 of file MultiPhase.cpp.
References MultiPhase::m_moleFractions, MultiPhase::m_moles, and MultiPhase::m_spphase.
## ◆ speciesIndex() [1/2]
size_t speciesIndex ( size_t k, size_t p ) const
inline
Return the global index of the species belonging to phase number p with local index k within the phase.
Parameters
k local index of the species within the phase p index of the phase
Definition at line 227 of file MultiPhase.h.
References MultiPhase::m_spstart.
## ◆ speciesIndex() [2/2]
size_t speciesIndex ( const std::string & speciesName, const std::string & phaseName )
Return the global index of the species belonging to phase name phaseName with species name speciesName.
Parameters
speciesName Species Name phaseName Phase Name
Returns
the global index
If the species or phase name is not recognized, this routine throws a CanteraError.
Definition at line 214 of file MultiPhase.cpp.
## ◆ minTemp()
doublereal minTemp ( ) const
inline
Minimum temperature for which all solution phases have valid thermo data.
Stoichiometric phases are not considered, since they may have thermo data only valid for conditions for which they are stable.
Definition at line 247 of file MultiPhase.h.
References MultiPhase::m_Tmin.
## ◆ maxTemp()
doublereal maxTemp ( ) const
inline
Maximum temperature for which all solution phases have valid thermo data.
Stoichiometric phases are not considered, since they may have thermo data only valid for conditions for which they are stable.
Definition at line 254 of file MultiPhase.h.
References MultiPhase::m_Tmax.
## ◆ charge()
doublereal charge ( ) const
Total charge summed over all phases (Coulombs).
Definition at line 205 of file MultiPhase.cpp.
References MultiPhase::nPhases(), and MultiPhase::phaseCharge().
## ◆ phaseCharge()
doublereal phaseCharge ( size_t p ) const
Charge (Coulombs) of phase with index p.
The net charge is computed as
$Q_p = N_p \sum_k F z_k X_k$
where the sum runs only over species in phase p.
Parameters
p index of the phase for which the charge is desired.
Definition at line 230 of file MultiPhase.cpp.
Referenced by MultiPhase::charge().
## ◆ elementMoles()
doublereal elementMoles ( size_t m ) const
Total moles of global element m, summed over all phases.
Parameters
m Index of the global element
Definition at line 190 of file MultiPhase.cpp.
Referenced by MultiPhaseEquil::MultiPhaseEquil().
## ◆ getChemPotentials()
void getChemPotentials ( doublereal * mu ) const
Returns a vector of Chemical potentials.
Write into array mu the chemical potentials of all species [J/kmol]. The chemical potentials are related to the activities by
$$\mu_k = \mu_k^0(T, P) + RT \ln a_k.$$.
Parameters
mu Chemical potential vector. Length = num global species. Units = J/kmol.
Definition at line 241 of file MultiPhase.cpp.
References MultiPhase::m_phase, MultiPhase::nPhases(), and MultiPhase::updatePhases().
## ◆ getValidChemPotentials()
void getValidChemPotentials ( doublereal not_mu, doublereal * mu, bool standard = false ) const
Returns a vector of Valid chemical potentials.
Write into array mu the chemical potentials of all species with thermo data valid for the current temperature [J/kmol]. For other species, set the chemical potential to the value not_mu. If standard is set to true, then the values returned are standard chemical potentials.
This method is designed for use in computing chemical equilibrium by Gibbs minimization. For solution phases (more than one species), this does the same thing as getChemPotentials. But for stoichiometric phases, this writes into array mu the user-specified value not_mu instead of the chemical potential if the temperature is outside the range for which the thermo data for the one species in the phase are valid. The need for this arises since many condensed phases have thermo data fit only for the temperature range for which they are stable. For example, in the NASA database, the fits for H2O(s) are only done up to 0 C, the fits for H2O(L) are only done from 0 C to 100 C, etc. Using the polynomial fits outside the range for which the fits were done can result in spurious chemical potentials, and can lead to condensed phases appearing when in fact they should be absent.
By setting not_mu to a large positive value, it is possible to force routines which seek to minimize the Gibbs free energy of the mixture to zero out any phases outside the temperature range for which their thermo data are valid.
Parameters
not_mu Value of the chemical potential to set species in phases, for which the thermo data is not valid mu Vector of chemical potentials. length = Global species, units = J kmol-1 standard If this method is called with standard set to true, then the composition-independent standard chemical potentials are returned instead of the composition- dependent chemical potentials.
Definition at line 251 of file MultiPhase.cpp.
## ◆ temperature()
doublereal temperature ( ) const
inline
Temperature [K].
Definition at line 329 of file MultiPhase.h.
References MultiPhase::m_temp.
## ◆ setTemperature()
void setTemperature ( const doublereal T )
Set the temperature [K].
Parameters
T value of the temperature (Kelvin)
Definition at line 707 of file MultiPhase.cpp.
## ◆ setState_TP()
void setState_TP ( const doublereal T, const doublereal Pres )
Set the state of the underlying ThermoPhase objects in one call.
Parameters
T Temperature of the system (kelvin) Pres pressure of the system (pascal)
Definition at line 424 of file MultiPhase.cpp.
## ◆ setState_TPMoles()
void setState_TPMoles ( const doublereal T, const doublereal Pres, const doublereal * Moles )
Set the state of the underlying ThermoPhase objects in one call.
Parameters
T Temperature of the system (kelvin) Pres pressure of the system (pascal) Moles Vector of mole numbers of all the species in all the phases (kmol)
Definition at line 434 of file MultiPhase.cpp.
## ◆ pressure()
doublereal pressure ( ) const
inline
Pressure [Pa].
Definition at line 389 of file MultiPhase.h.
References MultiPhase::m_press.
## ◆ volume()
doublereal volume ( ) const
The total mixture volume [m^3].
Returns the cumulative sum of the volumes of all the phases in the mixture.
Definition at line 472 of file MultiPhase.cpp.
References MultiPhase::m_moles, MultiPhase::m_phase, and MultiPhase::nPhases().
## ◆ setPressure()
void setPressure ( doublereal P )
inline
Set the pressure [Pa].
Parameters
P Set the pressure in the MultiPhase object (Pa)
Definition at line 404 of file MultiPhase.h.
Referenced by vcs_MultiPhaseEquil::equilibrate_TV().
## ◆ enthalpy()
doublereal enthalpy ( ) const
The enthalpy of the mixture [J].
Definition at line 292 of file MultiPhase.cpp.
## ◆ IntEnergy()
doublereal IntEnergy ( ) const
The internal energy of the mixture [J].
Definition at line 304 of file MultiPhase.cpp.
Referenced by vcs_MultiPhaseEquil::equilibrate(), and vcs_MultiPhaseEquil::equilibrate_HP().
## ◆ entropy()
doublereal entropy ( ) const
The entropy of the mixture [J/K].
Definition at line 316 of file MultiPhase.cpp.
Referenced by vcs_MultiPhaseEquil::equilibrate(), and vcs_MultiPhaseEquil::equilibrate_SP().
## ◆ gibbs()
doublereal gibbs ( ) const
The Gibbs function of the mixture [J].
Definition at line 280 of file MultiPhase.cpp.
## ◆ cp()
doublereal cp ( ) const
Heat capacity at constant pressure [J/K].
Note that this does not account for changes in composition of the mixture with temperature.
Definition at line 328 of file MultiPhase.cpp.
## ◆ nPhases()
size_t nPhases ( ) const
inline
## ◆ solutionSpecies()
bool solutionSpecies ( size_t kGlob ) const
Return true is species kGlob is a species in a multicomponent solution phase.
Parameters
kGlob index of the global species
Definition at line 271 of file MultiPhase.cpp.
References MultiPhase::m_phase, MultiPhase::m_spphase, and MultiPhase::nSpecies().
Referenced by MultiPhaseEquil::getComponents(), and MultiPhaseEquil::MultiPhaseEquil().
## ◆ speciesPhaseIndex()
size_t speciesPhaseIndex ( const size_t kGlob ) const
Returns the phase index of the Kth "global" species.
Parameters
kGlob Global species index.
Returns
the index of the owning phase.
Definition at line 800 of file MultiPhase.cpp.
References MultiPhase::m_spphase.
Referenced by MultiPhaseEquil::computeReactionSteps(), and MultiPhaseEquil::MultiPhaseEquil().
## ◆ moleFraction()
doublereal moleFraction ( const size_t kGlob ) const
Returns the mole fraction of global species k.
Parameters
kGlob Index of the global species.
Definition at line 805 of file MultiPhase.cpp.
References MultiPhase::m_moleFractions.
Referenced by Cantera::vcs_Cantera_update_vprob().
## ◆ setPhaseMoleFractions()
void setPhaseMoleFractions ( const size_t n, const doublereal *const x )
Set the Mole fractions of the nth phase.
This function sets the mole fractions of the nth phase. Note, the mole number of the phase stays constant
Parameters
n index of the phase x Vector of input mole fractions.
Definition at line 340 of file MultiPhase.cpp.
## ◆ setMolesByName() [1/2]
void setMolesByName ( const compositionMap & xMap )
Set the number of moles of species in the mixture.
Parameters
xMap CompositionMap of the species with nonzero mole numbers. Mole numbers that are less than or equal to zero will be set to zero. units = kmol.
Definition at line 353 of file MultiPhase.cpp.
Referenced by MultiPhase::setMolesByName().
## ◆ setMolesByName() [2/2]
void setMolesByName ( const std::string & x )
Set the moles via a string containing their names.
The string x is in the form of a composition map. Species which are not listed are set to zero.
Parameters
x string x in the form of a composition map where values are the moles of the species.
Definition at line 363 of file MultiPhase.cpp.
## ◆ getMoles()
void getMoles ( doublereal * molNum ) const
Get the mole numbers of all species in the multiphase object.
Parameters
[out] molNum Vector of doubles of length nSpecies containing the global mole numbers (kmol).
Definition at line 370 of file MultiPhase.cpp.
## ◆ setMoles()
void setMoles ( const doublereal * n )
Sets all of the global species mole numbers.
The state of each phase object is also updated to have the specified composition and the mixture temperature and pressure.
Parameters
n Vector of doubles of length nSpecies containing the global mole numbers (kmol).
Definition at line 385 of file MultiPhase.cpp.
Adds moles of a certain species to the mixture.
Parameters
indexS Index of the species in the MultiPhase object addedMoles Value of the moles that are added to the species.
Definition at line 415 of file MultiPhase.cpp.
References MultiPhase::getMoles(), MultiPhase::m_nsp, and MultiPhase::setMoles().
## ◆ getElemAbundances()
void getElemAbundances ( doublereal * elemAbundances ) const
Retrieves a vector of element abundances.
Parameters
elemAbundances Vector of element abundances Length = number of elements in the MultiPhase object. Index is the global element index. Units is in kmol.
Definition at line 442 of file MultiPhase.cpp.
## ◆ tempOK()
bool tempOK ( size_t p ) const
Return true if the phase p has valid thermo data for the current temperature.
Parameters
p Index of the phase.
Definition at line 810 of file MultiPhase.cpp.
References MultiPhase::m_temp_OK.
Referenced by MultiPhase::getValidChemPotentials(), and MultiPhaseEquil::MultiPhaseEquil().
Update the locally-stored composition within this object to match the current compositions of the phase objects.
Query the underlying ThermoPhase objects for their mole fractions and fill in the mole fraction vector of this current object. Adjust element compositions within this object to match.
This is an upload operation in the sense that we are taking downstream information (ThermoPhase object info) and applying it to an upstream object (MultiPhase object).
Definition at line 815 of file MultiPhase.cpp.
Referenced by vcs_MultiPhaseEquil::equilibrate_TP(), and MultiPhase::init().
## ◆ updatePhases()
void updatePhases ( ) const
Set the states of the phase objects to the locally-stored state within this MultiPhase object.
This method sets each phase to the mixture temperature and pressure, and sets the phase mole fractions based on the mixture mole numbers.
This is an download operation in the sense that we are taking upstream object information (MultiPhase object) and applying it to downstream objects (ThermoPhase object information)
Therefore, the term, "update", is appropriate for a downstream operation.
Definition at line 826 of file MultiPhase.cpp.
## ◆ calcElemAbundances()
void calcElemAbundances ( ) const
private
Calculate the element abundance vector.
Definition at line 450 of file MultiPhase.cpp.
## ◆ equilibrate_MultiPhaseEquil()
double equilibrate_MultiPhaseEquil ( int XY, doublereal err, int maxsteps, int maxiter, int loglevel )
private
Set the mixture to a state of chemical equilibrium using the MultiPhaseEquil solver.
Parameters
XY Integer flag specifying properties to hold fixed. err Error tolerance for $$\Delta \mu/RT$$ for all reactions. Also used as the relative error tolerance for the outer loop. maxsteps Maximum number of steps to take in solving the fixed TP problem. maxiter Maximum number of "outer" iterations for problems holding fixed something other than (T,P). loglevel Level of diagnostic output
Definition at line 482 of file MultiPhase.cpp.
Referenced by MultiPhase::equilibrate().
## ◆ m_moles
vector_fp m_moles
private
Vector of the number of moles in each phase.
Length = m_np, number of phases.
Definition at line 569 of file MultiPhase.h.
## ◆ m_phase
std::vector m_phase
private
## ◆ m_atoms
DenseMatrix m_atoms
private
Global Stoichiometric Coefficient array.
This is a two dimensional array m_atoms(m, k). The first index is the global element index. The second index, k, is the global species index. The value is the number of atoms of type m in species k.
Definition at line 580 of file MultiPhase.h.
## ◆ m_moleFractions
vector_fp m_moleFractions
private
Locally stored vector of mole fractions of all species comprising the MultiPhase object.
Definition at line 584 of file MultiPhase.h.
## ◆ m_spphase
std::vector m_spphase
private
Mapping between the global species number and the phase ID.
m_spphase[kGlobal] = iPhase Length = number of global species
Definition at line 591 of file MultiPhase.h.
## ◆ m_spstart
std::vector m_spstart
private
Vector of ints containing of first species index in the global list of species for each phase.
kfirst = m_spstart[ip], kfirst is the index of the first species in the ip'th phase.
Definition at line 599 of file MultiPhase.h.
## ◆ m_enames
std::vector m_enames
private
String names of the global elements.
This has a length equal to the number of global elements.
Definition at line 603 of file MultiPhase.h.
## ◆ m_atomicNumber
vector_int m_atomicNumber
private
Atomic number of each global element.
Definition at line 606 of file MultiPhase.h.
## ◆ m_snames
std::vector m_snames
private
Vector of species names in the problem.
Vector is over all species defined in the object, the global species index.
Definition at line 610 of file MultiPhase.h.
Referenced by MultiPhase::init(), MultiPhase::setMolesByName(), and MultiPhase::speciesName().
## ◆ m_enamemap
std::map m_enamemap
private
Returns the global element index, given the element string name.
-> used in the construction. However, wonder if it needs to be global.
Definition at line 616 of file MultiPhase.h.
## ◆ m_temp
doublereal m_temp
private
Current value of the temperature (kelvin)
Definition at line 619 of file MultiPhase.h.
## ◆ m_press
doublereal m_press
private
Current value of the pressure (Pa)
Definition at line 622 of file MultiPhase.h.
## ◆ m_nel
size_t m_nel
private
Number of distinct elements in all of the phases.
Definition at line 625 of file MultiPhase.h.
## ◆ m_nsp
size_t m_nsp
private
Number of distinct species in all of the phases.
Definition at line 628 of file MultiPhase.h.
## ◆ m_init
bool m_init
private
True if the init() routine has been called, and the MultiPhase frozen.
Definition at line 631 of file MultiPhase.h.
## ◆ m_eloc
size_t m_eloc
private
Global ID of the element corresponding to the electronic charge.
If there is none, then this is equal to -1
Definition at line 635 of file MultiPhase.h.
## ◆ m_temp_OK
std::vector m_temp_OK
mutableprivate
Vector of bools indicating whether temperatures are ok for phases.
If the current temperature is outside the range of valid temperatures for the phase thermodynamics, the phase flag is set to false.
Definition at line 642 of file MultiPhase.h.
Referenced by MultiPhase::addPhase(), MultiPhase::tempOK(), and MultiPhase::updatePhases().
## ◆ m_Tmin
doublereal m_Tmin
private
Minimum temperature for which thermo parameterizations are valid.
Stoichiometric phases are ignored in this determination. units Kelvin
Definition at line 646 of file MultiPhase.h.
## ◆ m_Tmax
doublereal m_Tmax
private
Minimum temperature for which thermo parameterizations are valid.
Stoichiometric phases are ignored in this determination. units Kelvin
Definition at line 650 of file MultiPhase.h.
## ◆ m_elemAbundances
vector_fp m_elemAbundances
mutableprivate
Vector of element abundances.
m_elemAbundances[mGlobal] = kmol of element mGlobal summed over all species in all phases.
Definition at line 657 of file MultiPhase.h.
The documentation for this class was generated from the following files:
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If every normal section of a surface is a geodesic, then for every point of the surface, the curvature is the same in any direction. [closed]
How should I show that if every normal section of a surface in $$\mathbb{R}^3$$ is a geodesic, then for every point of the surface, the curvature is the same in any direction?
I would like some hints. Thank you!
Sketch: Assume $$p=(0,0,0)$$ is a point on the surface $$S$$ and the unit normal vector to $$S$$ at $$p$$ is $$(0,0,1)$$. Take a ball of radius $$r>0$$ small enough on the tangent plane at $$S$$ (the $$X,Y$$-plane). Note that if you go $$r$$ distance from $$p$$, along $$S$$, in any direction, to say $$x$$, the normal vector at $$x$$ is in the plane passing through $$Z$$-axis and $$x$$ (uniqueness of geodesics given a direction and at a point).
Now consider the curve on $$S$$ which is the set of all points at Riemannian distance $$r$$ from $$p$$. Consider the function which assigns to each point on this curve its euclidean distance from the $$X,Y$$-plane. Note that the vectors tangent to this curve are perpendicular to planes containing the base point and the $$Z$$-axis by the observation above. This means that the function considered above is a constant. This is true for any $$r$$. Check that this gives local symmetry (rotations in the $$X,Y$$-plane send geodesics from $$p$$ to geodesics at $$p$$). The claim follows from this.
• When considering the curve which is the set of all points at distance $r$ from $p$ (second paragraph), do you mean Euclidean distance? or intrinsic distance? Nov 30, 2020 at 6:33
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# What are the oxidation numbers for CuSO_4?
For $\text{cupric sulfate}$? Well, there is a $C u \left(I I\right)$ ion.
And for the $\text{sulfate counterion}$, sulfur takes a $V I +$ oxidation state. And, as usual, the $O$ atoms takes a $- I I$ oxidation state. Again, as normal, the weighted sum of the oxidation numbers for a neutral species is $\text{ZERO} : 0 = 4 \times \left(2 -\right) + 2 + 6$.
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#### Scaling of Congestion in Small World Networks
##### Iraj Saniee, Gabriel H. Tucci
In this report we show that in a planar exponentially growing network consisting of $N$ nodes, congestion scales as $O(N^2/\log(N))$ independently of how flows may be routed. This is in contrast to the $O(N^{3/2})$ scaling of congestion in a flat polynomially growing network. We also show that without the planarity condition, congestion in a small world network could scale as low as $O(N^{1+\epsilon})$, for arbitrarily small $\epsilon$. These extreme results demonstrate that the small world property by itself cannot provide guidance on the level of congestion in a network and other characteristics are needed for better resolution. Finally, we investigate scaling of congestion under the geodesic flow, that is, when flows are routed on shortest paths based on a link metric. Here we prove that if the link weights are scaled by arbitrarily small or large multipliers then considerable changes in congestion may occur. However, if we constrain the link-weight multipliers to be bounded away from both zero and infinity, then variations in congestion due to such remetrization are negligible.
arrow_drop_up
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sdr_aux_module.f90 Source File
Source Code
! Copyright (c) 2011 Manuel Hasert <[email protected]>
! Copyright (c) 2011-2014 Kannan Masilamani <[email protected]>
! Copyright (c) 2011-2012, 2014 Harald Klimach <[email protected]>
! Copyright (c) 2012 Simon Zimny <[email protected]>
! Copyright (c) 2016 Tobias Girresser <[email protected]>
!
! Redistribution and use in source and binary forms, with or without
! modification, are permitted provided that the following conditions are met:
!
! 1. Redistributions of source code must retain the above copyright notice, this
! list of conditions and the following disclaimer.
!
! 2. Redistributions in binary form must reproduce the above copyright notice,
! this list of conditions and the following disclaimer in the documentation
! and/or other materials provided with the distribution.
!
! THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS "AS IS"
! AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
! IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE ARE
! DISCLAIMED. IN NO EVENT SHALL THE COPYRIGHT HOLDER OR CONTRIBUTORS BE LIABLE
! FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL
! DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS OR
! SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION) HOWEVER
! CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT LIABILITY,
! OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY OUT OF THE USE
! OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF SUCH DAMAGE.
! ******************************************************************************!
!> Some auxilary functionalities.
module sdr_aux_module
! treelm modules
use tem_aux_module, only: tem_print_execInfo
use tem_logging_module, only: logunit
implicit none
private
public :: sdr_init_global
contains
! ****************************************************************************!
!> Prominently let the user now, what he actually is running right now.
!!
!! Also set the solvername and version number in the solveHead.
!> contains solver header information
write(logunit(1),*) " "
write(logunit(1),*) " _ "
write(logunit(1),*) " ___ ___ ___ __| | ___ _ __ "
write(logunit(1),*) " / __|/ _ \/ _ \/ _ |/ _ \ '__| "
write(logunit(1),*) " \__ \ __/ __/ (_| | __/ | "
write(logunit(1),*) " |___/\___|\___|\__,_|\___|_" &
write(logunit(1),*) " "
write(logunit(1),*) &
& " (C) 2012 German Research School for Simulation Sciences"
write(logunit(1),*) " (C) 2013 University of Siegen "
write(logunit(1),*) " "
! Write the information about the executable, gathered at build time to
! the screen.
call tem_print_execInfo()
write(logunit(1),*) " "
write(logunit(1),*) " "
end subroutine sdr_init_global
! ****************************************************************************!
end module sdr_aux_module
! ******************************************************************************!
`
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# Some basic problems of group theory
1.Prove the every group G of order 4 is isomorphic to either Z4 or 4-group V,that is {1 (1,2)(3,4) (1,3)(2,4) (1,4)(2,3)}
2.If G is a group with $|G|\leq 5$ then G is abelian.
I have learned independently the chapter 1 of the book An introduction to group theory by Joseph Rotman . And can understand it ,can solve many exercises in it.I can understand the proofs in chapter 2 , however I totally don't have any intuition of the exercises. Does any suggestion?
-
What are the possible orders of the elements of the group? Try to fill the multiplication table. – Phira Apr 23 '11 at 14:20
Chandru1 Answered 2 pretty adequately. I'll focus on part 1 of your question.
Take $x\in G$ not the identity. By Lagrange's theorem the order of this element is either $2$ or $4$. If it is order four, you can map it to the generator $1$ of $\mathbb{Z}_4$. It is not hard to establish that this generates an isomorphism.
If $x$ has order $2$, you can look to the other elements. If none of these elements are of order $4$, you have that every square in this group is the identity. It is not hard to establish the required isomorphism to $V$ in this case.
-
Note:
• If $|G|=4$, then $G$ is either isomorphic $\mathbb{Z}_{4}$ or $\mathbb{Z}_{2} \times \mathbb{Z}_{2}$. Infact, if $|G|=p^{2}$, then either $G$ is isomorphic to $\mathbb{Z}_{p^2}$ or $\mathbb{Z}_{p} \times \mathbb{Z}_{p}$. For a proof of this you can see Dummit and Foote's Abstract Algebra book. ( I don't know where it's exactly given.)
• If $|G|=1$, then it's abelian, if $|G|=2$, it's cyclic, since $2$ is prime so it's abelian, if $|G|=3$, it's cyclic so abelian, if $|G|=4$, by above it's abelian, if $|G|=5$, it's cyclic so abelian.
-
Here is an alternate proof for (2), which is also a hint for 1:
We first prove the following:
(*) If any element in $G$ has order $2$, then $G$ is abelian.
Proof for (*): Let $x,y \in G$. Then
$$xxyy=ee=e=(xy)^2=xyxy \,.$$
Multiplying to the left by $x^{-1}$ and to the right by $y^{-1}$ yields
$$xy=yx \,.$$
Now (2) follows imediatelly from (*): By Lagrange Theorem any group of prime order is cyclic, thus abelian. For the case $|G| =4$, either there is an element of order $4$, or every element has order $2$; in both cases we get $G$ is abelian.
-
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# Numerical Problems , Work , Power , Energy
#### LEVEL – I
Q:1. An object of mass 5 kg falls from rest through a vertical distance of 20 m and attains a velocity of 10 m/s. How much work is done by the resistance of the air on the object? (g = 10m/s2).
Ans: -750 J
Q:2. A train of mass 100 metric tons is drawn up an incline of 1 in 49 at the rate of 36 km per hour by an engine. If the resistance due to friction be 10 N per metric ton, calculate the power of the engine. If the steam is shut off, how far will the train move before it comes to rest?
Ans: 238 m
Q:3. Shown in the figure is a smooth vertical frame of wire along which a small bead moves from the point A. Find its speed at the point B.
Ans : $\displaystyle v = \sqrt{2 g(H-h)}$
Q:4. A rubber ball after falling through a height h penetrates into the water through a distance x. Find the average force imparted by water on the rubber ball in ideal conditions.
Ans : $\displaystyle F = – mg (\frac{h}{x} + 1)$
Q:5. A bus of mass 1000 kg has an engine which produces a constant power of 50 kW. If the resistance to motion, assumed constant is 1000 N, find the maximum speed at which the bus can travel on level road and the acceleration when it’s travelling at 25 m/s.
Ans : v = 50 m/s ; a = 2 m/s2
Q:6. A particle of mass m is projected up the smooth inclined plane of inclination θ with a speed v0( > √(2glSinθ) ) as shown in the figure. Find the maximum height travelled by the particle.
Ans : $\displaystyle H = \frac{(v_0^2 – 2 g l sin\theta)sin^2 \theta}{2 g} + l sin\theta$
Q:7. A block of mass m collides with a horizontal weightless spring of force constant k. The block compresses the spring by x. Calculate the maximum momentum of the block.
Ans : $\displaystyle (\sqrt{k m} ) x$
Q:8. Prove that K. E. of two identical trains with respect to heliocentric frame of reference moving in opposite directions on equatorial line with same speed (w.r.t earth) are not equal.
Q:9. A small mass m starts from rest and slides down the smooth spherical surface of R. Assume zero potential energy at the top.
Find (a) the change in potential energy
(b) the kinetic energy
(c) the speed of the mass as a function of the angle made by the radius through the mass with the vertical.
Ans : (a) $\displaystyle -m g R(1-cos\theta)$
(b)$\displaystyle m g R(1-cos\theta)$
(c)$\displaystyle \sqrt{2 g R(1-cos\theta)}$
Q:10. An ideal massless spring can be compressed by 1 m by a force of 100 N. This same spring is placed at the bottom of a frictionless inclined plane which makes an angle θ = 30° with the horizontal. A 10 kg mass is released from rest at the top of the incline and is brought to rest momentarily after compressing the spring 2 meters.
(a) Through what distance does the mass slide before coming to rest?
(b) What is the speed of the mass just before it reaches the spring?
Ans: (a) s = 4 m (b) v = 2√5 m/s
Next Page → (Level-II)
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# Conductance and Conductivity – Definition, Unit, Formula and Examples
In this article, we shall discuss electrical conductance and conductivity, its definition, formula, unit of measurement, and the solved numerical examples.
## What is Electrical Conductance?
The measure of how easily the electric charge (or electrons or electric current) can flow through a material is called the conductance or electrical conductance of that material.
Therefore, electrical conductance defines the capability of a substance to conduct electricity. In order to understand the concept of electrical conductance, we have to recall the electrical resistance of materials.
Electrical resistance is defined as the measure of difficulty that a material offers in the flow of electric current or electric charge. The resistance is mainly the result of collisions of electric charges with ions of the atom within the material.
Thus, a material in which the number of collisions of electric charges with the ions of atoms is less, the conductance of the material will be more and vice-versa, and therefore, this material conducts the electric current more easily. Hence, this discussion shows that electrical conductance is the property of a material opposite to electrical resistance.
## Formula and Unit of Conductance
In electrical and electronic circuits, electrical conductance is defined as the reciprocal of resistance of a conductor. It is represented by the symbol ‘G’.
$$\mathrm{\therefore Conductance,\, G\, =\, \frac{1}{Resistance\left ( R \right )}\: \: \: \cdot \cdot \cdot \left ( 1 \right )}$$
According to Ohm’s law, the resistance of a conductor is given by,
$$\mathrm{Resistance,\, R\, =\, \frac{V}{I}}$$
Where, V is the voltage across and I is the current through the conductor.
$$\mathrm{\therefore G\, =\, \frac{I}{V}\: \: \cdot \cdot \cdot \left ( 2 \right )}$$
Also, the resistance of a conductor can be expressed in terms of the physical dimensions of the conductor as,
$$\mathrm{R\, =\, \frac{\rho l}{a}}$$
Therefore, the conductance of a conductor in terms of its physical dimensions will be,
$$\mathrm{G\, =\, \frac{a}{\rho l}\, =\, \sigma \frac{a}{ l}\: \: \cdot \cdot \cdot \left ( 3 \right )}$$
Where, σ is called conductivity or specific conductance of the conductor material.
$$\mathrm{\therefore \sigma \, =\, \frac{1}{\rho}}$$
### Unit of Conductance
As the conductance of a conductor is given by,
$$\mathrm{G=\frac{1}{R}=\frac{I}{V}=\frac{\sigma a}{l}}$$
$$\mathrm{\therefore Unit\: of\: G=\frac{1}{Ohm\, \left ( \Omega \right )}=\frac{Ampere}{Volt}=Mho}$$
Thus, the unit of conductance is Mho (Ω-1). But, in practice, we commonly use Siemen (S) as the unit of conductance. Where,
$$\mathrm{1\: Mho (\Omega ) = 1\: \: Siemen}$$
## Factors Affecting the Conductance
From equation (3), we have,
$$\mathrm{G=\frac{\sigma a}{l}}$$
Thus, the conductance of a conductor is −
• Directly proportional to the cross-sectional area of the conductor.
• Inversely proportional to the length of the conductor.
• Dependent upon the nature of the material (σ).
• Varied with the temperature.
## What is Conductivity?
The ability of a substance to provide ease in the flow of electric current or charge through it is called conductivity. It is denoted by the Greek letter sigma (σ).
Conductivity is one of the most important properties of substances that is used to select suitable material for making electrical conductors and insulators. Where, if a material has high conductivity, then it is best suited for making conducting wires, and if a material has low conductivity, then it is suitable for making insulators.
Basically, electrical conductivity explains the behavior of material for current flow on the application of an electric field.
Mathematically, the conductivity of a material is expressed as the reciprocal of its resistivity, i.e.,
$$\mathrm{Conductivity,\sigma =\frac{1}{Resistivity\left ( \rho \right )}\cdot \cdot \cdot \left ( 4 \right )}$$
The conductivity is also known as specific conductance. The conductivity is measured in Siemens per meter (S/m) or mho per meter (Ω-1m-1).
## Conductance of Series Circuits
Consider a series resistive circuit as shown in Figure1. The circuit consists of three resistors connected in series.
From the circuit, we have,
$$\mathrm{R_{t}=R_{1}\, +\,R_{2}\, +\,R_{3} }$$
$$\mathrm{ \because Conductance, G=\frac{1}{R}}$$
Therefore,
$$\mathrm{ \frac{1}{G_{t}}=\frac{1}{R_{1}}\, +\,\frac{1}{R_{2}}\, +\,\frac{1}{R_{3}}\, \, \cdot \cdot \cdot \left ( 5 \right ) }$$
Also,
$$\mathrm{G_{t}=\frac{1}{R_{t}}=\frac{1}{R_{1}\, +\,R_{2}\, +\,R_{3}} \, \, \cdot \cdot \cdot \left ( 6 \right )}$$
Hence, when a number of resistors are connected in series, then the reciprocal of total conductance is equal to the sum of the reciprocal of conductance of the individual resistors.
## Conductance of Parallel Circuits
Consider a parallel circuit as shown in Figure2. It consists of three resistors connected in parallel.
From the circuit, we can write,
$$\mathrm{ \frac{1}{R_{t}}=\frac{1}{R_{1}}\, +\,\frac{1}{R_{2}}\, +\,\frac{1}{R_{3}} }$$
From the definition of conductance, we have,
$$\mathrm{ G=\frac{1}{R}}$$
Therefore,
$$\mathrm{ G_{t}=G_{1}\, +\,G_{2}\, +\,G_{3}\: \: \cdot \cdot \cdot \left ( 7 \right ) }$$
Hence, when a number of resistors are connected in parallel, then the total conductance of the circuit is equal to the sum of the conductance of the individual resistors.
## Conductance vs. Conductivity
The conductance of a conductor is the measure of how easily the material of the conductor allows the flow of current through it. Whereas, conductivity is the property of a material by virtue of which the material offers ease in the flow of current.
## Numerical Example
Calculate the conductance of a wire of 915 m having a uniform cross-section of 0.88 cm2. The wire is made up of copper having a conductivity of 5.9 × 105 S/m.
## Solution
Given data,
$$\mathrm{length, l = 915 m }$$
$$\mathrm{Area\: of\: cross\: section, a = 0.88\: \: cm^{2} = 0.88 \times 10^{-4}\: m^{2} }$$
$$\mathrm{Conductivity, \sigma = 5.9\times 10^{5}\: Sm^{-1} }$$
Therefore, the conductance of the wire will be,
$$\mathrm{G=\frac{\sigma a}{l}=\frac{\left ( 5.9\times 10^{5} \right )\times \left ( 0.88\times 10^{-4} \right )}{\left ( 915 \right )}}$$
$$\mathrm{\therefore G=0.0867\: S}$$
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# Precalculus Examples
Find the value at .
Replace the variable with in the expression.
Simplify the result.
Simplify each term.
Anything raised to is .
Divide by .
Add and .
The final answer is .
The value at is .
Find the value at .
Replace the variable with in the expression.
Simplify the result.
Move to the numerator using the negative exponent rule .
Add and .
The final answer is .
The value at is .
Find the value at .
Replace the variable with in the expression.
Simplify the result.
Evaluate the exponent.
To write as a fraction with a common denominator, multiply by .
Write each expression with a common denominator of , by multiplying each by an appropriate factor of .
Combine.
Multiply by .
Combine the numerators over the common denominator.
Simplify the numerator.
Multiply by .
Add and .
The final answer is .
The value at is .
Find the value at .
Replace the variable with in the expression.
Simplify the result.
Raise to the power of .
To write as a fraction with a common denominator, multiply by .
Write each expression with a common denominator of , by multiplying each by an appropriate factor of .
Combine.
Multiply by .
Combine the numerators over the common denominator.
Simplify the numerator.
Multiply by .
Add and .
The final answer is .
The value at is .
List the points to graph.
Use the found points and asymptotes to graph .
Enter YOUR Problem
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# Bought Matthew Morris’ OCP 12c upgrade book for Key DBA Skills section
I haven’t even started on Sam Alapati’s OCP book but I realized today that Matthew Morris’s OCP 12c upgrade book has an outline of the Key DBA Skills section of the OCP exam and that is what I need most. I was two questions away on that section when I took the test the first time and I need to make sure I’m studying the right things. After I got back to the office after failing my first attempt (a $245 lesson) I made notes about what I remember being unsure of and that gave me some ideas what to study. But, it is nice to get some advice from someone else who has passed the exam. I saved a couple of bucks by getting the Kindle version for$9.99 so it seemed like a safe bet. Hopefully I can hit it hard on both books and my practice test software, which also covers the key DBA skills, and be able to take the test again by the end of the August.
I live in Chandler, Arizona with my wife and three daughters. I work for US Foods, the second largest food distribution company in the United States. I've been working as an Oracle database administrator and PeopleSoft administrator since 1994. I'm very interested in Oracle performance tuning.
This entry was posted in Uncategorized. Bookmark the permalink.
### 14 Responses to Bought Matthew Morris’ OCP 12c upgrade book for Key DBA Skills section
1. Alexandre Focante says:
Hey Bobby, your blog is very nice, simple and clear, and many times I get myself with the same thinkings of your posts.
I would like to see here you commenting a big about the Sam Alapati’s OCP book, if its really worthy for the 12c OCP upgrade test. Also, let us know about this book you just bought, from Matthew Morris.
• Bobby says:
Alexandre,
Thanks for your comment. I’ll definitely let people know how it goes. I feel good about my purchases. $45 for Sam Alapati’s book,$10 for Matthew Morris’ book and my employer bought me the testprep software which cost $100. Not bad when you consider that a week long class probably costs$3000.
I’m excited about Matthew’s book because it gives me some clarity on what to study on the second part of the test. I was starting to flounder because there is so much I could study and I would never get done. Now I’m planning to focus just on my three resources – the two books and the software – and let them drive any additional study of the manuals or testing of the software that is needed to clarify what the resources say.
– Bobby
2. praveen says:
Hi, during my Oracle 11g upgrade OCP, I used Sam’s book, I find lot of mistakes and incorrect information.
• Bobby says:
Praveen,
I had the same experience with his 11g OCP book, but then I passed the test so that’s the only thing that really matters. I used Sam Alapati’s 11g OCP book and the Kaplan Selftest software and I was able to pass the 11g OCP test on my first attempt. I figure that when I get to the point that I can find the errors in the book and the software then I know I’m ready. It worked on 11g. Time will tell with 12c. It’s a tough test.
– Bobby
3. fredjones95 says:
Odd, the “Key DBA Skills” section turned out to be the easiest but that was because I thoroughly studied and passed 1z0-052 and 1z0-053 late last year. With 11g under my belt, the upgrade section was all I had to tackle and I passed 1z0-060 at ease. Now all I need is an employer willing to accept my knowledge despite my only having a few months of hands on experience.
• Bobby says:
Thank you for your comment and I wish you well in your job search. Someone else said that they studied their 11g certification preparation materials and that helped them pass the Key DBA Skills section so that may be a good approach. I’m hoping to take another stab at the test by the end of next week.
It may not be of much use to you depending on where you live, but we do have an open DBA position. Also, I think that Pythian is hiring so you might try them. Companies need Oracle DBAs but the job search process can get in the way of connecting people to them.
– Bobby
p.s. Here is my post on our DBA position:
http://www.bobbydurrettdba.com/2014/05/30/senior-oracle-dba-position-in-tempe-arizona/
4. Hi Bobby,
I am too much worried regarding the “Key DBA Skills”. I am going to write exam on 13th Feb , please suggest me the topics what I should go through for the second section of exam.
• Bobby says:
I recommend the different books and programs that I listed on this blog post: http://www.bobbydurrettdba.com/2014/08/05/passed-ocp-12c-test/ Those are the ones I used. If you don’t pass the test on Feb 13 then after you get out of the test make notes for yourself on the things you need to study for the next attempt.
• Thanks Bobby for your quick response. I have to pass this exam otherwise I will loose my OCP 10G oracle certification validation. Anyways thanks again , I will try my level best.
• Bobby says:
I hope you do well on the test, but it may not be a big deal if you don’t pass it the first time. I did a quick check on the OCP expiration dates. I recommend that you check it but it seems like it is not that big of a deal if your 10G certification expires. If I am reading the web site correctly you can take your test Feb 13 and if you don’t pass your 10g OCP just becomes inactive March 1. But then you can retake the 12c OCP upgrade exam a few months later and then your 10g will be active again and your 12c. As long as they don’t make you pay for an expensive course or take a bunch of tests instead of the upgrade exam I don’t see it really mattering if your 10g certification goes inactive.
• Thanks Bobby, Please correct me If I am wrong. As I am getting you, you mean to say If I won’t pass this exam I will be able to take this after few months. I have one question that ” If I won’t update my certification before 1 March 2016, will Oracle allow me to write this exam as an OCP 10G certified Guy to upgrade my certification.”
What I thought they will not allow me to write this exam because after expiration of 10G certification I will not be OCP certified Guy any more so I will not be eligible for the 1Z0-060 exam. Please correct me if am wrong.
• Bobby says:
I don’t know for sure so don’t make any decisions based on my ideas here. You have to check for yourself. Look at this web page: https://education.oracle.com/pls/web_prod-plq-dad/db_pages.getpage?page_id=770 Here is the FAQ: “What happens if I allow my credential to become inactive? Can I follow an upgrade path and be reactivated, or do I need to start over?” Here is Oracle’s answer: “If you allow your credential to become inactive, simply upgrade to a current credential.” So, that sounds like you can still take the 1Z0-060 exam. But nothing is certain with Oracle so you have to check on this yourself and don’t blame me if their web site is misleading or wrong!
• Prabhakar Kumar says:
Hi Bobby,
Today I wrote the OCP 12c (1Z0-060) exam and passed the exam with 93%. Your guidance helped me a lot.
Thank you so much.
• Thanks a lot Bobby.
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# Given Below Are Two Statements, One Labelled As Assertion (A) and the Other Labelled As Reason (R):- Bzziii.com
Given below are two statements, one labelled as Assertion (A) and the other labelled as Reason (R): Assertion (A): A high operating ratio indicates
bzziii
Given below are two statements, one labelled as Assertion (A) and the other labelled as Reason (R):
Assertion (A): A high operating ratio indicates a favourable position.
Reasoning (R): A high operating ratio leaves a high margin to meet non-operating expenses. In the context of the above two statements, which of the following is correct?
Code:
(A) (A) and (R) both are correct and (R) correctly explains (A).
(B) Both (A) and (R) are correct but (R) does not explain (A).
(C) Both (A) and (R) are incorrect.
(D) (A) is correct but (R) is incorrect.
(C) Both (A) and (R) are incorrect.
Getting Info...
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# Maxwell equations and wave equation in a medium
#### ModusPwnd
1. Homework Statement
Consider an isotropic medium with constant conductivity $\sigma$. There is no free charge present, that is, $\rho = 0$.
a)What are the appropriate Maxwell equations for this medium?
b)Derive the damped wave equation for the electric field in the medium. Assume Ohm's law is of the form $\vec{J}=\sigma\vec{E}$.
2. Homework Equations
Maxwell equations and the curl
3. The Attempt at a Solution
a)
Maxwell equaitons with $\rho_f=0$ and $\vec{J}=\frac{\vec{E}}{\rho}$.
$$\nabla \cdot \vec{D} = 0$$
$$\nabla \times \vec{E} = - \frac{\partial \vec{B}}{\partial t}$$
$$\nabla \cdot \vec{B} = 0$$
$$\nabla \times \vec{H} = \sigma \vec{E} + \epsilon_0 \mu_0 \frac{\partial \vec{D}}{\partial t}$$
Its simply a matter of putting a $\sigma \vec{E}$ in place of the displacement current $\vec{J}$ right? hmmm...
b)
Here I am a little confused. I take the curl of the curl of $\vec{E}$,
$$\nabla \times (\nabla \times \vec{E}) = \nabla(\nabla \cdot \vec{E}) - \nabla^2 \vec{E} = -\nabla^2 \vec{E} = \nabla \times (-\frac{\partial \vec{B}}{\partial t}) = -\frac{\partial}{\partial t} (\nabla \times \vec{B})$$
Now here Im not sure if I am correct in assuming that $\nabla \cdot \vec{E} = 0$ and I'm not sure what $\nabla \times \vec{B}$ in this case, since its not in fee space...
Any ideas?
Related Advanced Physics Homework Help News on Phys.org
#### rude man
Homework Helper
Gold Member
First thing you do is assume sinusoids. It's pretty near impossible otherwise. So start with the equations for E and H assuming a sinusoidal plane wave. Use the exponetial form E = E0exp(jwt) and H = H0exp(jwt) if you're an engineer or substitute i for j if you're a physicist. :-)
Wind up eliminating H, and get a partial differential equation for E. Solve it.
#### ModusPwnd
Im not trying to solve the wave equation, I am trying to derive it.
#### ehild
Homework Helper
The fourth equation is correctly $\nabla \times \vec{H} = \sigma \vec{E} +\frac{\partial \vec{D}}{\partial t}$
and use also the "material equations" $\vec{D}=\epsilon \vec{E}$, $\vec{B}=\mu\vec{H}$
ehild
#### ModusPwnd
Can I get $\mu$ and $\epsilon$ from the conductivity I am given?
#### ehild
Homework Helper
No, they are also characteristics of the medium.
ehild
#### ModusPwnd
bleh, so the question does not provide enough for an answer? My profs. really suck at writing questions, this is not the first time this has happened...
#### ehild
Homework Helper
You have the appropriate Maxwell equations, and can write the damped wave equation replacing B=μH and D=εE. ε and μ are constants.
ehild
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# Question #4822a
Jun 18, 2015
Both $F e C {l}_{2}$ and $N {a}_{2} S {O}_{3}$ can act as reducing agents
#### Explanation:
Oxidation is loss of electrons. Reduction is gain of electrons.
An oxidising agent takes in electrons. A reducing agent gives them out.
Both ozone and chlorine are powerful oxidising agents.
Iron(II) choride can lose electrons to form iron(III):
$F {e}^{2 +} \rightarrow F {e}^{3 +} + e$
A solution of iron(II) chloride will slowly oxidise to iron(III) in the presence of air.
Sodium sulfite is also a reducing agent:
$S {O}_{3}^{2 -} + {H}_{2} O \rightarrow S {O}_{4}^{2 -} + 2 {H}^{+} + 2 e$
It is used as an anti - oxidant in food preservation.
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# Most “beautiful” presentations of the basic proofs for vector spaces?
I am familiar with the standard proofs presented in textbooks for stuff like linear independence/dependence, the dimensions of common vector spaces, any basis for a vector space V must be linearly independent and have at least n = dim V vectors, etc.
However, I am curious to know this: are there books that present these proofs in (the most?) an elegant way? By elegant here, I am alluding to some intangible sense of: "beautifully simple", "a proof that presents a new way of looking at things", "using non-standard methods to form a particularly straightforward argument", etc.
Perhaps these proofs have some quality akin to 'breathtaking' to students familiar only with the standard presentation, or perhaps they convincingly demonstrate the power of particular branch of mathematics?
In your answer, could you share a little as to why you consider the presentations you are advocating elegant?
• I would say that the subject matter you refer to as a whole is a marvel of efficiency and beauty: the basic material you describe can be done in about five pages. In my opinion the one result which makes it all beautiful and easy is the Steinitz Exchange Lemma. To me the basic dichotomy of exposition of this material is whether SEL is soft-pedaled/avoided (e.g. by row reduction arguments) or explicitly embraced, and when the latter is done you get something beautifully simple. It is also possible to treat this material in a more abstract context, leading to the notion of matroids. – Pete L. Clark Jan 28 '14 at 5:43
• On the other hand, I am not sure that I have seen any two presentations of this material that I would regard as "essentially different". – Pete L. Clark Jan 28 '14 at 5:45
• I think Lax's linear algebra book is mathematically beautiful. – littleO Jan 28 '14 at 5:48
• @PeteL.Clark Could you recommend a resource that presents the material using the Steinitz Exchange Lemma result? – user89 Jan 28 '14 at 5:52
• @twirlobite: I think that this is done in most of the more "theoretical" approaches to linear algebra. I don't really have a go-to linear algebra text, I'm afraid. But e.g. these notes give the approach that I have in mind:web.math.princeton.edu/~mdamron/teaching/S13/notes/…. – Pete L. Clark Jan 28 '14 at 6:41
I loved the part where he proved that an n-dimensional vector space is isomorphic to $F^n$ where $F$ is the base field, and then proceeds to explain why we still need to study finite dimensional vector spaces abstractly. He introduces all main concepts pretty easily, and early, like the concepts of dual space (and some of the aspects simplifies the proofs). I don't remember all the details, but I do remember that I loved following his proofs; all of them were elegant.
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# Changing the color of a specific curve in ContourPlot [closed]
How do I change the color of a specific curve in my ContourPlot? I want to change the color of the last curve (+1.5 additive factor) from purple to gray. Here's my code:
ContourPlot[{Sin[2 y] == -2/3 Cos[3 x] - 1.5,
Sin[2 y] == -2/3 Cos[3 x] - 1, Sin[2 y] == -2/3 Cos[3 x],
Sin[2 y] == -2/3 Cos[3 x] + 1,
Sin[2 y] == -2/3 Cos[3 x] + 1.5}, {x, -2 Pi, 2 Pi}, {y, -2 Pi,
2 Pi}]
## closed as off-topic by MarcoB, m_goldberg, Bob Hanlon, Carl Lange, ÖskåMar 7 at 21:58
This question appears to be off-topic. The users who voted to close gave this specific reason:
• "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – MarcoB, m_goldberg, Bob Hanlon, Carl Lange, Öskå
If this question can be reworded to fit the rules in the help center, please edit the question.
• Combine separate plots with Show is one simple way that comes to mind. – Michael E2 Mar 5 at 14:27
• That was really helpful. Thank you! @MichaelE2 – Levy Mar 5 at 14:35
• Or ContourPlot[ Evaluate[(Sin[2 y] == -2/3 Cos[3 x] - #) & /@ {-3/2, -1, 0, 1, 3/2}], {x, -2 Pi, 2 Pi}, {y, -2 Pi, 2 Pi}, ContourStyle -> {Automatic, Automatic, Automatic, Automatic, LightGray}, PlotLegends -> "Expressions"] where I have used LightGray rather than Gray to make the difference easier to see. – Bob Hanlon Mar 5 at 15:00
Two more methods in addition to the ones mentioned in comments by Michael E2 and Bob Hanlon:
1. Post-process Purple (actually, ColorData[97][5]) to Gray:
contours = {-3/2, -1, 0, 1, 3/2};
ContourPlot[Evaluate[Sin[2 y] + 2/3 Cos[3 x] == # & /@ contours],
{x, -2 Pi, 2 Pi}, {y, -2 Pi, 2 Pi}] /.
Directive[___, ColorData[97][5], ___] :> Directive[AbsoluteThickness[3], Gray]
1. Use a single function and style the contours using the {{contour1, style1}..} form for setting the option Contours:
styles = Append[ColorData[97] /@ Range[4], Directive[Thick, Gray]];
ContourPlot[Sin[2 y] - 2/3 Cos[3 x], {x, -2 Pi, 2 Pi}, {y, -2 Pi, 2 Pi},
Contours -> Thread[{contours, styles}], ContourShading -> None,
PlotLegends -> LineLegend[styles, Sin[2 y] - 2/3 Cos[3 x] == # & /@ contours]]
• Thank you very much! It worked great. – Levy Mar 6 at 12:14
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# 13. LO.1 & LO.2 (Relevant costs; writing) Because of a monumental error committed by its...
1 answer below »
13. LO.1 & LO.2 (Relevant costs; writing) Because of a monumental error committed by its purchasing department, Corner Grocery ordered 5,000 heads of lettuce rather than the 1,000 that should have been ordered. The company paid $0.65 per head for the lettuce. Although management is confident that it can sell 2,000 units through regular sales, the market is not large enough to absorb the other 3,000 heads. Management has identified two ways to dispose of the excess heads of lettuce. First, a wholesaler has offered to purchase them for$0.25 each. Second, a restaurant chain has offered to purchase the lettuce if Corner Grocery will agree to convert it into packaged lettuce for salads. This option would require Corner Grocery to incur $2,500 for conversion, and the heads of lettuce could then be sold for the equivalent of$1.05 each.
a. Which costs are sunk in this decision?
b. Actually, Corner Grocery can consider three alternatives in this decision. Describe the alternative that is not mentioned in the problem.
c. What are the relevant costs of each decision alternative, and what should the company do?
## 1 Approved Answer
Answer a Sunk Cost Sunk cost means a cost which already incurred and can not be recovered. In the given problem sunk cost will be Unit can not be sold in direct...
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Browse Questions
For the following half cell $P^+/CH_3COOH/H_2\;Ka_{CH_3COOH}=10^{-5}$ Find $E_{H^+/H_2}$
$\begin{array}{1 1}(a)\;-0.36\\(b)\;-0.18\\(c)\;-0.54\\(d)\;-0.09\end{array}$
$CH_3COOH(aq)\rightleftharpoons CH_3COO^-(aq)+H^+(aq)$
$Ka=\large\frac{C\alpha^2}{1-\alpha}$
$1-\alpha\approx 1$
So $\alpha=\sqrt{\large\frac{Ka}{C}}$
So $\alpha=\sqrt{\large\frac{10^{-3}}{10^{-1}}}$
$\Rightarrow 10^{-2}$
So $[H^+]=C\alpha=10^{-3}$
$P^H=-\log [H^+]=3$
$E_{H^+/H_2}=-0.06\times P^H=-0.18V$
Hence (b) is the correct answer.
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# How do you find the vertical, horizontal or slant asymptotes for (6x^2+2x-1) /( x^2-1)?
Sep 15, 2016
vertical asymptotes at $x = \pm 1$
horizontal asymptote at y = 6
#### Explanation:
The denominator of the function cannot be zero as this would make the function undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.
solve: ${x}^{2} - 1 = 0 \Rightarrow x = \pm 1$
$\Rightarrow x = - 1 \text{ and " x=1" are the asymptotes}$
Horizontal asymptotes occur as
${\lim}_{x \to \pm \infty} , f \left(x\right) \to c \text{ (a constant)}$
divide terms on numerator/denominator by the highest power of x that is ${x}^{2}$
$f \left(x\right) = \frac{\frac{6 {x}^{2}}{x} ^ 2 + \frac{2 x}{x} ^ 2 - \frac{1}{x} ^ 2}{{x}^{2} / {x}^{2} - \frac{1}{x} ^ 2} = \frac{6 + \frac{2}{x} - \frac{1}{x} ^ 2}{1 - \frac{1}{x} ^ 2}$
as $x \to \pm \infty , f \left(x\right) \to \frac{6 + 0 - 0}{1 - 0}$
$\Rightarrow y = 6 \text{ is the asymptote}$
Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (both of degree 2 ) Hence there are no slant asymptotes.
graph{(6x^2+2x-1)/(x^2-1) [-20, 20, -10, 10]}
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Martingale Decomposition
Is it possible to decompose a discrete-time martingale $(M_n)$ uniquely into two processes $$M_n=M_n^I+A_n$$ where $(M^I_n)$ is a martingale with independent increments and $(A_n)$ is a martingale? If no, under which condition(s) on $(M_n)$, do we have this kind of decomposition?
-
Yes. (Hint: Let $M_n^I$ be something very trivial; or, if you want something slightly less trivial, add and subtract.) – cardinal Oct 10 '12 at 0:19
For the first (very trivial) part, yes, $M_n^I = 0$ is valid. For the second (only slightly less trivial) part, add and subtract a martingale $(S_n)$ with independent increments that is independent of $(M_n)$. – cardinal Oct 10 '12 at 0:39
I edited my question. Actually what I mean is an unique decomposition similar to Doob decomposition. – Martungale Martun Oct 10 '12 at 0:44
Then, the answer is obviously "no", by considering examples in the same spirit as the ones I gave. – cardinal Oct 10 '12 at 0:46
thanks cardinal. I again edited the question. I would like to learn under which conditions it is possible to do. Any reference would also be helpful. – Martungale Martun Oct 10 '12 at 0:55
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## University Calculus: Early Transcendentals (3rd Edition)
$3 \ln |x|- \dfrac{x^{2}}{2}+c$
Calculate the anti-derivative. Since, we know $\int x^{n} dx=\dfrac{x^{n+1}}{n+1}+c$ and $\int \dfrac{1}{x} dx=\ln |x|+c$ where $c$ is a constant of proportionality. Then $\int (\dfrac{3}{x}-x) dx=3 \int \dfrac{1}{x} dx-\int x dx dx$ or, $=3 \ln |x|- \dfrac{x^{1+1}}{1+1}+c$ Thus, $=3 \ln |x|- \dfrac{x^{2}}{2}+c$
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# zbMATH — the first resource for mathematics
Phase space path integrals, coherent states and Wiener measure. (English) Zbl 0731.60107
Functional integration methods in stochastic quantum mechanics, Proc. Jt. Concordia-Sherbrooke Semin. Ser., Sherbrooke & Montréal/Québec (Can.) 1987, Suppl. Rend. Circ. Mat. Palermo, II. Ser. 25, 157-176 (1991).
Summary: [For the entire collection see Zbl 0722.00045.]
We define the phase space path integral for a one-dimensional quantum particle with Hamiltonian H, give its coherent state interpretation, and show how this path integral can be made rigorous by introducing a Wiener measure and taking the limit as the diffusion constant for the associated diffusion process tends to infinity.
##### MSC:
60K40 Other physical applications of random processes 81Q30 Feynman integrals and graphs; applications of algebraic topology and algebraic geometry
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# Why is high input impedance good?
Naive perhaps, but
• Why is high input impedance a good thing?
• Is high input impedance always a good thing?
It is a good thing for a voltage input, as if the input impedance is high compared to the source impedance then the voltage level will not drop too much due to the divider effect.
For example, say we have a $10V$ signal with $1k\Omega$ impedance.
We connect this to a $1M\Omega$ input, the input voltage will be $10V\cdot\frac{1M\Omega}{1M\Omega+1k\Omega} = 9.99V$.
If we reduce the input impedance to $10k\Omega$, we get $10V \cdot \frac{10k\Omega}{10k\Omega + 1k\Omega} = 9.09V$
Reduce it to 1k and we get $10V \cdot \frac{1k\Omega}{1k\Omega + 1k\Omega} = 5V$
Hopefully you get the picture - generally an input impedance of at least 10 times the source impedance is a good idea to prevent significant loading.
High input impedance is not always a good thing though, for example if you want to transfer as much power as possible then the source and load impedance should be equal. So in the above example the 1k input impedance would be the best choice.
For a current input a low input impedance (ideally zero) is desired, for example in a transimpedance (current to voltage) amplifier.
• People always use the matched load for max power transfer point. None of my high power equipment does that. You do not want to be dissipating a ton of power in your source, instead they use a higher voltage and you design for a high load impedance. Not to say I do not understand your point, but just to note for others. – Kortuk Nov 4 '11 at 6:33
• For maximum power transfer the source should have as low output impedance as possible. However, if the source has a relatively high output impedance and you can't modify it then the load should have the same impedance for maximum power. If the load impedance is higher, the power will be lower, if load impedance is lower, more power will be dissipated by the load. That's why vacuum tube amplifiers use output transformers to match the high impedance of the amp and low impedance of the speakers. – Pentium100 Apr 14 '12 at 3:43
The "best" value of Impedance depends on the situation and application.
When it is appropriate to have or need a high impedance it is because it is an approximation to an infinite impedance.
An input applied to a signal source acts as a voltage divider.
Vout = Vsignal x Zinput / (Zsource + Zinput)
To get no loading either Zsiganl is zero (low or no impeadance output) and / or Zinput = infinite.
"Suitably high" is the practical version of infinite would be nice."
How large "suitably" is depends on the application.
AC mains has an impedance well under 1 ohm (usually). A test meter with 1000 ohms impednace woul draw about 100 mA !!!! from 110 VAC mains but would only load it down my under 0.1 of a Volt in the process. A test meter of 1 megohm input impedance would draw about 100 uAmp which would be much more acceptable.
For high impedance sources "suitably) needs to be quite large.
A high impedance input places very little load on a signal that is applied to it.
It thus does not reduce it in level (or not much). A unity gain buffer usually has very high impedance and is often used as an input stage to an amplifier chain. A pH probe, used for measuring acidity and alkalinity of a solution, mat have an output impedance of 10's to 100's of megohms. It's voltage level is a direct measure of pH. So anything that seeks to measure the voltage must try not to alter it in the process. A voltage measuring probe will effectively act like a voltage divider. The probe impedance needs to be >> the measured impedance if loading is not to occur.
A probe which is 256 times the impedance of a circuit being measured will cause 1 bit error in an 8 bit system.
A probe which is 4096 times the impedance of a circuit being measured will cause 1 bit error in a 12 bit system.
So to measure with 1 bit in 256 = 1 bit in an 8 bit system with a 1 megohm source impedance you need a 256 Megohm input impedance. For a 10 Megohm source you need a 2.6 Gigohn input impedance. And for a 100 Megohm ource you need ... !!!
As per the formula above, for outputs, LOW impedance is good, with the ideal being zero impedance (a perfect voltage source).
Then there is the special case of matched impedances where source and input are the same. Half the signal is dissipated in the INPUT and half in the output (assuming otherwise lossless connection) BUT there are no reflections due to impedance mismatch. A whole new subject for another time.
The word "high input impedance" is always related to the amplifier (audio intermediate frequency power amplifier... etc.)
So let's consider the following circuit:
The input voltage $V_{in}$ has an internal impedance ($Z_{in}$) this voltage injected to the base of transistor to amplify the signal. We calculate the voltage across $Z_{in}$, called $v$ as follows:
$$v = \frac{V_{in} Z_{in}}{Z_{in} + Z.V_{in}}$$
If we take $V_{in}=5V$, $Z.V_{in}=2,000Ω$, $Z_{in}=10Ω$ we get:
$$V=\dfrac{5 \cdot 10}{2,000+10} = 0.02V$$
That's a very low voltage compared with the input voltage.
If we take $V_{in}=5V$ , $Z.V_{in}=2000Ω$ , $Z_{in}=1,000,000Ω=1MΩ$ we get:
$$V=\dfrac{5 \cdot 1,000,000}{2,000 + 1,000,000}=4.99V$$
That's a good voltage compared with the input voltage.
Let's see some value of the input impedance in the table below.
The answer is the high input impedance is good for the amplifier circuit to have a good amplification of the input signal other wise we get low voltage in, so low amplification.
I hope this can help, thank you.
To get all the voltage from a source to a target without loss.
you need high input impedance. This principle is called "voltage bridging" or "Impedance Bridging".
That is a relative low output impedance to a higher input impedance.
Usualy the input impedance is at least ten times higher then the output impedance.
Voltage Bridging
one which maximizes transfer of a voltage signal to the load.
The other typical configuration is an "Impedance matching connection",
which maximizes power delivered to the load.
The high impedance is not always good but it varies from application to application. In order for impedance matching with other circuits the designer will select the high input impedance using the theorem "Maximum Power transfer Thoerem"
• At high frequency impedance matching reduces reflected power ( look up transmission lines for more ). – russ_hensel Nov 4 '11 at 12:58
Infinite input impedance would allow one to feed any amount of voltage into a load without it absorbing any power. Zero input impedance would allow one to feed any amount of current into a load without it absorbing any power. In cases where one wants to sense voltage without absorbing power, infinite impedance is thus the ideal; conversely, if one wants to sense current, zero impedance is the ideal.
Although sometimes one wants a load that doesn't absorb any power, there are times one wants to feed power into the load. The amount of power fed into a load will be maximized when the input impedance of the load matches the output impedance of whatever is driving it. This situation does not imply maximal energy efficiency, however. Depending upon what's driving the load, a higher or lower input impedance may cause the driving device to waste more or less power internally.
An electrical signal has two components: (a) a voltage component (b) a current component.
To build a POWER amplifier requires equal amplification of both components and the "Maximum Power-transfer Theorem applies: i.e. a Load impedance must equal (the purely theoretical) Source impedance.
NOte that a soure impedance is not a true impedance - it cannot be measured but only calculated.
To drive an active component (valve or FET which has a high input impedance - large V/small I) a voltage amplifier must be driven from a low Source-impedance but deliver from a relatively low-impedance. (Thevenin's Theorem.)
To drive an active component (bipolar tansistor) which has a low input-impedance - small V/large I) a "current amplifier" must be driven from a high Source-impedance but deliver from a relatively high impedance. (Norton's Theorem.)
High Input means you only need the SIGNAL. Or lets call it the message of voltage. In this case low current is fine to drive the stuff.
High Input is NOT always a good thing. In case of not using the signal but driving an electronical part (for example for LED light) you need to calculate the current and you need to decrease the output resistance.
If you are using too high resistance while working with a signal message, the only point of view is the capacity to other parts.
If you are working in HF range of frequency modulation, it becomes more difficult. In any other case, yes, high input is a good thing to use to have less power consumption.
Regards
High impedance is not always good when a current must flow to achieve the desired result. For example, large area electrodes and conducting jelly are used to lower the impedance in Edison’s great invention, the electric chair.
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# Copycat Chess (Part 2/3)
The mandarin stared bewildered on the chess board after Li Chai finished his winning move:
1. d4 d5 2. Qd3 Qd6 3. Qf5 Qf4 4. Qxc8#
But it didn't last for long. He grabbed his queen and repeated Li Chais last move while announcing:
What kind of mate is that, if I can mate you as well?
This was of course against the rules, but Li Chai stayed calm and started the second game.
How did Li Chai win the second game?
The rules (same as in part 1):
• standard chess rules apply, if not stated otherwise
• you play white and your opponent copies all your moves (e.g 1. e4 would be followed by 1. ... e5)
• you are allowed to make "stupid" moves, your opponent will copy all moves regardless how bad they are
• you are not allowed to make moves which cannot be copied, of course except for the last move winning the game
New rule:
• make it impossible for the mandarin to repeat the mating move
The limit for this part is 7 moves.
• This could also be posted on the Chess Stack Exchange. – Mr Pie Jul 2 '18 at 3:18
This is new for me and I'm not sure but this seems to work:
1. d4 d5 2. e4 e5 3. d:e5 d:e4 4. Bg5 Bg4 5. Q:d8#
• @Sleafar, let me guess, now mandarin will eat his pawn with his king? – klm123 Apr 9 '16 at 12:58
• Nope. :) – Sleafar Apr 9 '16 at 13:01
• Oh, so the last move can be anything (legal)? – Anonymous Mar 20 '17 at 8:50
• @Anonymous, what do you mean? Last move must be checkmate. – klm123 Mar 20 '17 at 9:20
• I mean I thought the last move must be also copied. – Anonymous Mar 20 '17 at 9:35
Also, if you want to use exactly $7$ moves, then:
1. e4, e5. 2. Nf3, Nf6 3. Nxe5, Nxe4 4. d4, d5 5. Qh5, Qh4 6. Bg5, Bg4 7. Qf7#
• Sorry, but the mandarin would simply repeat the move: Qf3. Please read the question carefully, especially the new rule. – Sleafar Nov 18 '17 at 5:57
• @Sleafar : I suppose the point is that 7...Qxf2 wouldn't be a "counter-mate", because the queen can be taken by White's Qf7. – Evargalo Oct 4 '18 at 15:33
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## Stream: Is there code for X?
### Topic: Quotient thing
#### Adam Topaz (Feb 01 2021 at 16:02):
Do we have something like the following construction? If not, what should we call it?!
import algebra
variables (G : Type*) [group G] (X : Type*) [mul_action G X]
-- TODO: Rename this!
namespace quotient_thing
@[simps]
def setoid : setoid X := ⟨λ x y, ∃ g : G, g • x = y,
λ x, ⟨1, by simp⟩,
λ x y ⟨g,hg⟩, ⟨g⁻¹, by simp [← hg]⟩,
λ x y z ⟨g,hg⟩ ⟨h,hh⟩, ⟨h * g, by simp [mul_smul, hh, hg]⟩⟩
end quotient_thing
def quotient_thing := quotient (quotient_thing.setoid G X)
#### Eric Wieser (Feb 01 2021 at 16:07):
∃ g : G, g • x = y is y ∈ mul_action.orbit G x if that helps
#### Eric Wieser (Feb 01 2021 at 16:08):
Your quotient_thing.setoid is docs#mul_action.orbit_rel
#### Eric Wieser (Feb 01 2021 at 16:09):
And the quotient by it seems to be used here: https://github.com/leanprover-community/mathlib/blob/39ecd1abaabede897b2e5367efc2af8cdcadcc5b/src/group_theory/sylow.lean#L39
#### Adam Topaz (Feb 01 2021 at 16:13):
Any suggestions on naming the quotient?
#### Eric Wieser (Feb 01 2021 at 16:19):
None, I don't actually know this stuff, I just remembered finding it when moving stuff around to fix import cycles
#### Bryan Gin-ge Chen (Feb 01 2021 at 16:28):
mul_action.orbits?
Last updated: May 07 2021 at 22:14 UTC
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This is a closed-book, closed-notes exam. You have one hour to complete
the test. Write your answers in the space provided. You may use blank paper
for scratchwork or as additional space for solutions.
After you have completed your work, sign the statement below.
I accept full responsibility under the Binghamton University Academic
Honesty Code for my conduct on this examination.
Signed:
## Math 461, Fall 2021
Each problem is worth 5 points.
Problem 1. Definitions: Provide definitions of the following terms.
1. What properties must a set $X$ and a function $d: X \times X \rightarrow \mathbb{R}$ satisfy to be a metric space?
2. Given a topological space $X$ with an equivalence relation $\sim$, what is the quotient topology on $[X] ?$4
Problem 2.
True/False: Determine whether each statement below is True or False. Justify your answer as clearly and completely as you can.
1. Every metric space is Hausdorff.
Problem 3.
1. If $x$ is a limit point of $A \subset X$ then $x \notin A$
Problem 4.
1. If $f: X \rightarrow Y$ is continuous and $C \subset X$ is closed, then $f(C)$ is closed.
real analysis代写analysis 2, analysis 3请认准UprivateTA™. UprivateTA™为您的留学生涯保驾护航。
# Binghamton University Graduate Combinatorics, Algebra, and Topology Conference (BUGCAT)
The 14th Annual Binghamton University Graduate Combinatorics, Algebra, and Topology Conference (BUGCAT) is to be held online through Zoom, November 6th, 7th, 13th, and 14th 2021.
This year’s featured keynotes are Profs. Tara Holm from Cornell University, Dandrielle Lewis from High Point University, and Inna Zakharevich from Cornell University.
Visit the conference home page and the conference program page. There is also a Facebook Page, or you may e-mail [email protected] for more information.
Categories: 拓扑学数学代写
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# The model category structure on $\mathbf{TMon}$
I ask this question here since I asked it here on Math.SE, and got no answers after a week of a bounty offer.
I am trying to understand the homotopy colimit of a diagram of topological monoids, and whether there is an explicit construction of this object (even in the case of simple pushout diagrams).
Let TMon denote the category of well-pointed topological monoids which have the homotopy types of cell complexes.
TMon can be equipped with a model category structure where the fibrations and weak equivalences are the (Serre) fibrations and weak equivalences of the underlying topological spaces - this comes from section $$3$$ of the paper
R. Schwänzl and R.M. Vogt, The categories of $$A_\infty$$- and $$E_\infty$$-monoids and ring spaces as closed simplicial and topological model categories, Arch. Math 56 (1991) pp 405–411, doi:10.1007/BF01198229.
Cofibrations are the morphisms which have the appropriate lifting property.
I wish to understand what the homotopy colimit of a diagram of topological monoids is. One way of approaching this is to take the colimit of the cofibrant replacement of the diagram in question. This involves (firstly) understanding cofibrations in TMon.
I struggle with this. I really have no intuition for what a cofibration in this category is at all. This is the first thing preventing me from understanding homotopy colimits of diagrams in TMon.
Question:
Are there constructions of the homotopy colimit in TMon in the case of simple diagrams? For example, if a diagram has morphisms which are all inclusions on the level of topological spaces? Or if the diagram is Reedy? Or under any other sufficiently nice assumptions?
• Cofibrations in $TMon$ are going to be more specific: they will be retracts of "cellular maps", i.e., of maps built by iteratively attaching cells. A cell will be something of the form $F(S^{n-1})\to F(D^n)$, where $F$ is the free monoid on a space. Thus, the "simplest" examples of cofibrant objects are discrete free monoids. – Charles Rezk Aug 12 '19 at 16:30
• Thanks for your comment - I'm a bit of a beginner with this. Is it possible for you to elaborate a little on why cofibrations should be retracts of such maps, or why a "cell" is what you give above? – Matt Aug 13 '19 at 10:52
• It's an example of a general formalism for constructing model category structures by "lifting". Here we are lifting the usual model structure on spaces to $TMon$: so the we's and fib's of TMon are "detected" by the forgetful structure to spaces. Formally, the left adjoint $F$ to the forgetful functor will preserve cofibrations and trivial cofibrations. In particular, it must supply a large collection of cofibrations in TMon, which can be used to "build" all cofibrations. – Charles Rezk Aug 13 '19 at 16:52
• Key phrases here are "cofibrantly generated model category" and "tranferred model structure": ncatlab.org/nlab/show/cofibrantly+generated+model+category, ncatlab.org/nlab/show/transferred+model+structure – Charles Rezk Aug 13 '19 at 16:53
First of all, monoids have classifying spaces, and so a pushout diagram $$M_1\gets M_0 \to M_2$$ of monoids and monoid maps gives rise to a pushout diagram $$BM_1 \gets BM_0 \to BM_2$$, and the claim is that, essentially, $$\mathrm{hocolim}_{\mathbf{TMon}}( M_1\gets M_0 \to M_2 ) = \Omega ( \mathrm{colim}_{\mathbf{T}} (BM_1 \gets BM_0 \to BM_2 )) .$$
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## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)
We can find the energy $Q$ that is transferred to or from the system as heat as: $\Delta U = Q + W$ $Q = \Delta U - W$ $Q = (-200~J) - 500~J$ $Q = -700~J$ The negative sign means that 700 J of energy is transferred from the system as heat.
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# [OS X TeX] Intertext like command
Mon Jul 12 07:10:50 CEST 2004
On 12 Jul 2004, at 2:20 PM, Alain Schremmer wrote:
>
> As you typed it, I think Paragraph text … is part of Item 2 and lined
> up as such.
Oh, yeah. Makes sense now, what you want.
Besides manually outdenting the line, I can't think of anything:
\begin{enumerate}
\item item one
\item item two
\item[] \hspace{-20pt} The next items are boring but included for
completeness...
\item item three
\end{enumerate}
WARNING: I just guessed the amount of space to subtract. 20pt might not
be accurate!!!
You could then make your own command to you wouldn't have to deal with
the uglyness. Put this in your preamble:
\newcommand{\interitemtext}[1]{\item[] \hspace{-20pt} #1}
(You don't have to use \interitemtext, you can call it whatever you
like)
Then:
\begin{enumerate}
\item item one
\item item two
\interitemtext{The next items are boring but included for
completeness...}
\item item three
\end{enumerate}
W
-----------------------------------------------------
Post: <mailto:MacOSX-TeX at email.esm.psu.edu>
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String theory landscape
(Redirected from String landscape)
The string theory landscape or anthropic landscape refers to the large number of possible false vacua in string theory.[1] The "landscape" includes so many possible configurations that some physicists[who?] think that the known laws of physics, the standard model and general relativity with a positive cosmological constant, occur in at least one of them. The anthropic landscape refers to the collection of those portions of the landscape that are suitable for supporting human life, an application of the anthropic principle that selects a subset of the theoretically possible configurations.
In string theory the number of false vacua is thought to be somewhere between 1010 to 10100.[1] The large number of possibilities arises from different choices of Calabi–Yau manifolds and different values of generalized magnetic fluxes over different homology cycles. If one assumes that there is no structure in the space of vacua, the problem of finding one with a sufficiently small cosmological constant is NP complete,[2] being a version of the subset sum problem.
Anthropic principle
Main article: Anthropic principle
The idea of the string theory landscape has been used to propose a concrete implementation of the anthropic principle, the idea that fundamental constants may have the values they have not for fundamental physical reasons, but rather because such values are necessary for life (and hence intelligent observers to measure the constants). In 1987, Steven Weinberg proposed that the observed value of the cosmological constant was so small because it is not possible for life to occur in a universe with a much larger cosmological constant.[3] In order to implement this idea in a concrete physical theory, it is necessary to postulate a multiverse in which fundamental physical parameters can take different values. This has been realized in the context of eternal inflation.
Bayesian probability
Main article: Bayesian probability
Some physicists, starting with Weinberg, have proposed that Bayesian probability can be used to compute probability distributions for fundamental physical parameters, where the probability $P(x)$ of observing some fundamental parameters $x$ is given by,
$P(x)=P_{\mathrm{prior}}(x)\times P_{\mathrm{selection}}(x),$
where $P_\mathrm{prior}$ is the prior probability, from fundamental theory, of the parameters $x$ and $P_\mathrm{selection}$ is the anthropic selection function, determined by the number of "observers" that would occur in the universe with parameters $x$. These probabilistic arguments are the most controversial aspect of the landscape. Technical criticisms of these proposals have pointed out that:
• The function $P_\mathrm{prior}$ is completely unknown in string theory and may be impossible to define or interpret in any sensible probabilistic way.
• The function $P_\mathrm{selection}$ is completely unknown, since so little is known about the origin of life. Simplified criteria (such as the number of galaxies) must be used as a proxy for the number of observers. Moreover, it may never be possible to compute it for parameters radically different from those of the observable universe.
(Interpreting probability in a context where it is only possible to draw one sample from a distribution is problematic in frequentist probability but not in Bayesian probability, which is not defined in terms of the frequency of repeated events.)
Various physicists have tried to address these objections, and the ideas remain extremely controversial both within and outside the string theory community. These ideas have been reviewed by Carroll.[4]
Simplified approaches
Tegmark et al. have recently considered these objections and proposed a simplified anthropic scenario for axion dark matter in which they argue that the first two of these problems do not apply.[5]
Vilenkin and collaborators have proposed a consistent way to define the probabilities for a given vacuum.[6]
A problem with many of the simplified approaches people have tried is that they "predict" a cosmological constant that is too large by a factor of 10–1000 (depending on one's assumptions) and hence suggest that the cosmic acceleration should be much more rapid than is observed.[7][8][9]
Criticism
Although few dispute the idea that string theory appears to have an unimaginably large number of metastable vacua, the existence, meaning and scientific relevance of the anthropic landscape remain highly controversial. Prominent proponents of the idea include Andrei Linde, Sir Martin Rees and especially Leonard Susskind, who advocate it as a solution to the cosmological-constant problem. Opponents, such as David Gross, suggest that the idea is inherently unscientific, unfalsifiable or premature. A famous debate on the anthropic landscape of string theory is the Smolin–Susskind debate on the merits of the landscape.
The term "landscape" comes from evolutionary biology (see Fitness landscape) and was first applied to cosmology by Lee Smolin in his book.[10] It was first used in the context of string theory by Susskind.[11]
There are several popular books about the anthropic principle in cosmology.[12] Two popular physics blogs are opposed to this use of the anthropic principle.[13]
References
1. ^ a b The most commonly quoted number is of the order 10500. See M. Douglas, "The statistics of string / M theory vacua", JHEP 0305, 46 (2003). arXiv:hep-th/0303194; S. Ashok and M. Douglas, "Counting flux vacua", JHEP 0401, 060 (2004).
2. ^ Frederik Denef; Douglas, Michael R. (2006). "Computational complexity of the landscape". Annals of Physics 322 (5): 1096–1142. arXiv:hep-th/0602072. Bibcode:2007AnPhy.322.1096D. doi:10.1016/j.aop.2006.07.013.
3. ^ S. Weinberg, "Anthropic bound on the cosmological constant", Phys. Rev. Lett. 59, 2607 (1987).
4. ^ S. M. Carroll, "Is our universe natural?", arXiv:hep-th/0512148.
5. ^ M. Tegmark, A. Aguirre, M. Rees and F. Wilczek, "Dimensionless constants, cosmology and other dark matters", arXiv:astro-ph/0511774. F. Wilczek, "Enlightenment, knowledge, ignorance, temptation", arXiv:hep-ph/0512187. See also the discussion at [1].
6. ^ See, e.g. Alexander Vilenkin (2006). "A measure of the multiverse". Journal of Physics A: Mathematical and Theoretical 40 (25): 6777–6785. arXiv:hep-th/0609193. Bibcode:2007JPhA...40.6777V. doi:10.1088/1751-8113/40/25/S22.
7. ^ Abraham Loeb (2006). "An observational test for the anthropic origin of the cosmological constant" (subscription required). JCAP 0605: 009.
8. ^ Jaume Garriga and Alexander Vilenkin (2006). "Anthropic prediction for Lambda and the Q catastrophe" (subscription required). Prog. Theor.Phys. Suppl. 163: 245–57. arXiv:hep-th/0508005. Bibcode:2006PThPS.163..245G. doi:10.1143/PTPS.163.245.
9. ^ Delia Schwartz-Perlov and Alexander Vilenkin (2006). "Probabilities in the Bousso-Polchinski multiverse" (subscription required). JCAP 0606: 010.
10. ^ L. Smolin, "Did the universe evolve?", Classical and Quantum Gravity 9, 173–191 (1992). L. Smolin, The Life of the Cosmos (Oxford, 1997)
11. ^ L. Susskind, "The anthropic landscape of string theory", arXiv:hep-th/0302219.
12. ^ L. Susskind, The cosmic landscape: string theory and the illusion of intelligent design (Little, Brown, 2005). M. J. Rees, Just six numbers: the deep forces that shape the universe (Basic Books, 2001). R. Bousso and J. Polchinski, "The string theory landscape", Sci. Am. 291, 60–69 (2004).
13. ^ Lubos Motl's blog criticized the anthropic principle and Peter Woit's blog frequently attacks the anthropic string landscape.
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## Catalyst Example from lecture 3/12
Arrhenius Equation: $\ln k = - \frac{E_{a}}{RT} + \ln A$
KayleeMcCord1F
Posts: 31
Joined: Fri Sep 29, 2017 7:05 am
### Catalyst Example from lecture 3/12
In the ozone example about homogeneous catalysts in lecture today, which species is actually the catalyst?
Andres Reynoso 1J
Posts: 30
Joined: Fri Sep 29, 2017 7:06 am
### Re: Catalyst Example from lecture 3/12
I'm pretty sure the catalyst was NO (g).
aTirumalai-1I
Posts: 55
Joined: Thu Jul 27, 2017 3:00 am
Been upvoted: 1 time
### Re: Catalyst Example from lecture 3/12
Yes, NO(g) was the catalyst. It was used up in the first step of the mechanism, but then re-produced in the second step of the mechanism.
mayasinha1B
Posts: 70
Joined: Fri Sep 29, 2017 7:04 am
### Re: Catalyst Example from lecture 3/12
What was the purpose of the UV? Was it only to reach the activation energy?
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# Physics 11 Speed and acceleraton
[SOLVED] Physics 11 Speed and acceleraton
## Homework Statement
Linda sees an elephant dart into the road 50.0m ahead of her car while she is driving at 65km/hr. She slams on her brakes, which decelerate the car at the rate of -5.80m/s^2. Will she be able to avoid hitting the elephant? Find her stopping distance.
Answer is Yes, stopping distance is 28.1
## The Attempt at a Solution
65km/hr = 18.056m/s $$V_{avg}9.028$$
a = v / t -5.80m/s^2 = 9.028m/s / T = 1.56 seconds
using $$d = v_{0}t+.5at^{2}$$
d = .5(-5.80m/s^2) (2.43s^2) = 7meters
I am missing something here.
Thanks for the help on the previous question and thanks for any help for this question. Your quick response is appreciated.
Related Introductory Physics Homework Help News on Phys.org
I seem to have found the answer, but why is it that the rate is not averaged? The rate is going down from 65 to 0. Therefore, the rate should be an average before doing the calculations. I am confused.
V=0
U=18M/S
A=-5.80 M/S^2
V^2=U^2+2AS
>0=324-11.6S
>S=27m
why would ya average the rate???since the deceleration is uniform..u must not average
the rate.this shall lead to loss in generality of the problem.
Hootenanny
Staff Emeritus
Gold Member
V=0
U=18M/S
A=-5.80 M/S^2
V^2=U^2+2AS
>0=324-11.6S
>S=27m
Careful with rounding errors.
I seem to have found the answer, but why is it that the rate is not averaged? The rate is going down from 65 to 0. Therefore, the rate should be an average before doing the calculations. I am confused.
By rate, do you mean acceleration? For the case of constant acceleration, average acceleration = instantaneous acceleration.
Thanks guys.
why would ya average the rate???since the deceleration is uniform..u must not average
the rate.this shall lead to loss in generality of the problem.
This solved my question. I thought about it again and realized that I made an error. In any case, it will take me a while to wrap my mind around these questions. The deceleration is uniform and will go from 65 km/h to 0 km/h in a certain amount of time based on the rate of deceleration. The reason I was averaging the number was because of the formula sheet. It is written as
note: (change) = a triangle. I assume the triangle in front of the variable means change.
a=(change)v/(change)t
I assumed the change meant average.
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# SUPERCATEGORICAL ONTOLOGY OF COMPLEX SYSTEMS, META–SYSTEMS AND LEVELS: The Emergence of Life, Human Consciousness and Society.
acomared19-TAO1proceeds3.tex
August 19th, 2008
I. C. Baianu, R. Brown and J. F. Glazebrook
I. C. Baianu][email protected]
R. Brown][email protected]
J. F. Glazebrook][email protected]
## 0.1. Mathematical and Metaphysics Notes.
### 0.1.1. AN-1. On the Logical Foundations of Arithmetics
According to a website contributed entry (at $http://www.philosophypages.com/hy/6h.htm$): “The culmination of the new approach to logic lay in its capacity to illuminate the nature of the mathematical reasoning. While the idealists sought to reveal the internal coherence of absolute reality and the pragmatists offered to account for human inquiry as a loose pattern of investigation, the new logicians hoped to show that the most significant relations among things could be understood as purely formal and external. Mathematicians like Richard Dedekind realized that on this basis it might be possible to establish mathematics firmly on logical grounds. Giuseppe Peano had demonstrated in 1889 that all of arithmetic could be reduced to an axiomatic system with a carefully restricted set of preliminary postulates. Frege promptly sought to express these postulates in the symbolic notation of his own invention. By 1913, Russell and Whitehead had completed the monumental “Principia Mathematica” (1913), taking three massive volumes to move from a few logical axioms through a definition of number to a proof that “ 1 + 1 = 2 .” Although the work of Gödel (less than two decades later) made clear the inherent limitations of this approach, its significance for our understanding of logic and mathematics remains”.
AN-2.7:
### 0.1.2. Local–to–Global (LG) Construction Principles consistent with Quantum ‘Axiomatics’.
A novel approach to QST construction in Algebraic/Axiomatic QFT involves the use of generalized fundamental theorems of algebraic topology from specialized, ‘globally well-behaved’ topological spaces, to arbitrary ones (Baianu et al, 2007c). In this category, are the generalized, Higher Homotopy van Kampen theorems (HHvKT) of Algebraic Topology with novel and unique non-Abelian applications. Such theorems greatly aid the calculation of higher homotopy of topological spaces. R. Brown and coworkers (1999, 2004a,b,c) generalized the van Kampen theorem, at first to fundamental groupoids on a set of base points (Brown,1967), and then, to higher dimensional algebras involving, for example, homotopy double groupoids and 2-categories (Brown, 2004a). The more sensitive algebraic invariant of topological spaces seems to be, however, captured only by cohomology theory through an algebraic ring structure that is not accessible either in homology theory, or in the existing homotopy theory. Thus, two arbitrary topological spaces that have isomorphic homology groups may not have isomorphic cohomological ring structures, and may also not be homeomorphic, even if they are of the same homotopy type. Furthermore, several non-Abelian results in algebraic topology could only be derived from the Generalized van Kampen Theorem (cf. Brown, 2004a), so that one may find links of such results to the expected non-commutative geometrical’ structure of quantized space–time (Connes, 1994). In this context, the important algebraic–topological concept of a Fundamental Homotopy Groupoid (FHG) is applied to a Quantum Topological Space (QTS) as a “partial classifier” of the invariant topological properties of quantum spaces of any dimension; quantum topological spaces are then linked together in a crossed complex over a quantum groupoid (Baianu, Brown and Glazebrook, 2006), thus suggesting the construction of global topological structures from local ones with well-defined quantum homotopy groupoids. The latter theme is then further pursued through defining locally topological groupoids that can be globally characterized by applying the Globalization Theorem, which involves the unique construction of the Holonomy Groupoid. We are considering in a separate publication(Baianu et al 2007c) how such concepts might be applied in the context of Algebraic or Axiomatic Quantum Field Theory (AQFT) to provide a local-to-global construction of Quantum space-times which would still be valid in the presence of intense gravitational fields without generating singularities as in GR. The result of such a construction is a Quantum Holonomy Groupoid, (QHG) which is unique up to an isomorphism.
### 0.1.3. The Object-Based Approach vs Process-Based, Dynamic Ontology.
In classifications, such as those developed over time in Biology for organisms, or in Chemistry for chemical elements, the objects are the basic items being classified even if the ‘ultimate’ goal may be, for example, either evolutionary or mechanistic studies. Rutherford’s comment is pertinent in this context: “There are two major types of science: physics or stamp collecting.” An ontology based strictly on object classification may have little to offer from the point of view of its cognitive content. It is interesting that many psychologists, especially behavioural ones, emphasize the object-based approach rather than the process-based approach to the ultra-complex process of consciousness occurring ‘in the mind’ –with the latter thought as an ‘object’. Nevertheless, as early as the work of William James in 1850, consciousness was considered as a continuous stream that never repeats itself’–a Heraclitian concept that does also apply to super-complex systems and life, in general. We shall see more examples of the object-based approach to psychology in .
### 0.1.4. Procedures and Advantages of Poli’s Ontological Theory of Levels
According to Poli (2001), the ontological procedures provide:
• coordination between categories (for instance, the interactions and parallels between biological and ecological reproduction as in Poli, 2001);
• modes of dependence between levels (for instance, how the co-evolution/interaction of social and mental realms depend and impinge upon the material);
• the categorical closure (or completeness) of levels.
Already we can underscore a significant component of this essay that relates the ontology to geometry and topology; specifically, if a level is defined via ‘iterates of local procedures’ (viz. ‘items in iteration’, Poli, 2001), then we have some handle on describing its intrinsic governing dynamics (with feedback ) and, to quote Poli (2001), to ‘restrict the multi-dynamic frames to their linear fragments’. On each level of this ontological hierarchy there is a significant amount of connectivity through inter-dependence, interactions or general relations often giving rise to complex patterns that are not readily analyzed by partitioning or through stochastic methods as they are neither simple, nor are they random connections. But we claim that such complex patterns and processes have their logico-categorical representations quite apart from classical, Boolean mechanisms. This ontological situation gives rise to a wide variety of networks, graphs, and/or mathematical categories, all with different connectivity rules, different types of activities, and also a hierarchy of super-networks of networks of sub-networks. Then, the important question arises what types of basic symmetry or patterns such super-networks of items can have, and also how do the effects of their sub-networks ‘percolate’ through the various levels. From the categorical viewpoint, these are of two basic types: they are either commutative or non-commutative, where, at least at the quantum level, the latter takes precedence over the former.
It is often thought or taken for granted that the object-oriented approach can be readily converted into a process-based one. It would seem, however, that the answer to this question depends critically on the ontological level selected. For example, at the quantum level, object and process become inter-mingled. Either comparing or moving between levels, requires ultimately a process-based approach, especially in Categorical Ontology where relations and inter-process connections are essential to developing any valid theory. At the fundamental level of ‘elementary particle physics’ however the answer to this question of process-vs. object becomes quite difficult as a result of the ‘blurring’ between the particle and the wave concepts. Thus, it is well-known that any ‘elementary quantum object’ is considered by all accepted versions of quantum theory not just as a ‘particle’ or just a ‘wave’ but both: the quantum ‘object’ is both wave and particle, at the same-time, a proposition accepted since the time when it was proposed by de Broglie. At the quantum microscopic level, the object and process are inter-mingled, they are no longer separate items. Therefore, in the quantum view the ‘object-particle’ and the dynamic process-‘wave’ are united into a single dynamic entity or item, called the wave-particle quantum, which strangely enough is neither discrete nor continuous, but both at the same time, thus ‘refusing’ intrinsically to be an item consistent with Boolean logic. Ontologically, the quantum level is a fundamentally important starting point which needs to be taken into account by any theory of levels that aims at completeness. Such completeness may not be attainable, however, simply because an ‘extension’ of Gödel’s theorem may hold here also. The fundamental quantum level is generally accepted to be dynamically, or intrinsically non-commutative, in the sense of the non-commutative quantum logic and also in the sense of non-commuting quantum operators for the essential quantum observables such as position and momentum. Therefore, any comprehensive theory of levels, in the sense of incorporating the quantum level, is thus –mutatis mutandisnon-Abelian. Furthermore, as the non-Abelian case is the more general one, from a strictly formal viewpoint, a non-Abelian Categorical Ontology is the preferred choice. A paradigm-shift towards a non-Abelian Categorical Ontology has already started (Brown et al, 2007: ‘Non-Abelian Algebraic Topology’; Baianu, Brown and Glazebrook, 2006: NA-QAT; Baianu et al 2007a,b,c).
### 0.1.5. Fundamental Concepts of Algebraic Topology with Potential Application to Ontology Levels Theory and Space-Time Structures.
We shall consider briefly the potential impact of novel Algebraic Topology concepts, methods and results on the problems of defining and classifying rigorously Quantum space-times. With the advent of Quantum Groupoids–generalizing Quantum Groups, Quantum Algebra and Quantum Algebraic Topology, several fundamental concepts and new theorems of Algebraic Topology may also acquire an enhanced importance through their potential applications to current problems in theoretical and mathematical physics, such as those described in an available preprint (Baianu, Brown and Glazebrook, 2006), and also in several recent publications (Baianu et al 2007a,b; Brown et al 2007).
Now, if quantum mechanics is to reject the notion of a continuum, then it must also reject the notion of the real line and the notion of a path. How then is one to construct a homotopy theory? One possibility is to take the route signalled by Čech, and which later developed in the hands of Borsuk into ‘Shape Theory’ (see, Cordier and Porter, 1989). Thus a quite general space is studied by means of its approximation by open covers. Yet another possible approach is briefly pointed out in AN-2.6. A few fundamental concepts of Algebraic Topology and Category Theory are summarized in AN-2.6 that have an extremely wide range of applicability to the higher complexity levels of reality as well as to the fundamental, quantum level(s). We have omitted in this section the technical details in order to focus only on the ontologically-relevant aspects; full mathematical details are however also available in a recent paper by Brown et al (2007) that focuses on a mathematical/conceptual framework for a completely formal approach to categorical ontology and the theory of levels.
### 0.1.6. Towards Biological Postulates and Principles.
Often, Rashevsky considered in his Relational Biology papers, and indeed made comparisons, between established physical theories and principles. He was searching for new, more general relations in Biology and Sociology that were also compatible with the former. Furthermore, Rashevsky also proposed two biological principles that add to Darwin’s natural selection of species and the ‘survival of the fittest principle’, the emergent relational structure thus defining adaptive organisms:
1. The Principle of Optimal Design, and 2. The Principle of Relational Invariance (phrased by Rashevsky as “Biological Epimorphism).
In essence, the ‘Principle of Optimal Design’ defines the ‘fittest’ organism which survives in the natural selection process of competition between species, in terms of an extremal criterion, similar to that of Maupertuis; the optimally ‘designed’ organism is that which acquires maximum functionality essential to survival of the successful species at the lowest ‘cost’ possible. The ‘costs’ are defined in the context of the environmental niche in terms of material, energy, genetic and organismic processes required to produce/entail the pre-requisite biological function(s) and their supporting anatomical structure(s) needed for competitive survival in the selected niche. Further details were presented by Robert Rosen in his short but significant book on optimality (1970). The ‘Principle of Biological Epimorphism’ on the other hand states that the highly specialized biological functions of higher organisms can be mapped (through an epimorphism) onto those of the simpler organisms, and ultimately onto those of a (hypothetical) primordial organism (which was assumed to be unique up to an isomorphism or selection-equivalence). The latter proposition, as formulated by Rashevsky, is more akin to a postulate than a principle. However, it was then generalized and re-stated in the form of the existence of a limit in the category of living organisms and their functional genetic networks ($\textbf{GN}^{i}$), as a directed family of objects, $\textbf{GN}^{i}(-t)$ projected backwards in time (Baianu and Marinescu, 1968), or subsequently as a super-limit (Baianu, 1970 to 1987; Baianu, Brown, Georgescu and Glazebrook, 2006); then, it was re-phrased as the Postulate of Relational Invariance, represented by a colimit with the arrow of time pointing forward (Baianu, Brown, Georgescu and Glazebrook, 2006). Somewhat similarly, a dual principle and colimit construction was invoked for the ontogenetic development of organisms (Baianu, 1970), and also for populations evolving forward in time; this was subsequently applied to biological evolution although on a much longer time scale –that of evolution– also with the arrow of time pointing towards the future in a representation operating through Memory Evolutive Systems (MES) by A. Ehresmann and Vanbremeersch (2006).
### 0.1.7. Selective Boundaries and Homeostasis. Varying Boundaries vs Horizons.
Boundaries are especially relevant to closed systems. According to Poli (2008): “they serve to distinguish what is internal to the system from what is external to it”, thus defining the fixed, overall structural topology of a closed system. By virtue of possessing boundaries, a whole (entity) is something on the basis of which there is an interior and an exterior (viz. Baianu and Poli, 2007). One notes however that a boundary, or boundaries, may either change/vary or be quite selective/directional–in the sense of dynamic fluxes crossing such boundaries– if the system is open. In the case of an organism that grows and develops it will be therefore characterized by a variable topology that may also depend on the environment, and is thus context-dependent, as well. Perhaps one of the simplest example of a system that changes from closed to open, and thus has a variable topology, is that of a pipe equipped with a functional valve that allows flow in only one direction. On the other hand, a semi-permeable membrane such as a cellophane, thin-walled ’closed’ tube– that allows water and small molecule fluxes to go through but blocks the transport of large molecules such as polymers through its pores– is selective and may be considered as a primitive/’simple’ example of an open, selective system. Organisms, in general, are open systems with specific types or patterns of variable topology which incorporate both the valve and the selectively permeable membrane boundaries –albeit much more sophisticated and dynamic than the simple/fixed topology of the cellophane membrane; such variable structures are essential to maintaining their stability and also to the control of their internal structural order, of low microscopic entropy. (The formal definition of this important concept of ‘variable topology’ was introduced in our recent paper (Baianu et al 2007a) in the context of the space-time evolution of organisms, populations and species.)
As proposed by Baianu and Poli (2008), an essential feature of boundaries in open systems is that they can be crossed by matter; however, all boundaries may be crossed by either fields or by quantum wave-particles if the boundaries are sufficiently thin, even in ’closed’ systems. The boundaries of closed systems, however, cannot be crossed by molecules or larger particles. On the contrary, a horizon is something that one cannot reach. In other words, a horizon is not a boundary. This difference between horizon and boundary appears to be useful in distinguishing between systems and their environment (v. AN-4.1). Boundaries may be fixed, clear-cut, or they may be vague/blurred, mobile, varying/variable in time, or again they may be intermediate between these any of these cases, according to how the differentiation is structured. At the beginning of an organism’s ontogenetic development, there may be only a slightly asymmetric distribution in perhaps just one direction, but usually still maintaining certain symmetries along other directions or planes. Interestingly, for many multi-cellular organisms, including man, the overall symmetry retained from the beginning of ontogenetic development is bilateral–just one plane of mirror symmetry– from Planaria to humans. The presence of the head-to-tail asymmetry introduces increasingly marked differences among the various areas of the head, middle, or tail regions as the organism develops. The formation of additional borderline phenomena occurs later as cells divide and differentiate thus causing the organism to grow and develop
v. (AN-4.2.) AN4.2
Brown and Higgins, 1981a, showed that certain multiple groupoids equipped with an extra structure called connections were equivalent to another structure called a crossed complex which had already occurred in homotopy theory. such as double, or multiple groupoids (Brown, 2004; 2005). For example, the notion of an atlas of structures should, in principle, apply to a lot of interesting, topological and/or algebraic, structures: groupoids, multiple groupoids, Heyting algebras, $n$-valued logic algebras and $C^{*}$-convolution -algebras. One might incorporate a 3 or 4-valued logic to represent genetic dynamic networks in single-cell organisms such as bacteria. Another example from the ultra-complex system of the human mind is synaesthesia–the case of extreme communication processes between different types of ‘logics’ or different levels of ‘thoughts’/thought processes. The key point here is communication. Hearing has to communicate to sight/vision in some way; this seems to happen in the human brain in the audiovisual (neocortex) and in the Wernicke (W) integrating area in the left-side hemisphere of the brain, that also communicates with the speech centers or the Broca area, also in the left hemisphere. Because of this dual-functional, quasi-symmetry, or more precisely asymmetry of the human brain, it may be useful to represent all two-way communication/signalling pathways in the two brain hemispheres by a double groupoid as an over-simplified groupoid structure that may represent such quasi-symmetry of the two sides of the human brain. In this case, the 300 millions or so of neuronal interconnections in the corpus callosum that link up neural network pathways between the left and the right hemispheres of the brain would be represented by the geometrical connection in the double groupoid. The brain’s overall asymmetric distribution of functions and neural network structure between the two brain hemispheres may therefore require a non-commutative, double–groupoid structure for its relational representation. The potentially interesting question then arises how one would mathematically represent the split-brains that have been neurosurgically generated by cutting just the corpus callosum– some 300 million interconnections in the human brain (Sperry, 1992). It would seem that either a crossed complex of two, or several, groupoids, or indeed a direct product of two groupoids $G_{1}$ and $G_{2}$, $G_{1}\times G_{2}$ might provide some of the simplest representations of the human split-brain. The latter, direct product construction has a certain kind of built-in commutativity: $(a,b)(c,d)=(ac,bd)$, which is a form of the interchange law. In fact, from any two groupoids $G_{1}$ and $G_{2}$ one can construct a double groupoid $G_{1}\Join G_{2}$ whose objects are $\operatorname{Ob}(G_{1})\times\operatorname{Ob}(G_{2})$. The internal groupoid ‘connection’ present in the double groupoid would then represent the remaining basal/‘ancient’ brain connections between the two hemispheres, below the corpum callosum that has been removed by neurosurgery in the split-brain human patients.
The remarkable variability observed in such human subjects both between different subjects and also at different times after the split-brain (bridge-localized) surgery may very well be accounted for by the different possible groupoid representations. It may also be explained by the existence of other, older neural pathways that remain untouched by the neurosurgeon in the split-brain, and which re-learn gradually, in time, to at least partially re-connect the two sides of the human split-brain. The more common health problem –caused by the senescence of the brain– could be approached as a local-to-global, super-complex ageing process represented for example by the patching of a topological double groupoid atlas connecting up many local faulty dynamics in ‘small’ un–repairable regions of the brain neural network, caused for example by tangles, locally blocked arterioles and/or capillaries, and also low local oxygen or nutrient concentrations. The result, as correctly surmised by Rosen (1987), is a global, rather than local, senescence, super-complex dynamic process.
Social Autopoiesis Within a social system the autopoiesis of the various components is a necessary and sufficient condition for realization of the system itself. In this respect, the structure of a society as a particular instance of a social system is determined by the structural framework of the (autopoietic components) and the sum total of collective interactive relations. Consequently, the societal framework is based upon a selection of its component structures in providing a medium in which these components realize their ontogeny. It is just through participation alone that an autopoietic system determines a social system by realizing the relations that are characteristic of that system. The descriptive and causal notions are essentially as follows (Maturana and Varela, 1980, Chapter III):
• (1)
Relations of constitution that determine the components produced constitute the topology in which the autopoiesis is realized.
• (2)
Relations of specificity that determine that the components produced are the specific ones defined by their participation in autopoiesis.
• (3)
Relations of order that determine that the concatenation of the components in the relations of specification, constitution and order are the ones specified by the autopoiesis.
## 0.2. Propagation and Persistence of Organisms through Space and Time. Autopoiesis, Survival and Extinction of Species.
The autopoietic model of Maturana (1987) claims to explain the persistence of living systems in time as the consequence of their structural coupling or adaptation as structure determined systems, and also because of their existence as molecular autopoietic systems with a ‘closed’ network structure. As part of the autopoietic explanation is the ‘structural drift’, presumably facilitating evolutionary changes and speciation. One notes that autopoietic systems may be therefore considered as dynamic realizations of Rosen’s simple MR s. Similar arguments seem to be echoed more recently by Dawkins (2003) who claims to explain the remarkable persistence of biological organisms over geological timescales as the result of their intrinsic, (super-) complex adaptive capabilities.
The point is being often made that it is not the component atoms that are preserved in organisms (and indeed in ‘living fosils’ for geological periods of time), but the structure-function relational pattern, or indeed the associated organismic categories, higher order categories or supercategories. This is a very important point: only the functional organismic structure is ‘immortal’ as it is being conserved and transmitted from one generation to the next. Hence the relevance here, and indeed the great importance of the science of abstract structures and relations, i.e., Mathematics.
This was the feature that appeared paradoxical or puzzling to Erwin Schrödinger from a quantum theoretical point of view when he wrote his book “What is Life?” As individual molecules often interact through multiple quantum interactions, which are most of the time causing irreversible, molecular or energetic changes to occur, how can one then explain the hereditary stability over hundreds of years (or occasionally, a great deal longer, NAs) within the same genealogy of a family of men? The answer is that the ‘actors change but the play does not!’. The atoms and molecules turn-over, and not infrequently, but the structure-function patterns/organismic categories remain unchanged/are conserved over long periods of time through repeated repairs and replacements of the molecular parts that need repairing, as long as the organism lives. Such stable patterns of relations are, at least in principle, amenable to logical and mathematical representation without tearing apart the living system. In fact, looking at this remarkable persistence of certain gene subnetworks in time and space from the categorical ontology and Darwinian viewpoints, the existence of live ‘fossils’ (e.g., a coelacanth found alive in 1923 to have remained unchanged at great depths in the ocean as a species for 300 million years!) it is not so difficult to explain; one can attribute the rare examples of ‘live fossils’ to the lack of ‘selection pressure in a very stable niche’. Thus, one sees in such exceptions the lack of any adaptation apart from those which have already occurred some 300 million years ago. This is by no means the only long lived species: several species of marine, giant unicellular green algae with complex morphology from a family called the Dasycladales may have persisted as long as 600 million years (Goodwin, 1994), and so on. However, the situation of many other species that emerged through super-complex adaptations–such as the species of Homo sapiens–is quite the opposite, in the sense of marked, super-complex adaptive changes over much shorter time–scales than that of the exceptionally ‘lucky’ coelacanths. Clearly, some species, that were less adaptable, such as the Neanderthals or Homo erectus, became extinct even though many of their functional genes may be still conserved in Homo sapiens, as for example, through comparison with the more distant chimpanzee relative. When comparing the Homo erectus fossils with skeletal remains of modern men one is struck how much closer the former are to modern man than to either the Australopithecus or the chimpanzee (the last two species appear to have quite similar skeletons and skulls, and also their ‘reconstructed’ vocal chords/apparatus would not allow them to speak). Therefore, if the functional genomes of man and chimpanzee overlap by about $98\%$, then the overlap of modern man functional genome would have to be greater than $99\%$ with that of Homo erectus of 1 million years ago, if it somehow could be actually found and measured (but it cannot be, at least not at this point in time). Thus, one would also wonder if another more recent hominin than H. erectus, such as Homo floresiensis– which is estimated to have existed between 74,000 and 18,000 years ago on the now Indonesian island of Flores– may have been capable of human speech. One may thus consider another indicator of intelligence such as the size of region 10 of the dorsomedial prefrontal cortex, which is thought to be associated with the existence of self-awareness; this region 10 is about the same size in H. floresiensis as in modern humans, despite the much smaller overall size of the brain in the former (Falk, D. et al., 2005).
## 0.3. Neuro-Groupoids and Cat-Neurons
Categorical representations on nerve cells in the terminology of Ehresmann and Vanbremeersch (1987,2006) are called ‘categorical neurons’ (or cat–neurons for short). ‘Consciousness loops’ (Edelmann 1989, 1992) and the neuronal workspace of Baars (1988) (see also Baars and Franklin, 2003) are among an assortment of such local models that may be consistent with such local categorical representations. Among other notions, there were proposed several criteria for studying the overall integration of neuronal assemblies into an archetypal core: the cat–neuron resonates as an echo that propagates to target concepts through series of thalamocortical loops in response to the thalamus response to stimuli. Mimicking the neuron signalling through synaptic networks, cat–neurons would interact according tp certain linking procedures (Baianu, 1972), that could be then studied in the context of categorical logic, which in its turn may be applied either to semantic modelling of neural networks (Healy and Caudell, 2006) or possibly the schemata of adaptive resonance theory of Grossberg (1999). For such interactive network systems one would expect global actions and groupoid atlases to play more instrumental roles as possible realizations of various types of multi–agent systems (Bak et al, 2006). Let one be aware however, that such models tend to be reductionist in character, falling somewhere between simple and chaotic (‘complex’) systems. Although useful for the industry of higher level automata and robotics, they are unlikely to address at all the ontological problems of the human mind.
As regards to the role played by quantum events in mental processes the situation is different. Although there can be no reasonable doubt that quantum events do occur in the brain as elsewhere in the material world, there is no substantial, experimental evidence that quantum events are in any way efficacious or relevant for those aspects of brain activity that are correlates of mental activity. Bohm (1990), and Hiley and Pylkkännen (2005) have suggested theories of active information enabling ‘self’ to control brain functions without violating energy conservation laws. Such ideas are relevant to how quantum tunneling is instrumental in controlling the engagement of synaptic exocytosis (Beck and Eccles, 1992) and how the notion of a ‘(dendron) mind field’ (Eccles, 1986) could alter quantum transition probabilities as in the case of synaptic vesicular emission (nevertheless, there are criticisms to this approach as in Wilson, 1999).
Attempting to define consciousness runs into somewhat similar problems to those encountered in attempting to define Life, but in many ways far less ‘tangible’; one can make a long list of important attributes of human consciousness from which one must decide which ones are the essential or primary properties, and which ones are to be derived from such primary attributes in a rational manner.
Kant considered that the internal structure of reasoning, or the ‘pure reason’, was essential to human nature for knowledge of the world but the inexactness of empirical science amounted to limitations on the overall comprehension. At the same time,
Kozma et al. (2004) used network percolation models to analyze phase transitions of dynamic neural systems such as those embedded within segments of neuropil. This idea of neuro-percolation so provides a means of passage via transition states within a neurophysiological hierarchy (viz. levels). But the actual substance of the hierarchy cannot by itself explain the quality of intention. The constitution of the latter may be in part consciousness, but actual neural manifestations, such as for example pain, are clearly not products of a finite state Turing machine (Searle, 1983).
AN5-2 section 5 Point (5a) claims that a system should occupy either a macroscopic or a microscopic space-time region, but a system that comes into birth and dies off extremely rapidly may be considered either a short-lived process, or rather, a ‘resonance’ –an instability rather than a system, although it may have significant effects as in the case of ‘virtual particles’, ‘virtual photons’, etc., as in quantum electrodynamics and chromodynamics. Note also that there are many other, different mathematical definitions of systems, ranging from (systems of) coupled differential equations to operator formulations, semigroups, monoids, topological groupoid dynamic systems and dynamic categories. Clearly, the more useful system definitions include algebraic and/or topological structures rather than simple, discrete structure sets, classes or their categories (cf. Baianu, 1970, and Baianu et al., 2006).
It can be shown that such organizational order must either result in a stable attractor or else it should occupy a stable space-time domain, which is generally expressed in closed systems by the concept of equilibrium. On the other hand,
Quantum theories (QTs) were developed that are just as elegant mathematically as GR, and they were also physically ‘validated’ through numerous, extremely sensitive and carefully designed experiments. However, to date, quantum theories have not yet been extended, or generalized, to a form capable of recovering the results of Einstein’s GR as a quantum field theory over a GR-space-time altered by gravity is not yet available
AUTHORS’ AFFILIATIONS:
Title Supercategorical Approach to Complex Systems, Meta-systems and Ontology Multi-Levels SupercategoricalApproachToComplexSystemsMetasystemsAndOntologyMultiLevels 2013-03-11 19:53:05 2013-03-11 19:53:05 bci1 (20947) (0) 1 bci1 (0) Definition
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# Does the Laplace-Beltrami/surface gradient commute with orthogonal projection? (related to Galerkin method)
Let $\Gamma$ be a $C^k$ $(n-1)$-dimensional hypersurface embedded in $\mathbb{R}^n$. Let $H=L^2(\Gamma)$ and $V=H^1(\Gamma)$.
Suppose that $\{v_j\}$ is a basis for $H$ and $V$ (not necessarily orthogonal).
Let $V_m = \text{span}(v_1, ..., v_m)$.
Define a projection operator $P_m:H \to V_m$ satisfying $$(P_m h - h, v_m) = 0 \qquad\text{for all v_m \in V_m}.$$
Is it true that $$P_m (\nabla_{\Gamma}h) = \nabla_{\Gamma}(P_mh)$$ (on the LHS, the projection operator is applied to each element of the gradient vector) and $$P_m (\Delta_{\Gamma}h) = \Delta_{\Gamma}(P_mh)?$$ Where $\nabla_{\Gamma}$ is the surface gradient (defined by $\nabla_\Gamma f= \nabla f - \nabla f \cdot \nu \nu$ with $\nu$ the unit normal) and $\Delta_\Gamma$ is the Laplace-Beltrami operator. I.e. does the projection operator and the surface gradient operator commute??
-
The surface gradient is a vector field on $\Gamma$, so how do you make sense of it being in $V_m$, a space of functions? Do you use the embedding in some higher dimensional manifold? I guess you want $\Gamma$ to be a hypersurface in $\mathbb{R}^n$, but maybe you could make that clearer. – Ben McKay Feb 14 '14 at 13:25
Thanks for the comments. I edited my post. Yes $\Gamma$ is $(n-1)$-dimensional hypersurface in $\mathbb{R}^n$. So I think everything makes sense. – weasd Feb 14 '14 at 13:41
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# All Classes
Class
Description
A color scale using alpha and gray.
A 2D category, i.e., Key-Value pair, where Key is a String ID and value is a Number.
Definition of a function that maps a Double to a RandomVariable, with the possibility to throw an Exception.
A stile for a given graph specifying color, shape and stroke.
A color scale.
Some utilities for JFreeChart
A named object of type T.
A specification of the number axis, providing specific implementations.
Small convenient wrapper for JFreeChart line plot derived.
Small convenient wrapper for Java FX line plot.
Plots the value of an option under the Black-Scholes model as a function of strike and time-to-maturity.
Plots the value of an option under the Black-Scholes model as a function of strike and time-to-maturity.
Plots the regression estimation of a curve.
Plots the value of a finite difference approximation of the derivative of (exp(x)-x) at 0.
Plots a function and two scatters
Plots the value of an option under the Black-Scholes model as a function of strike and time-to-maturity.
Small convenient wrapper for Java FX line plot.
Plots the value of an option under the Black-Scholes model as a function of strike and time-to-maturity.
Small convenient wrapper for JZY3D derived from the JZY3D SurfaceDemo.
Axis wall
A two dimensional plotable.
Small convenient wrapper for JFreeChart line plot of a stochastic process.
Static factory methods for various plots (used to keep demos short).
Class representing a point in F^2 where F is the set of double floating point numbers.
Plots the regression estimation of a curve.
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Preprint Article Version 1 This version is not peer-reviewed
# Surface Tension and Viscosity of Blood
Version 1 : Received: 3 July 2019 / Approved: 5 July 2019 / Online: 5 July 2019 (05:08:09 CEST)
How to cite: Wesołowski, A.; Młynarczak, A. Surface Tension and Viscosity of Blood. Preprints 2019, 2019070090 (doi: 10.20944/preprints201907.0090.v1). Wesołowski, A.; Młynarczak, A. Surface Tension and Viscosity of Blood. Preprints 2019, 2019070090 (doi: 10.20944/preprints201907.0090.v1).
## Abstract
Blood has certain significant physical properties, which the authors investigated. This paper provides values of surface tension and viscosity of blood at 37 °C, as well as relationship between them plotted on a cross-plot. By applying Hagen- Poiseuille equation and Jurin's law to the experimentally obtained data, the authors calculated with great accuracy viscosity and surface tension of blood. Evaluating both of the properties using described methods required from the authors to know the density of investigated liquid, which was found to be on average $\rho=1063,56 \frac{kg}{m^3}$. The authors took blood samples from 30 healthy subjects and determined aforementioned physical properties. There has been made a distinction in regard of a sex of the blood donor. Results indicate that both surface tension and viscosity are independent from sex of a subject, as well as indicate that there is no correlation between viscosity and surface tension of blood at 37\degree C. Average values of surface tension and viscosity were found to be $\sigma=5,241\pm 0,262 \cdot 10^{-2}\frac{J}{m^2}$ and $\eta=3,352\pm0,360 \cdot 10^{-3} Pa \cdot s$, respectively.
## Subject Areas
surface tension; viscosity; blood
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# American Institute of Mathematical Sciences
July 2013, 33(7): 2829-2859. doi: 10.3934/dcds.2013.33.2829
## Splitting of separatrices in the resonances of nearly integrable Hamiltonian systems of one and a half degrees of freedom
1 School of Mathematics, Institute for Advanced Study, Einstein Drive, Simonyi Hall, Princeton, New Jersey, 08540, United States
Received March 2012 Revised April 2012 Published January 2013
In this paper we consider general nearly integrable analytic Hamiltonian systems of one and a half degrees of freedom which are a trigonometric polynomial in the angular state variable. In the resonances of these systems generically appear hyperbolic periodic orbits. We study the possible transversal intersections of their invariant manifolds, which is exponentially small, and we give an asymptotic formula for the measure of the splitting. We see that its asymptotic first order is of the form $K \varepsilon^{\beta} \text{e}^{-a/\varepsilon}$ and we identify the constants $K,\beta,a$ in terms of the system features. We compare our results with the classical Melnikov Theory and we show that, typically, in the resonances of nearly integrable systems Melnikov Theory fails to predict correctly the constants $K$ and $\beta$ involved in the formula.
Citation: Marcel Guardia. Splitting of separatrices in the resonances of nearly integrable Hamiltonian systems of one and a half degrees of freedom. Discrete & Continuous Dynamical Systems - A, 2013, 33 (7) : 2829-2859. doi: 10.3934/dcds.2013.33.2829
##### References:
show all references
##### References:
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2019 Impact Factor: 1.338
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# How do you factor 125 - 25h?
There is a $\textcolor{b l u e}{\text{ common factor " " of 25 in both terms }}$
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# Fight Finance
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If housing rents are constrained from growing more than the maximum target inflation rate, and houses can be priced as a perpetuity of growing net rental cash flows, then what is the implication for house prices, all things remaining equal? Select the most correct answer.
Background: Since 1990, many central banks across the world have become 'inflation targeters'. They have adopted a policy of trying to keep inflation in a predictable narrow range, with the hope of encouraging long-term lending to fund more investment and maintain higher GDP growth.
Australia's central bank, the Reserve Bank of Australia (RBA), has specifically stated their inflation target range is between 2 and 3% pa.
Some Australian residential property market commentators suggest that because rental costs comprise a large part of the Australian consumer price index (CPI), rent costs across the nation cannot significantly exceed the maximum inflation target range of 3% pa without the prices of other goods growing by less than the target range for long periods, which is unlikely.
An investor owns a whole level of an old office building which is currently worth $1 million. There are three mutually exclusive projects that can be started by the investor. The office building level can be: • Rented out to a tenant for one year at$0.1m paid immediately, and then sold for $0.99m in one year. • Refurbished into more modern commercial office rooms at a cost of$1m now, and then sold for $2.3m when the refurbishment is finished in one year. • Converted into residential apartments at a cost of$2m now, and then sold for $3.4m when the conversion is finished in one year. All of the development projects have the same risk so the required return of each is 10% pa. The table below shows the estimated cash flows and internal rates of returns (IRR's). Mutually Exclusive Projects Project Cash flownow ($) Cash flow inone year ($) IRR(% pa) Rent then sell as is -900,000 990,000 10 Refurbishment into modern offices -2,000,000 2,400,000 20 Conversion into residential apartments -3,000,000 3,400,000 13.33 Which project should the investor accept? Risk-free government bonds that have coupon rates greater than their yields: A 'fully amortising' loan can also be called a: For an asset price to double every 10 years, what must be the expected future capital return, given as an effective annual rate? Which of the following statements about the capital and income returns of a 25 year fully amortising loan asset is correct? Assume that the yield curve (which shows total returns over different maturities) is flat and is not expected to change. Over the 25 years from issuance to maturity, a fully amortising loan's expected annual effective: A firm pays out all of its earnings as dividends. Because of this, the firm has no real growth in earnings, dividends or stock price since there is no re-investment back into the firm to buy new assets and make higher earnings. The dividend discount model is suitable to value this company. The firm's revenues and costs are expected to increase by inflation in the foreseeable future. The firm has no debt. It operates in the services industry and has few physical assets so there is negligible depreciation expense and negligible net working capital required. Which of the following statements about this firm's PE ratio is NOT correct? The PE ratio should: Note: The inverse of x is 1/x. Suppose the Australian cash rate is expected to be 8.15% pa and the US federal funds rate is expected to be 3.00% pa over the next 2 years, both given as nominal effective annual rates. The current exchange rate is at parity, so 1 USD = 1 AUD. What is the implied 2 year forward foreign exchange rate? For a price of$95, Sherylanne will sell you a share which is expected to pay its first dividend of $10 in 7 years (t=7), and will continue to pay the same$10 dividend every year after that forever.
The required return of the stock is 10% pa.
Would you like to the share or politely ?
For a price of $95, Nicole will sell you a 10 year bond paying semi-annual coupons of 8% pa. The face value of the bond is$100. Other bonds with the same risk, maturity and coupon characteristics trade at a yield of 8% pa.
Would you like to the bond or politely ?
A person is thinking about borrowing $100 from the bank at 7% pa and investing it in shares with an expected return of 10% pa. One year later the person will sell the shares and pay back the loan in full. Both the loan and the shares are fairly priced. What is the Net Present Value (NPV) of this one year investment? Note that you are asked to find the present value ($V_0$), not the value in one year ($V_1$). The US government recently announced that subsidies for fresh milk producers will be gradually phased out over the next year. Newspapers say that there are expectations of a 40% increase in the spot price of fresh milk over the next year. Option prices on fresh milk trading on the Chicago Mercantile Exchange (CME) reflect expectations of this 40% increase in spot prices over the next year. Similarly to the rest of the market, you believe that prices will rise by 40% over the next year. What option trades are likely to be profitable, or to be more specific, result in a positive Net Present Value (NPV)? Assume that: • Only the spot price is expected to increase and there is no change in expected volatility or other variables that affect option prices. • No taxes, transaction costs, information asymmetry, bid-ask spreads or other market frictions. This annuity formula $\dfrac{C_1}{r}\left(1-\dfrac{1}{(1+r)^3} \right)$ is equivalent to which of the following formulas? Note the 3. In the below formulas, $C_t$ is a cash flow at time t. All of the cash flows are equal, but paid at different times. A stock just paid its annual dividend of$9. The share price is $60. The required return of the stock is 10% pa as an effective annual rate. What is the implied growth rate of the dividend per year? Your friend is trying to find the net present value of a project. The project is expected to last for just one year with: • a negative cash flow of -$1 million initially (t=0), and
• a positive cash flow of $1.1 million in one year (t=1). The project has a total required return of 10% pa due to its moderate level of undiversifiable risk. Your friend is aware of the importance of opportunity costs and the time value of money, but he is unsure of how to find the NPV of the project. He knows that the opportunity cost of investing the$1m in the project is the expected gain from investing the money in shares instead. Like the project, shares also have an expected return of 10% since they have moderate undiversifiable risk. This opportunity cost is $0.1m $(=1m \times 10\%)$ which occurs in one year (t=1). He knows that the time value of money should be accounted for, and this can be done by finding the present value of the cash flows in one year. Your friend has listed a few different ways to find the NPV which are written down below. (I) $-1m + \dfrac{1.1m}{(1+0.1)^1}$ (II) $-1m + \dfrac{1.1m}{(1+0.1)^1} - \dfrac{1m}{(1+0.1)^1} \times 0.1$ (III) $-1m + \dfrac{1.1m}{(1+0.1)^1} - \dfrac{1.1m}{(1+0.1)^1} \times 0.1$ (IV) $-1m + 1.1m - \dfrac{1.1m}{(1+0.1)^1} \times 0.1$ (V) $-1m + 1.1m - 1.1m \times 0.1$ Which of the above calculations give the correct NPV? Select the most correct answer. What is the net present value (NPV) of undertaking a full-time Australian undergraduate business degree as an Australian citizen? Only include the cash flows over the duration of the degree, ignore any benefits or costs of the degree after it's completed. Assume the following: • The degree takes 3 years to complete and all students pass all subjects. • There are 2 semesters per year and 4 subjects per semester. • University fees per subject per semester are$1,277, paid at the start of each semester. Fees are expected to stay constant for the next 3 years.
• There are 52 weeks per year.
• The first semester is just about to start (t=0). The first semester lasts for 19 weeks (t=0 to 19).
• The second semester starts immediately afterwards (t=19) and lasts for another 19 weeks (t=19 to 38).
• The summer holidays begin after the second semester ends and last for 14 weeks (t=38 to 52). Then the first semester begins the next year, and so on.
• Working full time at the grocery store instead of studying full-time pays $20/hr and you can work 35 hours per week. Wages are paid at the end of each week. • Full-time students can work full-time during the summer holiday at the grocery store for the same rate of$20/hr for 35 hours per week. Wages are paid at the end of each week.
• The discount rate is 9.8% pa. All rates and cash flows are real. Inflation is expected to be 3% pa. All rates are effective annual.
The NPV of costs from undertaking the university degree is:
Why is Capital Expenditure (CapEx) subtracted in the Cash Flow From Assets (CFFA) formula?
$$CFFA=NI+Depr-CapEx - \Delta NWC+IntExp$$
A company increases the proportion of debt funding it uses to finance its assets by issuing bonds and using the cash to repurchase stock, leaving assets unchanged.
Ignoring the costs of financial distress, which of the following statements is NOT correct:
Over the next year, the management of an unlevered company plans to:
• Achieve firm free cash flow (FFCF or CFFA) of $1m. • Pay dividends of$1.8m
• Complete a $1.3m share buy-back. • Spend$0.8m on new buildings without buying or selling any other fixed assets. This capital expenditure is included in the CFFA figure quoted above.
Assume that:
• All amounts are received and paid at the end of the year so you can ignore the time value of money.
• The firm has sufficient retained profits to pay the dividend and complete the buy back.
• The firm plans to run a very tight ship, with no excess cash above operating requirements currently or over the next year.
How much new equity financing will the company need? In other words, what is the value of new shares that will need to be issued?
Your friend just bought a house for $400,000. He financed it using a$320,000 mortgage loan and a deposit of $80,000. In the context of residential housing and mortgages, the 'equity' tied up in the value of a person's house is the value of the house less the value of the mortgage. So the initial equity your friend has in his house is$80,000. Let this amount be E, let the value of the mortgage be D and the value of the house be V. So $V=D+E$.
If house prices suddenly fall by 10%, what would be your friend's percentage change in equity (E)? Assume that the value of the mortgage is unchanged and that no income (rent) was received from the house during the short time over which house prices fell.
Remember:
$$r_{0\rightarrow1}=\frac{p_1-p_0+c_1}{p_0}$$
where $r_{0-1}$ is the return (percentage change) of an asset with price $p_0$ initially, $p_1$ one period later, and paying a cash flow of $c_1$ at time $t=1$.
One year ago you bought $100,000 of shares partly funded using a margin loan. The margin loan size was$70,000 and the other 30,000 was your own wealth or 'equity' in the share assets. The interest rate on the margin loan was 7.84% pa. Over the year, the shares produced a dividend yield of 4% pa and a capital gain of 5% pa. What was the total return on your wealth? Ignore taxes, assume that all cash flows (interest payments and dividends) were paid and received at the end of the year, and all rates above are effective annual rates. Hint: Remember that wealth in this context is your equity (E) in the house asset (V = D+E) which is funded by the loan (D) and your deposit or equity (E). There are many ways to calculate a firm's free cash flow (FFCF), also called cash flow from assets (CFFA). Some include the annual interest tax shield in the cash flow and some do not. Which of the below FFCF formulas include the interest tax shield in the cash flow? $$(1) \quad FFCF=NI + Depr - CapEx -ΔNWC + IntExp$$ $$(2) \quad FFCF=NI + Depr - CapEx -ΔNWC + IntExp.(1-t_c)$$ $$(3) \quad FFCF=EBIT.(1-t_c )+ Depr- CapEx -ΔNWC+IntExp.t_c$$ $$(4) \quad FFCF=EBIT.(1-t_c) + Depr- CapEx -ΔNWC$$ $$(5) \quad FFCF=EBITDA.(1-t_c )+Depr.t_c- CapEx -ΔNWC+IntExp.t_c$$ $$(6) \quad FFCF=EBITDA.(1-t_c )+Depr.t_c- CapEx -ΔNWC$$ $$(7) \quad FFCF=EBIT-Tax + Depr - CapEx -ΔNWC$$ $$(8) \quad FFCF=EBIT-Tax + Depr - CapEx -ΔNWC-IntExp.t_c$$ $$(9) \quad FFCF=EBITDA-Tax - CapEx -ΔNWC$$ $$(10) \quad FFCF=EBITDA-Tax - CapEx -ΔNWC-IntExp.t_c$$ The formulas for net income (NI also called earnings), EBIT and EBITDA are given below. Assume that depreciation and amortisation are both represented by 'Depr' and that 'FC' represents fixed costs such as rent. $$NI=(Rev - COGS - Depr - FC - IntExp).(1-t_c )$$ $$EBIT=Rev - COGS - FC-Depr$$ $$EBITDA=Rev - COGS - FC$$ $$Tax =(Rev - COGS - Depr - FC - IntExp).t_c= \dfrac{NI.t_c}{1-t_c}$$ One method for calculating a firm's free cash flow (FFCF, or CFFA) is to ignore interest expense. That is, pretend that interest expense $(IntExp)$ is zero: \begin{aligned} FFCF &= (Rev - COGS - Depr - FC - IntExp)(1-t_c) + Depr - CapEx -\Delta NWC + IntExp \\ &= (Rev - COGS - Depr - FC - 0)(1-t_c) + Depr - CapEx -\Delta NWC - 0\\ \end{aligned} Does this annual FFCF with zero interest expense or the annual interest tax shield? A company issues a large amount of bonds to raise money for new projects of similar risk to the company's existing projects. The net present value (NPV) of the new projects is positive but small. Assume a classical tax system. Which statement is NOT correct? Question 99 capital structure, interest tax shield, Miller and Modigliani, trade off theory of capital structure A firm changes its capital structure by issuing a large amount of debt and using the funds to repurchase shares. Its assets are unchanged. Assume that: • The firm and individual investors can borrow at the same rate and have the same tax rates. • The firm's debt and shares are fairly priced and the shares are repurchased at the market price, not at a premium. • There are no market frictions relating to debt such as asymmetric information or transaction costs. • Shareholders wealth is measured in terms of utiliity. Shareholders are wealth-maximising and risk-averse. They have a preferred level of overall leverage. Before the firm's capital restructure all shareholders were optimally levered. According to Miller and Modigliani's theory, which statement is correct? A firm has a debt-to-assets ratio of 50%. The firm then issues a large amount of debt to raise money for new projects of similar risk to the company's existing projects. Assume a classical tax system. Which statement is correct? Let the standard deviation of returns for a share per month be $\sigma_\text{monthly}$. What is the formula for the standard deviation of the share's returns per year $(\sigma_\text{yearly})$? Assume that returns are independently and identically distributed (iid) so they have zero auto correlation, meaning that if the return was higher than average today, it does not indicate that the return tomorrow will be higher or lower than average. The total return of any asset can be broken down in different ways. One possible way is to use the dividend discount model (or Gordon growth model): $$p_0 = \frac{c_1}{r_\text{total}-r_\text{capital}}$$ Which, since $c_1/p_0$ is the income return ($r_\text{income}$), can be expressed as: $$r_\text{total}=r_\text{income}+r_\text{capital}$$ So the total return of an asset is the income component plus the capital or price growth component. Another way to break up total return is to use the Capital Asset Pricing Model: $$r_\text{total}=r_\text{f}+β(r_\text{m}- r_\text{f})$$ $$r_\text{total}=r_\text{time value}+r_\text{risk premium}$$ So the risk free rate is the time value of money and the term $β(r_\text{m}- r_\text{f})$ is the compensation for taking on systematic risk. Using the above theory and your general knowledge, which of the below equations, if any, are correct? (I) $r_\text{income}=r_\text{time value}$ (II) $r_\text{income}=r_\text{risk premium}$ (III) $r_\text{capital}=r_\text{time value}$ (IV) $r_\text{capital}=r_\text{risk premium}$ (V) $r_\text{income}+r_\text{capital}=r_\text{time value}+r_\text{risk premium}$ Which of the equations are correct? Project Data Project life 1 year Initial investment in equipment8m Depreciation of equipment per year $8m Expected sale price of equipment at end of project 0 Unit sales per year 4m Sale price per unit$10 Variable cost per unit $5 Fixed costs per year, paid at the end of each year$2m Interest expense in first year (at t=1) $0.562m Corporate tax rate 30% Government treasury bond yield 5% Bank loan debt yield 9% Market portfolio return 10% Covariance of levered equity returns with market 0.32 Variance of market portfolio returns 0.16 Firm's and project's debt-to-equity ratio 50% Notes 1. Due to the project, current assets will increase by$6m now (t=0) and fall by $6m at the end (t=1). Current liabilities will not be affected. Assumptions • The debt-to-equity ratio will be kept constant throughout the life of the project. The amount of interest expense at the end of each period has been correctly calculated to maintain this constant debt-to-equity ratio. • Millions are represented by 'm'. • All cash flows occur at the start or end of the year as appropriate, not in the middle or throughout the year. • All rates and cash flows are real. The inflation rate is 2% pa. All rates are given as effective annual rates. • The project is undertaken by a firm, not an individual. What is the net present value (NPV) of the project? A man inherits$500,000 worth of shares.
He believes that by learning the secrets of trading, keeping up with the financial news and doing complex trend analysis with charts that he can quit his job and become a self-employed day trader in the equities markets.
What is the expected gain from doing this over the first year? Measure the net gain in wealth received at the end of this first year due to the decision to become a day trader. Assume the following:
• He earns $60,000 pa in his current job, paid in a lump sum at the end of each year. • He enjoys examining share price graphs and day trading just as much as he enjoys his current job. • Stock markets are weak form and semi-strong form efficient. • He has no inside information. • He makes 1 trade every day and there are 250 trading days in the year. Trading costs are$20 per trade. His broker invoices him for the trading costs at the end of the year.
• The shares that he currently owns and the shares that he intends to trade have the same level of systematic risk as the market portfolio.
• The market portfolio's expected return is 10% pa.
Measure the net gain over the first year as an expected wealth increase at the end of the year.
A company advertises an investment costing $1,000 which they say is underpriced. They say that it has an expected total return of 15% pa, but a required return of only 10% pa. Assume that there are no dividend payments so the entire 15% total return is all capital return. Assuming that the company's statements are correct, what is the NPV of buying the investment if the 15% return lasts for the next 100 years (t=0 to 100), then reverts to 10% pa after that time? Also, what is the NPV of the investment if the 15% return lasts forever? In both cases, assume that the required return of 10% remains constant. All returns are given as effective annual rates. The answer choices below are given in the same order (15% for 100 years, and 15% forever): An 'interest payment' is the same thing as a 'coupon payment'. or ? Currently, a mining company has a share price of$6 and pays constant annual dividends of $0.50. The next dividend will be paid in 1 year. Suddenly and unexpectedly the mining company announces that due to higher than expected profits, all of these windfall profits will be paid as a special dividend of$0.30 in 1 year.
If investors believe that the windfall profits and dividend is a one-off event, what will be the new share price? If investors believe that the additional dividend is actually permanent and will continue to be paid, what will be the new share price? Assume that the required return on equity is unchanged. Choose from the following, where the first share price includes the one-off increase in earnings and dividends for the first year only $(P_\text{0 one-off})$ , and the second assumes that the increase is permanent $(P_\text{0 permanent})$:
Note: When a firm makes excess profits they sometimes pay them out as special dividends. Special dividends are just like ordinary dividends but they are one-off and investors do not expect them to continue, unlike ordinary dividends which are expected to persist.
A fairly priced unlevered firm plans to pay a dividend of $1 next year (t=1) which is expected to grow by 3% pa every year after that. The firm's required return on equity is 8% pa. The firm is thinking about reducing its future dividend payments by 10% so that it can use the extra cash to invest in more projects which are expected to return 8% pa, and have the same risk as the existing projects. Therefore, next year's dividend will be$0.90.
What will be the stock's new annual capital return (proportional increase in price per year) if the change in payout policy goes ahead?
Assume that payout policy is irrelevant to firm value and that all rates are effective annual rates.
If the AUD appreciates against the USD, the European terms quote of the AUD will or ?
Investors expect the Reserve Bank of Australia (RBA) to keep the policy rate steady at their next meeting.
Then unexpectedly, the RBA announce that they will increase the policy rate by 25 basis points due to fears that the economy is growing too fast and that inflation will be above their target rate of 2 to 3 per cent.
What do you expect to happen to Australia's exchange rate in the short term? The Australian dollar is likely to:
In the 1997 Asian financial crisis many countries' exchange rates depreciated rapidly against the US dollar (USD). The Thai, Indonesian, Malaysian, Korean and Filipino currencies were severely affected. The below graph shows these Asian countries' currencies in USD per one unit of their currency, indexed to 100 in June 1997.
Of the statements below, which is NOT correct? The Asian countries':
The Chinese government attempts to fix its exchange rate against the US dollar and at the same time use monetary policy to fix its interest rate at a set level.
To be able to fix its exchange rate and interest rate in this way, what does the Chinese government actually do?
1. Adopts capital controls to prevent financial arbitrage by private firms and individuals.
2. Adopts the same interest rate (monetary policy) as the United States.
3. Fixes inflation so that the domestic real interest rate is equal to the United States' real interest rate.
Which of the above statements is or are true?
The accounting identity states that the book value of a company's assets (A) equals its liabilities (L) plus owners equity (OE), so A = L + OE.
The finance version states that the market value of a company's assets (V) equals the market value of its debt (D) plus equity (E), so V = D + E.
Therefore a business's assets can be seen as a portfolio of the debt and equity that fund the assets.
Let $\sigma_\text{V total}^2$ be the total variance of returns on assets, $\sigma_\text{V syst}^2$ be the systematic variance of returns on assets, and $\sigma_\text{V idio}^2$ be the idiosyncratic variance of returns on assets, and $\rho_\text{D idio, E idio}$ be the correlation between the idiosyncratic returns on debt and equity.
Which of the following equations is NOT correct?
Which of the following statements about risk free government bonds is NOT correct?
Hint: Total return can be broken into income and capital returns as follows:
\begin{aligned} r_\text{total} &= \frac{c_1}{p_0} + \frac{p_1-p_0}{p_0} \\ &= r_\text{income} + r_\text{capital} \end{aligned}
The capital return is the growth rate of the price.
The income return is the periodic cash flow. For a bond this is the coupon payment.
Bonds X and Y are issued by different companies, but they both pay a semi-annual coupon of 10% pa and they have the same face value ($100) and maturity (3 years). The only difference is that bond X and Y's yields are 8 and 12% pa respectively. Which of the following statements is true? A project's Profitability Index (PI) is less than 1. Select the most correct statement: An investor bought two fixed-coupon bonds issued by the same company, a zero-coupon bond and a 7% pa semi-annual coupon bond. Both bonds have a face value of$1,000, mature in 10 years, and had a yield at the time of purchase of 8% pa.
A few years later, yields fell to 6% pa. Which of the following statements is correct? Note that a capital gain is an increase in price.
The following equation is the Dividend Discount Model, also known as the 'Gordon Growth Model' or the 'Perpetuity with growth' equation.
$$p_0 = \frac{d_1}{r - g}$$
Which expression is NOT equal to the expected dividend yield?
The following is the Dividend Discount Model used to price stocks:
$$p_0=\frac{d_1}{r-g}$$
All rates are effective annual rates and the cash flows ($d_1$) are received every year. Note that the r and g terms in the above DDM could also be labelled as below: $$r = r_{\text{total, 0}\rightarrow\text{1yr, eff 1yr}}$$ $$g = r_{\text{capital, 0}\rightarrow\text{1yr, eff 1yr}}$$ Which of the following statements is NOT correct?
A bank grants a borrower an interest-only residential mortgage loan with a very large 50% deposit and a nominal interest rate of 6% that is not expected to change. Assume that inflation is expected to be a constant 2% pa over the life of the loan. Ignore credit risk.
From the bank's point of view, what is the long term expected nominal capital return of the loan asset?
Interest expense on debt is tax-deductible, but dividend payments on equity are not. or ?
A bathroom and plumbing supplies shop offers credit to its customers. Customers are given 60 days to pay for their goods, but if they pay within 7 days they will get a 2% discount.
What is the effective interest rate implicit in the discount being offered? Assume 365 days in a year and that all customers pay on either the 7th day or the 60th day. All rates given in this question are effective annual rates.
For certain shares, the forward-looking Price-Earnings Ratio ($P_0/EPS_1$) is equal to the inverse of the share's total expected return ($1/r_\text{total}$).
For what shares is this true?
Assume:
• The general accounting definition of 'payout ratio' which is dividends per share (DPS) divided by earnings per share (EPS).
• All cash flows, earnings and rates are real.
You own an apartment which you rent out as an investment property.
What is the price of the apartment using discounted cash flow (DCF, same as NPV) valuation?
Assume that:
• You just signed a contract to rent the apartment out to a tenant for the next 12 months at $2,000 per month, payable in advance (at the start of the month, t=0). The tenant is just about to pay you the first$2,000 payment.
• The contract states that monthly rental payments are fixed for 12 months. After the contract ends, you plan to sign another contract but with rental payment increases of 3%. You intend to do this every year.
So rental payments will increase at the start of the 13th month (t=12) to be $2,060 (=2,000(1+0.03)), and then they will be constant for the next 12 months. Rental payments will increase again at the start of the 25th month (t=24) to be$2,121.80 (=2,000(1+0.03)2), and then they will be constant for the next 12 months until the next year, and so on.
• The required return of the apartment is 8.732% pa, given as an effective annual rate.
• Ignore all taxes, maintenance, real estate agent, council and strata fees, periods of vacancy and other costs. Assume that the apartment will last forever and so will the rental payments.
A managed fund charges fees based on the amount of money that you keep with them. The fee is 2% of the end-of-year amount, paid at the end of every year.
This fee is charged regardless of whether the fund makes gains or losses on your money.
The fund offers to invest your money in shares which have an expected return of 10% pa before fees.
You are thinking of investing $100,000 in the fund and keeping it there for 40 years when you plan to retire. How much money do you expect to have in the fund in 40 years? Also, what is the future value of the fees that the fund expects to earn from you? Give both amounts as future values in 40 years. Assume that: • The fund has no private information. • Markets are weak and semi-strong form efficient. • The fund's transaction costs are negligible. • The cost and trouble of investing your money in shares by yourself, without the managed fund, is negligible. • The fund invests its fees in the same companies as it invests your funds in, but with no fees. The below answer choices list your expected wealth in 40 years and then the fund's expected wealth in 40 years. The perpetuity with growth equation is: $$P_0=\dfrac{C_1}{r-g}$$ Which of the following is NOT equal to the expected capital return as an effective annual rate? A share currently worth$100 is expected to pay a constant dividend of $4 for the next 5 years with the first dividend in one year (t=1) and the last in 5 years (t=5). The total required return is 10% pa. What do you expected the share price to be in 5 years, just after the dividend at that time has been paid? A very low-risk stock just paid its semi-annual dividend of$0.14, as it has for the last 5 years. You conservatively estimate that from now on the dividend will fall at a rate of 1% every 6 months.
If the stock currently sells for $3 per share, what must be its required total return as an effective annual rate? If risk free government bonds are trading at a yield of 4% pa, given as an effective annual rate, would you consider buying or selling the stock? The stock's required total return is: You have$100,000 in the bank. The bank pays interest at 10% pa, given as an effective annual rate.
You wish to consume an equal amount now (t=0), in one year (t=1) and in two years (t=2), and still have $50,000 in the bank after that (t=2). How much can you consume at each time? A levered firm has zero-coupon bonds which mature in one year and have a combined face value of$9.9m.
Investors are risk-neutral and therefore all debt and equity holders demand the same required return of 10% pa.
In one year the firm's assets will be worth:
• $13.2m with probability 0.5 in the good state of the world, or •$6.6m with probability 0.5 in the bad state of the world.
A new project presents itself which requires an investment of $2m and will provide a certain cash flow of$3.3m in one year.
The firm doesn't have any excess cash to make the initial $2m investment, but the funds can be raised from shareholders through a fairly priced rights issue. Ignore all transaction costs. Should shareholders vote to proceed with the project and equity raising? What will be the gain in shareholder wealth if they decide to proceed? On his 20th birthday, a man makes a resolution. He will deposit$30 into a bank account at the end of every month starting from now, which is the start of the month. So the first payment will be in one month. He will write in his will that when he dies the money in the account should be given to charity.
The bank account pays interest at 6% pa compounding monthly, which is not expected to change.
If the man lives for another 60 years, how much money will be in the bank account if he dies just after making his last (720th) payment?
A company has:
• 10 million common shares outstanding, each trading at a price of $90. • 1 million preferred shares which have a face (or par) value of$100 and pay a constant dividend of 9% of par. They currently trade at a price of $120 each. • Debentures that have a total face value of$60,000,000 and a yield to maturity of 6% per annum. They are publicly traded and their market price is equal to 90% of their face value.
• The risk-free rate is 5% and the market return is 10%.
• Market analysts estimate that the company's common stock has a beta of 1.2. The corporate tax rate is 30%.
What is the company's after-tax Weighted Average Cost of Capital (WACC)? Assume a classical tax system.
An industrial chicken farmer grows chickens for their meat. Chickens:
1. Cost $0.50 each to buy as chicks. They are bought on the day they’re born, at t=0. 2. Grow at a rate of$0.70 worth of meat per chicken per week for the first 6 weeks (t=0 to t=6).
3. Grow at a rate of $0.40 worth of meat per chicken per week for the next 4 weeks (t=6 to t=10) since they’re older and grow more slowly. 4. Feed costs are$0.30 per chicken per week for their whole life. Chicken feed is bought and fed to the chickens once per week at the beginning of the week. So the first amount of feed bought for a chicken at t=0 costs $0.30, and so on. 5. Can be slaughtered (killed for their meat) and sold at no cost at the end of the week. The price received for the chicken is their total value of meat (note that the chicken grows fast then slow, see above). The required return of the chicken farm is 0.5% given as an effective weekly rate. Ignore taxes and the fixed costs of the factory. Ignore the chicken’s welfare and other environmental and ethical concerns. Find the equivalent weekly cash flow of slaughtering a chicken at 6 weeks and at 10 weeks so the farmer can figure out the best time to slaughter his chickens. The choices below are given in the same order, 6 and 10 weeks. You're about to buy a car. These are the cash flows of the two different cars that you can buy: • You can buy an old car for$5,000 now, for which you will have to buy $90 of fuel at the end of each week from the date of purchase. The old car will last for 3 years, at which point you will sell the old car for$500.
• Or you can buy a new car for $14,000 now for which you will have to buy$50 of fuel at the end of each week from the date of purchase. The new car will last for 4 years, at which point you will sell the new car for $1,000. Bank interest rates are 10% pa, given as an effective annual rate. Assume that there are exactly 52 weeks in a year. Ignore taxes and environmental and pollution factors. Should you buy the or the ? You're trying to save enough money to buy your first car which costs$2,500. You can save $100 at the end of each month starting from now. You currently have no money at all. You just opened a bank account with an interest rate of 6% pa payable monthly. How many months will it take to save enough money to buy the car? Assume that the price of the car will stay the same over time. You just borrowed$400,000 in the form of a 25 year interest-only mortgage with monthly payments of $3,000 per month. The interest rate is 9% pa which is not expected to change. You actually plan to pay more than the required interest payment. You plan to pay$3,300 in mortgage payments every month, which your mortgage lender allows. These extra payments will reduce the principal and the minimum interest payment required each month.
At the maturity of the mortgage, what will be the principal? That is, after the last (300th) interest payment of $3,300 in 25 years, how much will be owing on the mortgage? A company has: • 50 million shares outstanding. • The market price of one share is currently$6.
• The risk-free rate is 5% and the market return is 10%.
• Market analysts believe that the company's ordinary shares have a beta of 2.
• The company has 1 million preferred stock which have a face (or par) value of $100 and pay a constant dividend of 10% of par. They currently trade for$80 each.
• The company's debentures are publicly traded and their market price is equal to 90% of their face value.
• The debentures have a total face value of $60,000,000 and the current yield to maturity of corporate debentures is 10% per annum. The corporate tax rate is 30%. What is the company's after-tax weighted average cost of capital (WACC)? Assume a classical tax system. Find Trademark Corporation's Cash Flow From Assets (CFFA), also known as Free Cash Flow to the Firm (FCFF), over the year ending 30th June 2013. Trademark Corp Income Statement for year ending 30th June 2013$m Sales 100 COGS 25 Operating expense 5 Depreciation 20 Interest expense 20 Income before tax 30 Tax at 30% 9 Net income 21
Trademark Corp Balance Sheet as at 30th June 2013 2012 $m$m Assets Current assets 120 80 PPE Cost 150 140 Accumul. depr. 60 40 Carrying amount 90 100 Total assets 210 180 Liabilities Current liabilities 75 65 Non-current liabilities 75 55 Owners' equity Retained earnings 10 10 Contributed equity 50 50 Total L and OE 210 180
Note: all figures are given in millions of dollars ($m). A new company's Firm Free Cash Flow (FFCF, same as CFFA) is forecast in the graph below. To value the firm's assets, the terminal value needs to be calculated using the perpetuity with growth formula: $$V_{\text{terminal, }t-1} = \dfrac{FFCF_{\text{terminal, }t}}{r-g}$$ Which point corresponds to the best time to calculate the terminal value? Project Data Project life 2 yrs Initial investment in equipment$600k Depreciation of equipment per year $250k Expected sale price of equipment at end of project$200k Revenue per job $12k Variable cost per job$4k Quantity of jobs per year 120 Fixed costs per year, paid at the end of each year $100k Interest expense in first year (at t=1)$16.091k Interest expense in second year (at t=2) $9.711k Tax rate 30% Government treasury bond yield 5% Bank loan debt yield 6% Levered cost of equity 12.5% Market portfolio return 10% Beta of assets 1.24 Beta of levered equity 1.5 Firm's and project's debt-to-equity ratio 25% Notes 1. The project will require an immediate purchase of$50k of inventory, which will all be sold at cost when the project ends. Current liabilities are negligible so they can be ignored.
Assumptions
• The debt-to-equity ratio will be kept constant throughout the life of the project. The amount of interest expense at the end of each period has been correctly calculated to maintain this constant debt-to-equity ratio. Note that interest expense is different in each year.
• Thousands are represented by 'k' (kilo).
• All cash flows occur at the start or end of the year as appropriate, not in the middle or throughout the year.
• All rates and cash flows are nominal. The inflation rate is 2% pa.
• All rates are given as effective annual rates.
• The 50% capital gains tax discount is not available since the project is undertaken by a firm, not an individual.
What is the net present value (NPV) of the project?
A levered firm has a market value of assets of $10m. Its debt is all comprised of zero-coupon bonds which mature in one year and have a combined face value of$9.9m.
Investors are risk-neutral and therefore all debt and equity holders demand the same required return of 10% pa.
Therefore the current market capitalisation of debt $(D_0)$ is $9m and equity $(E_0)$ is$1m.
A new project presents itself which requires an investment of $2m and will provide a: •$6.6m cash flow with probability 0.5 in the good state of the world, and a
• -$4.4m (notice the negative sign) cash flow with probability 0.5 in the bad state of the world. The project can be funded using the company's excess cash, no debt or equity raisings are required. What would be the new market capitalisation of equity $(E_\text{0, with project})$ if shareholders vote to proceed with the project, and therefore should shareholders proceed with the project? In Germany, nominal yields on semi-annual coupon paying Government Bonds with 2 years until maturity are currently 0.04% pa. The inflation rate is currently 1.4% pa, given as an APR compounding per quarter. The inflation rate is not expected to change over the next 2 years. What is the real yield on these bonds, given as an APR compounding every 6 months? Alice, Bob, Chris and Delta are traders in the futures market. The following trades occur over a single day in a newly-opened equity index future that matures in one year which the exchange just made available. 1. Alice buys a future from Bob. 2. Chris buys a future from Delta. 3. Delta buys a future from Bob. These were the only trades made in this equity index future. What was the trading volume and what is the open interest? A European put option will mature in $T$ years with a strike price of $K$ dollars. The underlying asset has a price of $S$ dollars. What is an expression for the payoff at maturity $(f_T)$ in dollars from owning (being long) the put option? Which of the following statements about Australian franking credits is NOT correct? Franking credits: A company conducts a 2 for 3 rights issue at a subscription price of$8 when the pre-announcement stock price was $9. Assume that all investors use their rights to buy those extra shares. What is the percentage increase in the stock price and the number of shares outstanding? The answers are given in the same order. Mr Blue, Miss Red and Mrs Green are people with different utility functions. Each person has$500 of initial wealth. A coin toss game is offered to each person at a casino where the player can win or lose $500. Each player can flip a coin and if they flip heads, they receive$500. If they flip tails then they will lose \$500. Which of the following statements is NOT correct?
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# Definition:Commutative Semiring
Let $\left({S, \circ, *}\right)$ be an additive semiring such that the distributor operation $*$ is a commutative operation.
Then $\left({S, \circ}\right)$ is a commutative semiring.
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# Figure Shows a Metallic Square Frame of Edge A In a Vertical Plane. a Uniform Magnetic Field B Exists in the Space in a Direction Perpendicular to the Plane of the Figure. - Physics
Sum
Figure shows a metallic square frame of edge a in a vertical plane. A uniform magnetic field B exists in the space in a direction perpendicular to the plane of the figure. Two boys pull the opposite corners of the square to deform it into a rhombus. They start pulling the corners at t = 0 and displace the corners at a uniform speed u. (a) Find the induced emf in the frame at the instant when the angles at these corners reduce to 60°. (b) Find the induced current in the frame at this instant if the total resistance of the frame is R. (c) Find the total charge which flows through a side of the frame by the time the square is deformed into a straight line.
#### Solution
(a) The effective length of each side is the length that is perpendicular to the velocity of the corners.
Thus, the effective length of each side is a sin θ.
Net effective length for four sides = 4 × a/2 = 2a
∴ Induced emf = Bvl = 2Bau
(b) Current in the frame is given by
$i= \frac{e}{R} = \frac{2auB}{R}$
(c) Total charge q, which flows through the sides of the frame, is given by
$q = \frac{∆ \phi}{R}$
Here,
ΔΦ = Change in the flux
R = Resistance of the coil
$\therefore q = \frac{∆ \phi}{R}$
$= \frac{B( a^2 - 0)}{R}$
$= \frac{B a^2}{R}$
Concept: Induced Emf and Current
Is there an error in this question or solution?
#### APPEARS IN
HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 16 Electromagnetic Induction
Q 13 | Page 306
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# Independence of 2 variables derived from dice rolls
A while ago, I was told that if you roll a d%, by rolling 2 ten sided dice and treating one as the tens digit and the other as the ones, then you can gain 2 independently distributed values by flipping which is 10s digit and which is ones.
I want to prove or disprove this, It feels like it shouldn't be true.
Ie: for X, Y independent and Uniformly discretely distributed, on 0,1,2..,9 Show, that for random variables A=10X+Y B=10Y+X A and B are Independent (or show the converse)
So it is clear that:
$P_X (x) = P_Y (y)=1/10$
$P_XY (x,y) = 1/100$ (from independence)
$P_A (a) = P_B(b) = 1/100$
Now I must show that $P_AB(a,b)=1/10000=P_A(a)\times P_B(b)$ to show independence. Just not to sure where to start with this, I tried to use the convolution rule for sums of random variables, but I must have made a mistake since i ened up with $P_AB(a,b)=1$
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Two events $A,B$ are independent if and only if $P(A|B) = P(A)$.
Clearly for a given number $x$, $P(A = x | B ) = 1$ if and only if $B$ is number that has the same (but reversed) digits as $x$. Otherwise, $P(A = x | B ) = 0$.
So, $A$ and $B$ cannot be independent.
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Ah, I was over thinking it. I should have been looking for a explicit counter example. Rather than trying to calculate a distribution – oxinabox Oct 10 '12 at 6:11
@oxinabox : Happens to us all the time. Glad I could help. Cheers~ :) – Legendre Oct 10 '12 at 6:16
I don't understand your notation, so I'll use my own.
You can calculate the probability that $A=42$, right?
You can also calculate the probability that $A=42$, given that $B=17$, right?
Doesn't that tell you what you need to know about independence of $A$ and $B$?
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mne.simulation.simulate_raw¶
mne.simulation.simulate_raw(info, stc=None, trans=None, src=None, bem=None, head_pos=None, mindist=1.0, interp='cos2', n_jobs=1, use_cps=True, forward=None, first_samp=0, max_iter=10000, verbose=None)[source]
Simulate raw data.
Head movements can optionally be simulated using the head_pos parameter.
Parameters
infoinstance of Info
The channel information to use for simulation.
Changed in version 0.18: Support for mne.Info.
stc
The source estimates to use to simulate data. Each must have the same sample rate as the raw data, and the vertices of all stcs in the iterable must match. Each entry in the iterable can also be a tuple of (SourceEstimate, ndarray) to allow specifying the stim channel (e.g., STI001) data accompany the source estimate. See Notes for details.
Changed in version 0.18: Support for tuple, and iterable of tuple or SourceEstimate.
trans
Either a transformation filename (usually made using mne_analyze) or an info dict (usually opened using read_trans()). If string, an ending of fif or fif.gz will be assumed to be in FIF format, any other ending will be assumed to be a text file with a 4x4 transformation matrix (like the –trans MNE-C option). If trans is None, an identity transform will be used.
srcstr | instance of SourceSpaces | None
Source space corresponding to the stc. If string, should be a source space filename. Can also be an instance of loaded or generated SourceSpaces. Can be None if forward is provided.
bem
BEM solution corresponding to the stc. If string, should be a BEM solution filename (e.g., “sample-5120-5120-5120-bem-sol.fif”). Can be None if forward is provided.
Name of the position estimates file. Should be in the format of the files produced by MaxFilter. If dict, keys should be the time points and entries should be 4x4 dev_head_t matrices. If None, the original head position (from info['dev_head_t']) will be used. If tuple, should have the same format as data returned by head_pos_to_trans_rot_t. If array, should be of the form returned by mne.chpi.read_head_pos(). See for example [1].
mindistfloat
Minimum distance between sources and the inner skull boundary to use during forward calculation.
interpstr
Either ‘hann’, ‘cos2’ (default), ‘linear’, or ‘zero’, the type of forward-solution interpolation to use between forward solutions at different head positions.
n_jobsint
The number of jobs to run in parallel (default 1). Requires the joblib package.
use_cpsbool
Whether to use cortical patch statistics to define normal orientations for surfaces (default True).
forwardinstance of Forward | None
The forward operator to use. If None (default) it will be computed using bem, trans, and src. If not None, bem, trans, and src are ignored.
New in version 0.17.
first_sampint
The first_samp property in the output Raw instance.
New in version 0.18.
max_iterint
The maximum number of STC iterations to allow. This is a sanity parameter to prevent accidental blowups.
New in version 0.18.
verbose
If not None, override default verbose level (see mne.verbose() and Logging documentation for more).
Returns
rawinstance of Raw
The simulated raw file.
Notes
Stim channel encoding
By default, the stimulus channel will have the head position number (starting at 1) stored in the trigger channel (if available) at the t=0 point in each repetition of the stc. If stc is a tuple of (SourceEstimate, ndarray) the array values will be placed in the stim channel aligned with the mne.SourceEstimate.
Data simulation
In the most advanced case where stc is an iterable of tuples the output will be concatenated in time as:
Data alignment and stim channel encoding
Channel
Data
M/EEG
fwd @ stc[0][0].data
fwd @ stc[1][0].data
...
STIM
stc[0][1]
stc[1][1]
...
time →
New in version 0.10.0.
References
1
Larson E, Taulu S (2017). “The Importance of Properly Compensating for Head Movements During MEG Acquisition Across Different Age Groups.” Brain Topogr 30:172–181
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Policies.Posterior.Beta module¶
Manipulate posteriors of Bernoulli/Beta experiments.
Rewards not in $${0, 1}$$ are handled with a trick, see bernoulliBinarization(), with a “random binarization”, cf., [Agrawal12] (algorithm 2). When reward $$r_t \in [0, 1]$$ is observed, the player receives the result of a Bernoulli sample of average $$r_t$$: $$r_t \sim \mathrm{Bernoulli}(r_t)$$ so it is well in $${0, 1}$$.
Agrawal12
http://jmlr.org/proceedings/papers/v23/agrawal12/agrawal12.pdf
Policies.Posterior.Beta.random() → x in the interval [0, 1).
Policies.Posterior.Beta.betavariate()
beta(a, b, size=None)
Draw samples from a Beta distribution.
The Beta distribution is a special case of the Dirichlet distribution, and is related to the Gamma distribution. It has the probability distribution function
$f(x; a,b) = \frac{1}{B(\alpha, \beta)} x^{\alpha - 1} (1 - x)^{\beta - 1},$
where the normalization, B, is the beta function,
$B(\alpha, \beta) = \int_0^1 t^{\alpha - 1} (1 - t)^{\beta - 1} dt.$
It is often seen in Bayesian inference and order statistics.
afloat or array_like of floats
Alpha, positive (>0).
bfloat or array_like of floats
Beta, positive (>0).
sizeint or tuple of ints, optional
Output shape. If the given shape is, e.g., (m, n, k), then m * n * k samples are drawn. If size is None (default), a single value is returned if a and b are both scalars. Otherwise, np.broadcast(a, b).size samples are drawn.
outndarray or scalar
Drawn samples from the parameterized beta distribution.
Policies.Posterior.Beta.bernoulliBinarization(r_t)[source]
Return a (random) binarization of a reward $$r_t$$, in the continuous interval $$[0, 1]$$ as an observation in discrete $${0, 1}$$.
• Useful to allow to use a Beta posterior for non-Bernoulli experiments,
• That way, Thompson sampling can be used for any continuous-valued bounded rewards.
Examples:
>>> import random
>>> random.seed(0)
>>> bernoulliBinarization(0.3)
1
>>> bernoulliBinarization(0.3)
0
>>> bernoulliBinarization(0.3)
0
>>> bernoulliBinarization(0.3)
0
>>> bernoulliBinarization(0.9)
1
>>> bernoulliBinarization(0.9)
1
>>> bernoulliBinarization(0.9)
1
>>> bernoulliBinarization(0.9)
0
class Policies.Posterior.Beta.Beta(a=1, b=1)[source]
Manipulate posteriors of Bernoulli/Beta experiments.
__init__(a=1, b=1)[source]
Create a Beta posterior $$\mathrm{Beta}(\alpha, \beta)$$ with no observation, i.e., $$\alpha = 1$$ and $$\beta = 1$$ by default.
N = None
List of two parameters [a, b]
__str__()[source]
Return str(self).
reset(a=None, b=None)[source]
Reset alpha and beta, both to 1 as when creating a new default Beta.
sample()[source]
Get a random sample from the Beta posterior (using numpy.random.betavariate()).
• Used only by Thompson Sampling and AdBandits so far.
quantile(p)[source]
Return the p quantile of the Beta posterior (using scipy.stats.btdtri()).
• Used only by BayesUCB and AdBandits so far.
mean()[source]
Compute the mean of the Beta posterior (should be useless).
forget(obs)[source]
Forget the last observation.
update(obs)[source]
• If obs is 1, update $$\alpha$$ the count of positive observations,
• If it is 0, update $$\beta$$ the count of negative observations.
Otherwise, a trick with bernoulliBinarization() has to be used.
__module__ = 'Policies.Posterior.Beta'
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# zbMATH — the first resource for mathematics
## Kateb, Djalil M.
Compute Distance To:
Author ID: kateb.djalil-m Published as: Kateb, D.; Kateb, Djalil; Kateb, Djalil M.
Documents Indexed: 29 Publications since 1993
all top 5
#### Co-Authors
2 single-authored 7 Dambrine, Marc 4 Drouiche, Karim 3 Afraites, Lekbir 3 Caubet, Fabien 3 Lemarié-Rieusset, Pierre Gilles 3 Seghier, Abdellatif 2 Le Louër, Frédérique 1 Ben Belgacem, Faker 1 Boichu, Daniel 1 Eppler, Karsten 1 Noiret, Céline 1 Robin, Vincent 1 Teyssière, Gilles 1 Timimoun, Chahnaz Zakia
all top 5
#### Serials
2 Inverse Problems 2 IEEE Transactions on Signal Processing 2 Comptes Rendus de l’Académie des Sciences. Série I 2 European Series in Applied and Industrial Mathematics (ESAIM): Control, Optimization and Calculus of Variations 1 Mathematische Nachrichten 1 Numerical Functional Analysis and Optimization 1 Proceedings of the American Mathematical Society 1 Revista Matemática Iberoamericana 1 Asymptotic Analysis 1 M$$^3$$AS. Mathematical Models & Methods in Applied Sciences 1 Journal of Elasticity 1 Applied and Computational Harmonic Analysis 1 Advances in Computational Mathematics 1 Journal of Mathematical Study 1 Discrete and Continuous Dynamical Systems. Series B 1 Inverse Problems and Imaging
all top 5
#### Fields
9 Calculus of variations and optimal control; optimization (49-XX) 7 Harmonic analysis on Euclidean spaces (42-XX) 6 Numerical analysis (65-XX) 4 Operator theory (47-XX) 3 Ordinary differential equations (34-XX) 3 Partial differential equations (35-XX) 3 Functional analysis (46-XX) 2 Functions of a complex variable (30-XX) 2 Approximations and expansions (41-XX) 2 Information and communication theory, circuits (94-XX) 1 Linear and multilinear algebra; matrix theory (15-XX) 1 Statistics (62-XX) 1 Mechanics of deformable solids (74-XX) 1 Fluid mechanics (76-XX) 1 Systems theory; control (93-XX)
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# ICHEP 2016 Chicago
3-10 August 2016
Chicago IL USA
US/Central timezone
There is a live webcast for this event.
## Identification of boosted hadronically decaying W bosons and top quarks using the ATLAS detector
Aug 6, 2016, 6:00 PM
2h
Riverwalk A/B
### Riverwalk A/B
Poster Top Quark and Electroweak Physics
### Speaker
ATLAS Collaboration (CERN)
### Description
At the LHC, massive hadronically decaying Standard Model particles, such as the W boson and the top quark, can be produced with high transverse momenta much larger than their mass. This will lead to increased collimation of the decay products in the direction of the boosted parent particle. ATLAS has commissioned and implemented jet substructure techniques to reconstruct and identify hadronically decaying W bosons and top quarks while rejecting backgrounds from light quarks and gluons. In Run1, the performance of these techniques, evaluated from Monte Carlo simulation, was validated with data. For Run2, these techniques are further optimised and their expected performance are studied. Several physics analyses with hadronically decaying W boson and top quark final states have employed these techniques to increase the sensitivity of their analyses.
### Primary author
ATLAS Collaboration (CERN)
### Presentation materials
There are no materials yet.
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# Hypervolume of n-d simplex in an n+1 space
Hello,
This is my first time asking a question on this site so please let me know if I'm doing it wrong.
I have been trying to find out how to compute the hypervolume of an n-d simplex in an n+1 space.
I have found how to find the hypervolume of an n-d simplex in an n-d sapce, but don't know how to do it if the simplex is in an n+1 d space link
That link would show me how to compute the area of a triangle described in a 2-d space. I would need to know how to compute the area of a triangle described in a 3-d space.
Thanks guys
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I voted to close, but then could not refrain from answering, so should somehow retract the close vote... – Igor Rivin Jun 15 '12 at 20:48
I am a bit conflicted, since this is not a research level question. However, the answer is nice:
1. Move one of the vertices to the origin.
2. if the remaining vertices are now $v_1, \dots, v_n$ find, by solving a linear system, the vector $v_0$ orthogonal to all of the $v_i.$ Normalize so that the norm of $v_0$ equals $1.$
3. Find the volume of the simplex with vertices $0, v_0, \dots, v_n.$
4. Multiply by $n+1.$
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I did not know about the research level bit(I've now read the faq). Thanks for answering anyway. Would you mind telling me where I can look up why this works (I'm not a mathematician as I'm sure you've already guessed)? – edgar Jun 16 '12 at 4:20
It works, because the volume of the $n$-simplex is $1/n$ the height times the base. In your case the simplex in step 3 has your degenerate simplex as the base, and has height $1$ by the normalization... – Igor Rivin Jun 16 '12 at 4:59
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## Testing ML applications
Image by Gordon Johnson from Pixabay
I’ve recently looked at the applications of different testing techniques for testing ML applications and got interested in the so called metamorphic testing. The idea is that we can check whether an output is within a specific range or set, which is called a metamorphic relation ( https://medium.com/trustableai/testing-ai-with-metamorphic-testing-61d690001f5c).
What is interesting about this paper is that it presents a framework for testing ML applications. I’ve not tried it yet, but I will as it seems very interesting to check how things work with this metamorphic testing and metamorphic relations. I’ve also interested in how to measure the quality of the software in this context.
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Maths
# Functions and graphics
## Basic theoretical information
### Coordinates and basic concepts of functions
The length of the segment on the coordinate axis is the formula:
The length of the segment on the coordinate plane is sought by the formula:
To find the length of a segment in a three-dimensional coordinate system, use the following formula:
The coordinates of the midpoint of the segment (only the first formula is used for the coordinate axis, the first two formulas for the coordinate plane, all three formulas for the three-dimensional coordinate system) are calculated using the formulas:
A function is a correspondence of the form = f ( x ) between variables, by virtue of which each considered value of some variable x (argument or independent variable) corresponds to a specific value of another variable, y (dependent variable, sometimes this value is simply called the value of the function ). Note that the function implies that only one value of the dependent variable y can correspond to one value of the argument x . At the same time, the same value of ycan be obtained for different x .
The domain of the function is all the values of the independent variable (the function argument, usually x ), in which the function is defined, i.e. its meaning exists. Denotes the domain of definition D ( y ). By and large, you are already familiar with this concept. The domain of definition of a function is differently called the domain of permissible values, or DHS, which you have long been able to find.
The range of the function is all possible values of the dependent variable of the function. Denoted by E ( y ).
The function increases on the interval where the larger value of the argument corresponds to the larger value of the function. The function decreases on the interval where the larger value of the argument corresponds to the smaller value of the function.
The intervals of the sign of the constancy of a function are the intervals of the independent variable at which the dependent variable retains its positive or negative sign.
The zeros of a function are those argument values for which the value of the function is zero. At these points, the graph of the function intersects the abscissa axis (OX axis). Very often, the need to find the zeros of a function means simply having to solve the equation. Also often, the need to find gaps in the sign of constancy means simply to solve inequalities.
The function y = f ( x ) is called even if it is defined on a symmetric set and for any x from the domain of definition the equality is true:
This means that for any opposite values of the argument, the values of the even function are equal. The graph of the even function is always symmetric about the ordinate axis of the OS.
The function y = f ( x ) is called odd if it is defined on a symmetric set and for any x from the domain of definition the equality is true:
This means that for any opposite values of the argument, the values of the odd function are also opposite. The graph of an odd function is always symmetric about the origin.
The sum of the roots of even and odd functions (the intersection points of the x-axis OX) is always zero, since for every positive root x there is a negative root – x .
It is important to note: some function does not have to be even or odd. There are many functions that are not even or odd. Such functions are called general functions , and none of the equalities or properties given above are satisfied for them.
### Linear function graph
A linear function is a function that can be given by the formula:
The graph of a linear function is straight and in the general case looks like this (an example is given for the case when k > 0, in this case the function is increasing; for the case of k <0, the function will be decreasing, that is, the line will be tilted in the other direction – from left to right):
### Graph of the quadratic function (Parabola)
The graph of a parabola is given by a quadratic function:
A quadratic function, like any other function, intersects the OX axis at the points that are its roots: ( 1 ; 0) and ( 2 ; 0). If there are no roots, then the quadratic function means that the OX axis does not intersect, if the root is one, then the quadratic function at this point ( 0 ; 0) only touches the OX axis, but does not intersect it. A quadratic function always intersects the OY axis at the point with the coordinates: (0; c ). The graph of a quadratic function (parabola) may look as follows (in the figure, examples that far from exhaust all possible types of parabolas):
Wherein:
• if the coefficient a > 0, in the function y = ax 2 + bx + c , then the branches of the parabola are directed upwards;
• if a <0, then the branches of the parabola are directed downwards.
The coordinates of the vertex of the parabola can be calculated by the following formulas. X vertices ( p – in the figures above) of a parabola (or the point at which the square trinomial reaches its highest or lowest value):
The vertex game ( q – in the figures above) of a parabola or maximum, if the branches of the parabola are directed down ( a <0), or minimal, if the branches of the parabola are directed up ( a > 0), the value of the square triple
### Graphs of other functions
The power function is the function given by the formula:
Here are some examples of graphs of power functions:
Inversely proportional dependence is called a function defined by the formula:
Depending on the sign of the number k, a graph of inversely proportional dependence can have two principal options:
An asymptote is a line to which the line of the graph of the function approaches infinitely close, but does not intersect. The asymptotes for inversely proportional plots shown in the figure above are the coordinate axes to which the function plots infinitely close, but do not intersect them.
The exponential function with the base а is the function given by the formula:
Depending on whether the number a is greater or less than unity, a graph of an exponential function can have two principal variants (we also give examples, see below):
A logarithmic function is a function defined by the formula:
Depending on whether the number a is greater or less than one, the graph of the logarithmic function can have two principal variants:
Graph of the function y = | x | as follows:
### Graphs of periodic (trigonometric) functions
The function y = f ( x ) is called periodic if there is such a non-zero, a number T such that f ( x + T ) = f ( x ), for any x in the domain of the function f ( x ). If the function f ( x ) is periodic with a period T , then the function:
where: A , k , b are constant numbers, and k is not equal to zero, also periodic with a period of 1 , which is defined by the formula:
Most examples of periodic functions are trigonometric functions. We present the graphs of the main trigonometric functions. The following figure shows a part of the graph of the function y = sin x (the entire graph continues indefinitely left and right), the graph of the function y = sin x is called a sinusoid :
The graph of the function y = cos x is called the cosine curve . This graph is shown in the following figure. Since the sine graph, it continues indefinitely along the OX axis to the left and to the right:
The graph of the function y = tg x is called the tangenzoid . This graph is shown in the following figure. Like the graphs of other periodic functions, this graph repeats indefinitely far along the OX axis to the left and to the right.
Finally, the graph of the function y = ctg x is called the cotangent . This graph is shown in the following figure. As well as graphs of other periodic and trigonometric functions, this graph repeats indefinitely far along the OX axis to the left and to the right.
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Volume 358 - 36th International Cosmic Ray Conference (ICRC2019) - GRI - Gamma Ray Indirect
Proving the outstanding capabilities of Imaging Atmospheric Cherenkov Telescopes in high time resolution optical astronomy
T. Hassan,* M. Daniel
*corresponding author
Full text: pdf
Pre-published on: 2019 July 22
Published on:
Abstract
Imaging Atmospheric Cherenkov Telescopes (IACTs) are very-large telescopes designed to detect the nanosecond-timescale flashes produced within extended air showers. Because IACTs are sensitive to the Cherenkov light (UV/blue) and use photodetectors with extremely fast time responses, they are also able to perform simultaneous optical observations. The large reflecting areas of these telescopes (larger than 100 m$^2$) makes them well-suited to studying fast optical transient phenomena with timescales ranging from seconds to milliseconds to nanoseconds, and the unique optical design provides a wide field of view monitoring capability with a modest point spread function. VERITAS, with its recently upgraded PMT current monitoring instrumentation, was able to provide the first detection of asteroid occultations with an IACT, resulting in the highest angular resolution measurements for stellar diameters ever taken in the visible band range. Here we explore the feasibility of using this technique to significantly expand the number of stars with directly measured stellar radii, usable for population studies to test stellar evolution modelling or transiting exoplanet radius measurements. A single observatory with a high-speed visible-band photometer with a sensitivity reaching the 13$^{th}$ magnitude could increase the number of directly measured K stars diameters by 50%.
Open Access
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Notes on Rational and Irrational Numbers | Grade 7 > Compulsory Maths > Rational and Irrational Numbers | KULLABS.COM
• Note
• Things to remember
• Videos
• Exercise
• Quiz
Rational Numbers
In general, A number that can be made by dividing two numbers is called rational number. Any numbers which can be expressed in the form of $$\frac{a}{b}$$ where a and b are integers and b ≠ 0 are called rational numbers. The set of rational numbers is denoted by the letter 'Q'. For examples Q = ( $$\frac{1}{2}$$, $$\frac{3}{4}$$, $$\frac{212}{100}$$, $$\frac{20}{12}$$, etc.. )
The set of the rational number is the set that includes the sets of natural numbers (N), whole numbers (W), and integers (Z). Therefore, the sets of natural numbers, whole numbers and integers are the proper subsets of the set of rational number.∴N < Q, W < Q and Z < Q.
Terminating and Non-Terminating rational numbers
When the rational number gets decimalised, the obtained decimal may be terminating or non-terminating decimal. Terminating rational number as decimal generally comes to an end and doesn't repeat after decimal. For example $$\frac{1}{2}$$ = 0.5, $$\frac{1}{4}$$ = 0.25, $$\frac{3}{8}$$ = 0.375 etc.) are the examples of terminating decimal.
Non-Terminating rational number as decimal goes on repeating in a pattern after the decimal. For examples ( $$\frac{1}{3}$$ = 0.333333..., $$\frac{1}{15}$$ = 0.0666666.... etc. are non-terminating rational number as decimal.
Irrational Numbers
The numbers which are not rational and that cannot be made by dividing two integers are called irrational numbers . If the irrational numbers are decimalised, the decimals are non-terminating non-recurring that goes on forever without repeating. For example ( $$\sqrt{5}$$ = 2.2360679775,( $$\sqrt{2}$$ = 1.41423562 etc. are the examples of irrational numbers.
Real Number System
The sets of number system i.e Natural numbers, Whole numbers, Integers, Rational numbers and Irrational numbers are defined under the set of real number system. The following chart shows the real number system :-
• A number that can be made by dividing two numbers is called rational number.
• The numbers which are not rational and that cannot be made by dividing two integers are called irrational numbers .
.
Click on the questions below to reveal the answers
Rational number at the middle of 2 and 3 = $$\frac{1}{2}$$(2+3)=$$\frac{5}{2}$$
Rational number at the middle of 2 and $$\frac{5}{2}$$ = $$\frac{1}{2}$$ (2+$$\frac{5}{2}$$ = $$\frac{1}{2}$$ x $$\frac{9}{2}$$ =$$\frac{9}{4}$$
So , $$\frac{5}{2}$$ and $$\frac{9}{4}$$ are any two rational numbers between 2 and 3.
0%
(sqrt{18})
(sqrt{8})
(sqrt{32})
(sqrt{42})
(sqrt{3})
3
(sqrt{2})
(sqrt{5})
(sqrt{16})
(sqrt{3})
(sqrt{4})
(sqrt{9})
2
16
4
8
• What will be the any two rational numbers between 2 and 3?
(frac{5}{4}), (frac{9}{4})
(frac{5}{2}), (frac{9}{2})
(frac{5}{2}), (frac{9}{4})
(frac{5}{3}), (frac{9}{4})
ASK ANY QUESTION ON Rational and Irrational Numbers
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Question
# If the tangent to the ellipse $$x^2+4y^2=16$$ at the point $$P(\phi)$$ is a normal to the circle $$x^2+y^2-8x-4y=0$$ then $$\phi$$ is equal to
A
π2
B
π4
C
tan112
D
π4
Solution
## The correct option is B $$tan^{-1}\frac{1}{2}$$$$y=mx+\sqrt {16m^2-4}$$$$c=\pm \sqrt {a^2m^2+b^2}$$$$\sqrt {16m^2+4}$$Now it is normal to circle so it passes through centre$$\therefore (4, 2)$$$$2=4m \pm \sqrt {16m^2-4}$$$$1=2m\pm \sqrt {4m^2-1}$$$$(1-2m)^2=(4m^2-1)$$$$4m^2-4m+1=4m^2-1$$$$4m=2, m=\frac {1}{2}$$$$\phi =tan^{-1}\frac {1}{2}$$Maths
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# The Sixth Annual Large Hadron Collider Physics conference LHCP 2018
Jun 4 – 9, 2018
Europe/Zurich timezone
## On the spin correlations of final leptons produced in the annihilation processes $e^+ e^- \rightarrow \mu^+ \mu^-, \tau^+ \tau^-$ and in the high-energy two-photon processes $\gamma \gamma \rightarrow e^+ e^-, \mu^+ \mu^-, \tau^+ \tau^-$
Jun 5, 2018, 4:00 PM
1h 30m
Library, Centro San Domenico
theoretical
### Speaker
Dr Valery Lyuboshitz (Joint Institute for Nuclear Research, Dubna )
### Description
The electromagnetic processes of annihilation of $(e^+ e^-)$ pairs, generated in high-energy nucleus-nucleus and hadron-nucleus collisions, into heavy lepton pairs are theoretically studied in the one-photon approximation, using the technique of helicity amplitudes . For the process $e^+e^- \rightarrow \mu^+\mu^-$, it is shown that -- in the case of the unpolarized electron and positron -- the final muons are also unpolarized but their spins are strongly correlated. For the final $(\mu^+ \mu^-)$ system, the structure of triplet states is analyzed and explicit expressions for the components of the spin density matrix and correlation tensor are derived.
It is demonstrated that here the spin correlations of muons have the purely quantum character, since one of the Bell-type incoherence inequalities for the correlation tensor components is always violated. In doing so, it is also established that, when involving the additional contribution of the weak interaction of lepton neutral currents through the virtual $Z^0$ boson, the qualitative character of the muon spin correlations does not change .
On the other hand, the theoretical investigation of spin structure for the processes of lepton pair production by pairs of photons ( which, in particular, may be emitted in relativistic heavy-ion and hadron-nucleus collisions ) is performed as well. For the two-photon process $\gamma \gamma \rightarrow e^+ e^-$, it is found that -- quite similarly to the process $e^+ e^- \rightarrow \mu^+ \mu^-$ -- in the case of unpolarized photons the final electron and positron remain unpolarized, but their spins prove to be strongly correlated. Explicit expressions for the components of the correlation tensor and for the relative fractions of singlet and triplet states of the final $(e^+ e^-)$ system are derived. Again, here one of the Bell-type incoherence inequalities for the correlation tensor components is always violated and, thus, spin correlations of the electron and positron have the strongly pronounced quantum character.
Analogous considerations can be wholly applied as well, respectively, to the annihilation process $e^+ e^- \rightarrow \tau^+ \tau^-$ and to the two-photon processes $\gamma \gamma \rightarrow \mu^+ \mu^-$, $\gamma \gamma \rightarrow \tau^+ \tau^-$, which become possible at much higher energies.
### Primary author
Dr Valery Lyuboshitz (Joint Institute for Nuclear Research, Dubna )
### Co-author
Dr Vladimir Lyuboshitz (Joint Institute for Nuclear Research, Dubna)
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Hello,
Nowadays, I think we have some classification of integral structure in semistable representation via Liu's $(\varphi, \hat{G})$-modules or via Caruso's $(\varphi, \tau)$-modules. I must say that because of lack of time and motivation, I didn't read their papers, nor the ones by Breuil or Kisin, so I know almost nothing about integral p-adic Hodge theory.
So my question is the following :
given a lattice $T$ in a semistable representation, is there a way to read the Hodge-Tate weights on the corresponding object (namely the associated $(\varphi, \hat{G})$-module) or the $(\varphi, \tau)$-module) ?
-
The Hodge-Tate weights make sense already without the integral structure, so it's unclear to me why you need integral p-adic Hodge theory. In any case, the answer is yes, because you can recover the weakly admissible Fontaine module associated with the semi-stable representation from Liu's $(\varphi,\hat{G})$-module (and probably Caruso's as well). In fact, you only need the $\varphi$ part of it, if I remember correctly. – Keerthi Madapusi Pera Nov 11 '11 at 19:33
I am mostly interested in torsion representations, that's why I only want to consider the integral structure. – user16131 Nov 11 '11 at 19:41
Actually, the Hodge-Tate weights can be read only from $\varphi$ as follows. Let $\mathfrak M$ be the corresponding $(\varphi,\hat G)$-module or $(\varphi,\tau)$-modul. Denote by $\varphi(\mathfrak M)$ the $\mathfrak S$-module generated by the image of $\varphi$. Then, since by definition $E(u)^r \mathfrak M \subset \varphi(\mathfrak M)$ (here $E(u)$ is the minimal polynomial of the fixed uniformizer $\pi$ of $K$ and $r$ is some nonnegative integer), the quotient $\mathfrak M[1/p]/\varphi(\mathfrak M)[1/p]$ is a direct sum of pieces of the form $\mathfrak S[1/p]/E(u)^{n_i}\mathfrak S[1/p]$ for some integers $n_i$. These integers are precisely the Hodge-Tate weights of the representation (or the opposite of them depending on your convention).
The point is that $\varphi^*\mathfrak{M}/E(u)\varphi^*\mathfrak{M}\left[\frac{1}{p}\right]$ is canonically identified with the $(\varphi,N)$-module over $K$ attached to the Galois representation. Under this identification, the Hodge filtration has the description given above. See Prop. 1.2.8 in Kisin's paper 'Crystalline representations and F-crystals'. – Keerthi Madapusi Pera Jun 18 '12 at 1:04
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# The existential theory of the reals
Some definitions of the existential theory of the reals (ETR) allow a real closed field and some definitions allow only rational numbers as coefficients of polynomials. Which one is correct? Will the answer to the question have an effect on the PSPACE proof by J. Canny? For example, ETR is not in PSPACE, if numbers in real closed field are allowed as coefficients?
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For each coefficient, you introduce one variable, an equation stating the variable is a root of the respective polynomial, and something to uniquely pick up the particular root: for example, if the coefficient is specified as the unique root of $f(x)$ in the interval $[a,b]$, you would include the inequalities $x-a\ge0$ and $b-x\ge0$ (which can be combined to $(x-a)(b-x)\ge0$). This is a polynomial-time reduction, the description of the new system has length polynomial in the length of the original input (in effect, it is the same input organized in a different way). ... – Emil Jeřábek Mar 8 '13 at 17:06
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# Generalised Hardy-Ramanujan Numbers
The number 1729 is famously the smallest positive integer expressible as the sum of two positive cubes in two different ways ($1729=1^3+12^3=9^3+10^3$). There is plenty of work on "taxicab numbers" - the smallest sums of cubes in $n$ different ways (which always exist) - Here's Ivars Peterson at MAA And here's another detailed analysis. (Does anyone know anything about the "Bill Butler" referred to in the second article)
However the sequence which caught my attention is OEIS A016078 - 4, 50, 1729, 635318657 which gives the smallest numbers which are sums of positive $n^{th}$ powers in two ways. Is there any more recent work or prospect of identifying such numbers for fifth powers and above? And should they be named as in the title of this post?
[This question arises from a much more frivolous one, which was closed, in which I learned why a $50^{th}$ birthday was special in this particular way].
-
Just a note, there is no upper bound on number of ways to write as the sum of two squares, and those with exactly two ways as $x^2 + y^2, \; 1 \leq x \leq y$ are in a small number of infinite families. For cubes, fourth and fifth powers, it is not clear that any examples with three or more ways are known...wait, P. Vojta 1983 $$15170836545 = 517^3 + 2468^3 = 709^3 + 2456^3 = 1733^3 + 2152^3.$$ Euler $$635318657 = 133^4 + 134^4 = 59^4 + 158^4,$$ no triples known. There are parametric families that give some of these pairs. – Will Jagy Oct 31 '12 at 21:45
oeis.org/A046881 has better comments. – SiliconCelery Oct 31 '12 at 21:46
Alrighty, the thing about the Vojta example with cubes is that $\gcd(x,y) = 1$ for each pair. Hardy and Wright, Theorem 412 on page 333 in the Fifth edition paperback, show that you can get at least $r$ different ways for any $r$ if you do not restrict $\gcd(x,y).$ – Will Jagy Oct 31 '12 at 21:54
@WillJagy Thanks for your comments on this and my previous. I have Hardy and Wright Fifth Edition, and will consult it. – Mark Bennet Oct 31 '12 at 22:08
@SiliconCelery Thanks for that - very helpful. – Mark Bennet Oct 31 '12 at 22:15
Guy, Unsolved Problems In Number Theory, 3rd edition, D1, writes, "... it is not known if there is any nontrivial solution of $a^5+b^5=c^5+d^5$. Dick Lehmer once thought that there might be a solution with a sum of about 25 decimal digits, but a search by Blair Kelly yielded no nontrivial solution with sum $\le1.02\times10^{26}$."
At F30, Guy writes, "... $x^5$ is a likely answer to the following unsolved problem of Erdos. Find a polynomial $P(x)$ such that all the sums $P(a)+P(b)$ ($0\le a\lt b$) are distinct."
The book was published in 2004. I don't know whether there has been any progress since.
-
Wikipedia is also not aware of any progress. – joriki Oct 31 '12 at 22:09
Guy's book is one I should have anyway, and it is missing from my collection. I have always been intrigued, because Hardy & Wright spend a fair amount of time on Waring's Problem (representation of integers as sums of $n^{th}$ powers) - but this seems to have been a bit of dead end - at least in the twentieth century - so far as deep mathematical insight and progress is involved. – Mark Bennet Oct 31 '12 at 22:13
@joriki, true --- and Wikipedia's only citation is Guy's book. – Gerry Myerson Nov 1 '12 at 0:12
Are $a$ and $b$ real and was there any recent progress? – draks ... Nov 26 '13 at 23:10
@draks..., in the Erdos question at F30 in Guy, $a$ and $b$ are meant to be whole numbers. I'm not aware of any recent progress. – Gerry Myerson Nov 27 '13 at 2:35
It is necessary to solve the equation:
$$x^5+y^5+z^5=q^5$$
For integers complex numbers solutions exist. $j=\sqrt{-1}$
Making this change.
$$a=p^2-2ps-s^2$$
$$b=p^2+2ps-s^2$$
$$c=p^2+s^2$$
You can write the solution.
$$x=jc+b$$
$$y=jc-b$$
$$z=a-jc$$
$$q=a+jc$$
$p,s$ - integers.
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# Thread: Stuck on a square root problem
1. ## Stuck on a square root problem
Hi! I am stuck on my homework. I came across a problem I didn't know how to do and there isn't a single example like it in my book or notes. The problem is-
Find where the square root of x-3 over -2x-8 is real
Ashley
2. Originally Posted by aperkin2
Hi! I am stuck on my homework. I came across a problem I didn't know how to do and there isn't a single example like it in my book or notes. The problem is-
Find where the square root of x-3 over -2x-8 is real
Ashley
Hi
$\sqrt{\frac{x-3}{-2x-8}}$ is real when $\frac{x-3}{-2x-8} \geq 0$ which means
( $x-3 \geq 0$ and $-2x-8 > 0$) or ( $x-3 \leq 0$ and $-2x-8 < 0$)
3. Thank you so much for your help! That is starting to sound a little more familiar. Would the next step involve a number line or is there another step before that? Sorry, I'm just very lost here. Thanks for your help!
4. You just need to find :
- the set of x where $x-3 \geq 0$ and $-2x-8 > 0$
- the set of x where $x-3 \leq 0$ and $-2x-8 < 0$
And take the union of both sets. This is the answer you are asked
5. Oh I see! Thank you so much! I never would have figured that out on my own
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July 08, 2020, 06:41:54 AM
Forum Rules: Read This Before Posting
### Topic: Work, dU and dH for vaporization of water (Read 3019 times)
0 Members and 1 Guest are viewing this topic.
#### cseil
• Full Member
• Posts: 131
• Mole Snacks: +4/-0
##### Work, dU and dH for vaporization of water
« on: August 14, 2015, 12:08:49 PM »
Hi,
I'd like to have a confirm about how I solved an exercise.
It asks me to find the work, the variation of H and U during the vaporization of 1 mol of water at 1atm and 100°C. I know that the adsorbed heat during the process (p is constant) is 9706 cal/mol and the book gives me the density of water at 100°C (0.958g/mL).
I firstly calculated the work. The pressure is constant so it's equal to:
$$W=-p \int_{V1}^{V2} dV$$
where V1 is the volume of the liquid water (18.78x10^-3 L) and V2 is the volume of the gas water (30.59L). The work is -740cal.
Talking about the enthalpy, ΔH = q at constant pressure so I've already got the value.
ΔU = ΔH - ΔnRT => ΔU = ΔH.
Is it right?
Thank you
#### mjc123
• Chemist
• Sr. Member
• Posts: 1752
• Mole Snacks: +245/-11
##### Re: Work, dU and dH for vaporization of water
« Reply #1 on: August 14, 2015, 12:32:47 PM »
Quote
ΔU = ΔH - ΔnRT => ΔU = ΔH.
Why does that make ΔU = ΔH? What is Δn in this context?
#### cseil
• Full Member
• Posts: 131
• Mole Snacks: +4/-0
##### Re: Work, dU and dH for vaporization of water
« Reply #2 on: August 14, 2015, 12:34:41 PM »
Isn't it 1? Because there were 0 moles of gas water, now there's one.
Edit: ops, it's ΔH - RT then!
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# Operations Management: Other Operations Management Problems – #230
Question: Consider a capital budgeting example with 5 projects from which to select. Let x1 = 1 if project a is selected, 0 if not, for a = 1, 2, 3, 4, 5. Conditions are independent. Projects cost $100,$200, $150,$75, and $300 respectively. The budget is$450. Write the appropriate constraint for the following condition. If project 3 is chosen, project 4 must be chosen.
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Geometric Mean Statistics, Business Law Subjects, A Bloody Feast Ac Odyssey Clues, Types Of Red Fish, East Fork Campground Best Sites, " />
# simplifying set theory expressions
The simplest expression for the set you're keeping, then, is $A^c\cup Z.$ Now, this assumes that it is possible for some of your people to be on no teams at all. These regions do not intersect--that is, their intersection is empty. It's been a long time since I did set theory and I don't remember my simplification rules. “Question closed” notifications experiment results and graduation, MAINTENANCE WARNING: Possible downtime early morning Dec 2/4/9 UTC (8:30PM…, Proving that the set of natural numbers is well-ordered. You are correct that $Z$ may contain some members of team $A$, but will include all who are members of team $A$ and at least one other team. The laws of set theory can be used to simplify complex expressions involving sets. Simplification of an algebraic expression can be defined as the process of writing an expression in the most efficient and compact form without affecting the value of the original expression. The simplest expression for the set you're keeping, then, is A c ∪ Z. By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. Problem 7TE from Chapter 2: Use Venn diagrams(a) to simplify the expressions (E F)(E F... Get solutions So, if you let $Z$ be the union of all teams except team $A,$ then you have $A\cap Z^c=\left(A^c\cup Z\right)^c$ as the set you're excluding. I found that a method I was hoping to publish is already known. Do other planets and moons share Earth’s mineral diversity? In the examples below, we will practice evaluating some of the expressions from previous examples; in part 1, we will evaluate the form with parentheses, and in part 2 we will evaluate the form we got after distributing. Search for: Simplifying and Evaluating Expressions Using the Distributive Property. Sorry, I think I'm being stupid here, I think $A^c \cup Z$ is actually what I want, I'm just being slow! To learn more, see our tips on writing great answers. Use MathJax to format equations. UNSOLVED! Is it not possible to express this in such a way that I can make: I want to avoid expressing anything in an exhaustive list like: (B ∪ C ∪ D ∪ E ∪ F ∪ G ∪ H) because this would break if a team 'I' were added (perhaps not possible? When we have to simplify algebraic expressions, we can often make the work easier by applying the Commutative or Associative Property first instead of automatically following the order of operations. Simplify algebraic expressions using identity, inverse and zero properties; Identify which property(ies) to use to simplify an algebraic expression; For "thinking that way", I like to consider this as a mirror of logic. Module 6: Set Theory and Logic. This website uses cookies to ensure you get the best experience. Choosing THHN colors when running 2 circuits together. So you are left with (A n B n C) u (A u B'). I have a set of staff members who are grouped into various teams (Team A, Team B, Team C etc ...up to Team H). Variation Theory Sequences and behaviour to enable mathematical thinking in the classroom - by Craig Barton @mrbartonmaths . MathJax reference. Why should they change? Free Set Theory calculator - calculate set theory logical expressions step by step. I have been given a maths problem to do by my lecturer but she hasn't explained how to do it. Pretty sure you just read it left to right. Is it possible to express “any set B composed of any two elements from set A” using set theory notation? ), Well, $B^c\cap C^c\cap\cdots\cap H^c=(B\cup C\cup\cdots\cup H)^c,$ so the set you're excluding is $$A\cap(B\cup C\cup D\cup E\cup F\cup G\cup H)^c.$$. So ABC = things in A, B, and C. ABc = things in A and B, but not C. A v B = AB + Ab + aB(A v B) ^ C = ABC + AbC + aBC[(A v B) ^ C] ^ B = ABBC + ABbC + aBBC = ABC + aBC{[(A v B) ^ C] ^ B} v C = ABC + aBC + ABC + AbC + aBC + abC{[(A v B) ^ C] ^ B} v C = ABC + AbC + aBC + abC = C, New comments cannot be posted and votes cannot be cast, More posts from the cheatatmathhomework community, Continue browsing in r/cheatatmathhomework, Press J to jump to the feed. Why did MacOS Classic choose the colon as a path separator? $$By using this website, you agree to our Cookie Policy. haha ok, I think I'm realising that no matter how well I word the question it not good enough... the issue with this is they could add extra teams in the future and I would have to update that to include e.g. Learning Outcomes. reply from a potential PhD advisor? A First Course in Probability (9th Edition) Edit edition. Now, this assumes that it is possible for some of your people to be on no teams at all. Z = (B \cap C \cap D \cap E \cap F \cap G \cap H) I need to exclude the set of staff that are ONLY in Team A. Like terms can be added or subtracted from one another. == 1 N A == A ... simplify. Module 6: Set Theory and Logic. For the more general question about manipulating set-theoretic expressions, it seems to me that Mathematica is limited to manipulating finite sets that can be represented as lists. To see why the set you tried is empty, make a Venn diagram with two overlapping circles, labeled A and Z. Then A\cap Z is the region contained in both circles, while A\cap Z^c is the region contained in A only. Is whatever I see on the internet temporarily present in the RAM? Please read! Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. A ∩ ( B ∪ C ∪ D ∪ E ∪ F ∪ G ∪ H) c. So, if you let Z be the union of all teams except team A, then you have A ∩ Z c = ( A c ∪ Z) c as the set you're excluding. Number. Given any n+2 integers, show that there exist two of them whose sum, or else whose difference, is divisible by 2n. What's the meaning of "Yo quería que nos pasara / y tú, y tú / lo dejaste pasar" in the song "Nada fue un error" by Coti? Set Theory \A set is a Many that allows itself to be thought of as a One."$$. site design / logo © 2020 Stack Exchange Inc; user contributions licensed under cc by-sa. (Georg Cantor) In the previous chapters, we have often encountered "sets", for example, prime numbers form a set, domains in predicate logic form sets as well. rev 2020.11.24.38066, The best answers are voted up and rise to the top, Mathematics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. Shouldn't some stars behave as black hole? Close. Still figuring out what I'm asking really... All are in at least one team, but since they can be in more than 1 team, going with just Z pulls through the people who are e.g. Search for: Simplifying Expressions Using the Properties of Identities, Inverses, and Zero. It only takes a minute to sign up. So the original expression is equivalent to just A. Regarding plotting Venn diagrams, take a look at the answers to this question: Hot to plot Venn diagrams. Evaluate Expressions Using the Distributive Property. Simplify arbitrary union expression in set theory. The following diagram shows some examples of like terms. Addition is OR, multiplication is AND. And then write the set I want to keep in terms of A and Z? Can the sum of the cardinalities of the intersections of all pairs of sets be expressed as a function of the cardinalities of the individual sets? This is my first time ever seeing these problems, thus finding it hard to understand without an explanation. Posted by 1 year ago. How can I deal with claims of technical difficulties for an online exam? The first part (A n B n C) n (A u B u C') simplifies to just (A n B n C). Each staff member can be in multiple teams (more often the case than not) and additional teams could be added in future. Asking for help, clarification, or responding to other answers. This is not really an answer, but it was too long for a comment. It is a question on simplifying a sets theory expression using the boolean laws. Therefore, you could have written: Example: Simplify the expressions: a) 14x + 5x b) 5y – 13y c) p – 3p. Find the course on: Wordpress: https://utsavm9.wordpress.com/courses/set-theory/ Udemy: Coming soon Eliademy: Coming soon OpenLearning: Coming soon UNSOLVED! I want to avoid that if possible, its better for me to create the equivalent of Z for all future cases by just grouping the teams excluding A. Learn more ... Identities Proving Identities Trig Equations Trig Inequalities Evaluate Functions Simplify.
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# Problems with SPI Management
Hello friends!
1. I use ODrive v3.5-48V
2. Encoders: AS5047P-TS_EK_AB connected by SPI.
3. Firmware: - https://github.com/TobinHall/ODrive/tree/Non-Blocking_Absolute_SPI
Set the settings:
odrv0.axis0.encoder.config.abs_spi_cs_gpio_pin = 4
odrv0.axis0.encoder.config.mode = 257
odrv0.axis0.encoder.config.cpr = 2**14
Encoders:
error = 0x0000 (int)
index_found = False (bool)
count_in_cpr = 4487 (int)
interpolation = 1.0 (float)
phase = 1.8239965438842773 (float)
pos_estimate = 20872.3125 (float)
pos_cpr = 4489.6875 (float)
hall_state = 7 (int)
vel_estimate = -625.0 (float)
pos_abs = 4487 (int)
config:
mode = 257 (int)
use_index = False (bool)
abs_spi_cs_gpio_pin = 4 (int)
pre_calibrated = False (bool)
idx_search_speed = 10.0 (float)
zero_count_on_find_idx = True (bool)
cpr = 16384 (int)
offset = 11282 (int)
offset_float = 0.7106562852859497 (float)
bandwidth = 1000.0 (float)
calib_range = 0.019999999552965164 (float)
ignore_illegal_hall_state = False (bool)
When turned on, the motor vibrates violently and jitter .
I use - move_to_pos and my engine shuts off with errors.
In [85]: hex(odrv0.axis0.motor.error)
Out[85]: '0x10'
In [86]: hex(odrv0.axis0.controller.error)
Out[86]: '0x1'
In [87]: hex(odrv0.axis0.error)
Out[87]: '0x200'
If disconnect <axis>.controller.config.vel_limit_tolerance = 0
Everything is working. But the vibration remains.
For some time the function even works move_to_pos.
But then everything turns off and an error pops up on the sensor.
In [94]: dump_errors(odrv0)
axis0
axis: Error(s):
ERROR_ENCODER_FAILED
motor: no error
encoder: Error(s):
controller: no error
axis1
axis: no error
motor: no error
encoder: no error
controller: no error
In [93]: hex(odrv0.axis0.encoder.error)
Out[93]: '0x80'
Help please. Can eat the ideas?
Can drive vibrations due to ADC rattle?
Maybe there are those who used SPI?
Have you tried tuning your gains?
Tried, it’s unreal. An error: 0x80
Tell me, is this normal?
I use: start_liveplotter(lambda:[odrv0.axis0.encoder.pos_estimate, odrv0.axis0.controller.pos_setpoint])
Oh, wow. I thought you meant the jittering, sorry. Yes, that is very strange behaviour. I have no idea what would cause that.
The sensor is sending the absolute position right? Such a big jump, looks like something is wrong with the signal, or the sensor is picking up noise.
From what I see in the specs you should have 0.063 degrees of noise = \frac{0.063}{(\frac{360}{2^{14}})} = 3 steps, so the jittering seems normal. But that jump of 200 steps is really weird.
See if you can electrically isolate your sensor. Using a plastic spacer or axle or something. To see if the noise goes away. In my case I changed my motor mounts to plastic, after that my other sensors picked up a lot less noise.
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# Ultra-lightweight superconducting wire based on Mg, B, Ti and Al
## Abstract
Actually, MgB2 is the lightest superconducting compound. Its connection with lightweight metals like Ti (as barrier) and Al (as outer sheath) would result in a superconducting wire with the minimal mass. However, pure Al is mechanically soft metal to be used in drawn or rolled composite wires, especially if applied for the outer sheath, where it cannot provide the required densification of the boron powder inside. This study reports on a lightweight MgB2 wire sheathed with aluminum stabilized by nano-sized γ-Al2O3 particles (named HITEMAL) and protected against the reaction with magnesium by Ti diffusion barrier. Electrical and mechanical properties of single-core MgB2/Ti/HITEMAL wire made by internal magnesium diffusion (IMD) into boron were studied at low temperatures. It was found that the ultra-lightweight MgB2 wire exhibited high critical current densities and also tolerances to mechanical stress. This predetermines the potential use of such lightweight superconducting wires for aviation and space applications, and for powerful offshore wind generators, where reducing the mass of the system is required.
## Introduction
Although many superconducting elements and compounds have been discovered1, only few of them can be used for thermally and mechanically stabilized long length wires with high current densities. Instead of high power cables and high field magnets, superconducting wires make possible the design of powerful and lightweight superconducting stators and rotors for aircraft engines and generators2,3. Lightweight superconducting wires are also attractive in other specific areas, where the total mass is extremely important, e.g. powerful wind turbines4,5,6 or any space applications7,8. Since the discovery of superconductivity in the lightest superconductive compound MgB29, extensive efforts have been expended in the development of practical composite wires made mostly by powder-in-tube (PIT) processes and in the enhancement of their superconductive properties, particularly the critical current density (Jc) and the upper critical field (Bc). Low cost MgB2 superconductor wires operated at 4–25 K can lower the upfront and ongoing operational costs of superconducting systems. It was found that sheath materials play an important role in the determination of transport properties of the wires made by powder in tube (PIT) technique10. Cu is an ideal thermally stabilizing metal for low-Tc superconducting wires. In the case of MgB2 wires, the Cu reaction with MgB2 has to be inhibited due to a possible radical reduction of the transport current density. Therefore, a protective (i.e. diffusion) barrier has to be used (e.g. Fe, Nb or Ta) in order to avoid any reaction, namely the one between Cu and Mg11. Ti sheathed MgB2 wires were tested initially by Allessadrini12, and Ti barriers were then successfully applied for multicore MgB2 wires stabilized by Cu13,14. Al may also be an appropriate sheath material for MgB2 superconductor due to its high electrical and thermal conductivity, low cost, magnetism, and good formability. However, pure Al is mechanically soft metal to be used in drawn or rolled composite wires, especially if applied for the outer sheath, where it cannot provide the required densification of the boron powder. While Al alloys can offer improved mechanical properties, the conductivities and formability are markedly deteriorated. Furthermore, the solidus temperatures of Al alloys are much lower compared with the melting point of pure Al, which makes even more difficult the formation of MgB2 by the heat treatment of Mg and B components at ≈650 °C. The first experiment with MgB2/Al tape superconductor was made by an ex-situ PIT method without final heat treatment, but it does not allow reaching high critical current density15. Several other experiments to stabilize MgB2 wire with pure Al were also performed16,17, but the stabilization was not effective enough. Also, another solution with Al bonding on the Ti sheathed wire was not successful due to intensively oxidized surfaces of both Al and Ti18. Thermally stable ultrafine-grained Al stabilized by a small content of nano-scale Al2O3 formed in situ in Al matrix, named HITEMAL (high temperature aluminum), was produced by a powder metallurgy approach19,20. HITEMAL shows attractive mechanical and recently also electrical properties at low temperatures21. The first attempt to make an Al-stabilized MgB2 wire was made using Ta diffusion barrier and HITEMAL outer sheath22, which demonstrated the possible production of Al-sheathed MgB2 wires. It allowed to verify the utilization of Al + Al2O3 outer sheath for MgB2 wire and to show what superconducting properties, especially current densities, can be reached in medium magnetic fields. Ta barrier is really heavy material (16.69 g/cm3) for lightweight composite wire, but it has been used due to minimal reaction with Al + Al2O3 during the final heat treatment.
In this work we present original manufacturing method and properties of ultra-lightweight superconducting wire prepared by internal Mg diffusion process (IMD) into the B, which utilizes the lightest superconducting compound (MgB2 with 2.5 gcm−3) combined with the lightweight composite sheath (Al + 1.37 vol.% Al2O3 with a density of ~2.7 gcm−3) and light metallic barrier material (Ti with a density of 4.5 gcm−3). Due to lowered melting point of Al + 1.37 vol.% Al2O3 ( 652 °C)22 in comparison to pure Al (660 °C) and close to melting of Mg (650 °C), really specific heat treatment is needed for MgB2/Ti/HITEMAL wire. It should allow: (i) the fast formation of dense MgB2 phase23, (ii) limited Ti/Al interaction and (iii) keeping the mechanical strength of Al + Al2O3 sheath.
## Results
It was shown that the melting point of Al + 1.37 vol.% Al2O3 sheath is relatively low ~652 °C22 while the temperature close to 650 °C is required for the fast formation of a dense MgB2 phase23. This could cause undesirable changes (e.g., melting or recrystallization) of the Al2O3 stabilized Al sheath. Therefore, fast ramp heat treatments (~25 °C/min) with the setting temperature 628–635 °C and overshoot up to 640–646.5 °C were applied for as-deformed Mg/B/Ti/HITEMAL wires named as wA, wB and wC, see Table 1 and Fig. 1. A transversal cross-section image of the wB wire is shown in Fig. 2(a), where the central hole (at the place of the original Mg core) and formed MgB2 layer of thickness 100 µm are well visible. It correlates with the kinetic of MgB2 layer formation presented by Li at al. who have calculated the time needed for the for Mg + B reaction24. Our previous experiments have confirmed this model and showed the optimal time of 8 minutes for HT at 635 °C and overshoot 654 °C23. Figure 2(b) shows a thin intermetallic reaction layer with a thickness of 1 µm at the Ti/Al interface of the wB wire subjected to heat treatment (HT) temperature 628 °C/10 min.
The local EDS analysis confirmed Al3Ti phase, which is in good agreement with other studies25. A comparable Al3Ti layer of a similar thickness was observed for the wA and wB wires that were heat treated for 10 min, while for the wC wire that was annealed for 30 min, the layer increased to 4 µm (see Table 1). Formation of the Al3Ti phase at the Ti/Al interface may significantly decrease thermal and electrical transport between the MgB2 core and outer Al + Al2O3 sheath. To minimize the Al3Ti phase formation, a short heat treatment regime with a very fast initial ramp is preferred.
The critical current densities of the compared wires wA-wC were determined from the magnetic loops by using Bean’s critical state model to establish a relationship between the width of the hysteresis loop Δm and the critical current density. Assuming a full penetration of the measured sample by a magnetic field, the particular form of the formulae relating Δm to Jc can be derived with regard to the current flow geometry. In the case of a cylindrical MgB2 core, the critical current density is obtained according to:
$$\begin{array}{c}{J}_{c}=\frac{3}{d}{\rm{\Delta }}M\,{\rm{for}}B\,\mathrm{applied}\,{\rm{parallel}}\,{\rm{to}}\,{\rm{the}}\,{\rm{wire}}\,{\rm{axis}}\,(B||)\,\,{\rm{and}}\\ \,{J}_{c}=\frac{3\pi }{4d}{\rm{\Delta }}M\,{\rm{for}}\,B\,{\rm{applied}}\,{\rm{perpendicular}}\,\mathrm{to}\,\mathrm{the}\,{\rm{wire}}\,{\rm{axis}}\,(B\,\perp \,)\end{array}$$
where ΔM is the width of the hysteresis loop divided by the volume of the MgB2 core, and d is the core diameter. For the wires made by the IMD process resulting in an annular MgB2 core shown in Fig. 1(a), all formulae must be multiplied by a factor considering the hollow core geometry26. The factor is $$(1-\frac{{d}_{I}^{2}}{{d}_{O}^{2}})/(1-\frac{{d}_{I}^{3}}{{d}_{O}^{3}})$$, where dI and do is the inner and outer diameter of the MgB2 core and the formula for Jc is:
$${J}_{c}=\frac{3{\rm{\Delta }}M}{{d}_{O}}\frac{1-\frac{{d}_{I}^{2}}{{d}_{O}^{2}}}{1-\frac{{d}_{I}^{3}}{{d}_{O}^{3}}}$$
Filament dimensions dI and do in wB wire are 0.00423 and 0.00625 cm, respectively. Figure 3(a) shows the critical current densities of the wB wire measured by a vibrating sample magnetometer at external magnetic fields 1–9 T (in B|| and B) and temperatures 5–25 K. Jc(B, T) values of the wire in the perpendicular field are identical with Jc of a single core MgB2/Ti/Cu wire made by the IMD process and annealed at 640 °C/60 min27, which is a result of a sufficiently dense boron powder deformed inside the Al + Al2O3 sheath. The dotted lines present Jc values at a parallel field, which are very similar to the perpendicular one for temperatures 5–15 K, and only slightly lowered at temperatures above 20 K. A small Jc difference between the perpendicular field direction (with currents flowing along the tubular core) and parallel one reflects an excellent homogeneity of the MgB2 compound created by the IMD process. An opposite behavior with large Jc differences between B|| and B was observed in MgB2 made by in-situ PIT process, which was attributed primarily to a texture caused by wire deformation and resulting to different ‘porosity’ or ‘connectivity’ in longitudinal and transversal direction28. Figure 3(b) shows the transport engineering current densities (calculated to the whole cross-section of the wire) at 4.2 K for wA, wB and wC samples. The highest Je is measured for the wA due to the Tmax = 646.5 °C, which is close to the melting point of Mg22. The mechanism of presented IMD process at temperatures below 650 °C considers fast Mg diffusion into boron powder and subsequent creation of MgB2 phase23,24. The resistive transitions of compared MgB2 layers are similar (see the insert in Fig. 1), but the systematic decrease of critical temperature (Tc = 37.00 K - wA, 36.89 K - wB and 36.87 K - wC) and small widening of R(T) transition (ΔTc = 1.90 K - wA, 2.03 K - wB and 2.15 K - wC) can be observed. It reflects the composition and purity of created MgB2 phase and correlates with Je values shown by Fig. 3(b).
Only slightly lowered peak temperature of 642.5 °C for the wB wire resulted in around 10% lower Je compared with that of the wA wire, but Je of wC wire is lowered by 37% at field 6 T in comparison to wA. Figure 3(b) shows also the relation between the transport and magnetic Je (from VSM), for the wB wire, where a falling off of the magnetic Je from the transport one was observed. It can be rationalized considering a different current flow combined with no fully identical connectivity along and across the core axis.
Changes in the critical currents of present wires subjected to axial tension at 4.2 K are shown in Fig. 4. Due to a larger thermal contraction of Ti and Al compared with that of MgB2, cooling down to 4.2 K results in compression stress which acts on the MgB2 layer and reduces the critical temperature and current1,29. Applied axial tension partially compensates the pressure stress and consequently critical current increases up to a level of irreversible strain (εirr), where the breaking of brittle MgB2 leads to a radical degradation. The εirr value defines the maximum strain at which the current still remains reversible29. One can see a considerable effect of applied heat treatment on the irreversible strain in Figure 4. The wA wire with the highest peak temperature of 646.5 °C behaves mechanically as the weakest, and consequently the lowest εirr = 0.166% is measured due to the apparent softening of the outer sheath.
However, the wC wire annealed at the peak temperature of 640 °C has the highest εirr = 0.342%, which is even comparable with the strain limit of a single-core IMD wire reinforced with a GlidCop sheath, see the filled circles in Fig. 4. GlidCop is dispersion strengthened copper which was already effectively used for some MgB2 wires23. Table 1 shows the irreversible strain εirr and irreversible stress σirr measured for wA-wC, which are correlating well with the sheath microhardness HV0.005 – the lowest for wA 43 GPa and the highest for wC 68 GPa. It was already shown that strain and stress tolerances (σirr and εirr) of MgB2 wires are dominantly affected by mechanical strength of the outer sheath30. Therefore, structural changes in outer sheath of wires wA-wC were examined by transmission electron microscopy.
The as-deformed Al + Al2O3 consists of Al grains intensively elongated in the wire drawing direction and transversal structure shows a randomly distributed nanometric Al2O3 dispersoids, see Fig. 5(a). The nano-dispersoids stemmed from native amorphous (am)-Al2O3 layers on as-atomised Al powders20. The induced shear deformation broke up the Al2O3 layers into am-Al2O3 platelets during the cold working steps21, and some remnants of the fractured am-Al2O3 platelets remained at high angle grain boundaries, see the white arrow in Fig. 5(a). However, a majority of the am-Al2O3 platelets transformed into nanometric crystalline Al2O3 dispersoids during the cold working were found at both, the Al grain boundaries and within the Al grain interiors, see the black arrow in Fig. 5(a). During the final heat treatment Al grains coarsening is observed. High angle grain boundaries are preferentially eliminated with increasing temperature, but low angle grain boundaries are still stabilized by Al2O3 dispersoids and are sustained even higher annealing temperatures, see Fig. 5(b). The black arrows show co-localization of low angle grain boundaries with Al2O3 dispersoids in the wA wire sheath.
Figure 6 shows TEM bright field images of transversal sections of the Al + Al2O3 sheath in the heat-treated wires wA - wC (a-c) in comparison to as-deformed one shown by Fig. 6(d). The TEM micrographs demonstrate the different microstructure upon annealing with the peak temperature between 640 °C and 646.5 °C. While Al grains of as-deformed wire have generally equiangular shape of averaged size dav 470 nm, enlarged and/or elongated (not equiangular) grains are visible in wB and wC wires due to grains coarsening. High angle grain boundaries of Al + Al2O3 sheaths are yet well stabilized by Al2O3 dispersoids at heat treatment temperature Tmax = 640 °C, see Fig. 6(c), where nearly doubled grain size dav 950 nm in comparison to as-deformed sheath is found. The outer sheath of wC wire stays polycrystalline with the structure similar to the as-deformed one, see Fig. 6(d). The grain size structure of wB sheath shown by Fig. 6(b) is more affected by annealing only 10 °C below the melting of Al + Al2O3 and dav 1380 nm was estimated for Tmax = 642.5 °C. Figure 6(a) shows that grain boundaries in wA (5.5 °C bellow the melting of Al + Al2O3) are not more stabilized and a big Al grains with sub-grains and low angle grain boundaries with localized Al2O3 dispersoids are present. Consequently, the correct estimation of dav for wA wire is not possible. Observed structural changes and grains coarsening leads to mechanical softening of heat treated Al + Al2O3 sheaths, which is accompanied by the decreased sheath micro-hardness (see Table 1) in comparison to not annealed Al + Al2O3 wire with HV0.005 70 GPa22.
## Discussion
The presented microstructural study clearly illustrates that the different Al grain structure of the Al + Al2O3 sheaths strongly affects the wire responses to axial tension. The Al + Al2O3 is a suitable material for a sufficiently strong superconducting wire, but conditions of the final heat treatment have to be controlled very precisely. Figure 4 shows the different strain tolerances, which are strongly affected by the applied annealing influencing the sheath microstructure (see Figs 5 and 6). While the apparent critical current degradation in the wA wire occurred at the tensile stress of 141 MPa due to not more stabilized grain boundaries by Al2O3 dispersoids, the wC wire is able to withstand much higher stress of 214 MPa. Due to polycrystalline structure and grain size dav 950 nm, the mechanical strength of the wC wire was by ~25% higher than determined for the wA wire, which is characterized by big grains with sub-grains and low angle grain boundaries. The averaged grain size of wB sheath (Tmax = 642.5 °C) is 1380 nm, which is larger than for wC and consequently σirr = 172 MPa is measured, see Table 1. Similar correlation (sheath softening) is observed by the micro-hardness (HV0.005) data, which decreased from HV0.005 = 68 to 43 as the peak temperature increased from 640 to 646.5 °C, respectively (see Table 1). Nevertheless, the achieved HV0.005 = 43 for the Al-Al2O3 sheath of the wA wire remains still much higher than that of pure Al (HV0.005 = 27)21 due to a dense net of low angle grain boundaries effectively stabilized by Al2O3 dispersoids. The observed differences are attributed only to structural changes in the Al + Al2O3 material (and formation of thicker Al3Ti layer) subjected to different heat treatment. Therefore, precisely chosen heat treatment has to be applied to form a high current density MgB2 core along with a high strength Al + Al2O3 sheath and Ti diffusion barrier with a limited interfacial reaction at the sheath interface.
Calculation of conductor mass based on the MgB2/Ti/Al + Al2O3 can lead to at least 2.5 times weight reduction when compared with a typical MgB2/Nb/Cu wire of the same cross-sectional dimensions. This clearly outlines the potential of the lightest MgB2/Ti/Al + Al2O3 superconducting wire, when compared with any other metallic or ceramic superconductors. Consequently, presented MgB2 wire meets demanding requirements on electrical and mechanical properties of superconductors for efficient superconductive and light-weight applications.
## Methods
A single-core MgB2 wire was fabricated by internal magnesium diffusion (IMD) into a boron process. Pure Mg99.99% rod 2.9 mm in diameter was precisely positioned in the central axis of a Ti99.99% tube with 5.5 mm inner diameter and 7.2 mm outer diameter. The free volume between the Ti tube and Mg rod was filled by a B99.8% powder (<1 µm size) in a glove-box under pure Argon atmosphere. The Mg/B/Ti composite was rotary swaged down to 6.2 mm diameter, cleaned, and inserted into a HITEMAL tube with 6.3 mm inner diameter and 9.1 mm outer diameter, that was machined from an as-extruded Al + 1.37 vol.% Al2O3 rod21. The Mg/B/Ti/Al + Al2O3 composite rod was rotary swaged down to 7.5 mm and then groove rolled to a rectangular wire with a cross-section of 1.02 × 1.02 mm2. A heat treatment process was applied at 300 oC for 30 min during the groove rolling process each time after reaching around 50% area reduction. The volume composition of the as-deformed wires corresponded to around 11% Mg, 12% B, 27% Ti, and 50% Al + Al2O3 outer sheath. The following final heat treatment was applied under Ar atmosphere at: (i) 632 °C for 10 min (wA wire); (ii) 628 °C for 10 min (wB wire); and (iii) 635 °C for 30 min (wC wire), with the peak temperatures of 646.5, 641.5 and 640 °C, respectively, see Fig. 1 and Table 1.
Hysteresis loops measured by a vibrating sample magnetometer (VSM) option in PPMS of Quantum Design system were recorded between −2 and +9 T with a constant field sweep of 6.3 mT/s in a temperature range of 5–25 K (at 5 K steps), and the field directed perpendicular and parallel to the wire axis. Using Bean’s critical state model, the critical current density Jc-mag was determined. Resistive transitions were measured by a standard four-probe method with DC current magnitude of 100 mA. Critical temperature (Tc) and the width of transition (ΔTc) were determined from R(T) dependences shown by the insert in Fig. 1. Transport critical currents were measured at liquid He temperature and an external magnetic field from 4.0 to 8.0 T using standard DC measurement with 1 μVcm−1 criterion for Ic values. A free-standing short sample (50 mm) configuration was used for the tensile load tests of the wires at 4.2 K29. The electro-mechanical characteristics: Ic versus the tensile strain (ε) and the stress-strain curves σ(ε) were measured at a constant external magnetic field of 6 T. Scanning electron microscopy (SEM, JOEL 7600 F) with energy dispersive spectrometry (EDS, Oxford Instruments X-Max 50) was used to characterize polished transversal-sections of the heat-treated wires. Transmission electron microscopy (TEM) observations were made using JEOL JEM 1200FX microscope. TEM specimens were prepared by mechanical grinding and polishing followed by Ar beam ion milling using GATAN PIPS II. The transversal Al grain size (dav) was determined by image analyse of multiple bright filed TEM micrographs.
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## Acknowledgements
This work was supported by the Slovak Scientific Agency under the APVV-14-0522 project and the VEGA 2/0129/16 project.
## Author information
P.K. and I.H. participated in technological verification. P.K. prepared the manuscript and Figures 2–3. A.R. performed SEM and TEM microscopy analyses and prepared Figures 1, 4–5. M.K., J.K., T.M. and L.K. performed electrical and mechanical characterization at low temperatures. M.B. and P.Kr. fabricated extruded Al + Al2O3 material for the outer sheath. All authors reviewed and approved the final manuscript.
Correspondence to P. Kováč.
## Ethics declarations
### Competing Interests
The authors declare no competing interests.
Publisher's note: Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.
## Rights and permissions
Reprints and Permissions
Kováč, P., Hušek, I., Rosová, A. et al. Ultra-lightweight superconducting wire based on Mg, B, Ti and Al. Sci Rep 8, 11229 (2018). https://doi.org/10.1038/s41598-018-29354-1
• Accepted:
• Published:
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• ### Small diameter wind and react coil made of anodised Al-sheathed MgB2 wire
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• , J Kováč
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• , I Hušek
• & D Berek
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• ### Thermal conductivities and thermal runaways of superconducting MgB2 wires stabilized by an Al + Al2O3 sheath
• P Kováč
• , M Bonura
• , S Santra
• , L Kopera
• , A Rosová
• , C Senatore
• & I Hušek
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• ### MgB2 Wires and Bulks With High Superconducting Performance
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• , Michael Eisterer
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• , Vitaliy V. Romaka
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• ### Comparison of interfacial and critical current behaviour of Al+Al2O3 sheathed MgB2 wires with Ta and Ti diffusion barriers
• S. Santra
• , C.R.M. Grovenor
• , S.C. Speller
• , P. Kováč
• , L. Kopera
• & I. Hušek
Journal of Alloys and Compounds (2019)
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Newtonian mechanics (elevator problem)
The question is to find the reading of the spring balance. I want to solve it without using pseudo forces. What should be the method to go about using relative acceleration from ground frame? It would be helpful if someone could help me with the equations from ground frame
So as you don't want to use the pseudo force concept we can without loss of generality assume acceleration of the blocks with respect to the pulley as $$a$$ and also assume that in the frame of the elevator the $$3.0 kg$$ block is going down then the acceleration with respect to the ground frame of both the blocks can be set up as $$a_{3kg} =a-\frac {g}{10}$$ and $$a_{1.5kg}=a+ \frac {g}{10}$$ and then you could use the Newton's second law to solve for $$a$$ and the tension in the string connecting the blocks $$T$$. The reading of the balance would be $$2T$$ if the pulley is massless.
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# Chain rule independent variables
Does the following identity $$I(X_1,X_2 ; Y_1, Y_2) = I(X_1; Y_1) + I(X_2; Y_2)$$ hold for mutual information for $$(X_1, Y_1)$$ and $$(X_2, Y_2)$$ independent?
Attempt: $$p(x_1, x_2, y_1, y_2) = p(x_1, y_1) \cdot p(x_2, y_2)$$ implies $$p(x_1, x_2) = p(x_1) \cdot p(x_2)$$ and $$p(y_1, y_2) = p(y_1) \cdot p(y_2)$$, where I use lower case $$p$$ to denote the probability of specific outcomes, e.g. $$p(x)$$ for $$\Pr\{X=x\}$$. Thus $$\sum_{x_1, x_2, y_1, y_2} p(x_1, x_2, y_1, y_2) \cdot \log \frac{p(x_1, x_2, y_1, y_2)}{p(x_1, x_2) \cdot p(y_1, y_2)} = \\ \sum_{x_1, x_2, y_1, y_2} p(x_1, x_2, y_1, y_2) \cdot \log \frac{p(x_1, y_1)}{p(x_1) \cdot p(y_1)} \cdot \frac{p(x_2, y_2)}{p(x_2) \cdot p(y_2)}$$
• Could you show what you tried? It would make it easier for us to guide you towards the answer. Also I believe on the right-hand side you have a typo since you dropped indices on the $Y$. Jun 23 at 8:22
• @adrien_vdb Thanks! I corrected the typo Jun 23 at 9:17
• Great thanks for adding it, let me give you two hints. 1) You can use the following property of the logarithm $\log(x\cdot y) = \log(x) +\log(y)$. and 2) Use the fact that probability distributions sum up to 1 --> $\sum_{x, y} p(x, y) = 1$ Jun 23 at 13:47
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# Why is LaTeX replacing 2010 for the math subject classification with 1991?
I am trying to compile a `.tex` document and convert it to PDF where I used the command:
``````\subjclass[2010]{***(primary), and ***(secondary)}
``````
(`***` correspond to some suitable number from the subject classification).
I used
``````\documentclass{amsart}
\usepackage{amscd,amsmath,amssymb,amsthm,amsfonts,epsfig,graphics}
``````
While compiling the document, the compiler replaces 2010 by 1991 automatically (it also tells me so) and in the PDF output, I also see 1991 subject classification, not 2010 subject classification!
Could you inform me why is this happening? What should I do to get the 2010 subject classification?
Finally the journal where the paper is accepted is asking to confirm that the authors use 2010 subject classification. By any chance, do you think they accept an older classification written at the footnote? (I know the last one is a bad question, but just in the worst case!)
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You're loading a bunch of useless packages; only `amssymb` among the AMS bundle is really necessary, as the others are automatically loaded by `amsart`. The package `epsfig` is obsolete and should never be used; prefer `graphicx` to `graphics`. Anyway, I get "2010", so you should try preparing a minimal example showing the problem. I suspect you have a very old TeX distribution. – egreg Jan 21 '13 at 16:38
Your title is a little confusing as it reads as converting .cpp to .exe . – percusse Jan 21 '13 at 16:38
The class checks that the classification is known by checking the internal macro
``````\@namedef{subjclassname@2010}{%
\textup{2010} Mathematics Subject Classification}
``````
If you are getting a warning that 2010 is not known, then this macro is presumably not defined in the class because you have an old copy of amsart, I copied the above from
``````\ProvidesClass{amsart}[2009/07/02 v2.20.1]
``````
Your log file (or output from `\listfiles` should show which version you have.
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# Why do we need perfect numbers?
Why are perfect numbers important? What is the best way of introducing these numbers to a first course on number theory? I could not find any application apart from the relation to Mersenne primes. Are there any other applications of perfect numbers?
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What is the specific question here? – Benjamin Dickman Jul 21 at 16:56
There is a premise of the question that seems not to be very well supported: that the perfect numbers are important enough that one has to include them in a first course on number theory. It feels difficult to answer the question without this somewhat dubious assumption. – Benoît Kloeckner Jul 21 at 18:56
I think it turns out that "perfect" numbers do not interact much with other parts of number theory. Some of these very old, elementary, very ad-hoc definitions of special classes of integers have proven (and will prove) to interact interestingly with other ideas, but some seem not to. It's not easy for a beginner to guess the significance or subtlety of one of these classes of integers from the innocuous elementary-sounding definitions, usually. But tradition itself lends some significance to such notions. And there is always the possibility that the future discloses some surprising connections. And sometimes the sheer long-standing irresolution of a question gives it considerable cachet, even without profound consequences.
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This answer is perfect, but it doesn't mention explicitly that the original definition of perfect numbers were just one of the many classes of numbers invented by the ancient Greeks. Their motivation is unknown, but was probably purely aesthetic. Maybe someone noticed that $6 = 2 + 3 + 1$ and got curious. – Jack M Jul 22 at 16:33
@JackM, ah, yes, there were/are also "amicable" numbers, "idoneal" numbers, and others I've forgotten. Some mysticism, too, I think. – paul garrett Jul 22 at 16:35
IMO "idoneal" does not really fit the list; the notion is some 2000 years more recent (Euler). – quid Jul 24 at 10:52
Mathematically, even perfect numbers give a good number theory example to the general idea of classification, i.e. all even perfects have a specific form.
I use perfect numbers in my number theory class for two or three pedagogical reasons: with some trial and error (and the help of some computational software), I have the students essentially discover Euclid's formula. There's a lot of good conjecturing and testing along the way. We then look at the proof which is one of the few that are long enough to be nontrivial from the student view, yet easy enough to follow that they can actually understand it (especially after the preliminary explorations). When it's all done, I talk a little about odd perfect numbers: no one knows of any nor can prove they don't exists. This ties into a common thread in my upper level courses: math is still alive, still being discovered, still full of the unknown.
All this being said, I surely wouldn't look down on a teacher or student who knew nothing about perfect numbers and didn't have them covered in class.
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Not sure why this was downvoted. Seems like a perfectly reasonable response. – Steven Gubkin Jul 24 at 4:02
We "need" the natural numbers to count, we need the real numbers to measure and we need the complex numbers to guarantee that every quadratic equation has solutions.
Whenever we have some collection of objects and we can create a new definition that makes it possible to make distinctions between the objects (some are the "same" and some are "different") the potential exists to get new insights into the objects involved, and to other related objects.
Thus, we can talk about the integers that are prime, those that are squares, cubes, the sum of two squares, etc. Some of these definitions create wonderful new theoretical and applied playgrounds. The primes are a good example. After noticing the primes then one can show the prime factorization theorem. Primes can be used to design a cryptographical system (RSA) powerful enough to protect many financial transactions at the current time.
Perfect numbers create a "playground" for the interested. One of my undergraduate professors, Leo Zippin, made the observation to me that if some mathematician can create a "nifty" bit of theoretical mathematics, eventually some other mathematician will find a "nifty" use for those ideas.
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I don't know of a practical application of perfect numbers per se (not saying they don't exist), but its historical status dating back to the Greeks makes them interesting for pedagogical purposes. The Greeks (the Pythagoreans in particular) put mystical significance on special sequences such as the primes, the figurate numbers (notably the triangular numbers), and of course the perfect numbers. If you find that injecting anecdotes into your teaching helps you grab the attention of your students, then perfect numbers are a well-documented example of the ongoing human fascination with number theory since antiquity.
For a very detailed history of perfect numbers, check out http://www-history.mcs.st-and.ac.uk/HistTopics/Perfect_numbers.html. (In fact entire MacTutor History of Mathematics Archive is quite fascinating, and makes for quite nice leisure reading.)
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# Is the way that Mathematicians have categorised the Linear Transformations ideal, or could we perhaps have done better?
The reason I asked these 4 questions regarding if all linear/affine transformations could be formed by some combination of:
• Rotation
• Stretching
• Projection
• Translation
Was that I was speculating if they could hence form some sort of basis for a space. However, in this speculation, I came across some problems — for example the non-commutativity of combinations of linear transformations, amongst other things.
1. If the answer to 1. is yes, then could the linear transformations in $\Bbb R^2$ be considered to be some sort of non-commutative vector space where the basis is:
$S_x$ - Stretching in the $x$ direction by a factor of 1
$S_y$ - Stretching in the $y$ direction by a factor of 1
$\Bbb R$ - Rotation by 360º
$T_x$ - Shearing in the $x$ direction by a factor of 1
$T_y$ - Shearing in the $y$ direction by a factor of 1
The non-commutativity of this space makes it seem weird… and I haven’t studied any spaces other than $\Bbb R^n$ so I don’t even know if non-commutative spaces can even exist since $a+b ≠ b+a$ generally…
And then if this was true, then it would imply linear transformations for $\Bbb R^2$ exist in 5 dimensions!? And for affine transformations in $\Bbb R^2$, 7 dimensions (since the $x$ and $y$ Translations would be added)?
2. Forgetting about the speculation, when we learn about the above types of Linear Transformations, could our teachers (or rather, mathematicians in general), have chosen a different way to categorise the linear transformations so that all linear transformations can be expressed as some combination of this new set of categories of linear transformations?
Is the set that mathematicians have chosen the best way to go about categorising linear/affine transformations, or is it arbitrary and there is no “best way”, or on the other hand, could there be improvements made in the way they are categorised?
3. Applying what I suggested in 5. to 6., since many different bases can be selected for spaces of the form $\Bbb R^n$ (and I would imagine this to be true for most, if not all vector spaces), perhaps the same can apply for the space formed by linear transformations (if they do form a space).
4. And lastly, could there be some way to select ‘better’ basis vectors for this space so that the space is commutative (and hence more ‘normal’)?
• A pretty natural structure with non-commutativity is the algebra of square matrices, it is a vector space with a product – janmarqz Oct 26 '15 at 18:38
• What do you mean by a "commutative space"? What do you mean "non-commutative vector space"? Without mentioning the binary operation, this doesn't have any meaning. – rschwieb Oct 26 '15 at 18:52
• however the en.wikipedia.org/wiki/Affine_group already is at ours disposal – janmarqz Oct 26 '15 at 19:10
• @rschwieb By "non-commutative vector space", I meant that the addition of vectors within this space wouldn't be commutative. (Sorry for bad use of terminology) – Shuri2060 Oct 27 '15 at 20:26
The question is better for invertible affine maps. Then you have the Lie group. But then the translations form a subgroup than you can handle later. In which case you work in the Lie group $GL(n,\Bbb R)$. This is not a vector space so it does not make sense to ask for a basis, but if you work infinitesimally you get the Lie algebra which has lots of convenient basis. These come exactly as stated from breaking up into rotations and the rest. The rotations are what is called the maximal compact. Then you put all these ingredients back together.
There are lots of ways to characterize linear transformations. You might look into Jordan canonical form, rational canonical form, polar decomposition, singular value decomposition, QR decomposition, LU decomposition, ...
If none of those fit your needs, you're welcome to define another one.
Choose a basis of $\mathbb R^2$. The basis $\{v_1 = (1,0), v_2 = (0,1)\}$ is good, but technically any basis will suffice as long as you specify what basis you have chosen.
Any linear transformation of $\mathbb R^2$ then can be uniquely specified by the pair of vectors $v_1', v_2'$ to which the transformation maps the pair of vectors $v_1, v_2$. Conversely, any pair of vectors $v_1', v_2'$ corresponds to a unique linear transformation that maps $v_1, v_2$ to $v_1', v_2'$.
As you can see from this description, two two-dimensional vectors are sufficient to identify any linear transformation of $\mathbb R^2$, so the linear transformations have four dimensions, not five.
Basically any approach to linear algebra is going to have to characterize all linear transformations somehow, and I do not think that beyond a very elementary level the primitives from which we choose to build all transformations will be any subset of the transformations you listed. The last time in my mathematical education that I recall those transformations being used as the "basis" for all linear transformations was in a "pre-calculus" type of curriculum during high school. (Of course in more advanced math it is still the case that we might remark that certain transformations fall into one of those categories, and sometimes we work with an interesting subset of transformations that can be characterized in such ways, for example, all the transformations are rotations of some kind.)
As for commutativity, you can get that only if you are willing to discard many possible transformations from the "space" that you are willing to consider. The reason for this is simply that some pairs of transformations do not commute. The only way to get a commutative algebra is to eliminate from your "space" at least one of the transformations in any such non-commuting pair. That's a lot of transformations you have to get rid of.
For example, the transformations consisting only of rotations and dilations (scaling uniformly in every direction) commute. But now you have lost stretching and shearing.
Rotation, stretching dilation, and translation are conformal, i.e. they do not change angles. Projection is singular, i.e. it maps regions of positive volume to regions of zero volume.
The linear transformation $(x,y) \mapsto (x, x+y)$, is a shear. It is non-singular, so it cannot have any projections among its factors. But it changes some angles (e.g. the angle between $(1,0)$ and $(0,1)$ becomes the angle between $(1,1)$ and $(0,1)$, which is different), so it cannot be made up only of transformations that never change angles.
Shearing happens whenever you do the elementary row operation of adding a multiple of one row of a matrix to another row.
POSTSCRIPT: It seems something other than multiplying every vector by the same scalar was meant by "stretching", so the question of the closure under composition, of the proposed set of transformations, is more than what the paragraphs above address.
• I agree that rotation/translation do not change angles... but surely stretching in a particular direction does? Also, an answer here disagrees with that.... who is correct? math.stackexchange.com/questions/1498833/… – Shuri2060 Oct 27 '15 at 20:29
• @MasterShuriken : It seems I misunderstood what was meant by "stretching". I took it to mean multiplication of every vector by the same scalar. ${}\qquad{}$ – Michael Hardy Oct 27 '15 at 20:33
• From my textbook, stretching occurs in the $x$ or $y$ directions by a certain factor (and hence my particular take on the meaning). – Shuri2060 Oct 27 '15 at 20:36
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# MATLAB: Fsolve result is not desirable even giving a close starting point
fsolvenonlinear equations
I have three nonlinear equations with three unknowns and I used fsolve function.
I know that x=[0.3173;0.3173;3.6590] is a good enough solution for the three equations. However, using fsolve, I could not get to this solution even for a very close starting point as shown below. Instead the result is (upon plugging these results into the 3 equations, the values are not close to 0):
x = 0.999999999999936 1.000000000000067 1.000000000000036
How can this be solved? Thanks!
CODE:
x0=[0.31; 0.31; 3.5];options=optimset('Display','iter'); [x,fval]=fsolve(@myfun1,x0,options)function F=myfun(x)F=[13*cos(x(3)) - 5*x(1)*x(3)^2 - 5*x(2)*x(3)^2 + 13*x(1)*x(3)*sin(x(3)) + 13*x(2)*x(3)*sin(x(3)) + 10*x(1)*x(2)*x(3)^2 - 13*x(1)*x(2)*x(3)^2*cos(x(3)) + 40; 5*x(3) - 13*sin(x(3)) + 30*x(1)*x(3) + 13*x(1)*x(3)*cos(x(3)) + 13*x(2)*x(3)*cos(x(3)) - 5*x(1)*x(2)*x(3)^3 + 13*x(1)*x(2)*x(3)^2*sin(x(3)); 390*x(3)^2*sin(x(3)) + 300*x(1)*x(3)^3 - 25*x(1)*x(3)^5 + 25*x(2)*x(3)^5 - 150*x(3)^3 - 260*x(1)*x(3)^3*cos(x(3)) - 130*x(2)*x(3)^3*cos(x(3)) + 130*x(1)*x(3)^4*sin(x(3)) - 130*x(2)*x(3)^4*sin(x(3)) - 169*x(1)*x(3)^3*cos(x(3))^2 + 169*x(2)*x(3)^3*cos(x(3))^2 - 169*x(1)*x(3)^3*sin(x(3))^2 + 169*x(2)*x(3)^3*sin(x(3))^2 ];
#### Best Answer
• Possibly because you're passing @myfun1 to fsolve instead of @myfun? When I make this change and run your code, I get the solution you expect.
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Models in set theory and continuum hypothesis
Some days ago I had the opportunity to listen to the talk about model theory and connections with algebra and geometry. I'm not at all specialist in this field so my question probably will be naive but nevertheless I'll try to explain my doubts. As far I understood: for example if we consider the theory of groups a model for such a theory will be a concrete group. For example the statement: "there is some $g \in G$ such that $g^2=e$" is a true statement in every model ($g=e$ is ok) therefore this statement is a theorem in a theory of groups. On the other hand the statement "for all $g \in G$ we have $g^2=e$" is no longer a theorem from the theory of groups: one can find examples (models) of the theory where this is satisfied but also one can construct counterexamples. In this sense this statement is independent from the axioms of group theory. So far everything looks clear. But I have a problem in understanding how the situation looks like in the context of set theory and ZFC axioms: for example I know that the statement "the cardinality of the reals is the next cardinality after the cardinality of the set of natural numbers" is independent from ZFC axioms. In other words one can construct two different models for set theory where in the one model this statement is true but in the second is false. What exactly does it mean "to construct model for set theory"? Let me return to the previous example about groups: for group theory a model is a concrete group so "to construct the model" means nothing more than "provide an example" but how to understand this in the context of whole set theory? What exactly do we have to construct? I'm sure that this question will be naive from the point of view of the specialist: from the other hand I suspect that there are at least few people which would like to know the answer for such "meta" question.
• Dude. Paragraphs. – Asaf Karagila Dec 6 '15 at 5:03
To construct a model of set theory means to produce a set $A$ and a relation $R$ on $A \times A$ such that all the axioms of ZFC are satisfied if we take "set" to mean "element of $A$" and take "$a \in b"$ to mean $aRb$.
Sometimes, in set theory, we allow a more general kind of model in which $A$ is a proper class and $R$ is a definable relation on pairs of elements of $A$. These are called "class models".
• "Inside" the model, there are sets of arbitrary cardinality. It will have a set it believes is $\mathbb{R}$, and the powerset of $\mathbb{R}$, etc. From the point of view of the model, the collection $A$ is a proper class. Of course, from outside the model, we see that the model only contains some sets, not all of them, and $A$ is a set. I think that one issue here is that we are used to looking at many different groups, so we don't have any qualms when we see different groups. But, outside of set theory, we don't normally look at different models of set theory, so it takes getting used to. – Carl Mummert Dec 5 '15 at 23:53
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### Browsed byCategory: Analysis
Topologies and Sigma-Algebras
## Topologies and Sigma-Algebras
Both topologies and $\sigma-$algebras are collections of subsets of a set $X$. What exactly is the difference between the two, and is there a relationship? We explore these notions by noting the definitions first. Let $X$ be any set.
### Topology
A topology $\tau$ is a collection of subsets of a set $X$ (also called a topology in $X$) that satisfies the following properties:
(1) $\emptyset \in \tau$, and $X \in \tau$
(2) for any finite collection of sets in $\tau$, $\{V_{i}\}_{i=1}^{n}$, $\cap_{i=1}^{n}V_{i} \in \tau$
(3) for any arbitrary collection of sets $\{V_{\alpha}: \alpha \in I\}$ in $\tau$ (countable or uncountable index set $I$), $\cup_{\alpha}V_{\alpha} \in \tau$
A topology is therefore a collection of subsets of a set $X$ that contains the empty set, the set $X$ itself, all possible finite intersections of the subsets in the topology, and all possible unions of subsets in the topology.
The simplest topology is called the trivial topology, where for a set $X$, $\tau = \{\emptyset, X\}$. Notice that (1) above is satisfied by design. All intersections we can make with the sets in $\tau$ are finite ones. (There’s just one, $X \cap \emptyset = \emptyset$.) Thus, (2) is satisfied. Any union here gives $X$, which is in $\tau$, so this is a topology.
This isn’t a very interesting topology, so let’s create another one.
Let’s take $X = \{1,2,3,4\}$ as the set. Let’s give a collection of subsets $\tau = \{\{1\},\{2\}, \{1,3\}, \{2,4\}, \{1,2,3\}, \{1,2,4\}, \emptyset, X\}.$ Notice that I didn’t include every single possible subset of $X$. There are two singleton sets, 4 pairs, and 1 set of triples missing. This example will illustrate that you can leave out subsets of a set and still have a topology. Notice that (1) is met. You can check all possible finite intersections of sets inside $\tau$, and notice that you either end up with $\emptyset$ or another of the sets in $\tau$. For example, $\{1,3\} \cap \{1,2,3\} = \{1,3\}$, $\{2\} \cap \{1\} = \emptyset$, etc. Lastly, we can only have finite unions here, since $\tau$ only has a finite number of things. You can check all possible unions, and notice that all of them result in a set already in $\tau$. For example, $\{1,3\} \cup \{2,4\} = X$, $\emptyset \cup \{1,2,4\} = \{1,2,4\}$, $\{1\} \cup \{1,3\} \cup \{2,4\} = X$, etc. Thus, $\tau$ is a topology on this set $X$.
We’ll look at one final example that’s a bit more abstract. Let’s take a totally ordered set $X$ (like the real line $\mathbb{R}$). Then the order topology on $X$ is the collection of subsets that look like one of the following:
• $\{x : a < x\}$ for all $a$ in $X$
• $\{x : b > x \}$ for all $b \in X$
• $\{x : a < b < x\}$ for all $a,b \in X$
• any union of sets that look like the above
To put something concrete to this, let $X = \{1,2,3,4\}$, the same set as above. This is a totally ordered set, since we can write these numbers in increasing order. Then
• The sets that have the structure $\{x : a < x\}$ for all $a \in X$ are
• $\{x : 1 < x\} = \{2,3,4\}$
• $\{x : 2 < x\} = \{3,4\}$
• $\{x : 3 < x\} = \{4\}$
• $\{x : 4 < x \} = \emptyset$
• The sets that have the structure $\{x : b > x\}$ for all $b \in X$ are
• $\{x : 1 > x\} = \emptyset$ (which we already handled)
• $\{x : 2 > x\} = \{1\}$
• $\{x : 3 > x \} = \{1,2\}$
• $\{x : 4 > x\} = \{1,2,3\}$
• The sets that have the structure $\{x : a < x < b\}$ for all $a,b \in X$ are
• $\{x : 1 < x < 3\} = \{2\}$
• $\{x : 1 < x < 4\} = \{2,3\}$
• $\{x : 2 < x < 4\} = \{3\}$
• The remaining combinations yield $\emptyset$
• The sets that are a union of the above sets (that aren’t already listed) are
• $X = \{x : 1 < x\} \cup \{x : 2 > x\}$
• $\{1,2,4\} = \{x : 3 > x\} \cup \{x : 3< x \}$
• $\{1,3,4\} = \{x : 2 < x\} \cup \{x : 2 > x\}$
• $\{1,3\} = \{x : 2 > x\} \cup \{x : 2< x < 4\}$
• $\{1,4\} = \{x : 2 > x\} \cup \{x : 3 < x\}$
• $\{2,4\} = \{x : 1 < x < 3\} \cup \{x : 3 < x \}$
The astute reader will note that in this case, the order topology on $X = \{1,2,3,4\}$ ends up being the collections of all subsets of $X$, called the power set.
### Sigma-Algebra
Let $X$ be a set. Then we define a $\sigma$-algebra.
A $\sigma$-algebra is a collection $\mathfrak{M}$ of subsets of a set $X$ such that the following properties hold:
(1) $X \in \mathfrak{M}$
(2) If $A \in \mathfrak{M}$, then $A^{c} \in \mathfrak{M}$, where $A^{c}$ is the complement taken relative to the set $X$.
(3) For a countable collection $\{A_{i}\}_{i=1}^{\infty}$ of sets that’s in $\mathfrak{M}$, $\cup_{i=1}^{\infty}A_{i} \in \mathfrak{M}$.
Let’s look at some explicit examples:
Again, take $X = \{1,2,3,4\}$. Let $$\mathfrak{M} = \{\emptyset, X, \{1,2\}, \{3,4\}\}.$$ We’ll verify that this is a $\sigma-$algebra. First, $X \in \mathfrak{M}$. Then, for each set in $\mathfrak{M}$, the complement is also present. (Remember that $X^{c} = \emptyset$.) Finally, any countable union will yield $X$, which is present in $\mathfrak{M}$, so we indeed have a $\sigma-$algebra.
Taking another example, we’ll generate a $\sigma$-algebra over a set from a single subset. Keep $X = \{1,2,3,4\}$. Let’s generate a $\sigma-$algebra from the set $\{2\}$. $$\mathfrak{M}(\{2\}) = \{X, \emptyset, \{2\},\{1,3,4\}\}$$ The singleton $\{2\}$ and its complement must be in $\mathfrak{M}(\{2\})$, and we also require $X$ and its complement $\emptyset$ to be present. Any countable union here results in the entire set $X$.
### What’s the difference between a topology and a $\sigma-$algebra?
Looking carefully at the definitions for each of a topology and a $\sigma-$algebra, we notice some similarities:
1. Both are collections of subsets of a given set $X$.
2. Both require the entire set $X$ and the empty set $\emptyset$ to be inside the collection. (The topology explicitly requires it, and the $\sigma-$algebra requires it implicitly by requiring the presence of $X$, and the presence of all complements.)
3. Both will hold all possible finite intersections. The topology explicitly requires this, and the $\sigma-$algebra requires this implicitly by requiring countable unions to be present (which includes finite ones), and their complements. (The complement of a finite union is a finite intersection.)
4. Both require countable unions. Here, the $\sigma-$algebra requires this explicitly, and the topology requires it implicitly, since all arbitrary unions–countable and uncountable–must be in the topology.
That seems to be a lot of similarities. Let’s look at the differences.
1. A $\sigma-$algebra requires only countable unions of elements of the collection be present. The topology puts a stricter requirement —all unions, even uncountable ones.
2. The $\sigma-$algebra requires that the complement of a set in the collection be present. The topology doesn’t require anything about complements.
3. The topology only requires the presence of all finite intersections of sets in the collection, whereas the $\sigma-$algebra requires all countable intersections (by combining the complement axiom and the countable union axiom).
It is with these differences we’ll exhibit examples of a topology that is not a $\sigma-$algebra, a $\sigma-$algebra that is not a topology, and a collection of subsets that is both a $\sigma-$algebra and a topology.
#### A topology that is not a $\sigma-$algebra
Let $X = \{1,2,3\}$, and $\tau = \{\emptyset, X, \{1,2\},\{2\}, \{2,3\}\}$. $\tau$ is a topology because
1. $\emptyset, X \in \tau$
2. Any finite intersection of elements in $\tau$ either yields the singleton $\{2\}$ or $\emptyset$.
3. Any union generates $X$, $\{1,2\}$, or $\{2,3\}$, all of which are already in $\tau$
However, because $\{2\}^{c} = \{1,3\} \not\in \tau$, we have a set in $\tau$ whose complement is not present, so $\tau$ cannot also be a $\sigma-$algebra. We used (2) in the list of differences to construct this example.
#### A $\sigma-$algebra that is not a topology
This example is a little trickier to construct. We need a $\sigma$-algebra, but not a topology, so we need to find a difference between the $\sigma-$algebra and the topology where the topology requirement is more strict than the $\sigma-$algebra’s version. We focus on difference (1) here.
Let $X = [0,1]$. Let $\mathfrak{M}$ be the collection of subsets of $X$ that are either themselves countable, or whose complements are countable. Some examples of things in $\mathfrak\{M\}:$
• all rational numbers between 0 and 1, represented as singleton sets. (countable)
• the entire collection of rational numbers between 0 and 1, represented as a set itself (countable)
• $\left\{\frac{1}{2^{x}}, x \in \mathbb{N}\right\}$. (countable)
• $[0,1] \setminus \{1/2, 1/4, 1/8\}$ (not countable, but its complement is $\{1/2, 1/4, 1/8\}$, which is countable)
• $\emptyset$ (countable)
• $X = [0,1]$ (not countable, but its complement $\emptyset$ is countable)
$\mathfrak{M}$ is a $\sigma-$algebra because:
• $X \in \mathfrak{M}$
• All complements of sets in $\mathfrak{M}$ are present, since we’ve designed the collection to be all pairs of countable sets with countable complements, and all uncountable sets with countable complements.
• Finally, all countable unions of countable sets are countable, so those are present. The countable union of uncountable sets with countable complements will have a countable complement1, and thus all countable unions of elements of $\mathfrak{M}$ are also in $\mathfrak{M}$, so we have a $\sigma-$algebra.
n particular, every single point of $[0,1]$ is in $\mathfrak{M}$ as a singleton set. To be a topology, any arbitrary union of elements of $\mathfrak{M}$ must also be in $\mathfrak{M}$. Take every real number between 0 and 1/2, inclusive. Then the union of all these singleton points is the interval $[0,1/2]$. However, $[0,1/2]^{c} = (1/2,1]$, which is uncountable. Thus, we have an uncountable set with an uncountable complement, so $[0,1/2] \notin \mathfrak{M}$. Since it can be represented as the arbitrary union of sets in $\mathfrak{M}$, $\mathfrak{M}$ is not a topology.
#### A collection that is both a $\sigma-$algebra and a topology
Take any set $X$ that is countable, and let $2^{X}$ be the power set on $X$ (the collection of all subsets of $X$). Then all subsets of $X$ are countable. We have that $\emptyset$ and $X$ are present, since both are subsets of $X$. Since the finite intersection of some subcollection of subsets of $X$ is a subset of $X$, it is in the collection. The arbitrary union of subsets of $X$ is either a proper subset of $X$ or $X$ itself. Thus $2^{X}$ is a topology (called the discrete topology).
The complement of a subset of $X$ is still a subset, and if all arbitrary unions are in $2^{X}$, then certainly countable unions are. Thus $2^{X}$ is also a $\sigma-$algebra.
To be explicit, return to the above where $X = \{1,2,3,4\}$. Write out all possible subsets of $X$, including singletons, $\emptyset$, and $X$ itself, and notice that all axioms in both the $\sigma-$algebra and the topology definitions are satisfied.
Simulating Soundscapes Using Convolutions
## Simulating Soundscapes Using Convolutions
One of the most powerful areas of electrical engineering that flourished in the 20th century is the field of signal processing. The field is broad and rich in some beautiful mathematics, but by way of introduction, here we’ll take a look at some basic properties of signals and how we can use these properties to find a nice compact representation of operations on them. As a motivating application, we’ll use what we study today to apply certain effects to audio signals. In particular, we’ll take a piece of audio, and be able to make it sound like it’s being played in a cathedral, or in a parking garage, or even through a metal spring.
First things first: what is a signal? For this discussion we’ll limit ourselves to looking at the space $\ell = \{x :\mathbb{Z} \rightarrow \mathbb{R}\}$ – the set of functions which take an integer and return a real number. Another way to think of a signal then is as an infinite sequence of real numbers. We’re limiting ourselves to functions where the domain is discrete (the integers), rather than continuous (the real numbers), since in many applications we’re looking at signals that represent some measurement taken at a bunch of different times1. It’s worth noting that any signal that’s been defined on a countable domain $\{..., t_{n-1}, t_n, t_{n+1},...\}$ can be converted to one defined on the integers via an isomorphism. We like to place one further restriction on the signals, in order to make certain operations possible. We restrict the space to so-called finite-energy signals:
$$\ell_2 = \left\{x \in \ell : \sum_{n = -\infty}^{\infty} |x(n)|^2 < \infty\right\}.$$
This restriction makes it much easier to study and prove things around these functions, while still giving us lots of useful signals to study, without having to deal with messy things like infinities. In practice, when dealing with audio we usually have a signal with a finite length and range, so this finite-energy property is trivially true.
Studying signals is only as useful if we can also define operations on them. We’ll study the interaction of signals with systems, which take one signal and transform it into another – essentially, a function operating on signals. Here, we’ll say that a system $H : \ell_2 \rightarrow \ell_2$ takes an input signal $x(t)$ and produces output signal $H\{x(t)\} = y(t)$.
## Linearity and Time Invariance
There are certain properties that are useful for systems to have. The first is linearity. A system $H$ is considered linear if for every pair of inputs $x_1, x_2 \in \ell_2$, and for any scalar values $\alpha, \beta \in R$, we have
$$H\{\alpha x_1 + \beta x_2\} = \alpha H\{x_1\} + \beta H\{x_2\}$$
This is very useful, because it allows us to break down a signal into simpler parts, study the response of the system to each of those parts, and understand the response to the more complex original signal.
The next property we’re going to impose on our systems is time-invariance:
$$\forall s \in \mathbb{Z}, H\{x(n)\} = y(n) \Rightarrow H\{x(n-s)\} = y(n-s)$$
This means that shifting the input by $s$ corresponds to a similar shift in the output. In our example of playing music in a cathedral, we expect our system to be time-invariant, since it shouldn’t matter whether we play our music at noon or at midnight, we’d expect it to sound the same. However, if we were playing in a building that, for example, lowered a bunch of sound-dampening curtains at 8pm every night, then the system would no longer be time-invariant.
So what are some more concrete examples of systems that are linear and time-invariant?
Let’s consider an audio effect which imitates an echo – it outputs the original signal, plus a quieter, delayed version of that signal. We might express such a system as
$$H_{\Delta, k}\{x(n)\} = x(n) + kx(n-\Delta)$$
where $\Delta \in \mathbb{Z}$ is the time delay of the echo (in terms of the number of samples), and $k \in \mathbb{R}$ is the relative volume of the echoed signal. We can see that this signal is time-invariant, because there is no appearance of the time variable outside of the input. If we replaced $k$ by a function $k(n) = \sin(n)$, for example, we would lose this time invariance. Additionally, the system is plainly linear:
\begin{aligned}H_{\Delta, k}\{\alpha x_1(n) + \beta x_2(n)\} &= \alpha x_1(n) + \beta x_2(n) + k x_1(n-\Delta) +k x_2(n-\Delta) \\&= H_{\Delta, k}\{x_1(n)\} + H_{\Delta, k}\{x_2(n)\}\end{aligned}
A common non-linearity in audio processing is called clipping — we limit the output to be between -1 and 1: $H\{x(n)\} = \max(\min(x(n), 1), -1)$. This is clearly non-linear since doubling the input will not generally double the output.
## The Kronecker delta signal
There is a very useful signal that I would be remiss not to mention here: the Kronecker delta signal. We define this signal as
$$\delta(n) = \begin{cases} 1 & n = 0 \\ 0 & n \neq 0 \end{cases}$$
The delta defines an impulse, and we can use it to come up with a nice compact description of linear, time-invariant systems. One property of the delta is that it can be used to “extract” a single element from another signal, by multiplying:
$$\forall s \in \mathbb{Z}, \delta(n-s)x(s) = \begin{cases} x(n) & n=s \\ 0 & n \neq s\end{cases}$$
Similarly, we can then write any signal as an infinite sum of these multiplications:
$$x(n) = \sum_{s=-\infty}^{\infty} \delta(n-s)x(s) = \sum_{s=-\infty}^{\infty}\delta(s)x(n-s)$$
Why would we want to do this? Let $H$ be a linear, time-invariant system, and let $h(t) = H\{\delta(t)\}$, the response of the system to the delta signal. Then we have
\begin{aligned} H\{x(n)\} &= H\left\{\sum_{s=-\infty}^{\infty} \delta(n-s)x(s)\right\}\\&=\sum_{s=-\infty}^{\infty}H\{\delta(n-s)\}x(s) \text{ by linearity}\\&=\sum_{s=-\infty}^{\infty}h(n-s)x(s) \text{ by time-invariance}\end{aligned}
We can write any linear, time-invariant system in this form. We call the function $h$ the impulse response of the system, and it fully describes the behaviour of the system. This operation where we’re summing up the product of shifted signals is called a convolution, and appears in lots of different fields of math 2.
## Firing a Gun in The Math Citadel
The power of this representation of a system is that if we want to understand how it will act on any arbitrary signal, it is sufficient to understand how it responds to an impulse. To demonstrate this, we’ll look at the example of how audio is affected by the environment it is played in. Say we were a sound engineer, and we wanted to get an instrument to sound like it was being played in a big, echoing cathedral. We could try to find such a place, and actually record the instrument, but that could be expensive, requiring a lot of setup and time. Instead, if we could record the impulse response of that space instead, we could apply the convolution to a recording we did back in a studio. How do we capture an impulse response? We just need a loud, very short audio source – firing a pistol or popping a balloon are common. To demonstrate, here are some example impulse responses, taken from OpenAirLib, and their effects on different audio signals.
First, here is the unprocessed input signal – a short piece of jazz guitar:
Here is the same clip, as if it were played in a stairwell at the University of York. First, the impulse response, then the processed audio.
That sounds a little different, but we can try a more extreme example: the gothic cathedral of York Minster. Again, here is the impulse response, followed by the processed signal.
In this case, we have a much more extreme reverberation effect, and we get the sound of a guitar in a large, ringing room. For our last example, we’ll note that impulse responses don’t have to be natural recordings, but instead could be entirely synthetic. Here, I’ve simply reversed the first impulse response from the stairwell, which creates this pre-echo effect, which doesn’t exist naturally.
This is just one of the most basic examples of what can be done with signal processing, but I think it’s a particularly good one – by defining some reasonable properties for signals and systems, we’re able to derive a nice compact representation that also makes a practical application very simple.
Sequences & Tendency: Topology Basics Pt. 2
## Introduction
In my previous post I presented abstract topological spaces by way of two special characteristics. These properties are enough to endow a given set with vast possibilities for analysis. Fundamental to mathematical analysis of all kinds (real, complex, functional, etc.) is the sequence.
We have covered the concept of sequences in some of our other posts here at The Math Citadel. As Rachel pointed out in her post on Cauchy sequences, one of the most important aspects of the character of a given sequence is convergence.
In spaces like the real numbers, there is convenient framework available to quantify closeness and proximity, and which allows naturally for a definition of limit or tendency for sequences. In a general topological space missing this skeletal feature, convergence must be defined.
This post will assume only some familiarity with sequences as mathematical objects and, of course, the concepts mentioned in Part 1. For a thorough treatment of sequences, I recommend Mathematical Analysis by Tom M. Apostol.
## Neighborhoods
Suppose $(X,\mathscr{T})$ is a given topological space, and nothing more is known. At our disposal so far are only open sets (elements of $\mathscr{T}$), and so it is on these a concept of vicinity relies.
Definition. Given a topological space $(X,\mathscr{T})$, a neighborhood of a point $x\in X$ is an open set which contains $x$.
That is, we say an element $T\in\mathscr{T}$ such that $x\in T$ is a neighborhood1 of $x$. To illustrate, take the examples from my previous post.
#### The Trivial Topology
When the topology in question is the trivial one: $\{\emptyset,X\}$, the only nonempty open set is $X$ itself, hence it is the only neighborhood of any point $x\in X$.
#### The Discrete Topology
Take $X=\{2,3,5\}$ and $\mathscr{T}$ to be the collection of all subsets of $X$:
$\emptyset$ $\{2\}$ $\{3\}$ $\{5\}$ $\{2,3\}$ $\{2,5\}$ $\{3,5\}$ $\{2,3,5\}$
Then, for, say $x=5$, neighborhoods include $\{5\}$, $\{2,5\}$, $\{3,5\}$, and $\{2,3,5\}$.
#### The Standard Topology on $\mathbb{R}$
The standard topology on $\mathbb{R}$ is defined to be the family of all sets of real numbers containing an open interval around each of its points. In this case, there are infinitely2 many neighborhoods of every real number. Taking $x=\pi$ for instance, then $(3,4)$, $(-2\pi,2\pi)$, and even
$$\bigcup_{n=1}^{\infty}\left(\pi-\frac{1}{n},\pi+\frac{1}{n}\right)$$
are all neighborhoods of $\pi$.
Remark. A special type of neighborhood in the standard topology is the symmetric open interval. Given a point $x$ and a radius $r>0$, the set
$$(x-r,x+r)=\{y\in\mathbb{R}\mathrel{:}|x-y|
is a neighborhood of $x$. These sets form what is referred to as a basis for the standard topology and are important to definition of convergence in $\mathbb{R}$ as a metric space.
## Convergence
“…the topology of a space can be described completely in terms of convergence.” —John L. Kelley, General Topology
At this point in our discussion of topological spaces, the only objects available for use are open sets and neighborhoods, and so it is with these that convergence of a sequence are built3.
Definition. A sequence $(\alpha_n)$ in a topological space $(X,\mathscr{T})$ converges to a point $L\in X$ if for every neighborhood $U$ of $L$, there exists an index $N\in\mathbb{N}$ such that $\alpha_n\in U$ whenever $n\geq N$. The point $L$ is referred to as the limit of the sequence $(\alpha_n)$.
Visually, this definition can be thought of as demanding the points of the sequence cluster around the limit point $L$. In order for the sequence $(\alpha_n)$ to converge to $L$, it must be the case that after finitely many terms, every one that follows is contained in the arbitrarily posed neighborhood $U$.
As you might expect, the class of neighborhoods available has a dramatic effect on which sequences converge, as well as where they tend. Just how close to $L$ are the neighborhoods built around it in the topology?
We will use the example topologies brought up so far to exhibit the key characteristics of this definition, and what these parameters permit of sequences.
#### The Trivial Topology
In case it was to this point hazy just how useful the trivial topology is, sequences illustrate the issue nicely. For the sake of this presentation, take the trivial topology on $\mathbb{R}$. There is precisely one neighborhood of any point, namely $\mathbb{R}$ itself. As a result, any sequence of real numbers converges, since every term belongs to $\mathbb{R}$. Moreover, every real number is a limit of any sequence. So, yes, the sequence $(5,5,5,\ldots)$ of all $5$‘s converges to $\sqrt{2}$ here.
#### The Discrete Topology
Whereas with the trivial topology a single neighborhood exists, the discrete topology is as packed with neighborhoods as can be. So, as the trivial topology allows every sequence to converge to everything, we can expect the discrete topology to be comparatively restrictive. Taking the set $\{2,3,5\}$ with the discrete topology as mentioned above, we can pinpoint the new limitation: every set containing exactly one point is a neighborhood of that point. Notice the sets4 $\{2\}$, $\{3\}$, and $\{5\}$ are all open sets.
What does this mean? Any sequence that converges to one of these points, say $3$, must eventually have all its terms in the neighborhood $\{3\}$. But that requires all convergent sequences to be eventually constant! This seems to be a minor issue with the finite set $\{2,3,5\}$, but it presents an undesirable, counter-intuitive problem in other sets.
Take $\mathbb{R}$ with the discrete topology, for example. Under these rules, the sequence
$$(\alpha_n)=\left(\frac{1}{n}\right)=\left(1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\ldots\right),$$
though expected to converge to $0$, does not converge at all.
So, the discrete topology is too restrictive, and the trivial topology lets us get away with anything. Fortunately, a happy middle ground exists by being a little more selective with neighborhoods.
#### The Standard Topology
By requiring an open set to contain an open interval around each of its points, it is impossible that a singleton be an open set. Therefore a singleton cannot be a neighborhood, and we eliminate the trouble posed by the discrete topology. Yet every open interval around a real number $L$ contains a smaller one, and each of these is a neighborhood.
This effectively corrals the points of any convergent sequence, requiring the distance between the terms and the limit to vanish as $n$ increases. Take again the sequence
$$(\alpha_n)=\left(\frac{1}{n}\right)=\left(1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\ldots\right).$$
We suspect $(\alpha_n)$ converges to $0$, but this requires proof. Therefore, we must consider an arbitrary neighborhood of $0$, and expose the index $N\in\mathbb{N}$ such that all terms, from the $N$th onward, exist in that neighborhood.
Suppose $U$ is a given neighborhood of $0$, so that $U$ contains an open interval surrounding $0$. Without loss of generality, we may assume this interval is symmetric; that is, the interval has the form $(-r,r)$ for some radius $r>0$. Take $N$ to be any integer greater than $\tfrac{1}{r}$. Then, whenever $n\geq N$,
$$\alpha_n = \frac{1}{n} \leq \frac{1}{N} < \frac{1}{1/r} = r.$$
But this means $\alpha_n\in(-r,r)\subset U$ so long as $n\geq N$. Since we chose $U$ arbitrarily, it follows $(\alpha_n)$ converges to $0$.
## Conclusion
The behavior of a sequence in a given set can change rather drastically depending on the network of neighborhoods the topology allows. However, with careful construction, it is possible to have all the sequential comforts of metric spaces covered under a briefly put definition.
My next post in this series will push the generalization of these concepts much further, by relaxing a single requirement. In order to define convergence in the preceding discussion, the set of indices $\mathbb{N}$ was important not for guaranteeing infinitely many terms, but instead for providing order. This allows us to speak of all terms subsequent to one in particular. It turns out that if we simply hold on to order, we can loosen the nature of the set on which it is defined. That is the key to Moore-Smith Convergence, to be visited next.
Paper Review: Active Queue Management with Non-Linear Packet Dropping Function
## Paper Review: Active Queue Management with Non-Linear Packet Dropping Function
As promised in the previous article, I plan to review Reference 2, Active Queue Management with Non-Linear Packet Dropping Function, by D. Augustyn, A. Domanski, and J. Domanska, published in HET-NETs 2010, which discusses a change in the structure of the packet drop probability function using the average queue length in a buffer. I mentioned previously that choosing a linear function of the average queue length can be viewed as a bit of an arbitrary choice, since we’re designing a control mechanism here, and this paper attempts to define a new form of this packet drop probability function.
In summary, the best lesson one can take from this paper is that publication in a journal or conference proceedings does not guarantee that the paper withstands scrutiny. The paper is linked above for the interested reader to peruse himself, and to investigate the claims.
## Summary
The paper intended to give a new function to calculate the probability of proactively dropping a packet in a queue in order to prevent a full buffer. It seemed to be presented as an alternative to RED, described in my previous article. The authors define this new function, then set up a simulation in order to examine the effects.
## When Orthogonal is Abused
The authors describe using a finite linear combination of orthogonal basis polynomials defined on a finite interval as the underlying mathematical structure.
First, we should discuss what we mean by orthogonal in context of functions. Orthogonal is most commonly understood in terms of vectors, and when we’re in two dimensions, orthogonal becomes our familiar perpendicular.
### Orthogonality
Beginning with the familiar notion of perpendicular, we can generalize this to understand orthogonality. The geometric interpretation of two vectors being perpendicular is that the angle between them is $90^{\circ}$. Once we leave two and three dimensions (or jump to the space of polynomials, as we’ll do soon), the concept of an angle isn’t as helpful.
Another way to define perpindicular is through an operation known as the dot productSuppose we take two 2D vectors, $\mathbf{x}$ and $\mathbf{y}$. Each vector will have coordinates: $\mathbf{x} = (x_{1},x_{2})$ and $\mathbf{y} = (y_{1}, y_{2})$. The dot product is a special type of multiplication defined on vectors, and denoted $\cdot$:
$$\mathbf{x}\cdot\mathbf{y} = x_{1}y_{1} + x_{2}y_{2}$$
The dot product can be described in words as the sum of the component-wise multiplication of the coordinates of the two vectors.
Now, we can say that two vectors are perpendicular if their dot product is 0. That is, $\mathbf{x}$ and $\mathbf{y}$ are perpindicular if $\mathbf{x}\cdot\mathbf{y} = 0$. (This article gives a nice overview of how we move from the algebraic to the geometric definition of perpendicular and orthogonal.)
Remember, perpendicular doesn’t make sense once we get into higher dimensions. Orthogonal is a more general notion of vectors being perpendicular, and is defined for two vectors (of any length) as their dot product equalling zero.
### From Dot Product to Inner Product
The dot product is used on vectors, and defines another type of product that is different from the scalar multiplication we know. In fact, we can generalize the notion of a dot product to something called an inner product, which can be defined on many different spaces than just vectors. We can define operations and products however we like, but for our definition to qualify as an inner product (denoted $\langle \cdot, \cdot\rangle$), it must meet certain criteria
For instance, on the set of real valued functions with domain $[a,b]$, we define the inner product of two functions $f(x)$ and $g(x)$ as
$$\langle f, g\rangle := \int_{a}^{b}f(x)g(x)dx$$
The concept of orthogonality generalizes to an inner product as well. If the inner product of two functions is zero (as defined above), we say the functions are orthogonal.
### Back to the paper
The authors claim to be using a set of orthogonal polynomials to define their drop probability function, and they give the structure of such functions. For a domain $[a,b]$, and for $\phi_{j}$ in the set of polynomials, they define $\phi_{j} = (x-a)^{j-1}(b-x)$. So, for example, $\phi_{1} = (b-x)$, and $\phi_{5} = (x-a)^{4}(b-x)$
Now, in order to be an orthogonal basis for a space1, the set of functions that are claimed to form the basis of the set must be pairwise orthogonal. That is, I need to be able to take the inner product of any two functions $\phi_{i}$ and $\phi_{j}$ and get 0. If that isn’t true for even one pair, then the set is not orthogonal.
As it turns out, if we take the inner product of any two functions in the basis set over the domain given, we find that there are no pairs that are orthogonal. To do this in general, we compute the integral
$$\int_{a}^{b}(x-a)^{i-1}(b-x)\cdot (x-a)^{j-1}(b-x)dx$$
The integral computation is one of simple polynomial integration, and can be done either by hand or using your favorite software (Mathematica) of choice. What we find here is that this set of functions defined in general this way is never orthogonal, yet the paper claims they are.
Applying to the particular situation of designing a drop probability function, they give the following for average queue length thresholds $T_{\min}$ and $T_{\max}$
$$p(x,a_{1},a_{2}) = \left\{\begin{array}{lr}0, &x < T_{\min}\\\phi_{0} + a_{1}\phi_{1}(x) + a_{2}\phi_{2}(x),&T_{\min}\leq x \leq T_{\max}\\1,&x > T_{\max}\end{array}\right.$$
where the basis functions are
\begin{aligned}\phi_{0}(x) &= p_{m}\frac{x-T_{\min}}{T_{\max}-T_{\min}}\\\phi_{1}(x) &= (x-T_{\min})(T_{\max}-x)\\\phi_{2}(x) &= (x-T_{\min})^{2}(T_{\max}-x)\end{aligned}
The reader will recognize $\phi_{0}$ as the original drop probability function from the RED algorithm. These functions are absolutely not orthogonal though (as the authors claim), and a simple check as we did above will verify it.
## Other issues
Another issue is that these mysterious coefficients $a_{1}$ and $a_{2}$ need to be determined. How, you ask? The authors do not say, other than to note that one can define “a functional” implicit on the unknown $p(x,a_{1},a_{2})$ that can be minimized to find the optimal values for those coefficients. They write that this mysterious functional2 can be based on either the average queue length or average waiting time, yet provide no details whatsoever as to the functional they have chosen for this purpose. They provide a figure with a sample function, but give no further details as to how it was obtained.
One other issue I have in their methodology is discussing the order of estimation. For those familiar with all sorts of ways to estimate unknown functions, from Taylor series, to splines, to Fourier series, we know that a function is exactly equal to an infinite sum of such functions. Any finite sum is an estimation, and the number of terms we use for estimation with a desired accuracy may change with the function being estimated.
For instance, if I want to use Taylor series (a linear combination of polynomials) to estimate a really icky function, how many terms should I include to ensure my accuracy at a point is within 0.001 of the real value? It depends on the function.3.
The authors simply claim that a second order term is appropriate in this case. The issue I take with that is these queueing management drop probability functions are designed. We’re not estimating a function describing a phenomenon, we are designing a control policy to seek some sort of optimal behavior. Fundamentally, the authors posing this as an estimation problem of some unknown drop probability function is incorrect. This isn’t a law of nature we seek to observe; it’s a control policy we seek to design and implement and optimize. Using language that implies the authors are estimating some “true function” is misleading.
Regarding the simulation itself, since the design was arbitrary, and not based on sound mathematical principles, I cannot give any real comment to the simulations and the results. The authors briefly discuss and cite some papers that explore the behavior of network traffic, and claim to take this into account in their simulations. To those, I cannot comment yet.
## Conclusion
Always verify a paper for yourself, and don’t accept anything at face value. Research and technical publications should be completely transparent as to choices and methodologies (and obviously free of mathematical inaccuracies), and derivations and proofs should be present when necessary, even if in an appendix.
Almost every textbook in probability or statistics will speak of classifying distributions into two different camps: discrete (singular in some older textbooks) and continuous. Discrete distributions have either a finite or a countable sample space (also known as a set of Lebesgue measure 0), such as the Poisson or binomial distribution, or simply rolling a die. The probability of each point in the sample space is nonzero. Continuous distributions have a continuous sample space, such as the normal distribution. A distribution in either of these classes is either characterized by a probability mass function (pmf) or probability distribution function (pdf) derived from the distribution function via taking a derivative. There is, however, a third kind.
One rarely talked about, or mentioned quickly and then discarded. This class of distributions is defined on a set of Lebesgue measure 0, yet the probability of any point in the set is 0, unlike discrete distributions. The distribution function is continuous, even uniformly continuous, but not absolutely continuous, meaning it’s not a continuous distribution. The pdf doesn’t exist, but one can still find moments of the distribution (e.g. mean, variance). They are almost never encountered in practice, and the only real example I’ve been able to find thus far is based on the Cantor set. This class is the set of red-headed step-distributions– the singular continuous distributions.
## Back up, what is Lebesgue measure?
Measure theory itself can get extremely complicated and abstract. The idea of measures is to give the “size” of subsets of a space. Lebesgue measure is one type of measure, and is actually something most people are familiar with: the “size” of subsets of Euclidean space in $n$ dimensions. For example, when $n=1$, we live in 1D space. Intervals. The Lebesgue measure of an interval $[a,b]$ on the real line is just the length of that interval: $b-a$. When we move to two dimensions, $\mathbb{R}\times \mathbb{R}$, the Cartesian product of 1D space with itself, our intervals combine to make rectangles. The Lebesgue measure in 2D space is area; so a rectangle built from $[a,b]\times [c,d]$ has Lebesgue measure $(b-a)(d-c)$. Lebesgue measure in 3D space is volume. And so forth.
Now, points are 0-dimensional in Euclidean space. They have no size, no mass. They have Lebesgue measure 01. Intuitively, we can simply see that Lebesgue measure helps us see how much “space” something takes up in the Euclidean world, and points take up no space, and hence should have measure 0.
In fact, any countable set of points has Lebesgue measure 0. Even an infinite but countable set. The union of disjoint Lebesgue measurable sets has a measure equal to the sum of the individual sets. Points are certainly disjoint, and they each have measure 0, and summing 0 forever still yields 0.2 So, the set $\{0,1,2\}$ has Lebesgue measure 0. But so do the natural numbers $\mathbb{N}$and the rational numbers $\mathbb{Q}$, even though the rational numbers contain the set of natural numbers.
It is actually possible to construct an uncountable infinite set that has Lebesgue measure 0, and we will need that in constructing our example of a singular continuous distribution. For now, we’ll examine discrete and continuous distributions briefly.
## Discrete (Singular) Distributions
These are the ones most probability textbooks begin with, and most of the examples that are familiar.
### Roll a fair die.
The sample space for a roll of a fair die $X$ is $S =\{1,2,3,4,5,6\}$. The PMF is $P(X = x) = 1/6$, where $x \in S$. The CDF is given by the function $P(X\leq x) = \sum_{j\leq x}P(X=j)$
Example:
$$P(X \leq 4) = \sum_{j\leq 4}\frac{1}{6} = \frac{2}{3}$$
### Binomial Distribution
A binomial random variable $X$ counts the number of “successes” or 1s in a binary sequence of $n$ Bernoulli random variables. Think a sequence of coin tosses, and counting the number of heads. In this case, the sample space is infinite, but countable: $S = \{0,1,2,\ldots\}$. If the probability of a 1, or “success” is $p$, then the PMF of $X$ is given by
$$P(X=x) = {n \choose x}p^{x}(1-p)^{n-x}$$
Note here again that the sample space is of Lebesgue measure 0, but the probability of any point in that space is a positive number.
## Continuous Distributions
Continuous distributions operate on a continuous sample space, usually an interval or Cartesian product of intervals or even a union of intervals. Continuous distribution functions $F$ are absolutely continuous, meaning that (in one equivalent definition), the distribution function has a derivative $f=F'$ almost everywhere that is Lebesgue integrable, and obeys the Fundamental Theorem of Calculus:
$$F(b)-F(a) = \int_{a}^{b}f(x)dx$$
for $a< b$. This $f$ is the probability distribution function (PDF), derived by differentiating the distribution function. Let’s mention some examples of these:
### The Continuous Uniform Distribution
Suppose we have a continuous interval $[a,b]$, and the probability mass is spread equally along this interval, meaning that the probability that our random variable $X$ lies in any subinterval of size $s$ has the same probability, regardless of location. Suppose we do not allow the random variable to take any values outside the interval. The sample space is continuous but over a finite interval. The distribution function for this $X$ is given by
$$F(x) = \left\{\begin{array}{lr}0&x< a\\\frac{x-a}{b-a}&a\leq x \leq b\\1&x > b\end{array}\right.$$
This is an absolutely continuous function. Then we may easily derive the PDF by differentiating $F$:
$$f(x) = \mathbb{1}_{x \in [a,b]}\frac{1}{b-a}$$
where $\mathbb{1}_{x \in [a,b]}$ is the indicator function that takes value 1 if $x$ is in the interval, and 0 otherwise.
This distribution is the continuous version of a die roll. The die roll is the discrete uniform distribution, and here we just allow for a die with uncountably many sides with values in $[a,b]$. The probability of any particular point is 0, however, even though it is possible to draw a random number from this interval. To see this, note that the probability that the random variable $X$ lies between two points in the interval, say $x_{1}$ and $x_{2}$ is given by multiplying the height of the PDF by the length (Lebesgue measure) of the subinterval. The Lebesgue measure of a point is 0, so even though a value for the PDF exists at that point, the probability is 0.
We don’t run into issues here mathematically because we are on a continuous interval.
### The Normal Distribution
Likely the most famous continuous distribution, the normal distribution is given by the famous “bell curve.” In this case, the sample space is the entire real line. The probability that a normally distributed random variable $X$ lies between any two points $a$ and $b$ is given by
$$P(a\leq X \leq b) = \int_{a}^{b}\frac{1}{\sqrt{2\pi\sigma^{2}}}\exp\left(-\frac{(x-\mu)^{2}}{2\sigma^{2}}\right)dx$$
where $\mu$ is the mean and $\sigma^{2}$ is the variance.
## Singular Continuous Distributions
We’re going to begin this section by discussing everyone’s favorite counterexample in mathematics: the Cantor set.
### The Cantor set
The Cantor set is given by the limit of the following construction:
1. Take the interval $[0,1]$
2. Remove the middle third: $(1/3, 2/3)$, so you’re left with $[0,1/3]\cup[2/3,1]$
3. Remove the middle third of each of the remaining intervals. So you remove $(1/9,2/9)$ from $[0,1/3]$ and $(7/9,8/9)$ from $[2/3,1]$, leaving you with the set $[0,1/9]\cup[2/9,1/3]\cup[2/3,7/9]\cup[8/9,1]$
Continue this process infinitely.
This is an example of a set that is uncountable, yet has Lebesgue measure 0. Earlier, when we discussed Lebesgue measure, we noted that all countable sets had measure 0. Thus we may conclude that only uncountable sets (like intervals) have nonzero Lebesgue measure. However, the Cantor set illustrates that not all uncountable sets have positive Lebesgue measure. To see why the Cantor set has Lebesgue measure 0, we will look at the measure of the sets that are removed (the complement of the Cantor set):
At the first step, we have removed one interval of size 1/3. At the second step, we remove two intervals of size 1/9. At the third step, we remove four intervals of size 1/27. Let’s call $S_{n}$ the subset removed from the interval [0,1] by the $n$th step. By the end of the third step, we have removed a set of size
$$m(S_{3}) = \frac{1}{3} + \frac{2}{3^{2}} + \frac{4}{3^{3}}$$
By the $n$th step,
$$m(S_{n}) = \sum_{j=0}^{n}\frac{2^{j}}{3^{j+1}}$$
This is the partial sum of a geometric series, so
$$m(S_{n}) = 1-\left(\frac{2}{3}\right)^{n}$$
Now, the Cantor set is formed when $n \to \infty$. The measure of the complement of the Cantor set, which we called $S_{\infty}$ then has measure
$$m(S_{\infty}) = \lim_{n \to \infty}m(S_{n}) = \lim_{n \to \infty}1-\left(\frac{2}{3}\right)^{n} = 1$$
But the original interval we started with had Lebesgue measure 1, and the union of the Cantor set with its complement $S_{\infty}$ is the interval [0,1]. That means that the measure of the Cantor set plus the measure of its complement must add to 1, which implies that the Cantor set is of measure 0. However, since we removed open intervals during the construction, there must be something left; in fact, there are uncountably many points left.
Now we have an uncountable set of Lebesgue measure 0. We’re going to use this set to construct the only example I could find of a singular continuous distribution. It is very important that the Cantor set is an uncountable set of Lebesgue measure 0.
### Building the Cantor distribution
Update: Following a correction from an earlier version, I’m going to show how to construct this distribution directly and via the complement of the Cantor set. The latter was used in a textbook I found, and is a bit convoluted in its construction, but I’m going to leave it.
The direct construction is to look at the intervals left behind at each stage $n$ of constructing the Cantor set. Assign a probability mass of $\frac{1}{2^{n}}$ to each of the $2^{n}$ intervals left behind, and this is your distribution function. It’s basically a continuous uniform distribution, but on stages of the Cantor set construction. Sending $n \to \infty$ yields the Cantor set, but the probability distribution moves to 0 on a set of measure 0. Thus, unlike the continuous uniform distribution, where the probability of any single point was 0, but the support has positive measure, we essentially have the continuous uniform distribution occurring on a set of measure 0, which means we have a continuous distribution function on a singular support of measure 0 that is uncountable and thus not discrete. This distribution is therefore neither continuous nor discrete.
Another way to construct this is by complement, via Kai Lai Chung’s A Course in Probability Theory.
(Note: after a second glance at this, I found this to be a relatively convoluted way of constructing this distribution, since it can be fairly easily constructed directly. However, I imagine the author’s purpose was to be very rigid and formal to cover all his bases, so I present a review of it here:)
Let’s go back to the construction of the Cantor set. At each step $n$ we have removed in total $2^{n}-1$ disjoint intervals. Let’s number those intervals, going from left to right as $J_{n,k}$, where $k = 1,2,\ldots, 2^{n}-1$
For example, at $n=2$ we have that $J_{2,1} = (1/9,2/9)$,$J_{2,2} = (1/3,2/3)$, and $J_{2,3} = (7/9,8/9)$
Now let the quantity $c_{n,k} = \frac{k}{2^{n}}$. This will be the probability mass assigned to interval $J_{n,k}$. So we define the distribution function as
$$F(x) = c_{n,k}, x \in J_{n,k}$$
Let $U_{n} = \cup_{k=1}^{2^{n}-1}J_{n,k}$, and $U = \lim_{n\to\infty}U_{n}$ The function $F$ is indeed a distribution function and can be shown to be uniformly continuous on the set $D = (-\infty,0)\cup U \cup (1,\infty)$. However, none of the points in $D$ is in the support of $F$, so the support of $F$ is contained in the Cantor set (and in fact is the Cantor set). The support (the Cantor set) has measure 0, so it is singular, but the distribution function is continuous, so it cannot be a discrete distribution. This distribution fits nowhere in our previous two classes, so we must now create a third class — the singular continuous distribution.
(By the way, even though the PDF doesn’t exist, the Cantor distribution still has mean of 1/2 and a variance of 1/8, but no mode. It does have a moment generating function.)
## Any other examples?
With some help, I spent some time poring through quite a few probability books to seek further study and other treatment of singular continuous distributions. Most said absolutely nothing at all, as if the class didn’t exist.
One book, Modern Probability Theory and Its Applications has a rather grumpy approach:
There also exists another kind of continuous distribution function, called singular continuous, whose derivative vanishes at almost all points. This is a somewhat difficult notion to picture, and examples have been constructed only by means of fairly involved analytic operations. From a practical point of view, one may act as if singular continuous distribution functions do not exist, since examples of these functions are rarely, if ever, encountered in practice.
This notion also has led me to a couple papers, which I intend to review and continue presenting my findings. I happen to have a great fondness for these “edge cases” and forgotten areas of mathematics. I believe they are the most ripe for groundbreaking new material.
Cauchy Sequences: the Importance of Getting Close
## Cauchy Sequences: the Importance of Getting Close
I am an analyst at heart, despite my recent set of algebra posts. Augustin Louis Cauchy can be argued as one of the most influential mathematicians in history, pioneering rigor in the study of calculus, almost singlehandedly inventing complex analysis and real analysis, though he also made contributions to number theory, algebra, and physics.
One of the fundamental areas he studied was sequences and their notion of convergence. Suppose I give you a sequence of numbers, and ask you what happens to this sequence if I kept appending terms forever? Would the path created by the sequence lead somewhere?
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# probability of rolling a die...
• April 16th 2011, 05:16 PM
slapmaxwell1
probability of rolling a die...
ok this problem is a little different from the others...
what is the probability of either a die roll of a six or flipping a coin once and getting a heads?
ok so this means i will use both laws together? first multiply and then add?
so the probability of a die roll is (1/6) times the probability of getting a heads (1/2)
so this is 1/12 + (1/6)(1/2)=1/6?
did i do this right?
• April 16th 2011, 06:15 PM
TKHunny
You must first decide what "or" means. Does it include both or is it exclusive?
p(6) = d
p(not six) = 1-d = f
p(not heads) = 1-h = j
Ponder this (d+f)*(h+j) = d*h + d*j + f*h + f*j
Four items to ponder. Which fit the given criteria?
• April 16th 2011, 06:22 PM
slapmaxwell1
the or means both, but im not understanding what you want me to ponder on? or i should ask, what 4 items do you want me to ponder on?
• April 16th 2011, 07:39 PM
slapmaxwell1
Quote:
Originally Posted by TKHunny
You must first decide what "or" means. Does it include both or is it exclusive?
p(6) = d
p(not six) = 1-d = f
p(not heads) = 1-h = j
Ponder this (d+f)*(h+j) = d*h + d*j + f*h + f*j
Four items to ponder. Which fit the given criteria?
---------------------------------------------------------------
ok i have to be honest, but i have no idea what you just said to me....but nonetheless i figured out the answer to the problem...here is my solution.
P(rolling a 6) = 1/6
P(flipping a head) = 1/2
P(rolling a 6 AND flipping a head) = 1/6 * 1/2 = 1/12
P(rolling a 6 OR flipping a head) = 1/6 + 1/2 - 1/12 = 7/12
or:
1H, 1T, 2H, 2T, 3H, 3T, 4H, 4T, 5H, 5T, 6H, 6T
Of those, the ones that are either a 6 or a head include
1H, 2H, 3H, 4H, 5H, 6H, and 6T
That's 7 out of 12
• April 17th 2011, 04:10 AM
Soroban
Hello, slapmaxwell1!
Quote:
What is the probability of either a die roll of a 6
or flipping a coin once and getting a Head?
You are expected to know this formula:
. . P(A $\cup$ B) .= .P(A) + P(B) - P(A $\cap$ B)
We have: .P(6) = 1/6
. . . . . . . .P(H) = 1/2
. . . . . P(6 $\cap$ H) = (1/6)(1/2) = 1/12
Therefore:
. . P(6 $\cup$ H) .= .P(6) + P(H) - P(6 $\cap$ H)
. . . . . . . . .= .(1/6) + (1/2) - (1/12)
. . . . . . . . .= . 7/12
• April 17th 2011, 06:52 AM
slapmaxwell1
Quote:
Originally Posted by Soroban
Hello, slapmaxwell1!
You are expected to know this formula:
. . P(A $\cup$ B) .= .P(A) + P(B) - P(A $\cap$ B)
We have: .P(6) = 1/6
. . . . . . . .P(H) = 1/2
. . . . . P(6 $\cap$ H) = (1/6)(1/2) = 1/12
Therefore:
. . P(6 $\cup$ H) .= .P(6) + P(H) - P(6 $\cap$ H)
. . . . . . . . .= .(1/6) + (1/2) - (1/12)
. . . . . . . . .= . 7/12
um thanks i know the formula now. :O)
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# Self Destruction, Final
## Lyrics
I am the lover in your bed (and I control you)
I am the sex that you provide (and I control you)
I am the hate you try to hide (and I control you)
I take you where you want to go
I give you all you need to know
I drag you down, I use you up
Mr. Self Destruct
Lyrics continue below...
I speak religion's message clear (and I control you)
I am denial, guilt, and fear (and I control you)
I am the prayers of the naive (and I control you)
I am the lie that you believe (and I control you)
I take you where you want to go
I give you all you need to know
I drag you down, I use you up
Mr. Self Destruct
You let me do this to you
You let me do this to you
You let me do this to you
You let me do this to you
You let me do this to you
(I am an exit)
You let me do this to you
(I am an exit)
You let me do this to you
(I am an exit)
You let me do this to you
(I am an exit)
I am the needle in your vein (and I control you)
I am the high you can't sustain (and I control you)
I am the pusher, I'm a whore (and I control you)
I am the need you have for more
"Falls wanking to the floor"
I am the bullet in the gun (and I control you)
I am the truth from which you run (and I control you)
I am the silencing machine (and I control you)
I am the end of all your dreams (and I control you)
I take you where you want to go
I give you all you need to know
I drag you down, I use you up
Mr. Self Destruct
"Wanking"
"Wanking"
"Wanking"
"Wanking"
I take you where you want to go
I give you all you need to know
I drag you down, I use you up
Mr. Self Destruct
I speak religion's message clear (and I control you)
I am denial, guilt, and fear (and I control you)
I am the prayers of the naive (and I control you)
I am the lie that you believe (and I control you)
I take you where you want to go
I give you all you need to know
I drag you down, I use you up
Mr. Self Destruct
"Wanking"
"Wanking"
"Wanking"
"Wanking"
"Wanking"
"Wanking"
"Wanking"
"Wanking"
"Wanking"
"Wanking"
"Wanking"
"Wanking"
"Wanking"
"Wanking"
"Wanking"
"Wanking"
I am the bullet in the gun (and I control you)
I am the truth from which you run (and I control you)
I am the silencing machine (and I control you)
I am the end of all your dreams (and I control you)
I take you where you want to go
I give you all you need to know
I drag you down, I use you up
Mr. Self Destruct
I take you where you want to go
I give you all you need to know
I drag you down, I use you up
Mr. Self Destruct
Writer(s): TRENT REZNOR
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# Relativity of Simultaneity
The classic example of demonstrating the relativity of simultaneity uses bolts of lightning striking two places "simultaneously". When you do the same experiment with sound then all observers can determine which event occured first because sound does not follow the 2nd postulate of SR.
Fredrik
Staff Emeritus
Gold Member
timetraveldude said:
The classic example of demonstrating the relativity of simultaneity uses bolts of lightning striking two places "simultaneously".
I have no idea what you're talking about.
timetraveldude said:
When you do the same experiment with sound...
What experiment?
timetraveldude said:
...then all observers can determine which event occured first because sound does not follow the 2nd postulate of SR.
What do you mean "determine which event occured first"? In SR, the only events that all observers agree happened in a certain order, are the ones that are timelike separated. This means that
$$-c^2t^2+x^2<0$$
where t is the time between the events (in some inertial frame) and x is the spatial distance between the events (in the same inertial frame).
jcsd
Gold Member
Yep I know the experiment, but timetrvaeldud your wrong as in relativty, the order in which an observer sees or even hears two events it not necessarily the order that they occur in his frame.
You need to read what I wrote
I know what SR says. What I am saying is that if you do the same experiemnt with sound all observers will agree on which occured first. What you are doing is taking what I am disputing and using it as evidence to refute my position.
Hurkyl
Staff Emeritus
Gold Member
What I am saying is that if you do the same experiemnt with sound all observers will agree on which occured first.
No they wouldn't.
Janus
Staff Emeritus
Gold Member
timetraveldude said:
I know what SR says. What I am saying is that if you do the same experiemnt with sound all observers will agree on which occured first.
Which is wrong. If you did the experiment with sound you would get the same results as you would with light. The speed of sound may not be invarient, but it is subject to the addition of velocities theorem.
Fredrik
Staff Emeritus
Gold Member
Timetraveldude, I take it you know SR pretty well, but it didn't seem that way to me when I read your first post in this thread. Jcsd and I both thought you meant something you didn't. Perhaps you should have told us more about that experiment, and explained your thoughts more carefully.
I think I understand (roughly) what you were trying to say, because I can imagine a thought experiment that uses light to define what events are simultaneous (to a certain observer) with a specified event. That would work with sound too. What you can't do with sound (in the same way) is determine what events are simultaneous to another observer.
Janus said:
Which is wrong. If you did the experiment with sound you would get the same results as you would with light. The speed of sound may not be invarient, but it is subject to the addition of velocities theorem.
Exactly, sound is subject to the addtion of velocities formula which light is not. I suggest you either do the though experiment or a proof.
Janus
Staff Emeritus
Gold Member
timetraveldude said:
Exactly, sound is subject to the addtion of velocities formula which light is not. I suggest you either do the though experiment or a proof.
Actually light and sound are subject to the same addition of velocities formula, namely:
$$w= \frac{u+v}{1+\frac{uv}{c^2}}$$
You are wrong again.
Janus said:
Actually light and sound are subject to the same addition of velocities formula, namely:
$$w= \frac{u+v}{1+\frac{uv}{c^2}}$$
If you want I can suggest some good books on basic relativity. This addition of velocity formula that you are stating is based on the fact that two intertial observers will never agree on events being simultaneous if they are moving relative to one another. Again you are taking the assumption that is under question and using its results to justify the assumption itself.
You need to stop regurgitating and start thinking.
Here is a very simple thought experiment. Stand at the midpoint between two machines A and B that launch tennis balls horizontally at the same speed. You begin moving towards machine B and the ball from B arrives before the ball from A. Since the speed of the balls are not the same in all interal reference frames you can determine that ball A must have traveled farther and at a slower speed. Working backwards you could determine that the balls were launched at the same time.
This is the same result that someone standing at the mid point would determine. Therefore, there is no relativity of simultaneity.
You need to be open to the possibility that what you think my be wrong.
Hurkyl
Staff Emeritus
Gold Member
You need to be open to the possibility that what you think my be wrong.
We are. Are you?
Hurkyl said:
We are. Are you?
Obviously I am open to it. What I am waiting for is someone to address the question. I have shot down all your comments. None of them have a made a dent in my contention.
Refute my thought experiment. Prove me wrong that is all you have to do.
Hurkyl
Staff Emeritus
Gold Member
If you've already made the assumption that special relativity is wrong and Gallilean relativity is right, it's fairly easy to disprove special relativity.
The trick is to do it without making the assumption, or equivalently, assume special relativity is right, and derive a contradiction.
Last edited:
anti_crank
This is the same result that someone standing at the mid point would determine. Therefore, there is no relativity of simultaneity.
This statement is extremely surprising from someone who understands relativity. The relativity of simultaneity is a mathematical consequence of the Lorentz tf. or the constancy of the speed of light principle; if that fails, so does all of SR. Thankfully, it does not.
timetraveldude said:
Here is a very simple thought experiment. Stand at the midpoint between two machines A and B that launch tennis balls horizontally at the same speed. You begin moving towards machine B and the ball from B arrives before the ball from A. Since the speed of the balls are not the same in all interal reference frames you can determine that ball A must have traveled farther and at a slower speed. Working backwards you could determine that the balls were launched at the same time.
Here's something that should settle this issue. Consider the following setup: observer A sits at the midpoint of a platform of length 2L, observer B sits somewhere else on the platform, and observer C moves at a constant speed u wrt the platform. Primed coordinates will correspond to C, unprimed to A. Choose the origin s.t. C passes by A when t=t'=0.
Now suppose that at t=0, *something* happens at both ends of the platform. It does not matter whether a light pulse is emitted, a sound wave is emitted, or a ball is sent towards A.
A will receive the two light pulses/sound waves/balls simultaneously at t=L/v. He knows where the events originated, and he assigns them spacetime coordinates (L,0) and (-L,0)
B will not receive the two light pulses/sound waves/balls simultaneously. To him, they arrive at t1=L1/v and t2=L2/v. However, once he accounts for the distance travelled, he concludes that they were emitted simultaneously and assigns them coordinates (L1, 0) and (-L2,0)
C is in a state of motion wrt A (his speed may be 1cm/s or 299000km/s, the principle still applies) so we can use the Lorentz tf. on the measurements of A. In particular, we are interested in the time components:
$$t{^'}_1 = \gamma t_1 - \frac{x_1 u}{c^2} = -\frac{Lu}{c^2}$$
Similarly $$t{^'}_2 = +Lu/c^2$$
I may have the signs inverted, but it should be clear that the times are different. This is not a distinction of whether the two light pulses/sound waves/balls *arrive* at C's position simultaneously; they obviously do not. However, unlike observer B, C still has a difference in the *emission* times after he accounts for their motion. To prevent misunderstandings, he is not allowed to explicitly refer to A or B when he makes his calculations, or to consider himself in any state of motion.
pervect
Staff Emeritus
timetraveldude said:
The classic example of demonstrating the relativity of simultaneity uses bolts of lightning striking two places "simultaneously". When you do the same experiment with sound then all observers can determine which event occured first because sound does not follow the 2nd postulate of SR.
You seem to me to be a bit confused. The fact that there is an atmosphere does appear at first glance to set up a specific coordinate system (it's not quite a frame because of various effects such as the rotation of the earth) that's at rest relative to the atmosphere.
In this coordinate system, it's possible to synchronize clocks. Unfortunately, the details of how this clock synchronization process works are going to depend on a large number of factors, including how the winds are blowing at that moment, because when you look closely, the atmosphere isn't at rest relative to itself.
So you haven't really defined a very useful or practical coordinate system at all so far.
If you manage to "fix up" your proposal to solve the problems of winds affecting how clocks are synchronized, you will still not have disproven relativity.
In fact, people already have a well defined system for synchronizing clocks on the Earth's surface. It's called, TAI time, which is a coordinate time. It's called a coordinate time because it sets up a coordinate system on the Earth's surface.
Now the next question is why you think that setting up a specific coodinate system with non-Einstenian clock synchronization somehow disproves relativity
A lot of people make the mistake (and I would guess that this is your error) that because they have set up some particular coordinate system, or used an existing system such as TAI time, Newton's laws must work in that coordinate system.
So the problem in these cases are not in relativity - it is in the assumption that Newton's laws must work in their coordiante system. This is a bad assumption.
Relativity points out otherwise - if you do not synchronize clocks according to Einstein's conventions, you will not have a coordinate system where you can apply Newton's laws, even at low velocities.
The "velocities" that one measures in a coordinate system that is not based on Einsteinian clock synchronization have an inherent spatial bias, a "preferred direction". They are not isotropic.
This means specifically that a mass m moving in one direction with a "coordinate velocity" v will not have the same momentum as a mass m moving in the opposite direction with a "coordinate velocity" v.
Another way of putting this is that if you were running an automobile race, automobiles would run faster in one direction than another.
The Einsteinan condition of using light to synchronize clocks is the same convention that's needed to run "fair" automobile races, ones in which no automobile has an advantage going in any particular direction - to use the technical term, an isotropic coordinate system.
Usually this gets simplified to the point of view of telling people that they "have to" use Einstein's method of synchronizing clocks. This is not a bad approach, however it's slightly over simplified. A more complete discussion includes the "or else", which is, as I described, the fact that your coordinate system won't be isotropic if you don't use Einstein's clock synchronization method.
I'd like to post some web references, but unfortunately I only have one which isn't very helpful - other than Einstein's original papers, I haven't seen much discussion about isotropy. Google finds a few crank pages, but nothing potentially useful, other than an abstract from a Physical review D paper by Clifford Will (and that's not very useful in abstract form, and I can't answer to how useful even the original would be not having seen it).
Fredrik
Staff Emeritus
Gold Member
Timetraveldude, don't you think the problem with the discussion in these two threads is that you haven't expressed yourself clearly enough? I read your thought experiment with the tennis balls, and unfortunately I must place it in the "not even wrong" category. You haven't stated what your postulates are, it's impossible to follow your reasoning, and even your conclusion is difficult to understand.
I can see that you're saying that "there is no relativity of simultaneity" (which is clearly in contradiction with SR), but what are you really saying? Are you saying that your conclusion is not in contradiction with SR, or that your thought experiment proves that SR is wrong?
Here is a very simple thought experiment. Stand at the midpoint between two machines A and B that launch tennis balls horizontally at the same speed. You begin moving towards machine B and the ball from B arrives before the ball from A. Since the speed of the balls are not the same in all interal reference frames you can determine that ball A must have traveled farther and at a slower speed. Working backwards you could determine that the balls were launched at the same time.
This is the same result that someone standing at the mid point would determine. Therefore, there is no relativity of simultaneity.
Let me see if you have your setup right:
Code:
A -->o O o<--- B
O' -->
O is an observer at rest with respect to A and B, at the midpoint between A and B.
A and B are launching projectiles with a certain speed, lets call it v.
O' is moving toward B at some speed v'.
Last edited:
anti_crank said:
This statement is extremely surprising from someone who understands relativity. The relativity of simultaneity is a mathematical consequence of the Lorentz tf. or the constancy of the speed of light principle; if that fails, so does all of SR. Thankfully, it does not.
Here's something that should settle this issue. Consider the following setup: observer A sits at the midpoint of a platform of length 2L, observer B sits somewhere else on the platform, and observer C moves at a constant speed u wrt the platform. Primed coordinates will correspond to C, unprimed to A. Choose the origin s.t. C passes by A when t=t'=0.
Now suppose that at t=0, *something* happens at both ends of the platform. It does not matter whether a light pulse is emitted, a sound wave is emitted, or a ball is sent towards A.
A will receive the two light pulses/sound waves/balls simultaneously at t=L/v. He knows where the events originated, and he assigns them spacetime coordinates (L,0) and (-L,0)
B will not receive the two light pulses/sound waves/balls simultaneously. To him, they arrive at t1=L1/v and t2=L2/v. However, once he accounts for the distance travelled, he concludes that they were emitted simultaneously and assigns them coordinates (L1, 0) and (-L2,0)
C is in a state of motion wrt A (his speed may be 1cm/s or 299000km/s, the principle still applies) so we can use the Lorentz tf. on the measurements of A. In particular, we are interested in the time components:
$$t{^'}_1 = \gamma t_1 - \frac{x_1 u}{c^2} = -\frac{Lu}{c^2}$$
Similarly $$t{^'}_2 = +Lu/c^2$$
I may have the signs inverted, but it should be clear that the times are different. This is not a distinction of whether the two light pulses/sound waves/balls *arrive* at C's position simultaneously; they obviously do not. However, unlike observer B, C still has a difference in the *emission* times after he accounts for their motion. To prevent misunderstandings, he is not allowed to explicitly refer to A or B when he makes his calculations, or to consider himself in any state of motion.
Again you have failed to see my motionvation for this post. It is clear from the Lorentz transformations that if one observer says two events we simultaneous another inertial observer moving relative to them will not say they were simultaneous. This is obvious.
The point I am making is that if the relativity of simultaneity is invalidated using material objects then there is something wrong with SR. This requires a new method of deriving the Lorentz transformations since the argument based on the constantcy of light in all inertial reference frames is the basis for the original derivation.
Fredrik said:
Timetraveldude, don't you think the problem with the discussion in these two threads is that you haven't expressed yourself clearly enough? I read your thought experiment with the tennis balls, and unfortunately I must place it in the "not even wrong" category. You haven't stated what your postulates are, it's impossible to follow your reasoning, and even your conclusion is difficult to understand.
I can see that you're saying that "there is no relativity of simultaneity" (which is clearly in contradiction with SR), but what are you really saying? Are you saying that your conclusion is not in contradiction with SR, or that your thought experiment proves that SR is wrong?
You inability to understand me is your problem.
Janus
Staff Emeritus
Gold Member
timetraveldude said:
Obviously I am open to it. What I am waiting for is someone to address the question. I have shot down all your comments. None of them have a made a dent in my contention.
Refute my thought experiment. Prove me wrong that is all you have to do.
All you done is set up a thought experiment were you assumed from the start that the Rules of Relativity don't hold. Then, suprise, surprise, you find that your conclusion doesn't agree with that of Relativity.
But that's exactly what you accused me of doing.
The only thing thought experiments can do is show you what results you should get from your starting postulates, and whether they are self consistant or not. ( Whether or not they lead to any internal contradictions.)
Both Gallilean relativity and Special Relativity are self consistant. Ergo, we cannot rely on thought experiment alone to choose between them.
We must actually perform a physical experiment and check the results against the predicitions of both.
To date, every actual experiment performed upholds the predictions of SR.
Fredrik
Staff Emeritus
Gold Member
timetraveldude said:
You inability to understand me is your problem.
You seem to have at least two problems yourself. One is your inability to express yourself clearly. The other is obviously an attitude problem.
You insult everyone who doesn't understand what you're trying to say. Have you ever considered that it might be your own fault? If nobody understands you, then maybe there's something wrong with the way you're writing.
The description of your thought experiment is not very good. You need to at least state clearly what your assumptions are.
timetraveldude said:
The point I am making is that if the relativity of simultaneity is invalidated using material objects then there is something wrong with SR.
Yes, but that's a pretty big if. Huge. Enormous.
timetraveldude said:
This requires a new method of deriving the Lorentz transformations ...
No, it doesn't. It would require that we abandon the Lorentz transformations and replace SR with a new theory.
timetraveldude said:
...since the argument based on the constantcy of light in all inertial reference frames is the basis for the original derivation.
You seem to be missing one important point. You don't have to derive the Lorentz transformations. That speed-of-light thing is how the Lorentz tranformations were discovered historically, but how they were discovered has no relevance to the validity of the theory.
SR only requires one postulate: "Space and time is represented mathematically by 4-dimensional Minkowski space". This is an extremely powerful statement that contains everything about Lorentz transformations and much more.
Janus said:
All you done is set up a thought experiment were you assumed from the start that the Rules of Relativity don't hold. Then, suprise, surprise, you find that your conclusion doesn't agree with that of Relativity.
But that's exactly what you accused me of doing.
The only thing thought experiments can do is show you what results you should get from your starting postulates, and whether they are self consistant or not. ( Whether or not they lead to any internal contradictions.)
Both Gallilean relativity and Special Relativity are self consistant. Ergo, we cannot rely on thought experiment alone to choose between them.
We must actually perform a physical experiment and check the results against the predicitions of both.
To date, every actual experiment performed upholds the predictions of SR.
All I assumed are the two postulates of SR.
quantumdude
Staff Emeritus
Gold Member
fixizrox said:
All I assumed are the two postulates of SR.
No, you assumed that all observers will agree on the priority of the two audible signals. That means you assumed that the postulates of SR are false.
jcsd
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# How do you solve 7/(x+3)<-5?
Jul 29, 2017
$\left(- \frac{22}{5} , - 3\right)$
#### Explanation:
$\text{add 5 to both sides}$
$\Rightarrow \frac{7}{x + 3} + 5 < 0$
$\text{express as a single rational function}$
$\frac{7 + 5 \left(x + 3\right)}{x + 3} < 0$
$\Rightarrow \frac{5 x + 22}{x + 3} < 0$
$\text{the zeros of the numerator/denominator are}$
$\text{numerator "x=-22/5," denominator } x = - 3$
These indicate where the rational function may change sign.
$\text{the intervals for consideration are}$
$x < - \frac{22}{5} , \textcolor{w h i t e}{x} - \frac{22}{5} < x < - 3 , \textcolor{w h i t e}{x} x > - 3$
$\text{consider a "color(blue)"test point "" in each interval}$
$\text{we want to find where the function is negative}$
$\text{substitute each test point into the function and consider}$
$\text{it's sign}$
$\textcolor{m a \ge n t a}{\text{x = - 5"to(-)/(-)tocolor(red)" positive}}$
$\textcolor{m a \ge n t a}{\text{x = - 4 "to(+)/(-)tocolor(blue)" negative}}$
$\textcolor{m a \ge n t a}{\text{x = 4"to(+)/(+)tocolor(red)" positive}}$
$\Rightarrow - \frac{22}{5} < x < - 3 \text{ or } \left(- \frac{22}{5} , - 3\right)$
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$$\renewcommand\AA{\unicode{x212B}}$$
# ISIS Powder Diffraction Scripts - OSIRIS Reference¶
## Creating OSIRIS Object¶
This method assumes you are familiar with A quick introduction to objects
To create a OSIRIS object the following parameters are required:
Optionally a configuration file may be specified if one exists using the following parameter:
See Using configuration files on YAML configuration files for more details
### Example¶
from isis_powder.osiris import Osiris
calibration_dir = r"C:\path\to\calibration_dir"
output_dir = r"C:\path\to\output_dir"
config_dir = r"C:\path\to\config.yaml"
osiris_example = Osiris(user_name="Calib",
calibration_directory=calibration_dir,
output_directory=output_dir,
config_file=config_dir)
## Methods¶
The following methods can be executed on a OSIRIS object:
For information on creating a OSIRIS object see: Creating OSIRIS Object
### run_diffraction_focusing¶
The diffraction_focusing method allows a user to process a series of runs into a focused dSpace workspace. Whilst processing the runs the scripts can apply any corrections the user enables.
On OSIRIS the following parameters are required when executing create_vanadium:
#### Example¶
# Notice how the filename ends with .yaml
cal_mapping_file = r"C:\path\to\cal_mapping.yaml"
osiris_example.run_diffraction_focusing(run_number="119977-119988",
merge_drange=True,
subtract_empty_can=True,
calibration_mapping_file=cal_mapping_file)
## Calibration Mapping File¶
The calibration mapping file holds the mapping between run numbers, current label, offset filename, empty run numbers, and vanadium run numbers.
For more details on the calibration mapping file see: Cycle mapping files
The layout on OSIRIS should look as follows for each block:
11-120:
2 label: "1_1"
3 offset_file_name: "offset_file.cal"
4 vanadium_drange1 : "13"
5 vanadium_drange2 : "14"
6 vanadium_drange3 : "15"
7 vanadium_drange4 : "16"
8 vanadium_drange5 : "17"
9 vanadium_drange6 : "18"
10 vanadium_drange7 : "19"
11 vanadium_drange8 : "20"
12 vanadium_drange9 : "21"
13 vanadium_drange10 : "22"
14 vanadium_drange11 : "23"
15 vanadium_drange12 : "24"
16 vanadium_run_numbers : "13-24"
17 empty_drange1 : "1"
18 empty_drange2 : "2"
19 empty_drange3 : "3"
20 empty_drange4 : "4"
21 empty_drange5 : "5"
22 empty_drange6 : "6"
23 empty_drange7 : "7"
24 empty_drange8 : "8"
25 empty_drange9 : "9"
26 empty_drange10 : "10"
27 empty_drange11 : "11"
28 empty_drange12 : "12"
29 empty_run_numbers : "1-12"
For each drange, any empty canisters and vanadium runs must be listed with the associated drange.
## Parameters¶
The following parameters for OSIRIS are intended for regular use when using the ISIS Powder scripts.
### calibration_directory¶
This parameter should be the full path to the calibration folder. Within the folder the following should be present:
• Grouping .cal file
• Folder(s) with the label name specified in mapping file (e.g. “1_1”) - Inside each folder should be the offset file with name specified in the mapping file
Example Input:
osiris_example = Osiris(calibration_directory=r"C:\path\to\calibration_dir", ...)
### calibration_mapping_file¶
This parameter gives the full path to the YAML file containing the calibration mapping. For more details on this file see: Calibration Mapping File
Note: This should be the full path to the file including extension
Example Input:
# Notice the filename always ends in .yaml
osiris_example = Osiris(calibration_mapping_file=r"C:\path\to\file\calibration_mapping.yaml", ...)
### config_file¶
The full path to the YAML configuration file. This file is described in detail here: Using configuration files It is recommended to set this parameter at object creation instead of on a method as it will warn if any parameters are overridden in the scripting window.
Note: This should be the full path to the file including extension
Example Input:
# Notice the filename always ends in .yaml
osiris_example = Osiris(config_file=r"C:\path\to\file\configuration.yaml", ...)
### do_van_normalisation¶
Indicates whether to divide the focused workspace within run_diffraction_focusing mode with an associated vanadium run.
Accepted values are: True or False
Example Input:
osiris_example = Osiris(do_van_normalisation=True, ...)
### file_ext¶
Optional
Specifies a file extension to use for the run_diffraction_focusing method.
This should be used to process partial runs. When processing full runs (i.e. completed runs) it should not be specified as Mantid will automatically determine the best extension to use.
Note: A leading dot (.) is not required but is preferred for readability
Example Input:
osiris_example = Osiris(file_ext=".s01", ...)
### merge_drange¶
Indicates whether to merge summed workspaces of different dranges after running the run_diffraction_focusing method.
Accepted values are: True or False
Example Input:
osiris_example = Osiris(merge_drange=True, ...)
### output_directory¶
Specifies the path to the output directory to save resulting files into. The script will automatically create a folder with the label determined from the calibration_mapping_file and within that create another folder for the current user_name.
Within this folder processed data will be saved out in several formats.
Example Input:
osiris_example = Osiris(output_directory=r"C:\path\to\output_dir", ...)
### run_number¶
Specifies the run number(s) to process when calling the run_diffraction_focusing method.
This parameter accepts a single value or a range of values with the following syntax:
- : Indicates an inclusive range of runs (e.g. 1-10 would process 1, 2, 3….8, 9, 10)
, : Indicates a gap between runs (e.g. 1, 3, 5, 7 would process run numbers 1, 3, 5, 7)
These can be combined like so: 1-3, 5, 8-10 would process run numbers 1, 2, 3, 5, 8, 9, 10.
In addition the input_mode parameter determines what effect a range of inputs has on the data to be processed
Example Input:
# Process run number 1, 3, 5, 6, 7
osiris_example = Osiris(run_number="1, 3, 5-7", ...)
# Or just a single run
osiris_example = Osiris(run_number=100, ...)
### user_name¶
Specifies the name of the current user when creating a new OSIRIS object. This is only used when saving data to sort data into respective user folders.
Example Input:
osiris_example = Osiris(user_name="Mantid", ...)
### subtract_empty_can¶
Provides the option to disable subtracting empty canister runs from the run being focused. Set to False to disable empty subtraction.
Example Input:
subtract_empty_can: True
Category: Techniques
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Mechanics Homework 14-3 1011 - Chain Puller Name_______________ Mr. Haynes Date_______________
Problem 14-3 A horizontal force P is applied to the handle of the puller. Determine, in terms of force P, the tension T in the chain and the magnitude and direction of the force acting on the handle at pin A.
$$\bar{T} = \text{5.36 P}$$ at 20° down and to the right
$$\bar{A} = \text{4.43 P}$$ at 24.4° up and to the left
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Journal of Advanced Studies in Topology http://m-sciences.com/index.php?journal=jast <p style="text-align: justify;">The Journal of Advanced Studies in Topology (<strong>JAST</strong>) is dedicated to rapid publication of the highest quality short papers, regular papers, and expository papers.<span class="Apple-converted-space"> </span><strong>JAST</strong><span class="Apple-converted-space"> </span>is a peer-reviewed international journal, which publishes original research papers and survey articles in all aspects of topology and its applications.</p> Modern Science Publishers en-US Journal of Advanced Studies in Topology 2090-8288 <p>No manuscript should be submitted which has previously been published, or which has been simultaneously submitted for publication elsewhere. The copyright in a published article rests solely with the Modern Science Publishers, and the paper may not be reproduced in whole in part by any means whatsoever without prior written permission.</p> Soft almost $$\alpha$$-continuous mappings http://m-sciences.com/index.php?journal=jast&page=article&op=view&path%5B%5D=1415 <p>In the present paper the concept of soft almost $$\alpha$$-continuous mappings and soft almost $$\alpha$$-open mappings in soft topological spaces have been introduced and studied.</p> S. S. Thakur A. S. Rajput ##submission.copyrightStatement## http://creativecommons.org/licenses/by-nc/4.0 2018-08-10 2018-08-10 9 2 94 99 10.20454/jast.2018.1415 Weak and strong forms of fuzzy $$\alpha$$-open (closed) sets and its applications http://m-sciences.com/index.php?journal=jast&page=article&op=view&path%5B%5D=1411 <p>In this paper, we generalize the concept of infra-$$\alpha$$-open (closed) and supra-$$\alpha$$-open (closed) sets to fuzzy topological spaces and basic properties of these new concepts have been introduced. Some applications on fuzzy (supra-) infra-$$\alpha$$-open (closed) sets, likely, fuzzy (supra-) infra-$$\alpha$$-continuous mappings, fuzzy (supra-) infra-$$\alpha$$-open (closed) mappings, fuzzy supra-$$\alpha$$- irresolute mapping and fuzzy supra-$$\alpha$$-connected space are introduced. The relations and converse relations between these new concepts and others kinds of fuzzy open sets and fuzzy continuous mappings are discussed. Special results about these new concepts are investigated and studied.</p> Hakeem A. Othman Alanod M. Sibih ##submission.copyrightStatement## http://creativecommons.org/licenses/by-nc/4.0 2018-08-10 2018-08-10 9 2 100–112 100–112 10.20454/jast.2018.1411 Separation axioms on function spaces defined on bitopological spaces http://m-sciences.com/index.php?journal=jast&page=article&op=view&path%5B%5D=1454 <p>In this paper, we introduce separation axioms on the function space <em>p</em><em>− </em><em>C</em><em>ω</em>(<em>Y, Z</em>) and study how they relate<br>to separation axioms defined on the spaces (<em>Z, δ</em><em>i</em>) for <em>i </em>= 1<em>, </em>2, (<em>Z, δ</em>1<em>, δ</em>2), 1 <em>− </em><em>C</em><em>ς</em>(<em>Y, Z</em>) and 2 <em>− </em><em>C</em><em>ζ</em>(<em>Y, Z</em>). It<br>is shown that the space <em>p </em><em>− </em><em>C</em><em>ω</em>(<em>Y, Z</em>) is <em>p</em><em>T</em><em>◦</em>, <em>p</em><em>T</em>1, <em>p</em><em>T</em>2 and <em>p</em>regular, if the spaces (<em>Z, δ</em>1) and (<em>Z, δ</em>2) are both<br><em>T0</em>, <em>T</em>1, <em>T</em>2 and regular respectively. The space <em>p </em><em>− </em><em>C</em><em>ω</em>(<em>Y, Z</em>) is also shown to be <em>p</em><em>T0</em>, <em>p</em><em>T</em>1, <em>p</em><em>T</em>2 and <em>p</em>regular,<br>if the space (<em>Z, δ</em>1<em>, δ</em>2) is <em>p </em><em>− </em><em>T0</em>, <em>p </em><em>− </em><em>T</em>1, <em>p </em><em>− </em><em>T</em>2 and <em>p</em>-regular respectively. Finally, the space <em>p </em><em>− </em><em>C</em><em>ω</em>(<em>Y, Z</em>) is<br>shown to be <em>p</em><em>T0</em>, <em>p</em><em>T</em>1, <em>p</em><em>T</em>2 and <em>p</em>regular, if and only if the spaces 1 <em>− </em><em>C</em><em>ς</em>(<em>Y, Z</em>) and 2 <em>− </em><em>C</em><em>ζ</em>(<em>Y, Z</em>) are both <em>T</em>0,<br><em>T</em>1, <em>T</em>2, and only if the spaces 1 <em>− </em><em>C</em><em>ς</em>(<em>Y, Z</em>) and 2 <em>− </em><em>C</em><em>ζ</em>(<em>Y, Z</em>) are both regular respectively.</p> N. E. Muturi J. M. Khalagai G. P. Pokhariyal ##submission.copyrightStatement## http://creativecommons.org/licenses/by-nc/4.0 2018-08-22 2018-08-22 9 2 113 118 10.20454/jast.2018.1454 $$\alpha_\beta$$-Connectedness as a characterization of connectedness http://m-sciences.com/index.php?journal=jast&page=article&op=view&path%5B%5D=1401 <p>In this paper, a new class of $$\alpha_\beta$$-open sets in a topological space $$X$$ is introduced which forms a topology on $$X$$. The connectedness of this new topology on $$X$$, called $$\alpha_\beta$$-connectedness, turns out to be equivalent to connectedness of $$X$$ and hence also to $$\alpha$$-connectedness of $$X$$. The $$\alpha_\beta$$-continuous and $$\alpha_\beta$$-irresolute mappings are defined and their relationship with other mappings such as continuous mappings and $$\alpha$$-continuous mappings are discussed. An intermediate value theorem is obtained. The hyperconnected spaces constitute a subclass of $$\alpha_\beta$$-connected spaces.</p> B. K. Tyagi Manoj Bhardwaj Sumit Singh ##submission.copyrightStatement## http://creativecommons.org/licenses/by-nc/4.0 2018-09-25 2018-09-25 9 2 119 129 10.20454/jast.2018.1401 On irresolute topological rings http://m-sciences.com/index.php?journal=jast&page=article&op=view&path%5B%5D=1474 <p>In this paper we introduce a new type of a topological ring which is an irresolute topological ring (semi topological ring). The relation among of them are studied. Several results are given. In particular, in a semi Hausdorff space, we show that if a subring is commutative, then its semi closure commutative subring. Furthermore, we show that the center of a ring is semi closed.</p> Haval M. Mohammed Salih ##submission.copyrightStatement## http://creativecommons.org/licenses/by-nc/4.0 2018-10-26 2018-10-26 9 2 130 134 10.20454/jast.2018.1474 Pre-$$(\omega)$$separation axioms http://m-sciences.com/index.php?journal=jast&page=article&op=view&path%5B%5D=1492 <p>In this paper we use the notion of $$(\omega)$$ preopen sets to introduce and study pre-separation axioms in an<br>$$(\omega)$$topological space.</p> Rupesh Tiwari ##submission.copyrightStatement## http://creativecommons.org/licenses/by-nc/4.0 2018-12-13 2018-12-13 9 2 135 138 10.20454/jast.2018.1492
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# 12.3 Stress, strain, and elastic modulus (Page 3/26)
Page 3 / 26
In either of these situations, we define stress as the ratio of the deforming force ${F}_{\perp }$ to the cross-sectional area A of the object being deformed. The symbol ${F}_{\perp }$ that we reserve for the deforming force means that this force acts perpendicularly to the cross-section of the object. Forces that act parallel to the cross-section do not change the length of an object. The definition of the tensile stress is
$\text{tensile stress}=\frac{{F}_{\perp }}{A}.$
Tensile strain is the measure of the deformation of an object under tensile stress and is defined as the fractional change of the object’s length when the object experiences tensile stress
$\text{tensile strain}=\frac{\text{Δ}L}{{L}_{0}}.$
Compressive stress and strain are defined by the same formulas, [link] and [link] , respectively. The only difference from the tensile situation is that for compressive stress and strain, we take absolute values of the right-hand sides in [link] and [link] .
Young’s modulus Y is the elastic modulus when deformation is caused by either tensile or compressive stress, and is defined by [link] . Dividing this equation by tensile strain, we obtain the expression for Young’s modulus:
$Y=\frac{\text{tensile stress}}{\text{tensile strain}}=\frac{{F}_{\perp }\text{/}\phantom{\rule{0.1em}{0ex}}A}{\text{Δ}L\phantom{\rule{0.1em}{0ex}}\text{/}\phantom{\rule{0.1em}{0ex}}{L}_{0}}=\frac{{F}_{\perp }}{A}\phantom{\rule{0.2em}{0ex}}\frac{{L}_{0}}{\text{Δ}L}.$
## Compressive stress in a pillar
A sculpture weighing 10,000 N rests on a horizontal surface at the top of a 6.0-m-tall vertical pillar [link] . The pillar’s cross-sectional area is $0{\text{.20 m}}^{2}$ and it is made of granite with a mass density of ${2700\phantom{\rule{0.2em}{0ex}}\text{kg/m}}^{3}.$ Find the compressive stress at the cross-section located 3.0 m below the top of the pillar and the value of the compressive strain of the top 3.0-m segment of the pillar.
## Strategy
First we find the weight of the 3.0-m-long top section of the pillar. The normal force that acts on the cross-section located 3.0 m down from the top is the sum of the pillar’s weight and the sculpture’s weight. Once we have the normal force, we use [link] to find the stress. To find the compressive strain, we find the value of Young’s modulus for granite in [link] and invert [link] .
## Solution
The volume of the pillar segment with height $h=3.0\phantom{\rule{0.2em}{0ex}}\text{m}$ and cross-sectional area $A=0.20\phantom{\rule{0.2em}{0ex}}{\text{m}}^{2}$ is
$V=Ah=\left(0.20\phantom{\rule{0.2em}{0ex}}{\text{m}}^{2}\right)\left(3.0\phantom{\rule{0.2em}{0ex}}\text{m}\right)=0.60\phantom{\rule{0.2em}{0ex}}{\text{m}}^{3}.$
With the density of granite $\rho =2.7\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}{\text{kg/m}}^{3},$ the mass of the pillar segment is
$m=\rho V=\left(2.7\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}{\text{kg/m}}^{3}\right)\left(0.60\phantom{\rule{0.2em}{0ex}}{\text{m}}^{3}\right)=1.60\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\text{kg}.$
The weight of the pillar segment is
${w}_{p}=mg=\left(1.60\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\text{kg}\right)\left(9.80\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}\right)=1.568\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\text{N.}$
The weight of the sculpture is ${w}_{s}=1.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\text{N},$ so the normal force on the cross-sectional surface located 3.0 m below the sculpture is
${F}_{\perp }={w}_{p}+{w}_{s}=\left(1.568+1.0\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\text{N}=2.568\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\text{N.}$
Therefore, the stress is
$\text{stress}=\frac{{F}_{\perp }}{A}=\frac{2.568\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\text{N}}{0.20\phantom{\rule{0.2em}{0ex}}{\text{m}}^{2}}=1.284\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}\text{Pa}=\text{128.4 kPa.}$
Young’s modulus for granite is $Y=4.5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{10}\text{Pa}=4.5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{7}\text{kPa}.$ Therefore, the compressive strain at this position is
$\text{strain}=\frac{\text{stress}}{Y}=\frac{128.4\phantom{\rule{0.2em}{0ex}}\text{kPa}}{4.5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{7}\text{kPa}}=2.85\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}.$
## Significance
Notice that the normal force acting on the cross-sectional area of the pillar is not constant along its length, but varies from its smallest value at the top to its largest value at the bottom of the pillar. Thus, if the pillar has a uniform cross-sectional area along its length, the stress is largest at its base.
what is vector
A quantity having both magnitude and direction
Zubair
quality having both magnitude and direction
ZINA
How to calculate for overall displacement
Well, take Final Position-Starting position
Grant
what is velocity
speed per unit time is called velocity. it is a vector quantity
Mukulika
velocity is distances overall time taking,it is a vector quantity, the units is metre per second.
Samuel
frequency is rate at which something happens or is repeated. it is a vector quantity
mawuo
what is the difference between resultant force and net force
net force is when you add forces numerically I.e. the total sum of all positive and negative or balanced and unbalanced forces. resultant force is a single vector which is the combination or addition of all x and y axes vector component forces in a system.
emmanuel
thanks
Ogali
resultant force is applied to hold or put together an object moving at the wrong direction. in other words it repairs.
Andrew
Damping is provided by tuning the turbulence levels in the moving water using baffles.How it happens? Give me a labelled diagram of it.
A 10kg ball travelling at 4meter per second collides elastically in a head-on collision with a 2kg ball.What are (a)the velocities and (b)the total momentum of the balls after collision?
a)v1 8/3s&v2 20/3s. b)in elastic collision total momentum is conserved.
Bala
multiply both weight which is 10*2 divided by the time give 4. and our answer will be 5.
Andrew
The displacement of the air molecules in sound wave is modeled with the wave function s(x,t)=5.00nmcos(91.54m−1x−3.14×104s−1t)s(x,t)=5.00nmcos(91.54m−1x−3.14×104s−1t) . (a) What is the wave speed of the sound wave? (b) What is the maximum speed of the air molecules as they oscillate in simple harmon
the question is wrong. if you need assistance with displacement I can help out.
Andrew
practical 1st year physics
huh
Luminous
allot of practicals, be specific with your topic and we can discuss.
Andrew
Whats the formular for newton law of motion
f=ma
F=m×a Where F=force M=mass of a body of an object a=acceleration due to gravity
Abubakar
what is speed
distance travelled per unit of time is speed.
distance travelled in a particular direction it is.
Andrew
Speed is define as the distance move per unit time. Mathematically is given as Speed = distance/time Speed = s/t
Abubakar
speed is a vector quantity. It is defined distance per unit time.It's unit in c.g.s cm/s and in S.I. m/s.It’s dimension is LT^-1
Mukulika
formula for velocity
v=ms^-1 velocity=distance time
Cleophas
(p-a/v)(v-b)=nrt what is the dimension of a
Amraketa
velocity= displacement time
Gold
Velocity = speed/time
Abubakar
what are evasive medical diagnosis?
If the block is displaced to a position y , the net force becomes Fnet=k(y−y0)−mg=0Fnet=k(y−y0)−mg=0 . But we found that at the equilibrium position, mg=kΔy=ky0−ky1mg=kΔy=ky0−ky1 . Substituting for the weight in the equation yields. Show me an equation of graph.
Shaina
where are you come from
Lida
samastipur Bihar
carrier
simple harmonic motion defination
how to easily memorize motion equation
Maharam
how speed destrog is uranium
where can we find practice problems?
I'm not well
YAZID
Sayed
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Then choose the lowest one. Add a new public comment to the blog: Cancel reply, The comments that you write here are moderated and can be seen by other users. In the fractions 3/8, 5/8, and 17/8, the numerators are 3, 5, and 17. Add Fractions with a Common Denominator The example above shows that to add the same-size pieces—meaning that the fractions have the same denominator—we just add the number of pieces. The sum 12/ 8 … In today's post, we're going to learn how to find prime numbers using the Sieve of Eratosthenes. In this example, the denominators are already the same. Then, convert each fraction to an equivalent fraction with denominator abcd. Multiply the numerator with the factor needed to come up with the least common denominator. Adding three fractions with the like denominators is a very simple process. If fractions with same denominators are to be added, we add the numerators only and keep the same denominator. Add these together. Now we are going to solve a problem that is a bit more difficult because the denominators are different prime numbers and we need to go through the process a bit more carefully: But as you can see, with the visual help it makes it easier as well! For a fraction problem that has a common denominator, the denominator is left as it is but the numerators are added as if they were integers. Now we are going to talk about adding fractions with different denominators. For example: 2 ¾ - 1 ¼ would give me 1 2/4 or 1 ½. That's why when you add fractions you first get all of them to have the same denominator, and then add them up. When the Denominators … Then, add the fractions together and simplify. Adding and Subtracting Fractions With Different Denominators. In this post, you are going to learn to add 3 fractions with different denominators with help from a visual aid. Check the fractions’ denominators are the same. He says that it is similar to adding two fractions with common denominator. I say that because when we come to sources like this we always have in mind that everything that is being posted was carefully revised and approved what can bring us to think that we can trust these info and use them to really learn with it. And in a situation like that a great explanation like you guys provided here, can be even more confusing than helpful. The answer is: Adding Numbers and a Fraction. Fractions with the same denominators (bottom number) have the same amount of equal parts.. You have already seen how easy adding fractions with the same or like denominators can be. Firstly, thanks so much for taking your time to tell us there was a mistake in the post. Fraction Addition If $a,b,\text{ and }c$ are numbers where $c\ne 0$, then What are Variables in Coding? He says that it is similar to adding two fractions with common denominator. The two things to remember is that before adding or subtracting, 1) The denominators of the fractions must be the same, and 2) The adding and subtracting is done with only the numerators of the fractions. Make sure the bottom numbers (denominators) are the same. Steps for Adding Fractions with Unlike Denominators. \frac{1}{3} + \frac{5}{12} = \frac{4}{12}+\frac{5}{12}=\frac{9}{12} He tells us first to add all the numbers in the numerator which gives us a new number which is the numerator in our answer and the denominator to our answer is the original common denominator. Add the top numbers (numerators): 3. Multiply the numerator with the factor needed to come up with the least common denominator. More information in. Worksheets include model and practice problems to teach students how to add fractions with the same denominator, and lesson plans incorporate hands-on activities. If you continue browsing you are agreeing to our use of cookies. 1/10 = multiply both by 7 = 7 / 70. Now, you’ll need to multiply the entire fraction to make the denominator become the least common multiple. We already took care of it. In this math tutorial the instructor shows us how to add three fractions with common denominators. How to Add Fractions. 65 / 70. make this as … The algebraic formula for adding fractions is: a/b + c/d = ad+bc/bd. How to add 3 fractions with different denominators (two of which are multiples) We are going to begin with the following addition problem: Simplify the fractions. Our adding fractions with the same denominator worksheets are unique, teacher-inspired assignments. To add fractions together when they have a common denominator, you simply add all the numerators together and rewrite the sum over the original denominator. For example, for 9/5 + 14/7, the multiples of 5 are 5, 10, 15, 20, 25, 30, and 35 while the multiples of 7 are 7, 14, 21, 28, and 35. It's as simple as that; the downside... What are geometric plane shapes? Simplify your result, if necessary. For example: 2. Simplify the sum if needed. First, we divide each rectangle into equal parts (adding the divisions to each rectangle): Now you can solve it easily! Then choose the lowest one. To add fractions there are Three Simple Steps: Step 1: Make sure the bottom numbers (the denominators) are the same, Step 2: Add the numerators, put that answer over the denominator, Step 3: Simplify the fraction (if needed) We also use third-party cookies that help us analyze and understand how you use this website. Step 3: Add or subtract the fractions. It is necessary to write the common denominator only once: It seems to me you’ve been taught the rules on how to add and multiply fractions with uncommon denominators without also being taught the logic behind the rules. Simplify the sum if needed. So, our LCD 6. For instance, in the equation 5/8 + 6/9, the denominators are 8 and 9. To add fractions, the denominators must be the same. In this example, the denominators are already the same. 5 combines the 1 red fifth with the 3 blue fifths. We will start by adding the first 2 addends. If you can simplify it, then do it! Firstly, thank you for all the info and help that you provide. Like before, we graphically represent each fraction: Since the sum of the first two is the same as before, we have already calculated that answer above: For an answer that is much easier to add, we represent 5/6 as a rectangle divided vertically and 1/5 as a rectangle divided horizontally: We divide both representations into equal parts: Now we have the 2 fractions in ”thirtieths” and the sum: I hope that it was easy to understand how to add 3 fractions. The denominators are the numbers on the bottoms of the fractions and they are the same in both fractions that we are adding. Now the denominators (the bottom numbers) are the same. 1/ 5 and 3/ 5 are like fractions because the denominators are the same. Detailed Help For Adding Fractions With Different Denominators. If you can, divide both the numerator and denominator … For example, suppose you want to add: 1 11 + 2 3 The LCM of 3 and 11 is 33 . Simplify the final result if possible. As the fractions have the same denominators, you can add them directly. The two are shown in ”sixths” and that way we get the following sum: We can represent the sum of this equation like this: Now all that’s left is to add the third addend, which we can do like this: And can be graphically represented like this: To be able to add them, we divide them into equal parts like before: We’ve converted the 2 fractions into ”twelfths” and now have the sum: That we can represent in the following way: Have you seen how easy it is to add fractions? Included is a PowerPoint with L.O and S.C as well as examples to work through. 3/7 = multiply both by 10 = 30 / 70. 35 is the least common multiple. Dimensions: Length, Width, and Height of an Object, Prime Numbers: How to Find Them with the Sieve of Eratosthenes, Geometric Plane Shapes: Circles, Triangles, Rectangles, Squares, and Trapezoids. The denominator remains the same with that of the given fractions. To add fractions with the same denominators, the denominator remains the same and we add the numerators together. 4 Any cookies that may not be particularly necessary for the website to function and is used specifically to collect user personal data via analytics, ads, other embedded contents are termed as non-necessary cookies. This fraction worksheet is great for practising adding 3 fractions with unlike denominators problems. That is, they must have a common denominator. In this tutorial you get to see just how easy it is to add up fractions once they have the same denominator! To add fractions there are Three Simple Steps: Step 1: Make sure the bottom numbers (the denominators) are the same, Step 2: Add the numerators, put that answer over the denominator, Step 3: Simplify the fraction (if needed) A lesson made looking at adding and subtracting fractions with the same denominator. In other words, multiply the numerator of each fraction by the denominator of the other: How to add and subtract fractions with different denominators. Step 2: Add the numerators directly. Find 5/8 + 7/8. For example: The only thing we need to do is add the numerators and leave the denominator alone. Since you are teaching and helping people, you should be more careful with your blog post review because it can confuse whoever is trying to use it as a source of learning or information. Learning about fractions works best when instruction is step-by-step, so for addition, start here with materials to help students add fractions with the same denominator. Step 3: Simplify the fraction with a common factor. When the addend denominators are the same, add the numerators to get the numerator of the sum. That way, you can understand the mathematical process behind the calculation. So 6/10 plus 3/10 would equal 9/10. That's why when you add fractions you first get all of them to have the same denominator, and then add them up. What characteristics do they have? Here is the online Adding Three Fractions with Same Denominators Calculator to add three fractions with same denominators, also called as the like denominators. You can add and subtract 3 fractions, 4 fractions, 5 fractions and up to 9 fractions at a time. Identify the least common denominator by finding the least common multiple for the denominators.. 2. We always revise our posts carefully before approving them, so this does not happen often. Cross-multiply the two fractions and add the results together to get the numerator of the answer. Step 3: Rewrite the fractions so they share the same denominator. The following video shows more examples of how to use models to add fractions with like denominators. If necessary we can simplify the fraction to lowest terms or a mixed number. That way, you can understand the mathematical process behind the calculation. This Adding and Subtracting Fractions with Unlike Denominators: Complete Guide includes several examples, a step-by-step tutorial, an animated video mini-lesson, and a free worksheet and answer key. While adding fractions can be hard, adding fractions with the same denominator is just as easy as adding numbers. In the fractions 3/5 and 2/5, the numerators are 3 and 2. Mini-Step 1.1: List the prime factors of both numbers: 9 = 1 × 3 × 3. The denominators of both fractions are ‘8’ and so the answer will also have a denominator of ‘8’. Page 1 of 3. The problems may be selected for different degrees of difficulty. Learn about fractions in a steady and entertaining way from the comfort of home or as an in-class activity. Adding mixed fractions? Here is the online Adding Three Fractions with Same Denominators Calculator to add three fractions with same denominators, also called as the like denominators. So, we need to find fractions equivalent to 1 11 and 2 3 which have 33 in Learning about fractions works best when instruction is step-by-step, so for addition, start here with materials to help students add fractions with the same denominator. How to add and subtract fractions with the same denominator. Fractions with the same denominators (bottom number) have the same amount of equal parts.. Step 4: Add the numerators. To start, locate the denominators in the fractions you’re dealing with. Continue reading >>. He tells us first to add all the numbers in the numerator which gives us a new number which is the numerator in our answer and the denominator to our answer is the original common denominator. Make sure the denominators (the bottom numbers) are the same. 5.NF.A.1 Add and subtract fractions with unlike denominators (including mixed numbers) by replacing given fractions with equivalent fractions in such a way as to produce an equivalent sum or difference of fractions with like denominators. Adding fractions is easy when the denominators are the same. While adding fractions can be hard, adding fractions with the same denominator is just as easy as adding numbers. This includes a practice sheet that children can work through for the questions on the board. There are three simple steps to add simple fractions: 1. The LCM of 2 and 3 is 6. These cookies do not store any personal information. On the Smartick blog, we have written a lot about fractions and the different operations we can do with them. Add the numerators and leave the denominator the same. If the denominators were not common, you could not add these fractions. You can see that the denominators are not the same, so let’s get to work on adjusting one or both fractions so their denominators “match”. The main rule of this game is that we can't do anything until the denominators are the same! In this tutorial you get to see just how easy it is to add up fractions once they have the same denominator! Worksheets include model and practice problems to teach students how to add fractions with the same denominator, and lesson plans incorporate hands-on activities. In order to add fractions with unlike denominators, there is a 3-step process. The denominator is 15. If necessary, we simplify the resulting fraction to lowest terms. For example: The only thing we need to do is add the numerators and leave the denominator alone. It includes adding fractions both with like and with, unlike denominators. While adding fractions can be hard, adding fractions with the same denominator is just as easy as adding numbers. Add 3 like fractions is just like adding decimals. Learn. Because you have to make the denominators the same before you add the fractions, find a common multiple that they share. When the denominators are the same, you are adding fractions that have the same amount of equal parts. This category only includes cookies that ensures basic functionalities and security features of the website. You also have the option to opt-out of these cookies. All rights reserved. 3. How to Add and Subtract Fractions When the Denominators are the Same. How to add and subtract fractions with the same denominator. The … The denominator remains the same with that of the given fractions. Necessary cookies are absolutely essential for the website to function properly. Adding fractions is easy when the denominators are the same. Cookies allow us to offer our services. Adding fractions is easy when the denominators (the numbers below the line) are the same. If they don't have common denominators, then find a common denominator and use it to rewrite each fraction. 2/5 = multiply both by 14 = 28 / 70. Improve your math knowledge with free questions in "Add 3 or more fractions with unlike denominators" and thousands of other math skills. Adding Fractions with the Same Denominator. If fractions with same denominators are to be added, we add the numerators only and keep the same denominator. Then add and simplify. © 2020 Smartick Intl. Add the numerators (the top numbers). Add three fractions with same denominators for the values a1 as 2, a2 as 3, a3 as 4 and b1,b2,b3 as 5, Sum of Like denominators = (a1 + b2 + a3) / b Choose abcd. We are going to begin with the following addition problem: To understand this problem correctly, we have graphically represented each addend: Using a rectangle as the unit, we divide them into 2, 3, and 4 parts, and in this case, each is a different color. Add or subtract the numerators. The first step is to make sure the denominators are the same. As you saw from the video, the concepts of adding and subtracting fractions are pretty much the same until the very last step. Adding and Subtracting Fractions: Why Do They Need the Same Denominator? To add fractions with unlike denominators, rename the fractions with a common denominator. Step 3: Rewrite the fractions so they share the same denominator. = (2 + 3 + 4) / 5 In this tutorial, take a look at adding together mixed fractions with unlike denominators! This Adding and Subtracting Fractions with Unlike Denominators: Complete Guide includes several examples, a step-by-step tutorial, an animated video mini-lesson, and a free worksheet and answer key. It's really just the LCM of our denominators, 2 and 3. Adding Fractions with the Visual Aid of Rectangles, Singapore Math Bar Model for Multiplying Fractions, Perimeter: What Is It and How to Find It for Any Polygon. In our example, this is what that looks like: Important: you multiply both the top and the bottom by the same amount to keep the value of the fraction the same. This tells us that we have 2 out of 3 equal parts. When one denominator is a multiple of the other, you can use a quick trick to find a common denominator: Increase only the terms of the fraction with the lower denominator to make both denominators the same. Adding three fractions with the like denominators is a very simple process. Access some of … Examples on adding fractions with common denominators. Visualizing everything makes it much easier! For example, if the fractions are 5/6 and 1/12, the LCM is 12, because 6 can be multiplied by 2 to create 12. Sum of the fractions = $\frac{a}{c}$ + $\frac{b}{c}$ = $\frac{(a + b)}{c}$, where a, b and c are any three real numbers. Add the fractions from Step 2. In order to add fractions, the denominators must be the same. With multiple activities to choose from, your child will develop personal goals and surpass them over the school year. HOW TO ADD AND SUBTRACT FRACTIONS. Before you read this post, you might want to check out a previous post where we explain how to add fractions step by step. Add the numerators. For example, 2/3 + 5/4 = 8/12 + … 1. These cookies will be stored in your browser only with your consent. It is mandatory to procure user consent prior to running these cookies on your website. This website uses cookies to improve your experience while you navigate through the website. Add the equivalent fractions that you wrote in step 2. Sample questions. So the first thing that might jump out at you is look, I have these fractions that I'm adding and subtracting, but they all have different denominators. Write equivalent fractions (making sure that each equivalent fraction contains the least common denominator (LCM)). 16 = 1 × 2 × 2 × 2 × 2 Add 3 like fractions is just like adding decimals. A fraction is a number that represents a part of a whole. That's why when you add fractions you first get all of them to have the same denominator, and then add them up. Changing fractions with unlike denominators but the same numerator requires you to just compare them and multiply them depending on the situation. A prime number is one which is only divisible by 1 and itself. Step 2: Multiply the top and bottom of the fraction by the same number to make the denominators the same. To add fractions you have to: Make the denominators the same using equivalent fractions. Now we have represented the 3 addends: 1/2, 1/3, and 1/4. While adding fractions can be hard, adding fractions with the same denominator is just as easy as adding numbers. In this tutorial you get to see just how easy it is to add up fractions once they have the same denominator! Fraction is nothing but the part of the whole number. Because you have to make the denominators the same before you add the fractions, find a common multiple that they share. You could first convert each to an improper fraction. Since the denominator are now alike, add the numerators and write LCD as the denominator to find the sum. So if you add these two fractions, your sum is going to have the same denominator, 15, and your numerator is just going to be the sum of the numerator, so it's going to be 3 plus 7, or it's going to be equal to 10/15. Delve into our collection of printable adding like fractions worksheets for grade 3 and grade 4 children to boast unquestionable competence in finding the sum of two proper fractions, two improper fractions, one proper and one improper fraction, and mixed numbers – all with like denominators. That's why when you add fractions you first get all of them to have the same denominator, and then add them up. Finally, since all three denominators are now the same, you simply add the numerators (6 + 8 + 9) while keeping the denominator (12) the same -- giving you a result of 23/12. These are the questions that we will answer in this post. In this math tutorial the instructor shows us how to add three fractions with common denominators. Procedure: To add two or more fractions that have the same denominators, add the numerators and place the resulting sum over the common denominator. In today’s blog post you can find much more information! Let’s begin with the simplest of examples: Adding Fractions with the Same Denominator. For private inquiries please write to [email protected]. This leads us to the following procedure for adding fractions with a common denominator. You have 7,5 and 10. the lowest number they will all go into is 70. If necessary, we simplify the resulting fraction to lowest terms. If you have a mixed number, you add the whole numbers then add the fractions. When the Denominators are the Same When fractions have the same denominators we simply add or subtract the numerators as indicated and place the result over the common denominator. And if you want more primary mathematics, sign up for Smartick and try it for free. You simply add the numerators and keep the same denominator, then simplify if needed. This can be done by finding the least common multiple (LCM) of the denominators. In this post, you are going to learn to add 3 fractions with different denominators with help from a visual aid. For example: The first thing that we need to do in this case is to convert the 2 into a fraction. You need to make the denominator the same on all. Learn. This leads us to the following procedure for adding fractions with a common denominator. For like fractions add only the numerators and not necessary to add the denominator because it remains the same. To add fractions, you have to create common (the same) denominators.You do this by multiplying the fractions times a fraction that is equivalent to 1 (such as 3/3). Page 1 of 3. That's why when you add fractions you first get all of them to have the same denominator, and then add them up. Out of these cookies, the cookies that are categorized as necessary are stored on your browser as they are essential for the working of basic functionalities of the website. Delve into our collection of printable adding like fractions worksheets for grade 3 and grade 4 children to boast unquestionable competence in finding the sum of two proper fractions, two improper fractions, one proper and one improper fraction, and mixed numbers – all with like denominators. The … How to add and subtract fractions with different denominators. Locate the denominators. In today’s entry, we’re going to talk about length, width, and height as tools to find the dimensions of an object. Step 5: Simplify fractions. Fractions don’t always have the same denominators, and in order to add or subtract those fractions, you must first find a common denominator. Add Fractions 2 . Featured here are exclusive resources on adding unit fractions, making a whole, finding the missing fractions, and finding the variables on fractions. Therefore, as a common denominator choose the LCM of the original denominators. = 9 / 5 Your personal details will not be shown publicly. Step 1: Determine the LCM of the denominators, 9 and 16. Robby’s New Laboratory. Here is the online Adding Three Fractions with Same Denominators Calculator to add three fractions with same denominators, also called as the like denominators. The easiest will have a set of denominators or 2,4,6,8. 35 is the least common multiple. Step 4: Add the numerators. Fraction is nothing but the part of the whole number. Same in both fractions that have the same denominator, and then add them up give me 2/4! Denominator is just as easy as adding numbers and a fraction is nothing the. Do with them hard, adding fractions is: adding fractions with common! ‘ 8 ’ need the same denominator and 2/5 unique, teacher-inspired assignments ×... Prime numbers using the Sieve of Eratosthenes math and coding in only 15 min protected ] the thing! Great explanation like you guys provided here, can be done by finding the common... Represents a part of a whole over the common denominators, you can solve it easily plans hands-on. Be added, we have represented the 3 addends: 1/2, 1/3 and... Of examples: adding numbers all go into is 70 can, both... Have already seen how easy adding fractions can be hard, adding fractions different. This example, the denominator are now alike, add the numerators keep. Then add them directly by 7 = 7 / 70 make this as … the! With, unlike how to add 3 fractions with same denominators work through find the sum 12/ 8 … to add fractions with the least denominator... Since the denominator alone this as … as the denominator the same and we add the same-size that. Well as examples to work through for the questions that we are adding fractions with the.! 1/3 and 2/5 you want to add 3 like fractions add only the numerators to get the numerator larger! Personal goals and surpass them over the school year and 9 today ’ s blog you... 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Credit Default Swaps
A credit default swap (CDS) is a contract that gives the buyer of the contract a right to receive compensation from the seller of the contract in the event of default of a third party. The buyer of the contract is typically a bondholder who is looking to transfer his credit exposure to another party. The seller is typically a bank which earns from the premiums it receives from the buyer.
Each CDS has a notional amount and it requires the buyer to pay a premium called CDS spread. Because the periodic premium rates are standardized, the buyer may also be required to pay an amount at the time 0 of the CDS seller. This amount is called upfront premium.
The seller of the CDS pays the buyer an amount equal to the loss incurred by the buyer on occurrence of a credit event. The credit event is binary in nature, i.e. it occurs, or it doesn’t. Typical credit events include (a) a filing for bankruptcy by the third party on whose bond the CDS was issued, (b) any failure by the third party to pay interest on its bonds and (c) any restructuring of the debt.
Formula
When it is established that a credit event has occurred, the amount paid by the CDS seller to the buyer is calculated using the following formula:
$$\text{Payout Amount}=\text{N}\times \text{Payout Ratio}=\text{N}\times(\text{1}\ -\ \text{Recovery Rate})$$
Where N is the notional amount and payout ratio is the loss incurred by a bondholder as a percentage of the bond’s par value. It equals 1 minus the recovery rate, which is the percentage of amount owed which is recovered by a bondholder during the bankruptcy proceedings.
During the life of the CDS, the profit (loss) that accrues to the buyer (seller) of the CDS can be approximated as follows:
$$\text{Profit to buyer of CDS}=\Delta \text{CDS}\ \times \text{N}\times \text{D}$$
ΔCDS is the basis point change in credit spread, N is the notional amount and D is the duration of the bond.
It follows that if the default spread increases over the life of the CDS, the buyer gains and if the spread shrinks the seller gains.
Example
A bank has loaned $40 million to a company for 5 years requiring periodic interest payments equal to LIBOR + 2.2%. The bank’s policy requires all loans to be backed by a credit default swap on the principal amount of loans made. In this case, the bank can buy a CDS with a notional amount of$40 million. The CDS costs 2%. The bank must pay an amount equal to 2% of the notional amount to the CDS seller each year. Annual premium amounts to $800,000 (2% ×$40 million).
If the borrower defaults on the final principal payment and the bank collects only 50% of its principal back, it can claim the differential from the seller of the CDS. The amount he will receive from the CDS sell is approximately equal to $20 million ($40 million × (1 – 50%)). If the borrower doesn’t default on the final principal amount, the bank doesn’t receive anything.
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## Archive for the 'Low dimensional topology' Category
### Knot Exteriors
Topologists often study knots by looking at what remains when the knot is removed from $S^3$. I suppose they’re looking (somewhat ironically) at everything but the knot. Hmm. In any case, it works, so why don’t we let $K$ be a knot in $S^3$ and define the knot exterior to be $E(K)=\overline{S^3\setminus N(K)}$, the closure of $S^3$ minus a regular neighborhood of the knot.
One way to visualize knot exteriors is to take the projection of $S^3$ to $\mathbb{R}^3$ where the point at infinity is chosen to lie inside $N(K)$. In this case, the knot exterior looks like a 3–ball minus a knotted arc inside. Two such exteriors are drawn at the right. I’ve added the knot (a neighborhood of which we removed, remember) in blue. The first is the exterior of the unknot, and the second is the exterior of a trefoil knot. It’s pretty easy to see in the first case that the exterior is a solid torus.
The converse of this fact, that if $E(K)$ is homeomorphic to a solid torus then $K$ is the unknot, is also true, because a compressing disk for the solid torus gives a Seifert surface for the knot.
### Some Simple Heegaard Splittings
Consider picture I, in which we have two balls with blue circles on their boundary. It’s not too hard to see that if we glue these two balls together so that the blue curves on their boundaries are identified, we obtain the space $S^3$. In picture II we have two solid tori with blue curves, and again, if we identify the boundaries of these two in such a way that the blue curves are identified, we obtain $S^3$.
This second one might be slightly harder to see, so let’s look at it another way: take the first solid torus, and glue a thickened disk along its edge to the blue curve. We consider the thickened disk to be a neighborhood of the disk bounding the blue curve in the second solid torus. What’s left after we’ve glued this disk to the first torus? A 3-ball. Glue that sucker in, you get $S^3$.
In picture III we ‘ve changed things up a bit. We still have to solid tori glued together, but this time the blue curves are different. It’s an exercise for the reader to determine this space. As a hint, it might be easier to see if you cut both solid tori along the disks bounding the blue curves.
An even more complicated (and difficult to visualize) example is pictured here. In the top line I’ve written it as a union of solid tori, and the second line it’s written as a thickened disk attached to the trefoil knot in the boundary of the second solid torus.
These are all examples of decompositions of 3-manifolds called Heegaard splittings. The first two are genus zero splittings of $S^3$, III is a genus one splitting of the mystery manifold $M$, and the last is a genus one splitting of a thingy called a “lens space.”
The point of all this is to see how a knot in bridge position gives a Heegaard splitting for a manifold associated to the knot, which I will write about next time.
### Bridge Number
Consider the pictured trefoil knot $K$ embedded in $\mathbb{R}^3$. The $xy$ plane is shown, and the intersections of the $xy$ plane with $K$ are marked in blue. The knot is drawn with a dashed line where it lies below the $xy$ plane; notice that the intersection of the lower half space with $K$ consists of three unknotted arcs. Similarly, the intersection of the upper half space with $K$ is three unknotted arcs. These arcs look a bit like bridges, and we say this is a three bridge presentation of $K$.
The smallest integer $n$ such that a knot $K$ has an $n$ bridge presentation is called the bridge number of $K$. The above picture shows that the bridge number of $K$ is less than or equal to three. We can do better though! Below is a two bridge presentation of the trefoil. Since a bridge number one knot is the unknot, this shows that the trefoil has bridge number two.
### An incompressible surface
Have a look at the this torus, embedded in a standard way in $S^3$. The shaded disk $D$ is inside the torus with its boundary on the torus itself. Consider $\partial D$, the boundary of $D$. It represents a nontrivial loop in the torus, but since it bounds a disk in $S^3$, the loop is trivial in $S^3$. Such a disk is called a compressing disk for the torus, and we can compress the torus by cutting along $\partial D$ and gluing in two copies of $D$ as shown in the second picture. This is also called surgering along $\partial D$. In this case, we obtain a sphere by compressing along $D$.
Notice that when we compress along $D$ we decrease the genus of our surface by one. With that in mind, consider the Seifert surface $F$ for the knot $K$ we looked at previously. Are there any compressing disks for $F$? Suppose there was a compressing disk $D$ and $\partial D$ didn’t separate $F$ into two components . Then we could compress $F$ along $D$, obtaining a new surface $F'$. Since we didn’t touch anything near the boundary of $F$, $F'$ is still a Seifert surface for $K$. But $F$ is a punctured torus, so $F'$ must be a disk, and the only knot bounding a disk is the unknot.
On the other hand, if there was a compressing disk $D$ such that $\partial D$ did separate $F$ into two components, then one of these components would have to be an annulus. As above, we would get a disk bounding the knot $K$. (Why? Working out these details is a fun exercise using Euler characteristic arguments.)
So if you believe that the trefoil $K$ is knotted, then the surface $F$ must be incompressible!
### Seifert Surfaces
Hello! I’m Sean, a graduate student here at the University of Texas and an aspiring topologist. Topologists seem to like pictures as a general rule, and I’m no exception, so I hope to post some nice pictures of interesting mathematical objects.
Speaking of, the picture on the right is a trefoil knot. (Isn’t it pretty?) We usually think of such knots as sitting inside $S^3$ (or $\mathbb{R}^3$ if you like).
Now it’s a fact that given any knot $K$, there’s an embedded orientable surface in $S^3$ whose boundary is $K$. Such a surface is called a Seifert surface or a spanning surface. The picture on the right shows a Seifert surface for the trefoil. As you can see, it consists of two regions on the left and right joined by three twisted bands. We can tell that it’s orientable because there’s no way to cross from the “top” to the “bottom” of the surface. In other words, there are no embedded Mobius bands in the surface.
What kind of surface is this? We know it has one boundary component and that it’s orientable, so from the classification of orientable surfaces we know that this must be a surface of genus $g$ with one puncture. To find out the genus of the surface, we can cut along the top two twists, being careful to keep track of how we cut. Then we can untwist at the bottom and straighten the surface out so that it lies flat. Finally, we can glue back the pieces marked with arrows as indicated. Hurray! Now we recognize the surface as a punctured torus!
(Cognoscenti will note that we could have seen immediately that a once punctured orientable surface with the homotopy type of the wedge of two circles must be a punctured torus. I think it’s more fun to draw pictures.)
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# 22. Operational amplifiers
Operational amplifiers, known as op-amps, are exceedingly useful integrated circuits. In fact, Horowitz and Hill refer to them as the “perfect part.” In this lesson, you will get an introduction to op-amps and learn how to use on as a non-inverting amplifier.
## Op-amps with feedback
An op-amp is an active (meaning externally powered with a DC power source) integrated circuit that has two inputs and one output. The difference between the two inputs is amplified with very high gain. Because of the high gain op-amps are almost always used with feedback.
Op-amps are powered by a positive and negative voltage source, $$V_+$$ and $$V_-$$, with $$V_- = -V_+$$. Because it requires negative voltage, we cannot use the Arduino Uno to power the op-amp, but need an external power supply. The Digilent PowerBRICKs can serve as a good power supply.
There are a few “rules” to keep in mind about op-amps used in feedback loops.
1. The gain is huge (typically $$10^5$$-$$10^6$$).
2. The inputs draw very small current (typically of order picoamps).
3. The output tends to make the voltage difference zero.
Rule (3) follows from rules (1) and (2), which we demonstrate below. To quickly think about op-amps, it is useful to keep these rules of thumb in mind.
We will apply op-amps to active filters, such as Sallen-Key filters, in a future lesson, and we will see other applications of op-amps in the next one. For now, we will explore the use of an op-amp in a non-inverting amplifier, which is simpler to analyze, and the analysis provides insight.
## A non-inverting amplifier
Consider the circuit below in which we connect an op-amp with feedback to a voltage divider.
Note that we have labeled a point $$a$$ in the circuit as a reference for our analysis. The symbol for an op-amp is a triangle with two inputs and a single output. The op-amp amplifies the difference between the input labeled with a “+” (referred to as the noninverting input) and that labeled with a “–” (referred to as the inverting input). Pay close attention when looking at a schematic; the inverting input can be on top or bottom. In some schematics, the power supplies ($$V_+$$ and $$V_-$$) for the op-amp are also included, given as vertical connections on the slanted sides of the triangle. In many schematics, including this one, we do not explicitly show those and note that they are in fact there.
Thinking intuitively about the circuit, imagine $$V_\mathrm{in}$$ goes up. Remembering rule 3 above, if $$V_\mathrm{in}$$ goes up, the output will be adjusted so that $$V_\mathrm{a}$$ also goes up in order to drive the difference between the inputs to the op-amp to be close to zero. In order to get $$V_a$$ to go up, $$V_\mathrm{out}$$ must also go up. Similarly, if $$V_\mathrm{in}$$ goes down, so too does $$V_\mathrm{out}$$.
To figure out how much $$V_\mathrm{out}$$ goes up, we need to find the value of $$V_\mathrm{out}$$ that will give $$V_\mathrm{a} = V_\mathrm{in}$$. Remembering what we have already worked out for a voltage divider, we have
\begin{align} V_\mathrm{a} = \frac{R_2}{R_1 + R_2}\,V_\mathrm{out}. \end{align}
Since the op-amp drives $$V_\mathrm{a} \approx V_\mathrm{in}$$, we have
\begin{align} V_\mathrm{out} = (1 + R_1/R_2)V_\mathrm{in}, \end{align}
which means that the op-amp amplifies the input by a factor of $$1 + R_1/R_2$$. Note that this amplification is independent of the gain or other properties of the op-amp. This is the beauty of op-amps. The nature of the amplification is entirely dependent on what circuit you put in the feedback and independent of the details of the op-amp itself.
Of course, this only works within a given range of input and output voltages. The maximum $$V_\mathrm{out}$$ for any op-amp is $$V_+$$ and the minimum $$V_\mathrm{out}$$ is $$V_-$$. In between these boundaries, good op-amps give amplification that is independent of $$V_+$$ and $$V_-$$.
### A more detailed look
It may seem unsatisfying that we used the “rules” to derive the above result. We could also do the analysis using Kirchhoff’s laws. The op-amp takes the differential input, $$V_\mathrm{in} - V_\mathrm{a}$$ and amplifies it with gain $$A$$. Then,
\begin{align} V_\mathrm{out} = A(V_\mathrm{in} - V_\mathrm{a}). \end{align}
Because negligible current goes through the op-amp, $$I_\mathrm{in} \approx I_\mathrm{a} \approx 0$$. From the voltage divider, as we have previously seen from Kirchhoff’s laws,
\begin{align} V_\mathrm{a} = \frac{R_2}{R_1 + R_2}\,V_\mathrm{out}. \end{align}
Thus, we have
\begin{align} V_\mathrm{out} = A\left(V_\mathrm{in} - \frac{R_2}{R_1 + R_2}\,V_\mathrm{out}\right). \end{align}
Rearranging this expression gives
\begin{align} \left(1 + \frac{A\,R_2}{R_1 + R_2}\right)V_\mathrm{out} = A\,V_\mathrm{in}. \end{align}
Unless we choose hugely disparate values of the resistors, e.g., $$R_1$$ to be in megaohms and $$R_2$$ to be one ohm, because $$A$$ is very large, the term in the parenthesis is
\begin{align} 1 + \frac{A\,R_2}{R_1 + R_2} \approx \frac{A\,R_2}{R_1 + R_2}. \end{align}
Thus,
\begin{align} \frac{A\,R_2}{R_1 + R_2}\,V_\mathrm{out} \approx A\,V_\mathrm{in}. \end{align}
The gains cancel! This means that the input-output relationship of the op-amp is independent of the gain. We get the same result as we did above by applying rule 3,
\begin{align} V_\mathrm{out} = (1 + R_1/R_2)V_\mathrm{in}. \end{align}
## A unit gain amplifier
Consider the following circuit.
This looks kind of funny. We take the output of the op-amp and put it right back in the inverting input without using a resister of capacitor. What is this doing? It makes more sense if we compare this circuit to the non-inverting amplifier. We see that our circuit is actually a non-inverting amplifier with $$R_1 \approx 0$$ and $$R_2 \approx \infty$$. Thus, we have $$V_\mathrm{out} \approx V_\mathrm{in}$$. The result is a unit-gain amplifier; the input signal is multiplied by unity.
This circuit is an example of an emitter follower (sometimes just called a follower), so named because the output is emitted, following an input. Importantly, the output has very low impedance, much lower than typical input impedance. As a result, the output is isolated from the input due to the large impedance difference. Unit gain amplifiers are quite useful to stabilize input signals prior to measuring them.
## Thinking exercise 7: Another amplifier
What is the relationship between $$V_\mathrm{in}$$ and $$V_\mathrm{out}$$ in the circuit below?
The circuit in this exercise is very important. Be sure you read the answer to this thinking exercise at the end of this lesson.
## Dual op-amps
We have LF412CP dual op-amps. A dual op-amp has two op-amps in the same integrated circuit. The pinout for this component is shown below.
Note that both of the op-amps share the same power source (Vcc+ and Vcc–).
When using this component, it is important to note the small circular indentation on the top. This is used to orient you so you know which pin is which.
The component is laid out as a dual inline package (DIP). The groove in the middle of your solderless breadboard is there for DIP support. You should mount the component across the groove. When you do so, none of the pins of the component are connected to each other, and you then can connect the pins to the necessary components of the circuit.
## Do-it-yourself exercise 10: Demonstration of noninverting amplification
Build the circuit below, taking into account the following considerations.
1. Power the dual op-amp with your 5 V PowerBRICK.
2. Do not connect the outputs of the op-amps to A0 and A2 right now. Do that after you upload your sketch.
Write an Arduino sketch to use the MCP4725 DAC to send one volt as input to the first op-amp and to send the outputs of the first and second op-amp to pins A0 and A2, respectively. Prior to plugging in pins A0 and A2 and uploading the sketch, what values to you think they will read?
Now plug in A0 and A2, upload the sketch, and see what values pins A0 and A2 read using the Serial Monitor. Hint: You may want to put a delay between opening the serial connection and reading and writing the results to make sure everything is ready to go for serial communication.
Answer to inverting op-amp thinking exercise
We will use the simple rules of op-amps. First, the voltage difference is close to zero, giving $$V_a \approx V_b$$. But, $$V_b = 0$$ because is ground. So, $$V_a = 0$$. Another rule of op-amps is that the input current is close to zero. Using Kirckhoff’s current law, the input current is the sum of the two currents coming in to node (a). The input current due to the input voltage is, by Ohm’s law, $$V_\mathrm{in} / R_1$$. The input current from the feedback is $$V_\mathrm{out} / R_2$$. These have to sum to zero, giving,
\begin{align} \frac{V_\mathrm{in}}{R_1} + \frac{V_\mathrm{out}}{R_2} = 0, \end{align}
from which we have
\begin{align} V_\mathrm{out} = -\frac{R_2}{R_1}\,V_\mathrm{in}. \end{align}
The minus sign means that the output voltage is of opposite sign as the input voltage. This circuit is therefore referred to as an inverting amplifier. This is one of the most widely used op-amp-based circuits. When $$R_1 = R_2$$, it is used to flip the sign of a voltage.
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# Algebra Difference in Roots Question.
Let D be the absolute value of the difference of the 2 roots of the equation 3x^2-10x-201=0. Find [D]. [x] denotes the greatest integer less than or equal to x.
I came across this question in a Math Competition and I am not sure how to solve it without using a calculator, since calculators are not allowed in the competition. Thanks.
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Can we assume familiarity with the quadratic formula? – abiessu Jun 3 '14 at 12:07
You mean, the general formula which allows you to find the 2 roots? – snivysteel Jun 3 '14 at 12:20
Yes. I answered the question assuming that this formula is available to those entering this challenge. – abiessu Jun 3 '14 at 12:26
Hint: If $a, b$ are the roots, $$|a-b|^2 = (a-b)^2 = (a+b)^2 - 4 ab = \frac{100}9+4\times \frac{201}3 = \frac{2512}9$$
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I've never seen this relationship between the roots of a quadratic... Is this a serious gap in my education, or a more minor note that I missed along the way? Also, I can't seem to see the relationship between this and the original quadratic... – abiessu Jun 3 '14 at 12:29
@abiessu $|a-b|^2=(a-b)^2=(a+b)^2-4ab$ holds true for all real $a, b$, not just roots. We use the fact that they are roots to quickly find the values of $a+b, ab$ using coefficients of the quadratic (see en.wikipedia.org/wiki/Quadratic_equation#Vieta.27s_formulas). Useful to know, IMHO. – Macavity Jun 3 '14 at 12:33
Got it, it just took me an extra moment to realize where the formulas came from. – abiessu Jun 3 '14 at 12:56
@abiessu The possible gap is the Fundamental Theorem of Symmetric Polynomials - see my answer. The essence of the matter will become clearer if you study Galois Theory. – Bill Dubuque Jun 3 '14 at 15:15
Hint $\$ By Vieta, $\,\ x^2 -\frac{10}3 x - 67\, =\, (x-a)(x-b)\iff \ \color{#0a0}{a+b} = 10/3,\ \color{#c00}{ab} = -67$
$(a-b)^2$ is symmetric in $\,a,b\,$ so by FTSP it can be written as a polynomial in $\,\color{#0a0}{a+b},\ \color{#c00}{ab}$
Indeed, applying Gauss's Algorithm we find that $\, (a-b)^2 = (\color{#0a0}{a+b})^2 -4\color{#c00}{ab}\, =\, \dfrac{16\cdot 157}9$
Remark $\$ The same algorithm works for polynomials in any number of variables. It reduces problem like this to rote mechanical computation, i.e. no guesswork is required to solve such problems, only simple polynomial arithmetic. The algorithm yields a constructive interpretation of the FTSP = Fundamental Theorem of Symmetric Polynomials, that every symmetric polynomial has a unique representation as a polynomial in the elementary symmetric polynomials.
Gauss's algorithm may be viewed as a special case of Gröbner basis methods (which may be viewed both as a multivariate generalization of the (Euclidean) polynomial division algorithm, as well as a nonlinear genralization of Gaussian elimination for linear systems of equation). Gauss's algorithm is the earliest known use of such a lexicographic order for term-rewriting (now mechanized by the Grobner basis algorithm and related methods).
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Very nice; I had missed symmetric polynomials in my undergraduate education, and while I have noticed them on occasion here on Math.SE and elsewhere, I haven't had much chance to incorporate them into what I understand of mathematics. – abiessu Jun 3 '14 at 16:04
@abiessu Symmetry plays a large role in solving many problems, so it is well-worth studying it (one will learn the basics in most courses on abstract agebra). Above it is a bit overkill to use Gauss's algorithm, but I wanted to make it clear that the process is completely mechanical/algorithmic - nothing needs to be pulled out of a hat like magic. Here are some other answers that exploit innate symmetry. – Bill Dubuque Jun 3 '14 at 16:11
$$3x^2-10x-201=0\\ \iff x^2-\frac{10}3x-67=0$$ Assuming the quadratic formula is available to use, $$x=\frac{10}6\pm\frac{\sqrt{\frac{100}9+4\cdot 67}}2\\=\frac 53\pm\sqrt{\frac{25}9+67}$$
So the square root term is greater than $\sqrt{64}$ but less than $\sqrt{81}$ and is therefore between $8$ and $9$ in value, and therefore the difference between the roots can be either be $16$ or $17$, but not $18$ since that term is less than $9$.
The difference can be determined this way because both roots have the offset fraction $\dfrac53$ which is removed upon subtraction.
To determine whether the difference is greater than $17$, consider whether the square root term is greater than $8.5:$
$$(8+0.5)^2=64+2\cdot8\cdot0.5+0.25=67+5+0.25$$
And our original square root term contains
$$67+\frac{25}9=67+2+\frac79$$
Therefore, half of the difference between the roots is less than $8.5$ but greater than $8$ and therefore the total difference is between $16$ and $17$, leaving $16$ as the value of $[D]$.
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This is a "brute force" way of attacking this problem, Macavity has listed a good shortcut in another answer. – abiessu Jun 3 '14 at 12:32
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 11.8: Box-and-Whisker Plots
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
At the end of this lesson, students will be able to:
• Make and interpret box-and-whisker plots.
• Analyze effects of outliers.
• Make box-and-whisker plots using a graphing calculator.
## Vocabulary
Terms introduced in this lesson:
first quartile, third quartile
five number summary
whiskers
inter-quartile range (IQR)
range
raw data
ordered list
outlier, mild outlier, extreme outlier
## Teaching Strategies and Tips
Use the introduction to point out that the median can be used to divide a data set into four quarters. See also Examples 1 and 2.
• After finding the quartiles, it is possible to construct the five-number summary and corresponding box-and-whisker plot.
• After finding the quartiles, it is possible to calculate the IQR.
Emphasize interpreting the box plot:
• 50%\begin{align*}50 \%\end{align*} of the data set lies between the first and third quartiles (IQR).
• 75%\begin{align*}75 \%\end{align*} of the data set lies above the first quartile; 75%\begin{align*}75 \%\end{align*} of the data set lies below the third quartile.
• The range is the distance from one whisker to the other.
• Compare the relative size of the box to the length of the whiskers: short whiskers indicate clustered data; long whiskers indicate a spread-out data set.
• If one whisker is shorter than another, then the distribution is skewed.
Construct box-and-whisker plots for two data sets and compare them side-by-side. Point out that this makes drawing inferences easy. See Example 3.
Compare the range and IQR for a data set. Ask:
• For what kind of distributions should the IQR be used? the range?
For the data set below, calculate the range and IQR. Which measure of dispersion do you think will give a better indication of the spread in the data?
2,5,8,8,8,9,9,10,100\begin{align*}2, 5, 8, 8, 8, 9, 9, 10, 100\end{align*}
## Error Troubleshooting
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## Finite simple groups with few Galois orbits on conjugacy classes. (arXiv:1810.07421v1 [math.GR])
All finite simple groups with at most 4 Galois orbits on conjugacy classes are determined. From this we list all finite simple groups G for which the group of central units of the integral group ring ZG is isomorphic to <\pm 1> \times Z. 查看全文>>
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SPHERICAL GEOMETRY
# Finding The Intersection Of Two Arcs That Lie On A Sphere
## Finding The Intersection Of Two Arcs That Lie On A Sphere
This is a common problem which usually arises because you want to find if two great circles (or segments of two great circles) on the earth’s surface intersect (obviously with are approximating the earth as a sphere).
We will choose to define each of the two arcs with two points on the surface of the sphere/earth. We could use either the geocentric (also called ECEF, coordinates are provided in terms of x, y, z) or geodetic (latitude/longitude) coordinate system to define these points. Let’s use geodetic for now:
Points for arc 1 ($$a1$$):
\begin{align} \b{P_{a11}} &= \left[ {\begin{array}{c} \lambda_{11} \\ \phi_{11} \end{array} } \right] \\ \b{P_{a12}} &= \left[ {\begin{array}{c} \lambda_{12} \\ \phi_{12} \end{array} } \right] \\ \end{align}
Points for arc 2 ($$a2$$):
\begin{align} \b{P_{a21}} &= \left[ {\begin{array}{c} \lambda_{21} \\ \phi_{21} \end{array} } \right] \\ \b{P_{a22}} &= \left[ {\begin{array}{c} \lambda_{22} \\ \phi_{22} \end{array} } \right] \\ \end{align}
where:
$$\lambda$$ = latitude, in degrees
$$\phi$$ = longitude, in degrees
We will then convert these into spherical coordinates with the following equations:
\begin{align} x = \cos(\lambda) \, \sin(\phi) \\ y = \cos(\lambda) \, \cos(\phi) \\ z = \sin(\lambda) \\ \end{align}
\begin{align} \b{P_{a11}} &= \left[ {\begin{array}{c} x_{11} \\ y_{11} \\ z_{11} \end{array} } \right] \\ \b{P_{a12}} &= \left[ {\begin{array}{c} x_{12} \\ y_{12} \\ z_{12} \end{array} } \right] \\ \b{P_{a21}} &= \left[ {\begin{array}{c} x_{21} \\ y_{21} \\ z_{21} \end{array} } \right] \\ \b{P_{a22}} &= \left[ {\begin{array}{c} x_{22} \\ y_{22} \\ z_{22} \end{array} } \right] \\ \end{align}
Calculate Cross-Products
We then need to calculate a cross-product for each arc, using the two vectors which define the arc as the inputs to the cross-product:
\begin{align} \b{N_1} = \b{P_{a11}} \times \b{P_{a12}} \\ \b{N_2} = \b{P_{a21}} \times \b{P_{a22}} \\ \end{align}
Note how we switched from using “point” to “vector” terminology here. Before we were talking about two “points” that lie on the sphere and define the arc. This is the easiest way to intuitively understand how two points can fully define an arc on the surface of a sphere. But now we can describe them as vectors. When we talk about them as vectors, we are referring to the vector which starts at $$[0, 0, 0]$$ and passes through this point.
These cross-products will be perpendicular to the plane that the arc lies in. Hence these cross-products are called plane normals.
Calculate The Cross-Product of the Cross-Product
Yup, you read that correctly. We now calculate the cross-product of the two cross-products we calculated above. Because the two arc cross-products define two plane normals, the cross-product of two plane normals gives us a vector that is in the same direction as the intersecting line.
$$\b{L} = \b{N_1} \times \b{N_2}$$
Because the line must pass through $$[0, 0, 0]^T$$, this vector is also the line itself!
Find The Intersecting Points
Now that we have the line of intersection, we can now find the two points of intersection between the arc planes by normalizing L (remember, we are using a unit sphere, so normalizing the line reduces it to a vector to one of the intersecting points):
$$\b{I_1} = \frac{\b{L}}{||\b{L}||} \\ \b{I_2} = -\b{I_1}$$
To get the other point of intersection, just take the negative of the first one.
Check If These Intersecting Points Are Within The Original Arc Segments
So we now have the two points of intersection between the arcs. If these arcs were complete great circles, then job done, you are guaranteed that the two above points lie within your great circles. But what if you had great-circle arc segments? These intersection points may not be within both arc segments.
We can determine if these intersection points lies within arc segments by doing a angle test, using the dot product:
$$\b{a} \cdot \b{b} = ||\b{a}|| ||\b{b}|| cos (\theta)$$
We will re-arrange for $$\theta$$:
$$\theta = \arccos \left( \frac{\b{a} \cdot \b{b}}{||\b{a}|| ||\b{b}|| } \right)$$
The angle from the start of arc 1 to intersecting point 1, and the angle from the end of arc 1 to intersecting point 1, and compare it against the angle between the start and end of arc 1.
The intersecting point is within arc 1 if:
$$\theta_{a11,i1} + \theta_{a12,i1} = \theta_{a11,a12}$$
Take note that if calculating this result using any data type that can lose precision (e.g. floats, doubles), you will have to check it is close to equal rather than exactly equal. This can be done by adding some epsilon. Usually $$1^{-10}$$ is sufficient.
This test has to be repeated between potential intersecting point 1 and arc 2. If both arcs intersect this potential intersecting point, then with have just confirmed that both arcs do intersect at this point!
The same process has to be applied to potential intersecting point 2 (remember, normalizing the line of intersection between the two planes gives us TWO potential points of intersection).
## Worked Example
Let’s define two arcs using two points each in geodetic (lat/lon) form (latitude and longitude are in degrees):
\begin{align} P_{a11} = \left[ {\begin{array}{c} 10 \\ 20 \end{array} } \right] P_{a12} = \left[ {\begin{array}{c} 60 \\ 90 \end{array} } \right] \\ P_{a21} = \left[ {\begin{array}{c} 50 \\ 10 \end{array} } \right] P_{a22} = \left[ {\begin{array}{c} 5 \\ 80 \end{array} } \right] \end{algin}
Then convert them to spherical coordinates:
$$\newcommand{\pAOneStart}{\left[ {\begin{array}{c} 5896 \\ 2146 \\ 1106 \end{array} } \right]} \b{a_{11}} = R \cdot \left[ {\begin{array}{c} x_{11} \\ y_{11} \\ z_{11} \end{array} } \right] \\ = \left[ {\begin{array}{c} \cos(\theta) \cos(\phi) \\ \cos(\theta) \sin(\phi) \\ \sin(\theta) \end{array} } \right] \\ = \left[ {\begin{array}{c} \cos(10) \cos(20) \\ \cos(10) \sin(20) \\ \sin(10) \end{array} } \right] \\ = \pAOneStart \\$$
And do the same for the other three points:
$$\b{a_{12}} = \left[ {\begin{array}{c} 0.0 \\ 3186 \\ 5517 \end{array} } \right] \b{a_{21}} = \left[ {\begin{array}{c} 4033 \\ 711 \\ 4880 \end{array} } \right] \b{a_{22}} = \left[ {\begin{array}{c} 1102 \\ 6250 \\ 555 \end{array} } \right]$$
Now calculate the normal vectors for arc 1 and 2 by taking the cross-product:
\begin{align} \b{N_1} &= \b{a_{11}} \times \b{a_{12}} \\ &= \left[ {\begin{array}{c} 5896 \\ 2146 \\ 1106 \end{array} } \right] \times \left[ {\begin{array}{c} 0.0 \\ 3186 \\ 5517 \end{array} } \right] \\ &= \left[ {\begin{array}{c} 8.316e6 \\ -3.253e7 \\ 1.878e7 \end{array} } \right] \\ \b{N_2} &= \b{a_{21}} \times \b{a_{22}} \\ &= \left[ {\begin{array}{c} 4033 \\ 711 \\ 4880 \end{array} } \right] \times \left[ {\begin{array}{c} 1102 \\ 6250 \\ 555 \end{array} } \right] \\ &= \left[ {\begin{array}{c} -3.011e7 \\ 3.139e6 \\ 2.442e7 \end{array} } \right] \end{align}
Now calculate the “normal of the normals”:
\begin{align} \b{L} &= \b{N_1} \times \b{N_2} \\ &= \left[ {\begin{array}{c} 8.316e6 \\ -3.253e7 \\ 1.878e7 \end{array} } \right] \times \left[ {\begin{array}{c} -3.011e7 \\ 3.139e6 \\ 2.442e7 \end{array} } \right] \\ &= \left[ {\begin{array}{c} -8.535e+14 \\ -7.686e+14 \\ -9.533e+14 \end{array} } \right] \end{align}
Now find the two points of intersection of the arc planes:
\begin{align} \newcommand{\iOne}{\left[ {\begin{array}{c} -0.5718 \\ -0.5149 \\ -0.6387 \end{array} } \right]} \b{I_1} &= \frac{ \b{L} }{ || \b{L}|| } \\ &= \left[ {\begin{array}{c} -8.535e+14 \\ -7.686e+14 \\ -9.533e+14 \end{array} } \right] \cdot \frac{1}{ 1.493e15 } \\ &= \iOne \\ \b{I_2} &= -\b{I_1} \\ &= \left[ {\begin{array}{c} 0.5717737 \\ 0.51491737 \\ 0.63869785 \end{array} } \right] \\ \end{align}
Check if these intersecting points are within the original arc segments. Let’s first check if $$\b{I_1}$$ intersects with the arc $$\b{a_1}$$ defined by the points $$\b{P_{a1\_start}}\quad$$ and $$\b{P_{a1\_end}}\quad$$.
\begin{align} \theta_{a1\_start,i1} + \theta_{a1\_end,i1} = \theta_{a1\_start,a1\_end} \end{align}
We need to calculate $$\theta_{a1\_start,i1}\quad$$, $$\theta_{a1\_end,i1}\quad$$, and $$\theta_{a1\_start,a1\_end}\quad$$. Lets find $$\theta_{a1\_start,i1}\quad$$ first:
\begin{align} \theta_{a1\_start,i1} &= \arccos \left( \frac{\b{P_{a1\_start}} \cdot \b{P_{i1}}}{||\b{P_{a1_start}}|| ||\b{P_{i1}}|| } \right) \\ &= \arccos \left( \frac{ \pAOneStart \cdot \iOne }{|| \pAOneStart || || \iOne || } \right) \\ &= 144.4 \end{align}
We can use the same equation to find $$\theta_{a1\_end,i1}\quad$$, and $$\theta_{a1\_start,a1\_end}\quad$$:
\begin{align} \theta_{a1\_end,i1} &= 144.2 \\ \theta_{a1\_start,a2\_start} &= 71.4 \\ \end{align}
Now we can check the equality:
\begin{align} \theta_{a1\_start,i1} + \theta_{a1\_end,i1} &= \theta_{a1\_start,a1\_end} \\ 144.4 + 144.2 &= 71.4 \\ 288.6 &= 71.4 \\ \end{align}
Obviouslty, this equality does not hold true! Therefore, potential intersection point $$\b{P_{i1}}$$ does not lie on the arc $$\b{a_1}$$ and we can rule it out as an intersection point (we do not need to test whether the potential intersection points lies on arc $$\b{a_2}$$, as if either of the equalities is false, we can immediately rule it out).
Now we test if potential intersection point $$\b{P_{i2}}$$ lies on both $$a_{1}$$ and $$a_{2}$$.
\begin{align} \theta_{a1\_start,i2} + \theta_{a1\_end,i2} &= \theta_{a1\_start,a1\_end} \\ 35.6 + 35.8 &= 71.4 \\ 71.4 &= 71.4 \\ \end{align}
The equality holds true, $$\b{P_{i2}}$$ lies on the first arc! Lets see if it lies on the second arc:
\begin{align} \theta_{a2\_start,i2} + \theta_{a2\_end,i2} &= \theta_{a2\_start,a2\_end} \\ 24.7 + 48.7 &= 73.4 \\ 73.4 &= 73.4 \\ \end{align}
This equality also holds true, $$\b{P_{i2}}$$ also lies on the second arc! Therefore we know that the point $$\b{P_{i2}}$$ is a valid intersection point between the two arcs. Problem solved!
## External Resources
https://stackoverflow.com/questions/2954337/great-circle-rhumb-line-intersection is a great page showing how to calculate the intersection between a arc and a Rhumb line (similar to above).
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# How would an attacker perform an exhaustive key search on a block cipher using ECB mode?
Do you always assume from Kerchoff's principle, that the attacker has access to everything but the decryption key?
That is, am I to assume that to perform an exhaustive key search, the attacker has a plaintext block and ciphertext block pair.
So an attacker would systematically try every possible key by decrypting the ciphertext? Then he would be able to identify when he has found the correct key if it matches the plaintext?
If he only has the ciphertext is an exhaustive key search still possible?
Should I be looking at what information a block cipher running in ECB mode provides?
A simple way to look at it is that they let your write equations involving the cipher parameters. For example, in $c = enc(p,k)$, you can fix two of the parameters and solve for the third. Clearly $c$ (ciphertext) must be obtainable from $\lbrace p,k \rbrace$ (encryption of plaintext $p$ under key $k$), and $p$ obtainable from $\lbrace c,k \rbrace$ (decryption), but $k$ should not be obtainable from $\lbrace c,p \rbrace$ (secure against known-plaintext attacks).
If all you have is ciphertext samples, then you have no means to verify a guess for the key. From an algebraic perspective, you cannot set up an equation using $c$ alone that will allow you to derive anything about $p$ or $k$. However, if you are running in ECB mode, then solving $k$ for one block under known-plaintext gives you $k$ for every other block, including blocks where only $c$ is known.
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# American Institute of Mathematical Sciences
September 2008, 7(5): 1225-1235. doi: 10.3934/cpaa.2008.7.1225
## Positive periodic solutions to a nonlinear fourth-order differential equation
1 Department of Mathematics, Jilin University, Changchun 130012, China, China 2 Department of Mathematics, and Key Laboratory of Symbolic Computation, and Knowledge Engineering of Ministry of Education, Jilin University, Changchun 130012
Received October 2007 Revised March 2008 Published June 2008
This paper is concerned with the existence of positive periodic solutions to a nonlinear fourth-order differential equation. By virtue of the first positive eigenvalue of the linear equation corresponding to the nonlinear fourth order equation, we establish the existence result by using the fixed point index theory in a cone.
Citation: Chunhua Jin, Jingxue Yin, Zejia Wang. Positive periodic solutions to a nonlinear fourth-order differential equation. Communications on Pure & Applied Analysis, 2008, 7 (5) : 1225-1235. doi: 10.3934/cpaa.2008.7.1225
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2019 Impact Factor: 1.105
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# UNC or Stern (full-time)
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11 Apr 2013, 14:38
Hi all- I'm a full-time admit for Fall 2013, trying to decide between 2 schools
GMAT: 770
GPA: 3.97 (at UNC)
NYU Stern: half tuition scholarship
UNC: full ride + generous stipend
Desired field: internal strategy or business development (the latter possibly within a rotational program) in financial sector
Current location: NYC
Desired geography: in/near a large east-coast city
Thoughts: very thankful to have two attractive options, having a hard time picking one. A bit uncertain/worried about getting interviews and having good networking opportunities in my desired field outside of the Southeast region if I attend UNC. Also gunshy about the extra cost of NYU (tuition & living) given that I won't be heading into ibanking or consulting.
thoughts??
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Re: UNC or Stern (full-time) [#permalink]
### Show Tags
11 Apr 2013, 16:29
You may not be headed to ibanking/consulting, but you still are thinking about the finance industry (at least enough to call it out here), so Stern will open up many doors for you. And you have half of your tuition covered, which puts you in a better spot than most people.
UNC recruiting will be very regional for mid-Atlantic/Southeast. Unless you want to end up in the Carolinas for life, I'd go with Stern. I think most people in this situation would lean Stern as well, but then again, since you went to UNC undergrad, you might have other ties to the Carolinas that make that full-ride there really compelling (withholding the obvious extra $and lower standard of living for expenses). _________________ aerien Note: I do not complete individual profile reviews. Please use the Admissions Consultant or Peer Review forums to get feedback on your profile. GMAT Club Premium Membership - big benefits and savings Director Joined: 07 Jan 2013 Posts: 758 Location: United States Concentration: Finance Followers: 5 Kudos [?]: 144 [0], given: 204 Re: UNC or Stern (full-time) [#permalink] ### Show Tags 11 Apr 2013, 17:28 I think you should go to Stern. Yes, it might be a financial burden right now, but in the long run, I think Stern will open lot more doors. By the way, you have stellar numbers. Did you apply anywhere else? _________________ Duke MBA Class of 2016 Intern Joined: 09 Nov 2011 Posts: 26 Location: India GMAT 1: 770 Q50 V44 Followers: 0 Kudos [?]: 8 [0], given: 1 Re: UNC or Stern (full-time) [#permalink] ### Show Tags 11 Apr 2013, 20:02 Are planning to return to your planet Krypton after the MBA ? Amazing stats! Pick Stern. The effort you'll put into recruiting while at UNC, if you put in the same while at NYU, you could land up pretty much where you want to. _________________ Ah well, may those few last steps that remain unfold. Re: UNC or Stern (full-time) [#permalink] 11 Apr 2013, 20:02 Similar topics Replies Last post Similar Topics: 1 UNC Kenan-Flagler($$) v/s NYU Stern? 8 03 Apr 2017, 13:42 UNC ($$) vs NYU Stern (No$) 2 28 Apr 2015, 19:57
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# Deriving logarithmic spiral equation from square corners
This is an interesting problem but I haven't been able to work it out (from Bender and Orszag, prob: 1.27) - any insight/assistance would be appreciated:
Four caterpillars, initially at rest at the four corners of a square centered at the origin, start walking clockwise, each caterpillar walking directly toward the one in front of it. If each caterpillar walks with unit velocity, show that the trajectories satisfy the differential equation
$$\frac{dy}{dx}=\frac{y-x}{y+x}$$
This is the equation for a logarithmic spiral in $$xy$$-coordinates. And I can see how the movement slowly traces points on rotating squares which get smaller and smaller in size, giving a logarithim spiral, but I cannot find a way to derive the equation from the given geometry in terms of dynamic parameters like $$dx, dy, dt, ...$$
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# Exponents
After addition, multiplication (serial addition), subtraction (addition in reverse), and division (inverse multiplication), comes exponentiation (serial multiplication). Most of us learned about the basic x² stuff in school — it just means x-times-x. Likewise, x³ just means x-times-x-times-x. Serial multiplication. No problem.
But sometimes it’s x-2 or x½, and it’s hard to see how those work. As it turns out, it can all be understood based on a single axiom:
$\displaystyle{x}^{n}={x_1}\times{x_2}\times\ldots\times{x_n}$
Which is the rule behind the basic understanding you already have. The exponent says how many times to multiply a value times itself. Put in formal math terms, it looks like this:
$\displaystyle\prod^{n}_{i=1}{x}_{i}$
From this axiom we can very easily derive the first theorem:
$\displaystyle{x}^{1}={x}$
Any number to the power of one is just that number because there is only one instance of it in the multiply chain.
We can also derive what is often presented as a second axiom (but is really another derived theorem):
$\displaystyle{x}^{a+b}={x^a}\times{x^b}$
Because:
$\displaystyle{x}^{n}={x}^{a+b}=\left({x_1}\cdot{x_2}\cdot\ldots{x_a}\right)\times\left({x_1}\cdot{x_2}\cdot\ldots{x_b}\right)$
The multiply chain can be broken into two parts consisting of acount instances and bcount instances (where a+b=n). Those two parts are obviously multiplied together, so we derive the second theorem:
$\displaystyle{x}^{a+b}={x^a}\times{x^b}$
Note that it’s possible to treat this equality as the axiom and derive the axiom above (along with everything else). In either case, a single axiom leads to all the theorems.
We derive another key equality by noting that (by addition and the first theorem):
$\displaystyle{x}^{1+0}={x}^{1}={x}$
And therefore (using the second theorem):
$\displaystyle{x}^{1+0}={x}^{1}\times{x}^{0}={x}$
If we divide the last two terms by x¹ (which, per the first theorem, is just x) we end up with the very important third theorem:
$\displaystyle{x}^{0}={1}$
Any number raised to the power of zero is just one.
The theorem for x0 lets us derive a rule that might be surprising. We start by noting that:
$\displaystyle{x}^{(n-n)}={x^0}={1}$
By subtraction and the third theorem. If we re-express this as:
$\displaystyle{x}^{\left(n+(-n)\right)}={x^0}={1}$
Then we then can invoke the second theorem to say:
$\displaystyle{x}^{\left(n+(-n)\right)}={x^n}\times{x^{-n}}={1}$
If we divide the last two terms by xn, we have the fourth theorem:
$\displaystyle{x}^{-n}=\frac{1}{x^n}$
Which says that negative exponents are the inverses of positive ones. A negative exponent just means “one-over” the positive version.
We can also derive what happens with non-integer exponents. We’ll start with a simple example by first noting that:
$\displaystyle{x}^{(\frac{1}{2}+\frac{1}{2})}={x}^{1}={x}$
By addition and the first theorem. As we’ve done above, we can invoke the second theorem to give us:
$\displaystyle{x}^{(\frac{1}{2}+\frac{1}{2})}={x}^{\frac{1}{2}}\times{x}^{\frac{1}{2}}=\left({x}^{\frac{1}{2}}\right)^2={x}$
And if we take the square root of the last two terms, we get a special case of the fifth theorem:
$\displaystyle{x}^{\frac{1}{2}}=\sqrt{x}$
We can make this more general by starting with:
$\displaystyle{x}^{(\frac{1}{n}+\frac{1}{n}+\cdots\frac{1}{n})}={x}^{1}={x}$
Where the fraction 1/n is repeated n times, and then, using the second theorem:
$\displaystyle{x}^{(\frac{1}{n}+\frac{1}{n}+\cdots\frac{1}{n})}=\left({x}^{\frac{1}{n}}\right)^n={x}$
Using the same logic as above (taking the nth root of last two terms) we get the general case of the fifth theorem:
$\displaystyle{x}^{\frac{1}{n}}=\sqrt[n]{x}$
So fractional exponents (with a numerator of one) give us roots.
Finally, recursive exponents such as:
$\displaystyle\left({x}^{a}\right)^{b}$
Note that, by the initial axiom:
$\displaystyle\left({x}^{a}\right)^{b}=(x^a)_{1}\times(x^a)_{2}\times\ldots(x^a)_{b}$
And each xa expands to:
$\displaystyle{x}^{a}={x}_{1}\times{x}_{2}\ldots{x}_{a}$
So, we have x times itself a times, times itself b times, which is just a times b in terms of total instances of x. This gives us the sixth theorem:
$\displaystyle\left({x}^{a}\right)^{b}={x}^{({a}\times{b})}$
Recursive exponents just multiply. This works nicely in reverse. For instance:
$\displaystyle{x}^{\frac{3}{5}}={x}^{(\frac{1}{5}\times{3})}=\left({x}^{\frac{1}{5}}\right)^{3}=\left(\sqrt[5]{x}\right)^3$
Which can be very helpful with fractional exponents that have numerators other than zero.
Amazin’ the theorems we can derive from that single axiom, eh?
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# A function that is not?
Calculus Level 5
$f(x) = \begin{cases} \sin x & \text{for } x \in \mathbb{Q} \\ (\ln x)^2 & \text{for } x \notin \mathbb{Q} \end{cases}$
Consider the function $$f$$ defined over the set of positive real numbers.
Which of the following statements about the function $$f$$ is true?
Notations:
×
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TeXShop:
A TeX previewer for Mac OS X
TeX-Soft:
An archive of TeX software
CTAN Archive
A host of the Comprehensive TeX Archive Network
/pub/tex
.
## LaTeX Margins
Margins in LaTeX are diificult to change and manipulate because all of the margins are interrelated. Please read the CTAN Faq on chnaging margins before attempting to use the commands below.
The most reliable and easiest way to change margins on our system is to use the geometry package and the commands it controls. Please see the document wide spacing page or below for those relatively simple commands.
Making a Style File
Another option, which is initially more difficult but will save time later, is to make a margins style file. Then, you could call that style file in any document you create with the command: \usepackage{margins}. This way, you would need to set your margins only once and you could call it for multiple documents and it would apply to the entire document.
Document Wide Spacing
Changing the spacing on a document-wide basis is done for different document settings using different packages. The sectsty package controls the spacing of section headings; the mdwlist package controls the spacing for lists; the geometry package controls the overall margins and text area of the document. All of these packages must be called in the preamble of the document and specified when called. For instance, the geometry package is called with the use package command, followed by the options, and then the package name, like this:
Another method to change the margins in the document is to use the package anysize and its command
\marginsize{left}{right}{top}{bottom}:
\documentclass[a4paper]{article}
\usepackage{anysize}
\marginsize{3cm}{2cm}{1cm}{1cm}
\begin{document}
Jawohl! Sommerlov!
\end{document}
Margins on the PageCommands to Change the Margins
Image from Peter Newbury's page on LaTeX Margins: http://www.iam.ubc.ca/~newbury/tex/page-set-up.html
%%%%%%%%%%%%%
\oddsidemargin 0.0in
\evensidemargin 1.0in
\textwidth 6.0in
\topmargin 0.5in
\textheight 9.0in
\footheight 1.0in
%%%%%%%%%%%%
(Please note that all of these are contingent on each other and that changing the margins this way is difficult.)
Margin Commands
All side margins on single-sided pages are controlled with the \oddsidemargin command. The distance from the left side of the page to the left side of the text is one inch + \oddsidemargin. To set margins of less than 1in, set the \oddsidemargin command with a negative length like -0.5in. The right margin is changed by combining the \oddsidemargin (default of 1) with the \textwidth. Thus, if you wanted to have 1inch margins on the left and the right sides of the page (on 8.5X11inch paper), you would use the commands:
\oddsidemargin 0.0in
%%this makes the odd side margin go to the default of 1inch
\textwidth 6.5
%%sets the textwidth to 6.5, which leaves 1 for the remaining right margin with 8 1/2X11inch paper
If you are writing a document with even and odd sided pages, then use the \evensidemargin command in addition to the \oddsidemargin command. Both commands are used within the same framework.
The vertical style parameters are complicated and setting \topmargin=0in will not negate the top space of the page, even with no headers.
C. Sean Bohun
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# Counting monomials and the Catalan numbers
Given a multi-variable polynomial $$F$$, denote the number of monomials by $$N(F)$$. Take for instance, \begin{align*}N(x(x+y)+(x+y)^2-(x-y)^2)=N(x^2+5xy)&=2 \qquad \text{and} \\ N((x+z)(x+y)^2)=N(x^3 + 2x^2y + x^2z + xy^2 + 2xyz + y^2z)&=6.\end{align*} Denote the symmetric group on $$n$$ letters $$\{1,2,\dots,n\}$$ by $$\mathfrak{S}_n$$. Define the action of $$\mathfrak{S}_n$$ on a function $$F(x_1,\dots,x_n)$$ in a natural way: given $$w\in\mathfrak{S}_n$$, then $$w\cdot F(x_1,\dots,x_n)=F(x_{w(1)},\dots,x_{w(n)})$$. Introduce the (multi-variable) rational functions $$G(\mathbf{x},\mathbf{z})=\prod_{k=1}^n\frac{x_1z_1+x_2z_2+\cdots+x_kz_k}{x_k-x_{k+1}}.$$
Assuming that the symmetric groups act only on the $$x$$-variables, let's compute the polynomial $$G_{\mathfrak{S}_{n+1}}=\sum_{w\in\frak{S}_{n+1}}w\cdot G.$$ I would like to ask:
QUESTION. Let $$C_n=\frac1{n+1}\binom{2n}n$$ be the Catalan numbers. Is there a combinatorial proof that $$N(G_{\mathfrak{S}_{n+1}})=C_n$$?
• Are you sure the $G(\mathbf{x},\mathbf{z})$ are polynomials (as opposed to rational functionals)? I guess the point is that after symmetrization the $G_{\mathfrak{S}_{n+1}}$ become polynomials... Apr 22 at 21:07
• Edited. Thank you. Apr 22 at 22:07
• I checked findstat for the distribution of coefficients, to no avail :-( Apr 23 at 9:18
Any monomial $$P:=\prod x_i^{c_i}$$ of degree $$\sum c_i=n$$ maps to a non-zero constant after symmetrization $$P\to \Phi(P):=G_{\mathfrak{S}_{n+1}}\frac{P}{(x_1-x_2)(x_2-x_3)\ldots (x_n-x_{n+1})}.$$ Indeed, $$\Phi(P)$$ is a constant by a degree consideration, to prove that this constant is non-zero you may use, for example, Theorem 2 here.
Thus any monomial $$\prod (x_iz_i)^{c_i}$$ maps to a non-zero constant times $$\prod z_i^{c_i}$$.
Thus, you simply count the number of monomials which may arise, when you multiply the linear forms $$\prod_{k=1}^n(x_1z_1+x_2z_2+\cdots+x_kz_k)$$. The only condition is that $$c_1+\ldots+c_i\geqslant i$$ for all $$i$$. Such sequences are indeed enumerated by Catalan numbers, a bijection with lattice paths below diagonal is straightforward.
• @MartinRubey the coefficient equals (upto sign) to the number of enumerations of $n+1$ vertices of the path which satisfy $n$ inequalities which in turn correspond to the $n$ edges. The inequality $\pi(x)<\pi(y)$ for an edge $xy$ corresponds to our monomial $C$ being $y$-biased in the following sense: remove the edge $xy$, let $k$ denote the number of vertices in the piece containing $y$; then the total degree of $C$ in these $k$ variables is at least $k$. So, the number of monomials with coefficient $\pm 1$ (happens when all inequalities have the same direction) is twice Catalan number. Apr 25 at 12:30
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## Songs as Another Kind of Parallel University
Meta Intelligence is a heterodox view of education where formal education (courses, diplomas, universities, fields) are incomplete and limited without adding informal education which is part of your life such as movies, songs, conversations and images (paintings, posters, etc). Your “lifeworld” (Edmund Husserl’s apt coinage) fuses all the kinds of education where the word education means thought-provoking and illuminating. Even personal experience counts such as illnesses or bad marriages! Only via this Meta Intelligence will you achieve a glimpsed “holism.” (Meta Intelligence is that meta-field outside fields, borders and boundaries.)
Take songs.
Think back to Jim Morrison’s classic tune, “Riders on the Storm” which begins:
“Riders on the storm
Riders on the storm
Into this house, we’re born
Into this world, we’re thrown
Like a dog without a bone
An actor out on loan
Riders on the storm”
This song (by the Doors), expresses in a simple way Heidegger’s notion of human existence as partly governed by “Geworfenheit” which derives from “werfen,” to throw. “Geworfenheit” means “thrownness.” Jim Morrison and his band the Doors are songphilosophers without (probably) being Heidegger’s acolytes. Max Weber, one of the fathers of modern sociology, uses the word “disenchantment” to describe the modern world, “Entzauberung” in German, where “zauber” means “magicality” and “ent” means “removal of,” and “ung” means “condition of being.” The magic here does not mean something like a card trick but rather sacred mysteries, perhaps like the feeling a medieval European felt on entering a cathedral.
Enchantment in the West survived in our notions of romantic love and was associated with the songs and outlook of the medieval troubadours. Such romantic enchantment which is fading from our culture in favor of sex is still celebrated in the classic Rogers and Hammerstein song, “Some Enchanted Evening” from the forties musical and fifties movie, South Pacific.
The song lyrics give you the philosophy of romantic love as the last stand of enchantment:
“Some enchanted evening, you may see a stranger,
You may see a stranger across a crowded room,
And somehow you know, you know even then,
That somehow you’ll see here again and again.
Some enchanted evening, someone may be laughing,
You may hear her laughing across a crowded room,
And night after night, as strange as it seems,
The sound of her laughter will sing in your dreams.
“Who can explain it, who can tell you why?
Fools give you reasons, wise men never try.
“Some enchanted evening, when you find your true love,
When you hear her call you across a crowded room,
Then fly to her side and make her your own,
Or all through your life you may dream all alone.
“Once you have found her, never let her go,
Once you have found her, never let her go.”
Notice that “chant” is a component of enchantment.
One could say that conventional enchantment has been transferred to the world of science and mathematics where a deep beauty is intuited. Professor Frank Wilczek of MIT (Nobel Prize) wrote several books on this intersection of science and the quest for beauty whereas Sabine Hossenfelder of Germany has argued, per contra, that this will be a “bum steer.”
You should sense that like movies, songs give you a “side window” or back door into thinking and knowledge, which should be center stage and not depreciated.
## Education and “Intuition Pumps”
Professor Daniel Dennett of Tufts uses the word “intuition pumps” in discussing intuitive understanding and its tweaking.
Let’s do a simple example, avoiding as always “rocket science,” where the intricacies weigh you down in advance. We make a U-turn and go back by choice to elementary notions and examples.
Think of the basic statistics curve. It’s called the Bell Curve, the Gaussian, the Normal Curve.
The first name is sort of intuitive based on appearance unless of course it’s shifted or squeezed and then it’s less obvious. The second name must be based on either the discoverer or the “name-giver” or both, if the same person. The third is a bit vague.
Already one’s intuitions and hunches are not fool-proof.
The formula for the Bell Curve is:
$$y = \frac{1}{\sqrt{2\pi}}e^{\frac{-x^2}{2}}$$
We immediately see the two key constants: π (pi) and e. These are: 22/7 and 2.71823 (base of natural logs).
The first captures something about circularity, the second continuous growth as in continuous compounding of interest.
You would not necessarily anticipate seeing these two “irrational numbers” (they “go on” forever) in a statistics graph. Does that mean your intuition is poor or untutored or does it mean that “mathworld” is surprising?
It’s far from obvious.
For openers, why should π (pi) be everywhere in math and physics?
Remember Euler’s identity: e + 1 = 0
That the two key integers (1 and 0) should relate to π (pi), e, and i ($\sqrt{\mathrm{-1}}$) is completely unexpected and exotic.
Our relationship to “mathworld” is quite enigmatic and this raises the question whether Professor Max Tegmark of MIT who proposes to explain “ultimate reality” through the “math fabric” of all reality might be combining undoubted brilliance with quixotism. We don’t know.
## Education and Finality Claims
Stephen Hawking kept saying he wanted to discover the ultimate world-equation. This would be the final “triumph of the rational human mind.”
This would presumably imply that if one had such a world-equation, one could infer or deduce all the formalisms in a university physics book with its thousand pages of equations, puzzles and conundrums, footnotes and names and dates.
While hypothetically imaginable, this seems very unlikely because too many phenomena are included, too many topics, too many rules and laws.
There’s another deep problem with such Hawking-type “final equation” quests. Think of the fact that a Henri Poincaré (died in 1912) suddenly appears and writes hundreds of excellent science papers. Think of Paul Erdős (died in 1996) and his hundreds of number theory papers. Since the appearance of such geniuses and powerhouses is not knowable in advance, the production of new knowledge is unpredictable and would “overwhelm” any move towards some world-equation which was formulated without the new knowledge since it was not known at the time that the world-equation was formalized.
Furthermore, if the universe is mathematical as MIT’s Professor Max Tegmark claims, then a Hawking-type “world-equation” would cover all mathematics without which parts of Tegmark’s universe would be “unaccounted for.”
In other words, history and the historical experience, cast doubt on the Stephen Hawking “finality” project. It’s not just that parts of physics don’t fit together. (General relativity and quantum mechanics, gravity and the other three fundamental forces.) Finality would also imply that there would be no new Stephen Hawking who would refute the world-equation as it stands at a certain point in time. In other words, if you choose, as scientists like Freeman Dyson claim that the universe is a “vast evolutionary” process, then the mathematical thinking about it is also evolving or co-evolving and there’s no end.
There are no final works in poetry, novels, jokes, language, movies or songs and there’s perhaps also no end to science.
Thus a Hawking-type quest for the final world-equation seems enchanting but quixotic.
## Science and Its Limits
The outstanding physics theoretician Max Tegmark of MIT tells the story of how Ernest Rutherford’s 1933 prediction about atomic energy (i.e., that is was “moonshine”)—was refuted before 24 hours had passed when Szilard (the Hungarian genius) realized that a nuclear chain reaction could be set in motion getting around Rutherford’s pessimistic prediction of only a few hours before:
“In London, where Southampton Row passes Russell Square, across from the British Museum in Bloomsbury, Leo Szilard waited irritably one gray Depression morning for the stoplight to change. A trace of rain had fallen during the night; Tuesday, September 12, 1933, dawned cool, humid and dull. Drizzling rain would begin again in early afternoon. When Szilard told the story later he never mentioned his destination that morning. He may have had none; he often walked to think. In any case another destination intervened. The stoplight changed to green. Szilard stepped off the curb. As he crossed the street time cracked open before him and he saw a way to the future, death into the world and all our woes, the shape of things to come…”
(Richard Rhodes, The Making of the Atomic Bomb)
This Tegmark/Szilard “refutation” of Rutherford in our times reminds one of MIT’s AI pioneer, Prof. Marvin Minsky’s limitless and perhaps too rosy predictions for AI and human intelligence in the sixties and seventies.
A student pursuing education has to live with the paradox and puzzle that unpredicted surprises and leaps do occur in the world of science and they are astonishing. It is true at the same time, that the realm of science (i.e., “how” questions) cannot address “why” questions. The question “how was I born?” cannot replace “why was I born?”
Both of these questions have possible answers at various levels and are subject to hierarchies.
Steven Jay Gould, the late Harvard biologist, had a felicitous phrase, “separate magisteria” (i.e., separate realms or domains) to describe this gap between the pursuit of personal meaning (human quest) and the pursuit of (tentative) accuracy (scientific quest).
## Essay 40: Movies as a “Backdoor Into Financial History”
“Financial history” (the Professor Niall Ferguson PBS miniseries, The Ascent of Money, of recent years tries to “flag” this domain) can be exciting and eye-opening if the student fins the kind of “backdoor” into it that makes it all enchanting and not a tiresome slog through opaque textbooks. Movies are a good way to “parachute” into fields, domains, areas of study:
The 1963 movie Mary Poppins is partly about bank runs and the “Tuppence” song in the movie communicates the centrality of London finance in the world of 1910, the setting of the movie:
“You see, Michael, you’ll be part of railways through Africa
Dams across the Nile, fleets of ocean Greyhounds
Majestic, self-amortizing canals
Plantations of ripening tea
All from tuppence, prudently fruitfully, frugally invested
In the, to be specific
In the Dawes, Tomes, Mousely, Grubbs
Fidelity fiduciary bank
Now Michael, when you deposit tuppence in a bank account
Soon you’ll see
That it blooms into credit of a generous amount
Semiannually
And you’ll achieve that sense of stature
To the high financial strata
That established credit, now commands
You can purchase first and second trust deeds
Think of the foreclosures
Bonds! Chattels! Dividends! Shares
Bankruptcies! Debtor sales! Opportunities
All manner of private enterprise
Shipyards! The mercantile
Collieries! Tanneries
Incorporations! Amalgamations! Banks”
The current U.S. Treasury Secretary Mnuchin was a foreclosure king of the Great Recession of 2008. An American movie on “bank runs” is of course the classic It’s a Wonderful Life (with James Stewart as the local banker.)
The 1910 world of London finance show in the movie Mary Poppins can be now contextualized by realizing all of this crashed down in August 1914 which represents the beginning of post-WW 1deglobalization.” Thus, finance and globalization issues haunt the present.
Walter Bagehot’s masterpiece of 1873, Lombard Street, is a kind of anticipation of this syndrome and Charles Kindleberger‘s (MIT) Manias, Panics and Crashes gives the sense of the underlying instability.
Kevin Phillips’s book Bad Money of 2008 outlines the dangers of “over financialization.”
The movie and the fun song can help a student find his or her way in to these areas and domains.
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