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# Differences

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 phy131studiof15:lectures:chapter1 [2015/08/24 08:46]mdawber [Accuracy and precision] phy131studiof15:lectures:chapter1 [2015/08/24 08:49]mdawber [Stating errors] Both sides previous revision Previous revision 2015/08/24 09:19 mdawber [Scientific Notation] 2015/08/24 08:51 mdawber [Visual comparison of different types of eror] 2015/08/24 08:49 mdawber [Stating errors] 2015/08/24 08:46 mdawber [Accuracy and precision] 2015/08/24 08:45 mdawber [Scientific Notation] 2015/08/21 14:48 mdawber [Scientific Notation] 2015/08/21 14:32 mdawber [Measurement Exercises] 2015/08/21 14:30 mdawber [Some propagation rules] 2015/08/21 13:46 mdawber [Some propagation rules] 2015/07/27 16:04 mdawber [Stating errors] 2015/07/27 16:02 mdawber [Stating errors] 2015/07/27 16:02 mdawber [Error and Uncertainty] 2015/07/27 15:58 mdawber [1.P.005] 2015/07/27 15:55 mdawber [Scientific Notation] 2015/07/20 11:06 mdawber [Example of systematic error] 2015/07/20 11:05 mdawber [Class 1 - Units, Measurement and Error] 2015/07/20 11:04 mdawber created 2015/08/24 09:19 mdawber [Scientific Notation] 2015/08/24 08:51 mdawber [Visual comparison of different types of eror] 2015/08/24 08:49 mdawber [Stating errors] 2015/08/24 08:46 mdawber [Accuracy and precision] 2015/08/24 08:45 mdawber [Scientific Notation] 2015/08/21 14:48 mdawber [Scientific Notation] 2015/08/21 14:32 mdawber [Measurement Exercises] 2015/08/21 14:30 mdawber [Some propagation rules] 2015/08/21 13:46 mdawber [Some propagation rules] 2015/07/27 16:04 mdawber [Stating errors] 2015/07/27 16:02 mdawber [Stating errors] 2015/07/27 16:02 mdawber [Error and Uncertainty] 2015/07/27 15:58 mdawber [1.P.005] 2015/07/27 15:55 mdawber [Scientific Notation] 2015/07/20 11:06 mdawber [Example of systematic error] 2015/07/20 11:05 mdawber [Class 1 - Units, Measurement and Error] 2015/07/20 11:04 mdawber created Next revision Both sides next revision Line 93: Line 93: But here is where it gets confusing... When you estimate the uncertainty of your measurement, as you will do frequently in the lab component of this course, you should consider the possible sources of error that contribute to the uncertainty. This way if there are large sources of error in your experiment, you will have a large uncertainty which will not exclude the accurate value of the quantity you are trying to measure. But here is where it gets confusing... When you estimate the uncertainty of your measurement, as you will do frequently in the lab component of this course, you should consider the possible sources of error that contribute to the uncertainty. This way if there are large sources of error in your experiment, you will have a large uncertainty which will not exclude the accurate value of the quantity you are trying to measure. - ===== Stating errors ===== + ===== Stating uncertainty ===== Because we cannot know any experimentally measured value to absolute certainty we state such values in the format Because we cannot know any experimentally measured value to absolute certainty we state such values in the format - $\mathrm{value}\pm\mathrm{error\,in\,value}$ + $\mathrm{value}\pm\mathrm{uncertainty\,in\,value}$ - This is equivalent to saying that the actual value of the quantity could be anywhere between the value minus the error and the value plus the error. + This is equivalent to saying that the actual value of the quantity could be anywhere between the value minus the uncertainty and the value plus the uncertainty. Much of our activity in the lab will be about the best way to realistically determine and estimate for the uncertainty in any value we obtain. Much of our activity in the lab will be about the best way to realistically determine and estimate for the uncertainty in any value we obtain.


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# The VB program CASVB

CASVB is a general program for valence bond calculations
written by T. Thorsteinsson and D. L. Cooper (1996–2005).

This program can be used in two basic modes:

• variational optimization of quite general types of nonorthogonal MCSCF or modern valence bond wavefunctions
• representation of CASSCF wavefunctions in modern valence form, using overlap- (relatively inexpensive) or energy-based criteria.

Bibliography:

T. Thorsteinsson, D. L. Cooper, J. Gerratt, P. B. Karadakov and M. Raimondi, Theor. Chim. Acta 93, 343–66 (1996).
T. Thorsteinsson and D. L. Cooper, in Quantum Systems in Chemistry and Physics. Volume 1: Basic problems and models systems, eds. A. Hernández-Laguna, J. Maruani, R. McWeeny, and S. Wilson (Kluwer, Dordrecht, 2000); pp 303–26. [Please contact David L. Cooper if you have problems accessing this key reference]
All publications resulting from use of this program should acknowledge relevant publications. There is a more complete bibliography at http:%%//%%www.liv.ac.uk/ dlc/CASVB.html

All CASVB sub-commands may be abbreviated by four letters. The general input structure can be summarized as follows:

• For generating representations of CASSCF wavefunctions, the program is invoked by the command CASVB. For variational optimization of wavefunctions it is normally invoked inside MULTI by the sub-command VB (see optimizing valence bond wavefunctions).
• Definition of the CASSCF wavefunction (not generally required).
• Definition of the valence bond wavefunction.
• Recovery and/or storage of orbitals and vectors.
• Manual input of starting guess (optional).
• Optimization control.
• Definition of molecular symmetry and possible constraints on the VB wavefunction.
• Wavefunction analysis.
• Further general options.

Items a) and b) should precede everything else in the input; apart from this, commands may come in any order.

CASVB is interfaced with the determinant part of MULTI (i.e., CONFIG,CSF; must not be specified). When this program is run prior to CASVB, the CI vector must dumped using one of the directives SAVE, NATORB, CANONICAL, or LOCALI (see section saving the CI vectors and information for a gradient calculation). The three latter are recommended.

VBDUMP[,vbdump];[casvb:vbdump]

If present, the VBDUMP card must occur first in the CASVB input. It is not required for variational calculations.

Note that in the majority of cases (e.g., if a CASVB run occurs immediately after MULTI, or for variational calculations), explicit specification of dump records with vbdump is not required.

Wavefunction definitions may be restored here using VBDUMP cards (see also Section saving wavefunction information for CASVB). The default record name (vbdump) is 4299.2. If a  VBDUMP card is not present and record 4299.2 does not exist, then CASVB will attempt to generate the wavefunction information automatically based on the latest MCSCF calculation (however, STATE and WEIGHT information will not be restored in such a case).

The definitions of the CASSCF wavefunction may also be specified manually using some or all of the directives:

• OCC Occupied orbitals.
• CLOSED Closed-shell orbitals.
• FROZEN Frozen-core orbitals.
• WF Wavefunction card.
• STATE Number of states for this wavefunction symmetry.
• WEIGHT Weights of states.

For the exact definition of these cards see sections defining the orbital subspaces and defining the optimized states. These commands may also be used to modify the values defined in VBDUMP. The information given on these cards should correspond to the CI vector saved in the CASSCF calculation. The cards, and their ordering, should therefore coincide with those used in MULTI, except for the WEIGHT cards which may differ. At present, the VB wavefunction must correspond to a well-defined number of electrons and total spin. Other states may be present, but an error condition will occur if non-zero weights are specified for wavefunction symmetries with varying values of elec or spin.

The number of core and active orbitals ($mcore$, $mact$), active electrons ($Nact$), and the value of the total spin will be identical to that defined for the CASSCF wavefunction. The spatial VB configurations are defined in terms of the active orbitals only, and may be specified using one or more CON cards (note that the RESTRICT and SELECT keywords are not used in CASVB):

CON,$n_1,n_2,n_3,n_4,\ldots$;

The configurations can be specified by occupation numbers, exactly as in MULTI (see section specifying orbital configurations), so that $n_i$ is the occupation of the $i$th valence bond orbital. Alternatively a list of $Nact$ orbital numbers (in any order) may be provided – the program determines which definition applies. The two cards CON,1,0,1,2; and CON,1,3,4,4; are thus equivalent.

If no configurations are specified the single covalent configuration $\phi_1\phi_2\cdots\phi_{Nact}$ is assumed.

SPINBASIS,key;

key may be chosen from KOTANI (default), RUMER, PROJECT or LTRUMER, specifying the basis of spin eigenfunctions used in the definition of valence bond structures. PROJECT refers to spin functions generated using a spin projection operator, LTRUMER to Rumer functions with the so-called “leading term“ phase convention.

The appropriate Molpro records may be specified explicitly using the START directive (an alternative is the vbdump mechanism described in section the VBDUMP directive):

START,ci,vb,orb,trnint;[casvb:trnint]

ci: record name for the CASSCF CI vector. The CI vector must have been dumped previously using either of the SAVE, NATORB, CANONICAL, or LOCALI directives (see section saving the CI vectors and information for a gradient calculation). A default value for ci is determined from the most recent vbdump record(s).

Note that if the ci record is not found, only an energy-based optimization of the VB wavefunction can be carried out.

vb: record name for the valence bond orbitals and structure coefficients, as saved by a previous CASVB calculation. If the VB wavefunction was previously saved in the AO basis the orbitals will be projected onto the present active space (note that it is necessary to specify a record name for the molecular orbitals (orb below) for this to be possible).

orb: record name for the molecular orbitals defining the CASSCF wavefunction. This information is necessary if one wants to output the valence bond orbitals in the atomic orbital basis.

trnint: record name for the transformed CASSCF integrals. These are required for the energy-based criteria (i.e., if CRIT,ENERGY is specified), and can be saved inside MULTI by the TRNINT sub-command (see saving transformed integrals). The default record name, both here and in MULTI, is 1900.1.

SAVE,vb,civb;

vb: record name for VB wavefunction (default is first available record after 3200.2), i.e., orbitals and structure coefficients.

civb: record name for valence bond full CI vector defined in terms of the CASSCF MOs (default is 3300.2). Saving this vector is necessary for the calculation of further properties, geometry optimization, etc.

It is normally advisable to use records on file 2 for vb and civb.

GUESS;key-1,…;key-2,…;…

The GUESS keyword initiates the input of a guess for the valence bond orbitals and structure coefficients. key-i can be either ORB, STRUC or READ. These keywords modify the guess provided by the program, or specified by the START directive. It is thus possible to modify individual orbitals in a previous solution to construct the starting guess.

ORB,i, c$_1$, c$_2$,…c$_{mact}$;

Specifies a starting guess for valence bond orbital number $i$. The guess is specified in terms of the $mact$ active MOs defining the CASSCF wavefunction. (Note that the definition of these MOs will depend on how the CI vector was dumped – i.e. which of the SAVE, NATORB, CANONICAL, or LOCALI directives was used (see section saving the CI vectors and information for a gradient calculation). Use of one of the three latter keywords is recommended.)

STRUC,c$_1$, c$_2$,…c$_{NVB}$;

Specifies a starting guess for the $NVB$ structure coefficients. If this card is not provided, and no guess specified by START, the perfect-pairing mode of spin coupling is assumed for the spatial configuration having the least number of doubly occupied orbitals. Note that the definition of structures depends on the value of SPINBASIS. Doubly occupied orbitals occur first in all configurations, and the spin eigenfunctions are based on the singly occupied orbitals being in ascending order.

The READ keyword can take one of the following forms:

READ,ORB,iorb1[,TO,iorb2] [,AS,jorb1[,TO,jorb2]] [,FROM,record];

READ,STRUC,istruc1[,TO,istruc2] [,AS,jstruc1[,TO,jstruc2]] [,FROM,record];

READ,ALL [,FROM,record];

In this way a subset of orbitals and/or structure coefficients may be picked out from a previous calculation. Renumbering of orbitals or structures can be done using the “AS” construct as outlined above. If the VB wavefunction was previously saved in the AO basis, the orbitals will be projected onto the present active space (note that it is necessary to specify a record name for the molecular orbitals (orb in the START commmand) for this to be possible).

Default for record is the vb record name specified in keyword START (if applicable).

ORBPERM,$i_1$,…,$i_{mact}$;

Permutes the orbitals in the valence bond wavefunction and changes their phases according to $\phi_j'={\rm sign}(i_j)\phi_{{\rm abs}(i_j)}$. The guess may be further modified using the GUESS keyword. Also the structure coefficients will be transformed according to the given permutation (note that the configuration list must be closed under the orbital permutation for this to be possible).

CRIT,method;

Specifies the criterion for the optimization. method can be OVERLAP or ENERGY (OVERLAP is default). The former maximizes the normalized overlap with the CASSCF wavefunction: $${\rm max}\left(\frac{\langle\Psi_{CAS}|\Psi_{VB}\rangle}{(\langle\Psi_{VB}|\Psi_{VB}\rangle)^{1/2}}\right)$$ and the latter simply minimizes the energy: $${\rm min}\left(\frac{\langle\Psi_{VB}|\hat{H}|\Psi_{VB}\rangle}{\langle\Psi_{VB}|\Psi_{VB}\rangle}\right).$$

MAXITER,$N_{iter}$;

Specifies the maximum number of iterations in the second order optimizations. Default is $N_{iter}$=50.

(NO)CASPROJ;

With this keyword the structure coefficients are picked from the transformed CASSCF CI vector, leaving only the orbital variational parameters. For further details see the bibliography. This option may be useful to aid convergence.

SADDLE,n;

Defines optimization onto an n$^{\rm th}$-order saddle point. See also T. Thorsteinsson and D. L. Cooper, Int. J. Quant. Chem. 70, 637–50 (1998).

More than one optimization may be performed in the same CASVB deck, by the use of OPTIM keywords:

OPTIM[;;FINOPTIM];

The subcommands may be any optimization declarations defined in this section, as well as any symmetry or constraints specifications described in section point group symmetry and constraints. Commands given as arguments to OPTIM will be particular to this optimization step, whereas commands specified outside will act as default definitions for all subsequent OPTIM keywords.

If only one optimization step is required, the OPTIM keyword need not be specified.

When only a machine-generated guess is available, CASVB will attempt to define a sequence of optimization steps chosen such as to maximize the likelihood of successful convergence and to minimize CPU usage. To override this behaviour, simply specify one or more OPTIM cards.

A loop over two or more optimization steps may be specified using:

ALTERN,Niter;;FINALTER

With this specification the program will repeat the enclosed optimization steps until either all optimizations have converged, or the maximum iteration count, Niter, has been reached.

The problems associated with symmetry-adapting valence bond wavefunctions are considered, for example, in: T. Thorsteinsson, D. L. Cooper, J. Gerratt and M. Raimondi, Theor. Chim. Acta 95, 131 (1997).

SYMELM,label,sign;

Initiates the definition of a symmetry operation referred to by label (any three characters). sign can be + or $-$; it specifies whether the total wavefunction is symmetric or antisymmetric under this operation, respectively. A value for sign is not always necessary but, if provided, constraints will be put on the structure coefficients to ensure that the wavefunction has the correct overall symmetry (note that the configuration list must be closed under the orbital permutation induced by label for this to be possible).

The operator is defined in terms of its action on the active MOs as specified by one or more of the keywords IRREPS, COEFFS, or TRANS (any other keyword will terminate the definition of this symmetry operator). If no further keyword is supplied, the identity is assumed for label. The alternative format SYMELM,label,sign;key-1,…; key-2,…;…may also be used.

IRREPS,i$_1$, i$_2$,…;

The list i$_1$, i$_2$,… specifies which irreducible representations (as defined in the CASSCF wavefunction) are antisymmetric with respect to the label operation. If an irreducible representation is not otherwise specified it is assumed to be symmetric under the symmetry operation.

COEFFS,i$_1$, i$_2$,…;

The list i$_1$, i$_2$,… specifies which individual CASSCF MOs are antisymmetric with respect to the label operation. If an MO is not otherwise specified, it is assumed to be symmetric under the symmetry operation. This specification may be useful if, for example, the molecule possesses symmetry higher than that exploited in the CASSCF calculation.

TRANS,n$_{dim}$, i$_1$, …i$_{n_{dim}}$, c$_{11}$, c$_{12}$, …c$_{n_{dim}n_{dim}}$;

Specifies a general n$_{dim}\times n_{dim}$ transformation involving the MOs i$_1$, …i$_{n_{dim}}$, specified by the $c$ coefficients. This may be useful for systems with a two- or three-dimensional irreducible representation, or if localized orbitals define the CASSCF wavefunction. Note that the specified transformation must always be orthogonal.

In general, for a VB wavefunction to be symmetry-pure, the orbitals must form a representation (not necessarily irreducible) of the symmetry group. Relations between orbitals under the symmetry operations defined by SYMELM may be specified according to:

ORBREL,i$_1$, i$_2$, label1, label2,…;

Orbital $i_1$ is related to orbital $i_2$ by the sequence of operations defined by the label specifications (defined previously using SYMELM). The operators operate right to left. Note that $i_1$ and $i_2$ may coincide. Only the minimum number of relations required to define all the orbitals should be provided; an error exit will occur if redundant ORBREL specifications are found.

As an alternative to incorporating constraints, one may also ensure correct symmetry of the wavefunction by use of a projection operator:

(NO)SYMPROJ[,irrep$_1$,irrep$_2$,…];

The effect of this keyword is to set to zero coefficients in unwanted irreducible representations. For this purpose the symmetry group defined for the CASSCF wavefunction is used (always a subgroup of D$_{2h}$). The list of irreps in the command specifies which components of the wavefunction should be kept. If no irreducible representations are given, the current wavefunction symmetry is assumed. In a state-averaged calculation, all irreps are retained for which a non-zero weight has been specified in the wavefunction definition. The SYMPROJ keyword may also be used in combination with constraints.

FIXORB,i$_1$, i$_2$,…;

This command freezes the orbitals specified in the list i$_1$, i$_2$,… to that of the starting guess. Alternatively the special keywords ALL or NONE may be used. These orbitals are eliminated from the optimization procedure, but will still be normalized and symmetry-adapted according to any ORBREL keywords given.

FIXSTRUC,i$_1$, i$_2$,…;

Freezes the coefficients for structures i$_1$, i$_2$,… . Alternatively the special keywords ALL or NONE may be used. The structures are eliminated from the optimization procedure, but may still be affected by normalization or any symmetry keywords present.

DELSTRUC,i$_1$, i$_2$,…,[ALL],[NONE];

Deletes the specified structures from the wavefunction. The special keywords ALL or NONE may be used. A structure coefficient may already be zero by symmetry (as defined by SYMELM and ORBREL), in which case deleting it has no effect.

ORTHCON;key-1,…;key-2,…;…

The ORTHCON keyword initiates the input of orthogonality constraints between pairs of valence bond orbitals. The sub-keywords key-i can be one of ORTH, PAIRS, GROUP, STRONG or FULL as described below. Orthogonality constraints should be used with discretion. Note that orthogonality constraints for an orbital generated from another by symmetry operations (using the ORBREL keyword) cannot in general be satisfied.

ORTH,i$_1$, i$_2$, …;

Specifies a list of orbitals to be orthogonalized. All overlaps between pairs of orbitals in the list are set to zero.

PAIRS,i$_1$, i$_2$, …;

Specifies a simple list of orthogonalization pairs. Orbital $i_1$ is made orthogonal to $i_2$, $i_3$ to $i_4$, etc.

GROUP,label,i$_1$, i$_2$, …;

Defines an orbital group to be used with the ORTH or PAIRS keyword. The group is referred to by label which can be any three characters beginning with a letter a–z. Labels defining different groups can be used together or in combination with orbital numbers in ORTH or PAIRS. i$_1$, i$_2$, … specifies the list of orbitals in the group. Thus the combination GROUP,AZZ,1,2; GROUP,BZZ,3,4; ORTH,AZZ,BZZ; will orthogonalize the pairs of orbitals 1-3, 1-4, 2-3 and 2-4.

STRONG;

This keyword is short-hand for strong orthogonality. The only allowed non-zero overlaps are between pairs of orbitals ($2n$$-$$1$, $2n$).

FULL;

This keyword is short-hand for full orthogonality. This is mainly likely to be useful for testing purposes.

(NO)SCORR;

With this option, expectation values of the spin operators $(\hat{s}_\mu+\hat{s}_\nu)^2$ are evaluated for all pairs of $\mu$ and $\nu$. Default is NOSCORR. The procedure is described by: G. Raos, J. Gerratt, D. L. Cooper and M. Raimondi, Chem. Phys. 186, 233–250 (1994); ibid, 251–273 (1994); D. L. Cooper, R. Ponec, T. Thorsteinsson and G. Raos, Int. J. Quant. Chem. 57, 501–518 (1996).

At present this analysis is only implemented for spin-coupled wavefunctions.

For further details regarding the calculation of weights in CASVB, see T. Thorsteinsson and D. L. Cooper, J. Math. Chem. 23, 105-26 (1998).

VBWEIGHTS,key1,key2,…

Calculates and outputs weights of the structures in the valence bond wavefunction $\Psi_{VB}$. key specifies the definition of nonorthogonal weights to be used, and can be one of:

• CHIRGWIN Evaluates Chirgwin-Coulson weights (see: B. H. Chirgwin and C. A. Coulson, Proc. Roy. Soc. Lond. A201, 196 (1950)).
• LOWDIN Performs a symmetric orthogonalization of the structures and outputs the corresponding weights.
• INVERSE Outputs “inverse overlap populations“ as in G. A. Gallup and J. M. Norbeck, Chem. Phys. Lett. 21, 495–500 (1973).
• ALL All of the above.
• NONE Suspends calculation of structure weights.

The commands LOWDIN and INVERSE require the overlap matrix between valence bond structures, and some computational overhead is thus involved.

For further details regarding the calculation of weights in CASVB, see T. Thorsteinsson and D. L. Cooper, J. Math. Chem. 23, 105-26 (1998).

CIWEIGHTS,key1,key2,…[,$N_{\rm conf}$];

Prints weights of the CASSCF wavefunction transformed to the basis of nonorthogonal VB structures. For the key options see VBWEIGHTS above. Note that the evaluation of inverse overlap weights involves an extensive computational overhead for large active spaces. Weights are given for the total CASSCF wavefunction, as well as the orthogonal complement to $\Psi_{VB}$. The default for the number of configurations requested, $N_{\rm conf}$, is 10. If $N_{\rm conf}$=$-1$ all configurations are included.

PRINT,i$_1$, i$_2$,…;

Each number specifies the level of output required at various stages of the execution, according to the following convention:

• -1 No output except serious, or fatal, error messages.
• 0 Minimal output.
• 1 Standard level of output.
• 2 Extra output.

The areas for which output can be controlled are:

• $i_1$ Print of input parameters, wavefunction definitions, etc.
• $i_2$ Print of information associated with symmetry constraints.
• $i_3$ General convergence progress.
• $i_4$ Progress of the 2nd order optimization procedure.
• $i_5$ Print of converged solution and analysis.
• $i_6$ Progress of variational optimization.
• $i_7$ Usage of record numbers on file 2.

For all, the default output level is +1. If $i_5\geq$2 VB orbitals will be printed in the AO basis (provided that the definition of MOs is available); such output may be especially useful for plotting of orbitals.

Calculations can also be performed for various types of direct product wavefunctions and/or with strictly localized orbitals. Details are available from the authors. These facilities will be documented in a later release.

SERVICE;
This keyword takes precedence over any others previously defined to CASVB. It provides simple facilities for retrieving orbital coefficients and VB structure coefficients. It should not be used during a run of CASVB that has been invoked from inside MULTI.

START,record.file;
Coefficients are taken from record.file. The default value is 2100.2.

WRITE,iwrite;
Vectors in the symmetry orbital basis are written to channel iabs(iwrite). The default action is to write these vectors to the standard output. If iwrite is negative, then the vectors are instead written to a binary file as a single record.

SPECIAL,idim1,idim2,idim3,idim4;
If present, this keyword must come last. The program attempts to retrieve from record.file a vector of length idim1*idim2+idim3, after first skipping idim4 elements. The vector is written according to the setting of iwrite. (Default idim values are zero.)

***, ch2                      ! A1 singlet state
geometry={angstrom
c
h1,c,1.117
h2,c,1.117,h1,102.4}
int
hf
{multi;occ,4,1,2;closed,1      ! 6 in 6 CASSCF
natorb,,ci,save=3500.2;vbdump}
{casvb                         ! Overlap-based VB using
save,3200.2}                   ! the spin-coupled wavefunction
{casvb                         ! Energy-based VB calculation
start,,3200.2;save,3220.2
crit,energy}
{multi;occ,4,1,2;closed,1      ! Fully variational VB calculation
{vb;start,,3220.2;save,3240.2;print,,,,,2}}
---
***,n2s2 (model a)            ! Variational calculation for N2S2.
v=2.210137753 bohr
geometry={ n, -v, 0, 0;       ! NOTE: other choices of active space
n, +v, 0, 0;       ! give alternative (competing) models.
s, 0, -v, 0;
s, 0, +v, 0}
basis=VTZ
hf
{multi; occ,7,4,5,2,4,2,2,0; closed,7,4,5,2,1,0,1,0; natorb,,ci,save=3500.2}
{multi; occ,7,4,5,2,4,2,2,0; closed,7,4,5,2,1,0,1,0;
{vb; start,3500.2; scorr}}
---
***, lih                      ! Fully-variational VB calculation
r=2.8 bohr                    ! and geometry optimization.
geometry={li;h,li,r}
basis={
s,1,921.300000,138.700000,31.940000,9.353000,3.158000,1.157000;
c,1.6,0.001367,0.010425,0.049859,0.160701,0.344604,0.425197;
s,1,0.444600,0.076660,0.028640;
p,1,1.488000,0.266700,0.072010,0.023700;
c,1.2,0.038770,0.236257;
s,2,13.36,2.013,0.4538,.1233;
c,1.2,0.032828,0.231204}
hf
{multi; occ,4,0,0,0; closed,0,0,0,0; natorb,,ci,save=3500.2}
{multi; occ,4,0,0,0; closed,0,0,0,0;
{vb; start,3500.2}}
optg
---

format: plain


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SSC CGL Tier 1 :: EC Test 77
Home » EC Test 77 » General Questions

SSC CGL Tier 1 English Comprehension Questions and Answers Online Test 77

Get free ssc cgl tier 1 free English Comprehension questions and answers free online test mock tests quiz sample questions ssc online test ssc practice test freeonline Test & Mock Test Quiz 77

Exercise

1 . In Question out of the four alternatives, choose the one which best expresses the meaning of the given word and mark it in the Answer Sheet
$Heretic$
 Offender Conformable Religious Believer
2 . In Question out of the four alternatives, choose the one which best expresses the meaning of the given word and mark it in the Answer Sheet
$Lassitude$
 Vivacity Agility Lethargy Lively
3 . In question, choose the word opposite in meaning to the given word and mark it in the Answer Sheet
$Miscreant$
 Dishonest Vagabond Innocent Scoundrel
4 . In question, choose the word opposite in meaning to the given word and mark it in the Answer Sheet
$Giggle$
5 . In question, choose the word opposite in meaning to the given word and mark it in the Answer Sheet
$Rout$
 Defeat Achievement Vanquish Subjugate
6 . In question, four words are given in each question, out of which only one word is correctly spelt. Find the correctly spelt word and mark your answer in the Answer Sheet.
 Quarantine Qarantine Quaerantene Quaerantene
7 . In question, four words are given in each question, out of which only one word is correctly spelt. Find the correctly spelt word and mark your answer in the Answer Sheet.
 Ostentasion Austentaesion Ostentation Austantation
8 . In question, four words are given in each question, out of which only one word is correctly spelt. Find the correctly spelt word and mark your answer in the Answer Sheet.
 Enocuous Inocuouse Enocuuous Innocuous


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# Box Filtering Height Maps for Smooth Rolling Hills

By Graham Wihlidal | Published Oct 23 2004 02:03 PM in Game Programming

width height filtering value heightmap terrain bounds box convolution
 If you find this article contains errors or problems rendering it unreadable (missing images or files, mangled code, improper text formatting, etc) please contact the editor so corrections can be made. Thank you for helping us improve this resource

# Introduction

Real-time terrain is an extremely common element within most 3D engines, and generally serves as the building block for outdoor scenes.

One of the most common methods of terrain rendering is using height values to perform displacement mapping on a flat mesh grid; a method of visualizing geometry based on a source image, and
scaling factors for vertex positioning. Using this technique for terrain results in sharp edges, and is generally something to avoid when developing the visuals for a cutting-edge game, especially if
the environment for the game is best suited with smooth hills.

The sharp edges produced when working with height maps generally result from noise in the source image, and can be removed using one of many convolution filters. The easiest type of convolution
filter to implement is Box Filtering, and will be utilized in this example to remove noise present in terrain height maps prior to rendering.

# Box Filtering Technique

Box filtering, also known as average or mean filtering, is a method of reducing the intensity variation between pixels in an image, and is a commonly used technique to reduce noise.

This type of filtering is accomplished by replacing each pixel value with the average value of its surrounding neighbors, including itself. Doing so removes large intensity variations between
pixels, and produces a much smoother displacement map.

A convolution filter is a simple mathematical operation which is commonly used in image processing operations. Convolution filters provide a way of multiplying two numerical arrays together,
generally of different sizes, but of the same dimensionality, to produce a third numerical array which is the result of the filter operation. Of the two input arrays, one is generally the source
image to process, and the second array is commonly referred to as a filtering or convolution kernel. The resultant array has the same dimensionality as the source image array.

The convolution is performed by sliding the filtering kernel over the image through all positions where the kernel fits within the boundaries of the image. This is not always the case in regards
to edge smoothing though, as some implementations handle bounds checking so that the edges may be filtered as well.

Box filtering can be thought of as a convolution filter, because it too is based around a filtering kernel. This kernel represents the size of the surrounding area that will be sampled. The most
common convolution filter for the box technique is a 3x3 cell size, as shown in figure 1. A larger kernel can be used for severe smoothing, or several iterations of a smaller kernel can be used to
achieve similar results.

Figure 1

Using this technique, we can box filter any terrain height map, and produce smooth rolling hills, with hardly any noticeable edges.

# Algorithm Example

void BoxFilterHeightMap(unsigned long width, unsigned long height,

float*& heightMap, bool smoothEdges)

{

//     width: Width of the height map in bytes

//    height: Height of the height map in bytes

// heightMap: Pointer to your height map data

// Temporary values for traversing single dimensional arrays

long x = 0;

long z = 0;

long  widthClamp = (smoothEdges) ?  width : width  - 1;

long heightClamp = (smoothEdges) ? height : height - 1;

// [Optimization] Calculate bounds ahead of time

unsigned int bounds = width * height;

// Validate requirements

if (!heightMap)

return;

// Allocate the result

float* result = new float[bounds];

// Make sure memory was allocated

if (!result)

return;

for (z = (smoothEdges) ? 0 : 1; z < heightClamp; ++z)

{

for (x = (smoothEdges) ? 0 : 1; x < widthClamp; ++x)

{

// Sample a 3x3 filtering grid based on surrounding neighbors

float value = 0.0f;

float cellAverage = 1.0f;

// Sample top row

if (((x - 1) + (z - 1) * width) >= 0 &&

((x - 1) + (z - 1) * width) < bounds)

{

value += heightMap[(x - 1) + (z - 1) * width];

++cellAverage;

}

if (((x - 0) + (z - 1) * width) >= 0 &&

((x - 0) + (z - 1) * width) < bounds)

{

value += heightMap[(x    ) + (z - 1) * width];

++cellAverage;

}

if (((x + 1) + (z - 1) * width) >= 0 &&

((x + 1) + (z - 1) * width) < bounds)

{

value += heightMap[(x + 1) + (z - 1) * width];

++cellAverage;

}

// Sample middle row

if (((x - 1) + (z - 0) * width) >= 0 &&

((x - 1) + (z - 0) * width) < bounds)

{

value += heightMap[(x - 1) + (z    ) * width];

++cellAverage;

}

// Sample center point (will always be in bounds)

value += heightMap[x + z * width];

if (((x + 1) + (z - 0) * width) >= 0 &&

((x + 1) + (z - 0) * width) < bounds)

{

value += heightMap[(x + 1) + (z    ) * width];

++cellAverage;

}

// Sample bottom row

if (((x - 1) + (z + 1) * width) >= 0 &&

((x - 1) + (z + 1) * width) < bounds)

{

value += heightMap[(x - 1) + (z + 1) * width];

++cellAverage;

}

if (((x - 0) + (z + 1) * width) >= 0 &&

((x - 0) + (z + 1) * width) < bounds)

{

value += heightMap[(x    ) + (z + 1) * width];

++cellAverage;

}

if (((x + 1) + (z + 1) * width) >= 0 &&

((x + 1) + (z + 1) * width) < bounds)

{

value += heightMap[(x + 1) + (z + 1) * width];

++cellAverage;

}

// Store the result

result[x + z * width] = value / cellAverage;

}

}

// Release the old array

delete [] heightMap;

// Store the new one

heightMap = result;

}

# Screenshots

height="379" alt="" border="0">
Unfiltered

"379" alt="" border="0">
Filtered

# Conclusion

I would like to point out that box filtering is not the only solution to smooth sharp edges in terrain, and generally will not be a decent convolution filter to use in certain situations. For
example; if you want to have sharp cliffs and canyons, box filtering will smooth your cliffs into a gentle slope. In this case, what you would want to add to your filtering operation is edge
detection; also known as a “Smart Blur”, which smoothes noise patterns without adversely affecting sharpness or fine details in the height map. Using this filter method will smooth noise,
but leave your cliffs and canyons recognizable as such.

The example provided with this article is simplistic, and even though the algorithm is fairly optimized, a couple optimizations were neglected to preserve readability. The out of bounds testing
could be simplified further, and the loop conditionals could also be evaluated prior to processing.

The ability to specify whether or not to perform edge smoothing is dependent upon your implementation. Having the option to disable edge smoothing will allow terrain blocks to be tileable, as long
as the height map was designed properly. With edge smoothing, and multiple blocks of terrain, cracks would appear between the blocks, and skirting would be required to remove them.

Terrain rendering is a popular and important topic for 3D graphics, and height mapping is just one method of visualizing terrain geometry. It is my intent with this article to help improve the
quality of your terrain engine using box filtering as a simple, yet effective way of removing sharp edges.


text:
Question

# Sound of frequency 1000 Hz from a stationary source is reflected from an object approaching the source at $$30\, ms^{-1}$$, back to a stationary observer located at the source.The speed of sound in air is $$330 ms^{-1}$$.The frequency of the sound heard by the observer is

A
1200Hz
B
1000Hz
C
1090Hz
D
1100Hz

Solution

## The correct option is A $$1200 Hz$$Frequency recieved by approaching car$$f_1=f_0(1+\cfrac{30}{330})Hz$$where $$f_0=1000Hz$$$$f_1=1000(1+\cfrac{1}{11})Hz$$Frequency recieved by source again$$f_2=\cfrac{f_1}{1-\cfrac{30}{330}}=\cfrac{1000(1+\cfrac{1}{11})}{1-\cfrac{1}{11}}\\ \quad\quad=1200Hz$$Physics

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## Duke Mathematical Journal

### Regularly related lattices in Lie groups

Marc A. Rieffel

#### Article information

Source
Duke Math. J., Volume 45, Number 3 (1978), 691-699.

Dates
First available in Project Euclid: 20 February 2004

https://projecteuclid.org/euclid.dmj/1077312957

Digital Object Identifier
doi:10.1215/S0012-7094-78-04531-3

Mathematical Reviews number (MathSciNet)
MR507464

Zentralblatt MATH identifier
0405.22009

Subjects

#### Citation

Rieffel, Marc A. Regularly related lattices in Lie groups. Duke Math. J. 45 (1978), no. 3, 691--699. doi:10.1215/S0012-7094-78-04531-3. https://projecteuclid.org/euclid.dmj/1077312957

#### References

• [1] A. Borel, Density properties for certain subgroups of semi-simple groups without compact components, Ann. of Math. (2) 72 (1960), 179–188.
• [2] C. W. Curtis and I. Reiner, Representation theory of finite groups and associative algebras, Pure and Applied Mathematics, Vol. XI, Interscience Publishers, a division of John Wiley & Sons, New York-London, 1962.
• [3] E. G. Effros, Transformation groups and $C\sp\ast$-algebras, Ann. of Math. (2) 81 (1965), 38–55.
• [4] J. Glimm, Locally compact transformation groups, Trans. Amer. Math. Soc. 101 (1961), 124–138.
• [5] W. H. Gottschalk and G. A. Hedlund, Topological dynamics, American Mathematical Society Colloquium Publications, Vol. 36, American Mathematical Society, Providence, R. I., 1955.
• [6] G. Hochschild, The structure of Lie groups, Holden-Day Inc., San Francisco, 1965.
• [7] G. W. Mackey, Induced representations of locally compact groups. I, Ann. of Math. (2) 55 (1952), 101–139.
• [8] G. W. Mackey, The theory of unitary group representations, University of Chicago Press, Chicago, Ill., 1976.
• [9] M. S. Raghunathan, Discrete subgroups of Lie groups, Springer-Verlag, New York, 1972.
• [10] M. A. Rieffel, Commutation theorems and generalized commutation relations, Bull. Soc. Math. France 104 (1976), no. 2, 205–224.
• [11] M. A. Rieffel, The type of group measure space von Neumann algebras, Monatshefte Math. to appear.
• [12] M. A. Rieffel, Von Neumann algebras associated with some discrete co-compact subgroups, in preparation.
• [13] A. Selberg, On discontinuous groups in higher-dimensional symmetric spaces, Contributions to function theory (internat. Colloq. Function Theory, Bombay, 1960), Tata Institute of Fundamental Research, Bombay, 1960, pp. 147–164.
• [14] C. L. Siegel, Discontinuous groups, Ann. of Math. (2) 44 (1943), 674–689.
• [15] S. P. Wang, On density properties of $S$-subgroups of locally compact groups, Ann. of Math. (2) 94 (1971), 325–329.
• [16] S. P. Wang, On density properties of certain subgroups of locally compact groups, Duke Math. J. 43 (1976), no. 3, 561–578.


text:
# Thread: Request explanation of the formula

1. ## Request explanation of the formula

Each edge of a cube is equally divided into n parts, thus there are total n3 smaller cubes.

1) Let N0 be the number of smaller cubes with no exposed surfaces.

2) Let N1 be the number of smaller cubes with one exposed surfaces.

3) Let N2 be the number of smaller cubes with two exposed surfaces.

For (1) i.e the number of smaller cubes with no exposed surface N0 is given by the formula (n-2)3. Why is 2 subtracted from n ? And why is (n-2) raised to the power 3 ?

For (2) i.e the number of smaller cubes with one exposed surfaces N1 is given by the formula 6(n-2)2 . Again why is 2 subtracted from n ? Why are we multiplying (n-2) with 6 ? And why is (n-2) raised to the power 2 and not 3 ?

For (3) i.e
the number of smaller cubes with two exposed surfaces N2 is given by the formula 12(n-2). Again why are we multiplying (n-2) with 12 ? And why are we subtracting 2 from n ?

I will be very grateful if a detailed explanation is provided for the above formulae showing the dimensions clearly. For example Is 2 subtracted from n because it is the side layer or is there any other reason or is there any other way the cube is getting affected dimensionally? Thanks in advance!

2. ## Re: Request explanation of the formula

Originally Posted by Robocop

Each edge of a cube is equally divided into n parts, thus there are total n3 smaller cubes.

1) Let N0 be the number of smaller cubes with no exposed surfaces.

2) Let N1 be the number of smaller cubes with one exposed surfaces.

3) Let N2 be the number of smaller cubes with two exposed surfaces.

For (1) i.e the number of smaller cubes with no exposed surface N0 is given by the formula (n-2)3. Why is 2 subtracted from n ? And why is (n-2) raised to the power 3 ?

For (2) i.e the number of smaller cubes with one exposed surfaces N1 is given by the formula 6(n-2)2 . Again why is 2 subtracted from n ? Why are we multiplying (n-2) with 6 ? And why is (n-2) raised to the power 2 and not 3 ?

For (3) i.e
the number of smaller cubes with two exposed surfaces N2 is given by the formula 12(n-2). Again why are we multiplying (n-2) with 12 ? And why are we subtracting 2 from n ?

I will be very grateful if a detailed explanation is provided for the above formulae showing the dimensions clearly. For example Is 2 subtracted from n because it is the side layer or is there any other reason or is there any other way the cube is getting affected dimensionally? Thanks in advance!

There is a single layer of cubes on the outside that has > 0 faces showing.

Thus the interior "cube" of cubes with faces not showing is going to be of dimensions $(n-2)(n-2)(n-2)$, i.e. n, minus the right and left layer, and minus the top and bottom layer, and minus the front and back layers. So altogether there are $(n-2)^3$ cubes w/no faces showing.

Cubes with one face showing are on the outside layers but not along the intersection of two layers.

There are going to be $(n-2)(n-2)$ cubes per face having a single face showing thus a total of $6(n-2)^2$ as there are six faces.

Cubes with 2 faces showing are along the intersections of two large faces but not at the corners.

There will be $(n-2)$ cubes along each of these intersections. There are 12 intersections. 4 at the top, 4 on the bottom, and the 4 sides so there are

$12(n-2)$ cubes with 2 faces showing.

It should be clear that there are always 8 cubes with 3 faces showing. These are the corner cubes.


text:
# Need hint on 'simple' differentiation problem

1. Dec 1, 2004

### redshift

I need to differentiate z = artan(y/x) with respect to x. Somehow z' = y/(1 +(1/x)^2) doesn't seem right.

2. Dec 1, 2004

### Nylex

The derivative of arctan x = 1/(1 + x^2). There's a proof of that here.

3. Dec 1, 2004

### dextercioby

You know that
$$[\arctan(f(x))]'=\frac{f'(x)}{1+f^2 (x)}$$
,so apply the formula correctly.This is of course if "y" and "x" are independent variables.

4. Dec 1, 2004

### HallsofIvy

$$z'= \frac{1}{1+(\frac{y}{x})^2} $$-\frac{y}{x^2}$$$$
by the chain rule.

Now multiply both numerator and denominator by x2.

5. Dec 1, 2004

### dextercioby

Apparently the sofware didn't read the paranthesis you've written,so it should be:
$$z'= \frac{1}{1+(\frac{y}{x})^2} (-\frac{y}{x^2}) =-\frac{y}{x^2+y^2}$$


text:
## Rydberg Constant

H-Atom ($E_{n}=-\frac{hR}{n^{2}}$)

Jessica Li 4F
Posts: 115
Joined: Fri Aug 09, 2019 12:16 am

### Rydberg Constant

For the Rydberg equation, does the negative sign cancel out?

Tai Metzger 3K
Posts: 103
Joined: Sat Aug 24, 2019 12:17 am

### Re: Rydberg Constant

Not necessarily. If the energy is negative, it means that energy has been released and the e- has fallen from a higher energy level to a lower one. If energy is positive, it means that energy has been absorbed and the e- has jumped to a higher energy level.


text:
# Math Help - One to one

1. ## One to one

Let A be a nonempty set and let f: A-->B be a function. Prove that f is one-to-one iff there exists a function g: B-->A such that gof = 1(subA).

Can someone please show me how this is solved.

2. Originally Posted by hayter221
Let A be a nonempty set and let f: A-->B be a function. Prove that f is one-to-one iff there exists a function g: B-->A such that gof = 1(subA).
There is a well known theorem: There is an injection from A to B iff there is a surjection from B to A.

3. Are you sure, I thought it had something to do with inverses.

4. Originally Posted by hayter221
Are you sure, I thought it had something to do with inverses.
Do you understand this problem?
What I posted is all about 'inverses'!


text:
OGRE  13.4 Object-Oriented Graphics Rendering Engine
Ogre::ProfileFrame Struct Reference

Represents the total timing information of a profile since profiles can be called more than once each frame. More...

#include <OgreProfiler.h>

## Public Attributes

uint calls
The number of times this profile was called this frame. More...

ulong frameTime
The total time this profile has taken this frame. More...

uint hierarchicalLvl
The hierarchical level of this profile, 0 being the main loop. More...

## Detailed Description

Represents the total timing information of a profile since profiles can be called more than once each frame.

## ◆ frameTime

 ulong Ogre::ProfileFrame::frameTime

The total time this profile has taken this frame.

## ◆ calls

 uint Ogre::ProfileFrame::calls

The number of times this profile was called this frame.

## ◆ hierarchicalLvl

 uint Ogre::ProfileFrame::hierarchicalLvl

The hierarchical level of this profile, 0 being the main loop.

The documentation for this struct was generated from the following file:


text:
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# définition - Signal-to-noise ratio

signal-to-noise ratio (n.)

1.the ratio of signal intensity to noise intensity

## définition (complément)

voir la définition de Wikipedia

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# synonymes - Signal-to-noise ratio

signal-to-noise ratio (n.)

## dictionnaire analogique

acoustics[Domaine]

RationalNumber[Domaine]

rate, ratio[Hyper.]

signal-to-noise ratio (n.)

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Wikipedia

# Signal-to-noise ratio

Signal-to-noise ratio (often abbreviated SNR or S/N) is a measure used in science and engineering that compares the level of a desired signal to the level of background noise. It is defined as the ratio of signal power to the noise power. A ratio higher than 1:1 indicates more signal than noise. While SNR is commonly quoted for electrical signals, it can be applied to any form of signal (such as isotope levels in an ice core or biochemical signaling between cells).

The signal-to-noise ratio, the bandwidth, and the channel capacity of a communication channel are connected by the Shannon–Hartley theorem.

Signal-to-noise ratio is sometimes used informally to refer to the ratio of useful information to false or irrelevant data in a conversation or exchange. For example, in online discussion forums and other online communities, off-topic posts and spam are regarded as "noise" that interferes with the "signal" of appropriate discussion.

## Definition

Signal-to-noise ratio is defined as the power ratio between a signal (meaningful information) and the background noise (unwanted signal):

$\mathrm{SNR} = \frac{P_\mathrm{signal}}{P_\mathrm{noise}},$

where P is average power. Both signal and noise power must be measured at the same or equivalent points in a system, and within the same system bandwidth. If the signal and the noise are measured across the same impedance, then the SNR can be obtained by calculating the square of the amplitude ratio:

$\mathrm{SNR} = \frac{P_\mathrm{signal}}{P_\mathrm{noise}} = \left ( \frac{A_\mathrm{signal}}{A_\mathrm{noise} } \right )^2,$

where A is root mean square (RMS) amplitude (for example, RMS voltage). Because many signals have a very wide dynamic range, SNRs are often expressed using the logarithmic decibel scale. In decibels, the SNR is defined as

$\mathrm{SNR_{dB}} = 10 \log_{10} \left ( \frac{P_\mathrm{signal}}{P_\mathrm{noise}} \right ) = {P_\mathrm{signal,dB} - P_\mathrm{noise,dB}},$

which may equivalently be written using amplitude ratios as

$\mathrm{SNR_{dB}} = 10 \log_{10} \left ( \frac{A_\mathrm{signal}}{A_\mathrm{noise}} \right )^2 = 20 \log_{10} \left ( \frac{A_\mathrm{signal}}{A_\mathrm{noise}} \right ).$

The concepts of signal-to-noise ratio and dynamic range are closely related. Dynamic range measures the ratio between the strongest un-distorted signal on a channel and the minimum discernable signal, which for most purposes is the noise level. SNR measures the ratio between an arbitrary signal level (not necessarily the most powerful signal possible) and noise. Measuring signal-to-noise ratios requires the selection of a representative or reference signal. In audio engineering, the reference signal is usually a sine wave at a standardized nominal or alignment level, such as 1 kHz at +4 dBu (1.228 VRMS).

SNR is usually taken to indicate an average signal-to-noise ratio, as it is possible that (near) instantaneous signal-to-noise ratios will be considerably different. The concept can be understood as normalizing the noise level to 1 (0 dB) and measuring how far the signal 'stands out'.

[note 1]

## Alternative definition

An alternative definition of SNR is as the reciprocal of the coefficient of variation, i.e., the ratio of mean to standard deviation of a signal or measurement:[2][3]

$\mathrm{SNR} = \frac{\mu}{\sigma}$

where $\mu$ is the signal mean or expected value and $\sigma$ is the standard deviation of the noise, or an estimate thereof.[note 2] Notice that such an alternative definition is only useful for variables that are always non-negative (such as photon counts and luminance). Thus it is commonly used in image processing,[4][5][6][7] where the SNR of an image is usually calculated as the ratio of the mean pixel value to the standard deviation of the pixel values over a given neighborhood. Sometimes SNR is defined as the square of the alternative definition above.

The Rose criterion (named after Albert Rose) states that an SNR of at least 5 is needed to be able to distinguish image features at 100% certainty. An SNR less than 5 means less than 100% certainty in identifying image details.[8]

Yet another alternative, very specific and distinct definition of SNR is employed to characterize sensitivity of imaging systems; see signal to noise ratio (imaging).

Related measures are the "contrast ratio" and the "contrast-to-noise ratio".

## SNR for various modulation systems

### Amplitude modulation

Channel signal-to-noise ratio is given by

$\mathrm{(SNR)_{C,AM}} = \frac{A_c^2 (1 + k_a^2 P)} {2 W N_0}$

where W is the bandwidth and ka is modulation index

Output signal-to-noise ratio (of AM receiver) is given by

$\mathrm{(SNR)_{O,AM}} = \frac{A_c^2 k_a^2 P} {2 W N_0}$

### Frequency modulation

Channel signal-to-noise ratio is given by

$\mathrm{(SNR)_{C,FM}} = \frac{A_c^2} {2 W N_0}$

Output signal-to-noise ratio is given by

$\mathrm{(SNR)_{O,FM}} = \frac{A_c^2 k_f^2 P} {2 N_0 W^3}$

## Improving SNR in practice

Recording of the noise of a thermogravimetric analysis device that is poorly isolated from a mechanical point of view; the middle of the curve shows a lower noise, due to a lesser surrounding human activity at night.

All real measurements are disturbed by noise. This includes electronic noise, but can also include external events that affect the measured phenomenon — wind, vibrations, gravitational attraction of the moon, variations of temperature, variations of humidity, etc., depending on what is measured and of the sensitivity of the device. It is often possible to reduce the noise by controlling the environment. Otherwise, when the characteristics of the noise are known and are different from the signals, it is possible to filter it or to process the signal. When the signal is constant or periodic and the noise is random, it is possible to enhance the SNR by averaging the measurement.

## Digital signals

When a measurement is digitised, the number of bits used to represent the measurement determines the maximum possible signal-to-noise ratio. This is because the minimum possible noise level is the error caused by the quantization of the signal, sometimes called Quantization noise. This noise level is non-linear and signal-dependent; different calculations exist for different signal models. Quantization noise is modeled as an analog error signal summed with the signal before quantization ("additive noise").

This theoretical maximum SNR assumes a perfect input signal. If the input signal is already noisy (as is usually the case), the signal's noise may be larger than the quantization noise. Real analog-to-digital converters also have other sources of noise that further decrease the SNR compared to the theoretical maximum from the idealized quantization noise, including the intentional addition of dither.

Although noise levels in a digital system can be expressed using SNR, it is more common to use Eb/No, the energy per bit per noise power spectral density.

The modulation error ratio (MER) is a measure of the SNR in a digitally modulated signal.

### Fixed point

For n-bit integers with equal distance between quantization levels (uniform quantization) the dynamic range (DR) is also determined.

Assuming a uniform distribution of input signal values, the quantization noise is a uniformly distributed random signal with a peak-to-peak amplitude of one quantization level, making the amplitude ratio 2n/1. The formula is then:

$\mathrm{DR_{dB}} = \mathrm{SNR_{dB}} = 20 \log_{10}(2^n) \approx 6.02 \cdot n$

This relationship is the origin of statements like "16-bit audio has a dynamic range of 96 dB". Each extra quantization bit increases the dynamic range by roughly 6 dB.

Assuming a full-scale sine wave signal (that is, the quantizer is designed such that it has the same minimum and maximum values as the input signal), the quantization noise approximates a sawtooth wave with peak-to-peak amplitude of one quantization level[9] and uniform distribution. In this case, the SNR is approximately

$\mathrm{SNR_{dB}} \approx 20 \log_{10} (2^n \sqrt {3/2}) \approx 6.02 \cdot n + 1.761$

### Floating point

Floating-point numbers provide a way to trade off signal-to-noise ratio for an increase in dynamic range. For n bit floating-point numbers, with n-m bits in the mantissa and m bits in the exponent:

$\mathrm{DR_{dB}} = 6.02 \cdot 2^m$
$\mathrm{SNR_{dB}} = 6.02 \cdot (n-m)$

Note that the dynamic range is much larger than fixed-point, but at a cost of a worse signal-to-noise ratio. This makes floating-point preferable in situations where the dynamic range is large or unpredictable. Fixed-point's simpler implementations can be used with no signal quality disadvantage in systems where dynamic range is less than 6.02m. The very large dynamic range of floating-point can be a disadvantage, since it requires more forethought in designing algorithms.[10]

## Optical SNR

Optical signals have a carrier frequency that is much higher than the modulation frequency (about 200 THz and more). This way the noise covers a bandwidth that is much wider than the signal itself. The resulting signal influence relies mainly on the filtering of the noise. To describe the signal quality without taking the receiver into account, the optical SNR (OSNR) is used. The OSNR is the ratio between the signal power and the noise power in a given bandwidth. Most commonly a reference bandwidth of 0.1 nm is used. This bandwidth is independent of the modulation format, the frequency and the receiver. For instance an OSNR of 20dB/0.1 nm could be given, even the signal of 40 GBit DPSK would not fit in this bandwidth. OSNR is measured with an optical spectrum analyzer

## Notes

1. ^ The connection between optical power and voltage in an imaging system is linear. This usually means that the SNR of the electrical signal is calculated by the 10 log rule. With an interferometric system, however, where interest lies in the signal from one arm only, the field of the electromagnetic wave is proportional to the voltage (assuming that the intensity in the second, the reference arm is constant). Therefore the optical power of the measurement arm is directly proportional to the electrical power and electrical signals from optical interferometry are following the 20 log rule.[1]
2. ^ The exact methods may vary between fields. For example, if the signal data are known to be constant, then $\sigma$ can be calculated using the standard deviation of the signal. If the signal data are not constant, then $\sigma$ can be calculated from data where the signal is zero or relatively constant.
3. ^ Often special filters are used to weight the noise: DIN-A, DIN-B, DIN-C, DIN-D, CCIR-601; for video, special filters such as comb filters may be used.
4. ^ Maximum possible full scale signal can be charged as peak-to-peak or as RMS. Audio uses RMS, Video P-P, which gave +9 dB more SNR for video.

## References

1. ^ Michael A. Choma, Marinko V. Sarunic, Changhuei Yang, Joseph A. Izatt. Sensitivity advantage of swept source and Fourier domain optical coherence tomography. Optics Express, 11(18). Sept 2003.
2. ^ D. J. Schroeder (1999). Astronomical optics (2nd ed.). Academic Press. p. 433. ISBN 978-0-12-629810-9.
3. ^ Bushberg, J. T., et al., The Essential Physics of Medical Imaging, (2e). Philadelphia: Lippincott Williams & Wilkins, 2006, p.280.
4. ^ Rafael C. González, Richard Eugene Woods (2008). Digital image processing. Prentice Hall. p. 354. ISBN 0-13-168728-X.
5. ^ Tania Stathaki (2008). Image fusion: algorithms and applications. Academic Press. p. 471. ISBN 0-12-372529-1.
6. ^
7. ^
8. ^ Bushberg, J. T., et al., The Essential Physics of Medical Imaging, (2e). Philadelphia: Lippincott Williams & Wilkins, 2006, p.280. ISBN 0-683-30118-7
9. ^ Defining and Testing Dynamic Parameters in High-Speed ADCsMaxim Integrated Products Application note 728
10. ^ Fixed-Point vs. Floating-Point DSP for Superior AudioRane Corporation technical library

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# Ring of functions of generic fiber of affine special formal schemes

Fix $R$ a complete DVR. Recall from Berkovich's Vanishing Cycles for Formal Schemes II paper that we have a class of special formal schemes which are not topologically of finite type over $\operatorname{spf} R$, but modeled on special $R$-algebras which are controlled enough to work with. An example is the completion of $R[x]$ along the ideal $(\pi,x)$ where $\pi$ is a uniformizer of $R$. It turns out special $R$-algebras are always quotients for some $n,m \geq 0$ of the adic ring

$$R^{n,m} := R[[T_1,\ldots,T_n]]\{ S_1,\ldots,S_m \}.$$

To construct the generic fiber of a special $R$-algebra $A$, we write it as a quotient

$$I \subset R^{n,m} \to R^{n,m}/I \simeq A$$

Now to construct the generic fiber of $\operatorname{spf}A$, we consider $R^{n,m}$ to be a formal model for a partially-closed polydisc $E^m \times D^n$, and we define the generic fiber of $A$ to be the closed subspace of $E^m \times D^n$ determined by the coherent ideal $I \cdot \mathscr{O}_{E^m\times D^n}$

My question is how does the generic fiber constructed above compare to the naive generic fiber $\operatorname{sp} A \otimes_R K$? If $A$ is topologically of finite type they agree. In general the rings $\mathscr{O}_{E^m\times D^n} / I \cdot \mathscr{O}_{E^m\times D^n}$ and $A \otimes_R K$ are different, but is the former the completion of the latter with respect to some norm? Are the Berkovich spectra isomorphic?

It seems to be the case that $\mathscr{O}_{E^m\times D^n} / I \cdot \mathscr{O}_{E^m\times D^n}$ is a decreasing intersection of Banach rings $C_n$, and further each $C_n$ is a $A \otimes_R K$ algebra which further satisfies that the image of $A \otimes_R K \to C_n$ is topologically everywhere dense. Does this imply that $A\otimes_R K \to \mathscr{O}_{E^m\times D^n} / I \cdot \mathscr{O}_{E^m\times D^n}$ is topologically everywhere dense?

• What is $E^m\times D^n$? – Hang Aug 27 '18 at 1:34


text:
# 澳门新葡8455最新网站,www.8455.com,新葡京最新官网

Pole assignment in the regular row-by-row decoupling problem

In this talk we discuss pole assignment in the regular row-by-row decoupling problem by applying the canonical decomposition of the right invertible system $\{C, A, B\}$ obtained in Wei, Cheng and Wang (2010a). When the regular row-by-row decoupling problem is solvable, we first derive the formulas of all solutions of the regular row-by-row decoupling problem, then derive a matrix $\mathcal{D}$ and prove that the system is controllable, if and only if $\rank(\mathcal{D})=l$, where $l$ is a parameter obtained in the canonical decomposition. We then define so called decoupling controllable vectors, with different cases of $\rank(\mathcal{D})$ and the dimension of the decoupling controllable vector subspace, characterize all attainable transfer function matrices for the decoupling and pole assignment problem.


text:
# Ratio and Proportion Quiz for SSC CGL & Railways

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Here is a ratio and proportion quiz for upcoming exams like SSC CGL & Railways. This quiz contains important questions matching the exact pattern and syllabus of upcoming exams. Make sure you attempt today’s Quant Quiz for Upcoming Exams to check your preparation level and also read other SSC CGL articles below.

 SSC CGl Admit Card SSC CGL Exam Pattern SSC CGL Result SSC CGL Exam Date SSC CGL Cut off SSC CGL Books SSC CGL Application Status

Ratio & Proportion Quiz for SSC CGL & Railways
Que. 1

The ratio of weekly incomes of A and B is 9 : 7 and the ratio of their expenditures is 4 : 3. If each saves Rs. 200 per week, then the sum of their weekly incomes is

1.
Rs. 3,200
2.
Rs. 4,200
3.
Rs. 4,800
4.
Rs. 5,600

Que. 2

a : b for the greatest possible value of 5 – (3a – b)2 is

1.
3 : 5
2.
1 : 5
3.
1 : 3
4.
3 : 1

Que. 3

If 60% of A = 3/4 of B, then the ratio A:B is

1.
9 : 20
2.
20 : 9
3.
4 : 5
4.
5 : 4

Que. 4

By decreasing 15° of each angle of a triangle, the ratios of their angles are 2:3:5, The radian measure of greatest angle is:

1.
$$\frac{{11\pi }}{{24}}$$
2.
$$\frac{\pi }{{12}}$$
3.
$$\frac{\pi }{{24}}$$
4.
$$\frac{{5\pi }}{{24}}$$

Que. 5

Two numbers are in the ratio 4 : 7. If 30% of the first number is 24, then 60% of the second number is –

1.
84
2.
140
3.
64
4.
118

Que. 6

If 2r = h + √(r2 + h2) , (r ≠ 0) then find r : h.

1.
1 : 2
2.
3 : 1
3.
1 : 1
4.
4 : 3

Que. 7

The cost of making an article is divided between materials, labor and overheads in the ratio 3 : 4 : 2. If the cost of material is Rs.33.60 the cost of article is

1.
Rs.100.80
2.
Rs.11.20
3.
Rs.44.80
4.
Rs.22.40

Que. 8

A sum of Rs. 300 is divided among P, Q and R in such a way that Q gets Rs. 30 more than P and R gets Rs. 60 more than Q. The ratio of their share is

1.
5 : 3 : 2
2.
2 : 3 : 5
3.
3 : 2 : 5
4.
2 : 5 : 3

Que. 9

What must be added to each term of the ratio 2 : 5 so that it may be equal to 5 : 6 ?

1.
12
2.
78
3.
65
4.
13

Que. 10

The salaries of A, B and C are in the ratio 1 : 3 : 4 . If their salaries are increased by 5%, 10% and 15 % respectively, then their salaries will be in the ratio

1.
20 : 66 : 95
2.
21 : 66 : 95
3.
21 : 66 : 92
4.
19 : 66 : 92

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## More Quizzes from Testbook.com

Hopefully, this article on the SSC CGL was informative for all the SSC CGL aspirants. The candidates can also download this syllabus article as PDF and use it for future reference. For the preparation of such government job exams, the candidates can download the SSC CGL App and can also make use of exciting offers and you should also have a look at the other articles on SSC CGL that will help you in the application and preparation of coming SSC CGL Exam.

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text:
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# If x > y, x < 6, and y > -3, what is the largest prime number that

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Director
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If x > y, x < 6, and y > -3, what is the largest prime number that  [#permalink]

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22 Feb 2008, 13:01
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Difficulty:

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Question Stats:

27% (01:32) correct 73% (01:37) wrong based on 122 sessions

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If x > y, x < 6, and y > -3, what is the largest prime number that could be equal to x + y?

A. 11
B. 13
C. 7
D. 5
E. 2
Intern
Joined: 03 Mar 2007
Posts: 28
Re: If x > y, x < 6, and y > -3, what is the largest prime number that  [#permalink]

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22 Feb 2008, 14:56
Im going with 7

X>Y
X<6
Y>-3

Possible X Possible Y
5 2 = 7
5 0

Primes: 3, 5, 7, 11, 13, etc.
Director
Joined: 12 Jul 2007
Posts: 661
Re: If x > y, x < 6, and y > -3, what is the largest prime number that  [#permalink]

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22 Feb 2008, 15:18
2
jimmyjamesdonkey wrote:
If x>y, x<6, and y> -3, what is the largest prime number that could be equal to x+y?

Primes: 2, 3, 5, 7, 11, 13, 17, etc

If X is less than 6 and y is less than X, then 6+6 = 12 would be impossible. We know 13 and all the primes > 13 are out. Let's try 11.

5.9 + 5.1 = 11

5.9 < 6
5.1 < 5.9

everything checks out

Only trick is to remember there aren't any restrictions on the numbers, ie they don't have to be integers.
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Joined: 05 Mar 2008
Posts: 12
Re: If x > y, x < 6, and y > -3, what is the largest prime number that  [#permalink]

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24 Jun 2008, 10:47
Question:

If x>y, x<6 and y > -3, what is the largest prime number that could be equal to x+y ?

sorting out inequalities here, simplified form is -3 < y < x < 6

So as per this x not equals y and is less greater than x. As x is <6, largest possible value is 5. Hence largest possible value for y is 4 (as it should be < x). Maximum SUM (x+y) =10 and largest prime possible is 7

As per MGMAT explanation answer is 11. Here is the explanation from the book which doesn't make sense to me:

"The upper extreme for x is less than 6. The upper extreme for y is also < 6 as long as it is less than x. Therefor, x+y must be less than 12.The larget prime number less than 12 is 11".

The part I am not clear here is "The upper extreme for y is also < 6 as long as it is less than x". How it can be ever > 4 as it should be always less than x which in turn should be < 6 ?
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Re: If x > y, x < 6, and y > -3, what is the largest prime number that  [#permalink]

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24 Jun 2008, 10:54
1
The question does not state that x and y are integers. So x must be less than 6

Make x = 5.999999
Make y = 5.111111

Now x + y = 11.

pbvmba wrote:
Question:

If x>y, x<6 and y > -3, what is the largest prime number that could be equal to x+y ?

sorting out inequalities here, simplified form is -3 < y < x < 6

So as per this x not equals y and is less greater than x. As x is <6, largest possible value is 5. Hence largest possible value for y is 4 (as it should be < x). Maximum SUM (x+y) =10 and largest prime possible is 7

As per MGMAT explanation answer is 11. Here is the explanation from the book which doesn't make sense to me:

"The upper extreme for x is less than 6. The upper extreme for y is also < 6 as long as it is less than x. Therefor, x+y must be less than 12.The larget prime number less than 12 is 11".

The part I am not clear here is "The upper extreme for y is also < 6 as long as it is less than x". How it can be ever > 4 as it should be always less than x which in turn should be < 6 ?
Intern
Joined: 30 Jul 2007
Posts: 27
Re: If x > y, x < 6, and y > -3, what is the largest prime number that  [#permalink]

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24 Jun 2008, 11:03
Rewrite -3 < y < x < 6. We know that if x > y, then you can assign the highest possible value for x, and the next highest value possible value for y. It didn't say y or x had to be a negative number, so I allowed y and x to be positive.

I plugged in 5.99 for x and 5.98 for y = 11.97. Since that's not a prime, the next possible number down is 11.
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Re: If x > y, x < 6, and y > -3, what is the largest prime number that  [#permalink]

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21 Nov 2010, 09:45
Guys,
i have a question here from the MGMAT but I can't understand the answer to the following question :
If x>y, x<6, and y>-3, what is the largest prime number that could be equal to x + y ?
could someone help ?
Math Expert
Joined: 02 Sep 2009
Posts: 59587
Re: If x > y, x < 6, and y > -3, what is the largest prime number that  [#permalink]

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21 Nov 2010, 09:53
2
whichscore wrote:
Guys,
i have a question here from the MGMAT but I can't understand the answer to the following question :
If x>y, x<6, and y>-3, what is the largest prime number that could be equal to x + y ?
could someone help ?

So we have: $$-3<y<x<6$$.

Note that we are not told that $$x$$ and $$y$$ are integers. Obviously as $$x<6$$ and $$y<x$$ then $$x+y$$ must be less than 6+6=12. Let's check whether $$x+y$$ could be equal to the first prime less then 12, so to 11. Now, if $$y=5.1<5.9=x<6$$ then $$x+y=11=prime$$.

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Re: If x > y, x < 6, and y > -3, what is the largest prime number that  [#permalink]

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21 Nov 2010, 10:01
In fact the answer is 11.
I think that the trap in this question was assuming that x and y are integers.

Thank you bunnel !
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Posts: 1546
Re: If x > y, x < 6, and y > -3, what is the largest prime number that  [#permalink]

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31 Mar 2011, 10:40
kamalkicks wrote:
Q1. if x>y, x<6, and y>-3
what is the largest prime number that could be equal to x+y

kindly solve... and check what is the answer -- 11 or 7

x=5.6
y=5.4

x>y
5.6>5.4

x<6
5.6<6

y>-3
5.4>-3

x+y=5.6+5.4=11
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Re: If x > y, x < 6, and y > -3, what is the largest prime number that  [#permalink]

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31 Mar 2011, 19:15
>>>> if x>y, x<6, and y>-3

These conditions just tell you that on the number line, both x and y lie between -3 and 6 and that x is to the right of y:

Attachment:

Ques2.jpg [ 2.86 KiB | Viewed 5007 times ]

>>>>>what is the largest prime number that could be equal to x+y

The maximum value x and y can take is a little less than 6 so their sum must be a little less than 12. The largest prime less than 12 is 11. You can make 11 in a number of ways: (5.6 = x, 5.4 = y), (5.8 = x, 5.2 = y) etc.

They don't say that x and y must be integers. Had they said that, then the greatest sum would have been 7 (4 = x, 3 = y)
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Re: If x > y, x < 6, and y > -3, what is the largest prime number that  [#permalink]

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31 Mar 2011, 21:34
1
If y > -3, then x > -3

So we have -3 < x < 6 and -3 < y < 6

=> x + y < 12

To get the maximum value of x+y we have to maximize both x and y in way that y < x

so x can be max like 5.9 and y = 5.1, and this results in (x+y) becoming a prime number as 11 (11 is the biggest prime number < 12)
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Re: If x > y, x < 6, and y > -3, what is the largest prime number that  [#permalink]

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02 Apr 2011, 14:03
x,y need not be integers.

Given
x<6
y>-3
x>y => -3<x<6
-3<y<x

sum cannot be greater than 12
maximum value of x+y = maximum of x + maximum of y
=5.9 +5.1

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Re: If x > y, x < 6, and y > -3, what is the largest prime number that  [#permalink]

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07 Apr 2011, 04:06
1) If x>y, x<6 and y>-3, what is the largest prime number that could be equal to x+y?

as x<6 max x can be 5

y>-3 to get the largest prime number equal to x+y

y= -2

x+y = 5-2 = 3

Which is not correct. i want to know why?

Second method if we line up
If we line up the inequalities we get -3<y<x<6

But how to proceed further and how we get OA 11

Pls help
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Re: If x > y, x < 6, and y > -3, what is the largest prime number that  [#permalink]

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04 Sep 2011, 09:00
info given:

-3 < y < 6
-3 < x < 6

-3-3 = -6
-3+6 = +3
+6-3 = +3
+6+6 = +12

-6 < x+y < 12

hence, 11 is the largest prime
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Re: If x > y, x < 6, and y > -3, what is the largest prime number that  [#permalink]

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04 Sep 2011, 10:04
excellent question.... my answer was 7
i did reach till -3<y<x<6, then i "assumed" that to get a prime number(which is an integer) , we need to add two integers :O
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Re: If x > y, x < 6, and y > -3, what is the largest prime number that  [#permalink]

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22 Sep 2013, 22:42
atewari wrote:
If x>y, x<6 and y>-3, what is the largest prime number that could be equal to x+y?

I solve the question to have the result as 7 with the inequality expression being -3<y<x<6. Whereas, the Manhattan guide gives a solution as 11 with the same inequality expression. Am I missing something?

Yes, you are missing something.

Firstly, both x & y are not necessarily integers. Again, y>-3, thus, y can take any value as long as y<x and x<6. Had x=6, then y could be any value between {-3,6}[not including both the end points]. Thus, Max value of (x+y) <12. The nearest prime =11.

For better clarity, assume x=5.9, y=11-5.9 = 5.1

or x= 5.8, y = 11-5.8 = 5.2

and so on.

Hope this is clear.
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Re: If x > y, x < 6, and y > -3, what is the largest prime number that  [#permalink]

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29 Oct 2018, 22:00
From question stem,

-3<y<x<6

And we have to find out largest prime no. after adding x+y

now first thing is that x+y is an integer because all primes are positive integers.
Second thing is we have maximize the value so take maximum of x and to meet the condition x>y, take value of y just below of x.

lets take x= 5.99
y= 5.98
(Here I am taking two digit after decimal.)
now X+Y = 11.97 ...
but we already know that primes are integers then only possible prime just below this 11.97 is 11.
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Re: If x > y, x < 6, and y > -3, what is the largest prime number that  [#permalink]

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Re: If x > y, x < 6, and y > -3, what is the largest prime number that   [#permalink] 23 Nov 2019, 09:40
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# zbMATH — the first resource for mathematics

## Grujić, Zoran

Compute Distance To:
 Author ID: grujic.zoran Published as: Grujić, Z.; Grujić, Zoran External Links: MGP
 Documents Indexed: 47 Publications since 1998
all top 5

#### Co-Authors

 10 single-authored 9 Bradshaw, Zachary 7 Dascaliuc, Radu 7 Kalisch, Henrik 5 Kukavica, Igor 4 Bona, Jerry L. 4 Farhat, Aseel 2 Barbu, Viorel 2 Guberović, Rafaela 2 Lasiecka, Irena 2 Leitmeyer, Keith 2 Ruzmaikina, Anastasia A. 2 Tuffaha, Amjad M. 1 Jolly, Michael S. 1 Zhang, Qi-Shuhuason
all top 5

#### Serials

 7 Communications in Mathematical Physics 5 Indiana University Mathematics Journal 4 Journal of Mathematical Fluid Mechanics 3 Nonlinearity 3 Journal of Differential Equations 2 Archive for Rational Mechanics and Analysis 2 Journal of Mathematical Physics 2 Nonlinear Analysis. Theory, Methods & Applications. Series A: Theory and Methods 2 Annales de l’Institut Henri Poincaré. Analyse Non Linéaire 2 Journal of Nonlinear Science 2 Discrete and Continuous Dynamical Systems 1 Applicable Analysis 1 Studia Mathematica 1 Journal of Functional Analysis 1 Differential and Integral Equations 1 M$^3$AS. Mathematical Models & Methods in Applied Sciences 1 Journal of Dynamics and Differential Equations 1 Electronic Journal of Differential Equations (EJDE) 1 Journal of Nonlinear Mathematical Physics 1 Comptes Rendus. Mathématique. Académie des Sciences, Paris 1 Communications in Mathematical Sciences
all top 5

#### Fields

 45 Partial differential equations (35-XX) 33 Fluid mechanics (76-XX) 2 Functions of a complex variable (30-XX) 2 Dynamical systems and ergodic theory (37-XX) 2 Mechanics of deformable solids (74-XX) 1 Special functions (33-XX) 1 Harmonic analysis on Euclidean spaces (42-XX)


text:
Question: In one of the greatest disasters in the history of collective planning, all 330M residents of the U.S. join the same Zoom call by picking two random times between 8:00 and 9:00. They dial in at the earlier time and hang up at the later time. What is the probability that there is at least one person who is on for some portion of everyone else’s call, a so-called Superzoomer? Further, what is the chance of there being two Superzoomers on the same call?

## Solution

### Insights

Under the surface fact that the zoom start and end times are random between 8:00 and 9:00 is the fact that they impose an order. The spacing between consecutive points can be scaled however we like — but each ordering of the points has the same set of possible spacings. Finally, if we reassigned the labels of the points at random then we’d have another valid call, so all orders are equally likely.

In sum, the spacings don’t affect the problem, and all possible call schedules are equally probable.

If a Superzoomer exists they have to call in before the first person hangs up and they can’t hang up until the last person starts. This means that they have one of the following relationships with every other person on the call:

1. Superzoomer calls in before them and hangs up after them
2. Superzoomer calls in before them and hangs up before them
3. Superzoomer calls in after them and hangs up after them

Some possible Superzoomers look like this:

Two exemplary Superzoomers. In the top row we have a Superzoomer with long span, they get on the call first and hang up as soon as the last person calls in. The bottom row shows that a Superzoomer doesn’t necessarily need to be on the call for long. Note that both rows have multiple Superzoomers beyond the one highlighted in blue.

### Walking through the back door

Thinking about the forward problem got us very little — we couldn’t see a recursion and we couldn’t organize the combinatorics. This got us thinking, what would it take to dismantle a Superzoomer, e.g., what is the minimal surgery we’d have to do to erase a Superzoomer?

Necessarily, we have to snip the edge of the Superzoomer. But if that’s all we do, we can only reconnect that edge. So we have to cut another edge. If you do this enough times you start to see that it’s always possible to pick the two edges to cut. Moreover, when you reconnect those edges, there are $3$ possible pairings, and $2$ of them make a Superzoomer.

Similarly, if you start with a call schedule that has no Superzoomer, you can trivially create one. Pick two slots such that if they were connected, they’d make a Superzoomer, connect them (making the Superzoomer), and then connect their former connected slots to form a new pair.

Superzoomer rises Two edges are targeted for removal, then a long two of the open slots are used to make a Superzoomer (who starts before the first person hangs up, and hangs up after the last person calls in). The remaining two edges are connected as well. In this case, both new callers are superzoomers.

Keeping on with this game, you can do the same with $2$ Superzoomers by cutting $3$ edges, with $3$ Superzoomers by cutting $4$ edges, and so on.

This is an exhilarating realization, but is it a mirage?

Suppose these surgeries are always possible: aren’t we cherry picking the points, spoiling the chance that this has anything to do with organically formed call schedules?

No. Every call schedule is equally likely, and it doesn’t matter whether we set a particular caller’s interval first or last.

### But how do we pick the edges for surgery?

The essential insight is that the Superzoomer needs to span, at a minimum, $1$ call endpoint (call-in or hang-up) for $(n-1)$ unique callers. If there are more endpoints, which there probably are, it means that some callers started and ended within the Superzoomer’s calling window. So, the condition for forming a Superzoomer is a contiguous stretch of points that contain at least a call-in or a hang-up from $(n-1)$ distinct callers, with empty slots on either side.

Common Core This core contains at least one terminus for each of the $(n-1)$ callers whose times have already been picked. On either side of the core are open slots.

Notice that the zone of contiguous points can be anywhere along the line, but has to contain the middle point (the zone contains at least $(n - 1)$ points, and the earliest it can end is at position $1 + (n - 1) = n$):

Early to bed, and early to rise The earliest possible Superzoomer is the one who is first to join, leaves right after the last person joins, and noone else hangs up in between..

Start at one of the center slots and pick another at random. This is our first calling interval. Our goal is A. to build up a core of lone call-ins, lone hang-ups, or paired call-in/hang-ups, and B. to leave empty slots on either side.

The trouble is, depending on where our next slot leads, we could start to build up an imbalance, e.g. so that the holes are only on one side. For instance, our first slot connects to another slot on the left, so we have $2$ points on the left and $0$ on the right. If we pick our next slot to the left of the first, and it connects to a point on the left, then we have $4$ on the left and $0$ on the right. Clearly, this can result in a situation where we’re ready for our $2$ surgery edges. and they’re both on one side of the core.

Happily, this is easy to avoid. If the first slot connects to a point on the left, then make the next slot the one directly to the right of the first, if this connects to a point on the right, then the imbalance will become zero. If it connects to a point on the left, it will simply maintain the current imbalance. So, if we follow this rule, we can ensure that we always end up with a $(3,1)$ or $(2,2)$ balance of open slots when we’re ready for surgery.

A tale of two cores Two possible constructions of the core — one leading to an imbalanced distribution of the $4$ remaining calling slots about the core, and one balanced.

By construction, the empty slots we have when we’re ready for the surgery edges are always on either side of the core, since we are obsessively picking new slots exactly adjacent to the last slot we picked. So, the core contains one or both endpoints for every other of the $(n - 2)$ callers.

If we have a $(3,1)$ balance, the potential pairings are like this, and $2$ of the $3$ form a Superzoomer.

Decisions decisions Since there is only $1$ open slot on the left, and there are $3$ on the right, $2$ of the slots on the right will endpoints for the same caller. If we join the lonesome empty slot to the right empty slot on the right, then that caller won’t overlap with the caller formed by pairing the two rightmost slots. So, the core-spanning arc cannot land in the first slot on the right.

If we have a $(2,2)$ balance, then unless we pair the ones that are on the same side of the core, we’ll get a Superzoomer, again $2$ out of $3$.

### One Superzoomer is never enough.

Like we said in the hands-on feel good section, a surgery to make two Superzoomer means snipping $3$ edges. So, stop the core assembly when there are $6$ empty slots. Again, since we make sure to balance the holes, we can either have $(2,4)$ or $(3,3)$.

If we have $(2,4)$ then we’re going to be immediately forced into three distinct $(2,2)$ scenarios. The simple reason is that two of the slots on the right of the core have to form their own one-sided pair. If we pair the first slot (underlined blue) with the left-most point on the right, it means that this will not overlap with the one-sided pair, ruling the caller out from being a second Superzoomer. So, we have $3$ ways to pick the destination for the $1^\text{st}$ point, which leaves us with a balanced $(2,2)$ experience, which has $2$ ways of producing a Superzoomer. In all, there are $3\cdot 2$ ways to pick them.

One more time We can’t pair the underlined blue slot with the first open slot on the right because the resulting call wouldn’t intersect the path made by thetwo leftover points on the right side.

In the balanced $(3,3)$ case it’s clear that we have $3\cdot 2$ ways to pair them (moreover, a few of the scenarios result in $3$ Superzoomers).

### Let’s have $s$ Superzoomers

When there are an equal number of open slots on either side we immediately get $(s+1)!$ ways to pair them up.

In the case when they’re off by $2$, one side has $(s-1)$ open slots and the other side has $(s+1)$ of them. There are $(s+1)$ choices for how to pair off the one-sided pair (on the side with $(s+1)$ open slots), followed by $s!$ ways to pair up the remaining $2s$ endpoints that cross the core.

The pairing choices are made randomly of course, and there are $(2s+1)(2s-1)(2s-3)\cdots 3\cdot 1 = (2s+1)!!$ total ways to pair off a collection of $2(s+1)$ points (e.g. including the core-crossing Superzoomer arcs and not, as well as the cases that don’t meet our Superzoom needs).

So, the probability of finding $s$ Superzoomers (or more) is

$P(SZ \geq s) = \dfrac{(s+1)!}{(2s+1)!!}.$

Which matches the simulation

import numpy as np

def round():
people = [];
max_start = 0;
min_end = 1;
for i in range (1000):
x, y = np.random.random(), np.random.random()
temp_start = min(x,y)
temp_end = max(x,y)
if temp_start > max_start:
max_start = temp_start
if temp_end < min_end:
min_end = temp_end
people.append([temp_start, temp_end])

successes=0
for person in people:
if person[0] <= min_end and person[1] >= max_start:
successes +=  1
return successes

data = np.unique([round() for i in range(10000)], return_counts=True)


text:
CERN-PH-TH-2015-310
MCNET-15-31
arXiv:1512.08107
Eur.Phys.J. C76 (2016) 288

by: Milhano, José Guilherme (CERN) et al.

Abstract:
The di-jet asymmetry --- the measure of the momentum imbalance in a di-jet system --- is a key jet quenching observable. Using the event generator \jewel we show that the di-jet asymmetry is dominated by fluctuations both in proton-proton and in heavy ion collisions. We discuss how in proton-proton collisions the asymmetry is generated through recoil and out-of-cone radiation. In heavy ion collisions two additional sources contribute to the asymmetry, namely energy loss fluctuations and differences in path length. The latter is shown to be a sub-leading effect. We discuss the implications of our results for the interpretation of this observable.


text:
# what does positive Gaussian copula dependency describe

Suppose I have two variables and that their dependency structure is Gaussian copula. Suppose that the parameter of the Gaussian copula is 0.8 and the corresponding Kendall's tau is 0.7. How can I interpret the result? How can I describe this dependency from the plot?

Can I said that there are strong correlation at the middle values while these two variables are not correlated at the tails. Also, can I said the relationship is positive which means that if the first variable increases then the second variables increases as well. If yes, why this increase is not apply for extremes. That is, saying the variables are increase together that means they also should increased even at the extremes.

How rigorous do you want the exposition to be? In simple terms, even though we often talk about "dependence" and "correlation" in intuitive terms, they have formal definitions. In particular, Kendall's tau has a formal mathematical definition, and it measures the probability of concordance minus the probability of disconcordance, where concordance between $$(X,Y)$$ is defined as the probability that $$X_1 > X_2$$ and $$Y_1 > Y_2$$ jointly, where $$(X_1,Y_1)$$, $$(X_2,Y_2)$$ are independent random variables with the same distribution of $$(X,Y)$$. As such, even if $$(X,Y)$$ have a high value for kendall's tau, they can certainly have little concordance at the tails (or any other part of the distribution) as long as the mass of being at those parts are low enough. If I have no association at a certain part of the distribution, but the density is really low, then an overall measure of kendall's tau may say little about the dependence at certain areas.

To answer your question more explicitly, if you know that the copula is a Gaussian copula, then knowing the parameter $$.8$$ is in some sense all you need to know. A commonly studied feature of copulas is the upper (lower) tail dependence, which precisely measures the 'relationship' of the copula at the tails. It can be shown that for gaussian copulas, the tail dependence is $$0$$. So in some sense, this is why the correlation in the middle does not apply at the extremes--it has to do with your choice of copula. For a rigorous proof of this, and also an intuitive visualization, check out this post:

Why is Gaussian Copula's Tail Dependence Zero?

Generally, copulas measure arbitrarily complicated and intricate relationships, and there's no reason why correlation in the middle implies anything about correlation at the tails, and measures such as Kendall's tau provide information about the overall 'relationship.' Maybe not the best comparison, but it's sort of similar to how for a random variable, the mean and variance provide information about the variable, but it says little about tail behavior, and so just how tail dependence exits as a concept for copulas, so does concepts like kurtosis for a random variable.

• Thank you so much for your great help. One point still confuse me and do not know how interpret it. So, can I said that " $X_1$ and $X_2$ has positive gaussian dependency at a short period (at the middle), while they are independence at their extreme. That is, if $X_1$ increase then this independent from $X_2$. In other words, can I said that $X_1$ does not affected by $X_2$ at extreme. My main problem still in how to interpret this dependency in more details. This is the second part of my question. – Maryam Jun 14 '20 at 17:58
• @Maryam Yes, you can say such a statement at the limit, but for any finite interval (ie the xth upper percentile of the joint distribution), there will still be a dependence, but at the limit this disappears. It's not clear what it means to "describe the dependence," but typically in your case there is a positive dependence between $X_1,X_2$, but this dependence tapers off at the extreme, and at the limit, there is tail independence. – doubled Jun 14 '20 at 20:21
• I really do not know how to thank you for your great help. I just have two variables and would like to see is there a relationship between them. Mainly I would like to find does $X_1$ affect $X_2$. That is, if $X_1$ increases what happen to $X_2$. I fit copula and that the relationship between them is Gaussian. So, I need to say what will happen to $X_2$ if $X_1$ increases. – Maryam Jun 15 '20 at 17:00
• @Maryam It seems like you have all the tools you need, what in particular you still stuck on? – doubled Jun 15 '20 at 19:04
• @Maryam Not sure what "time" really means here, but if $(X_1,X_2)$ have a corresponding gaussian copula, then you can say that $X_1,X_2$ are related, but that the relationship decreases as you move to the limits, and at the limit, they are unrelated. – doubled Jun 18 '20 at 2:33