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# MullikenPopulation¶

class MullikenPopulation(configuration)

Class for calculating the Mulliken population for a configuration.

Parameters: configuration (DeviceConfiguration | BulkConfiguration | MoleculeConfiguration.) – The Configuration which the Mulliken population should be calculated for.
atomicAngles()
Returns: The atomic polarization angles as a list of ($$\theta$$, $$\phi$$) tuples representing spherical coordinates. list of tuple
atomicCharge()

Calculate the charge of each atom.

Returns: The charge of each atom in a length n array. PhysicalQuantity of type charge
atoms(spin=None)

Return an array with the Mulliken population of each atom.

Parameters: spin (Spin.All | Spin.Sum | Spin.X | Spin.Y | Spin.Z | Spin.Up | Spin.Down | Spin.RealUpDown | Spin.ImagUpDown) – The mulliken population is calculated for this spin. Default: Spin.All
bond(atom_i, atom_j)

Return a float which is the Mulliken population projected onto the bond between atom_i and atom_j.

Parameters: atom_i (int) – Index of the first atom in the bond projection. atom_i should satisfy 0 <= atom_i < number_of_atoms. atom_j (int) – Index of the second atom in the bond projection. atom_j should satisfy 0 <= atom_j < number_of_atoms.
metatext()
Returns: The metatext of the object or None if no metatext is present. str | unicode | None
nlprint(stream=None)

Print a string containing an ASCII table useful for plotting the AnalysisSpin object.

Parameters: stream (python stream) – The stream the table should be written to. Default: NLPrintLogger()
orbitals(spin=None)

Calculate the Mulliken population of all orbitals.

Parameters: spin (Spin.All | Spin.Sum | Spin.X | Spin.Y | Spin.Z | Spin.Up | Spin.Down | Spin.RealUpDown | Spin.ImagUpDown) – The mulliken population is calculated for this spin. Default: Spin.All One array for each atom. Each array holds the Mulliken population of each orbital of that atom. list of numpy.array
setMetatext(metatext)

Set a given metatext string on the object.

Parameters: metatext (str | unicode | None) – The metatext string that should be set. A value of “None” can be given to remove the current metatext.

## Usage Examples¶

Calculate the Mulliken population of an ammonia molecule and print projections of orbitals and bonds:

# Set up configuration
molecule_configuration = MoleculeConfiguration(
elements=[Nitrogen, Hydrogen, Hydrogen, Hydrogen],
cartesian_coordinates=[[ 0.,       0.,        0.124001],
[ 0.,       0.941173, -0.289336],
[ 0.81508, -0.470587, -0.289336],
[-0.81508, -0.470587, -0.289336]] * Angstrom)

# Define the calculator
calculator = HuckelCalculator()
molecule_configuration.setCalculator(calculator)

# Calculate and save the mulliken population
mulliken_population = MullikenPopulation(
configuration=molecule_configuration)

nlsave('mulliken.hdf5', mulliken_population)

# print all occupations
nlprint(mulliken_population)

# print partial occupations of N
print('N ', mulliken_population.atoms(spin=Spin.Sum)[0])
print(' | ', mulliken_population.orbitals(spin=Spin.Sum)[0])
print('')

# print partial occupations of first H
print('H ', mulliken_population.atoms(spin=Spin.Sum)[1])
print(' | ', mulliken_population.orbitals(spin=Spin.Sum)[1])
print('')

# print Mulliken population of N-H bond
print('N-H bond ', mulliken_population.bond(0, 1))

mulliken_nh3.py

## Notes¶

The total number of electrons, $$N$$, is given by

$N = \sum_{ij} D_{ij} S_{ij},$

where $$D$$ is the density matrix, $$S$$ the overlap matrix, and the sum is over all orbitals in the system.

The Mulliken population is defined by different partitions of this sum into orbitals, atoms and bonds.

• Mulliken population of orbitals

The Mulliken Population of orbitals, $$M_i$$ , is defined by restricting one of the sum indexes to the orbital, i.e.

$M_i = \sum_{j} D_{ij} S_{ji}.$
• Mulliken population of atoms

The Mulliken Population of atoms, $$M_{\mu}$$ , is defined by adding up all the orbital contributions on atom number $$\mu$$

$M_\mu = \sum_{i \in \mu} \sum_{j} D_{ij} S_{ji}.$
• Mulliken population of bonds

The Mulliken Population of bonds, $$M_{\mu \nu}$$ , is defined by restricting the sum indexes to the orbitals on atom number $$\mu$$ and $$\nu$$ , i.e.

$M_{\mu \nu} = (2-\delta_{\mu \nu}) \sum_{i \in \mu} \sum_{j \in \nu} D_{ij} S_{ji}.$

Note the factor two, which ensures that

$N = \sum_{\mu \ge \nu} M_{\mu \nu}.$

### Noncollinear spin¶

For noncollinear systems, the Mulliken population of atoms is a four component spin tensor:

$M_\mu^{\sigma \sigma'} = \sum_{i \in \mu} \sum_{j} D_{ij}^{\sigma \sigma'} S_{ji}.$

The Mulliken Population of atoms can be diagonalized to give a local spin direction. This direction is reported by the nlprint command for noncollinear systems.

The nlprint report for noncollinear spin also shows the orbital Mulliken populations in the local spin direction.


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Here's the question you clicked on:

## Moneeka96 Group Title Pleaseeee help me.. What is the electric force between a +2 µC point charge and a –2 µC point charge if they are separated by a distance of 5.0 cm? (µC = 1.0 × 10–6 C) one year ago one year ago

• This Question is Closed
1. electrokid Group Title

$\Huge \color{Lime}{~\:\:\:\:\:\:\:\:\:\:\:\:\frak Welcome\:\:\frak To\:\:\frak Open\frak Study}$

2. Yahoo! Group Title

F = Kq1q2/r^2

3. electrokid Group Title

Coulomb's Law: $F=K{q_1q_2\over d^2}$ $K = const.\\ q_1 = 2\times10^{-6}C\\ q_2= 2\times10^{-6}C\\ d=5{\rm cm}\times {1{\rm m}\over 100{\rm cm}}=0.05{\rm m}$

4. electrokid Group Title

the important thing is to have all the units in SI system

5. Moneeka96 Group Title

So i had to put cm into m like i believe you did above?

6. electrokid Group Title

yep

7. Moneeka96 Group Title

now from here where do I go?

8. Moneeka96 Group Title

wouldnt q2 be negative?

9. electrokid Group Title

we plug em in the equation for the force. $F=8.99\times10^{9}\times \frac{2\times10^{-6}\times2\times10^{-6}}{(0.05)^2}$

10. electrokid Group Title

the negative sign can be neglected since we are only looking for the magnitude (i.e., the amount) of force. since the two charges are oppositely charged, the above force will be a force of attraction. this is the information we get from the charges

11. Moneeka96 Group Title

Okay cool so then Id just plug the above into my calculator and thats the answer. right?

format: plain


text:
Sachin chakote

does anyone have an idea about DACL on windows file system
i used "GetNamedSecurityInfo" for getting the dacl info but every time i get null value for sidOwner , sidGroup, DACL.
its shows some value for the security descriptor but when i try to get DACL info by
GetSecurityDescriptorDacl() it shows NULL DACL value.

even for files shared manually with user defined permission shows same result.

I dont have my machine over DOMAIN.

### Re: Visual C++ General DACL for Windows XP File System

Simple Samples

Read the documentation of ACLs and such.


text:
## Image of the optimized structures, representations of the frontier MOs, computed positions of the electronic transitions, oscillator strength (f), major contributions and the calculated absorption spectra for all molecules are given from Theoretical investigation of the structural and spectroscopic properties of expanded metalloporphyrin complexes

2019-01-09T08:47:09Z (GMT) by
The frontier molecular orbitals, UV–Vis absorption spectra, charge transfer (CT) and triplet excited states of 12 expanded D–A porphyrin/benzoporphyrin complexes were investigated using the density functional theory (DFT) method and time-dependent DFT in this work. The results showed that thiophene was an effective fragment for absorption of ‘long wavelength’, while the benzoporphyrin worked on the ‘short wavelength’, which was derived from its saddle-shaped structure; this expanded D–A conjugated system had a mild CT process with anthraquinone/isoindigo as acceptors and a strong CT process with naphtoquinone as acceptor. In addition, based on the simulation of the triplet state, the theoretical phosphorescence wavelength range of this series of derivatives was between 1000 and 1200 nm. This study is expected to assist the design of conjugated porphyrin for the field of porphyrin chemistry.


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# Optical Transceiver using LCML

#1

In this example, we will demonstrate how to use the compact models in the CML to build an optical transceiver.

## Associated files

lcml_optical_transceiver.icp (1.5 MB)

Download and open the file lcml_optical_transceiver.icp. The optical transceiver circuit in lcml_optical_transceiver.icp is composed of the following elements:

The optical transceiver is measured by an Eye Diagram analyzer and Oscilloscopes. The following figures show the eye diagram and time domain waveform measured from the circuit.

Circuit design example using CMLs


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## Friday, February 12, 2010

### Warming by 20 °C in Vancouver

I was checking whether January 2010 was the warmest ever January in Vancouver, as you can read in some news outlets. Using the WeatherData function in Mathematica, the answer is No. For example, the average temperature on the airport was 7.8 °C in January 1994 but only 7.63 °C in January 2010. So as far as I can say, the media are not telling us the truth.

But I was intrigued by a supercool January 2003 which was apparently 20 °C cooler than most other Januaries. I immediately decided it had to be a bug. The WeatherData bug is shared by Wolfram Alpha. If you ask Wolfram Alpha about the temperature in Vancouver in February 2003, you get the following cute graph:

The temperature suddenly jumped by 20 °C. Similarly, you will find a 20 °C drop in December 2002. It's really not hard to become sure that the discontinuities are errors. Just look at Bellingham, WA, which is just 76 km away. You won't find any discontinuities in February 2003.

My cherished contact at Wolfram wrote me that I should trust the temperatures - they really jumped by 20 °C in a few hours, probably because of a "major warm front" - because the data were taken directly from the weather station and because some nearby stations show the same behavior. No kidding. ;-) I hope that he has gotten sane since that time but I still have no proof of it.

Of course, it wasn't hard to find out what was the cause of the mistake - which may be repeated at many other places of the WeatherData database. (This error could be easily looked for systematically.)

It was a simple Fahrenheit-Celsius confusion. The actual temperature was around +5 °C. It was incorrectly interpreted as +5 °F, which is around -15 °C, which is what you see in the bogus graph. Also, the fluctuations during the "ice age" are suppressed by a factor of two - well, 1.8, as you know very well.

So the Al Gore Rhythm producing the error is known. What is not known yet is who is actually responsible for this error.

Robbert Dijkgraaf vs Henk Tennekes

As Roger Pielke Sr analyzes, climate skeptic Henk Tennekes resigned from the Dutch Academy of Sciences. The current president of the Academy, string theorist Robbert Dijkgraaf, had to deal with this case, and he did so diplomatically and bureaucratically - without having said anything nontrivial.

My impressions are mixed.

Henk Tennekes is certainly bothered by many of the same things as your humble correspondent. However, a big portion of his complaints is the fact that physics is treated as the umbrella of all natural sciences. Well, it simply is. That's how the world - and causal relationships between various disciplines - work. Physics is the most fundamental natural science and others are its special examples and approximations for more specific situations.


text:
# Replanting forests:has no effect on the water supplyhelps put water into underground reservoirs

###### Question:

Replanting forests: has no effect on the water supply
helps put water into underground reservoirs

### HURRY PLEASE!!what is the rate of change for (-1,-6) and (-3,-2) ?

HURRY PLEASE!! what is the rate of change for (-1,-6) and (-3,-2) ?...

### PLEASE HELP ASAP I'M DOING AN EXAM!!A majority of the land in the Middle East us unsuitable for crops because A. Ocean water cannot be converted to

PLEASE HELP ASAP I'M DOING AN EXAM!! A majority of the land in the Middle East us unsuitable for crops because A. Ocean water cannot be converted to freshwater B. Freshwater sources are scarce in the region C. People cannot live in these desert areas D. Mountains block water from coming inland. ...

### Complete a frase abaixo com o Simple Past. We David in town a few days ago. (see) * Did sawed Was saw Seen Saw Did see Seen

Complete a frase abaixo com o Simple Past. We David in town a few days ago. (see) * Did sawed Was saw Seen Saw Did see Seen...

### Look at the map of europe after world war ii. how did poland change after the war

Look at the map of europe after world war ii. how did poland change after the war...

### Make the following statement negative: Nosotros siempre vamos al café.Nosotros nunca vamos al café.Nosotros vamos al café nunca.Nosotros vamos

Make the following statement negative: Nosotros siempre vamos al café. Nosotros nunca vamos al café. Nosotros vamos al café nunca. Nosotros vamos nunca al café. Nunca nosotros vamos al café....

### All of the following cases involve due process EXCEPT...O Gideon v. WainwrightO McCulloch v. MarylandO Miranda v. ArizonaO New

All of the following cases involve due process EXCEPT... O Gideon v. Wainwright O McCulloch v. Maryland O Miranda v. Arizona O New Jersey v. T. L.O....

### Choose the correct figure of speech. the lining of john's new jacket felt like a kitten's fur. 1.simile

Choose the correct figure of speech. the lining of john's new jacket felt like a kitten's fur. 1.simile 2.metaphor 3.personification...

### What is the best interpretation of the political cartoonA. Southern states welcomed the passage of the Reconstruction Amendments.B. Many white voters

What is the best interpretation of the political cartoon A. Southern states welcomed the passage of the Reconstruction Amendments. B. Many white voters who could not pass a literacy test were still able to vote in the South. C. Voters of all races who could not read could pay a pool tax to avoid ta...

Tyler’s brother works in a shoe store. a. Tyler’s brother earns a commission. He makes 2.5% of the amount he sells. Last week, he sold $900 worth of shoes. How much was his commission? b. The store bought a pair of shoes for$50, and sold it for $80. What percentage was the markup? c. Tyler’... 4 answers ### About 5000 tigers live in captivity in the united states. about 10% of these tigers live in reputable About 5000 tigers live in captivity in the united states. about 10% of these tigers live in reputable zoos. around how many of these tigers do not live in reputable zoos?... 4 answers ### What pros and cons would you weigh if you were a sailor trying to decide whether to sign on with da what pros and cons would you weigh if you were a sailor trying to decide whether to sign on with da gama, columbus, or magellan?... 4 answers ### How do interest groups use their bias to try and shape public opinion How do interest groups use their bias to try and shape public opinion... 1 answer ### List one example that uses the different subsets of real numbers in real-lifesituations. An example would be using rational numbers when dividing a List one example that uses the different subsets of real numbers in real-life situations. An example would be using rational numbers when dividing a pie into six equal pieces. Each piece is of the whole pie. Where would you use counting numbers? Whole numbers? Integers?... 4 answers ### Noah solved an equation as shown below and found that the equation has infinitely many solutions. Noah solved an equation as shown below and found that the equation has infinitely many solutions.... 3 answers ### Jack's big shop had a huge back to college special for$1000 you get even desk chairs, five tables and

Jack's big shop had a huge back to college special for \$1000 you get even desk chairs, five tables and two laptops. write an equation to determine the number of items jack's big shop sells...

### Where do costa ricans and panamanian bioh go to see cultural performances

Where do costa ricans and panamanian bioh go to see cultural performances...

### Create a query calculating the profit margin of bike models (use Production. ProductCategory. Name to determine if the product is a bike.)

Create a query calculating the profit margin of bike models (use Production. ProductCategory. Name to determine if the product is a bike.) Profit margin is the difference between ListPrice and StandardCost divided by the ListPrice. Only consider bike models currently sold (Use DiscontinuedDate in Pr...


text:
Kategoriler

arrow_right. Computer Versioning Instead, QDA assumes that each class has its own covariance matrix. Quadratic discriminant analysis is attractive if the Quadratic Discriminant Analysis. Course Material: Walmart Challenge . Javascript Number Quadratic Discriminant Analysis. Data Concurrency, Data Science Lexical Parser Unlike LDA however, in QDA there is no assumption that the covariance of each of the classes is identical. 33 Comparison of LDA and QDA boundaries ¶ The assumption that the inputs of every class have the same covariance $$\mathbf{\Sigma}$$ can be … Linear Algebra Graph If we assume data comes from multivariate Gaussian distribution, i.e. This quadratic discriminant function is very much like the linear discriminant function except that because Σ k, the covariance matrix, is not identical, you cannot throw away the quadratic terms. This tutorial explains Linear Discriminant Analysis (LDA) and Quadratic Discriminant Analysis (QDA) as two fundamental classification methods in statistical and probabilistic learning. The classification rule is similar as well. Like, LDA, it seeks to estimate some coefficients, plug those coefficients into an equation as means of making predictions. A distribution-based Bayesian classiﬁer is derived using information geometry. The assumption of groups with matrices having equal covariance is not present in Quadratic Discriminant Analysis. Quadratic discriminant analysis performed exactly as in linear discriminant analysis except that we use the following functions based on the covariance matrices for each category: This quadratic discriminant function is very much like the linear discriminant function except that because Σ k, the covariance matrix, is not identical, you cannot throw away the quadratic terms. Because, with QDA, you will have a separate covariance matrix for every class. This discriminant function is a quadratic function and will contain second order terms. 217. close. $$\hat{\mu}_0=(-0.4038, -0.1937)^T, \hat{\mu}_1=(0.7533, 0.3613)^T$$, \(\hat{\Sigma_0}= \begin{pmatrix} You just find the class k which maximizes the quadratic discriminant function. We can also use the Discriminant Analysis data analysis tool for Example 1 of Quadratic Discriminant Analysis, where quadratic discriminant analysis is employed. The model fits a Gaussian density to each class. number of variables is small. Order Quadratic discriminant analysis (QDA) is a variant of LDA that allows for non-linear separation of data. Quadratic discriminant analysis uses a different covariance matrix for each class. Tree


text:
American Institute of Mathematical Sciences

eISSN:
2163-2480

All Issues

Evolution Equations & Control Theory

June 2017 , Volume 6 , Issue 2

Select all articles

Export/Reference:

2017, 6(2): 155-175 doi: 10.3934/eect.2017009 +[Abstract](3174) +[HTML](64) +[PDF](510.6KB)
Abstract:

In this paper we study the one-dimensional logarithmic Schrödin-\break ger equation perturbed by an attractive \begin{document}$δ^{\prime}$\end{document}-interaction

where $γ>0$. We establish the existence and uniqueness of the solutions of the associated Cauchy problem in a suitable functional framework. In the attractive $δ^{\prime}$-interaction case, the set of the ground state is completely determined. More precisely: if $0 < γ≤ 2$, then there is a single ground state and it is an odd function; if $γ>2$, then there exist two non-symmetric ground states. Finally, we show that the ground states are orbitally stable via a variational approach.

2017, 6(2): 177-186 doi: 10.3934/eect.2017010 +[Abstract](2591) +[HTML](58) +[PDF](356.9KB)
Abstract:

We investigate the long time behavior of solutions to the differential equation:

where \begin{document}$c$\end{document} is nonnegative constant, \begin{document} $α∈\lbrack0,1[,Φ \ \ {\rm{is \ \ a}}\ \ C^{1}$\end{document} convex function defined on a Hilbert space \begin{document}$\mathcal{H}$\end{document} and \begin{document} $g∈ L^{1}(0,+∞;\mathcal{H}).$\end{document} We obtain sufficient conditions on the source term \begin{document}$g(t)$\end{document} ensuring the weak or the strong convergence of any trajectory \begin{document} $x(t)$\end{document} of (1) as \begin{document}$t\to ∞$\end{document} to a minimizer of the function \begin{document} $Φ$\end{document} if one exists.

2017, 6(2): 187-217 doi: 10.3934/eect.2017011 +[Abstract](2390) +[HTML](52) +[PDF](633.1KB)
Abstract:

We study an optimal control problem for certain evolution equations in two component composites with \begin{document}$\varepsilon$\end{document}-periodic disconnected inclusions of size \begin{document}$\varepsilon$\end{document} in presence of a jump of the solution on the interface that varies according to a parameter \begin{document}$γ$\end{document}. In particular the case \begin{document}$γ=1$\end{document} is examinated.

2017, 6(2): 219-237 doi: 10.3934/eect.2017012 +[Abstract](3120) +[HTML](69) +[PDF](368.5KB)
Abstract:

In this paper, we discuss the existence of the periodic solutions of a Cahn-Hillard/Allen-Cahn equation which is introduced as a simplification of multiple microscopic mechanisms model in cluster interface evolution. Based on the Schauder type a priori estimates, which here will be obtained by means of a modified Campanato space, we prove the existence of time-periodic solutions in two space dimensions. The uniqueness of solutions is also discussed.

2017, 6(2): 239-260 doi: 10.3934/eect.2017013 +[Abstract](3720) +[HTML](78) +[PDF](447.2KB)
Abstract:

In this paper, we consider a viscoelastic Kirchhoff equation with Balakrishnan-Taylor damping, dynamic boundary conditions and a time-varying delay term acting on the boundary. By using the Faedo-Galerkin approximation method, we first prove the well-posedness of the solutions. By introducing suitable energy and perturbed Lyapunov functionals, we then prove the general decay results, from which the usual exponential and polynomial decay rates are only special cases. To achieve these results, we consider the following two cases according to the coefficient α of the strong damping term: for the presence of the strong damping term (α>0), we use the strong damping term to control the time-varying delay term, under a restriction of the size between the time-varying delay term and the strong damping term; for the absence of the strong damping term (α=0), we use the viscoelasticity term to control the time-varying delay term, under a restriction of the size between the time-varying delay term and the kernel function.

2017, 6(2): 261-276 doi: 10.3934/eect.2017014 +[Abstract](3014) +[HTML](51) +[PDF](410.7KB)
Abstract:

In this paper we consider a viscoelastic plate equation with a nonlinear weakly dissipative feedback localized on a part of the boundary. Without imposing restrictive assumptions on the boundary frictional damping, we establish an explicit and general decay rate result that allows a wider class of relaxation functions and generalizes previous results existing in the literature.

2017, 6(2): 277-297 doi: 10.3934/eect.2017015 +[Abstract](2679) +[HTML](51) +[PDF](422.2KB)
Abstract:

We consider evolution inclusions driven by a time dependent subdifferential plus a multivalued perturbation. We look for periodic solutions. We prove existence results for the convex problem (convex valued perturbation), for the nonconvex problem (nonconvex valued perturbation) and for extremal trajectories (solutions passing from the extreme points of the multivalued perturbation). We also prove a strong relaxation theorem showing that each solution of the convex problem can be approximated in the supremum norm by extremal solutions. Finally we present some examples illustrating these results.

2017, 6(2): 299-318 doi: 10.3934/eect.2017016 +[Abstract](2445) +[HTML](55) +[PDF](438.0KB)
Abstract:

A 3D-2D dimension reduction for a nonhomogeneous constrained energy is performed in the realm of $Γ$-convergence, and two-scale convergence for slender domains, providing an integral representation for the limit functional. Applications to supremal functionals are also given.

2020 Impact Factor: 1.081
5 Year Impact Factor: 1.269
2020 CiteScore: 1.6


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1. ## Disjoint Cycles

P=
(1234567
3125746)

q=
(1234567
5641723)

r=
(1234567
7631425)

s=
(1234567
5713264)

Find pq(^-1)sr(^2) write the answer as a product of disjoint cylcles. Find its order Parity and Inverse.

I understand parts of the question but for the majority I am unable to do it.

2. ## Re: Disjoint Cycles

Which parts, specifically, are you unable to do?

3. ## Re: Disjoint Cycles

(132)(4576)*(15734)(26)
(1
q sends 1 to 5 and p sends 5 to 7
(17- because p.q(1) = p(5) = 7
so the composite permutation begins (17...)
then see what p.q(7) is and so on
q sends 7 to 3 and p sends 3 to 2
(172
q sends 2 to 6 and p sends 6 to 4
(1724
q sends 4 to 1 and p sends 1 to 3
(17243
q sends 3 to 4 and p sends 4 to 5
(172435
q sends 5 to 7 and p sends 7 to 6
(1724356
just to check, q sends 6 to 2 and p sends 2 to 1
(1724356)-7 digits so completed as pq

To find sr we do the same
Sr
1
R takes 1 to 7 and s takes 7 to 4
14
R takes 4 to 1 and s takes 1 to 5
145
R takes 5 to 4 and s takes 4 to 3
1453
R takes 3 to 3 and s takes 3 to 1
14531

this is as far as ive got

4. ## Re: Disjoint Cycles

ok, well, i do not know if for you:

pq means first do q, and then do p, or vice-versa (different authors use different conventions). i will assume pq(x) = p(q(x)) (do q first).

so first we have to calculate pq-1sr2. this is tedious.

first we do r twice:

1→7→5
2→6→2
3→3→3
4→1→7
5→4→1
6→2→6
7→5→4

then we apply s:

1→5→2
2→2→7
3→3→1
4→7→4
5→1→5
6→6→6
7→4→3

then we apply q-1, which is:

1→4
2→6
3→7
4→3
5→1
6→2
7→5, to get:

1→2→6
2→7→5
3→1→4
4→4→3
5→5→1
6→6→2
7→3→7

finally, we apply p to get:

1→6→4
2→5→7
3→4→5
4→3→2
5→1→3
6→2→1
7→7→6, so that pq-1sr2 = (1 4 2 7 6)(3 5)

this is the product of a disjoint 5-cycle and 2-cycle, so has order lcm(2,5) = 10.

alternatively, we have pq-1sr2 = (1 3 2)(4 5 7 6)(1 4 3 7 5)(2 6)(1 5 2 7 4 3)(1 7 5 4)(2 6)(1 7 5 4)(2 6).

thus pq-1sr2(1) = p(q-1(s(r(r(1)))))

= p(q-1(s(r(7)))) = p(q-1(s(5))) = p(q-1(2)) = p(6) = 4 so we start:

(1 4......

then we calculate pq-1sr2(4) = p(q-1(s(r(r(4)))))

= p(q-1(s(r(1)))) = p(q-1(s(7))) = p(q-1(4)) = p(3) = 2, so we continue:

(1 4 2.....

next, we want to know pq-1sr2(2) = p(q-1(s(r(r(2)))))

= p(q-1(s(r(6)))) = p(q-1(s(2))) = p(q-1(7)) = p(5) = 7, so we are up to:

(1 4 2 7....

AND.....pq-1sr2(7) = p(q-1(s(r(r(7)))))

= p(q-1(s(r(5)))) = p(q-1(s(4))) = p(q-1(3)) = p(7) = 6, so we have:

(1 4 2 7 6.....

and quickly, now! pq-1sr2(6) = p(q-1(s(r(r(6)))))

= p(q-1(s(r(2)))) = p(q-1(s(6))) = p(q-1(6)) = p(2) = 1, so our first cycle closes:

(1 4 2 7 6)(....

the next number not in the first cycle is 3, so we calculate:

pq-1sr2(3) = p(q-1(s(r(r(3)))))

= p(q-1(s(r(3)))) = p(q-1(s(3))) = p(q-1(1)) = p(4) = 5, and by bijectivity,

we must have pq-1sr2(5) = 3, so we get (1 4 2 7 6)(3 5), as we did above.

5-cycles are even, and (3 5) is a transposition, therefore pq-1sr2 is odd.


text:
# Grating projections FDTD

#1

Dear Sir/Mrs,

Concerning grating projections.
1- for this command
?G; # grating results
?G*T; # normalized to source power

when we normalize the transmitted power from certain grating order why do we multiply by the T not dividing by T?

#2

G=grating(mname);   # power to each order (fraction of transmitted power)
T=transmission(mname); # total power transmitted through monitor (fraction of source power)

G returns the fraction of power in each order assuming transmission is 1. This result might be multiplied by transmission if we want to obtain actual power in each order.

Hope this was clear.

#3

yes it is.
Thank you.


text:
## Introductory Statistics 9th Edition

The mean of the sampling distribution is equal to the mean of the population. The standard deviation of the sampling distribution of $\bar{x}\$ is equal to the standard deviation of the population divided by the square root of the sample size.
The mean of the sampling distribution is equal to the mean of the population. The standard deviation of the sampling distribution of $\bar{x}\$ is equal to the standard deviation of the population divided by the square root of the sample size.


text:
I have an assignment that has me reading in a file and creating a token for each string and special characters in it (which isn't the part I need help with). I currently have part of the program working 98% but am getting an error when I execute the program. I've looked around and couldn't solve it the way I need it. Anyway here is the error I'm getting:

"Debug Assertion Failed!

Expression: string subscript out of range "

Now from what I've gathered it has to do with the size of the string & the container size not matching up.

There was an old thread that gave a solution which partially worked but it was skipping the last word of my line (which could be due to the way I currently have the program.

Anyway here is what the line of txt from the file reads: main is awesome

Here is my code at the current moment (which displays all the text as tokens but generates the error

#include <iostream>
#include <fstream>
#include <string>

int main()
{
std::ifstream source;  //create variable for input file

std::string myToken;   //create string variable to hold data from file
std::string txtToken;  //variable to house each token as it's created
bool done = false;  //variable to end do/while once finished
int k = 0;   //variable to keep program moving through file grabbing valid characters

source.open("file\\input.txt");  //open file

if(!source.is_open()) //check to see if file exists if not proceed with statement within {}
{
std::cout << "File not found, or cannot be opened!"<<std::endl;  //generate error message for missing file
return 0;  //end program
}

getline(source,myToken);  //read file into memory and store in string variable myToken for seperation

do
{

char s = myToken[k++]; //each character is stored within "s"

if( (s>='a' && s<='z') || (s>='A' && s<='Z') )  //checks if found character is valid or invalid
{
txtToken = txtToken + s; //if character is valid it is stored within txtToken
}
else
{
std::cout << "Token is: " << txtToken << std::endl; //txtToken is now displayed to screen
txtToken = ""; //clear txtToken to input next string set
}

}while (!done); //finish do/while

source.close(); //close file

return 0; //end program
}

I also found that if I remove the myToken[k++] and make it myToken[k] the error goes away but I get no text.

4
Contributors
7
Replies
8
Views
7 Years
Discussion Span
Last Post by RayvenHawk

Instead of [k++] why don't you try incrementing k outside of the [] so you know for sure when it is happening. Then output k and compare it to yourString.size() and make sure everything seems kosher. This error definitely means you are accessing the 5th character in a string of length 4, or something equivalent.

Dave

You never set done to True. Your do {} while(!done) is a nice infinite loop. Of course it runs out of the string limits.

david - I tried placing the k++ outside the [] before and it still ran into that error. I ran the debug and step-over to watch what value "s" takes and it goes through like it's suppose to but no mention of the error when I do it that way.

Nez - Hmm good point. I'll load up the project and add that line and see what happens.

I'll post back what I found.

Change your do...while loop to a while loop and use k as your tracker for your exit condition.

while( k != myToken.size())
{
char s = myToken[k];  // no ++ here
// rest of your code here
k++
}

The reason for this is that you are running to infinity with your current loop and as soon as k is larger then myToken.size() you get a failed assert. Having the loop as I presented will ensure that you do not go beyond the range of the string.

Nez & Nathan - Ok I did both suggestions (not together but on their own) along with a few other tweaks and what is strange is while my original error is gone (which is good), it seems to avoid the last word for some reason.

What I got with the error is this output to the screen:

Token is: main
Token is: is
Token is: awesome

Now doing it either of your suggestions the error is gone but now I get this output:

Token is: main
Token is: is

it skips the last word for whatever reason.

Could you post what you have now?

Here is what I currently have

getline(source,myToken);
while(k != myToken.size())
{
//getline(source,myToken);  //read file into memory and store in string variable myToken for seperation

char s = myToken[k]; //each character is stored within "s"
k++;

if( (s>='a' && s<='z') || (s>='A' && s<='Z') )  //checks if found character is valid or invalid
{
txtToken = txtToken + s; //if character is valid it is stored within txtToken
}
else
{
std::cout << "Token is: " << txtToken << std::endl; //txtToken is now displayed to screen
txtToken = ""; //clear txtToken to input next string set
}

}
source.close(); //close file

return 0; //end program
}

Now if I put k++ prior to the char s = myToken[k], I get everything from "a" on meaning:

Token is: ain
Token is: is
Token is: awesome

I know the reason it skips m is because it jumps to the next spot before it even puts the value in txtToken. I've placed k++ at the end of the while loop but it gives me the error I reported last.

Now another weird thing is as it is currently if I change what the contents are of my input.txt I get no display. For example if I put "main isawesome" (excluding the "") all I get is main nothing else.

Also if I change the line all together to anything other than main I get nothing displayed.

This topic has been dead for over six months. Start a new discussion instead.
Have something to contribute to this discussion? Please be thoughtful, detailed and courteous, and be sure to adhere to our posting rules.


text:
# Math Help - Determining whether a sequence of partial sums is convergent or divergent

1. ## Determining whether a sequence of partial sums is convergent or divergent

Given the following sequence of partial sums:
$S_N=\frac{1}{N}cos(N\pi)$

My attempt:
If I can find the series for this sequence of partial sums, then I can test the series to see if it's convergent or divergent.

$a_N=S_N-S_{N-1}=\sum^N_{n=1}a_n-\sum^{N-1}_{n=1}a_n$

Thus:
$a_N=\frac{1}{N}cos(N\pi)-\frac{1}{N-1}cos((N-1)\pi)$

Since $cos(N\pi)=(-1)^N$:

$a_N=\frac{1}{N}(-1)^N-\frac{1}{N-1}(-1)^{N-1}$

But this is as far as I got.
I would like to express this with a factor $(-1)^N$ so that I can use the alternating series test, but I'm not sure how exactly.

$a_N=(-1)^N\cdot \left ( \frac{1}{N}-\frac{1}{N-1}\right )$
Would this be correct?

2. ## Re: Determining whether a sequence of partial sums is convergent or divergent

Originally Posted by MathIsOhSoHard
Given the following sequence of partial sums:
$S_N=\frac{1}{N}cos(N\pi)$
If the real question is contained in the title of this thread, then you have wasted a lot effort.
Because the partial sums are given as $S_N=\frac{1}{N}cos(N\pi)$.
You know $(S_N)\to 0$ so the series converges to zero.

3. ## Re: Determining whether a sequence of partial sums is convergent or divergent

This is the question in its entirety:

For a series $\sum^{/infty}_{n=1}a_n$ the sequence of partial sums is $S_N=\frac{1}{N}cos(N\pi)$. Determine whether the series is absolutely convergent, conditionally convergent, or divergent.
Hint: Find the series n'th term $a_n$ and express it as an alternating series.

(It's supposed to be an infinity sign above the sum, but for some reason LATEX doesn't want to display it)

4. ## Re: Determining whether a sequence of partial sums is convergent or divergent

Originally Posted by MathIsOhSoHard
This is the question in its entirety:
For a series $\sum^{/infty}_{n=1}a_n$ the sequence of partial sums is $S_N=\frac{1}{N}cos(N\pi)$. Determine whether the series is absolutely convergent, conditionally convergent, or divergent.
Hint: Find the series n'th term $a_n$ and express it as an alternating series.
Well the only difficult part is "Determine whether the series is absolutely convergent,"
For that reason you may want to find $a_n$.
$a_N=S_{N}-S_{N-1}=(-1)^N\left[\frac{2N-1}{N(N-1)}\right]$.

$$\sum\limits_{n = 0}^\infty {ar^n }$$ gives $\sum\limits_{n = 0}^\infty {ar^n }$

5. ## Re: Determining whether a sequence of partial sums is convergent or divergent

Originally Posted by Plato
Well the only difficult part is "Determine whether the series is absolutely convergent,"
For that reason you may want to find $a_n$.
$a_N=S_{N}-S_{N-1}=(-1)^N\left[\frac{2N-1}{N(N-1)}\right]$.

$$\sum\limits_{n = 0}^\infty {ar^n }$$ gives $\sum\limits_{n = 0}^\infty {ar^n }$
Are you sure $a_n$ can't be written as $a_N=(-1)^N\cdot \left ( \frac{1}{N}-\frac{1}{N-1}\right )$

6. ## Re: Determining whether a sequence of partial sums is convergent or divergent

Originally Posted by MathIsOhSoHard
Are you sure $a_n$ can't be written as $a_N=(-1)^N\cdot \left ( \frac{1}{N}-\frac{1}{N-1}\right )$
YES

It is $a_N=(-1)^N\cdot \left ( \frac{1}{N}+\frac{1}{N-1}\right )$

7. ## Re: Determining whether a sequence of partial sums is convergent or divergent

Originally Posted by Plato
YES

It is $a_N=(-1)^N\cdot \left ( \frac{1}{N}+\frac{1}{N-1}\right )$

How do you rewrite $(-1)^{N-1}$ into $(-1)^N$?

8. ## Re: Determining whether a sequence of partial sums is convergent or divergent

Originally Posted by mathisohsohard
how do you rewrite $(-1)^{n-1}$ into $(-1)^n$?
$(-1)^{n-1}=(-1)^n(-1)^{-1}=-(-1)^n$

You see $(-1)^{-1}=\frac{1}{-1}=-1$

9. ## Re: Determining whether a sequence of partial sums is convergent or divergent

Great! Now I get it

So by using the alternating series test, it can be shown that the series is convergent.

And to test for absolute convergence, using the comparison test:
$\left | a_n \right | \le \left | b_n \right |$
$\sum^\infty_{n=1}\frac{1}{n}\le \sum^\infty_{n=1}(1)^n \left ( \frac{1}{n}+\frac{1}{n-1} \right )$

Since $\sum^\infty_{n=1}\frac{1}{n}$ is divergent, then the absolute of $b_n$ is divergent too, thus it is not absolutely convergent.

So conclusion would be that it's conditionally convergent?
Would the absolute value of the fraction be $\frac{1}{n-1}$ or $\frac{1}{n+1}$? I wasn't quite sure about that.

10. ## Re: Determining whether a sequence of partial sums is convergent or divergent

Originally Posted by MathIsOhSoHard
Great! Now I get it

So by using the alternating series test, it can be shown that the series is convergent.

And to test for absolute convergence, using the comparison test:
$\left | a_n \right | \le \left | b_n \right |$
$\sum^\infty_{n=1}\frac{1}{n}\le \sum^\infty_{n=1}(1)^n \left ( \frac{1}{n}+\frac{1}{n-1} \right )$

Since $\sum^\infty_{n=1}\frac{1}{n}$ is divergent, then the absolute of $b_n$ is divergent too, thus it is not absolutely convergent.

So conclusion would be that it's conditionally convergent?
Would the absolute value of the fraction be $\frac{1}{n-1}$ or $\frac{1}{n+1}$? I wasn't quite sure about that.
Very good.


text:
# Changeset 1790 for Deliverables

Ignore:
Timestamp:
Feb 28, 2012, 4:35:26 PM (8 years ago)
Message:

edited the report so that all the spelling is consistent (british english, naturally), fixed italianisms, typos, etc. etc.

File:
1 edited

### Legend:

Unmodified


text:
# Check Out Parity

Calculus Level 3

For any real valued function $$f: [-a,a] \to \Re$$ $$\forall a>0$$ we have the relation: $\displaystyle \int_{-a}^{a} {f(x)}^2dx=2\displaystyle \int_{0}^{a}{f(x)}^2dx$ Is this True of False?  This is part of this set

×


text:
How much power is produced if a voltage of 5 V is applied to a circuit with a resistance of 96 Omega?

Apr 7, 2018

The power is approximately $0.26$ Watts

Explanation:

Let $I =$ current, $P =$ power, $V =$ voltage, and $R =$ resistance.

Ohm's law states

$V = I R$

We can solve this for current.

$I = \frac{V}{R}$

We also know that

$P = I V$.

We can eliminate the current with Ohm's Law.

$P = {V}^{2} / R = {5}^{2} / 96 \approx 0.26$ Watts

Apr 8, 2018

Around $0.26$ watts

Explanation:

Ohm's law states that,

$V = I R$

• $I$ is the current in amperes

• $R$ is the resistance in ohms

$\iff I = \frac{V}{R}$

Meanwhile, power is measured through,

$P = I V$

• $V$ is the voltage in volts

$\iff I = \frac{P}{V}$

Through the two equations, we get a relationship:

$\frac{V}{R} = \frac{P}{V}$

$P = {V}^{2} / R$

Now, we can simply plug in the values, and get:

$P = {\left(5 \setminus \text{V}\right)}^{2} / \left(96 \setminus \Omega\right)$

$= \frac{25 \setminus {\text{V}}^{2}}{96 \setminus \Omega}$

$\approx 0.26 \setminus \text{W}$


text:
The quadratic formula is a formula that helps us to find the solutions of an equation of degree 2. First of all, we need to turn our equation of degree 2 to standard form $ax^2+bx+c=0$.

The quadratic formula is: $\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$.

How to use the quadratic formula?

Let’s see some examples of the quadratic formula.

Example 1: Solve the quadratic equation $x^2+5x+6=0$.
Solution: In this quadratic equation a=1, b=5 and c=6. We put these numbers into the quadratic formula and we get:
$\dfrac{-5\pm \sqrt{25-24}}{2}=\dfrac{-5 \pm 1}{2}=-2,-3$.

Example 2: Solve the quadratic equation $x^2+5x+7=0$.
Solution: In this quadratic equation a=1, b=5 and c=7. We put these numbers into the quadratic formula and we get:
$\dfrac{-5\pm \sqrt{25-28}}{2}=$. As you see, under the square root we have a negative number, so there we have no solution for this quadratic equation.

When do we use the quadratic formula?

The quadratic formula is the most general formula to find solutions to quadratic equations. You can use it when you have a quadratic equation.


text:
# Environment with mdframed

The following mwe showcases the environment I am currently working with:

\documentclass{article}
\usepackage[tikz]{mdframed}

\definecolor{greentitle}{RGB}{165,224,168}

\newenvironment{notation}[1]
{\begin{mdframed}[
bottomline=true,
leftline=true,
linecolor=greentitle,
innerrightmargin=25pt,
singleextra={
\node[overlay,anchor=east,fill=greentitle,rotate=90,font=\footnotesize\scshape] at ([yshift=0.4pt]O|-P) {#1};
},
firstextra={
\node[overlay,anchor=east,fill=greentitle,rotate=90,font=\footnotesize\scshape] at ([yshift=0.4pt]O|-P) {#1};
},
]
}
{\end{mdframed}}
\begin{document}
\begin{notation}{Notation}
Sample Text
\end{notation}
\begin{notation}{Notation}
\begin{enumerate}
\item sample
\item sample
\item sample
\item sample
\item sample
\item sample
\end{enumerate}
\end{notation}
\begin{notation}{Notation}
\begin{enumerate}
\item sample
\item sample
\end{enumerate}
\end{notation}
\begin{notation}{Notation}
\begin{enumerate}
\item sample
\item sample
\end{enumerate}
\end{notation}
\end{document}

The issue I have is that I would like the environment to expand to the height of any word, in this case, notation by default. Therefore, the minimum height of the framed environment should be the height of the boxed word "notation." In the last examples you can see that the bottom line is not aligned with the boxed word.

Also I would like the boxed word to anchored outside of the framed box at the top right corner, not as is.

Note also that there is not enough space after each environment and as the first and second example show, they overlap.

• Is a solution with tcolorbox acceptable to you? – CarLaTeX Jul 16 '19 at 4:34
• @CarLaTeX Sure. If that is another way of doing it yes. I would like to compare both ways if that is the case; tcolorbox and mdframed. – azetina Jul 16 '19 at 4:44

This is what I managed to do with tcolorbox:

\documentclass{article}
\usepackage{graphicx}
\usepackage[most]{tcolorbox}
\definecolor{greentitle}{RGB}{165,224,168}
\newcommand{\mytitle}[1]{\rotatebox[origin=c]{90}{#1}}
\newtcolorbox{notation}{
sharp corners,
bicolor,
sidebyside,
colback=greentitle,
colframe=greentitle,
colbacklower=white,
lefthand width=.2cm,
sidebyside align=top seam,
left=0pt,
middle=0pt,
sidebyside gap=10pt,
fontupper=\footnotesize\scshape,
}

\begin{document}
\begin{notation}
\mytitle{Notation}
\tcblower
Sample Text
\end{notation}
\begin{notation}
\mytitle{Notation}
\tcblower
\begin{enumerate}
\item sample
\item sample
\item sample
\item sample
\item sample
\item sample
\end{enumerate}
\end{notation}
\begin{notation}
\mytitle{Notation}
\tcblower
\begin{enumerate}
\item sample
\item sample
\end{enumerate}
\end{notation}
\begin{notation}
\mytitle{Notation}
\tcblower
\begin{enumerate}
\item sample
\item sample
\end{enumerate}
\end{notation}
\end{document}


text:
Defining multiplication on a Koszul complex

Let $R$ be a Noetherian commutative ring and $x$ and $y$ two elements in $R$. We construct the Koszul complex on $x$ and $y$. We start by the following two chain complexes:

$$C_2=0\to C_1=R\xrightarrow{\ x\ } C_0= R\to C_{-1}=0$$ $$D_2=0\to D_1=R\xrightarrow{\ y\ } D_0= R\to D_{-1}=0$$ Now we construct the tensor product chain complex which we denote $CD:=C\otimes D$, $$CD_2=C_1\otimes D_1$$ $$CD_1=C_1\otimes D_0 \oplus C_0\otimes D_1$$ $$CD_0=C_0\otimes D_0$$ and we get the chain complex $$CD_3=0 \to CD_2\xrightarrow{\ \partial_2\ } CD_1\xrightarrow{\ \partial_1\ } CD_0 \to CD_{-1} =0$$ where

$$\partial_2 (c_1\otimes d_1)=(-c_1\otimes (yd_1)\,,\,(xc_1)\otimes d_1)$$ and $$\partial_1 (c_1\otimes d_0\,,\,c_0\otimes d_1)=(xc_1)\otimes d_0+c_0\otimes (yd_1)$$ It is clear that $\partial_1\circ\partial_2=0$

Question: How do we define the multiplication $CD_i\times CD_j\to CD_{i+j}$ on $(CD,\partial)$ that will make it a graded differential algebra; for example if we take two elements in $CD_1=C_1\otimes D_0 \oplus C_0\otimes D_1$, $(c_1\otimes d_0,c_0\otimes d_1)$ and $(c'_1\otimes d'_0,c'_0\otimes d'_1)$ how do we multiply them to obtain an element in $CD_2=C_1\otimes D_1$ ?

• This doesn't exactly answer your question, but it is maybe worth mentioning here that it is perhaps easier to understand the algebraic structure of the Koszul complex if one thinks of it as an exterior algebra equipped with a differential (as in section 1.6 of "Cohen-Macaulay Rings" by Bruns and Herzog).
– user55407
Mar 2 '13 at 17:58
• Well, given $A, B$ two dga over $R$, can you see that $A\otimes_R B$ is then a naturally a dga? In your case, $C_\bullet = R[x]/(x^2)$, $D_\bullet = R[y]/(y^2)$ are dga with differential $x, y$ respectively. Mar 4 '13 at 0:47
• can you please be more specific and tell me how do we define the multiplication $CD_i\times CD_j\rightarrow CD_{i+j}$? Thanks Mar 4 '13 at 15:03
• suppose $a_1 , a_2 \in C_\bullet; b_1,b_2\in D_\bullet$, then $(a_1\otimes b_1), (a_2\otimes b_2) \in C_\bullet\otimes D_\bullet = CD$, now multiplication is just $a_1a_2\otimes b_1b_2$. You can put all the degree back by assuming all the elements are homogeneous Mar 4 '13 at 23:06
• If i understand, the multiplication of two elements of $CD_1=C_1\otimes D_0\oplus C_0\otimes D_1$ is $(c_1\otimes d_0\,,\,c_0\otimes d_1)*(c'_1\otimes d'_0\,,\,c'_0\otimes d'_1)=(c_1c'_1\otimes d_0d'_0\,,\, c_0c'_0\otimes d_1d'_1)$ but this is not in $CD_2=C_1\otimes D_1$ as is supposed to be a gradation on $CD$, am i missing something? Mar 5 '13 at 5:19

To my understanding you are trying to consider $2$ differential graded algebras (dgas) $(C,d_C)$ and $(D,d_D)$ and to find the associative product on the differential graded algebra $(C\otimes D, d_{C\otimes D})$. In this case, the formula written by Aaron in his comment is correct.
Your complexes $C$ and $D$ are no dgas, though for any associative ring $R$, though. The Leibniz rule fails. Check this example in $C$:
$d_C( t \star w)= d_C(t)\star w - t \star d_C(w)$, with $t,w\in C_1=R$ and associative product $\star: C \otimes C \rightarrow C$ on $C$ induced by the multiplication on $R$. Note that $t \star w=0=C_2$, though. The r.h.s. of the above relation is $(xt)w-t(xw) \in C_1$, instead.
Consider that the Koszul complex is in general a free resolution of the given ring $R$ as left $R$-module; what is important in the applications is the module structure. You can even consider the above Koszul complex as an $R$-$R$-bimodule.


text:
Research Output

Now showing 1 - 10 of 35
• Publication
Breadth first search on massive graphs
(American Mathematical Society, 2009)
We consider the problem of Breadth First Search (BFS) traversal on massive sparse undirected graphs. Despite the existence of simple linear time algorithms in the RAM model, it was considered non-viable for massive graphs because of the I/O cost it incurs. Munagala and Ranade [29] and later Mehlhorn and Meyer [27] gave efficient algorithms (refered to as MR BFS and MM BFS, respectively) for computing BFS level decompositions in an external memory model. Ajwani et al. [3] implemented MR BFS and the randomized variant of MM BFS using the external memory library STXXL and gave a comparative study of the two algorithms on various graph classes. In this paper, we review and extend that result demonstrating the effectiveness and viability of the BFS implementations on various other synthetic and real world benchmarks. Furthermore, we present the implementation of the deterministic variant of MM BFS and show that in most cases, it outperforms the randomized variant.
188
• Publication
Automated Knowledge Hierarchy Assessment
(CEUR Workshop Proceedings, 2018-07-12)
Automated construction of knowledge hierarchies is gaining increasing attention to tackle the infeasibility of manually extracting and semantically linking millions of concepts. With the evolution of knowledge hierarchies, there is a need for measures to assess its temporal evolution, quantifying the similarities between different versions and identifying the relative growth of different subgraphs in the knowledge hierarchy. This work proposes a principled and scalable similarity measure, based on Katz similarity between concept nodes, for comparing knowledge hierarchies, modeled as generic Directed Acyclic Graphs (DAGs).
109
• Publication
A Geometric Distance Oracle for Large Real-World Graphs
(2018)
Many graph processing algorithms require determination of shortest-path distances between arbitrary numbers of node pairs. Since computation of exact distances between all node-pairs of a large graph, e.g., 10M nodes and up, is prohibitively expensive both in computational time and storage space, distance approximation is often used in place of exact computation. In this paper, we present a novel and scalable distance oracle that leverages the hyperbolic core of real-world large graphs for fast and scalable distance approximation. We show empirically that the proposed oracle significantly outperforms prior oracles on a random set of test cases drawn from public domain graph libraries. There are two sets of prior work against which we benchmark our approach. The first set, which often outperforms other oracles, employs embedding of the graph into low dimensional Euclidean spaces with carefully constructed hyperbolic distances, but provides no guarantees on the distance estimation error. The second set leverages Gromov-type tree contraction of the graph with the additive error guaranteed not to exceed $2\delta\log{n}$, where $\delta$ is the hyperbolic constant of the graph. We show that our proposed oracle 1) is significantly faster than those oracles that use hyperbolic embedding (first set) with similar approximation error and, perhaps surprisingly, 2) exhibits substantially lower average estimation error compared to Gromov-like tree contractions (second set). We substantiate our claims through numerical computations on a collection of a dozen real world networks and synthetic test cases from multiple domains, ranging in size from 10s of thousand to 10s of millions of nodes.
94
• Publication
Characterizing the Performance of Flash Memory Storage Devices and Its Impact on Algorithm Design
(Springer, 2008-06-25)
Initially used in digital audio players, digital cameras, mobile phones, and USB memory sticks, flash memory may become the dominant form of end-user storage in mobile computing, either completely replacing the magnetic hard disks or being an additional secondary storage. We study the design of algorithms and data structures that can exploit the flash memory devices better. For this, we characterize the performance of NAND flash based storage devices, including many solid state disks. We show that these devices have better random read performance than hard disks, but much worse random write performance. We also analyze the effect of misalignments, aging and past I/O patterns etc. on the performance obtained on these devices. We show that despite the similarities between flash memory and RAM (fast random reads) and between flash disk and hard disk (both are block based devices), the algorithms designed in the RAM model or the external memory model do not realize the full potential of the flash memory devices. We later give some broad guidelines for designing algorithms which can exploit the comparative advantages of both a flash memory device and a hard disk, when used together.
• Publication
Creating and Characterising Electricity Load Profiles of Residential Buildings
(Springer, 2020-09-18)
Intelligent planning, control and forecasting of electricity usage has become a vitally important element of the modern conception of the energy grid. Electricity smart-meters permit the sequential measurement of electricity usage at an aggregate level within a dwelling at regular time intervals. Electricity distributors or suppliers are interested in making general decisions that apply to large groups of customers, making it necessary to determine an appropriate electricity usage behaviour-based clustering of these data to determine appropriate aggregate load profiles. We perform a clustering of time series data associated with 3670 residential smart meters from an Irish customer behaviour trial and attempt to establish the relationship between the characteristics of each cluster based upon responses provided in an accompanying survey. Our analysis provides interesting insights into general electricity usage behaviours of residential consumers and the salient characteristics that affect those behaviours. Our characterisation of the usage profiles at a fine-granularity level and the resultant insights have the potential to improve the decisions made by distribution and supply companies, policy makers and other stakeholders, allowing them, for example, to optimise pricing, electricity usage, network investment strategies and to plan policies to best affect social behavior.
• Publication
Co-optimizing application partitioning and network topology for a reconfigurable interconnect
(Elsevier, 2016-10)
To realize the full potential of a high-performance computing system with a reconfigurable interconnect, there is a need to design algorithms for computing a topology that will allow for a high-throughput load distribution, while simultaneously partitioning the computational task graph of the application for the computed topology. In this paper, we propose a new framework that exploits such reconfigurable interconnects to achieve these interdependent goals, i.e., to iteratively co-optimize the network topology configuration, application partitioning and network flow routing to maximize throughput for a given application. We also present a novel way of computing a high-throughput initial topology based on the structural properties of the application to seed our co-optimizing framework. We show the value of our approach on synthetic graphs that emulate the key characteristics of a class of stream computing applications that require high throughput. Our experiments show that the proposed technique is fast and computes high-quality partitions of such graphs for a broad range of hardware parameters that varies the bottleneck from computation to communication. Finally, we show how using a particular topology as a seed to our framework significantly reduces the time to compute the final topology.
• Publication
Generating synthetic task graphs for simulating stream computing systems
(Elsevier, 2013-10)
Stream-computing is an emerging computational model for performing complex operations on and across multi-source, high-volume data flows. The pool of mature publicly available applications employing this model is fairly small, and therefore the availability of workloads for various types of applications is scarce. Thus, there is a need for synthetic generation of large-scale workloads to drive simulations and estimate the performance of stream-computing applications at scale. We identify the key properties shared by most task graphs of stream-computing applications and use them to extend known random graph generation concepts with stream computing specific features, providing researchers with realistic input stream graphs. Our graph generation techniques serve the purpose of covering a disparity of potential applications and user input. Our first "domain-specific" framework exhibits high user-controlled configurability while the second "application- agnostic" framework focuses solely on emulating the key properties of general stream-computing systems, at the loss of domain-specific fine-tuning. © 2013 Elsevier Inc. All rights reserved.
• Publication
Geometric Algorithms for Private-Cache Chip Multiprocessors
(Springer, 2010-09-08)
We study techniques for obtaining efficient algorithms for geometric problems on private-cache chip multiprocessors.


text:
?

# Econometrics

Free Version
Easy

EMETRC-YJ4EAZ

Let's suppose you are estimating the following model, where monthly saving depends upon income:

$$savings = \beta_{0} + \beta_{1} income + u$$

If u = education, then the assumption $E(u|x) = E(u)$ can be written as:

A

$E(edu|income) = 0$

B

$E(income|edu) = E(income)$

C

$E(edu|income) = E(edu)$

D

$E(income|edu) = E(edu)$


text:
# Homework Help: Needing help with calculating currents

1. Jul 16, 2010

### astenroo

1. The problem statement, all variables and given/known data
The currents in a DC circuit is to be calculated, and I end up with a system of 5 equations and 5 unknowns. This is where my headache starts.

2. Relevant equations

(1) I$$_{1}$$=I$$_{2}$$ + I$$_{3}$$
(2) I$$_{2}$$=I$$_{4}$$ + I$$_{5}$$
(3) 6 - 2I$$_{1}$$ - I$$_{3}$$ = 0
(4) 4.5 - 5I$$_{2}$$ - 3I$$_{5}$$ + I$$_{3}$$ = 0
(5) 3I$$_{5}$$ - 3I$$_{4}$$ -3 = 0

3. The attempt at a solution

Now, I have no problem solving equations with two unknowns. But 5?
Anyway, I gave it a try. I tried to substitute (2) in (1) and substituting the result into (3) and then substituting (2) into (4), resulting in a new system with three unknowns

(3') -3I$$_{3}$$ - 2I$$_{4}$$ - 2I$$_{5}$$ = -6
(4') I$$_{3}$$ - 5I$$_{2}$$ - 8I$$_{5}$$ = -4,5
(5') -3I$$_{5}$$ + 3I$$_{5}$$ = 3

I've been on this problem for the last 6 hours and I still haven't figured out how to solve it.
I've even tried using a matrix (never done one before today), and it all goes wrong, since I apparently don't know how to use a matrix :P

2. Jul 16, 2010

### phyzguy

You are on the right track with matrices. Write the current as a vector C=(I1,I2,I3,I4,I5), and then you can write the equations as a matrix equation M * C = B, where B is a constant vector which has the constant coefficients in it. Then you invert the matrix M, and can write the current C as C = Inverse(M) * B. I carried it out and got an answer for the currents using Mathematica. Can you try to write down the matrix M and the constant vector B?

3. Jul 16, 2010

### astenroo

i have to get back on that one as soon as I find instructions on how to set up said matrices... The one I tried was using it on (3'), (4') and (5'). Apparently that is a no go ;)

4. Jul 16, 2010

### dulrich

Matricies are cleaner and if you have the time to figure it out, that is the preferable way to go. However, this "brute-force" way will work. The best part is that if you keep clean notes, simply plug your answers back in and you will know for sure if you have the right answer regardless how you get there.

Except for a couple of typos in your subscripts, your equations look right. If you substitute out I5 from (3') and (4') using (5'), you will end up with two equations (3'') and (4'') with two unknowns, I3 and I4. Solve for those and then plug them back into (5') and you have I5. Plug these back into (2) then (1) and you'll be done.

5. Jul 19, 2010

### astenroo

Oh yes! Indeed, it is exactly as solving systems with two equations and 2 unknowns. Now I noticed the typos :) I might stick around using the brute-force method for now, but I will eventually have to learn matrices also :) Thank you for the help.


text:
Ronald Ross.

# The prevention of malaria online

. (page 14 of 55)
Online LibraryRonald RossThe prevention of malaria → online text (page 14 of 55)
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definition is more correct.

27] CONDITIONS OF INFECTION 153

(6). By the malarial case mortality we mean, of course, the
proportion of infected persons who die of the disease.

(7). By the malarial mortality we mean the proportion of
the total population who die of the disease.

(8). I use the word ratio to denote a proportion, and the
word rate to denote a percentage. The latter is
obtained by multiplying the former by one hundred.

27. Conditions required for the Production of New
Infections in a Locality. — From what has been written already
it will appear that new infections can occur in a locality only
if all the following conditions exist : —

(i). That a person whose blood contains a sufiflcient number
of gametids (sexual forms) is living in or near the
locality.
(2). That an Anopheline capable of carrying the parasites

sucks enough of that person's blood.
(3). That this Anopheline lives for a week or more after-
wards under suitable conditions — long enough to
allow the parasites to mature in it.
(4). That it next succeeds in biting another person who is
not immune against the disease or is not protected
by quinine.
To these we must now add the following principle : —
(5). That few or no new infections will occur in a com-
munity unless the persons with gametids in their
blood and the carrying Anophelines are sufficiently
numerous.
This last proposition is the basis of the public prevention
of malaria. Although tacitly admitted, it is never properly
discussed in the monographs, and the reader must therefore
examine it here in detail and as precisely as possible.

Let us suppose that we have to do with a population of
1,000 people living over an area in which indigenous malaria
does not exist ; and suppose that one of these people is an

154 MALARIA IN THE COMMUNITY [Sect.

imported case with suitable gametids in his blood. Next,
suppose that a single suitable Anopheline is liberated within
the area. What are the chances that this insect will ever cause
a new infection ?

First, we observe that not every mosquito can succeed in
biting human beings at all : suppose that the chances are
4 to I against this happening. Next, we observe that as there
is only one infected person among the i,ooo people in the place,
and as the particular Anopheline liberated in the area may
bite any one of these people, the chances are i,ooo to i against
its happening to bite the patient, even if it succeeds in biting
at all. That is, altogether, the chances against its biting the
patient are 4,000 to i. But suppose that this has happened.
It must now live for a week or more afterwards, and not all
mosquitos live so long. Suppose that the chances are 3 to i
against this particular Anopheline living long enough to mature
the parasites in it — so that the chances are 12,000 to i against
the Anopheline reaching the infective stage. But it must now
bite a second person. Suppose that the chances against this
are, again, 4 to i. Thus the total chances against this
Anopheline inoculating another person will be something like
48,000 to I. In other words, if instead of liberating a single
Anopheline within the area of observation we had liberated
48,000, the chances are that only one of these would succeed
in biting the patient, and would also live long enough to become
infective and to bite a second person. Of course, by bad luck,
so to speak, a larger proportion of the insects might succeed
in this ; but, on the other hand, by good luck, none at all might
succeed. Such calculations, though obviously based on con-
jectural data, yet suffice to show the absurdity of supposing
that the presence of a few Anophelines must cause an epidemic
of malaria.

Or we may consider the subject as follows : — Suppose that
48 Anophelines on the average live near the infected person.
Then, if the conjectural data are sound, 1/4 of these, or 12,

27] CONDITIONS OF INFECTION 155

will succeed in biting him ; 1/3 of these, or 4, will live for a
week or more ; and only 1/4 of these, or i, will succeed in biting
another person. If 48 Anophelines live near each one of the
999 healthy people, there will be 47,952 Anophelines" which
can bite only healthy people, and which cannot therefore cause
new infections. Thus, on the average, only i out of 48,000
Anophelines will succeed in infecting another individual.

If instead of only one person with gametids in the blood
there are 2, 3, 4 ... or more in the locality, then if there are
48 Anophelines on the average to each person, 2, 3, 4 ... or
more new infections may be caused. Thus, obviously, the
number of new infections in a locality, that is, the inoculation
rate, depends on two factors (other things being equal), namely,
the number of Anophelines and the number of previously
infected persons in the locality.

It is useful to put these thoughts into simple symbolic
language. Let/ denote the human population of the locality ;
nip the number of malaria - infected persons ; and imp the
number of these with gametids in the blood. Here vi and /
are fractions, since 7np is less than /, and imp less than mp.
Also m may vary from o to i, since the number of infected
persons may be anything from zero to the whole population.
The fraction i, being the proportion of infected persons with
gametids in their blood at the moment of enquiry, may be
put at 1/4, or may be much less.

Again, let a denote the number of Anophelines (of some
malaria - bearing species) to each human being — so that ap
denotes the total number of Anophelines in the locality, and
aimp the number of Anophelines compared with the number
of persons with gametids. Let b be the proportion of these
(say 1/4) which succeed in biting ; s the proportion (say 1/3)
which succeed in maturing the parasites ; and b the proportion
which succeed in biting another person. Then bsbaimp gives
the number of Anophelines which succeed in infecting persons.

156 MALARIA IN THE COMMUNITY [Sect.

We have assumed that b^ij^, •\$'=i/3. 2= 1/4; so that if each
of these bites a different person, we shall have —
No. of \noc\x\2A.\ons = b'^saiinp = amp I ig2

Inoculation rate % = — ^- • — = about h am.
'° 192 p

That is to say, the inoculation rate per icx) of population equals

about half the malaria ratio (w) multiplied by the mosquito

ratio {a).

For example, in a village containing 1,250 people, 750
infected people, and 3,000 Anophelines,/'= 1,250, m = o-b, rt = 2"4;
and we calculate roughly that the number of infecting Anophe-
lines, and also of inoculations, will be about 9*4. The inoculation
rate per cent, will be about 1/2 (2*4XO"6) = o72 ; that is, the
chances of being inoculated in the village will be as 72 is
to 10,000.

Or suppose that in another village half the people are in-
fected, and there are about twenty Anophelines to each person.
Then the chances of becoming infected there will be about 5/100.

Such calculations may appear far - fetched to many ; but
they are useful, not so much for the numerical estimates yielded
by them, but because they give more precision to our ideas,
and a guide for future investigations.

28. Laws which Regfulate the Amount of Malaria in a
Locality. — The number of infecting mosquitos which succeed
in biting again is b^saiinp. If all of these bite different people,
and all these people are healthy, and all become infected, this
expression will also denote the number oinew infections occurring
in the locality. But, of course, the infecting mosquitos may
often happen to bite on the second occasion, not healthy
persons, but persons already infected, especially if the pro-
portion of the latter is large.

If, as before, mp denotes the number of persons already
infected at the beginning of the enquiry, then/ — ?;// or (i —m)p
denotes the number of healthy people. Hence by proportion

28] LAWS OF DIFFUSION 157

the number of infecting mosquitos which bite healthy persons
(on the second biting) will be x, where

X : U^saimp :: {i—m)p : p

or X = b'^sai {i — m) mp . . . . (i)

and if each bites a different person and each person becomes
infected, the same expression will denote the number of new
infections which occur in the locality — that is, will denote the
addition to the number of malaria cases there.

But this is not the whole change which may occur. While,
during the period of observation, new infections are being pro-
duced, it may happen that some of the old cases may have
recovered. The number of those old cases was originally mp :
let rmp denote the number of those who have recovered during
the period of observation — so that r is a fraction. Hence the
whole number of cases in the locality will have increased or
decreased at the end of the period of observation, according
to whether b'^sai (i — ;;/) mp, the number of new cases, is greater
or less than rmp, the number of recoveries. Thus (neglecting
common factors) the change depends upon whether b-sai {\ —in)
is greater or less than r.

Suppose that no change occurs — that the recoveries exactly
equal the new infections. Then

b'^saz {i—in) = r .... (2)
From this equation we can calculate the values of a and m
when the amount of malaria in the locality remains constant —
that is, if we know the values of b, s, z, and r.

Suppose that the period of observation is one month. Now,
in section 21, I estimated roughly that only about half the cases
of malaria remain infected after three months. If this rate
holds for smaller periods, we may suppose that the ratio of
people who remain infected after only one month will be given
by the cube root of 1/2 — that is, by 07937 ; so that the pro-
portion of those who recover in one month will be 1—07937.
That is, we may write r = 0*2062. Let us take the values of
b, s, t as suggested in the previous section — so that b-si= 1/192.

158 MALARIA IN THE COMMUNITY [Sect.

Hence the new infections will equal the recoveries if

a{i- m) = 192 X 0-2062 = 39-59 = 40 {say) ... (3)
and the malaria will increase or diminish in the locality accord-
ing to whether a{\ —in) is greater or less than 40. Thus, if in,
the original malaria rate, is very small, the malaria will not
increase unless a, the monthly number of Anophelines per head
of human population, is greater than 40. On the other hand,
if ;;; is a larger fraction, say 1/2, the malaria will not increase
unless a is greater than 80. If in is still larger, say 3/4, the
malaria will not increase unless a is greater than 160. This
suggests that the malaria rate is not likely to increase
indefinitely unless the number of mosquitos is enormous. On
the other hand, if the number of Anophelines is below the
figure given by the equation, the malaria rate ought to begin
to fall, because the new infections can no longer keep pace with
the recoveries.

It should be noted that by the number of Anophelines we
here mean the number of different Anophelines which may bite
each person during a whole month, and not the insects which
may be, so to speak, allotted to each person at any one moment.
We say one month because this is the period we have selected
for observation — the period during which we suppose that
0*2 or I /5th of the cases recover. If we had selected one week
for the period of observation, the proportion of recoveries would,
of course, be lower (about 0*056 142), and the number of different
Anophelines to each person, required to compensate for the
recoveries during the week, would be correspondingly less (as
shown by the equation).

Now, what will happen if the malaria rate, instead of
remaining constant, increases or decreases ? On the one hand,
will it increase until every one becomes infected ; or, on the
other hand, will it decrease until it vanishes?

We have supposed that inp denotes the number of cases
at the beginning of the enquiry, which lasts, let us say, for a
month. Let ni^, tn^ip, w^p , . . denote the number of cases

28] LAWS OF DIFFUSION 159

at the end of i, 2, 3 . . . months respectively. Then if the

other figures remain constant, we have,

m^p = original cases -\- new infections — recoveries

= vip + b'^sai ( I — w) mp — rmp . . . . . (4)

and as the same process repeats itself month after month we

continue to have,

m^p = Wj/ + d'^sai ( i — m■^) m^p — rni^p
m^p = m^J) + b'^sai ( i — vi^ w^p — rm^J)

and so on. We can calculate m^ from the first equation.
Substituting its value in the second equation we calculate m^ ;
and substituting this in the third equation we calculate m.^ ;
and so on indefinitely.

If r=o'2 and b^si='00% the value of the 7ith term of the
series m^p, m^ . . . m„..,p, m„p may be written

in „p = {160 -\- a — am „..,) m„..,px o-oo^ • • • (S)

Let us now consider for examples the case of a village
containing 1,000 people of whom half are infected at the
beginning of the enquiry. By equation (3) of this section, and
also by this equation, if the number of different Anophelines
per person during one month {a) is 80, the malaria will neither
increase nor decrease, so that w„ will always be the same as w,
namely, 0*5. But if a is greater or less than this, the malaria
will increase or decrease accordingly.

(i). First, suppose that there are 100 different Anophelines
per person during a month — so that the malaria should increase.
Thus we calculate,

m^p = {i6o+ 100— iooxo"5)x 5x0*5 = 525

W2^ = (i6o+ 100— iooxo'525)x5xo-525 = 544'5.

Proceeding in this way we find that the number of cases
should increase every month as follows : —

Increase of Cases with 100 Anophelines

Months

I

2

3

4

. finally

Cases

500

525

544

560

571

600

i6o MALARIA IN THE COMMUNITY [Sect.

But how do we reach the last figure ? As we calculate the
number of cases month after month we observe that they always
increase, but by a constantly decreasing increment. Finally,
this increment becomes very small, so that the number of
cases approaches a fixed limit. It is easy to calculate this
limit mathematically (as will be done presently), but we can
also calculate it very simply as follows. If the number of cases
does ever arrive at a fixed limit, so that it no longer increases,
then equation (3) of this section must hold ; that is, the number
of cases and the number of mosquitos will be exactly balanced
according to the formula a (i— ;//) = 40. From this we have,
when this exact balance is reached,

m=\— Afi\a (6)

Here

Online LibraryRonald RossThe prevention of malaria → online text (page 14 of 55)


text:
chrono::ChContinuumDistribution Class Reference

## Description

Class that can be used to generate sample numbers according to a probability distribution.

Probability distribution is defined with x,y points, at least a dozen of pairs to be more precise in the reconstruction of probability. It uses the "Smirnov transform" (inverse probability integral transform)

#include <ChDistribution.h>

Inheritance diagram for chrono::ChContinuumDistribution:
Collaboration diagram for chrono::ChContinuumDistribution:

## Public Member Functions

ChContinuumDistribution (ChVectorDynamic<> &mx, ChVectorDynamic<> &my)
Create an object that can be used to generate sample numbers according to a generic probability distribution. More...

virtual double GetRandom () override
Compute a random value whose probability is the probability curve that has been entered with x,y points during the creation of this object.

const ChVectorDynamicGetProbabilityXpoints () const

const ChVectorDynamicGetProbabilityYpoints () const

const ChVectorDynamicGetProbabilityCDFcumulativeX () const

const ChVectorDynamicGetProbabilityCDFcumulativeY () const

Public Member Functions inherited from chrono::ChDistribution
virtual ~ChDistribution ()
Default destructor for distribution object.

## ◆ ChContinuumDistribution()

 chrono::ChContinuumDistribution::ChContinuumDistribution ( ChVectorDynamic<> & mx, ChVectorDynamic<> & my )

Create an object that can be used to generate sample numbers according to a generic probability distribution.

The probability distribution is a curve represented by simplified x,y pairs of points. The integral of the probability curve must be unit, i.e normalized (but if not, a normalization will be enforced) Note: too few points means approximate results, but too many points might give a small performance overhead when calling GetRandom().

The documentation for this class was generated from the following files:
• /builds/uwsbel/chrono/src/chrono/core/ChDistribution.h
• /builds/uwsbel/chrono/src/chrono/core/ChDistribution.cpp


text:
# Ifthenelse fails if is part of argument

I have followed very simple tutorial to create own listof. Link is https://texblog.org/2008/07/13/define-your-own-list-of/

It worked great, I just wanted to add optional parameter. If this optional is given, to the custom ToC it will send this optional parametr. But I get error, which I don't understand.

\usepackage{tocloft}
...
\newcommand{\listTMname}{List of TMs}
\newlistof{TM}{tms}{\listTMname}

\newcommand{\TM}[2][]{%
\refstepcounter{TM}
\par\noindent\textbf{TM \theTM. #2}
}\par}

I wanted to use it as follows: \TM{To text and ToC} and \TM[to ToC]{To text} But I get error calling is on line 94:

Illegal parameter number in definition of \reserved@a. l.94 \TM{Test...}

What I did wrong? If in \addcontentsline... command I remove ifthenelse and put it into \textbf 2 line above it works, but not opposite way.

• Have you tried enclosing the whole \ifthenelse in braces? – TeXnician Apr 30 '17 at 15:10
• Welcome to TeX - LaTeX! To good to get useful answers it usually best to collect your code snippets in to minimal compilable document demonstrating the problem. – Andrew Swann Apr 30 '17 at 15:28
• @TeXnician yes I tried, did not help, Andrew Swann thanks, next time I will know – Arxeiss Apr 30 '17 at 18:22

The command \ifthenelse is fragile and so is the whole construction

\ifthenelse{\equal{#1}{}}{#2}{#1}

Use an expandable test, such as

\newcommand{\TM}[2][]{%
\par\noindent
\refstepcounter{TM}% <--- don't forget %
\protect\numberline{\theTM}%
\if\relax\detokenize{#1}\relax#2\else#1\fi
}%
\textbf{TM \theTM. #2}% <--- don't forget %
\par\nopagebreak
}

Alternatively, with \usepackage{xparse}:

\NewDocumentCommand{\TM}{om}{%
\par\noindent
\refstepcounter{TM}% <--- don't forget %

I changed the position of the various items for placing the anchor (in case you use hyperref) in the right place.
• Works perfectly, thank you. Glad that you pointed version with and withou package xparse – Arxeiss Apr 30 '17 at 18:35


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Solution - Areas of Sector and Segment of a Circle

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ConceptAreas of Sector and Segment of a Circle

Question

Find the area of the minor segment of a circle of radius 14 cm, when its central angle is 60˚. Also find the area of the corresponding major segment.[use π=22/7]

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Reference Material

Solution for concept: Areas of Sector and Segment of a Circle. For the course 8th-10th CBSE
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# zbMATH — the first resource for mathematics

Serre’s conjectures. (English) Zbl 0848.11019
Murty, V. Kumar (ed.), Seminar on Fermat’s last theorem. The Fields Institute for Research in Mathematical Sciences, 1993-1994, Toronto, Ontario, Canada. Proceedings. Providence, RI: American Mathematical Society (publ. for the Canadian Mathematical Society). CMS Conf. Proc. 17, 135-153 (1995).
Let $$F$$ be a finite field of characteristic $$p$$ and write $$G_{\mathbb{Q}}= \text{Gal}(\overline{\mathbb{Q}}, \mathbb{Q})$$. Serre’s conjectures are concerned with irreducible two-dimensional representations $$\rho: G_{\mathbb{Q}}\to \text{GL}_2(\mathbb{F})$$ which are assumed to be odd, i.e. for complex conjugation $$c$$, $$\rho(c)$$ has eigenvalues $$1$$ and $$- 1$$. If $$\ell$$ is a prime at which $$\rho$$ is unramified, the characteristic polynomial of $$\rho(\text{Frob}_\ell)$$ of the Frobenius element $$\text{Frob}_\ell$$ at $$\ell$$ is well defined. With the usual notation, one has the notion of normalized eigenforms for Hecke operators acting on spaces of cusp forms (and on spaces of newforms) $$S_k(N, \varepsilon, R)$$ of weight $$k$$, level $$N$$ and character $$\varepsilon$$, for any subring $$R$$ of $$\mathbb{C}$$. The representation $$\rho$$ is said to be modular if there exists a normalized eigenform $$f$$ (of some weight $$k\geq 2$$, level $$N$$, and character $$\varepsilon$$) with Fourier coefficients in $$\mathbb{F}$$, such that for all $$\ell\nmid Np$$, where $$\rho$$ is unramified, $$\rho(\text{Frob}_\ell)$$ has characteristic polynomial $$x^2- a_\ell x+ \ell^{k- 1} \varepsilon(\ell)$$, where $$a_\ell$$ is the $$\ell$$th Fourier coefficient of $$f$$.
In this case $$\rho$$ and $$f$$ are called associated. Serre’s conjectures are stated in two forms: (S1, the vague form) Any odd irreducible $$\rho$$ as above is modular; (S2, the precise form) There exists a normalized $$\text{mod } p$$ eigenform of level $$N(\rho)$$, weight $$k(\rho)$$, and $$\mathbb{F}$$-valued character $$\varepsilon(\rho)$$ which is associated to $$\rho$$. Here the $$N(\rho)$$, $$k(\rho)$$ and $$\varepsilon(\rho)$$ are given by a precise recipe of Serre. (The $$\text{char } 2$$ or 3 cases need special treatment.) Of course, with respect to Wiles’ work on the Shimura-Taniyama conjecture and Fermat’s last theorem, a case of particular importance is the one of the Galois representations $$\rho_{E, p}: G_{\mathbb{Q}}\to \operatorname{Aut}(E_p)\simeq \text{GL}_2(\mathbb{F}_2)$$ associated to a semi-stable elliptic curve $$E$$ defined over $$\mathbb{Q}$$ with reduction $$\text{mod } p$$, $$E_p$$. If $$N_E$$ denotes the conductor of $$E$$ (i.e. the product of the primes of bad reduction of $$E$$) and $$\Delta_E$$ is the minimal discriminant of $$E$$, one has: (i) $$N(\rho_{E, p})$$ is the product of all primes $$\ell\neq p$$ such that $$\text{ord}_p(\Delta_E)\neq 0\pmod p$$; (ii) $$k(\rho_{E, p})= 2$$, if $$\text{ord}_p(\Delta_E)\equiv 0\pmod p$$, and is equal to $$p+ 1$$ otherwise; (iii) $$\varepsilon(\rho_{E, p})= 1$$. Unfortunately, S2 remains very difficult to attack in most cases, but one may try to prove Serre’s epsilon conjecture: (Assume $$\text{char}(\mathbb{F})> 3$$) S1 implies S2. In this direction one has results due to Mazur and Ribet.
In particular, Ribet’s result is enough to prove that the Shimura-Taniyama conjecture implies Fermat’s last theorem. Several examples of elliptic curves illustrating Mazur’s and Ribet’s results are discussed in some detail. The latest result is due to F. Diamond: Assume $$F$$ has odd characteristic. If $$\rho: G_\mathbb{Q}\to \text{GL}_2(\mathbb{F})$$ is a representation arising from an eigenform, then $$\rho$$ is associated to an eigenform of level $$N(\rho)$$, weight $$k(\rho)$$, and, if $$\rho$$ is not induced from a character of $$\text{Gal}(\overline{\mathbb{Q}}/ \mathbb{Q}(\sqrt{- 3}))$$ in case $$\text{char}(\mathbb{F})= 3$$, character $$\varepsilon(\rho)$$.
There are a few cases where S1 can be shown to be true, e.g. if the image of $$\rho$$ is isomorphic to a dihedral group $$D_{2n}$$, $$(n, p)= 1$$. This may then be combined with Diamond’s result. The case $$\mathbb{F}= \mathbb{F}_2$$ remains unclear.
One may also try to attack the author’s and Granville’s generalized Fermat conjectures on equations of the form $$x^p+ y^q= z^r$$, $${1\over p}+ {1\over q}+ {1\over r}< 1$$, $$\text{gcd}(x, y, z)= 1$$, $$xyz\neq 0$$. Assuming S2 one has some results in this direction. The idea is to associate an appropriate Frey curve (an elliptic curve) to the equation, and use Serre’s conjecture to show that the associated $$\text{ mod } p$$ representation does not exist, and, via Ribet’s result, apply the Shimura-Taniyama conjecture for this situation. Frey made the following conjecture: (F) Let $$A$$ be an elliptic curve defined over a number field $$K$$. There are only finitely many pairs $$(E, p)$$ consisting of an elliptic curve $$E$$ over $$K$$, not isogenous to $$A$$, and a prime number $$p> 5$$, such that $$E_p \simeq A_p$$ as $$G_K$$-modules. If both S2 and F hold, then the generalized Fermat equation of the form $$Ax^n+ By^n= Cz^n$$, $$n\geq 3$$, $$\text{gcd}(x, y, z)= 1$$, admits only finitely many integer solutions $$(x, y, z, n)$$. Apart from the fact that S2 implies the Shimura-Taniyama conjecture, Ribet proved the following: If S2 is true, then an elliptic curve over $$\overline\mathbb{Q}$$ is modular (i.e. is the quotient of some $$J_0(N)$$) if and only if $$E$$ is a $$\mathbb{Q}$$-curve, i.e. $$E$$ is an elliptic curve defined over a number field which is isogenous to all of its Galois conjugates.
The fifth (and last) section relates Wiles’ work to a generalized Shimura-Taniyama conjecture, essentially due to Mazur. This conjecture can be broken into two parts: (i) Serre’s conjecture on $$\text{mod } p$$ representations of $$G_{\mathbb{Q}}$$; (ii) A lifting conjecture of Serre’s $$\text{mod } p$$ representations to so-called admissible $$p$$-adic representations of $$G_{\mathbb{Q}}$$. Here is where Wiles’ work has led to substantial inroads. The paper closes with the remark that Serre’s conjecture (in general), as a first step in the direction of a ‘Langlands philosophy $$\text{mod } p$$’, remains wide open and that it will keep number theorists busy in years to come, perhaps long after the Shimura-Taniyama conjecture has been completely proved.
For the entire collection see [Zbl 0828.00033].

##### MSC:
 11F11 Holomorphic modular forms of integral weight 11F33 Congruences for modular and $$p$$-adic modular forms 11R32 Galois theory 20C11 $$p$$-adic representations of finite groups 11G05 Elliptic curves over global fields 11F70 Representation-theoretic methods; automorphic representations over local and global fields


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# A box of chocolates contains 20 identically shaped chocolates. 5 of them are filled jelly, 3 are filled with caramel and 12 are filled with nuts. What is the probability that one chocolate chosen at random is filled with jelly, caramel or nuts?

Dec 14, 2017

100%

#### Explanation:

$P \left(J e l l y\right) \mathmr{and} P \left(C a r a m e l\right) \mathmr{and} P \left(N u t s\right)$

$= P \left(J e l l y\right) + P \left(C a r a m e l\right) + P \left(N u t s\right)$ (Here, I used the or rule)

$= \frac{5}{20} + \frac{3}{20} + \frac{12}{20}$

$= 1$

Also, if you think about it, the only options are jelly, caramel and nuts.

But, the question is asking what is the probability for all of the only options.

Therefore, it is certain (i.e. 100%) that either a jelly, caramel or nuts would be chosen.

Hope that makes sense!