text:
# Circles class 9 Notes Mathematics

## myCBSEguide App

CBSE, NCERT, JEE Main, NEET-UG, NDA, Exam Papers, Question Bank, NCERT Solutions, Exemplars, Revision Notes, Free Videos, MCQ Tests & more.

CBSE class 9 Mathematics Chapter 10 Circles notes in PDF are available for free download in myCBSEguide mobile app. The best app for CBSE students now provides Circles class 9 Notes latest chapter wise notes for quick preparation of CBSE board exams and school based annual examinations. Class 9 Mathematics notes on Chapter 10 Circles are also available for download in CBSE Guide website.

## CBSE Guide Circles class 9 Notes

CBSE guide notes are the comprehensive notes which covers the latest syllabus of CBSE and NCERT. It includes all the topics given in NCERT class 9 Mathematics text book. Users can download CBSE guide quick revision notes from myCBSEguide mobile app and my CBSE guide website.

## 9 Mathematics notes Chapter 10 Circles

Download CBSE class 9th revision notes for Chapter 10 Circles in PDF format for free. Download revision notes for Circles class 9 Notes and score high in exams. These are the Circles class 9 Notes prepared by team of expert teachers. The revision notes help you revise the whole chapter in minutes. Revising notes in exam days is on of the best tips recommended by teachers during exam days.

CBSE Class 09 Mathematics
Revison Notes
CHAPTER 10
CIRCLES

• Circles and its Related Terms : A Review
• Angle Subtended by the chord at a Point
• Perpendicular from the centre to a Chord
• Circle through three given points.
• Equal Chords and their distances from the Centre
• Angle Subtended by an Arc of a Circle
• Circle – Circle is a locus of such points which are at equidistant from a fixed point in a plane. Also, a circle is the colleciton of those points in a plane that are at a given constant distance from a given fixed-point in the plane. The fixed point is called the centre and the given distance is called the radius of the circle.
• Concentric circles – Circles having same centre and different radii area called concentric circles.
• Arc – A continuous piece of a circle is called an arc of the circle.
• Chord – A line segment joining any two points on a circle is called the chord of the circle.
• A chord passing through the centre of a circle is called the diameter of the circle.
• A diameter of a circle divides it into two equal parts which are arcs. Each of these two arcs is called a semi-circle.
• Two arcs of a circle  are called congruent if they have the same degree of  measure.
• If two arcs are equal,then their corresponding chords are also equal.
• The perpendicular drawn from centre to  the chord of circle bisects the chord and vice-versa.
• There is one and only one circle passing through three non-collinear points.
• Equal chords of circle are equidistant from centre.
• If two circles intersect in two points, then the line through the centres is perpendicular to the common chord.
• The angle subtended by an arc at the centre of circle is twice the angle  subtended at remaining part of circumference.
• Any two angles in the same segment of the circle are equal.
• Equal chords of a circle subtend equal angle at the centre.
• Out of two chords of a circle, the larger chord is nearer to the centre.
• Angle of semicircle is right angle.
• Equal chords of circle subtend equal angle at the centre of circle.
• If  all the vertices of a quadrilateral lies on the circumference of circle, then quadrilateral  is called cyclic.
• In a cyclic quadrilateral the sum of opposite angles is ${180^o}$ and vice-versa.
• The exterior angle of a cyclic quadrilateral is equal to the  interior opposite angle.


text:
## Wednesday, May 7, 2014

### Hopping Robots and Reinforcement Learning

All too often, when we deal with data the outcome needed is a strategy or an algorithm itself. To arrive at that strategy we may have historic data or some model on how entities in system respond to various situations. In this write up, I'll go over the method of reinforcement learning. The general idea behind reinforcement learning is to come up with a strategy to maximize some measurable goal. For example, if you are modelling a robot that learns to navigate around obstacles, you want the learning process to come back with a strategy that minimizes collisions (say) with other entities in the environment.

Pattern Recognition and Machine Learning (Information Science and Statistics)

For the sake of simplicity, lets assume the following scenario. A robot is placed (at random) on flat plank of wood which has some sticky glue in the center. To its left there is a hole which damages the robot a bit and to its right is a reward which is its destination, as shown in the figure below

The robot has the following capabilities
• It can sense what state $${S_1,S_2,...,S_7}$$ it is in
• It can detect a reward or damage that happens to it while in a particular state
• It can move one space left or right
• It can hop a space on to the right
If the robot ends up in either of the terminal states of $$S_1$$ or $$S_7$$ it is taken and placed in another state at random. State $$S_1$$ does damage to the robot while state $$S_7$$ rewards it. $$S_4$$ is not exactly a damage causing state but it is an avoidable state and we want to learn that over time. All other states are neither harmful nor rewarding. The robot's goal is to "learn" to get to state $$S_7$$ in as few steps as possible.

In order to get reinforcement learning to work, you need to know what the reward values are for each of the states the robot can be in. In this particular example, we will assume the following reward structure for each of the states.
Note, the numbers are fairly arbitrary. In addition to this we need a function or a table, mapping out actions/states pairs leading to new states. Given the robot's movement description above, we can use a table as follows
Zero is being used to designate the scenario when the robot is reset to a random state. With the above two sets of information we are good to start the learning process.

Reinforcement learning works in the following way. You maintain a matrix of values for each state/action pair. This is the reward that can be attained by arriving at a particular state by taking a particular action. But the trick is to also account for what possible future state you can get to, given that you arrive at a given state. For example it may be beneficial to be in a certain state $$X$$, but all downstream states from $$X$$ may be bad to be in. You want to avoid such states. This way, the algorithm tries to ensure that future favourable and achievable states are taken into account. The algorithm does not immediately update itself based on whatever it learns, it preserves old values and learns gradually. The above methodology can be stated as follows.

If $$Q^{*}(s_t,a_t)$$ represents the pay off received by being in state then

$$Q^{*}(s_t,a_t) = R(s_{t+1},s_t) + \alpha max_{a_t}Q^{*}(s_{t+1},a_{t+1})$$

To slow down learning a bit, we stick with whatever prior estimate of $$Q^{*}(s_t,a_t)$$ we have by a fraction of $$\beta$$ as shown below

$$Q^{*}_{new}(s_t,a_t) = \beta Q^{*}_{prior}(s_t,a_t) + (1 - \beta)Q^{*}(s_t,a_t)$$
That's it! We now let the system take on various initial states, and let the device play around moving over to different states while we constantly update our $$Q$$ matrix. After several iterations, the $$Q$$ matrix will end with some values which reflect what strategy to take.

To give a swirl, here is an R code that walks through the entire process for this particular example

A thing to note in the code is how the reward function is encoded. There is a penalty imposed on moving from a higher state to a lower state. This is the simple way to ensure that whatever strategy it comes up with does not involve  going backwards from rightward positional gains that have been made. If you try it without the penalty, you will see cases where the strategy does not care going left or right in some states. The above code when run, creates the following output (screen grab below)

The rows represent the states, the columns represent the three possible actions move-left, move-right and hop-right that the robot can take. Notice what it's trying to say:
• State 1, do nothing
• State 2, move right, but don't hop
• State 3, don't move right, hop
• State 4, move right or hop
• State 5, move right (small chance), hop (definitely)
• State 6, move right, move left (small chance)
This is exactly what you would expect and want! More on this subject can be read from the following books
Reinforcement Learning: State-of-the-Art (Adaptation, Learning, and Optimization)
an
Reinforcement Learning: An Introduction (Adaptive Computation and Machine Learning)
If you are looking to buy some books on probability theory here is a good list to own.

format: plain


text:
# Informatics1-2017/Practice7

## Latex mathematical formulas

### Formulas

Reproduce the following in latex:

\begin{align} a_{11}x_1 + a_{12}x_2 + \ldots + a_{1n}x_n &\leq b_1\\ a_{21}x_1 + a_{22}x_2 + \ldots + a_{2n}x_n &\leq b_2\\ &\vdots\\ a_{m1}x_1 + a_{m2}x_2 + \ldots + a_{mn}x_n &\leq b_m \end{align}

where $x_i \geq 0 \, \forall i=1,2,\ldots,n$. The form of the objective function:

$z = c_1x_1 + c_2x_2 + \ldots + c_nx_n \Rightarrow \mathop{max} \text{ or } \mathop{min}$

The same with vectors and matrices:

\begin{align} \overrightarrow{x} &\geq 0\\ A\overrightarrow{x} &\leq \overrightarrow{b}\\ z=\overrightarrow{c}^T\overrightarrow{x}&\Rightarrow\mathop{max} \text{ or } \mathop{min} \end{align}

### Matrices, tables

Try the tabular environment with different aligns! Create a 3x3 matrix with all kinds of brackets. Try to do an nxn matrix as well for example, try to create this formula from wikipedia: matrix.

### Theorems, definitions

Let's do some theorems. For that append this to the preamble:

\newtheorem{mydef}{Definition}
• Create a new threorem style environment!
• Try the different styles (remark, theorem, definition)!

### Labels, references

\begin{theorem}\label{thm:sample_thm}
Theorem text
\end{theorem}

In Theorem \ref{thm:sample_thm} we...

#### Floating pictures

\begin{figure}[p]
\centering
\includegraphics[width=0.8\textwidth]{image.png}
\caption{Awesome Image}
\label{fig:awesome_image}
\end{figure}

Change the placement (h,t,p,b,!,H)!

## BibTeX

BibTeX is a package to create nice looking bibliographies. Create a test bibliography using these sites: http://www.bibtex.org/Using/, https://en.wikibooks.org/wiki/LaTeX/Bibliography_Management.


text:
Unique factorisation of prime geodesics?

In T. Sunada's 1985 paper Riemannian coverings and isospectral manifolds'', it is noted that for a compact Riemannian manifold $M$, prime (closed and non self-intersecting) geodesics behave like primes in a number field.

That is, given a finite normal (i.e. Galois'') cover $M\to N$ of Riemannian manifolds, the preimage of prime geodesic $\mathfrak{p}$ in $N$ is a disjoint union $\mathfrak{P}_1,...,\mathfrak{P}_k$ of prime geodesics in $M$. The degree $f_i$ of the covering map $\mathfrak{P}_i\to \mathfrak{p}$ of circles should be thought of as the inertia degree of $\mathfrak{P}_i/\mathfrak{p}$, and the Frobenius $\sigma_{\mathfrak{P}_i}\in \text{Deck}(M/N)$ acts on $\mathfrak{P}_i$ by a $1/f_i$ rotation. Note $[M:N]=\sum f_i$. Other similar properties also hold, for instance there is also a Chebatorev density theorem.

My question is: is there an analogue of unique factorisation of ideals in this context?

I don't know what the right analogue of a non-prime ideal is here: perhaps $\mathfrak{p}^2$ should be $\mathfrak{p}$, but the geodesic loops round twice, but I can't think of the right definition for $I$ in general.

• Thank you. For those not wanting to look through the paper: I think the paper hasn't found any geometric object corresponding to ideals, instead talking in terms of the formal product of prime ideals. The exception to this is prime power ideals, which correspond to prime geodesics looped round $n$ times. – Meow Jul 28 '18 at 18:28


text:
# American Institute of Mathematical Sciences

December  2011, 31(4): 1039-1051. doi: 10.3934/dcds.2011.31.1039

## On some frictional contact problems with velocity condition for elastic and visco-elastic materials

 1 University of La Réunion, PIMENT EA4518, 97715 Saint-Denis Messag cedex 9 La Réunion, France, France, France

Received  October 2009 Revised  May 2010 Published  September 2011

We study the evolution of a class of quasistatic problems, which describe frictional contact between a body and a foundation. The constitutive law of the materials is elastic, or visco-elastic: with short or long memory, and the contact is modelled by a general subdifferential condition on the velocity. We derive weak formulations for the models and establish existence and uniqueness results. The proofs are based on evolution variational inequalities, in the framework of monotone operators and $fi$xed point methods. We show the approach of the viscoelastic solutions to the corresponding elastic solutions, when the viscosity tends to zero. Finally we also study the approach to short memory visco-elasticity when the long memory relaxation coefficients vanish.
Citation: Khalid Addi, Oanh Chau, Daniel Goeleven. On some frictional contact problems with velocity condition for elastic and visco-elastic materials. Discrete & Continuous Dynamical Systems - A, 2011, 31 (4) : 1039-1051. doi: 10.3934/dcds.2011.31.1039
##### References:

show all references

##### References:
 [1] Toyohiko Aiki. A free boundary problem for an elastic material. Conference Publications, 2007, 2007 (Special) : 10-17. doi: 10.3934/proc.2007.2007.10 [2] Linglong Du. Long time behavior for the visco-elastic damped wave equation in $\mathbb{R}^n_+$ and the boundary effect. Networks & Heterogeneous Media, 2018, 13 (4) : 549-565. doi: 10.3934/nhm.2018025 [3] Maria-Magdalena Boureanu, Andaluzia Matei, Mircea Sofonea. Analysis of a contact problem for electro-elastic-visco-plastic materials. Communications on Pure & Applied Analysis, 2012, 11 (3) : 1185-1203. doi: 10.3934/cpaa.2012.11.1185 [4] Grigory Panasenko, Ruxandra Stavre. Asymptotic analysis of a non-periodic flow in a thin channel with visco-elastic wall. Networks & Heterogeneous Media, 2008, 3 (3) : 651-673. doi: 10.3934/nhm.2008.3.651 [5] Daniele Davino, Ciro Visone. Rate-independent memory in magneto-elastic materials. Discrete & Continuous Dynamical Systems - S, 2015, 8 (4) : 649-691. doi: 10.3934/dcdss.2015.8.649 [6] Alessia Berti, Claudio Giorgi, Elena Vuk. Free energies and pseudo-elastic transitions for shape memory alloys. Discrete & Continuous Dynamical Systems - S, 2013, 6 (2) : 293-316. doi: 10.3934/dcdss.2013.6.293 [7] Sergei A. Avdonin, Boris P. Belinskiy. On controllability of a linear elastic beam with memory under longitudinal load. Evolution Equations & Control Theory, 2014, 3 (2) : 231-245. doi: 10.3934/eect.2014.3.231 [8] Stanisław Migórski, Anna Ochal, Mircea Sofonea. Analysis of a frictional contact problem for viscoelastic materials with long memory. Discrete & Continuous Dynamical Systems - B, 2011, 15 (3) : 687-705. doi: 10.3934/dcdsb.2011.15.687 [9] Romain Aimino, Huyi Hu, Matthew Nicol, Andrei Török, Sandro Vaienti. Polynomial loss of memory for maps of the interval with a neutral fixed point. Discrete & Continuous Dynamical Systems - A, 2015, 35 (3) : 793-806. doi: 10.3934/dcds.2015.35.793 [10] Valeria Danese, Pelin G. Geredeli, Vittorino Pata. Exponential attractors for abstract equations with memory and applications to viscoelasticity. Discrete & Continuous Dynamical Systems - A, 2015, 35 (7) : 2881-2904. doi: 10.3934/dcds.2015.35.2881 [11] Shujuan Lü, Zeting Liu, Zhaosheng Feng. Hermite spectral method for Long-Short wave equations. Discrete & Continuous Dynamical Systems - B, 2019, 24 (2) : 941-964. doi: 10.3934/dcdsb.2018255 [12] Francesco Maddalena, Danilo Percivale, Franco Tomarelli. Adhesive flexible material structures. Discrete & Continuous Dynamical Systems - B, 2012, 17 (2) : 553-574. doi: 10.3934/dcdsb.2012.17.553 [13] Vittorino Pata. Exponential stability in linear viscoelasticity with almost flat memory kernels. Communications on Pure & Applied Analysis, 2010, 9 (3) : 721-730. doi: 10.3934/cpaa.2010.9.721 [14] Toyohiko Aiki. On the existence of a weak solution to a free boundary problem for a model of a shape memory alloy spring. Discrete & Continuous Dynamical Systems - S, 2012, 5 (1) : 1-13. doi: 10.3934/dcdss.2012.5.1 [15] G. Idone, A. Maugeri. Variational inequalities and a transport planning for an elastic and continuum model. Journal of Industrial & Management Optimization, 2005, 1 (1) : 81-86. doi: 10.3934/jimo.2005.1.81 [16] Stanislaw Migórski, Anna Ochal, Mircea Sofonea. Analysis of a dynamic Elastic-Viscoplastic contact problem with friction. Discrete & Continuous Dynamical Systems - B, 2008, 10 (4) : 887-902. doi: 10.3934/dcdsb.2008.10.887 [17] C. R. Chen, S. J. Li. Semicontinuity of the solution set map to a set-valued weak vector variational inequality. Journal of Industrial & Management Optimization, 2007, 3 (3) : 519-528. doi: 10.3934/jimo.2007.3.519 [18] Mark S. Gockenbach, Akhtar A. Khan. Identification of Lamé parameters in linear elasticity: a fixed point approach. Journal of Industrial & Management Optimization, 2005, 1 (4) : 487-497. doi: 10.3934/jimo.2005.1.487 [19] Jáuber Cavalcante Oliveira, Jardel Morais Pereira, Gustavo Perla Menzala. Long time dynamics of a multidimensional nonlinear lattice with memory. Discrete & Continuous Dynamical Systems - B, 2015, 20 (8) : 2715-2732. doi: 10.3934/dcdsb.2015.20.2715 [20] Mircea Sofonea, Cezar Avramescu, Andaluzia Matei. A fixed point result with applications in the study of viscoplastic frictionless contact problems. Communications on Pure & Applied Analysis, 2008, 7 (3) : 645-658. doi: 10.3934/cpaa.2008.7.645

2018 Impact Factor: 1.143


text:
# Inclined plane projectile

1. Jun 10, 2013

1. The problem statement, all variables and given/known data
A particle of mass m is projected with velocity of magnitude u at an angle of θ, measured anticlockwise from the line parallel to the plane, from a point (sx,0,sy,0) on a plane inclined at an angle of α measured anticlockwise from the line parallel to the ground. The particle coordinates refer to the particle's initial position (it starts from the plane) relative to the ground (i.e. sx is on an x-axis parallel to the ground, sy is on a y-axis perpendicular to the ground).
Find an expression each for sx and sy in times of time t, given that the particle undergoes no horizontal acceleration and that its only vertical acceleration is -g ms-2.

2. Relevant equations
I don't know. We have to find the equation.

3. The attempt at a solution

Well if the plane were not inclined this would be pretty easy.

$${s_y} = u \cdot \sin{θ} \cdot t + \frac{1}{2} \cdot {a_y} \cdot t^2$$

$${s_x} = u \cdot \cos{θ} \cdot t$$

Where θ is the angle of projection, and ay happens to be -g=-9.8 ms-2 in this case.

But now that the plane is inclined, I really am not sure!

2. Jun 10, 2013

### rude man

Gives me a headache just to figure the problem.

Would be good if the initial velocity were given in terms of v = vx i + vy j + vz k as well as its initial position (x0, y0, z0). Or if it's 2-dimensional (I can't tell from the wording), leave out the z0 and vz k. Then we could compute x(t), y(t) and (if appropriate) z(t).


text:
Tuesday, September 16, 2014

Fullerene Van der Waals Oligomers as Electron Traps

Shubina, T. E.; Sharapa, D. I.; Schubert, C.; Zahn, D.; Halik, M.; Keller, P. A.; Pyne, S. G.; Jennepalli, S.; Guldi, D. M.; Clark, T. J. Am. Chem. Soc. 2014, 136, 10890-10893
Contributed by Steven Bachrach.
Reposted from Computational Organic Chemistry with permission

Clark and co-workers have examined small fullerene clusters for their ability to capture electrons.1 They first looked at the fullerene dimer, comparing the electron affinity of the dimer having a C-C bond between the two cages (about 1.6-1.7 Å between the two cages) 1 and where the two cages are interacting only through van der Waals attractions (around 2.6 Å) 2. The structures and their radical anions were computed at RI-BP86/TZV. The structures of the two radical anions are shown in Figure 1. Interestingly, the radical anion of 2 is actually lower in energy that the radical anion of 1. Comparisons with some other methods are discussed, including a CASSPT2(5,4)/ANO-L-VDZ, computation, that support this result.
 1 2 3 4
Figure 1. RI-BP86/TZV optimized geometries of the radical anions of 1-4.
(Be sure to click on these images to be able to manipulate these structures in 3-D!)

This suggests that the added electron is being held between the cages, in an interstitial region. That suggested looking at the trimer and tetramer structures 3 and 4. The radical anions of these two oligomers are also shown in Figure 1. These oligomers show electron affinities of 1 eV greater than for fullerene itself, along with the ability to stabilize the dianion and even the trianion, what the authors call “deep electron traps”.

References

(1) Shubina, T. E.; Sharapa, D. I.; Schubert, C.; Zahn, D.; Halik, M.; Keller, P. A.; Pyne, S. G.; Jennepalli, S.; Guldi, D. M.; Clark, T. "Fullerene Van der Waals Oligomers as Electron Traps," J. Am. Chem. Soc. 2014,136, 10890-10893, DOI: 10.1021/ja505949m.


text:
# Garland

How much tinsel do you need to decorate a Christmas tree?

Our idealised tree will be a cone with opening angle $2\varphi$. The tinsel lies on the surface of the cone, and I’ll state that the curve described by the tinsel can be expressed parametrically by the equations

$\bold{r}(\theta)=\begin{pmatrix}\alpha\theta\cos\theta \\ \alpha\theta\sin\theta \\ \beta\theta\end{pmatrix},\quad\,\alpha,\beta\in\left[0,\infty\right)$

where the $z$ axis coincides with the tree’s axis of rotational symmetry, and $\bold{r}(0)$ corresponds to the tip of the tree. This vector describes the locus of a conical spiral, for which the distance between successive loops is constant.

To make sure this spiral lies on the surface of the cone, we have to make sure the constants $\alpha$ and $\beta$ are chosen appropriately. To do so, it’s useful to define the tinsel’s radial distance $\rho$ from the tree’s axis of symmetry:

$\rho(\theta)=\sqrt{x^2(\theta)+y^2(\theta)}$

$\rho(\theta)=\alpha\theta$

Then if $\theta\to\theta+2\pi$,

$\rho\to\rho+2\pi\alpha$

$z\to z+2\pi\beta$

So after wrapping around the tree once, the tinsel has moved outwards by an amount $2\pi\alpha$ and downwards by an amount $2\pi\beta$. I hope you’ll agree that this means the angle made by the tinsel’s spiral path with the $z$ axis satisfies

$\displaystyle\tan\varphi'=\frac{2\pi\alpha}{2\pi\beta}$

Hence for the tinsel to lie on the surface of the tree, we require that $\alpha$ and $\beta$ satisfy

$\displaystyle \frac{\alpha}{\beta}=\tan\varphi$

which we can also express as

$\displaystyle\frac{\alpha}{\sqrt{\alpha^2+\beta^2}}=\sin\varphi$

We can also introduce a second property of the tinsel, defined by

$\displaystyle \lambda=2\pi\sqrt{\alpha^2+\beta^2}$

This is the distance between successive loops, as measured along the surface of the tree. It’s a bit like the tinsel’s ‘wavelength’, which is why I’ve called the quantity $\lambda$.

For now we’ll continue to work with the parameters $\alpha$ and $\beta$, and switch to the physical quantities $\varphi$ and $\lambda$ when it’s appropriate.

$\$

To measure the length of the tinsel, we use an integral of small line elements:

$\displaystyle S=\int dS$

The length of each element can be written in terms of the differentials $dx, dy, dz$ using the equation

$dS=\sqrt{(dx)^2+(dy)^2+(dz)^2}$

which can conveniently be expressed using the parameter $\theta$:

$\displaystyle dS=\sqrt{\left(\frac{dx}{d\theta}\right)^2+\left(\frac{dy}{d\theta}\right)^2+\left(\frac{dz}{d\theta}\right)^2}d\theta$

Using the equation for $\bold{r}(\theta)$, we can work out the necessary derivatives:

$\displaystyle \frac{dx}{d\theta}=\frac{d}{d\theta}(\alpha\theta\cos\theta)=\alpha\cos\theta-\alpha\theta\sin\theta$

$\displaystyle \frac{dy}{d\theta}=\frac{d}{d\theta}(\alpha\theta\sin\theta)=\alpha\sin\theta+\alpha\theta\cos\theta$

$\displaystyle \frac{dz}{d\theta}=\frac{d}{d\theta}(\beta\theta)=\beta$

Substituting these in gives

$dS=\sqrt{(\alpha\cos\theta-\alpha\theta\sin\theta)^2+(\alpha\sin\theta+\alpha\theta\cos\theta)^2+(\beta)^2}$

$dS=\sqrt{\alpha^2+\beta^2+\alpha^2\theta^2}$

$\displaystyle dS=\sqrt{\alpha^2+\beta^2}\sqrt{1+\frac{\alpha^2}{\alpha^2+\beta^2}\theta^2}$

Now would be a good time to change to the parameters we introduced before:

$\displaystyle dS=\frac{1}{2\pi}\lambda\sqrt{1+\sin^2\varphi\,\theta^2}$

So we need to solve the integral

$\displaystyle S=\frac{1}{2\pi}\lambda\int\sqrt{1+\sin^2\varphi\,\theta^2}\,d\theta$

By exploiting the identity

$\cosh^2 y-\sinh^2 y=1$

the integrand can be simplified using the substitution

$\sinh y=\sin\varphi\,\theta$

$\cosh y\,dy=\sin\varphi\,d\theta$

This gives

$\displaystyle S=\frac{1}{2\pi}\lambda\csc\varphi\int\sqrt{1+\sinh^2y}\cosh y\,dy$

$\displaystyle S=\frac{1}{2\pi}\lambda\csc\varphi\int\cosh^2 y\,dy$

The next few steps require the ‘double-angle’ formulae

$\cosh 2y=2\cosh^2 y -1$

$\sinh 2y=2\sinh y\cosh y$

allowing us to say

$\displaystyle S=\frac{1}{4\pi}\lambda\csc\varphi\int \left(\cosh 2y+1\right)\,dy$

$\displaystyle S=\frac{1}{4\pi}\lambda\csc\varphi\left(\frac{1}{2}\sinh 2y+y\right)$

$\displaystyle S=\frac{1}{4\pi}\lambda\csc\varphi\left(\sinh y\cosh y+y\right)$

$\displaystyle S=\frac{1}{4\pi}\lambda\csc\varphi\left(\sin\varphi\,\theta\sqrt{1+\sin^2\varphi\,\theta^2}+\text{arcsinh}\left(\sin\varphi\,\theta\right)\right)$

which is equivalent to

$\displaystyle S(\theta|\lambda,\varphi)=\frac{1}{4\pi}\lambda\csc\varphi\left(\sin\varphi\,\theta\sqrt{1+\sin^2\varphi\,\theta^2}+\ln\left(\sin\varphi\,\theta+\sqrt{1+\sin^2\varphi\,\theta^2}\right)\right)$

$\$

Let’s try a few values. For a Christmas tree with opening angle 35 degrees adorned with 5 turns of tinsel separated by a length of 40 cm, the total length of tinsel needed is

$S(10\pi|0.4,17.5^\circ)\approx 10\,\text{m}$

That’s just for a modest tree! Try some for yourself. For very long lengths of tinsel, the expression above can be approximated as

$\displaystyle S(\theta|\lambda,\varphi)\approx\frac{1}{4\pi}\lambda\sin\varphi\,\theta^2$

since the logarithmic term grows much more slowly than the quadratic term.


text:
To install click the Add extension button. That's it.

The source code for the WIKI 2 extension is being checked by specialists of the Mozilla Foundation, Google, and Apple. You could also do it yourself at any point in time.

4,5
Kelly Slayton
Congratulations on this excellent venture… what a great idea!
Alexander Grigorievskiy
I use WIKI 2 every day and almost forgot how the original Wikipedia looks like.
Live Statistics
English Articles
Improved in 24 Hours
What we do. Every page goes through several hundred of perfecting techniques; in live mode. Quite the same Wikipedia. Just better.
.
Leo
Newton
Brights
Milds

# Residence time (statistics)

In statistics, the residence time is the average amount of time it takes for a random process to reach a certain boundary value, usually a boundary far from the mean.

## Definition

Suppose y(t) is a real, scalar stochastic process with initial value y(t0) = y0, mean yavg and two critical values {yavgymin, yavg + ymax}, where ymin > 0 and ymax > 0. Define the first passage time of y(t) from within the interval (−ymin, ymax) as

${\displaystyle \tau (y_{0})=\inf\{t\geq t_{0}:y(t)\in \{y_{\operatorname {avg} }-y_{\min },\ y_{\operatorname {avg} }+y_{\max }\}\},}$

where "inf" is the infimum. This is the smallest time after the initial time t0 that y(t) is equal to one of the critical values forming the boundary of the interval, assuming y0 is within the interval.

Because y(t) proceeds randomly from its initial value to the boundary, τ(y0) is itself a random variable. The mean of τ(y0) is the residence time,[1][2]

${\displaystyle {\bar {\tau }}(y_{0})=E[\tau (y_{0})\mid y_{0}].}$

For a Gaussian process and a boundary far from the mean, the residence time equals the inverse of the frequency of exceedance of the smaller critical value,[2]

${\displaystyle {\bar {\tau }}=N^{-1}(\min(y_{\min },\ y_{\max })),}$

where the frequency of exceedance N is

${\displaystyle N(y_{\max })=N_{0}e^{-y_{\max }^{2}/2\sigma _{y}^{2}},}$

(1)

σy2 is the variance of the Gaussian distribution,

${\displaystyle N_{0}={\sqrt {\frac {\int _{0}^{\infty }{f^{2}\Phi _{y}(f)\,df}}{\int _{0}^{\infty }{\Phi _{y}(f)\,df}}}},}$

and Φy(f) is the power spectral density of the Gaussian distribution over a frequency f.

### Generalization to multiple dimensions

Suppose that instead of being scalar, y(t) has dimension p, or y(t) ∈ ℝp. Define a domain Ψ ⊂ ℝp that contains yavg and has a smooth boundary ∂Ψ. In this case, define the first passage time of y(t) from within the domain Ψ as

${\displaystyle \tau (y_{0})=\inf\{t\geq t_{0}:y(t)\in \partial \Psi \mid y_{0}\in \Psi \}.}$

In this case, this infimum is the smallest time at which y(t) is on the boundary of Ψ rather than being equal to one of two discrete values, assuming y0 is within Ψ. The mean of this time is the residence time,[3][4]

${\displaystyle {\bar {\tau }}(y_{0})=\operatorname {E} [\tau (y_{0})\mid y_{0}].}$

### Logarithmic residence time

The logarithmic residence time is a dimensionless variation of the residence time. It is proportional to the natural log of a normalized residence time. Noting the exponential in Equation (1), the logarithmic residence time of a Gaussian process is defined as[5][6]

${\displaystyle {\hat {\mu }}=\ln \left(N_{0}{\bar {\tau }}\right)={\frac {\min(y_{\min },\ y_{\max })^{2}}{2\sigma _{y}^{2}}}.}$

This is closely related to another dimensionless descriptor of this system, the number of standard deviations between the boundary and the mean, min(ymin, ymax)/σy.

In general, the normalization factor N0 can be difficult or impossible to compute, so the dimensionless quantities can be more useful in applications.

## Notes

1. ^ Meerkov 1987, pp. 1734–1735.
2. ^ a b Richardson 2014, p. 2027.
3. ^ Meerkov 1986, p. 494.
4. ^ Meerkov 1987, p. 1734.
5. ^ Richardson 2014, p. 2028.
6. ^ Meerkov 1986, p. 495, an alternate approach to defining the logarithmic residence time and computing N0

## References

• Meerkov, S. M.; Runolfsson, T. (1986). Aiming Control. Proceedings of 25th Conference on Decision and Control. Athens: IEEE. pp. 494–498.CS1 maint: ref=harv (link)
• Meerkov, S. M.; Runolfsson, T. (1987). Output Aiming Control. Proceedings of 26th Conference on Decision and Control. Los Angeles: IEEE. pp. 1734–1739.CS1 maint: ref=harv (link)
• Richardson, Johnhenri R.; Atkins, Ella M.; Kabamba, Pierre T.; Girard, Anouck R. (2014). "Safety Margins for Flight Through Stochastic Gusts". Journal of Guidance, Control, and Dynamics. AIAA. 37 (6): 2026–2030. doi:10.2514/1.G000299. hdl:2027.42/140648.CS1 maint: ref=harv (link)


text:
ABC is an isosceles triangle with AC = BC. If AB^2 = 2 AC^2, prove that ABC is a right triangle.

# Ask a Question

### Question Asked by a Student from EXXAMM.com Team

Q 3061580425.     ABC is an isosceles triangle with AC = BC. If AB^2 = 2 AC^2, prove that ABC is a right triangle.

#### HINT

(Provided By a Student and Checked/Corrected by EXXAMM.com Team)

#### Access free resources including

• 100% free video lectures with detailed notes and examples
• Previous Year Papers
• Mock Tests
• Practices question categorized in topics and 4 levels with detailed solutions
• Syllabus & Pattern Analysis


text:
# Blank output from TextRecognize

Why am I getting blank output from TextRecognize for the following image? Is it because of the size of the image?

Neither am I getting output for this code:

x = Import["http://i.stack.imgur.com/RbqrD.jpg"];
TextRecognize[x]

nor for this one:

x = Import["http://i.stack.imgur.com/RbqrD.jpg"];
img = ImagePartition[x, {50, 69}];
MatrixForm[img];
p=img[[4,1]];
TextRecognize[p]

• The way TextRecognize works is that it looks for groups of letters and guesses what the word is. So it's better for complete sentences, it's not good for individual letters. In this image especially the letters have a spatial arrangement TextRecognize cannot make sense of, so you are better off using ImageCorrelate and some sample letters (there are other Q&As about this). Jul 9, 2014 at 9:55
• This is what you need. Jul 9, 2014 at 10:04
• @Pickett This was also my first thought, but please see the section recognizing random characters from my answer. It's far from a complete answer though. Jul 9, 2014 at 17:48
• @JacobAkkerboom I've seen TextRecognize being used like this before, for example here, but it's just not very good at it. For varying spatial distributions it is a much safer bet to use image correlation. Jul 9, 2014 at 18:25
• If talking about spatial distribution then my second code should work properly in there this whole image is partitioned in 228 images of size 50x69 even there its not working Jul 10, 2014 at 3:22

It does not seem to recognise a single character in isolation. Even when I make it easier

y1 = ImageCrop[
ImageCrop[img1, {100, 225}, {Left, Bottom}], {20, 25}, {Left, Top}];
y2 = ImageResize[y1, Scaled[10]];
y3 = Sharpen[y2];
y4 = Binarize[y3]

We still have that

TextRecognize@y4

gives "" . But there is hope, see the next section.

Recognizing multiple random characters

TextRecognize can find characters that do not make up words (I'm not sure if sentences matter here). A key element here is that the characters are regularly spaced. Let's get an example from the docs

wolframImg =
ImageCrop[
ImageCrop[
ImageResize[
Import[
"http://reference.wolfram.com/language/ref/Files/TextRecognize.\
en/I_20.gif"]
,
Scaled[5]
]
,
{434, 240}
,
{Left, Bottom}
]
,
{408, Full}
,
{Right, Top}
]

![enter image description here][2]

TextRecognize works reasonably well on this, as advertised.

TextRecognize[wolframImg]

ZSI-IC
HSKRN
CHKRVD

TextRecognize actually often gives quite a few false positives, when it is confident about a lot of (other) characters. This gave me the (silly) idea to compose this image with characters from your example.

y4Copies = Table[y4, {2}, {5}];
y5 = ImageAssemble[y4Copies];
y6 = ImageResize[y5, First@ImageDimensions@wolframImg];
compoImg = ImageAssemble[{{y6}, {wolframImg}}]

Now we can do

TextRecognize[compoImg]

which gives

EEEEE
EEEEE
ZSHC
HSKRN
CHKRVD

• Yes Recognizes it successfully but tesseract in mathematica does not recognize single character it requires a whole word be it a random or logical..Still on my way to make it recognize single character Jul 10, 2014 at 3:56

Here is a shorter implementation of the idea of Jacob Akkerboom to assemble the characters in a line:

\$Version
i = Import["https://i.stack.imgur.com/RbqrD.jpg"];
c = ComponentMeasurements[ColorNegate@Binarize@i, "Image", #"ConvexCoverage" < .9 &];
row = ImageAssemble[{ImagePad[ImageResize[#, {Automatic, 200}], 10] & /@ c[[;; , 2]]}]
TextRecognize[row]

"11.1.0 for Microsoft Windows (64-bit) (March 13, 2017)"

"ESEEEEEEEEEEEEEAAAAAAHHHHA"

The parameters 200 and 10 were carefully chosen by trial and error for achieving best possible recognition quality. And the latter is still leaves much to be desired. :(

Such poor quality of TextRecognize is especially surprising considering that Tesseract (which TextRecognize uses under the hood) shows excellent quality and does not require any assistance at all, as it can be checked for example using its Windows GUI gImageReader.

• Actually I have given a report last time,but don't get any response till now.
– yode
Apr 3, 2017 at 7:04
• @yode Please inform us if you receive an answer from the tech support. I was playing with the image from this question in the hope that improvements of TextRecognize in version 11.1 make recognition of individual characters straightforward. But I see that it is not the case: RecognitionPrior -> "Character" improves nothing as well as the other option choices! It is a shame that WRI can't properly integrate Tesseract into Mathematica... Apr 3, 2017 at 7:52


text:
`ISRN Astronomy and AstrophysicsVolume 2013 (2013), Article ID 174741, 6 pageshttp://dx.doi.org/10.1155/2013/174741`
Research Article

LRS Bianchi Type-I Dark Energy Cosmological Models in General Scalar Tensor Theory of Gravitation

Department of Applied Mathematics, Andhra University, Visakhapatnam 530003, India

Received 19 April 2013; Accepted 29 May 2013

Academic Editors: H. Bushouse, M. Grewing, D. Kieda, V. Pierrard, K. P. Rauch, and W. W. Zeilinger

Copyright © 2013 V. U. M. Rao and D. Neelima. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

Locally rotationally symmetric (LRS) Bianchi type-I dark energy cosmological model with variable equation of state (EoS) parameter in (Nordtvedt 1970) general scalar tensor theory of gravitation with the help of a special case proposed by (Schwinger 1970) is obtained. It is observed that these anisotropic and isotropic dark energy cosmological models always represent an accelerated universe and are consistent with the recent observations of type-Ia supernovae. Some important features of the models, thus obtained, have been discussed.

1. Introduction

Nordtvedt [1] proposed a general class of scalar tensor gravitational theories in which the parameter of the Brans-Dicke (BD) theory is allowed to be an arbitrary (positive definite) function of the scalar field (). Considering the static spherically symmetric solution for a point mass source, Nordtvedt [1] found a variety of experimental consequences of , including a contribution to the rate of precession of Mercury’s perihelion.

Several investigations have been made in higher dimensional cosmology in the framework of different scalar tensor theories of gravitation. Barker [2], Ruban and Finkelstein [3], Banerjee and Santos [4, 5], and Shanti and Rao [6, 7] are some of the authors who have investigated several aspects of the Nordtvedt general scalar tensor theory in four dimensions. Rao and Sreedevi Kumari [8] have discussed a cosmological model with negative constant deceleration parameter in a general scalar tensor theory of gravitation. Rao et al. [9] have obtained the Kaluza-Klein radiating model in a general scalar tensor theory of gravitation. Rao et al. [10] have discussed LRS Bianchi type-I dark energy cosmological model in the Brans-Dicke theory of gravitation. Rao et al. [11] have discussed Bianchi type-II, -VIII, and -IX dark energy cosmological models in the Saez-Ballester theory of gravitation. Recently, Rao et al. [12] have obtained perfect fluid dark energy cosmological models in the Saez-Ballester and general theory of gravitation.

Recently, there has been considerable interest in cosmological models with dark energy in general relativity because of the fact that our universe is currently undergoing an accelerated expansion which has been confirmed by a host of observations, such as type Ia supernovae (Reiss et al. [13]; Perlmutter et al. [14]; and Tegmark et al. [15]). Based on these observations, cosmologists have accepted the idea of dark energy, which is a fluid with negative presence making up around 70% of the present universe energy content to be responsible for this acceleration due to repulsive gravitation. Cosmologists have proposed many candidates for dark energy to fit the current observations such as cosmological constant, tachyon, quintessence, and phantom. Current studies to extract the properties of a dark energy component of the universe from observational data focus on the determination of its equation of state , which is the ratio of the dark energy’s pressure to its energy density , which is not necessarily constant. The methods for restoration of the quantity from experimental data have been developed (Sahni and Starobinsky [16]), and an analysis of the experimental data has been conducted to determine this parameter as a function of cosmological time (Sahni et al. [17]). The simplest dark energy candidate is the vacuum energy , which is mathematically equivalent to the cosmological constant . The other conventional alternatives, which can be described by minimally coupled scalar fields, are quintessence , phantom energy , and quintom (that can cross from phantom region to quintessence region as evolved) and have time dependent EoS parameter. Due to lack of observational evidence in making a distinction between constant and variable , usually the equation of state parameter is considered as a constant (Kujat et al. [18]; Bartelmann et al. [19]) with phase wise value −1, 0, +1/3, and +1 for vacuum fluid, dust fluid, radiation, and stiff dominated universe, respectively. But in general, is a function of time or redshift (Jimenez [20]; Das et al. [21]). Ray et al. [22], A. K. Yadav and L. Yadav [23], Kumar [24], and Pradhan et al. [25] are some of the authors who have investigated dark energy models in general relativity with variable EoS parameters in different contexts. Yadav and Saha [26] have obtained an LRS Bianchi-I anisotropic cosmological model with dominance of dark energy.

In this paper, we will study LRS Bianchi type-I dark energy cosmological models in the Nordtvedt [1] general scalar tensor theory with the help of a special case proposed by Schwinger [27], that is, , where is a constant.

2. Metric and Field Equations

We consider the LRS Bianchi type-I metric in the following form: where and are functions of cosmic time only.

The field equations of general scalar tensor theory proposed by Nordtvedt (using geometrized units with , ) are where is the Ricci tensor, is the curvature invariant, is the stress energy of the matter, and comma and semicolon denote partial and covariant differentiation, respectively.

Also, we have which is a consequence of the field equations (2).

The energy momentum tensor components of the fluid can be written in anisotropic diagonal form as We can parameterize the components of the energy momentum tensor as follows: where is the energy density of the fluid. , , and are the pressures, , , and are the directional equation of state (EoS) parameters of the fluid along -, -, and -axes, respectively, and is the deviation-free EoS parameter of the fluid.

Here, we have parameterized the deviation from isotropy by setting . Also is the skewness parameter, which is a deviation from along - and -axes. The parameters , are not necessarily constants and can be functions of the cosmic time .

Using commoving coordinates, the field equations (2)-(3) for the metric (1) with the help of (5) can be written as where the overhead dot denotes derivative with respect to the cosmic time .

By using the transformation , , and , the field equations (6) can be written as where the overhead dash denotes derivative with respect to .

3. Dark Energy Cosmological Models in General Scalar Tensor Theory of Gravitation

The field equations (7) to (10) are four independent equations with seven unknowns , , , , , , and . Here, we obtain dark energy cosmological model in Nordtvedt’s general scalar tensor theory in a special case proposed by Schwinger [27] in the following form: From (7) to (10), we get From (12) and (13), we get Then, let us construct a physically meaningful model by considering Equations (15) will satisfy (14) provided that the arbitrary constants , , and are related by . For various values of , , and , we will get different cosmological models.

From (9), we get the energy density From (7), we get the EoS parameter From (7) and (8), we get the skewness parameter Then, the metric (1) can be written in the following form: Thus, the metric (19) together with (16) to (18) constitutes LRS Bianchi type-I dark energy cosmological models in general scalar tensor theory of gravitation. Since , , and are arbitrary constants, for different values of , , and , we will get different cosmological models. But in this paper, we will present the following anisotropic as well as isotropic cosmological models.

3.1. Anisotropic Dark Energy Cosmological Model in General Scalar Tensor Theory of Gravitation

Spatially homogeneous cosmological models play an important role in attempts to understand the structure and properties of the space of all cosmological solutions of Einstein field equations. Moreover, from the theoretical point of view, anisotropic universe has a greater generality than isotropic models. Although the present day universe is satisfactorily described by homogeneous and isotropic models given by the Friedmann- Robertson-Walker (FRW) space-time, as we know the universe in a smaller scale is neither homogeneous nor isotropic nor do we expect the universe in its early stages to have these properties. In fact, to get a physically realistic description of the universe, one has to consider inhomogeneous models. In this case, the solutions of Einstein’s field equations become more complicated or may be impossible. Therefore, many theoretical cosmologists are trying to use the spatially homogeneous and anisotropic Bianchi type models instead of inhomogeneous models. These types of space-times present a “middle way” between FRW models and inhomogeneous and anisotropic universes and hence play an important role in modern cosmology.

If , , and , from (15) to (18), we get Then, the model (19) can be written as Thus, the metric (21) together with , , , and , as given above, constitutes LRS Bianchi type-I anisotropic dark energy cosmological model in general scalar tensor theory of gravitation.

3.2. Isotropic Dark Energy Cosmological Model in General Scalar Tensor Theory of Gravitation

If and , from (15) to (18), we get Then, the model (19) can be written as Thus, the metric (23) together with , , , and as given above constitutes LRS Bianchi type-I dark energy cosmological model in the isotropic form in general scalar tensor theory of gravitation.

4. Some Important Features of the Models

4.1. Anisotropic Dark Energy Cosmological Model

The volume element of the model (21) is given by Mean Hubble’s parameter is given by The deceleration parameter is given by Average anisotropy parameter is given by The expansion scalar is given by The shear scalar is defined as , The overall density parameter is given by The tensor of rotation is identically zero, and hence, this universe is nonrotational.

4.2. Isotropic Dark Energy Cosmological Model

The volume element of the model (23) is given by Mean Hubble’s parameter is given by The deceleration parameter is given by Average anisotropy parameter is given by The expansion scalar is given by The shear scalar is defined as , The overall density parameter is given by The tensor of rotation is identically zero, and hence, this universe is nonrotational.

5. Conclusions

In this paper, we have presented a spatially homogeneous LRS Bianchi type-I anisotropic as well as isotropic dark energy cosmological models in the Nordtvedt [1] general scalar tensor theory of gravitation with the help of a special case proposed by Schwinger [27]. For both the models, the spatial volume is constant at and increases exponentially with time. This shows that at the initial epoch, the universe starts with constant volume and expands exponentially approaching infinite volume. It is observed that the model (19) is free from singularities. For both the models, the expansion scalar exhibits the constant value. This shows that the universe expands homogeneously. We observe that the shear scalar and the Hubble parameter are also constants for both the models. From (21), we can observe that the energy density and the EoS parameter will become zero as , while skewness parameter is constant. From (23), we can observe the energy density and skewness parameter will become zero as , while the EoS parameter is constant. For all the previous models, the overall density parameter vanishes as . Also, the deceleration parameter that appears with negative sign implies accelerating expansion of the universe as one can expect for exponential volumetric expansion, which is consistent with the present day observations. By recent observations of type Ia supernovae (SN Ia), Perlmutter et al. [14] and Riess et al. ([13, 28, 29]) proved that the decelerating parameter of the universe is in the range , and the present day universe is undergoing accelerated expansion. For the model (21), we can observe that which indicates that the model is anisotropic and represents the early stages of the universe. Recent experiments show that there is a certain amount of anisotropy in the universe. Hence, anisotropic space-times are important. For the model (23), we can see that , which indicates that the model is isotropic and represents the present stage of the universe. Thus, the anisotropic as well as isotropic dark energy cosmological models presented here are expanding, nonrotating, and accelerating in the standard way.

Acknowledgment

The second author (D. Neelima) is grateful to the Department of Science and Technology (DST), New Delhi, India, for providing INSPIRE fellowship.

References

1. K. Nordtvedt Jr., “Post-Newtonian metric for a general class of scalar-tensor gravitational theories and observational consequences,” The Astrophysical Journal, vol. 161, pp. 1059–1067, 1970.
2. B. M. Barker, “General scalar-tensor theory of gravity with constant G,” Astrophysical Journal, vol. 219, no. 1, pp. 5–11, 1978.
3. V. A. Ruban and A. M. Finkelstein, “Generalization of the Taub-Kazner cosmological metric in the scalar-tensor gravitation theory,” Lettere Al Nuovo Cimento Series 2, vol. 5, no. 3, pp. 289–293, 1972.
4. A. Banerjee and N. O. Santos, “Homogeneous radiation-filled Universe in general scalar tensor theory,” Journal of Physics A, vol. 14, no. 10, article 2829, 1981.
5. A. Banerjee and N. O. Santos, “Homogeneous cosmological model in general scalar-tensor theory,” Physical Review D, vol. 23, pp. 2111–2115, 1981.
6. K. Shanti and V. U. M. Rao, “Conformally flat static space-time in the general scalar-tensor theory of gravitation,” Astrophysics and Space Science, vol. 162, no. 1, pp. 163–165, 1989.
7. K. Shanti and V. U. M. Rao, “Bianchi type-VI0 cosmological model in the general scalar-tensor theory of gravitation,” Astrophysics and Space Science, vol. 172, no. 1, pp. 83–88, 1990.
8. V. U. M. Rao and G. Sreedevi Kumari, “A cosmological model with negative constant deceleration parameter in a general scalar-tensor theory of gravitation,” International Journal of Theoretical Physics, vol. 51, no. 1, pp. 311–315, 2012.
9. V. U. M. Rao, G. S. Kumari, and B. J. M. Rao, “Kaluza-Klein radiating model in a general scalar-tensor theory,” Astrophysics and Space Science, vol. 337, no. 2, pp. 795–798, 2012.
10. V. U. M. Rao, M. Vijaya Santhi, T. Vinutha, and G. Sreedevi Kumari, “LRS Bianchi type-I dark energy cosmological model in Brans-Dicke theory of gravitation,” International Journal of Theoretical Physics, vol. 51, no. 10, pp. 3303–3310, 2012.
11. V. U. M. Rao, K. V. S. Sireesha, and P. Suneetha, “Bianchi type-II, VIII and IX dark energy cosmological models in saez-ballester theory of gravitation,” The African Review of Physics, vol. 7, p. 0054, 2012.
12. V. U. M. Rao, K. V. S. Sireesha, and D. Neelima, “Bianchi type II, VIII, and IX perfect fluid dark energy cosmological models in Saez—Ballester and general theory of gravitation,” ISRN Astronomy and Astrophysics, vol. 2013, Article ID 924834, 11 pages, 2013.
13. A. G. Reiss, A. V. Filippenko, P. Challis, et al., “Observational evidence from supernovae for an accelerating universe and a cosmological constant,” The Astrophysical Journal, vol. 116, article 1009, 1998.
14. S. Perlmutter, G. Aldering, G. Goldhaber, et al., “Measurements of Ω and Λ from 42 High-Redshift Supernovae,” The Astrophysical Journal, vol. 517, no. 2, article 565, 1999.
15. M. Tegmark, M. R. Blanton, M. A. Strauss, et al., “The three-dimensional power spectrum of galaxies from the sloan digital sky survey,” The Astrophysical Journal, vol. 606, no. 2, article 702, 2004.
16. V. Sahni and A. Starobinsky, “Reconstructing dark energy,” International Journal of Modern Physics D, vol. 15, no. 12, pp. 2105–2132, 2006.
17. V. Sahni, A. Shafielooa, and A. Starobinsky, “Two new diagnostics of dark energy,” Physical Review D, vol. 78, Article ID 103502, 11 pages, 2008.
18. J. Kujat, A. M. Linn, R. J. Scherrer, and D. H. Weinberg, “Prospects for determining the equation of state of the dark energy: what can be learned from multiple observables?” The Astrophysical Journal, vol. 572, no. 1, 2002.
19. M. Bartelmann, K. Dolagb, F. Perrotta, et al., “Evolution of dark-matter haloes in a variety of dark-energy cosmologies,” New Astronomy Reviews, vol. 49, no. 2–6, pp. 199–203, 2005.
20. R. Jimenez, “The value of the equation of state of dark energy,” New Astronomy Reviews, vol. 47, no. 8–10, pp. 761–767, 2003.
21. A. Das, S. Gupta, T. D. Saini, and S. Kar, “Cosmology with decaying tachyon matter,” Physical Review D, vol. 72, no. 4, Article ID 043528, 6 pages, 2005.
22. S. Ray, F. Rahaman, U. Mukhopadhyay, and R. Sarkar, “Variable equation of state for generalized dark energy model,” International Journal of Theoretical Physics, vol. 50, no. 9, pp. 2687–2696, 2010.
23. A. K. Yadav and L. Yadav, “Bianchi type III anisotropic dark energy models with constant deceleration parameter,” International Journal of Theoretical Physics, vol. 50, no. 1, pp. 218–227, 2010.
24. S. Kumar, “Some FRW models of accelerating universe with dark energy,” Astrophysics and Space Science, vol. 332, no. 2, pp. 449–454, 2010.
25. A. Pradhan, H. Amirhashchi, and B. Saha, “Bianchi type-I anisotropic dark energy model with constant deceleration parameter,” International Journal of Theoretical Physics, vol. 50, no. 9, pp. 2923–2938, 2011.
26. A. K. Yadav and B. Saha, “LRS Bianchi-I anisotropic cosmological model with dominance of dark energy,” Astrophysics and Space Science, vol. 337, no. 2, pp. 759–765, 2012.
27. J. Schwinger, Particles, Sources and Fields, Addison-Wesley, Reading, Mass, USA, 1970.
28. A. G. Riess, R. P. Kirshner, B. P. Schmidt, et al., “BVRI light curves for 22 type Ia supernovae,” The Astronomical Journal, vol. 117, no. 2, article 707, 1999.
29. A. G. Riess, L.-G. Strolger, J. Tonry, et al., “Type Ia supernova discoveries at $z>1$ from the hubble space telescope: evidence for past deceleration and constraints on dark energy evolution,” The Astrophysical Journal, vol. 607, no. 2, p. 665, 2004.


text:
It is currently 18 Jan 2018, 06:17

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# What is the remainder when (47)(49) is divided by 8?

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 43320

Kudos [?]: 139387 [0], given: 12789

What is the remainder when (47)(49) is divided by 8? [#permalink]

### Show Tags

11 Jun 2015, 02:55
Expert's post
7
This post was
BOOKMARKED
00:00

Difficulty:

15% (low)

Question Stats:

72% (00:39) correct 28% (00:47) wrong based on 160 sessions

### HideShow timer Statistics

What is the remainder when 47*49 is divided by 8?

(A) 1
(B) 3
(C) 4
(D) 5
(E) 7

Kudos for a correct solution.
[Reveal] Spoiler: OA

_________________

Kudos [?]: 139387 [0], given: 12789

Manager
Joined: 12 Nov 2014
Posts: 61

Kudos [?]: 103 [1], given: 3

Re: What is the remainder when (47)(49) is divided by 8? [#permalink]

### Show Tags

11 Jun 2015, 03:38
1
KUDOS
3
This post was
BOOKMARKED
we can make use of the rule :
Remainder of { (a * b)/n}} = Remainder of (a/n) * Remainder of (b/n)

Here
Remainder of { 47 * 49)/8}} = Remainder of (47/8) * Remainder of (49/8) = 7 * 1 = 7

_________________

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Kindly press Kudos if the explanation is clear.
Thank you
Ambarish

Kudos [?]: 103 [1], given: 3

SVP
Joined: 08 Jul 2010
Posts: 1911

Kudos [?]: 2473 [1], given: 51

Location: India
GMAT: INSIGHT
WE: Education (Education)
Re: What is the remainder when (47)(49) is divided by 8? [#permalink]

### Show Tags

11 Jun 2015, 03:40
1
KUDOS
Expert's post
1
This post was
BOOKMARKED
Bunuel wrote:
What is the remainder when 47*49 is divided by 8?

(A) 1
(B) 3
(C) 4
(D) 5
(E) 7

Kudos for a correct solution.

CONCEPT: Remainder Theorem states that Remainders can be calculated for each term multiplied or added together in Numerator and then the operation (multiplication or addition) can be performed on remainders thereafter and finally remainder can be calculated

i.e. Remainder when 47*49 is divided by 8 = Remainder [47/8] * Remainder [49/8]

i.e. Remainder when 47*49 is divided by 8 = 7 * 1 = 7

[Reveal] Spoiler:
E

_________________

Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION

Kudos [?]: 2473 [1], given: 51

Intern
Joined: 04 Apr 2015
Posts: 44

Kudos [?]: 13 [1], given: 31

Re: What is the remainder when (47)(49) is divided by 8? [#permalink]

### Show Tags

11 Jun 2015, 04:07
1
KUDOS
2
This post was
BOOKMARKED
One way to solve this

(8x6-1)(8x6+1)/8
remaninder -1x1=-1 ie 7

Another way

(48-1)(48+1)/8
(48^2-1^2)/8
remainder when 48^2/8 is 0
remainder when -1^2/8 is -1 ie 7

Ans is E

jimmy

Bunuel wrote:
What is the remainder when 47*49 is divided by 8?

(A) 1
(B) 3
(C) 4
(D) 5
(E) 7

Kudos for a correct solution.

Kudos [?]: 13 [1], given: 31

Senior Manager
Status: Math is psycho-logical
Joined: 07 Apr 2014
Posts: 432

Kudos [?]: 145 [0], given: 169

Location: Netherlands
GMAT Date: 02-11-2015
WE: Psychology and Counseling (Other)
Re: What is the remainder when (47)(49) is divided by 8? [#permalink]

### Show Tags

13 Jun 2015, 06:53
I think we could also use the last digits of the numbers like this:

47 * 49 = 7*9 = 63.
63/8 = 7,... and leaves a remainder of 7.

One question on one of the solutions above:
Another way

(48-1)(48+1)/8
(48^2-1^2)/8
remainder when 48^2/8 is 0
remainder when -1^2/8 is -1 ie 7

Why does this tell us that the remainder is 7? I read the post about remainder tricks and tips, but it is not that clear to me...

Kudos [?]: 145 [0], given: 169

SVP
Joined: 08 Jul 2010
Posts: 1911

Kudos [?]: 2473 [0], given: 51

Location: India
GMAT: INSIGHT
WE: Education (Education)
Re: What is the remainder when (47)(49) is divided by 8? [#permalink]

### Show Tags

13 Jun 2015, 07:06
pacifist85 wrote:
I think we could also use the last digits of the numbers like this:

47 * 49 = 7*9 = 63.
63/8 = 7,... and leaves a remainder of 7.

One question on one of the solutions above:
Another way

(48-1)(48+1)/8
(48^2-1^2)/8
remainder when 48^2/8 is 0
remainder when -1^2/8 is -1 ie 7

Why does this tell us that the remainder is 7? I read the post about remainder tricks and tips, but it is not that clear to me...

There are two things for you to note

1) the method of calculating remainder by calculating unit digits and dividing by divisor is fundamentally wrong if the divisors are not 2 or 5 or 10.
Just luckily the answer is matching so please don't use this method because it fits to one question by chance.

2) remainder can be written in two forms either positive or negative. The understanding of positive and negative remainders are as follows

Remainder of 7 when a number divided by 8 means the number has 7 extra else the no. Would have been divisible by 8

Remainder of -1 when a number divided by 8 means the number is short by 1 else the no. Would have been divisible by 8

So technically they are sane and remainder of -1 is same as remainder of 7 for a divisor 8

Similarly, remainder of -2 is same as remainder of 6 for a divisor 8

Similarly, remainder of -4 is same as remainder of 5 for a divisor 9

Similarly, remainder of -5 is same as remainder of 7 for a divisor 12

I hope it clears your doubt!
_________________

Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION

Kudos [?]: 2473 [0], given: 51

Senior Manager
Status: Math is psycho-logical
Joined: 07 Apr 2014
Posts: 432

Kudos [?]: 145 [0], given: 169

Location: Netherlands
GMAT Date: 02-11-2015
WE: Psychology and Counseling (Other)
Re: What is the remainder when (47)(49) is divided by 8? [#permalink]

### Show Tags

13 Jun 2015, 07:27
Hello,

Thank you for the explanations.

OK I can now understand what remainders for a divisor mean.
However, I still do not understand why when you divide -1^2 by 8 you get a remainder of 7. Isn't it the same as 1/8? This is the 0.125 and I cannot see how you end up with 7 as a remainder..

Kudos [?]: 145 [0], given: 169

SVP
Joined: 08 Jul 2010
Posts: 1911

Kudos [?]: 2473 [1], given: 51

Location: India
GMAT: INSIGHT
WE: Education (Education)
Re: What is the remainder when (47)(49) is divided by 8? [#permalink]

### Show Tags

13 Jun 2015, 07:59
1
KUDOS
Expert's post
1
This post was
BOOKMARKED
pacifist85 wrote:
Hello,

Thank you for the explanations.

OK I can now understand what remainders for a divisor mean.
However, I still do not understand why when you divide -1^2 by 8 you get a remainder of 7. Isn't it the same as 1/8? This is the 0.125 and I cannot see how you end up with 7 as a remainder..

Point 1: You are making a mistake again and to identify check the explanation below

Remainder [(47 x 49)/8] = Remainder [(48-1)(48+1)/8] = Remainder [(48^2 - 1^2)/8]
i.e. Remainder [(47 x 49)/8] = Remainder [48^2/8] - Remainder [1^2/8] = Remainder [0] - Remainder [1] = 0-1 = -1

Remainder (-1) means Remainder (7) for Divisor = 8

Please Note: $$-1^2$$ is not same as $$(-1)^2$$

$$-1^2$$ = $$-1$$ which is the case here AND $$(-1)^2$$ = $$+1$$

Point 2: The tradition of saying "Thank you" here on GMAT CLUB is by pressingthe button "+1Kudos"
_________________

Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION

Kudos [?]: 2473 [1], given: 51

Director
Joined: 21 May 2013
Posts: 566

Kudos [?]: 57 [0], given: 510

Re: What is the remainder when (47)(49) is divided by 8? [#permalink]

### Show Tags

13 Jun 2015, 09:08
Bunuel wrote:
What is the remainder when 47*49 is divided by 8?

(A) 1
(B) 3
(C) 4
(D) 5
(E) 7

Kudos for a correct solution.

Since the product can be easily calculated here, 47*49=2303
Simply divide this by 8
Remainder=7

Kudos [?]: 57 [0], given: 510

Intern
Joined: 04 Apr 2015
Posts: 44

Kudos [?]: 13 [1], given: 31

Re: What is the remainder when (47)(49) is divided by 8? [#permalink]

### Show Tags

13 Jun 2015, 19:51
1
KUDOS
Further to what GMATinsight wrote:

1. When you divide an integer by a positive integer N, the possible remainders range from 0 to (N- 1). There are thus N possible remainders.

2. You can add and subtract remainders directly, as long as you correct excess or negative remainders. so to correct the negative remainder, which is unacceptable remainder, add and extra 8. you get 7 as the answer.

Please note here that excess and negative remainders are not acceptable. for more, please refer chapter 10 of manhattan math book 1.

regards

jimmy

GMATinsight wrote:
pacifist85 wrote:
I think we could also use the last digits of the numbers like this:

47 * 49 = 7*9 = 63.
63/8 = 7,... and leaves a remainder of 7.

One question on one of the solutions above:
Another way

(48-1)(48+1)/8
(48^2-1^2)/8
remainder when 48^2/8 is 0
remainder when -1^2/8 is -1 ie 7

Why does this tell us that the remainder is 7? I read the post about remainder tricks and tips, but it is not that clear to me...

There are two things for you to note

1) the method of calculating remainder by calculating unit digits and dividing by divisor is fundamentally wrong if the divisors are not 2 or 5 or 10.
Just luckily the answer is matching so please don't use this method because it fits to one question by chance.

2) remainder can be written in two forms either positive or negative. The understanding of positive and negative remainders are as follows

Remainder of 7 when a number divided by 8 means the number has 7 extra else the no. Would have been divisible by 8

Remainder of -1 when a number divided by 8 means the number is short by 1 else the no. Would have been divisible by 8

So technically they are sane and remainder of -1 is same as remainder of 7 for a divisor 8

Similarly, remainder of -2 is same as remainder of 6 for a divisor 8

Similarly, remainder of -4 is same as remainder of 5 for a divisor 9

Similarly, remainder of -5 is same as remainder of 7 for a divisor 12

I hope it clears your doubt!

Kudos [?]: 13 [1], given: 31

Manager
Joined: 01 Jan 2015
Posts: 56

Kudos [?]: 3 [1], given: 7

Re: What is the remainder when (47)(49) is divided by 8? [#permalink]

### Show Tags

14 Jun 2015, 08:52
1
KUDOS
47/8=R(-1)
49/8=R(+1)
-1*+1=-1
R(-1) OR R(7)

Kudos [?]: 3 [1], given: 7

Math Expert
Joined: 02 Sep 2009
Posts: 43320

Kudos [?]: 139387 [0], given: 12789

Re: What is the remainder when (47)(49) is divided by 8? [#permalink]

### Show Tags

15 Jun 2015, 03:39
Bunuel wrote:
What is the remainder when 47*49 is divided by 8?

(A) 1
(B) 3
(C) 4
(D) 5
(E) 7

Kudos for a correct solution.

MANHATTAN GMAT OFFICIAL SOLUTION:

One way to solve would be to multiply (47)(49), then either divide the result by 8 or repeatedly subtract known multiples of 8 from the result until we are left with a remainder smaller than 8.

An alternative is to rewrite the given product as an equivalent yet easier-to-manipulate product. Note that 47 and 49 are equidistant from 48, a multiple of 8. We can write each of the original factors as terms in the form (a + b) or (a – b).

(47)(49) = (48 + 1)(48 – 1)

Recognizing the “difference of two squares” special product, (a + b)(a – b) = a^2 – b^2, we can quickly manipulate again:

(48 + 1)(48 – 1) = (48^2 – 1^2)

48 is a multiple of 8, and therefore so is 48^2. Thus, (48^2 – 1^2) is 1 less than a multiple of 8. All such numbers (e.g. 7, 15, 23, 31, etc.) have a remainder of 7 when divided by 8.

_________________

Kudos [?]: 139387 [0], given: 12789

Intern
Joined: 09 Jun 2015
Posts: 4

Kudos [?]: 5 [1], given: 0

Re: What is the remainder when (47)(49) is divided by 8? [#permalink]

### Show Tags

15 Jun 2015, 04:24
1
KUDOS
Because we know that 8 divides into 48 evenly, we just need to find the remainder for 47/8:

49 x 47 = (48 x 47) + 47. We know that 8 divides into 48 x 47 evenly because it divides into 48 evenly so you can remove this portion entirely.

Kudos [?]: 5 [1], given: 0

Manager
Joined: 21 Jan 2015
Posts: 149

Kudos [?]: 121 [0], given: 24

Location: India
Concentration: Strategy, Marketing
WE: Marketing (Other)
Re: What is the remainder when (47)(49) is divided by 8? [#permalink]

### Show Tags

15 Jun 2015, 20:06
Bunuel wrote:
What is the remainder when 47*49 is divided by 8?

(A) 1
(B) 3
(C) 4
(D) 5
(E) 7

Ans: E

Solution: 47*49 = (48-1)*(48+1) = 48^2 -1^2
divisor = 8
48/8 = Reminder 0
1/8 reminder 1 put this in the equation = 0-1= -1..as remainder can not be negative so (8-1)= 7 is the remainder.
_________________

--------------------------------------------------------------------
The Mind is Everything, What we Think we Become.
Kudos will encourage many others, like me.
Thanks

Kudos [?]: 121 [0], given: 24

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7863

Kudos [?]: 18468 [1], given: 237

Location: Pune, India
Re: What is the remainder when (47)(49) is divided by 8? [#permalink]

### Show Tags

15 Jun 2015, 22:03
1
KUDOS
Expert's post
pacifist85 wrote:
Hello,

Thank you for the explanations.

OK I can now understand what remainders for a divisor mean.
However, I still do not understand why when you divide -1^2 by 8 you get a remainder of 7. Isn't it the same as 1/8? This is the 0.125 and I cannot see how you end up with 7 as a remainder..

Here is a post on negative remainders that might interest you: http://www.veritasprep.com/blog/2014/03 ... -the-gmat/
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for \$199

Veritas Prep Reviews

Kudos [?]: 18468 [1], given: 237

Non-Human User
Joined: 09 Sep 2013
Posts: 14249

Kudos [?]: 291 [0], given: 0

Re: What is the remainder when (47)(49) is divided by 8? [#permalink]

### Show Tags

05 Dec 2017, 05:29
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Kudos [?]: 291 [0], given: 0

Re: What is the remainder when (47)(49) is divided by 8?   [#permalink] 05 Dec 2017, 05:29
Display posts from previous: Sort by


text:
# Hypergeometric Distribution

-

The hypergeometric distribution is a discrete probability distribution that arises when we try to draw a random sample without replacement from a given population. For example, suppose there are N balls in a bag out of which M are white and the remaining N-M are black. Suppose we choose a sample of size n from the bag. Then the probability that the sample has k white balls can be calculated using the pdf of the hypergeometric distribution.

Example: Suppose that a bag contains 10 balls out of which 6 are white and 4 are black. If we choose 4 balls randomly from the bad then calculate the probability that two of these balls are white.

Solution: Let X denote the number of white balls in our sample.

Given N=10, M=6, n=4, and k=2.

Substituting all this in the above formula we get,

P(X=2) = (62)*(42)/(104) = 15*6/210 = 90/210 = 0.429

Mean and Variance of Hypergeometric Distribution:

Example: In the above example, calculate the mean number of white balls selected and the variance.

Solution: Given N=10, M=6, n=4 and k=2.

Substituting all this in the above formula we get,

Mean = 4*6/10 = 2.4 ,that is, 2.4 white balls will be chosen on average.

Variance = (10*6*4*6)/(100*9) = 1.6

Use of hypergeometric distribution:

1. It is used in the theory of Quality control to calculate the number of lots to be selected for inspection of quality of goods.
2. It is used in sampling theory to calculate the probabilities of selecting particular kinds of samples without replacement.

Binomial approximation to Hypergeometric Distribution:

As the value of N tends toward infinity and the value M/N approaches a finite number then the hypergeometric distribution can be approximated using the pdf of the binomial distribution with parameters n and p=M/N.

Hey 👋

I have always been passionate about statistics and mathematics education.

I created this website to explain mathematical and statistical concepts in the simplest possible manner.

If you've found value from reading my content, feel free to support me in even the smallest way you can.


text:
# [算法竞赛入门经典]Crossword Answers ACM/ICPC World Finals 1994,UVa232

## Description

A crossword puzzle consists of a rectangular grid of black and white squares and two lists of definitions (or descriptions).
One list of definitions is for “words” to be written left to right across white squares in the rows and the other list is for words to be written down white squares in the columns. (A word is a sequence of alphabetic characters.)
To solve a crossword puzzle, one writes the words corre- sponding to the definitions on the white squares of the grid.
The definitions correspond to the rectangular grid by
means of sequential integers on “eligible” white squares.
White squares with black squares immediately to the left or
above them are “eligible.” White squares with no squares ei-
ther immediately to the left or above are also “eligible.” No other squares are numbered. All of the squares on the first row are numbered.
The numbering starts with 1 and continues consecutively across white squares of the first row, then across the eligible white squares of the second row, then across the eligible white squares of the third row and so on across all of the rest of the rows of the puzzle. The picture below illustrates a rectangular crossword puzzle grid with appropriate numbering.
An “across” word for a definition is written on a sequence of white squares in a row starting on a numbered square that does not follow another white square in the same row.
The sequence of white squares for that word goes across the row of the numbered square, ending immediately before the next black square in the row or in the rightmost square of the row.
A “down” word for a definition is written on a sequence of white squares in a column starting on a numbered square that does not follow another white square in the same column.
The sequence of white squares for that word goes down the column of the numbered square, ending immediately before the next black square in the column or in the bottom square of the column.
Every white square in a correctly solved puzzle contains a letter.
You must write a program that takes several solved crossword puzzles as input and outputs the lists of across and down words which constitute the solutions.

## Input

Each puzzle solution in the input starts with a line containing two integers r and c (1 ≤ r ≤ 10 and 1 ≤ c ≤ 10), where r (the first number) is the number of rows in the puzzle and c (the second number) is the number of columns.
The r rows of input which follow each contain c characters (excluding the end-of-line) which describe the solution. Each of those c characters is an alphabetic character which is part of a word or the character ‘*’, which indicates a black square.
The end of input is indicated by a line consisting of the single number

## Output

Output for each puzzle consists of an identifier for the puzzle (puzzle #1:, puzzle #2:, etc.) and the list of across words followed by the list of down words. Words in each list must be output one-per-line in increasing order of the number of their corresponding definitions.
The heading for the list of across words is ‘Across’. The heading for the list of down words is ‘Down’.
In the case where the lists are empty (all squares in the grid are black), the ‘Across’ and ‘Down’ headings should still appear.
Separate output for successive input puzzles by a blank line.

## Sample Input

22
AT
*O
67
AIM*DEN
*ME*ONE
UPON*TO
SO*ERIN
*SA*OR*
IES*DEA
0

## Sample Output

puzzle #1:
Across
1.AT
3.O
Down
1.A
2.TO
puzzle #2:
Across
1.AIM
4.DEN
7.ME
8.ONE
9.UPON
11.TO
12.SO
13.ERIN
15.SA
17.OR
18.IES
19.DEA
Down 1.A
2.IMPOSE
3.MEO
4.DO
5.ENTIRE
6.NEON
9.US
10.NE
14.ROD
16.AS
18.I
20.A
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <cctype>
#define MAX 1005
using namespace std;

char s[MAX][MAX];       // 储存网格
int flag[MAX][MAX];     // 记录起始点
char in[5];

int main()
{
int cnt = 0;
while(~scanf("%[^\n]%*c",in) && (in[0]-'0')){
memset(flag,0,sizeof(flag)); memset(s,0,sizeof(s)); // 清除内容
int r = in[0]-'0',c = in[2]-'0';
for(int i = 0;i < r;i ++) scanf("%s",s[i]);         // 输入网格
int flagCNT = 0;
for(int i = 0;i < r;i ++)                           // 判断起始点
for(int j = 0;j < c;j ++)
if((s[i][j]!='*') &&
(i == 0 || j == 0 || s[i-1][j]=='*' || s[i][j-1]=='*'))
flag[i][j] = ++flagCNT;
if(cnt) printf("\n");
printf("puzzle #%d:\nAcross\n",++cnt);
bool next = true; bool fis = true;
for(int i = 0;i < r;i ++)
for(int j = 0;j < c;j ++){
if(next && s[i][j]!='*'){
fis ? printf("%3d.",flag[i][j]) :
printf("\n%3d.",flag[i][j]);
}
if(s[i][j]!='*') {
putchar(s[i][j]); next = false;
if(j==c-1 && s[i+1][0]!='*') next = true;
}
else { fis = false; next = true; }
}
printf("\nDown\n");
next = true; fis = true;
for(int i = 0;i < r;i ++)
for(int j = 0;j < c;j ++){
if(next && s[i][j]!='*' && flag[i][j] && (i==0 || s[i-1][j]=='*')){
fis ? printf("%3d.",flag[i][j]) :
printf("\n%3d.",flag[i][j]);
}
if(flag[i][j] && (i==0 || s[i-1][j]=='*')){
int down = i;
while(s[down][j]!='*' && down < r) {
putchar(s[down++][j]); next = false;
}
fis = false; next = true;
}
}
printf("\n");
getchar();          // 吞掉输入后的回车
}
return 0;
}
posted @ 2018-09-21 12:37  1Kasshole  阅读(...)  评论(... 编辑 收藏


text:
Solve the following and write the quotient and remainder.8543 ÷ 41 - Mathematics

Sum

Solve the following and write the quotient and remainder.
8543 ÷ 41

Solution

208

41)overline8543
-82
034
- 00
343
-    328
15
Quotient = 208
Remainder = 15

Concept: Concept of Division
Is there an error in this question or solution?

APPEARS IN

Balbharati Mathematics 5th Standard Maharashtra State Board
Chapter 4 Multiplication and Division
Problem Set 15 | Q 1.4 | Page 19


text:
# Steady solution of ODE

1. May 17, 2009

### KFC

I have a set of ODE of the following form

$$\begin{cases} \displaystype{\frac{dx(t)}{dt}} = F(x, y, z; \delta e^{i\omega t}, \Delta e^{-i\omega t})\\[4mm] \displaystype{\frac{dy(t)}{dt}} = G(x, y, z; \delta e^{i\omega t}, \Delta e^{-i\omega t})\\[4mm] \displaystype{\frac{dz(t)}{dt}} = H(z, y, z; \delta e^{i\omega t}, \Delta e^{-i\omega t}) \end{cases}$$

where $$\delta, \Delta, \omega$$ are constants.

If only concern about the steady solution, can I conclue that the solution must be time-independent?

The equations is quite complicate so one must consider the small pertubration ($$\delta, \Delta$$ are very small number. So when $$\delta \to 0$$ and $$\Delta \to 0$$, the steady solutions are $$x^{(0)}, y^{(0)}, z^{(0)}$$. Take x as example, the first order corrections of the steady solution is of the form

$$x = x^{(0)} + y^{(1)} \delta e^{i\omega t} + z^{(1)} \Delta e^{-i\omega t}$$

I wonder why the above steady solution is time dependent? In this sense, can I conclude that $$y^{(1)}, z^{(1)}$$ are time independent?

2. May 18, 2009

### HallsofIvy

A "steady state solution" is by definition a solution constant in time. Yes, it is independent of time.

Again, the definition of "steady state solution" is that it is time independent!

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook


text:
# Homework Help: L'Hopital's Rule

1. May 15, 2010

### kd001

1. The problem statement, all variables and given/known data

Evalutate the limit, as x approaches 0, log(coshx)/x2

2. Relevant equations

L'Hopital's rule

3. The attempt at a solution

I can get as far as (sinhx.cosh-1x)/2x by differentiating the top and bottom separately. I'm not sure how to do the next differentiation.

Thanks for any help.

2. May 15, 2010

### hgfalling

Be careful about writing cosh-1 x; that means something quite different from what you mean.

So you have quite rightly differentiated top and bottom, leaving you with:

$$\frac{\frac{sinh x}{cosh x}}{2x}$$

Fix up that fraction so that it is in a more usual form (ie, move the cosh x to the denominator), and then apply L'Hopital's rule again.

3. May 15, 2010

### Staff: Mentor

By cosh-1x, you really mean 1/cosh x, but this notation suggests the inverse cosh function. Don't write cosh-1x if you mean 1/cosh x.

Your numerator is sinh(x)/cosh(x) = tanh(x). Since the numerator approaches zero, and 2x approaches zero, use L'Hopital's Rule another time.

4. May 15, 2010

### kd001

Here's what I've done. I've treated the numerator as being sinhx/coshx and differentiated it to get (1/cosh2x). I've treated denominator as being 2x and differentiated it to get 2. Therefore I end up with (1/cosh2x)/2 which gives 1/2 as x approaches 0. Is this the correct answer?

5. May 15, 2010

Yup.


text:
1. ## Re: NEW QUESTION about Fitch

Originally Posted by rai009
Given ¬(p∧q), use the Fitch system to prove (¬p∨¬q)
This is a little trickier and requires the use of negation elimination, which is properly called double-negation elimination. It is easier to derive (¬p ∨ ¬q) if you can derive (A ∨¬A) for any formula A. But assuming this is not done, we proceed as follows.

Assume ¬(¬p ∨ ¬q). From this in the end we will derive a contradiction and thus ¬¬(¬p ∨ ¬q), and then use double-negation elimination to derive (¬p ∨ ¬q). Now assume p and q. Then p ∧ q, which contradicts ¬(p ∧ q). Therefore, ¬p, which implies (¬p ∨ ¬q). This in turn contradicts ¬(¬p ∨ ¬q), so we conclude ¬q. Then we repeat by deriving (¬p ∨ ¬q), which again contradicts ¬(¬p ∨ ¬q). Thus, ¬¬(¬p ∨ ¬q) and (¬p ∨ ¬q).

2. ## Re: NEW QUESTION about Fitch

Thanks for the help but i can't show contradiction as i mentioned before my program doesn't allow contradiction introduction.....how do i do it without using contradiction introduction rule?

3. ## Re: NEW QUESTION about Fitch

You can read the word "contradiction" informally. In the description in post #16, contradiction is derived from A and ~A for some formula A, and this contradiction is immediately used to close one of the open assumption and derive the negation of that assumption. In your case, you derive the negation of the assumption directly from A and ~A.

Page 2 of 2 First 12


text:
Back to Course

0% Complete
0/0 Steps

• ## Do you like this content?

Lesson 7, Topic 4
In Progress

# Metre Bridge

Lesson Progress
0% Complete

The metre bridge uses principle of Wheatstone bridge to measure an unknown resistance.

It consists of a uniform wire AB 100cm long; the ends are soldered to two copper or brass strips with the wire stretched over a calibrated scale. The third strip or plate is placed parallel to the wire.

The unknown resistance r1 and known resistance R2, are connected across the gaps. By closing the key, the jockey is moved along the wire until a balanced point is obtained and the galvanometer reads zero i.e. show no deflection.

Using Wheatstone bridge relation,

$$\frac{R_1}{R_2} = \frac{R_3}{R_4}$$

R3 = L1,  R4 = L2 , then,

$$\frac{R_1}{R_2} = \frac{L_1}{L_2}$$

R1 =$$\frac{L_1}{L_2} \scriptsize \; \times \; R_2$$

L1  and L2 can be read directly from the calibrated scale attached to the wire

Precaution

1. Jockey should not be dragged but tapped lightly on the wire so as not to change the cross-sectional area of the wire.
2. All connections must be tight to avoid leakage of current
3. Switch is made open when readings are not beings taken so as not to run down battery
4. The battery key should be closed before the jockey is tapped on the wire for maximum result/ completely of circuit.

error:


text:
In the last article we looked at how Komposition automatically classifies moving and still segments in imported video media, how I went from ineffective testing by eye to finding curious bugs using property-based testing (PBT). If you haven’t read it, or its preceding posts, I encourage you to check them out first:

This is the final case study in the “Property-Based Testing in a Screencast Editor” series. It covers property-based integration testing and its value during aggressive refactoring work within Komposition.

## A History of Two Stacks

In Komposition, a project’s state is represented using an in-memory data structure. It contains the hierarchical timeline, the focus, import and render settings, project storage file paths, and more. To let users navigate backwards and forwards in their history of project edits, for example when they have made a mistake, Komposition supplies undo and redo commands.

The undo/redo history was previously implemented as a data structure recording project states, compromised of:

• a current state variable
• a stack of previous states
• a stack of possible future states

The undo/redo history data structure held entire project state values. Each undoable and redoable user action created a new state value. Let’s look a bit closer at how this worked.

### Performing Actions

When a user performed an undoable/redoable action, the undo/redo history would:

• push the previous state onto the undo stack
• perform the action and replace the current state
• discard all states in the redo stack

This can be visualized as in the following diagram, where the state d is being replaced with a new state h, and d being pushed onto the undo stack. The undo/redo history to the left of the dividing line is the original, and the one to the right is the resulting history.

Again, note that performing new actions discarded all states in the redo stack.

### Undoing Actions

When the user chose to undo an action, the undo/redo history would:

• pop the undo stack and use that state as the current state
• push the previous state onto the redo stack

The following diagram shows how undoing the last performed action’s resulting state, d, pushes d onto the redo stack, and pops c from the undo stack to use that as the current state.

### Redoing Actions

When the user chose to redo an action, the undo/redo history would:

• pop the redo stack and use that state as the current state
• push the previous state onto the undo stack

The last diagram shows how redoing, recovering a previously undone state, pops g from the redo stack to use that as the current state, and pushes the previous state d onto the undo stack.

Note that not all user actions in Komposition are undoable/redoable. Actions like navigating the focus or zooming are not recorded in the history.

### Dealing With Performance Problems

While the “two stacks of states” algorithm was easy to understand and implement, it failed to meet my non-functional requirements. A screencast project compromised of hundreds or thousands of small edits would consume gigabytes of disk space when stored, take tens of seconds to load from disk, and consume many gigabytes of RAM when in memory.

Now, you might think that my implementation was incredibly naive, and that the performance problems could be fixed with careful profiling and optimization. And you’d probably be right! I did consider going down that route, optimizing the code, time-windowing edits to compact history on the fly, and capping the history at some fixed size. Those would all be interesting pursuits, but in the end I decided to try something else.

## Refactoring with Property-Based Integration Tests

Instead of optimizing the current stack-based implementation, I decided to implement the undo/redo history in terms of inverse actions. In this model, actions not only modify the project state, they also return another action, its inverse, that reverses the effects of the original action. Instead of recording a new project state data structure for each edit, the history only records descriptions of the actions themselves.

I realized early that introducing the new undo/redo history implementation in Komposition was not going to be a small task. It would touch the majority of command implementation code, large parts of the main application logic, and the project binary serialization format. What it wouldn’t affect, though, was the module describing user commands in abstract.

To provide a safety net for the refactoring, I decided to cover the undo/redo functionality with tests. As the user commands would stay the same throughout my modifications, I chose to test at that level, which can be characterized as integration-level testing. The tests run Komposition, including its top-level application control flow, but with the user interface and some other effects stubbed out. Making your application testable at this level is hard work, but the payoff can be huge.

With Komposition featuring close to twenty types of user commands, combined with a complex hierarchical timeline and navigation model, the combinatory explosion of possible states was daunting. Relying on example-based tests to safeguard my work was not satisfactory. While PBT couldn’t cover the entire state space either, I was confident it would improve my chances of finding actual bugs.

## Undo/Redo Tests

Before I began refactoring, I added tests for the inverse property of undoable/redoable actions. The first test focuses on undoing actions, and is structured as follows:

1. Generate an initial project and application state
2. Generate a sequence of undoable/redoable commands (wrapped in events)
3. Run the application with the initial state and the generated events
4. Run an undo command for each original command
5. Assert that final timeline is equal to the initial timeline

Let’s look at the Haskell Hedgehog property test:

hprop_undo_actions_are_undoable = property $do -- 1. Generate initial timeline and focus timelineAndFocus <- forAllWith showTimelineAndFocus$
Gen.timelineWithFocus (Range.linear 0 10) Gen.parallel

-- ... and initial application state
initialState <- forAll (initializeState timelineAndFocus)

-- 2. Generate a sequence of undoable/redoable commands
events <- forAll $Gen.list (Range.exponential 1 100) genUndoableTimelineEvent -- 3. Run 'events' on the original state beforeUndos <- runTimelineStubbedWithExit events initialState -- 4. Run as many undo commands as undoable commands afterUndos <- runTimelineStubbedWithExit (undoEvent <$ events) beforeUndos

-- 5. That should result in a timeline equal to the one we started
-- with
timelineToTree (initialState ^. currentTimeline)
=== timelineToTree (afterUndos ^. currentTimeline)

The second test, focusing on redoing actions, is structured very similarly to the previous test:

1. Generate an initial project and application state
2. Generate a sequence of undoable commands (wrapped in events)
3. Run the application with the initial state and the generated events
4. Run an undo commands for each original command
5. Run an redo commands for each original command
6. Assert that final timeline is equal to the timeline before undoing actions

The test code is also very similar:

hprop_undo_actions_are_redoable = property $do -- 1. Generate the initial timeline and focus timelineAndFocus <- forAllWith showTimelineAndFocus$
Gen.timelineWithFocus (Range.linear 0 10) Gen.parallel

-- ... and the initial application state
initialState <- forAll (initializeState timelineAndFocus)

-- 2. Generate a sequence of undoable/redoable commands
events <- forAll $Gen.list (Range.exponential 1 100) genUndoableTimelineEvent -- 3. Run 'events' on the original state beforeUndos <- runTimelineStubbedWithExit events initialState -- 4. Run undo commands corresponding to all original commands afterRedos <- runTimelineStubbedWithExit (undoEvent <$ events) beforeUndos
-- 5. Run redo commands corresponding to all original commands
>>= runTimelineStubbedWithExit (redoEvent <\$ events)

-- 6. That should result in a timeline equal to the one we had
-- before undoing actions
timelineToTree (beforeUndos ^. currentTimeline)
=== timelineToTree (afterRedos ^. currentTimeline)

Note that these tests only assert on the equality of timelines, not entire project states, as undoable commands only operate on the timeline.

### All Tests Passing, Everything Works

The undo/redo tests were written and run on the original stack-based implementation, kept around during a refactoring that took me two weeks of hacking during late nights and weekends, and finally run and passing with the new implementation based on inverse actions. Except for a few minimal adjustments to data types, these tests stayed untouched during the entire process.

The confidence I had when refactoring felt like a super power. Two simple property tests made the undertaking possible. They found numerous bugs, including:

• Off-by-one index errors in actions modifying the timeline
• Inconsistent timeline focus:
• focus was incorrectly restored on undoing an action
• focus was outside of the timeline bounds
• Non-inverse actions:
• actions returning incorrectly constructed inverses
• the inverse of splitting a sequence is joining sequences, and joining them back up didn’t always work

After all tests passed, I ran the application with its GUI, edited a screencast project, and it all worked flawlessly. It’s almost too good to be true, right?

Property testing is not a silver bullet, and there might still be bugs lurking in my undo/redo history implementation. The tests I run are never going to be exhaustive and my generators might be flawed. That being said, they gave me a confidence in refactoring that I’ve never had before. Or maybe I just haven’t hit that disastrous edge case yet?

## Why Test With Properties?

This was the last case study in the “Property-Based Testing in a Screencast Editor” series. I’ve had a great time writing these articles and giving talks on the subject. Before I wrap up, I’ll summarize my thoughts on PBT in general and my experience with it in Komposition.

Property-based testing is not only for pure functions; you can use it to test effectful actions. It is not only for unit testing; you can write integration tests using properties. It’s not only for functional programming languages; there are good frameworks for most popular programming languages.

Properties describe the general behavior of the system under test, and verify its correctness using a variety of inputs. Not only is this an effective way of finding errors, it’s also a concise way of documenting the system.

The iterative process in property-based testing, in my experience, comes down to the following steps:

1. Think about the specification of your system under test
2. Think about how generators and tests should work
3. Write or modify generators, tests, and implementation code, based on steps 1 and 2
4. Get minimal examples of failing tests
5. Repeat

Using PBT within Komposition has made it possible to confidently refactor large parts of the application. It has found errors in my thinking, my generators, my tests, and in my implementation code. Testing video scene classification went from a time consuming, repetitive, and manual verification process to a fast, effective, and automated task.

In short, it’s been a joy, and I look forward to continue using PBT in my work and in my own projects. I hope I’ve convinced you of its value, and inspired you to try it out, no matter what kind of project you’re working on and what programming language you are using. Involve your colleagues, practice writing property tests together, and enjoy finding complicated bugs before your users do!


text:
# Lease 1 answer below »

Metro Industries manufacture s equipment for sale or lease. Federated Corp (lessee) leased a tooling machine from Metro Industries (lessor) on 12/31/2010, for a three-year period ending December 31, 2013. On 12/31/2013, the equipment will revert back to the lessor. The equipment has a fair value or “selling price” of $128,872 and it costs the lessor$90,000. The lease agreement specified annual payments (note: you need to figure out the annual payment with the given info) to be paid with the first payment at the inception of the lease, and each December 31 through 2012. The machine’s useful life is four years. The residual value at the end of 12/31/2013 is expected to be $35,000. The lessee does guarantee the residual value. The lessee has incremental borrowing rate of 14% and is aware of the lessor’s implicit rate of return of 12%. Assume the fair value of the equipment at the end of lease term (12/31/2013) will be$10,000. Do the followings: 1) Show how the lessor determines the lease payment; 2) Prepare the appropriate entries for both the lessee and the lessor from the inception of the lease through the return of the equipment back to the lessor.


text:
# scipy.sparse.linalg.SuperLU.perm_r¶

SuperLU.perm_r

Permutation Pr represented as an array of indices.

The row permutation matrix can be reconstructed via:

>>> Pr = np.zeros((n, n))
>>> Pr[perm_r, np.arange(n)] = 1