text:
# American Institute of Mathematical Sciences

October  2011, 16(3): 739-766. doi: 10.3934/dcdsb.2011.16.739

## Existence of radial stationary solutions for a system in combustion theory

 1 INRA, Equipe BIOSP, Centre de Recherche d'Avignon, Domaine Saint Paul, Site Agroparc, 84914 Avignon cedex 9, France 2 Departamento de Ingeniería Matemática and CMM (UMI 2807 CNRS), Universidad de Chile, Blanco Encalada 2120 - 5 Piso, Santiago, Chile

Received  June 2010 Revised  February 2011 Published  June 2011

In this paper, we construct radially symmetric solutions of a nonlinear non-cooperative elliptic system derived from a model for flame balls with radiation losses. This model is based on a one step kinetic reaction and our system is obtained by approximating the standard Arrehnius law by an ignition nonlinearity, and by simplifying the term that models radiation. We prove the existence of 2 solutions using degree theory.
Citation: Jérôme Coville, Juan Dávila. Existence of radial stationary solutions for a system in combustion theory. Discrete & Continuous Dynamical Systems - B, 2011, 16 (3) : 739-766. doi: 10.3934/dcdsb.2011.16.739
##### References:
 [1] R. Aris, "The Mathematical Theory of Diffusion and Reaction in Permeable Catalysts,", Clarendon Press, (1975).   Google Scholar [2] J. B. van den Berg, V. Guyonne and J. Hulshof, Flame balls for a free boundary combustion model with radiative transfer,, SIAM J. Appl. Math., 67 (2006), 116.  doi: 10.1137/050636516.  Google Scholar [3] H. Berestycki and B. Larrouturou, Quelques aspects mathématiques de la propagation des flammes prémélangées,, Nonlinear Partial Differential Equations and Their Applications, (1991), 1987.   Google Scholar [4] H. Berestycki, B. Larrouturou and J.-M. Roquejoffre, Mathematical investigation of the cold boundary difficulty in flame propagation theory,, Dynamical Issues in Combustion Theory (Minneapolis, (1989), 37.   Google Scholar [5] H. Berestycki and P.-L. Lions, Nonlinear scalar field equations. I. Existence of a ground state,, Arch. Rational Mech. Anal., 82 (1983), 313.  doi: 10.1007/BF00250555.  Google Scholar [6] H. Berestycki, B. Nicolaenko and B. Scheurer, Traveling wave solutions to combustion models and their singular limits,, SIAM J. Math. Anal., 16 (1985), 1207.  doi: 10.1137/0516088.  Google Scholar [7] H. Bockhorn, J. Fröhlich and K. Schneider, An adaptive two-dimensional wavelet-vaguelette algorithm for the computation of flame balls,, Combustion Theory and Modelling, 3 (1999), 177.  doi: 10.1088/1364-7830/3/1/010.  Google Scholar [8] I. Brailovsky and G. I. Sivashinsky, On stationay and travelling flameballs,, Combustion and Flame, 110 (1997), 524.  doi: 10.1016/S0010-2180(97)00001-1.  Google Scholar [9] J. Brindley, N. A. Jivraj, J. H. Merkin and S. K. Scott, Stationary-state solutions for coupled reaction-diffusion and temperature-conduction equations. II. Spherical geometry with Dirichlet boundary conditions,, Proc. Roy. Soc. London Ser. A, 430 (1990), 479.  doi: 10.1098/rspa.1990.0102.  Google Scholar [10] J. Buckmaster, G. Joulin and P. Ronney, The structure and stability of nonadiabatic flame balls,, Combustion and Flame, 79 (1990), 381.  doi: 10.1016/0010-2180(90)90147-J.  Google Scholar [11] J. Buckmaster, G. Joulin and P. D. Ronney, The structure and stability of nonadiabatic flame balls: II. Effects of far-field losses,, Combustion and Flame, 84 (1991), 411.  doi: 10.1016/0010-2180(91)90015-4.  Google Scholar [12] C. J. Lee and J. Buckmaster, The structure and stability of flame balls: a near-equidiffusional flame analysis,, SIAM J. Appl. Math., 51 (1991), 1315.  doi: 10.1137/0151066.  Google Scholar [13] R. S. Cantrell and C. Cosner, "Spatial Ecology Via Reaction-Diffusion Equations,", Wiley Series in Mathematical and Computational Biology, (2003).   Google Scholar [14] P. Clément and G. Sweers, Existence and multiplicity results for a semilinear elliptic eigenvalue problem,, Ann. Scuola Norm. Sup. Pisa Cl. Sci., 14 (1987), 97.   Google Scholar [15] Y. Du and Y. Lou, Proof of a conjecture for the perturbed Gelfand equation from combustion theory,, J. Differential Equations, 173 (2001), 213.  doi: 10.1006/jdeq.2000.3932.  Google Scholar [16] V. Giovangigli, Nonadiabatic plane laminar flames and their singular limits,, SIAM J. Math. Anal., 21 (1990), 1305.  doi: 10.1137/0521072.  Google Scholar [17] L. Glangetas and J.-M. Roquejoffre, Rigorous derivation of the dispersion relation in a combustion model with heat losses,, Preprint Publications du Laboratoire d'Analyse Numérique R95032, (1995).   Google Scholar [18] V. Guyonne and L. Lorenzi, Instability in a flame ball problem,, Discrete Contin. Dyn. Syst. Ser. B, 7 (2007), 315.   Google Scholar [19] V. Guyonne and P. Noble, On a model of flame ball with radiative transfer,, SIAM J. Appl. Math., 67 (2007), 854.  doi: 10.1137/060659612.  Google Scholar [20] V. Hutson, J. López-Gómez, K. Mischaikow and G. Vickers, Limit behaviour for a competing species problem with diffusion,, Dynamical Systems and Applications, (1995), 343.   Google Scholar [21] L. Kagan and G. I. Sivashinsky, Self-fragmentation of nonadiabatic cellular flames,, Combust. Flame, 108 (1997), 220.  doi: 10.1016/S0010-2180(96)00108-3.  Google Scholar [22] L. Kagan, S. Minaev and G. I. Sivashinsky, On self-drifting flame balls,, Math. Comput. Simulation, 65 (2004), 511.  doi: 10.1016/j.matcom.2004.01.013.  Google Scholar [23] A. K. Kapila, B. J. Matkowsky and J. Vega, Reactive-diffusive system with Arrhenius kinetics: peculiarities of the spherical geometry,, SIAM J. Appl. Math., 38 (1980), 382.  doi: 10.1137/0138032.  Google Scholar [24] M. K. Kwong, Uniqueness of positive solutions of $\Delta u-u+u^p=0$ in $R^n$,, Arch. Rational Mech. Anal., 105 (1989), 243.  doi: 10.1007/BF00251502.  Google Scholar [25] C. Lederman, J.-M. Roquejoffre and N. Wolanski, Mathematical justification of a nonlinear integrodifferential equation for the propagation of spherical flames,, Ann. Mat. Pura Appl., 183 (2004), 173.  doi: 10.1007/s10231-003-0085-1.  Google Scholar [26] C. J. Lee and J. Buckmaster, The structure and stability of flame balls: a near-equidiffusional flame analysis,, SIAM J. Appl. Math., 51 (1991), 1315.  doi: 10.1137/0151066.  Google Scholar [27] B. Lewis and G. von Elbe, "Combustion, Flames and Explosion of Gases,", 3rd ed., (1987).   Google Scholar [28] M. Marion, Qualitative properties of a nonlinear system for laminar flames without ignition temperature,, Nonlinear Anal., 9 (1985), 1269.  doi: 10.1016/0362-546X(85)90035-5.  Google Scholar [29] S. Minaev, L. Kagan, G. Joulin and G. I. Sivashinsky, On self-drifting flame balls,, Combust. Theory Model., 5 (2001), 609.  doi: 10.1088/1364-7830/5/4/306.  Google Scholar [30] P. Ronney, Near-Limit Flame Structures at Low Lewis Number,, Combustion and Flame, 82 (1990), 1.  doi: 10.1016/0010-2180(90)90074-2.  Google Scholar [31] L. Roques, Study of the premixed flame model with heat losses. The existence of two solutions,, European J. Appl. Math., 16 (2005), 741.  doi: 10.1017/S0956792505006431.  Google Scholar [32] G. Sagon, Steady fronts solutions of semilinear elliptic equations in exterior domains arising in flame propagation,, Ann. Mat. Pura Appl., 185 (2006), 273.   Google Scholar [33] A. A. Shah, R. W. Thatcher and J. W. Dold, Stability of a spherical flame ball in a porous medium,, Combustion Theory and Modelling, 4 (2000), 511.  doi: 10.1088/1364-7830/4/4/308.  Google Scholar [34] Y. B. Zeldovich, A theory of the limit of slow flame propagation,, Zh. Prikl. Mekh. i Tekhn. Fiz., 1 (1941), 159.   Google Scholar

show all references

##### References:
 [1] R. Aris, "The Mathematical Theory of Diffusion and Reaction in Permeable Catalysts,", Clarendon Press, (1975).   Google Scholar [2] J. B. van den Berg, V. Guyonne and J. Hulshof, Flame balls for a free boundary combustion model with radiative transfer,, SIAM J. Appl. Math., 67 (2006), 116.  doi: 10.1137/050636516.  Google Scholar [3] H. Berestycki and B. Larrouturou, Quelques aspects mathématiques de la propagation des flammes prémélangées,, Nonlinear Partial Differential Equations and Their Applications, (1991), 1987.   Google Scholar [4] H. Berestycki, B. Larrouturou and J.-M. Roquejoffre, Mathematical investigation of the cold boundary difficulty in flame propagation theory,, Dynamical Issues in Combustion Theory (Minneapolis, (1989), 37.   Google Scholar [5] H. Berestycki and P.-L. Lions, Nonlinear scalar field equations. I. Existence of a ground state,, Arch. Rational Mech. Anal., 82 (1983), 313.  doi: 10.1007/BF00250555.  Google Scholar [6] H. Berestycki, B. Nicolaenko and B. Scheurer, Traveling wave solutions to combustion models and their singular limits,, SIAM J. Math. Anal., 16 (1985), 1207.  doi: 10.1137/0516088.  Google Scholar [7] H. Bockhorn, J. Fröhlich and K. Schneider, An adaptive two-dimensional wavelet-vaguelette algorithm for the computation of flame balls,, Combustion Theory and Modelling, 3 (1999), 177.  doi: 10.1088/1364-7830/3/1/010.  Google Scholar [8] I. Brailovsky and G. I. Sivashinsky, On stationay and travelling flameballs,, Combustion and Flame, 110 (1997), 524.  doi: 10.1016/S0010-2180(97)00001-1.  Google Scholar [9] J. Brindley, N. A. Jivraj, J. H. Merkin and S. K. Scott, Stationary-state solutions for coupled reaction-diffusion and temperature-conduction equations. II. Spherical geometry with Dirichlet boundary conditions,, Proc. Roy. Soc. London Ser. A, 430 (1990), 479.  doi: 10.1098/rspa.1990.0102.  Google Scholar [10] J. Buckmaster, G. Joulin and P. Ronney, The structure and stability of nonadiabatic flame balls,, Combustion and Flame, 79 (1990), 381.  doi: 10.1016/0010-2180(90)90147-J.  Google Scholar [11] J. Buckmaster, G. Joulin and P. D. Ronney, The structure and stability of nonadiabatic flame balls: II. Effects of far-field losses,, Combustion and Flame, 84 (1991), 411.  doi: 10.1016/0010-2180(91)90015-4.  Google Scholar [12] C. J. Lee and J. Buckmaster, The structure and stability of flame balls: a near-equidiffusional flame analysis,, SIAM J. Appl. Math., 51 (1991), 1315.  doi: 10.1137/0151066.  Google Scholar [13] R. S. Cantrell and C. Cosner, "Spatial Ecology Via Reaction-Diffusion Equations,", Wiley Series in Mathematical and Computational Biology, (2003).   Google Scholar [14] P. Clément and G. Sweers, Existence and multiplicity results for a semilinear elliptic eigenvalue problem,, Ann. Scuola Norm. Sup. Pisa Cl. Sci., 14 (1987), 97.   Google Scholar [15] Y. Du and Y. Lou, Proof of a conjecture for the perturbed Gelfand equation from combustion theory,, J. Differential Equations, 173 (2001), 213.  doi: 10.1006/jdeq.2000.3932.  Google Scholar [16] V. Giovangigli, Nonadiabatic plane laminar flames and their singular limits,, SIAM J. Math. Anal., 21 (1990), 1305.  doi: 10.1137/0521072.  Google Scholar [17] L. Glangetas and J.-M. Roquejoffre, Rigorous derivation of the dispersion relation in a combustion model with heat losses,, Preprint Publications du Laboratoire d'Analyse Numérique R95032, (1995).   Google Scholar [18] V. Guyonne and L. Lorenzi, Instability in a flame ball problem,, Discrete Contin. Dyn. Syst. Ser. B, 7 (2007), 315.   Google Scholar [19] V. Guyonne and P. Noble, On a model of flame ball with radiative transfer,, SIAM J. Appl. Math., 67 (2007), 854.  doi: 10.1137/060659612.  Google Scholar [20] V. Hutson, J. López-Gómez, K. Mischaikow and G. Vickers, Limit behaviour for a competing species problem with diffusion,, Dynamical Systems and Applications, (1995), 343.   Google Scholar [21] L. Kagan and G. I. Sivashinsky, Self-fragmentation of nonadiabatic cellular flames,, Combust. Flame, 108 (1997), 220.  doi: 10.1016/S0010-2180(96)00108-3.  Google Scholar [22] L. Kagan, S. Minaev and G. I. Sivashinsky, On self-drifting flame balls,, Math. Comput. Simulation, 65 (2004), 511.  doi: 10.1016/j.matcom.2004.01.013.  Google Scholar [23] A. K. Kapila, B. J. Matkowsky and J. Vega, Reactive-diffusive system with Arrhenius kinetics: peculiarities of the spherical geometry,, SIAM J. Appl. Math., 38 (1980), 382.  doi: 10.1137/0138032.  Google Scholar [24] M. K. Kwong, Uniqueness of positive solutions of $\Delta u-u+u^p=0$ in $R^n$,, Arch. Rational Mech. Anal., 105 (1989), 243.  doi: 10.1007/BF00251502.  Google Scholar [25] C. Lederman, J.-M. Roquejoffre and N. Wolanski, Mathematical justification of a nonlinear integrodifferential equation for the propagation of spherical flames,, Ann. Mat. Pura Appl., 183 (2004), 173.  doi: 10.1007/s10231-003-0085-1.  Google Scholar [26] C. J. Lee and J. Buckmaster, The structure and stability of flame balls: a near-equidiffusional flame analysis,, SIAM J. Appl. Math., 51 (1991), 1315.  doi: 10.1137/0151066.  Google Scholar [27] B. Lewis and G. von Elbe, "Combustion, Flames and Explosion of Gases,", 3rd ed., (1987).   Google Scholar [28] M. Marion, Qualitative properties of a nonlinear system for laminar flames without ignition temperature,, Nonlinear Anal., 9 (1985), 1269.  doi: 10.1016/0362-546X(85)90035-5.  Google Scholar [29] S. Minaev, L. Kagan, G. Joulin and G. I. Sivashinsky, On self-drifting flame balls,, Combust. Theory Model., 5 (2001), 609.  doi: 10.1088/1364-7830/5/4/306.  Google Scholar [30] P. Ronney, Near-Limit Flame Structures at Low Lewis Number,, Combustion and Flame, 82 (1990), 1.  doi: 10.1016/0010-2180(90)90074-2.  Google Scholar [31] L. Roques, Study of the premixed flame model with heat losses. The existence of two solutions,, European J. Appl. Math., 16 (2005), 741.  doi: 10.1017/S0956792505006431.  Google Scholar [32] G. Sagon, Steady fronts solutions of semilinear elliptic equations in exterior domains arising in flame propagation,, Ann. Mat. Pura Appl., 185 (2006), 273.   Google Scholar [33] A. A. Shah, R. W. Thatcher and J. W. Dold, Stability of a spherical flame ball in a porous medium,, Combustion Theory and Modelling, 4 (2000), 511.  doi: 10.1088/1364-7830/4/4/308.  Google Scholar [34] Y. B. Zeldovich, A theory of the limit of slow flame propagation,, Zh. Prikl. Mekh. i Tekhn. Fiz., 1 (1941), 159.   Google Scholar
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# Use of Windows resource scheduled_task

I am trying to use the windows resource scheduled_task. I don't want to create the task in the default location but instead I want to put it in it's own library name. No matter what I try, I get the following error:

Error: Failed to call #<Win32::TaskScheduler::COM::TaskScheduler:0x000000093d2580>::NewWorkItem with HRESULT: -2147024893.:  The system cannot find the path specified.
Error: /Stage[main]/Puppet_agent::Init_windows/Scheduled_task[\Task Scheduler\Task Scheduler Library\Puppet\PuppetKickoff]/ensure: change from absent to present failed: Failed to call #<Win32::TaskScheduler::COM::TaskScheduler:0x000000093d2580>::NewWorkItem with HRESULT: -2147024893.:  The system cannot find the path specified.

Here is my code:

scheduled_task { '\Task Scheduler\Task Scheduler Library\Puppet\PuppetKickoff':
ensure    => present,
enabled   => true,
command   => 'C:\\Windows\\System32\\WindowsPowerShell\\v1.0\\powershell.exe',
arguments => ' E:\\Program Files\\PuppetLabs\\bin\\run_puppet_shell.bat -Verb runas',
trigger   => {
schedule   => daily,
every      => 1,            # Specifies every other day. Defaults to 1 (every day).
start_date => '2018-08-15', # Defaults to 'today'

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# If P + |k| > |P| + k then which of the following is true of Inqual

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Re: If P + |k| > |P| + k then which of the following is true of Inqual  [#permalink]

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29 Oct 2016, 20:43
sunny3011 wrote:
If P + |k| > |P| + k then which of the following is true of Inequalities ?

A. P=K
B. P<=k
C. P<k
D. P>k
E. Can't determined

Again strucked with Inequalities ? Need Help.

Bunuel and VeritasPrepKarishma ?

P + |k| > |P| + k

Square both sides

P^2 + k^2 +2*P*|k| > P^2 + k^2 + 2*|P|*k

P*|k| > |P|*k

P/|P| > k/|k|

We all know that x/|x| can be 1 or -1.

So in the above inequality LHS will be +1 and RHS will be -1.

Which means P is positive and k is negative.

P>k

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If P + |k| > |P| + k then which of the following is true of Inqual  [#permalink]

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30 Nov 2016, 04:01
akumar5 wrote:
given that P + |k| > |P| + k.................1

lets square both sides : p^2+k^2+2p|k| > p^2+k^2+2k|p|

so, p|k|>k|p| ;
Clearly, the above statement is valid only if p>k

I like your answer, but can you really square both sides of inequality when you dont know whether either side is +ve or -ve?
I've heard you shouldnt, if you don't know
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Re: If P + |k| > |P| + k then which of the following is true of Inqual  [#permalink]

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30 Nov 2016, 18:55
Krage17 wrote:
I am neither of two you asked to help, but I am retaking the GMAT soon, so here is how I approached this problem:

Assuming that P and k and both sides have the absolute value, the only way we will have an inequality is when when one or more numbers have negative signs, and thus change their value when an absolute value is taken. We already know that P on the left side equals |P| on the right side because the left side of the inequality is larger (and therefore, there is no way P is negative). So, what truly shifts the equation here is the sign flip when we take an absolute value of k. In other words, k must be negative, otherwise |k| would be equal to k and P+|k| would be equal to |P|+k. Hence, P (which is positive) must be larger than k (which is negative).
I know it's somewhat conceptual and may not be easy to follow, but that's how I looked at it.

This approach is interesting, and i think the most efficient way to approach this problem. But your assumption here is wrong my friend-

Here's how P can be negative:
p = -1, k = -2
p + |k| > |p| + k
-1 + |-2| > |-1| + (-2)
1 > -1

This is true for any negative value of p when k<p.

So we cannot conclude that p is non-negative. However, what we can certainly conclude from the given inequality is- k has to be negitive

|Any number| $$\geq$$ That number

Now, the given inequality:
p + |k| > |p| + k
The |k| on LHS can be greater than or equal to k on RHS

Case 1:
|k| = k, which means that k is positive
in this case, to hold the ineq. true p > |p|. But this cant be true. Hence K is negative.

Case 2:
|k| > k, which means that k is negative
in this case, to hold the ineq. true, p on the LHS can be equal to (p is positive) or less than (p is negative) |p| on RHS.

So we can safely conclude from the given ineq. that
p can be +ve or -ve, but
k will always be -ve

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Re: If P + |k| > |P| + k then which of the following is true of Inqual  [#permalink]

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Re: If P + |k| > |P| + k then which of the following is true of Inqual   [#permalink] 07 Feb 2019, 08:28

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The Rational Roots Theorem

# The Rational Roots Theorem

 Theorem 1 (The Rational Roots Theorem): If $p \in \mathbb{Z}[x]$ with $p(x) = c_0 + c_1x + ... + c_nx^n$ and $\frac{a}{b} \in \mathbb{Q}$ is a root of $p$ with $(a, b) = 1$ then $a | c_0$ and $b | c_n$.
• Proof: Suppose that $\frac{a}{b} \in \mathbb{Q}$ with $(a, b) = 1$ is a root of $p$. Then $p \left ( \frac{a}{b} \right ) = 0$, so:
(1)
\begin{align} \quad 0 &= c_0 + c_1 \frac{a}{b} + ... + c_n \frac{a^n}{b^n} \\ &= c_0b^n + c_1ab^{n-1} + ... + c_na^n \end{align}
• Since $b | c_0b^n, c_1ab^{n-1}, ..., c_{n-1}a^{n-1}b, 0$ we have that $b | c_na^n$. Similarly, since $a | c_1ab^{n-1}, ..., c_na^n, 0$ we have that $a | c_0b^n$. Since $(a, b) = 1$ and since $b | c_na^n$ and [$a | c_0b^n$ we have that $b | c_n$ and $a | c_0$. $\blacksquare$

The rational roots theorem says that if $\frac{a}{b}$ is a rational root in lowest terms of $p(x) = c_0 + c_1x + ... + c_nx^n$ then it must be that the numerator $a$ divides $c_0$ and that the denominator $b$ divides $c_n$. This is extremely useful in determining if a rational number is a root of a particular $p \in \mathbb{Z}[x]$.

For example, let $p(x) = 4 + 4x^2 + 5x^3 + 5x^4$. Suppose that $\frac{a}{b}$ is a rational root of $p(x)$. Then by the theorem above we must have that $a|4$ and $b|5$. So $a = 0, \pm 1, \pm 2, \pm 4$ and $b = \pm 1, \pm 5$. So the possible rational roots of $p$ are:

(2)
\begin{align} \quad 0, \pm 1, \pm 2, \pm 4, \pm \frac{1}{5}, \pm \frac{2}{5}, \pm \frac{4}{5} \end{align}
 Corollary 1: The number $\sqrt{2}$ is irrational.
• Proof: Suppose instead that $\sqrt{2} = \frac{a}{b}$ is rational. Consider the polynomial $p(x) = x^2 - 2$. Clearly $p(\sqrt{2}) = 0$. By the rational roots theorem, we must have that $a | 2$ and $b | 1$. So $a = \pm 1, \pm 2$ and $b = \pm 1$. So $\frac{a}{b} = \pm 1, \pm 2$. However, $p(\pm 1) \neq 0$ and $p(\pm 2) \neq 0$ which is a contradiction. Therefore $\sqrt{2}$ must instead be irrational. $\blacksquare$


text:
### Home > INT2 > Chapter 9 > Lesson 9.4.1 > Problem9-138

9-138.

Graph the following points on graph paper and then connect them in the order given. Then find the area and perimeter of the shape. Show all work. 9-138 HW eTool (Desmos).

$A(-2,-3)$, $B(-6,5)$, $C(11,5)$, $D(7,2)$,$E(7,-3)$

Try using the Pythagorean Theorem to find the lengths of different sides. Also try breaking up this shape into easier shapes to find the area.

$\text{Perimeter } = 44.9 \; \text{units}$
$\text{Area} = 94 \; \text{square units}$

Use the eTool below to plot the points correctly.
Click the link at right for the full version of the eTool: INT2 9-138 HW eTool


text:
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# If w, x, y and z are integers such that w/x and y/z are

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Updated on: 13 Mar 2012, 14:00
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If w, x, y and z are integers such that w/x and y/z are integers, is w/x + y/z odd?

(1) wx + yz is odd
(2) wz + yx is odd

Originally posted by japped187 on 01 Jun 2008, 05:41.
Last edited by Bunuel on 13 Mar 2012, 14:00, edited 1 time in total.
Edited the question and added the OA
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Re: If w, x, y and z are integers such that w/x and y/z are  [#permalink]

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13 Mar 2012, 14:24
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If w, x, y and z are integers such that w/x and y/z are integers, is w/x + y/z odd?

Given: $$\frac{w}{x}=integer$$ and $$\frac{y}{z}=integer$$. Hence, $$\frac{w}{x}+\frac{y}{z}=\frac{wz+yx}{xz}=integer$$ and the question is whether this integer is odd.

(1) wx + yz is odd --> if $$w=x=1$$ and $$y=z=2$$ then $$\frac{w}{x}+\frac{y}{z}=2=even$$ but if $$w=x=1$$ and $$y=2$$, $$z=1$$ then $$\frac{w}{x}+\frac{y}{z}=3=odd$$. Not sufficient.

(2) wz + yx is odd --> $$\frac{wz+yx}{xz}=\frac{odd}{xz}=integer$$ --> $$odd=(xz)*integer$$ --> all multiple must be odd in order the product to be odd, hence $$integer =odd$$. Sufficient.

Hope it's clear.
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Re: if w,x,y,and z are integers...DS  [#permalink]

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01 Jun 2008, 09:00
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I think question is asking w/x + y/z odd and not w/x + w/z odd or not? If this is the case then here is my explanation.
Rephrased question is wz+xy/xz is odd or not?

Statement 1:
wx + yz is odd, this implies one pair is odd and other pair is even. As even + odd = odd. But we are not sure which pair is even or odd. So this is not sufficient and answer cannot be A or D.

Statement 2;
wz + xy is odd, this numerator of question is odd. As given in question w/x + y/z is integer so wz+xy/xz is not a fraction and this implies wz+xy is divisible by xz. Only way a odd number divisible by another number is that divisor has to be odd as well. So odd divided by odd will yield odd. So question is answered.

##### General Discussion
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Re: If w, x, y, and z are integers such that w/x and y/z are  [#permalink]

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13 Mar 2012, 12:06
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japped187 wrote:
If w, x, y, and z are integers such that w/x and y/z are integers, is w/x + w/z odd?

(1) wx + yz is odd
(2) wz + xy is odd

IMO A is also satisfactory ....
wx + yz is odd

Case 1
wx is odd & yz is even
wx is odd ==> w & x are odd (odd*odd = odd) ==> w/x is odd
yz is even ==> There can be 3 cases
y is even and z is odd = Not possible as we are given y/z is integer
y is odd and z is even = Not possible as we are given y/z is integer
y is even and z is even = Possible ==> y/z is even
w/x (odd) + y/z (even) = Odd

Case 2
wx is even & yz is odd can be proved in a similar manner.

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Re: If w, x, y and z are integers such that w/x and y/z are  [#permalink]

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14 Mar 2012, 01:02
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Hi Bunuel,

I agree with ur exp .. but is there a problem with my algebraic method ?
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Re: If w, x, y and z are integers such that w/x and y/z are  [#permalink]

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20 Sep 2012, 06:33
5
Other way to look at the problem

As w/x is integer we can say w=xa+0 , where a= any integer------> xa/x = a
Same way y/z is integer we can say y=zb+0 , where b= any integer------->zb/z = b

So the basically the question is, "Is a+b Odd"

1) wx + yz ----> x^2a + z^2b = odd------> a+b may be or may not be odd ---> Insufficient
2) wz + yx -----> xza + xzb ----> xz(a+b) = odd--->it means that both xz & (a+b) are odd ---->Sufficient

Hope it helps.
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Re: If w, x, y and z are integers such that w/x and y/z are  [#permalink]

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20 Sep 2012, 07:16
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rohitgoel15 wrote:
IMO A is also satisfactory ....
wx + yz is odd

Case 1
wx is odd & yz is even
wx is odd ==> w & x are odd (odd*odd = odd) ==> w/x is odd
yz is even ==> There can be 3 cases
y is even and z is odd = Not possible as we are given y/z is integer
y is odd and z is even = Not possible as we are given y/z is integer
y is even and z is even = Possible ==> y/z is even
w/x (odd) + y/z (even) = Odd

Case 2
wx is even & yz is odd can be proved in a similar manner.

rohitgoel15 wrote:
Hi Bunuel,

I agree with ur exp .. but is there a problem with my algebraic method ?

Yes there is a problem in statement:
y is even and z is odd = Not possible as we are given y/z is integer
Even/Odd can be integer, Consider eg Y=6 , z=3.. Hence that solution is incorrect.
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Re: If w, x, y and z are integers such that w/x and y/z are  [#permalink]

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19 Jul 2013, 01:22
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From 100 hardest questions.
Bumping for review and further discussion.
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Re: If w, x, y and z are integers such that w/x and y/z are  [#permalink]

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31 Jul 2013, 07:33
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St 2 can also be proved in the following way.

w/x + y/z = something

=> (wz+xy)/xz = something
=> wz+xy = A -----------------------> Is xz*something odd ??

As per st 2 , wz+xy = odd .

Hence B

Hope there are no loopholes in my understanding.

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Re: If w, x, y and z are integers such that w/x and y/z are  [#permalink]

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02 Aug 2013, 08:33
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Hi All,

Problem helps us to understand the following:
For $$\frac{w}{x}$$ to be an integer, both w and x need to be either even or odd. Same is the case with y and z, both either need to be even or odd.
We need, w/x + y/z to be of the following format:

ODD + EVEN, so of the two terms, one of them has to be even and the other to be odd.

Now, lets look the statements..

Statement (1): $$wx+yz$$ is odd

From this statement we can deduce that either wx is odd and yz is even OR wx is even and yz is odd. Lets assume that wx is odd and yz is even. For the term wx to be odd and w/x to be an integer, both w and x needs to be odd.
However, even though yz is even since both y and z are even, it doesn't guarantee that y/z will be even. (6/2 = 3)

So from statement 1, what we get is ODD + EVEN/ODD, hence not sufficient.

Statement (2): $$wz+yx$$ is odd

We can simply multiply and divide by xz as below:

(w/x + y/z) * xz = ODD

Now since we know that only ODD * ODD = ODD, we can be certain that w/x + y/z is odd, hence sufficient.

So my response, B
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Re: If w, x, y and z are integers such that w/x and y/z are  [#permalink]

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10 Oct 2013, 07:16
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Statement 1 is indeed tricky... Statement 2 is easy.
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Re: If w, x, y and z are integers such that w/x and y/z are  [#permalink]

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27 Dec 2013, 07:48
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HI,

Just for further discussion,

if AB is odd, is A/B also odd? What is the rule on this type of construction?

Thanks
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Re: If w, x, y and z are integers such that w/x and y/z are  [#permalink]

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29 Dec 2013, 04:43
Paris75 wrote:
HI,

Just for further discussion,

if AB is odd, is A/B also odd? What is the rule on this type of construction?

Thanks

For integers a and b, ab is odd only if both are odd. Now, odd/odd can be odd or not an integer at all. For example, 3/1=3=odd but 1/3 is not an integer at all.
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Re: If w, x, y and z are integers such that w/x and y/z are  [#permalink]

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29 Dec 2013, 20:27
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japped187 wrote:
If w, x, y and z are integers such that w/x and y/z are integers, is w/x + y/z odd?

(1) wx + yz is odd
(2) wz + yx is odd

Plug in approach requiring minimal thinking:

Take the case that is to be proved i.e, w/x + y/z is odd and then take the contrary case i.e, w/x + y/z is even for both the statements

(i) Case to be proved : wx+ yz is odd and w/x + y/z is odd.

The former is satisfied by odd + even and the latter again by odd+ even

Thus this will be satisfied say, if the following is satisfied: wx and w/x are both odd and y/z and yz are both even. an example is w=15 , x=3 and y=4 and z=2

How about the contrary case for (i). i.e., wx+ yz is odd but w/x + y/z is even.

The former is satisfied by odd+ even and the latter by odd+odd or even +even

This will be satisfied say, if wx and w/x are both odd and yz is even and y/z is odd. An example is w=9 x=3 and y=6 z=2

since the case to be proved and the contrary are both satisfied (i) is not sufficient

(ii) Case to be proved: wz+yx is odd and w/x and y/z is odd

The former is satisfied by odd+ even and the latter again by odd+ even

Thus this will be satisfied say, if the following is satisfied: wz and w/x are both odd and y/z and yx are both even. an example is w=15 , x=3 and y=12 and z=1

How about the contrary case for (i). i.e., wx+ yz is odd but w/x + y/z is even.

The former is satisfied by odd+ even and the latter by odd+odd or even +even

This will be satisfied say, if wz and w/x are both odd and yx is even and y/z is odd.

We find this cannot be satisfied

Since the contrary case cannot be proved for (ii) , this alone is sufficient and the answer is B.
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Re: If w, x, y and z are integers such that w/x and y/z are  [#permalink]

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japped187 wrote:
If w, x, y and z are integers such that w/x and y/z are integers, is w/x + y/z odd?

(1) wx + yz is odd
(2) wz + yx is odd

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# Pincell Depletion¶

This notebook is intended to introduce the reader to the depletion interface contained in OpenMC. It is recommended that you are moderately familiar with building models using the OpenMC Python API. The earlier examples are excellent starting points, as this notebook will not focus heavily on model building.

If you have a real power reactor, the fuel composition is constantly changing as fission events produce energy, remove some fissile isotopes, and produce fission products. Other reactions, like $$(n, \alpha)$$ and $$(n, \gamma)$$ will alter the composition as well. Furthermore, some nuclides undergo spontaneous decay with widely ranging frequencies. Depletion is the process of modeling this behavior.

In this notebook, we will model a simple fuel pin in an infinite lattice using the Python API. We will then build and examine some of the necessary components for performing depletion analysis. Then, we will use the depletion interface in OpenMC to simulate the fuel pin producing power over several months. Lastly, we will wrap up with some helpful tips to improve the fidelity of depletion simulations.

[1]:

%matplotlib inline
import math
import openmc

## Build the Geometry¶

Much of this section is borrowed from the “Modeling a Pin-Cell” example. If you find yourself not understanding some aspects of this section, feel free to refer to that example, as some details may be glossed over for brevity.

First, we will create our fuel, cladding, and water materials to represent a typical PWR.

[2]:

fuel = openmc.Material(name="uo2")

[3]:

fuel.add_element("U", 1, percent_type="ao", enrichment=4.25)
fuel.set_density("g/cc", 10.4)

[4]:

clad = openmc.Material(name="clad")

[5]:

clad.add_element("Zr", 1)

[6]:

water = openmc.Material(name="water")

[7]:

water.add_element("O", 1)
water.set_density("g/cc", 1.0)

[8]:

water.add_s_alpha_beta("c_H_in_H2O")

[9]:

materials = openmc.Materials([fuel, clad, water])

Here, we are going to use the openmc.model.pin function to build our pin cell. The pin function anticipates concentric cylinders and materials to fill the inner regions. One additional material is needed than the number of cylinders to cover the domain outside the final ring.

[10]:

radii = [0.42, 0.45]

Using these radii, we define concentric ZCylinder objects. So long as the cylinders are concentric and increasing in radius, any orientation can be used. We also take advantage of the fact that the openmc.Materials object is a subclass of the list object to assign materials to the regions defined by the surfaces.

[11]:

pin_surfaces = [openmc.ZCylinder(r=r) for r in radii]

[12]:

pin_univ = openmc.model.pin(pin_surfaces, materials)

The first material, in our case fuel, is placed inside the first cylinder in the inner-most region. The second material, clad, fills the space between our cylinders, while water is placed outside the last ring. The pin function returns an openmc.Universe object, and has some additional features we will mention later.

[13]:

pin_univ.plot()

[13]:

<matplotlib.image.AxesImage at 0x7f9aea1a50d0>

[14]:

bound_box = openmc.rectangular_prism(0.62, 0.62, boundary_type="reflective")

[15]:

root_cell = openmc.Cell(fill=pin_univ, region=bound_box)

[16]:

root_univ = openmc.Universe(cells=[root_cell])

[17]:

geometry = openmc.Geometry(root_univ)

Lastly we construct our settings. For the sake of time, a relatively low number of particles will be used.

[18]:

settings = openmc.Settings()

[19]:

settings.particles = 100
settings.inactive = 10
settings.batches = 50

The depletion interface relies on OpenMC to perform the transport simulation and obtain reaction rates and other important information. We then have to create the xml input files that openmc expects, specifically geometry.xml, settings.xml, and materials.xml.

[20]:

geometry.export_to_xml()

[21]:

settings.export_to_xml()

Before we write the material file, we must add one bit of information: the volume of our fuel. In order to translate the reaction rates obtained by openmc to meaningful units for depletion, we have to normalize them to a correct power. This requires us to know, or be able to calculate, how much fuel is in our problem. Correctly setting the volumes is a critical step, and can lead to incorrect answers, as the fuel is over- or under-depleted due to poor normalization.

For our problem, we can assign the “volume” to be the cross-sectional area of our fuel. This is identical to modeling our fuel pin inside a box with height of 1 cm.

[23]:

fuel.volume = math.pi * radii[0] ** 2

[24]:

materials.export_to_xml()

## Setting up for depletion¶

The OpenMC depletion interface can be accessed from the openmc.deplete module, and has a variety of classes that will help us.

[25]:

import openmc.deplete

In order to run the depletion calculation we need the following information:

1. Nuclide decay, fission yield, and reaction data
2. Operational power or power density
3. Desired depletion schedule
4. Desired time integration scheme

The first item is necessary to determine the paths by which nuclides transmute over the depletion simulation. This includes spontaneous decay, fission product yield distributions, and nuclides produced through neutron-reactions. For example, * Te129 decays to I129 with a half life of ~70 minutes * A fission event for U-235 produces fission products like Xe135 according to a distribution * For thermal problems, Am241 will produce metastable Am242 about 8% of the time during an $$(n,\gamma)$$ reaction. The other 92% of capture reactions will produce ground state Am242

These data are often distributed with other nuclear data, like incident neutron cross sections with ENDF/B-VII. OpenMC uses the openmc.deplete.Chain <https://docs.openmc.org/en/latest/pythonapi/generated/openmc.deplete.Chain.html#openmc.deplete.Chain>__ to collect represent the various decay and transmutation pathways in a single object. While a complete Chain can be created using nuclear data files, users may prefer to download pre-generated XML-representations instead. Such files can be found at https://openmc.org/depletion-chains/ and include full and compressed chains, with capture branching ratios derived using PWR- or SFR-spectra.

For this problem, we will be using a much smaller depletion chain that contains very few nuclides. In a realistic problem, over 1000 isotopes may be included in the depletion chain.

[26]:

chain = openmc.deplete.Chain.from_xml("./chain_simple.xml")

[27]:

chain.nuclide_dict

[27]:

OrderedDict([('I135', 0),
('Xe135', 1),
('Xe136', 2),
('Cs135', 3),
('Gd157', 4),
('Gd156', 5),
('U234', 6),
('U235', 7),
('U238', 8)])

The primary entry point for depletion is the openmc.deplete.Operator. It relies on the openmc.deplete.Chain and helper classes to run openmc, retrieve and normalize reaction rates, and other perform other tasks. For a thorough description, please see the full API documentation.

We will create our Operator using the geometry and settings from above, and our simple chain file. The materials are read in automatically using the materials.xml file.

[28]:

operator = openmc.deplete.Operator(geometry, settings, "./chain_simple.xml")

We will then simulate our fuel pin operating at linear power of 174 W/cm, or 174 W given a unit height for our problem.

[29]:

power = 174

For this problem, we will take depletion step sizes of 30 days, and instruct OpenMC to re-run a transport simulation every 30 days until we have modeled the problem over a six month cycle. The depletion interface expects the time to be given in seconds, so we will have to convert. Note that these values are not cumulative.

[30]:

time_steps = [30 * 24 * 60 * 60] * 6

And lastly, we will use the basic predictor, or forward Euler, time integration scheme. Other, more advanced methods are provided to the user through openmc.deplete

[31]:

integrator = openmc.deplete.PredictorIntegrator(operator, time_steps, power)

To perform the simulation, we use the integrate method, and let openmc take care of the rest.

[32]:

integrator.integrate()

## Processing the outputs¶

The depletion simulation produces a few output files. First, the statepoint files from each individual transport simulation are written to openmc_simulation_n<N>.h5, where <N> indicates the current depletion step. Any tallies that we defined in tallies.xml will be included in these files across our simulations. We have 7 such files, one for each our of 6 depletion steps and the initial state.

[33]:

!ls *.h5

c5g7.h5                  openmc_simulation_n2.h5  openmc_simulation_n6.h5
depletion_results.h5     openmc_simulation_n3.h5  statepoint.50.h5
openmc_simulation_n0.h5  openmc_simulation_n4.h5  summary.h5
openmc_simulation_n1.h5  openmc_simulation_n5.h5

The depletion_results.h5 file contains information that is aggregated over all time steps through depletion. This includes the multiplication factor, as well as concentrations. We can process this file using the openmc.deplete.ResultsList object

[34]:

results = openmc.deplete.ResultsList.from_hdf5("./depletion_results.h5")

[35]:

time, k = results.get_eigenvalue()

[36]:

time /= (24 * 60 * 60)  # convert back to days from seconds

[37]:

k

[37]:

array([[0.76882937, 0.00982155],
[0.75724033, 0.00827689],
[0.75532242, 0.01031746],
[0.74796855, 0.00919769],
[0.74066561, 0.01157708],
[0.73184492, 0.00971504],
[0.7207293 , 0.00703074]])

The first column of k is the value of k-combined at each point in our simulation, while the second column contains the associated uncertainty. We can plot this using matplotlib

[38]:

from matplotlib import pyplot

[39]:

pyplot.errorbar(time, k[:, 0], yerr=k[:, 1])
pyplot.xlabel("Time [d]")
pyplot.ylabel("$k_{eff}\pm \sigma$");

Due to the low number of particles selected, we have not only a very high uncertainty, but likely a horrendously poor fission source. This pin cell should have $$k>1$$, but we can still see the decline over time due to fuel consumption.

We can then examine concentrations of atoms in each of our materials. This requires knowing the material ID, which can be obtained from the materials.xml file.

[40]:

_time, u5 = results.get_atoms("1", "U235")
_time, xe135 = results.get_atoms("1", "Xe135")

[41]:

pyplot.plot(time, u5, label="U235")
pyplot.xlabel("Time [d]")
pyplot.ylabel("Number of atoms - U235");

[42]:

pyplot.plot(time, xe135, label="Xe135")
pyplot.xlabel("Time [d]")
pyplot.ylabel("Number of atoms - Xe135");

We can also examine reaction rates over time using the ResultsList

[43]:

_time, u5_fission = results.get_reaction_rate("1", "U235", "fission")

[44]:

pyplot.plot(time, u5_fission)
pyplot.xlabel("Time [d]")
pyplot.ylabel("Fission reactions / s");

Depletion is a tricky task to get correct. Use too short of time steps and you may never get your results due to running many transport simulations. Use long of time steps and you may get incorrect answers. Consider the xenon plot from above. Xenon-135 is a fission product with a thermal absorption cross section on the order of millions of barns, but has a half life of ~9 hours. Taking smaller time steps at the beginning of your simulation to build up some equilibrium in your fission products is highly recommended.

When possible, differentiate materials that reappear in multiple places. If we had built an entire core with the single fuel material, every pin would be depleted using the same averaged spectrum and reaction rates which is incorrect. The Operator can differentiate these materials using the diff_burnable_mats argument, but note that the volumes will be copied from the original material.

Using higher-order integrators, like the CECMIntegrator, EPCRK4Integrator with a fourth order Runge-Kutta, or the LEQIIntegrator, can improve the accuracy of a simulation, or at least allow you to take longer depletion steps between transport simulations with similar accuracy.

Fuel pins with integrated burnable absorbers, like gadolinia, experience strong flux gradients until the absorbers are mostly burned away. This means that the spectrum and magnitude of the flux at the edge of the fuel pin can be vastly different than that in the interior. The helper pin function can be used to subdivide regions into equal volume segments, as follows.

[45]:

div_surfs_1 = [openmc.ZCylinder(r=1)]
div_1 = openmc.model.pin(div_surfs_1, [fuel, water], subdivisions={0: 10})

[46]:

div_1.plot(width=(2.0, 2.0))

[46]:

<matplotlib.image.AxesImage at 0x7f9ae8572a90>

The innermost region has been divided into 10 equal volume regions. We can pass additional arguments to divide multiple regions, except for the region outside the last cylinder.

## Register depletion chain¶

The depletion chain we created can be registered into the OpenMC cross_sections.xml file, so we don’t have to always pass the chain_file argument to the Operator. To do this, we create a DataLibrary using openmc.data. Without any arguments, the from_xml method will look for the file located at OPENMC_CROSS_SECTIONS. For this example, we will just create a bare library.

[47]:

data_lib = openmc.data.DataLibrary()

[48]:

data_lib.register_file("./chain_simple.xml")

[49]:

data_lib.export_to_xml()

[50]:

!cat cross_sections.xml

<?xml version='1.0' encoding='utf-8'?>
<cross_sections>
<depletion_chain path="chain_simple.xml" type="depletion_chain" />
</cross_sections>

This allows us to make an Operator simply with the geometry and settings arguments, provided we exported our library to OPENMC_CROSS_SECTIONS. For a problem where we built and registered a Chain using all the available nuclear data, we might see something like the following.

[51]:

new_op = openmc.deplete.Operator(geometry, settings)

[52]:

len(new_op.chain.nuclide_dict)

[52]:

3820

[53]:

[nuc.name for nuc in new_op.chain.nuclides[:10]]

[53]:

['H1', 'H2', 'H3', 'H4', 'H5', 'H6', 'H7', 'He3', 'He4', 'He5']

[54]:

[nuc.name for nuc in new_op.chain.nuclides[-10:]]

[54]:

['Ds268',
'Ds269',
'Ds270',
'Ds270_m1',
'Ds271',
'Ds271_m1',
'Ds272',
'Ds273',
'Ds279_m1',
'Rg272']

## Choice of depletion step size¶

A general rule of thumb is to use depletion step sizes around 2 MWd/kgHM, where kgHM is really the initial heavy metal mass in kg. If your problem includes integral burnable absorbers, these typically require shorter time steps at or below 1 MWd/kgHM. These are typically valid for the predictor scheme, as the point of recent schemes is to extend this step size. A good convergence study, where the step size is decreased until some convergence metric is satisfied, is a beneficial exercise.

We can use the Operator to determine our maximum step size using this recommendation. The heavy_metal attribute returns the mass of initial heavy metal in g, which, using our power, can be used to compute this step size.

$\frac{2\,MWd}{kgHM} = \frac{P\times\Delta}{hm_{op}}$
[55]:

operator.heavy_metal

[55]:

5.080339195584719

[56]:

max_step = 2 * operator.heavy_metal / power * 1E3

[57]:

print("\"Maximum\" depletion step: {:5.3} [d]".format(max_step))

"Maximum" depletion step:  58.4 [d]

Alternatively, if we were provided the power density of our problem, we can provide this directly with openmc.deplete.PredictorIntegrator(operator, time_steps, power_density=pdens). The values of power and power_density do not have to be scalars. For problems with variable power, we can provide an iterable with the same number of elements as time_steps.


text:
# zbMATH — the first resource for mathematics

A note on a semilinear elliptic problem. (English) Zbl 0786.35060
The author studies the fourth order problem $\Delta^ 2u+c \Delta u=b [(u+1)_ +-1] \text{ in } \Omega,\quad u=\Delta u=0 \text{ on }\partial \Omega,$ for bounded domains $$\Omega \subset \mathbb{R}^ n$$ and for $$c<\lambda_ 1$$, the first Dirichlet eigenvalue of the Laplace operator. It is shown that $$b \geq \lambda_ 1$$ $$(\lambda_ 1-c)$$ is necessary and sufficient for the existence of nontrivial solutions, and that $$b \geq \lambda_ 1$$ $$(\lambda_ 1-c)>0$$ is sufficient for the existence of negative solutions.
(Reviewer’s remark: In fact $$b \geq \lambda_ 1$$ $$(\lambda_ 1-c)>0$$ is also necessary for the existence of negative solutions, as can be seen from the inequality preceding (1.3).) There are some connections to second order problems with similar nonlinearities, see [P. J. Mc Kenna, Arch. Ration. Mech. Anal. 98, 167-190 (1987; Zbl 0676.35003)].

##### MSC:
 35J65 Nonlinear boundary value problems for linear elliptic equations 35J40 Boundary value problems for higher-order elliptic equations
##### Keywords:
biharmonic operator; semilinear elliptic problem


text:
329 views

At first sight, it looks as though panchayati raj, the lower layer of federalism in our polity, is as firmly entrenched in our system as is the older and higher layer comprising the Union Government and the States. Like the democratic institutions at the higher level, those at the panchayat level, the panchayati raj institutions (PRIs), are written into and protected by the Constitution. All the essential features, which distinguish a unitary system from a federal one, are as much enshrined at the lower as at the upper level of our federal system. But look closely and you will discover a fatal flaw. The letter of the Constitution as well as the spirit of the present polity have exposed the intra-State level of our federal system to a dilemma of which the inter-State and Union-State layers are free. The flaw has many causes. But all of them are rooted in an historical anomaly, that while the dynamics of federalism and democracy have given added strength to the rights given to the States in the Constitution, they have worked against the rights of panchayats.

At both levels of our federal system there is the same tussle between those who have certain rights and those who try to encroach upon them if they believe they can. Thus the Union Government was able to encroach upon certain rights given to the States by the Constitution. It got away with that because the single dominant party system, which characterised Centre-State relations for close upon two decades, gave the party in power at the Union level many extra-constitutional political levers. Second, the Supreme Court had not yet begun to extend the limits of its power. But all that has changed in recent times. The spurt given to a multi-party democracy by the overthrow of the Emergency in $1977$ became a long-term trend later on because of the ways in which a vigorously democratic multi-party system works in a political society which is as assertively pluralistic as Indian society is. It gives political clout to all the various segments which constitute that society. Secondly, because of the linguistic reorganisation of States in the $1950$s, many of the most assertive segments have found their most assertive expression as States. Thirdly, with single-party dominance becoming a thing of the past at the Union level, governments can be formed at that level only by multi-party coalitions in which State-level parties are major players. This has made it impossible for the Union Government to do much about anything unless it also carries a sufficient number of State-level parties with it. Indian federalism is now more real than it used to be, but an unfortunate side-effect is that India's panchayati raj system, inaugurated with such fanfare in the early $1980$s, has become less real.

By the time the PRIs came on the scene, most of the political space in our federal system had been occupied by the Centre in the first $30$ years of Independence, and most of what was still left after that was occupied by the States in the next $20.$ PRIs might have hoped to wrest some space from their immediate neighbour, the States, just as the States had wrested some from the Centre. But having at last managed to checkmate the Centre's encroachments on their rights, the States were not about to allow the PRIs to do some encroaching of their own.

By the $1980$s and early $1990$s, the only national party left, the Congress, had gone deeper into a siege mentality. Finding itself surrounded by State-level parties, it had built walls against them instead of winning them over. Next, the States retaliated by blocking Congress proposals for panchayati raj in Parliament, suspecting that the Centre would try to use panchayats to bypass State Governments. The suspicion fed on the fact that the powers proposed by the Congress for panchayats were very similar to many of the more lucrative powers of State Governments. State-level leaders also feared, perhaps, that if panchayat-level leaders captured some of the larger PRIs, such as district-level panchayats, they would exert pressure on State-level leaders through intra-State multi-party federalism.

It soon became obvious to Congress leaders that there was no way the panchayati raj amendments they wanted to write into the Constitution would pass muster unless State-level parties were given their pound of flesh. The amendments were allowed only after it was agreed that the powers of panchayats could be listed in the Constitution. Illustratively, they would be defined and endowed on PRIs by the State Legislature acting at its discretion.

This left the door wide open for the States to exert the power of the new political fact that while the Union and State Governments could afford to ignore panchayats as long as the MLAs were happy, the Union Government had to be sensitive to the demands of State-level parties. This has given State-level actors strong beachheads on the shores of both inter-State and intra-State federalism. By using various administrative devices and nonelected parallel structures, State Governments have subordinated their PRIs to the State administration and given the upper hand to State Government officials against the elected heads of PRIs. Panchayats have become local agencies for implementing schemes drawn up in distant State capitals. And their own volition has been further circumscribed by a plethora of “Centrally-sponsored schemes”. These are drawn up by even more distant Central authorities but at the same time tie up local staff and resources on pain of the schemes being switched off in the absence of matching local contribution. The "foreign aid" syndrome can be clearly seen at work behind this kind of "grass roots development".

What is the "dilemma" at the intra-State level mentioned in the first paragraph of the passage?

1. Should the state governments wrest more space from the Union, before considering the panchayati system?
2. Should rights similar to those that the States managed to get be extended to panchayats as well?
3. Should the single party system which has withered away be brought back at the level of the States?
4. Should the States get "their pound of flesh" before allowing the Union government to pass any more laws?


text:
the relation between Laplace transform and Z-transform?

Who?
Deniz Genlik (TU Berlin)
When?
2013/11/08, 16:00
Where?
TU Berlin, at room MA 742
About what?

The Laplace transform plays an important role in the analysis of continuous linear time invariant systems and the Z-transform does the same for Discrete Linear Shift Invariant Systems. In this talk, these systems will be briefly characterized and these transformations will be related to the characterizations. The relation between the Laplace transform and the Z-Transform will then be introduced, which is important for the analysis of mixed-type CLTI and DLSI Systems.


text:
# Would going first in Egg Russian Roulette increase or decrease your chances of winning? [duplicate]

Egg Russian Roulette:

6 eggs, 5 of which are hard boiled & 1 of which is raw. Players (1v1) each choose eggs in a turn based system whereby the egg is smashed against the forehead. The first player to smash the raw egg against his/her forehead is the loser.

Should you have the option of choosing who goes first, would going first increase or decrease your chances of winning?

## marked as duplicate by Nate Eldredge, user147263, Aditya Hase, Henry probability StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Dec 3 '14 at 7:05

• Sounds like copious amounts of vodka would be a pre-requisite for this. – Tom Collinge Dec 3 '14 at 6:54

Let's list all the probabilities shall we?

If you went first there is a:
$$\frac16$$ chance of you losing on the first turn,
$$\frac56\times\frac45\times\frac14=\frac16$$ chance of you losing on the third turn, $$\frac56\times\frac45\times\frac34\times\frac23\times\frac12=\frac16$$ chance of you losing on the fifth turn,

Therefore there is a $$\frac16+\frac16+\frac16=\frac12$$ chance of you losing, thus your chances will stay the same.

• Actually ended up running a Monte Carlo simulation and saw the same results on 1,000,000 tests. (50.0275% choosing first would lose, 49.9725% choosing second would lose). So +1, thanks. – Moose Dec 3 '14 at 8:20
• @moose No problems – Abraham Zhang Dec 3 '14 at 9:06

If you went first you'd have a $1\over6$ chance of getting a raw one and the sequence of moves would look like this. $$\text{you}={1\over6}$$$$\text{them}={1\over5}$$$$\text{you}={1\over4}$$$$\text{them}={1\over3}$$$$\text{you}={1\over2}$$$$\text{them}={1\over1}$$ Which you'd have the best odds in the beginning and if it did happen to go all the way to the end your opponent would be the one to lose.

If you went second though your opponent would go twice in the time you only go once. So just within three moves you would've had a ${1\over5}$ chance of getting the raw one, but your opponent would have had a ${1\over6}+{1\over4}={5\over 12}$ which is a much higher percentage that yours.

So I could easily justify either way. That's just my logic. Don't know if it helps or not!

If you forced me to pick, I'd say go second. Just so if you happened to make it to the end and you knew you were going to get the raw egg you could use it as a weapon against your opponent.

• Adding up these probabilities gives a number greater than $1$. Instead consider: you lose first turn $\dfrac16$, they lose second turn $\dfrac56 \times \dfrac15$, you lose third turn $\dfrac56 \times \dfrac45 \times \dfrac14$, etc. – Henry Dec 3 '14 at 7:07


text:
# Function to match the case of a text as per another text

I am translating a text into another language using Google APIs.

Sometimes the translated text returned from the API does not match the case of the text provided - e.g. when translating the text "Now Activated" to German, I get the translated text response as "jetzt aktiviert" instead of "Jetzt Aktiviert".

I wrote a function which can change the case of translated text to match the input text:

public string MaintainCase(string text, string translatedText)
{
// Split the inputs provided into string arrays
var textStrArray = text.Split(' ');
var translatedTextStrArray = translatedText.Split(' ');

// Count the no of words of both the texts provided
var noOfWordsInText = textStrArray.Length;
var noOfWordsInTranslatedText = translatedTextStrArray.Length;

// Check if the word count matches
bool wordCountMatches =
noOfWordsInText == noOfWordsInTranslatedText
? true
: false;

// Start changing the case of translated text as per text
for (int i = 0; i < noOfWordsInText; i++)
{
// Take the word no i of text and translated text
var textWord = textStrArray[i];
var translatedTextWord = translatedTextStrArray[i];

// Using string builder to modify the first character of translated text
// based on the case identified of the first character of text
var strBuilder = new StringBuilder(translatedTextWord);
var changedChar =
char.IsUpper(textWord[0])
? char.ToUpper(translatedTextWord[0])
: char.ToLower(translatedTextWord[0]);

strBuilder[0] = changedChar;
translatedTextStrArray[i] = strBuilder.ToString();

// If the word count does not match, after changing the case of only first word, break the loop
if (!wordCountMatches)
{
break;
}
}

// Return the modified translated text
return string.Join(" ", translatedTextStrArray);
}

Kindly suggest if this can be made more efficient (usage of string builders etc) and if I need to add some additional checks which I may have missed.

• Frame challenge: is the result really what we want? E.g. translating German mein Buch to English, we normally want my book, not my Book. Similarly, when going the other way, we must respect the capitalisation that the rules of the language demand, rather than slavishly following the source language. – Toby Speight Oct 1 at 7:46
• In the application this can be made optional by providing a default bool parameter 'maintainCase = false'. However is the code fine for the scenario where we do want to utilize this functionality? – Akshunya Oct 1 at 7:52
• This looks like trouble to me and as a method that will never be good enough because of all strange possibilities in real world texts. Have you considered the functionality provided by CultureInfo.TextInfo for instance ToTitleCase() and the like? – Henrik Hansen Oct 1 at 10:03


text:
# Fixed-point iterations for quadratic function $x\mapsto x^2-2$

Let $$f(x)$$ be $$x^2-x-2$$. I want to find the root using FPI in an interval where it will converge. I have chosen $$g(x)=x^2-2$$ and so $$g'(x)=2x$$. The convergence condition, $$|g'(x)|<1$$ is obviously satisfied in $$-0.5.

Problem: I have failed to find a consistent interval outside of this convergence interval up to $$\pm 1$$ for which the iteration consistently diverges.

Question: Does this convergence condition only guarantee convergence (in the bounds) but not divergence (outside the bounds)?

• Is that fixed-point iteration fixed? From $x^2=2+x$ one finds the better iteration $x_{n+1}=\sqrt{2+x_n}$ for the positive root. – LutzL Apr 16 at 16:25
• Yes, but I thought the reason it’s ‘better’ is because it satisfies abs(g’(x))<1 in some interval. But g(x) in op works just fine up to -+1. – AKubilay Apr 16 at 18:10
• $g(x)=\sqrt{2+x}$ has $g'(x)=\frac1{2\sqrt{2+x}}\le\frac25$ for all $x>0$, $g(x)=1+\frac2x$ has $g'(x)=-\frac{2}{x^2}>-1$ for $x>\frac32$, so you get intervals with $|g'(x)|<1$ for many fixed-point functions. – LutzL Apr 16 at 18:53

Let us consider the fixed point iterations associated to the function $$g: x \mapsto x^2-2$$, defined by the quadratic map $$x_{n+1} = {x_n}^2 - 2, \qquad x_0 \in \Bbb R .$$ This map has many periodic points, even with large period. The period-one fixed points $$-1$$, $$2$$ are both repelling fixed points (indices $$2>1$$ and $$4>1$$, respectively). Thus, fixed-point iterations will not converge towards these values unless the starting value $$x_0$$ is exactly equal to $$-1$$ or $$2$$. Setting $$y_n = -\frac{1}{4} x_n + \frac{1}{2}$$, the logistic map $$y_{n+1} = r y_n (1-y_n)$$ with parameter $$r=4$$ is obtained, which exact solution is bounded and exhibits chaotic behavior (see also this article). Therefore, $$x_n = 2\cos\left(2^n\cos^{-1}(x_0/2)\right)$$ is bounded and exhibits chaotic behavior too (the sequence does not diverge to infinity).


text:
# Are there good resources explaining mean field approximation?

I am a computer science master student. In a statistical learning theory course I am taking, mean field approximation was introduced to approximately solve non-factorizable Gibbs distributions that were derived using maximum entropy inference. Our professor has a strong background in physics and often uses terms from statistical physics. Unfortunately, I lack that background. So,

1. Are there any resources explaining mean field approximations from a non-physical/computer science perspective? I couldn't find any.

2. Or alternatively, are there "crash course"-like resources that would allow me to understand one of the more physically motivated explanations without looking up tons of terms?

EDIT: Here are some details that hopefully help to narrow it down: In this course, we are trying to sample from intractable, non-factorizable Gibbs-distributions (mainly in the context of clustering). Apparently, this cannot be done (efficiently), therefore we retreat to a mean-field approximation.

-

## migrated from stats.stackexchange.comMay 22 '12 at 13:24

This question came from our site for people interested in statistics, machine learning, data analysis, data mining, and data visualization.

There is a better site for that. –  mbq May 22 '12 at 13:24
Your best bet is to ask your professor for more details, but actually aside from its historical development within physics, mean-field approximations basically are defined by replacing non-factorizable probability distributions with factorizable ones that have some correlations put in by hand (the "mean fields" or "Weiss fields" of stat.phys.) The extra correlations should be predicted by the distribution, though, so they are calculated self-consistently in the factorizable one. –  wsc May 22 '12 at 13:51
Try > W. D. McComb, Renormalization Methods: a guide for beginners, 2004. The first three chapters give you what you want. –  user9365 May 22 '12 at 20:45
@Mbq: Given that the OP wants a "non-physical" description this might have been better moved to Computation Science, but that's a beta site. –  dmckee May 22 '12 at 21:03
@dmckee You're right, I've... missed. –  mbq May 22 '12 at 22:14

In your case, the algebra is commutative, so you don't need to understand all the fine points related to noncommutative observables in the quantum case, and the mean field will simply be a high-dimensional Gaussian, where the cumulant generating function $W(f)$ is just the exponential of a quadratic form.


text:
Also inspect the Derivative Calculator
!
en español
Integralrechner
auf Deutsch
Калькулятор Интегралов
на Русском

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Maxima takes care of actually computing the integral the the mathematical function. Maxima"s output is revolutionized to LaTeX again and is then presented to the user. The antiderivative is computed making use of the Risch algorithm, i beg your pardon is tough to know for humans. That"s why reflecting the measures of calculate is very an overwhelming for integrals.

In bespeak to display the steps, the calculator applies the very same integration techniques that a human would apply. The regimen that does this has been arisen over several years and is composed in Maxima"s own programming language. It is composed of much more than 17000 present of code. When the integrand matches a well-known form, it uses fixed rule to resolve the integral (e.g. Partial fraction decomposition for rational functions, trigonometric substitution because that integrands including the square root of a quadratic polynomial or integration by components for commodities of details functions). Otherwise, that tries different substitutions and also transformations till either the integral is solved, time runs the end or there is nothing left come try. The calculator lacks the math intuition the is an extremely useful for finding an antiderivative, but on the various other hand the can shot a large number of possibilities within a quick amount that time. The step by step antiderivatives are regularly much shorter and an ext elegant 보다 those uncovered by Maxima.

The "Checkanswer" feature has to deal with the an overwhelming task of determining whether two mathematical expressions space equivalent. Their difference is computed and also simplified as far as feasible using Maxima. For example, this entails writing trigonometric/hyperbolic attributes in your exponential forms. If it have the right to be presented that the distinction simplifies to zero, the job is solved. Otherwise, a probability algorithm is applied that evaluates and also compares both functions at randomly liked places. In the instance of antiderivatives, the entire procedure is repeated with every function"s derivative, because antiderivatives are enabled to differ by a constant.

The interactive duty graphs room computed in the browser and also displayed in ~ a canvas element (HTML5). For each role to it is in graphed, the calculator creates a JavaScript function, i m sorry is climate evaluated in tiny steps in stimulate to attract the graph. When graphing, singularities (e.g. Poles) room detected and also treated specially. The gesture control is implemented using Hammer.js.

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oil producing machinery company

Established in 2001, Puyang Zhong Yuan Restar Petroleum Equipment Co.,Ltd, “RSD” for short, is Henan’s high-tech enterprise with intellectual property advantages and independent legal person qualification. With registered capital of RMB 50 million, the Company has two subsidiaries-Henan Restar Separation Equipment Technology Co., Ltd We are mainly specialized in R&D, production and service of various intelligent separation and control systems in oil&gas drilling,engineering environmental protection and mining industries.We always take the lead in Chinese market shares of drilling fluid shale shaker for many years. Our products have been exported more than 20 countries and always extensively praised by customers. We are Class I network supplier of Sinopec,CNPC and CNOOC and registered supplier of ONGC, OIL India,KOC. High quality and international standard products make us gain many Large-scale drilling fluids recycling systems for Saudi Aramco and Gazprom projects.

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The Calculation of Coal Rock Fracture Conductivity with ...

By the definition of fracture diversion capacity, the fracture conductivity of coal rock KW is. With ( 8) ~ ( 11) and ( 20) ~ ( 23) substituted into the above ,formula,, a model for calculating the fracture conductivity of coal rock considering the factors of proppant embedment and deformation in Model 4-1 is given.

Search Injection Rate: Fracturing fluids and additives | SPE

Fault activation by hydraulic fracturing in western Canada ...

This event occurred during ,flowback,, a postinjection process during which ,fracturing fluid, is partially recovered in a controlled manner . This timing invites speculation that the earthquake was triggered by ,fluid, withdrawal; however, only ~7% of the injected fluids at well 1 flowed back to the wellhead, an unusually low recovery level compared with typical values of ~50% in other parts of western Canada ( …

(PDF) Computer simulation of hydraulic fractures | Jean ...

The conversion ,formula, between concentration in ppa and volumetric concentration c is (footnote continued) c ppa ¼ ðc=ð1 À cÞÞG s r w , where G s is the proppant solid's specific gravity, and r w is the water density in pounds per gallon (r w ' 8:34 lb=gal). 5 In reality, part of the gel fraction of the ,fluid,, plus the ,fluid, trapped in the interstitial pores of the proppant pack remains ...

Fluids | Free Full-Text | Discrete Element Simulation of ...

During the fluid flow calculation, the increment of fluid pressure (∆ P) in a reservoir is computed from the bulk modulus of fluid (K f), the volume of the domain (V d), the sum of the flow volume for one time step (∆ t), and the volume change of the domain due to mechanical loading according to Equation (12) :

Selection of Diameter of Spherical Tagin Fracturing ...

rate, of fracture ,fluid, at well head. The following are force ... ,Fracturing fluid flowback, is an important part of hydraulic ,fracturing, technique, ... and the ,formula, of air drag showing as 2 1 2 f C Sv , we can think that the thrustapply to tag from ,fracturing fluid

International Journal of Oil Gas and Coal Technology ...

Abstract: Hydraulic ,fracturing, is considered as an effective approach for the development of tight oil reservoirs. Unlike conventional reservoirs, after a large volume injection of the ,fracturing fluid,, the ,flowback rate, in tight oil reservoirs is considerably lower whereas the underlying mechanism explaining water retention remains unclear.

Study of the Potential Impacts of Hydraulic Fracturing on ...

For this study, ",flowback," is the ,fluid, returned to the surface after hydraulic ,fracturing, has occurred, but before the well is placed into production, while "produced water" is the ,fluid, returned to the surface after the well has been placed into production.10 They are collectively referred to as "hydraulic ,fracturing, wastewater" and may contain chemicals injected as part of the hydraulic ...

Study of Hydraulic Fracturing Real-time Evaluation Technology

Dali Guo 1, Lei Xu 1, Xiaohui Zeng 2, Xijun Ke 1 and Bo Tao 1. 1 State Key Laboratory of Oil and Gas Reservoir Geology and Exploitation, School of Sciences, Southwest Petroleum University, Chengdu, Sichuan, 610500, China; 2 School of Petroleum Engineering, Southwest Petroleum University, Chengdu, Sichuan, 610500, China. Received: March 31, 2014. Revised: July 18, 2014.

A semi‐analytical model for capturing dynamic behavior of ...

Abstract Hydraulic ,fracturing, has been successfully employed for unconventional oil and gas recovery for decades. During ,flowback,, the closure of the fracture may exhibit with the pressure drop of ...

Comparison of analytical models for hydraulic fracture ...

The calculation formula for the fracture conductivity is as follows: $${F}_{\mathrm{RCD}}=\frac{74.8{f}_1\left({f}_1-0.74\right)\left(D{f}_1-{D}_1{f}_2-\frac{2{D}_2p}{E_2}\right)}{0.5{f_1}^2+1}{D_1}^2$$

The effect of N-ethyl-N-hydroxyethyl perfluorooctanoamide ...

So ,fracturing fluid flowback, plays a vital role in shale gas development. Its purpose is to reduce the reflow of the proppant and to maintain the reservoir cracks with a strong ability for diversion.

(PDF) Computer simulation of hydraulic fractures | Jean ...

The conversion ,formula, between concentration in ppa and volumetric concentration c is (footnote continued) c ppa ¼ ðc=ð1 À cÞÞG s r w , where G s is the proppant solid's specific gravity, and r w is the water density in pounds per gallon (r w ' 8:34 lb=gal). 5 In reality, part of the gel fraction of the ,fluid,, plus the ,fluid, trapped in the interstitial pores of the proppant pack remains ...

Study of Hydraulic Fracturing Real-time Evaluation Technology

Dali Guo 1, Lei Xu 1, Xiaohui Zeng 2, Xijun Ke 1 and Bo Tao 1. 1 State Key Laboratory of Oil and Gas Reservoir Geology and Exploitation, School of Sciences, Southwest Petroleum University, Chengdu, Sichuan, 610500, China; 2 School of Petroleum Engineering, Southwest Petroleum University, Chengdu, Sichuan, 610500, China. Received: March 31, 2014. Revised: July 18, 2014.

Study of Hydraulic Fracturing Real-time Evaluation Technology

Dali Guo 1, Lei Xu 1, Xiaohui Zeng 2, Xijun Ke 1 and Bo Tao 1. 1 State Key Laboratory of Oil and Gas Reservoir Geology and Exploitation, School of Sciences, Southwest Petroleum University, Chengdu, Sichuan, 610500, China; 2 School of Petroleum Engineering, Southwest Petroleum University, Chengdu, Sichuan, 610500, China. Received: March 31, 2014. Revised: July 18, 2014.

The effect of N-ethyl-N-hydroxyethyl perfluorooctanoamide ...

So ,fracturing fluid flowback, plays a vital role in shale gas development. Its purpose is to reduce the reflow of the proppant and to maintain the reservoir cracks with a strong ability for diversion.

Selection of Diameter of Spherical Tagin Fracturing ...

rate, of fracture ,fluid, at well head. The following are force ... ,Fracturing fluid flowback, is an important part of hydraulic ,fracturing, technique, ... and the ,formula, of air drag showing as 2 1 2 f C Sv , we can think that the thrustapply to tag from ,fracturing fluid

Testing device and method for simulating the flowback in ...

The invention discloses a testing device and method for simulating the ,flowback, in the shut-in period of fractured wells, comprising a ,fracturing fluid, tank, a ,flowback fracturing fluid, tank, a hydraulic ,fracturing, pump, a ,flowback fluid, collection beaker, and a shut-in ,flowback, testing system; the shut-in ,flowback, testing system is composed of a box body and an acoustic wave testing device, a ...

A semi‐analytical model for capturing dynamic behavior of ...

Abstract Hydraulic ,fracturing, has been successfully employed for unconventional oil and gas recovery for decades. During ,flowback,, the closure of the fracture may exhibit with the pressure drop of ...

The effect of N-ethyl-N-hydroxyethyl perfluorooctanoamide ...

So ,fracturing fluid flowback, plays a vital role in shale gas development. Its purpose is to reduce the reflow of the proppant and to maintain the reservoir cracks with a strong ability for diversion.