text:
Van't Hoff Equation

Van't Hoff equation - effect on temperature

Equilibrium constant Kp of a reaction is constant at a given temperature but when the temperature is changed, the value of equilibrium constant also changed.
The quantitative relation, relation, known as Van't Hoff equation is derived by using the Gibbs - Helmholtz equation.
The Gibbs - Helmholtz equation is,
ΔG⁰ = ΔH⁰ + T[d(ΔG⁰)/dT]p
Zero superscripts are indicating stranded values

or,- (ΔH⁰/T² )= -(ΔG⁰/T² )+(1/T)[d(ΔG⁰)/dT]p
or, - (ΔH⁰/T² ) = [d/dT(ΔG⁰/T)]p

Again Van't Hoff isotherm is,

- RT lnKp = ΔG⁰

or, - R lnKp = ΔG⁰/T

Differentiating with respect to T at constant P,
- R [dlnKp/dT]p = [d/dT(ΔG⁰/T)]p

Comparing the above two equation we have,

dlnKp/dT = ΔH⁰/T²
This is the differential form of Van't Hoff reaction equation.
The greater the value of ΔH⁰, the faster the equilibrium constant (Kp) changes with temperature (T).
Separating the variables and integrating,
∫ dlnKp = (ΔH⁰/R)∫ (dT/T²)

Assuming that ΔH⁰ is independent of temperature.
or, lnKp = - (ΔH⁰/R)(1/T)+C
where C is integrating constant

The integration constant can be evaluated and identified as ΔS⁰/R using the relation,
ΔG⁰ = ΔH⁰ - TΔS⁰

Thus the Van't hoff Equation is,

lnKp = - (ΔH⁰/RT) + (ΔS⁰/R)

Problem
For a reaction 2A + B ⇆ 2C, ΔG⁰(500 K) = 2 KJ mol⁻¹ find the Kp at 500 K for the reaction A + ½B ⇆ C.
Solution
ΔG⁰(500 K) for the reaction,
A + ½B ⇆ C; (2 KJ mol⁻¹)/2 = 1 KJ mol⁻¹.
The relation is ΔG⁰ = - RT lnKp
or, 1 = - (8.31 × 10-3 ) × (500) lnKP
or, lnKP = 1/(8.31 × 0.5)
or, lnKP = 0.2406

∴ KP = 1.27

Plot of lnKP vs 1/T from Van't Hoff equation

• For exothermic reaction, ΔH⁰ = (-) ve.
Examples are the formation of ammonia from H₂ and N₂.
 N₂ + H₂ ⇆ 2NH₃ ΔH⁰ = (-) ve
• For endothermic reaction, ΔH⁰ = (+) ve.
Examples are the dissociation of HI into H₂ and I₂.
 HI ⇆ H₂ + I₂ ΔH⁰ = (+) ve
• For the reaction, ΔH⁰ = 0.
lnKp is independent of T. Provided ΔS⁰ does not change much with T.
 Van't Hoff equation

Since for ideal system, H is not a function of P and ΔH⁰ = ΔH and Van't Hoff equation is,
dlnKp/dt = ΔH/RT²

The integrated equation is,
lnKp = (ΔH/RT) + ΔS/R
However, S of an ideal gas depends strongly on P, so ΔS and ΔG per mole of reaction in the mixture differ quite substantially from ΔS⁰ and ΔG⁰.
If we consider the integrated Van't Hoff equation at two temperature then, it becomes,
ln(Kp₂/Kp₁) = (ΔH/R){(T₂ - T₁)/T₁T₂)}
Where Kp₁ and Kp₂ are the equilibrium constants of the reaction at two different temperatures T₁ and T₂ respectively.
Thus the determination of Kp₁ and Kp₂ at two temperatures helps to calculate the value of ΔH of the reaction.
The above relation is called Van't Hoff reaction isobar since P remains constant during the change of temperature.
Problem
Use Gibbs - Helmholtz equation to derive the Van't Hoff reaction isochore. What condition do you expect a linear relationship between log k and 1/T?
Solution
Again Kp = Kc (RT)ΔƔ
or, lnKp = lnKc + ΔƔ lnR + ΔƔ lnT
Differentiating with respect to T,
dlnKp/dT = dlnKc/dT + ΔƔ/T
But dlnKp/dT = ΔH⁰/RT²
Hence, dlnKc/dT = ΔH⁰/RT² - ΔƔ/T
or, dlnKc/dT = (ΔH⁰ - ΔƔRT)/RT²

∴ dlnKc/dT = ΔU⁰/RT²
ΔU⁰ is the standers heat of reaction at constant volume. This is really the Van't Hoff reaction isochore in differential form.
The integrated form of the equation at two temperature is,
ln(KC₂/KC₁) = (ΔU⁰/R){(T₂ - T₁)/T₁T₂)}
For ideal system ΔU⁰ = ΔU. Since the reaction occurs in an equilibrium box at constant Volume this is called Van't Hoff reaction isochore.
• Two important assumptions are
1. The reacting system is assumed to behave ideally.
2. ΔH is taken independent of temperature for a small range of temperature change.
Due to the assumption involved ΔH and ΔU do not produce the precise value of these reactions. As ΔG is independent of pressure for ideal system ΔH and ΔU also independent of pressure.
Hence, ΔG⁰ = - RT lnKp
or, [dlnKp/dT]T = 0
Problem
Show that the equilibrium condition for any chemical reaction is given by ΔG = 0.
Solution
Van't Hoff reaction isotherm is,
ΔG = - RT lnKa + RT lnQa
But when the reaction attains an equilibrium,
Qa = Ka

∴ ΔG = 0

Le -Chatelier principle from Van't Hoff equation

Van't Hoff equations give a quantitative expression of the Le-Chatelier Principle.
lnKp = - (ΔH/R)(1/T) + C
It is evident that for endothermic reaction (ΔH〉0), an increase of T increases the value of any of the reactions.
But for exothermic reaction (ΔHㄑ 0), with rising in temperature, a is decreased. This change of Kp also provides the calculation of the quantitative change of equilibrium yield of products.
This above statement is in accordance with the Le - Chatelier Principle which states that whenever stress is placed on any system in a state of equilibrium, the system always reacts in a direction to reduce the applied stress.
Thus when T is increased, the system in equilibrium will try to move in a direction in which heat is absorbed that is endothermic reaction is favors.

Effect of temperature on equilibrium constant, Van't Hoff equation, expression of Le -Chatelier principle from Van't Hoff equation, problem solution

Name

Email *

Message *


text:
## CryptoDB

### Xianhui Lu

#### Publications

Year
Venue
Title
2018
ASIACRYPT
Motivated by abstracting the common idea behind several implicitly authenticated key exchange (AKE) protocols, we introduce a primitive that we call double-key key encapsulation mechanism (2-key KEM). It is a special type of KEM involving two pairs of secret-public keys and satisfying some function and security property. Such 2-key KEM serves as the core building block and provides alternative approaches to simplify the constructions of AKE. To see the usefulness of 2-key KEM, we show how several existing constructions of AKE can be captured as 2-key KEM and understood in a unified framework, including widely used HMQV, NAXOS, Okamoto-AKE, and FSXY12-13 schemes. Then, we show (1) how to construct 2-key KEM from concrete assumptions, (2) how to adapt the classical Fujisaki-Okamoto transformation and KEM combiner to achieve the security requirement of 2-key KEM, (3) an elegant Kyber-AKE over lattice using the improved Fujisaki-Okamoto technique.
2015
EUROCRYPT
2014
EPRINT
2011
PKC
2010
EPRINT
In this paper we propose two public key encryption schemes over the semi-smooth subgroup introduced by Groth05. Both the schemes are proved secure against chosen ciphertext attacks under the factoring assumption. Since the domain of exponents is much smaller, both our schemes are significantly more efficient than Hofheiz-Kiltz 2009 encryption.
2008
EPRINT
Kiltz proposed a practical key encapsulation mechanism(Kiltz07-KEM) which is secure against adaptive chosen ciphertext attacks(IND-CCA2) under the gap hashed Diffie-Hellman(GHDH) assumption\cite{Kiltz2007}. We show a variant of Kiltz07-KEM which is more efficient than Kiltz07-KEM in encryption. The new scheme can be proved to be IND-CCA2 secure under the same assumption, GHDH.
2008
EPRINT
Concurrent signature provided a novel idea for fair exchange protocol without trusted third party. Perfect Concurrent Signature is proposed to strengthen theambiguity of the concurrent signature. Wang et al, pointed out there exist an attack against the fairness of Perfect Concurrent Signature and proposed the improved perfect concurrent signature. This paper find that in proposed (perfect) concurrent signature protocol, no matter two party or multi-party, the signer could bind multiple messages with one keystone set but let the other signers know only one of the messages. This is a new unfair case in the application of concurrent signature. Based on this observation, we propose that accountability should be one of the security properties of (perfect) concurrent signature and we give the definition of accountability of concurrent signature. To illustrate this idea, we give an attack scene against the accountability of improved perfect concurrent signature proposed by Wang et al, and propose an update version of perfect concurrent signature to avoid such attack.
2008
EPRINT
An efficient variant of the ElGamal public key encryption scheme is proposed which is provably secure against adaptive chosen ciphertext attacks(IND-CCA2) under the decisional Diffie-Hellman(DDH) assumption. Compared to the previously most efficient scheme under DDH assumption by Kurosawa and Desmedt [Crypto 2004] it has one group element shorter ciphertexts, 50\% shorter secret keys, 25\% shorter public keys and it is 28.6\% more efficient in terms of encryption speed, 33.3\% more efficient in terms of decryption speed. A new security proof logic is used, which shows directly that the decryption oracle will not help the adversary in the IND-CCA2 game. Compared to the previous security proof, the decryption simulation is not needed in the new logic. This makes the security proof simple and easy to understand.
2007
EPRINT
Kurosawa and Matsuo\cite{Kurosawa20042} showed that MAC can be removed from DHIES while the underlying symmetric-key encryption(SKE) scheme is secure against adaptive chosen ciphertext attacks(IND-CCA). We construct a variant of DHIES which eliminate the MAC while the SKE scheme is secure against passive attacks(IND-PA). Since IND-PA is the basic requirement of SKE schemes, the new scheme is more flexible than \cite{Kurosawa20042}. Our new scheme can be seen as a combination of a tag-KEM \cite{Abe2005} and a DEM. Our construction offers the first tag-KEM with single element. When the hash function $H$ in the ODH assumption is a non-malleable hash function we can prove that the new scheme is IND-CCA secure under the ODH assumption.
2007
EPRINT
For all current adaptive chosen ciphertext(CCA) secure public key encryption schemes in standard model there are two operations in the decryption algorithm, validity check" and decryption. The decryption algorithm returns the corresponding plaintext if the ciphertext is valid otherwise it returns a rejection symbol $\perp$. We call this paradigm invalid ciphertext rejection". However the validity check" is not necessary for an encryption scheme. Also in this case the adversary will get the information that the ciphertext is "invalid" which he may not know before the decryption query. We propose a new paradigm for constructing CCA secure public key encryption schemes which combines validity check" and decryption together. The decryption algorithm will execute the same operation regardless of the ciphertext's validity. We call this new paradigm uniform decryption". Compared with the "invalid ciphertext rejection" paradigm, the decryption oracle of schemes in the new paradigm will reveal less information. The attacker even can not get whether the queried ciphertext is valid" or not. Moreover the combination of validity check" and the decryption will yield more efficient schemes. Using the new paradigm we construct an efficient public key encryption scheme. Our scheme is more efficient than CS98 in both computation and bandwidth. Compered with KD04 and HK07 the new scheme is more efficient in bandwidth and the same efficient in computation. The new scheme is as efficient as Kiltz07 both in computation and bandwidth. However the new scheme is CCA secure based on DDH assumption which is more flexible than GHDH assumption that Kiltz07 based on. Kurosawa and Desmedt proposed an efficient hybrid scheme named as KD04\cite{Kurosawa2004}. Although the key encapsulation part of KD04(KD04-KEM) is not CCA secure \cite{Hofheinz2006}, the whole scheme can be proved to be CCA secure. We show that if the key derivation function(KDF) of KD04-KEM is a non-malleable hash function it will be a CCA secure KEM in the new paradigm.
2007
EPRINT
In this paper, by elaborately choosing the parameters of Waters Hash function, we propose a new efficient signature scheme. It is shown that the scheme is secure against strongly unforgeable chosen-message attacks in the standard model under Computational Diffie-Hellman (CDH) assumption. Further, among all the known secure signatures in the standard model, our scheme is the shortest one and has the efficient security reduction as well.
2007
EPRINT
We propose a security notion named as weak adaptive chosen ciphertext security(IND-WCCA) for hybrid encryption schemes. Although it is weaker than adaptive chosen ciphertext security(IND-CCA), a IND-WCCA secure hybrid encryption scheme can be used in any situations that a IND-CCA secure hybrid encryption scheme used in. We show that IND-WCCA secure hybrid encryption scheme can be constructed from IND-CCA secure KEM and IND-PA secure DEM. Since IND-PA is the basic requirement of symmetric key encryption schemes, IND-WCCA hybrid encryption scheme is very flexible and can use most of the stream ciphers and block ciphers as the DEM part of the scheme. Use the new secure notion we can refine current IND-CCA secure hybrid encryption schemes and get more efficient IND-WCCA secure hybrid encryption schemes.
2007
EPRINT
M. Abdalla, M. Bellare and P. Rogaway proposed a variation of Diffie-Hellman assumption named as oracle Diffie-Hellman(ODH) assumption. They recommend to use a one-way cryptographic hash function for the ODH assumption. We notice that if the hash function is just one-way then there will be an attack. We show that if the the hash function is non-malleable then the computational version of ODH assumption can be reduced to the computational Diffie-Hellman(CDH) assumption. But we can not reduce the ODH assumption to the decisional Diffie-Hellman(DDH) even if the hash function is non-malleable. It seems that we need a random oracle hash function to reduce the ODH assumption to the DDH assumption.
2006
EPRINT
KD04 proposed by K. Kurosawa and Y. Desmedt is the most efficient public key encryption scheme provably secure against adaptive chosen ciphertext attack in standard model based on decision diffie-hellman problem. We proposed a simplify version of KD04 which is more efficient than KD04 while still can be proved to be secure against adaptive chosen ciphertext attack in standard model based on decision diffie-hellman problem.
2006
EPRINT
Cramer and Shoup proposed the first provably secure practical public-key encryption scheme in the standard model (CS98). We find new way to construct the secure reduction in which the decryption oracle is not needed yet. Thus we get a simplified version of CS98 which is more efficient than the original scheme, and also provably secure against chosen ciphertext attack in standard model.
2006
EPRINT
We construct an efficient identity based encryption scheme from pairing. The basic version of the new scheme is provably secure against chosen plaintext attack, and the full version of the new scheme is provably secure against adaptive chosen ciphertext attack. Our scheme is based on a new assumption (decision weak bilinear Diffie-Hellman assumption ) which is no stronger than decision bilinear Diffie-Hellman assumption.

Jia Fan (1)
Dake He (10)
Jingnan He (1)
Dingding Jia (4)
Li Kang (1)
Xuejia Lai (8)
Guomin Li (6)
Yunfeng Li (1)
Bao Li (5)
Bei Liang (1)
Qixiang Mei (3)
Xiaohu Tang (1)
Haiyang Xue (1)

format: plain


text:
Subjects -> MATHEMATICS (Total: 1074 journals)     - APPLIED MATHEMATICS (86 journals)    - GEOMETRY AND TOPOLOGY (23 journals)    - MATHEMATICS (793 journals)    - MATHEMATICS (GENERAL) (43 journals)    - NUMERICAL ANALYSIS (23 journals)    - PROBABILITIES AND MATH STATISTICS (106 journals) MATHEMATICS (793 journals)                  1 2 3 4 | Last

1 2 3 4 | Last

Similar Journals
 Acta Applicandae MathematicaeJournal Prestige (SJR): 0.675 Citation Impact (citeScore): 1Number of Followers: 1      Hybrid journal (It can contain Open Access articles) ISSN (Print) 0167-8019 - ISSN (Online) 1572-9036 Published by Springer-Verlag  [2570 journals]
• Symmetry and Localization for Magnetic Schrödinger Operators: Landau
Levels, Gabor Frames and All That
• Abstract: We investigate the relation between broken time-reversal symmetry and localization of the electronic states, in the explicitly tractable case of the Landau model. We first review, for the reader’s convenience, the symmetries of the Landau Hamiltonian and the relation of the latter with the Segal-Bargmann representation of Quantum Mechanics. We then study the localization properties of the Landau eigenstates by applying an abstract version of the Balian-Low Theorem to the operators corresponding to the coordinates of the centre of the cyclotron orbit in the classical theory. Our proof of the Balian-Low Theorem, although based on Battle’s main argument, has the advantage of being representation-independent.
PubDate: 2019-03-18

• Iterative Regularization Methods for the Multiple-Sets Split Feasibility
Problem in Hilbert Spaces
• Abstract: Abstract In this paper, we introduce iterative regularization methods for solving the multiple-sets split feasibility problem, that is to find a point closest to a family of closed convex subsets in one space such that its image under a bounded linear mapping will be closest to another family of closed convex subsets in the image space. We consider the cases, when the families are either finite or infinite. We also give two numerical examples for illustrating our main method.
PubDate: 2019-03-07

• Decay Estimates for Solutions of Porous Medium Equations with Advection
• Abstract: Abstract In this paper, we show that bounded weak solutions of the Cauchy problem for general degenerate parabolic equations of the form $$u_{t} + \operatorname{div}f(x,t,u) = \operatorname{div}\bigl( u ^{\alpha } \nabla u\bigr), \quad x \in \mathbb{R}^{n} , \ t > 0,$$ where $$\alpha > 0$$ is constant, decrease to zero, under fairly broad conditions on the advection flux $$f$$ . Besides that, we derive a time decay rate for these solutions.
PubDate: 2019-03-05

• Determine a Space-Dependent Source Term in a Time Fractional
Diffusion-Wave Equation
• Abstract: Abstract This paper is devoted to identify a space-dependent source term in a multi-dimensional time fractional diffusion-wave equation from a part of noisy boundary data. Based on the series expression of solution for the direct problem, we improve the regularity of the weak solution for the direct problem under strong conditions. And we obtain the uniqueness of inverse space-dependent source term problem by the Titchmarsh convolution theorem and the Duhamel principle. Further, we use a non-stationary iterative Tikhonov regularization method combined with a finite dimensional approximation to find a stable source term. Numerical examples are provided to show the effectiveness of the proposed method.
PubDate: 2019-03-05

• The Dual Kaczmarz Algorithm
• Abstract: Abstract The Kaczmarz algorithm is an iterative method for solving a system of linear equations. It can be extended so as to reconstruct a vector $$x$$ in a (separable) Hilbert space from the inner-products $$\{\langle x, \phi _{n} \rangle \}$$ . The Kaczmarz algorithm defines a sequence of approximations from the sequence $$\{\langle x, \phi _{n} \rangle \}$$ ; these approximations only converge to $$x$$ when $$\{\phi _{n}\}$$ is effective. We dualize the Kaczmarz algorithm so that $$x$$ can be obtained from $$\{\langle x, \phi _{n} \rangle \}$$ by using a second sequence $$\{\psi _{n}\}$$ in the reconstruction. This allows for the recovery of $$x$$ even when the sequence $$\{\phi _{n}\}$$ is not effective; in particular, our dualization yields a reconstruction when the sequence $$\{\phi _{n}\}$$ is almost effective. We also obtain some partial results characterizing when the sequence of approximations from $$\{\langle x, \phi _{n} \rangle \}$$ using $$\{\psi _{n}\}$$ converges to $$x$$ , in which case $$\{(\phi _{n}, \psi _{n})\}$$ is called an effective pair.
PubDate: 2019-02-14

• Existence of Continuous Eigenvalues for a Class of Parametric Problems
Involving the ( p , 2 ) $(p,2)$ -Laplacian Operator
• Abstract: Abstract We discuss a parametric eigenvalue problem, where the differential operator is of $$(p,2)$$ -Laplacian type. We show that, when $$p\neq 2$$ , the spectrum of the operator is a half line, with the end point formulated in terms of the parameter and the principal eigenvalue of the Laplacian with zero Dirichlet boundary conditions. Two cases are considered corresponding to $$p>2$$ and $$p<2$$ , and the methods that are applied are variational. In the former case, the direct method is applied, whereas in the latter case, the fibering method of Pohozaev is used. We will also discuss a priori bounds and regularity of the eigenfunctions. In particular, we will show that, when the eigenvalue tends towards the end point of the half line, the supremum norm of the corresponding eigenfunction tends to zero in the case of $$p>2$$ , and to infinity in the case of $$p < 2$$ .
PubDate: 2019-02-07

• Spine Decompositions and Limit Theorems for a Class of Critical
Superprocesses
• Abstract: Abstract In this paper we first establish a decomposition theorem for size-biased Poisson random measures. As consequences of this decomposition theorem, we get a spine decomposition theorem and a 2-spine decomposition theorem for some critical superprocesses. Then we use these spine decomposition theorems to give probabilistic proofs of the asymptotic behavior of the survival probability and Yaglom’s exponential limit law for critical superprocesses.
PubDate: 2019-02-07

• Existence and Convergence Theorems for Global Minimization of Best
Proximity Points in Hilbert Spaces
• Abstract: Abstract In order to solve global minimization problems involving best proximity points, we introduce general Mann algorithm for nonself nonexpansive mappings and then prove weak and strong convergence of the proposed algorithm under some suitable conditions in real Hilbert spaces. Furthermore, we also provide numerical experiment to illustrate the convergence behavior of our proposed algorithm.
PubDate: 2019-02-06

• Existence and Multiplicity Solutions for the p $p$ -Fractional
Schrödinger–Kirchhoff Equations with Electromagnetic Fields and
Critical Nonlinearity
• Abstract: Abstract This paper is devoted to the study of the $$p$$ -fractional Schrödinger–Kirchhoff equations with electromagnetic fields and critical nonlinearity. By using the variational methods, we obtain the existence of mountain pass solutions $$u_{\varepsilon }$$ which tend to the trivial solutions as $$\varepsilon \rightarrow 0$$ . Moreover, we get $$m^{\ast }$$ pairs of solutions for the problem in absence of magnetic effects under some extra assumptions.
PubDate: 2019-02-04

• Bernstein Fractal Trigonometric Approximation
• Abstract: Abstract Fractal interpolation and approximation received a lot of attention in the last thirty years. The main aim of the current article is to study a fractal trigonometric approximants which converge to the given continuous function even if the magnitude of the scaling factors does not approach zero. In this paper, we first introduce a new class of fractal approximants, namely, Bernstein $$\alpha$$ -fractal functions using the theory of fractal approximation and Bernstein polynomial. Using the proposed class of fractal approximants and imposing no condition on corresponding scaling factors, we establish that the set of Bernstein $$\alpha$$ -fractal trigonometric functions is fundamental in the space of continuous periodic functions. Fractal version of Gauss formula of trigonometric interpolation is obtained by means of Bernstein trigonometric fractal polynomials. We study the Bernstein fractal Fourier series of a continuous periodic function $$f$$ defined on $$[-l,l]$$ . The Bernstein fractal Fourier series converges to $$f$$ even if the magnitude of the scaling factors does not approach zero. Existence of the $$\mathcal{C}^{r}$$ -Bernstein fractal functions is investigated, and Bernstein cubic spline fractal interpolation functions are proposed based on the theory of $$\mathcal{C}^{r}$$ -Bernstein fractal functions.
PubDate: 2019-02-01

• Nonlinear Elliptic Equations Without Sign Condition and L 1 $L^{1}$ -Data
in Musielak-Orlicz-Sobolev Spaces
• Abstract: Abstract In this research we give the existence of solutions to a elliptic problem containing two lower order terms, the first nonlinear term satisfying the growth conditions and without sign conditions and the second is a continuous function on ℝ. Not also that for right hand side, it is assumed that to be merely integrable. This results in formulation of the problem in Musielak-Orlicz-Sobolev spaces.
PubDate: 2019-02-01

• Study of a Tritrophic Food Chain Model with Non-differentiable Functional
Response
• Abstract: Abstract We study a model of three interacting species in a food chain composed by a prey, an specific predator and a generalist predator. The capture of the prey by the specific predator is modelled as a modified Holling-type II non-differentiable functional response. The other predatory interactions are both modelled as Holling-type I. Moreover, our model follows a Leslie-Gower approach, in which the function that models the growth of each predator is of logistic type, and the corresponding carrying capacities depend on the sizes of their associated available preys. The resulting model has the form of a set of nonlinear ordinary differential equations which includes a non-differentiable term. By means of topological equivalences and suitable changes of parameters, we find that there exists an Allee threshold for the survival of the prey population in the food chain, given, effectively, as a critical level for the generalist predator. The dynamics of the model is studied with analytical and computational tools for bifurcation theory. We present two-parameter bifurcation diagrams that contain both local phenomena (Hopf, saddle-node transcritical, cusp, Bogdanov-Takens bifurcations) and global events (homoclinic and heteroclinic connections). In particular, we find that two types of heteroclinic cycles can be formed, both of them containing connections to the origin. One of these cycles is planar involving the absence of the specific predator. In turn, the other heteroclinic cycle is formed by connections in the full three-dimensional phase space.
PubDate: 2019-01-24

• Multiple Solutions for a Non-cooperative Elliptic System of Kirchhoff Type
Involving p $p$ -Biharmonic Operator and Critical Growth
• Abstract: Abstract In this paper, we consider a class of non-cooperative elliptic systems of Kirchhoff type involving $$p$$ -biharmonic operator and critical growth. With the help of the Limit index theory due to Li (Nonlinear Anal. TMA 30(7):4619–4627, 1997) and the concentration compactness principle, we establish the existence of infinitely many solutions for the problem under the suitable conditions on the nonlinearity.
PubDate: 2019-01-23

• Monopoles, Dipoles, and Harmonic Functions on Bratteli Diagrams
• Authors: Sergey Bezuglyi; Palle E. T. Jorgensen
Abstract: Abstract In our study of electrical networks we develop two themes: finding explicit formulas for special classes of functions defined on the vertices of a transient network, namely monopoles, dipoles, and harmonic functions. Secondly, our interest is focused on the properties of electrical networks supported on Bratteli diagrams. We show that the structure of Bratteli diagrams allows one to describe algorithmically harmonic functions as well as monopoles and dipoles. We also discuss some special classes of Bratteli diagrams (stationary, Pascal, trees), and we give conditions under which the harmonic functions defined on these diagrams have finite energy.
PubDate: 2018-05-31
DOI: 10.1007/s10440-018-0189-7

• Skeleton Decomposition and Law of Large Numbers for Supercritical
Superprocesses
• Authors: Zhen-Qing Chen; Yan-Xia Ren; Ting Yang
Abstract: Abstract The goal of this paper is twofold. First, we establish skeleton and spine decompositions for superprocesses whose underlying processes are general symmetric Hunt processes. Second, we use these decompositions to obtain weak and strong law of large numbers for supercritical superprocesses where the spatial motion is a symmetric Hunt process on a locally compact metric space $$E$$ and the branching mechanism takes the form $$\psi _{\beta }(x,\lambda )=-\beta (x)\lambda +\alpha (x)\lambda ^{2}+ \int _{(0,{\infty })}\bigl(e^{-\lambda y}-1+\lambda y\bigr)\pi (x,dy)$$ with $$\beta \in \mathcal{B}_{b}(E)$$ , $$\alpha \in \mathcal{B}^{+}_{b}(E)$$ and $$\pi$$ being a kernel from $$E$$ to $$(0,{\infty })$$ satisfying $$\sup_{x\in E}\int _{(0,{\infty })} (y\wedge y^{2}) \pi (x,dy)<{\infty }$$ . The limit theorems are established under the assumption that an associated Schrödinger operator has a spectral gap. Our results cover many interesting examples of superprocesses, including super Ornstein-Uhlenbeck process and super stable-like process. The strong law of large numbers for supercritical superprocesses are obtained under the assumption that the strong law of large numbers for an associated supercritical branching Markov process holds along a discrete sequence of times, extending an earlier result of Eckhoff et al. (Ann. Probab. 43(5):2594–2659, 2015) for superdiffusions to a large class of superprocesses. The key for such a result is due to the skeleton decomposition of superprocess, which represents a superprocess as an immigration process along a supercritical branching Markov process.
PubDate: 2018-05-29
DOI: 10.1007/s10440-018-0190-1

• Dynamics for a Nonlocal Reaction-Diffusion Population Model with a Free
Boundary
• Authors: Yaling Zhao; Zuhan Liu; Ling Zhou
Abstract: Abstract In this paper we focus on a nonlocal reaction-diffusion population model. Such a model can be used to describe a single species which is diffusing, aggregating, reproducing and competing for space and resources, with the free boundary representing the expanding front. The main objective is to understand the influence of the nonlocal term in the form of an integral convolution on the dynamics of the species. Precisely, when the species successfully spreads into infinity as $$t\rightarrow \infty$$ , it is proved that the species stabilizes at a positive equilibrium state under rather mild conditions. Furthermore, we obtain a upper bound for the spreading of the expanding front.
PubDate: 2018-05-22
DOI: 10.1007/s10440-018-0188-8

• Localization Results for Impulsive Second Order Coupled Systems on the
Half-Line and Application to Logging Timber by Helicopter
• Authors: Feliz Minhós; Robert de Sousa
Abstract: Abstract This work considers two types of a second order impulsive coupled system of differential equations with generalized jump conditions in half-line. For both problems it will be presented two localization results with different monotonicity assumptions on the nonlinearities and on the impulsive conditions. The arguments apply lower and upper solutions method combined with Nagumo type condition and truncature techniques. Last section contains an applications of one of this impulsive problems to logging timber by helicopter.
PubDate: 2018-05-08
DOI: 10.1007/s10440-018-0187-9

• Weak Type Estimates of the Fractional Integral Operators on Morrey Spaces
with Variable Exponents
• Authors: Kwok-Pun Ho
Abstract: Abstract We show that when the infimum of the exponent function equals to 1, the fractional integral operator is a bounded operator from the Morrey space with variable exponent to the weak Morrey space with variable exponent.
PubDate: 2018-04-26
DOI: 10.1007/s10440-018-0181-2

• Computing Optimal Distances to Pareto Sets of Multi-Objective Optimization
Problems in Asymmetric Normed Lattices
• Authors: X. Blasco; G. Reynoso-Meza; E. A. Sánchez-Pérez; J. V. Sánchez-Pérez
Abstract: Given a finite dimensional asymmetric normed lattice, we provide explicit formulae for the optimization of the associated (non-Hausdorff) asymmetric “distance” among a subset and a point. Our analysis has its roots and finds its applications in the current development of effective algorithms for multi-objective optimization programs. We are interested in providing the fundamental theoretical results for the associated convex analysis, fixing in this way the framework for this new optimization tool. The fact that the associated topology is not Hausdorff forces us to define a new setting and to use a new point of view for this analysis. Existence and uniqueness theorems for this optimization are shown. Our main result is the translation of the original abstract optimal distance problem to a clear optimization scheme. Actually, this justifies the algorithms and shows new aspects of the numerical and computational methods that have been already used in visualization of multi-objective optimization problems.
PubDate: 2018-04-26
DOI: 10.1007/s10440-018-0184-z

• Authors: Rasul Ganikhodzhaev; Farrukh Mukhamedov; Mansoor Saburov
Abstract: Abstract The paper is devoted to the study of elliptic quadratic operator equations over the finite dimensional Euclidean space. We provide necessary and sufficient conditions for the existence of solutions of elliptic quadratic operator equations. The iterative Newton-Kantorovich method for stable solutions is also presented.
PubDate: 2018-04-26
DOI: 10.1007/s10440-018-0183-0

JournalTOCs
School of Mathematical and Computer Sciences
Heriot-Watt University
Edinburgh, EH14 4AS, UK
Email: journaltocs@hw.ac.uk
Tel: +00 44 (0)131 4513762


text:
Math Help - First order exact diff. equation.

2. Re: First order exact diff. equation.

I've never seen the solution method that you have presented so I can't find if there's an error in it. On the other hand if we look at the solution method I know I do not get your book answer either. The fix is trivial and I suspect a typo. I'll run through how I know how to do it and maybe you can compare the two methods to find where your problem is.

I won't bother to prove that your original equation is exact. I'm assuming you already checked that. So I will start with the assumption that there is a function F(x, y) such that
$\frac{\partial F}{\partial x} = P(x, y) = e^x sin(y) + e^{-y}$

which means that
$F(x, y) = \int P(x, y) \partial x + \phi (y)$
(The $\partial x$ in the integrand is merely to specify we are keeping the value of y fixed during the integration.)

$F(x, y) = e^x sin(y) + xe^{-y} + \phi (y)$

We also know that
$\frac{\partial F}{\partial y} = Q(x, y)$

Integrating this out we get that $\phi (y) = C$

So the final solution will be
$F(x, y) = e^x sin(y) + xe^{-y} + C$

whereas the book claims $F(x, y) = C = e^x sin(y) + xe^{-y}$. Hence the suspicion of a typo.

-Dan

3. Re: First order exact diff. equation.

thank you, it makes sense.


text:
# Precalculus Examples

,
Use the distance formula to determine the distance between the two points.
Substitute the actual values of the points into the distance formula.
Simplify.
Subtract from .
One to any power is one.
Subtract from .
Raise to the power of .


text:
# Kvalor k de multiplicando por 2 de 105

###### Question:

Kvalor k de multiplicando por 2 de 105

### Ito ang nag dadagdag kariktan sa isang tula o sulatin at maaari ring tawaging metaporikal na pahayag. Ano ito?

Ito ang nag dadagdag kariktan sa isang tula o sulatin at maaari ring tawaging metaporikal na pahayag. Ano ito?...

### Research the Minnesota Importance Questionnaire (MIQ) and write one paragraph explaining the purpose

Research the Minnesota Importance Questionnaire (MIQ) and write one paragraph explaining the purpose and benefits of taking this questionnaire....

### What usually comes in our mind when we think about art, answer?

What usually comes in our mind when we think about art, answer?...

### Is Gause’s Principle accurate? Explain why or why not.

Is Gause’s Principle accurate? Explain why or why not....

### Match each term with its correct definition.Balance sheet?What someone ownsLiabilities?What someone

Match each term with its correct definition. Balance sheet ? What someone owns Liabilities ? What someone owes Assets ? Statement of financial position...

### Lori wants to distribute 35 peaches equally into a basket she will use more than one but fewer than 10 baskets how many

Lori wants to distribute 35 peaches equally into a basket she will use more than one but fewer than 10 baskets how many baskets does lori need...

### Use the figure below to determine the length of sides b and c. options: b = 5√3, c = 10 b =10, c =

Use the figure below to determine the length of sides b and c. options: b = 5√3, c = 10 b =10, c = 5√3 b = 5√2, c = 10√3 b = 5, c = 10√2 $Use the figure below to determine the length of sides b and c. options: b = 5√3, c = 10 b =10, c =$...

### Fill in the blanks ''The Good Samaritan.'

Fill in the blanks ''The Good Samaritan."...

### The fossils of human ancestors show an increase in complexity as we move forward in the geological time

The fossils of human ancestors show an increase in complexity as we move forward in the geological time scale. what does this complexity indicate?...


text:
pure chemical eqaution when a piece of calcium

Give the chemical equation for th following.Piece of …

When a piece of Calcium is placed in water, then it form Calcium Hydroxide [C a (O H) 2 ] and hydrogen gas (H 2 ).

Chemical Reactions | Encyclopedia

11/6/2018· CHEMICAL REACTIONS CONCEPT If chemistry were compared to a sport, then the study of atomic and molecular properties, along with learning about the elements and how they relate on the periodic table [1], would be like going to practice.

Chemical Reactions: An Introduction | Introductor…

Calcium oxide is sometimes very challenging to store in the chemistry laboratory. This compound reacts with moisture in the air and is converted to calcium hydroxide. If a bottle of calcium oxide is left on the shelf too long, it gradually absorbs moisture from the

/a>

The relative formula mass can be determined experimentally by reacting a measured mass of the pure solid with an excess of hydrochloric acid. The equation for this reaction is +2HCl caC12 + 21-120 The unreacted acid can then be determined by titration with a standard sodium //c …

Chemical Reactions | Chemistry for Non-Majors

A chemical equation is a representation of a chemical reaction that displays the reactants and products with chemical formulas. The chemical equation for the reaction of methane with oxygen is shown: The equation above, called a skeleton equation, is an equation that shows only the formulas of the reactants and products with nothing to indie the relative amounts.

Calcium | Ca (Element) - PubChem

Calcium, a metallic element, is fifth in abundance in the earth''s crust, of which it forms more than 3%. It is an essential constituent of leaves, bones, teeth, and shells. Never found in nature uncoined, it occurs abundantly as limestone, gypsum, and fluorite. Apatite is the fluorophosphate or chlorophosphate of calcium.

Balance Chemical Equation - Online Balancer

calcium hydroxide + carbon dioxide = calcium carbonate + water sulfur + ozone = sulfur dioxide Examples of the chemical equations reagents (a complete equation will be suggested): H 2 SO 4 + K 4 Fe(CN) 6 + KMnO 4 Ca(OH) 2 + H 3 PO 4 Na 2 S 2

Write chemical equation for the reaction a piece of …

19/1/2018· Calcium reacts slowly with water. The reaction forms calcium hydroxide, Ca(OH)2and hydrogen gas (H2). The calcium metal sinks in water and after an hour or so bubbles of hydrogen are evident, stuck to the surface of the metal. Ca+2H2O---->Ca(OH)2+H2

A piece of calcium metal is dropped in water write a …

2/1/2018· A piece of calcium metal is dropped in water write a chemical equation. beenafaisalhb7485 is waiting for your help. Add your answer and earn points.

A small piece of calcium metal is put into a trough …

A small piece of calcium metal is put into a trough containing water. There is effervescence and white turbidity is formed. Write an equation for the reaction. SOLUTION. C a + 2 H 2 O → C a ( O H) 2 + H 2. View Full Answer. Its FREE, you''re just one step away. Subjective Medium Published on …

Calcium | Ca (Element) - PubChem

Calcium, a metallic element, is fifth in abundance in the earth''s crust, of which it forms more than 3%. It is an essential constituent of leaves, bones, teeth, and shells. Never found in nature uncoined, it occurs abundantly as limestone, gypsum, and fluorite. Apatite is the fluorophosphate or chlorophosphate of calcium.

5.9: Classifying Chemical Reactions - Chemistry LibreTexts

23/5/2020· After exchanging partners, the balanced equation is: $\ce{NaCN} \left( aq \right) + \ce{HBr} \left( aq \right) \rightarrow \ce{NaBr} \left( aq \right) + \ce{HCN} \left( g \right)$ Think about your result. This is a double replacement reaction. All formulas are correct and

What happens when calcium reacts with water? | Study

Calcium hydroxide and hydrogen gas are generated when calcium reacts with water. This is a single displacement reaction. The equation is: Ca + 2H 2 O See full answer below.

Caridge International Examinations Caridge International …

6 Calcium chlorate(V), Ca(Cl O 3) 2, is made by reacting calcium hydroxide with chlorine gas. 6Ca(OH) 2 + 6Cl 2 Ca(Cl O 3) 2 + 5CaCl 2 + 6H 2 O 3(a) 8.88 g of calcium hydroxide and 7200 cm of chlorine gas are mixed together. (i) How many moles is 8.88

5.9: Classifying Chemical Reactions - Chemistry LibreTexts

23/5/2020· After exchanging partners, the balanced equation is: $\ce{NaCN} \left( aq \right) + \ce{HBr} \left( aq \right) \rightarrow \ce{NaBr} \left( aq \right) + \ce{HCN} \left( g \right)$ Think about your result. This is a double replacement reaction. All formulas are correct and

Experiment 3: PROPERTIES & CHANGES: PHYSICAL vs. CHEMICAL

Part I Physical versus Chemical Change: Separation of a Mixture In the procedure outlined below you will be observing some chemical as well as some physical changes. Part of the procedure will involve the separation of a mixture of solid calcium chloride 2 3

Chemical Reactions and Equations Class 10 Extra …

17/9/2020· When quicklime or calcium oxide (Cao) reacts with water, slaked lime (Ca(OH) 2] is formed. During this reaction a large amount of heat is released. So, this reaction is an exothermic reaction.

Balance Chemical Equation - Online Balancer

calcium hydroxide + carbon dioxide = calcium carbonate + water sulfur + ozone = sulfur dioxide Examples of the chemical equations reagents (a complete equation will be suggested): H 2 SO 4 + K 4 Fe(CN) 6 + KMnO 4 Ca(OH) 2 + H 3 PO 4 Na 2 S 2

Name: Unit 7- Chemical Equations

Unit 7: Chemical Equations Page 6 3. Solid calcium will react with liquid water to produce aqueous calcium hydroxide and hydrogen gas. 4. Butane gas (C 4 H 10) will react with oxygen gas to produce carbon dioxide gas and water vapor. 5. Solid aluminum 6.

Word Equations for Chemical Reactions Tutorial

A word equation shows the reactant reacting to produce the product: reactant → product The arrow in the general form of the equation above shows the direction in which the chemical reaction proceeds: that is, we start with "reactant" and we end up with

Question #a5e9d + Example - Socratic

2/7/2016· Calcium metal, Ca, will react with bromine, Br2, to form calcium bromide, CaBr2, a white solid. This is a redox reaction in which calcium is being oxidized, its oxidation state going from 0 to +2, and bromine is being reduced, its oxidation state going from 0 to −1.

Welcome to CK-12 Foundation | CK-12 Foundation

Write the balanced chemical equation for the reaction that occurs when a piece of aluminum metal is placed in a solution of silver nitrate. Using the solubility rules given above, predict whether or not the following compounds are soluble or insoluble in water.

7.10: Classifying Chemical Reactions - Chemistry LibreTexts

2/8/2019· Most chemical reactions can be classified into one or more of five basic types: acid–base reactions, exchange reactions, condensation reactions (and the reverse, cleavage reactions), and oxidation–reduction reactions. The general forms of these five kinds of reactions are summarized in Table 7.10. 1, along with examples of each.

What is the chemical equation for calcium and water? - …

Calcium is a metal and a typical reaction of a metal with water will be as follows: Metal + Water → Metal Hydroxide + Hydrogen Gas Thus, the chemical reaction for calcium and water will be: [Word Equation] Calcium + Water → Calcium hydroxide + Hyd

Demonstrations - Calcium + Water

Ca (s) + 2H 2 O (l) ——> Ca (OH) 2 (aq) + H 2 (g) In the following demonstration, a chunk of calcium metal is dropped into a beaker of distilled water. After a second or so, the calcium metal begins to bubble vigorously as it reacts with the water, producing …

A small piece of calcium metal is put into a trough …

A small piece of calcium metal is put into a trough containing water. There is effervescence and white turbidity is formed. Write an equation for the reaction. SOLUTION. C a + 2 H 2 O → C a ( O H) 2 + H 2. View Full Answer. Its FREE, you''re just one step away. Subjective Medium Published on …

Caridge International Examinations Caridge International …

6 Calcium chlorate(V), Ca(Cl O 3) 2, is made by reacting calcium hydroxide with chlorine gas. 6Ca(OH) 2 + 6Cl 2 Ca(Cl O 3) 2 + 5CaCl 2 + 6H 2 O 3(a) 8.88 g of calcium hydroxide and 7200 cm of chlorine gas are mixed together. (i) How many moles is 8.88


text:
## Thursday, April 23, 2015

### 4/21 Purple Comet Online Meet

This week of math club we tried out the online Purple Comet Math Contest. http://purplecomet.org. Running the contest required a few modifications to our normal format. We worked in the library so we could have access to the computers and skipped any snacks or warmups since the lasted the entire 60 minutes. I've also been delaying posting of this entry until after the contest ends so I can discuss a few of the problems.

Overall the contest worked out fairly well. Of 20 questions most of the first 15 were potentially doable with their knowledge base .  A few in between I think were most likely to need algebra and then their were the remainder that made me raise my eyebrows. Each team of 4-5 kids solved 4-6 problems out of 20 which was about what I expected although I was hoping for a little closer to 8. The hardest part for the kids was keeping focus after about 30-40 minutes. Its really hard to keep going and not just give up and start chatting after you've tilted at problems that need a clever solution.  I'm going to debrief everyone next time and see how they liked the experience. In the future we could do some of the problems as individual items on a week to week basis like I sometime use the five triangle problems instead of actually competing.   If done that way they could be potentially scaffolded a bit.  On the other hand sometimes the element of competition is motivating.  The other logistical problem I have is that more kids were out than normal and I plan to go over the problems as a group next session. I'm going to email out a pdf of the contest so those who were absent can look at them. The dilemma is that those who were here deserve a group discussion / understanding of solutions  to the problems they tried but talking about a large group of  problems that  were unseen tends to be counter productive for the rest. Also I don't plan to discuss most of the back problems unless prompted by the kids since I think they were mostly out of scope.

#### Selected Problem Discussion

6Find the least positive integer whose digits add to a multiple of 27 yet the number itself is not a multiple  of 27. For example, 87999921 is one such number.

My first thought was if the digits add to 27 then its definitely a multiple of 9 and that the example is more a red herring.  This is actually quite easy to find via a guess and check strategy. First there are no 1 or 2 digit numbers whose digits sum to 27 at all and there is only one 3 digit number which is 999 but its also divisible by 27 and not the answer. So then onto the 4 digit cases. Since we want to start with the smallest candidate set the 1000th place to 1 and maximize the lower places i.e. the next number is 1899.  1899 / 9 is 211 which is not divisible by 3 using the divisibility test and thus the answer.  As I remember both groups arrived at the solution.

12.
Right triangle ABC with a right angle at A has AB = 20 and AC = 15. Point D is on AB with BD  = 2
Points E and F are placed on ray CA and ray CB respectively such that CD is a median of triangle CEF,  Find the area of triangle CEF.

The hardest part of this problem is that the drawing was incomplete and I completely misinterpreted what they were asking on my first read. I assumed E was in the middle of CA and F in the middle of CB and that CD went along the median as so:

This figure as I found out is unconstrained and has multiple solutions. What I completely missed was they really meant ray not segment where it was written and CD is the exact median.

That means point F has to be out beyond the original triangle like the above picture and this is quite a different problem. Luckily this version is much simpler than my first interpretation and very solvable using similar triangles. Some initial observations: triangle CEF is made up of 3 triangles  one of which we already can calculate: CBD. Also because CD is the median of the triangle CBD + BDF is one half of the median and CDE is the other and they both have an equal area. So our goal is to to find either one of the remaining triangles and the area will fall out.

Now comes the  use of similar triangles. Since CD is a median that means ED is the same length as DF and that triangle EDA is similar to EFA'  scaled up 2x. So since DA is 18 FA' is 36 in length. Then we also know triangle CBA is similar to CFA' and we can scale again CA' / CA = FA'/BA or CA'/15 = 36/20 and CA' = 27.    That gives us all the dimensions needed to find requested triangle. CBD is  1/2 * 15 * 2 = 15 based on the original dimension. BDF 1/2 * 2 * (AA')  AA' = (27-15) = 12. so BDF is 12. That means CFD the sum of the two previous triangles is 15 + 12 = 27 and since that's also one half of the median of CEF, the final area must be 54.  Overall doable but a full diagram would improve the question making it less about what I found a subtle visualization.

19.  Find the real solution for the following:
$a^2 + 3b^2 + \frac{c^2 + 3d^2}2 = a + b + c + d - 1$

At first glance this looks ill-defined given the four variables. A little bit of experimentation shows that the left hand side is almost always larger than right and that there are no real solutions if you zero out any of the terms (although that requires alg2 analysis and some understanding of the quadratic formula)

The key here is realizing that this is actually the sum of four parabolas:

$a^2 -a, 3b^2 - b, \frac{c^2}2 - c$ and  $\frac{3d^2}2 - d$ each of which must have a minimum value.  The problem as stated strongly suggests that the sum of the minimums is likely to be the solution or otherwise there would be multiple answers but in any case its interesting to determine the floor for the total value.

Now either you know the formula for the vertex of a parabola $ax^2 + bx + c$ is  $-b/2a$ or set the derivative of each parabola function to zero to solve (Advanced Geometry or 1st year Calculus) for each term and you derive

$a^2 -a$ has a minimum of -1/4 at a =1/2
$3b^2 - b$ has a minimum of  -1/12 at b = 1/6
$\frac{c^2}2 - c$ has a minimum of -1/2 at c  = 1
$\frac{3d^2}2 - d$ has a minimum of -1/6  at d = 1/3

Lo and behold -1/4 + -1/12 + -1/2 + -1/6  =  -1!  and its done.  I'm waiting to see the official contest solutions but I don't see any obvious shortcuts for the geometry/quadratic analysis besides maybe graphing the parabolas and guessing the vertex from the graph. And even that would still require insight to split the function apart and that look at the sub functions. All of which I think makes this a great maybe 10-11th grade problem in my book rather than even a MS level question.


text:
# Math Help - [SOLVED] asymptotes of a tan function

1. ## [SOLVED] asymptotes of a tan function

what are the equations of the asymptotes for the function y=tan((2pi)/4)x where 0<x<4

I have trouble with this style of question whether it be find the x intercepts etc.

I got the normal asymptotes (pi/2, 3pi/2, etc.)
Edit-b=pi/2
but if i multiply pi/2 by pi/2, 3pi/2 It gives me the wrong answer.
HOw do I solve these kind of questions?

Thanks.

2. Originally Posted by brentwoodbc
what are the equations of the asymptotes for the function y=tan((2pi)/4)x where 0<x<4

I have trouble with this style of question whether it be find the x intercepts etc.

I got the normal asymptotes (pi/2, 3pi/2, etc.)
Edit-b=pi/2
but if i multiply pi/2 by pi/2, 3pi/2 It gives me the wrong answer.
HOw do I solve these kind of questions?

Thanks.
Since $\tan \left( \frac{2 \pi}{4} x \right) = \frac{\sin \left( \frac{2 \pi}{4} x\right)}{\cos \left( \frac{2 \pi}{4} x\right)}$, you just need to solve $\cos \left( \frac{2 \pi}{4} x\right) = 0$.

Note that you should first simplify $\frac{2 \pi}{4} x$.

3. so that would give you "one" value for x when y=0 correct, what's that have to do with the asymptotes?
I still dont understand why you cant multiply the original asymptotes by the b value.

4. Originally Posted by brentwoodbc
so that would give you "one" value for x when y=0 correct, what's that have to do with the asymptotes?
I still dont understand why you cant multiply the original asymptotes by the b value.
Do you understand the definition of a vertical asymptote? Do you see why I said to solve the equation I gave you? Did you try solving that equation?

5. Originally Posted by mr fantastic
Do you understand the definition of a vertical asymptote? Do you see why I said to solve the equation I gave you? Did you try solving that equation?

I do understand what an asymptote is
I "kind of" understand why you told me to solve it. maybe to find the first 0, then multiply that by the b value?
I got x=1?

6. Originally Posted by brentwoodbc
I do understand what an asymptote is
I dont understand why you told me to solve it. maybe to find the 0's?
I got x=1?
A vertical asymptote is a vertical line that passes through the value of x for which the function is undefined. When you're function has the form of a fraction, it will obviously be undefined when the denominator is equal to zero. tan can be written in the form of a fraction. I told you to solve when the denominator of this fraction is equal to zero.

x = 1 is one solution to the equation I told you to solve. There are others lying in the domain [0, 4].

7. Originally Posted by mr fantastic
A vertical asymptote is a vertical line that passes through the value of x for which the function is undefined. When you're function has the form of a fraction, it will obviously be undefined when the denominator is equal to zero. tan can be written in the form of a fraction. I told you to solve when the denominator of this fraction is equal to zero.

x = 1 is one solution to the equation I told you to solve. There are others lying in the domain [0, 4].
wow ok I think I understand, so to find the asymptotes do I multiply 1 by the b value? ie
pi/2, pi, 3pi/2, 2pi?

I have graphed it and the 1'st asymptote appears to be 1, then every 2, like 1,3,5...

so I guess since the period is 2, it's the 1 "first asymptote" plus the period (2) = 1,3,5 but since x is 0<x<4 its 1,3.

So when you solve for x like you got me to do it will always give the first asymptote/zero?

8. Originally Posted by brentwoodbc
wow ok I think I understand, so to find the asymptotes do I multiply 1 by the b value? ie
pi/2, pi, 3pi/2, 2pi?

I have graphed it and the 1'st asymptote appears to be 1, then every 2, like 1,3,5...

so I guess since the period is 4, it's the 1 "first asymptote" times 4/2, 2 times 4/2, = 1,3,5 but since x is 0<x<4 its 1,3.
Forget about recipes. Understand and apply the basic principles. Solve $\cos \left( \frac{\pi}{2} x \right) = 0$:

$\frac{\pi}{2} x = \frac{\pi}{2}$, or $\frac{\pi}{2} x = \frac{3\pi}{2}$

9. Originally Posted by mr fantastic
Forget about recipes. Understand and apply the basic principles. Solve $\cos \left( \frac{\pi}{2} x \right) = 0$:

$\frac{\pi}{2} x = \frac{\pi}{2}$, or $\frac{\pi}{2} x = \frac{3\pi}{2}$
I edited that post.
I forget, when you solve for x when y = 0 what x is given if there are multiple x's, the first one, or the one closest to 0?

10. Originally Posted by brentwoodbc
I edited that post.
I forget, when you solve for x when y = 0 what x is given if there are multiple x's, the first one, or the one closest to 0?
I don't understand what you're asking. I've given you the equation to solve and I've explained why you have to solve it. And I've all but solved that equation in my previous post.

The vertical asymptotes are x = 1 and x = 3. Try to see why.

11. Originally Posted by mr fantastic
I don't understand what you're asking. I've given you the equation to solve and I've explained why you have to solve it. And I've all but solved that equation in my previous post.

The vertical asymptotes are x = 1 and x = 3. Try to see why.
I 100% understand how to solve this now, (thanks very much.)
I am just asking if when you solve an equation for an x value like this equation "or any other equation" that has multiple x values, which x value will you get, the first, the closest to y=0 etc.

this is just purley out of curiosity.

an example could be (x-5)^2-4=0
x=8
why does solving for x give you 8 and not 4?

12. Originally Posted by brentwoodbc
....
an example could be (x-5)^2-4=0
x=8
why does solving for x give you 8 and not 4?
$(x - 5)^2 = 4 \Rightarrow x - 5 = \pm 2$ and so on.


text:
Expressing power sum symmetric polynomials in terms of lower degree power sums - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T20:28:25Z http://mathoverflow.net/feeds/question/36771 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/36771/expressing-power-sum-symmetric-polynomials-in-terms-of-lower-degree-power-sums Expressing power sum symmetric polynomials in terms of lower degree power sums Peter Erskin 2010-08-26T15:11:47Z 2010-09-01T07:35:04Z
Is there an explicit formula expressing the power sum symmetric polynomials $$p_k(x_1,\ldots,x_N)=\sum\nolimits_{i=1}^N x_i^k = x_1^k+\cdots+x_N^k$$ of degree $k$ in $N < k$ variables entirely through the power sum symmetric polynomials $p_j(x_1,\ldots,x_N)$ of degrees $j \le N$?

Examples: $$N=1,\ k=2: \quad p_2=x^2=x\times x=p_1^2$$

$$N=2,\ k=3: \quad p_3 = x^3 + y^3 = [3(x^2+y^2)(x+y)-(x+y)^3]/2 = (3 p_2 p_1-p_1^3)/2$$

What is the general formula?

I am looking for a formula similar to that for the expansion of the Schur functions $s_\lambda$ in terms of the symmetric power sums:

$$s_\lambda=\sum_{\rho=(1^{r_1},2^{r_2},3^{r_3},\dots)}\chi^\lambda_\rho \prod_j \frac{p^{r_j}_j}{r_j!},$$ where the coefficients $\chi^\lambda_\rho$ are the characters of the representation of the symmetric group indexed by the partition $\lambda$ evaluated at elements of cycle type indexed by the partition $\rho=(1^{r_1},2^{r_2},3^{r_3},\dots)$, which contains $r_j$ parts of length $j$.

Clearly, the power sums of degree higher than $N$ can be expanded in a similar manner: $$p_k=\sum_{\rho}a_{k;\rho}\prod_{j=1}^N p_j^{r_j},$$ where $\rho=(1^{r_1},2^{r_2},\dots,N^{r_N})$ is the partition of $k$ such that $k=r_1+2r_2+3r_3+...+Nr_N$.

In the above example for $N=2,\ k=3$ one has $a_{3;\ (1^{1},2^{1}) }=3/2$ and $a_{3;\ (1^{3},2^{0})}=-1/2$.

My question can be thus reformulated as follows: given $r_1,...,r_N$ what is the explicit formula for $a_{k;\rho}$?

Note Added

Actually, Wikipedia tells us how to construct a certain explicit formula for $p_k$. It gives the following expressions for $p_n$ with $n=N$ in terms of $e_j,$

$$p_n = \begin{vmatrix} e_1 & 1 & 0 & \cdots & \\ 2e_2 & e_1 & 1 & 0 & \cdots & \\ 3e_3 & e_2 & e_1 & 1 & \cdots & \\ \vdots &&& \ddots & \ddots & \\ ne_n & e_{n-1} & \cdots & & e_1 & \end{vmatrix},$$

and for $e_n$ with $n=N$ in terms of $p_j,$

$$e_n=\frac1{n!} \begin{vmatrix}p_1 & 1 & 0 & \cdots\\ p_2 & p_1 & 2 & 0 & \cdots \\ \vdots&& \ddots & \ddots \\ p_{n-1} & p_{n-2} & \cdots & p_1 & n-1 \\ p_n & p_{n-1} & \cdots & p_2 & p_1 \end{vmatrix}.$$

As far as I can see from the derivation described in Wikipedia, these determinant expressions are also valid for $p_n$ with $n > N$ and for $e_n$ with $n < N$.

For $p_n$ with $n>N$ one should take into account that all $e_k=0$ for $k > N$, so that the resulting matrix has zeros in both right-upper and left-lower corners.

Substituting the determinants for $e_j$ into the determinant for $p_k$, one gets the explicit formula which seems to solve the problem.

However, I still don't know how to obtain the coefficients $a_{k;\rho}$ in the expansion of $p_k$ in terms of the first $N$ power sums which would be the desired (really explicit) formula.
 http://mathoverflow.net/questions/36771/expressing-power-sum-symmetric-polynomials-in-terms-of-lower-degree-power-sums/36779#36779 Answer by Robin Chapman for Expressing power sum symmetric polynomials in terms of lower degree power sums Robin Chapman 2010-08-26T16:41:05Z 2010-08-26T18:15:43Z
If $e_1,\ldots,e_N$ are the elementary symmetric functions of $x_1,\ldots,x_N$ then for $k\ge N$ one has $$p_k=\sum_{j=1}^N(-1)^{j-1}e_j p_{k-j}.$$ This formula uses the elementary symmetric functions, which I presume you want to avoid, but it means that for $k\ge 2N$ the $(N+1)$ by $(N+1)$ matrix $$M_k=(p_{k-i-j})_{i,j=0}^N$$ has the null-vector $(1,-e_1,e_2,-e_3,\ldots,\pm e_N)$ and so $\det(M_k)=0$. Expanding this out gives an explicit formula for $p_k$ as a rational function (alas!) of $p_{k-1},\ldots,p_{k-2N}$.

Added I suppose one can express the $e_j$ in terms of $p_1,\ldots,p_n$ and put them into the above linear recurrence for $p_k$.
 http://mathoverflow.net/questions/36771/expressing-power-sum-symmetric-polynomials-in-terms-of-lower-degree-power-sums/36873#36873 Answer by Gjergji Zaimi for Expressing power sum symmetric polynomials in terms of lower degree power sums Gjergji Zaimi 2010-08-27T11:16:51Z 2010-08-27T12:06:21Z
Assuming you have $n$ variables then for $k\geq n$, Robin Chapman's identity above can be written as $$(p_n,p_{n-1},\dots, p_1)\begin{pmatrix} e_1 & 1 & \cdots & 0 \\ -e_2 & 0 & \ddots & \vdots \\ \vdots & \vdots & \ddots & 1 \\ (-1)^{n-1}e_n & 0 & \cdots & 0 \end{pmatrix}^{k-n}=(p_k,p_{k-1},\dots, p_{k-n+1})$$

Now to finish the job you need to express the $e_i$'s in terms of the power sum symmetric functions too. This is given by $$e_n=\sum_{|\lambda|=n}(-1)^{|\lambda|-l(\lambda)} z_{\lambda}^{-1}p_{\lambda}$$ where $|\lambda|$ is the size of the partition $\lambda$ and $l(\lambda)$ is its length, $p_{\lambda}=p_{\lambda_1}p_{\lambda_2}\cdots$ and $$z_{\lambda}=\prod_{i\geq 1}\left(i^{m_i}\cdot m_i!\right)$$ where $m_i$ is the number of parts of $\lambda$ equal to $i$.

I thought I'd remark that the formulas you are quoting are all valid in $\Lambda_{\mathbb{Q}}$, the ring of symmetric functions in infinitely many variables while the one you are searching for is not, because the $p_\lambda$'s are an orthogonal basis in this ring with $\langle p_{\lambda},p_{\mu}\rangle =\delta_{\lambda \mu}z_{\lambda}$. This is also the same reason why the formula for Schur polynomials may contain arbitrary $p_{\lambda}$'s in it. In fact the reason why that formula is important is because it gives you the transition from the basis of Schur polynomials to that of power sum polynomials in $\Lambda_{\mathbb{Q}}$.
 http://mathoverflow.net/questions/36771/expressing-power-sum-symmetric-polynomials-in-terms-of-lower-degree-power-sums/37235#37235 Answer by Akkerman for Expressing power sum symmetric polynomials in terms of lower degree power sums Akkerman 2010-08-31T08:59:39Z 2010-09-01T07:35:04Z
Combining the trace formula proposed by Gjergji Zaimi and Qiaochu Yuan, $$p_k={\rm Tr}\begin{pmatrix} e_1 & 1 & \cdots & 0 \\ -e_2 & 0 & \ddots & \vdots \\ \vdots & \vdots & \ddots & 1 \\ (-1)^{N-1}e_N & 0 & \cdots & 0 \end{pmatrix}^{k},$$ with the formula quoted by Peter Erskin, $$e_n=\frac1{n!} \begin{vmatrix}p_1 & 1 & 0 & \cdots\\ p_2 & p_1 & 2 & 0 & \cdots \\ \vdots&& \ddots & \ddots \\ p_{n-1} & p_{n-2} & \cdots & p_1 & n-1 \\ p_n & p_{n-1} & \cdots & p_2 & p_1 \end{vmatrix},$$ Mathematica produces the following expansions of $p_k$:

$$N=2$$

$$p_3=-\frac{1}{2}\ p_1^3+\frac{3}{2}\ p_1p_2$$

$$p_4=-\frac{1}{2}\ p_1^4+p_1^2p_2+\frac{1}{2}\ p_2^2$$

$$p_5=-\frac{1}{4}\ p_1^5+\frac{5}{4}\ p_1p_2 ^2$$

$$p_6=-\frac{3}{4}\ p_1^4p_2+\frac{3}{2}\ p_1^2p_2^2+\frac{1}{4}\ p_2^3$$

$$p_7=\frac{1}{8}\ p_1^7-\frac{7}{8}\ p_1^5p_2+\frac{7}{8}\ p_1^3p_2^2+\frac{7}{8}\ p_1p_2^3$$

$$p_8=\frac{1}{8}\ p_1^8-\frac{1}{2}\ p_1^6p_2-\frac{1}{4}\ p_1^4p_2^2+\frac{3}{2}\ p_1^2p_2^3+\frac{1}{8}\ p_2^4$$

$$p_9=\frac{1}{16}\ p_1^9-\frac{9}{8}\ p_1^5p_2^2+\frac{3}{2}\ p_1^3p_2^3 +\frac{9}{16}\ p_1p_2^4$$

$$p_{10}=\frac{5}{16}\ p_1^8p_2-\frac{5}{4}\ p_1^6p_2^2+\frac{5}{8}\ p_1^4p_2^3 +\frac{5}{4}\ p_1^2p_2^4+\frac{1}{16}\ p_2^5$$

$$p_{11}=-\frac{1}{32}\ p_1^{11}+\frac{11}{32}\ p_1^9p_2-\frac{11}{16}\ p_1^7p_2^2-\frac{11}{16}\ p_1^5p_2^3 +\frac{55}{32}\ p_1^3p_2^4+\frac{11}{32}\ p_1p_2^5$$

$$N=3$$

$$p_4=\frac{1}{6}\ p_1^4-p_1^2p_2+\frac{1}{2}\ p_2^2+ \frac{4}{3}\ p_1p_3$$

$$p_5=\frac{1}{6}\ p_1^5-\frac{5}{6}\ p_1^3p_2+\frac{5}{6}\ p_1^2p_3+\frac{5}{6}\ p_2p_3$$

$$p_6=\frac{1}{12}\ p_1^6-\frac{1}{4}\ p_1^4p_2-\frac{3}{4}\ p_1^2p_2^2+\frac{1}{4}\ p_2^3+\frac{1}{3}\ p_1^3p_2^3+p_1 p_2 p_3 +\frac{1}{3}\ p_3^2$$

$$p_7=\frac{1}{36}\ p_1^7-\frac{7}{12}\ p_1^3p_2^2+\frac{7}{36}\ p_1^4p_3+\frac{7}{12}\ p_2^2p_3+\frac{7}{9}\ p_1p_3^2$$

$$p_8=\frac{1}{72}\ p_1^8-\frac{1}{18}\ p_1^6p_2+\frac{1}{12}\ p_1^4p_2^2-\frac{1}{2}\ p_1^2p_2^3+\frac{1}{8}\ p_2^4+\frac{2}{9}\ p_1^5p_3$$ $$-\frac{8}{9}\ p_1^3p_2p_3+\frac{2}{3}\ p_1p_2^2p_3+\frac{8}{9}\ p_1^2p_3^2+\frac{4}{9}\ p_2p_3^2$$

$$N=4$$

$$p_5=-\frac{1}{24}\ p_1^5+\frac{5}{12}\ p_1^3p_2-\frac{5}{8}\ p_1p_2^2-\frac{5}{6}\ p_1^2p_3+\frac{5}{6}\ p_2p_3+\frac{5}{4}\ p_1p_4$$

$$p_6=-\frac{1}{24}\ p_1^6+\frac{3}{8}\ p_1^4p_2-\frac{3}{8}\ p_1^2p_2^2-\frac{1}{8}\ p_2^3-\frac{2}{3}\ p_1^3p_3+\frac{1}{3}\ p_3^2+\frac{3}{4}\ p_1^2p_4+\frac{3}{4}\ p_2p_4$$

$$p_7=-\frac{1}{48}\ p_1^7+\frac{7}{48}\ p_1^5p_2+\frac{7}{48}\ p_1^3p_2^2-\frac{7}{16}\ p_1p_2^2-\frac{7}{24}\ p_1^4p_3-\frac{7}{12}\ p_1^2p_2p_3$$ $$+\frac{7}{24}\ p_2^2p_3 +\frac{7}{24}\ p_1^3p_4+\frac{7}{8}\ p_1p_2p_4+\frac{7}{12}\ p_3p_4$$

$$p_8=-\frac{1}{144}\ p_1^8+\frac{1}{36}\ p_1^6p_2+\frac{5}{24}\ p_1^4p_2^2 -\frac{1}{4}\ p_1^2p_2^2-\frac{1}{16}\ p_2^4$$ $$-\frac{1}{9}\ p_1^5p_3-\frac{2}{9}\ p_1^3p_2p_3 -\frac{1}{3}\ p_1p_2^2p_3-\frac{4}{9}\ p_1^2p_3^2+\frac{4}{9}\ p_2p_3^2$$ $$+\frac{1}{12}\ p_1^4p_4+\frac{1}{2}\ p_1^2p_2p_4+\frac{1}{4}\ p_2^2p_4 +\frac{2}{3}\ p_1p_3p_4+\frac{1}{4}\ p_4^2.$$

It seems to me that a nice and compact formula for $a_{k,\rho}$ does exist. Indeed, the coefficients in the above examples are extremely simple.

In particular, I observe that the last terms in each of $p_k$ for $N=8$ have the form $$k\ \prod_j \frac{1}{j^{r_j}r_j!}p_j^{r_j},$$ which corresponds to $$a_{k,\rho}=k\prod_j \frac{1}{j^{r_j}r_j!}.$$ This formula (whose structure resembles the coefficients in the expansion of Schur functions quoted by Peter Erskin) also works for all terms of the type $p_jp_{k-j}$ at arbitrary $N$.

Apparently, this is not a general formula, as can be seen from the coefficients in front of $p_1^k$ which do depend on $N$. I believe, however, that the general formula for $a_{k,\rho}$ with $N$ properly included should not be much more complex than the empirical one above.

Hope this helps.


text:
# Indefinite Integral Question; e^x

1. Mar 3, 2012

1. The problem statement, all variables and given/known data
Find the indefinite integral (preferably using u-substitution):
∫(ex-e-x)2 dx

2. Relevant equations
N/A

3. The attempt at a solution
To be honest, I'm slightly confused as to which path I'm supposed to take with this, especially since I'm not sure what I should be using u to substitute here. This is one of my attempts at solving the question:

∫[(ex-e-x)2 dx]
= ∫[e2x dx] - ∫[e-2x dx]
= (e2x / 2) + (1 / 2e2x) + C

Could someone please tell me if I have the right answer / I'm on the right track / I'm completely wrong? If the latter, I'd appreciate it if you could help get me on track!

2. Mar 3, 2012

### BruceW

You almost got it. But you've forgotten the cross-terms when you did the square.

There is a good substitution you could use instead. But your method is just as good.

Edit: p.s. welcome to physicsforums, your name is good. reminds me of the town in pokemon

3. Mar 3, 2012

### rock.freak667

If you need the u-substitution, try just putting u=ex.

(Remember that 1/a is the same as a-1)

EDIT: Also in your attempt you should also have

∫e2x dx + ∫e-2x dx +∫2 dx

4. Mar 3, 2012

Before I continue asking questions, thanks for all the help! I really do appreciate it and I'm sorry if I'm a bit slow, haha.

Now then...
@BruceW: Thanks for the help, the welcome, and the name compliment! I'm actually a fan of the colour itself and the city in Pokemon, heh.
@rock.freak667: Ugh, how could I forget 1/a = a-1? Thanks for the reminder. I have a question about the +∫2 dx, though. How exactly did it get there?

I think I'm actually a bit more confused than I initially thought I was. Let's say I went with this method:

u=ex
du=exdx ; dx=du/ex

...and then what? This is what I'm leaning towards:
=∫(u2 dx) - ∫(1/u2 dx)

Do I substitute dx for du/ex or am I over-complicating things? Oh man, my head's gonna burst. @@

5. Mar 3, 2012

### SammyS

Staff Emeritus
(ex-e-x)2 = (ex-e-x)(ex-e-x)

Multiply it out. (Use FOIL, if that's what you're used to.)

6. Mar 3, 2012

### Dick

(e^x-e^(-x))^2=(e^x-e^(-x))*(e^x-e^(-x)). Multiply that out and you'll see where the '2' came from. rock.freak667 had the sign wrong on it.

7. Mar 3, 2012

Okay, do I tried expanding it and then went through this:

(ex-e-x)2 = (ex-e-x)(ex-e-x) = e2x + e-2x

∫(e2x + e-2x) dx
u = e2x
du = (e2x / 2) dx

Final solution:
(u2/e2x) + (2 / e2x)ln|e2x|

Last edited: Mar 3, 2012
8. Mar 3, 2012

### Staff: Mentor

No, this is incorrect, as Sammy and Dick pointed out. (a - b)2 ≠ a2 - b2, which is essentially what you are doing.
When you expand (ex - e-x)2 you should get three terms, not two.

Also, there is an error below - you shouldn't get any log terms.

9. Mar 3, 2012

Oh I see it now.
(ex - e-x)(ex - e-x) = e2x - 2 + e-2x

I somehow wrote "+ e0 - e0" at some point, oops. Thanks!
Time to try again. (:

10. Mar 3, 2012

### Staff: Mentor

For you last error, I think this is what you did:

∫dx/e2x = ln|e2x|

That is incorrect. Don't write the integrand as a fraction - use the negative exponents.

11. Mar 3, 2012

Okay, another attempt. Hope I fixed more than I broke, this time. :/
Thanks again everyone, I'm sorry for bothering you all. D:

∫(ex-e-x)2dx = ∫(e2x-2+e-2x)dx
= ∫e2xdx + ∫e-2xdx - ∫2dx

Let u = ex
du = exdx

∫e2xdx + ∫e-2xdx - ∫2dx = (1/ex)(u3/3)+(1/ex)(u-1/-1)-2x+C
= (1/ex)[(u3/3)+(u-1/-1)]-2x+C
= (1/ex)[(e3x/3) - (1/ex)]-2x+C
= (1/ex)[(e4x-3)/(3ex)]-2x+C
= [(e4x-3)/(3e2x)]-2x+C

Last edited: Mar 3, 2012
12. Mar 3, 2012

### SammyS

Staff Emeritus
What is $\displaystyle\int du \,?$

13. Mar 3, 2012

ex+C?

14. Mar 3, 2012

### SammyS

Staff Emeritus
Ignore my last post.

If u = ex and du = exdx,

then $\displaystyle\int e^{2x}dx$ becomes what integral in terms of u ?

15. Mar 3, 2012

∫u2dx? :x

[EDIT] Another attempt:
2e2x-2x-2e-2x+C

x_x

Last edited: Mar 3, 2012
16. Mar 3, 2012

### SammyS

Staff Emeritus
Yes. You appear to be confused.

You can't take 1/ex out of the integral. In fact you should leave it inside the integral as 1/u .

What is u2/u ?

17. Mar 3, 2012

u1 or just u? xD

(also, I edited my previous post with another answer in case you missed it.)

18. Mar 3, 2012

### SammyS

Staff Emeritus
OK, so what is $\displaystyle\int u \,du \,?$

19. Mar 3, 2012


text:
# Dual Energy Tests (Part 2)

## September 30, 2015

This post focuses on when the dual energy formalism should have an effect. From Bryan et al., 2014 (B14): “It should be stressed that as long as the parameter $\eta_1$ is small enough the dual energy method will have no dynamical effect.”

Here, $\eta_1$ refers to the ratio that determines whether a given cell updates its internal energy using the conservative update vs. the non-conservative update. As given in B14, Equation 44:

For the following tests, we use the Teyssier 2015 (T15) nonconservative update (Equation 161), with a centered-difference estimate for the divergence of the velocity:

We are using the B14 switch described above to calculate the internal energy for a given cell. We are also using the B14 method to decide when to synchronize the internal energy for a cell with the total energy. This is done again according to a ratio of internal to total energy, but this ratio takes into account the maximum energy of nearby cells, as well:

Note that in the tests shown below, the bulk velocity is high enough that the total energy is synced for every cell each timestep. That is, the ratio of conservatively calculated internal energy to maximum nearby total energy is always much less than $\eta_2$.

The tests

Is the statement above regarding dynamical impact true for Cholla?

The tests shown below consist of a shock tube with periodic boundaries, to allow for high mach number bulk flow.

For the high mach number test, $v_L = v_R = 120$. The tests are run until $t = 0.8$, on a domain from -3 to 3.

Without Dual Energy

Figure 1: PPMP, exact solver, no dual energy, $v_{bulk} = 100 c_s$

With Dual Energy

Figure 2: Same as figure one but with dual energy, $\eta_1 = 10^{-6}$

As expected, the dual energy formalism has no effect in this case. Practically speaking, this is because the internal energy for each cell is reset to the conservatively calculated value at every timestep, NOT because the hydro update is unaffected by a non-conservatively calculated internal energy.

So what happens if we increase $\eta_1$?

With $\eta_1 = 10^{-5}$ one cell every so often (not every time step) chooses the non-conservative internal energy update. The non-conservative $e$ in these cases is always slightly larger than the conservatively calculated $e$. Nevertheless, there is not a discernable dynamic impact.

With $\eta_1 = 10^{-4}$ all cells to the right of the discontinuity (where $e = 0.075$ initially) are using the non-conservative update. As the test begins, the two energies are the same for the majority of these, because there is no energy flux or velocity gradient. For the cells within the shock, however, the two values differ. At first, the non-conservative update has higher internal energy values, but as the test continues the values are sometimes lower. It seems that the non-conservative internal energy values are suffering from osciallations. Perhaps the instability is a result of the centered-difference approximation?

After time $t = 0.1863830$ the conservatively calculated internal energy is occasionally negative. The test does run to completion without the non-conservative energy update producing negative pressures, but the results are a mess in the regions where the non-conservative update was used.

Figure 3: Same as Figure 1 but with dual energy, $\eta_1 = 10^{-4}$

With $\eta_1 = 10^{-3}$ (the value given in B14), all cells are using the non-conservative energy update, and the results are a bit better (but still inferior to the non dual energy case).

Figure 4: Same as Figure 1 but with dual energy, $\eta_1 = 10^{-3}$

Is it the way the divergence is approximated?

B14 have a slightly different way of estimating the $P \nabla \cdot \mathbf{v}$ term. They use the velocities at the interface returned by the Riemann solver to approximate the derivative. Does using this version help?

Figure 5: Same as Figure 4 but with the B14 internal energy update

The results are strikingly similar to Figure 4. Evidently the Pdv approximation is not the problem. Perhaps there is an issue somewhere else in the calculation. The next post will use a high mach advection problem to test this.


text:
# Ubuntu – not working as writing 3rd and 4th level characters

keyboardkeyboard-layoutshortcut-keys

I originally asked a question here: Is it possible to write special characters using the alt key? but was asked to submit a new question including more info

I want to be able to type special characters such as ellipsis, en dash and em dash using the same key combinations as in osx (that is, alt+period for ellipsis, alt+hyphen for en dash, alt+shift+hyphen for em dash, etc.) Apparently, Right Alt is supposed to function this way natively. Both of my Alt keys work as normal Alt keys in keyboard shortcuts, but I'm not able to use them to create special characters.

My keyboard looks like this:

(with eject instead of power button). Ubuntu's original Norwegian (macbook) layout has a lot of errors ( | instead of ' etc.), so I use a custom layout file that I found on github: github.com/gulrotkake/Ubuntu-MacBook-Norwegian-keyboard I checked the file, and the 3rd and 4th level characters are included:

key <AB09>  { [    period,      colon, ellipsis,   periodcentered ] };
key <AB10>  { [     minus, underscore,       endash,       emdash ] };

I tried changing back to the original layout file (which I had stored as a backup), but I am still not able to produce any of the special characters.

I followed the advice of this post: Setup 3rd/4th level characters (typographic layout) by going into org.gnome.desktop.input-sources and adding 'lv3:ralt_switch', 'misc:typo'. gsettings get org.gnome.desktop.input-sources xkb-options now outputs ['ctrl:swap_lwin_lctl', 'compose:rwin', 'lv3:ralt_switch', 'misc:typo']

This lets me write 3rd and 4th level characters with my Right Alt key, but they aren't mapped correctly (and I'm also no longer able to use the Right Alt key in my keyboard shortcuts). When I type alt+period I get » instead of ellipsis, alt+hyphen gives ellipsis instead of en dash, and so on.

I'm using Ubuntu 20.04 LTS

Edit:
cat /etc/default/keyboard outputs:

BACKSPACE=guess
XKBLAYOUT=no
XKBVARIANT=mac

gsettings get org.gnome.desktop.input-sources sources outputs:

[('xkb', 'no+mac'), ('ibus', 'mozc-jp')]

I|ve tried using the custom keyboard layout from github as well as the default one. The behavior of the two keyboard layouts is the same: i normally cant write any special characters using Right Alt. Adding 'lv3:ralt_switch', I can write the characters but they are mapped incorrectly. Pressing the period key gives me . : · ÷ and pressing the hyphhen key gives me – _ ̣ ˙
Adding 'lv3:ralt_switch', 'misc:typo' the mapping is different but still incorrect. pressing the period key gives me . : » > and pressing the hyphen key gives me – _ … ˙

Also, this might or might not be relevant, but before I switched to the github layout I tried to fix the wrong bindings on the default layout using sudo dpkg-reconfigure keyboard-configuration – choosing MacBook Pro international and norwegian. This made æ å ø and several other keys disappear. And this persisted on reboot. I fixed it by switching to the github layout. I|m unsure whether the Right Alt key functioned properly before doing this.

There is a simple explanation: The symbols you want to be able to type (ellipsis, endash and emdash) are not present in the Norwegian (Macintosh) keyboard layout, neither with the original Norwegian symbols file nor with the file you picked from the github project. The lines from the no file you included in your question are not made use of in Norwegian (Macintosh).


text:
Thread: Triple integral View Single Post
 HW Helper P: 6,189 "Cartesian" form in a triple integral means x, y, and z. "Polar" is another form (meaning r, theta, and phi). So I would conclude that you're not supposed to use polar coordinates.


text:
# Unable to start JAZZ server 4.0.3

Rosa Naranjo (2.9k11422) | answered Aug 17 '13, 4:07 p.m.
FORUM MODERATOR / JAZZ DEVELOPER
Hello
Whenever I see this problem, it has resolved itself by reprovisioning the WAR files associated with the applications.  With the Tomcat version, I delete the work folder and sometimes also the directory under the webapps directory that corresponds to the WAR file.  For WAS, there is another cache directory structure that can be reset by delete. I think the folder is called wscache.

commented Aug 19 '13, 1:05 p.m.

Thanks Rosa.  Just to clarify.  You indicated that I should delete:
1) work directory under Tomcat (i.e. C:\Program Files\IBM\JazzTeamServer_4.0\server\tomcat\work)
2) the directories under Tomcat webapps directory (i.e. C:\Program Files\IBM\JazzTeamServer_4.0\server\tomcat\webapps)
Then, restart JAZZ 4.0.3 server.  Sound about right?

commented Aug 19 '13, 2:25 p.m.
FORUM MODERATOR / JAZZ DEVELOPER

# 2 - you should only delete the folders under webapps that correspond to the WAR files deployed by the Jazz applications.  for ex., jts, ccm, qm. Don't delete the folder called ROOT.

commented Aug 20 '13, 12:55 a.m.

Thanks Rosa.  JAZZ server is starting now.  Regarding #2, I deleted the files (war files) directly under the Tomcat webapps directory, but none of the sub folders or files under the sub folders.  That seemed to do the trick.  Thanks for your help.

 0 SEC Servizi (971●2●25●49) | answered May 07 '15, 9:47 a.m. We are on RTC v4.0.7 and we are facing the same exception: ```CRJAZ0291E Non è stato possibile avviare il seguente bundle: com.ibm.team.repository.service 1.3.0.v20140626_0050. Potrebbe essere necessario fornire o installare nuovamente il server.org.osgi.framework.BundleException: State change in progress for bundle "file:/D:/IBMJAZZ407/JazzTeamServer/server/conf/jazz/sites/update-site/plugins/com.ibm.team.repository.service_1.3.0.v20140626_0050.jar" by thread "Start Level Event Dispatcher".at org.eclipse.osgi.framework.internal.core.AbstractBundle.beginStateChange(AbstractBundle.java:1077)at org.eclipse.osgi.framework.internal.core.AbstractBundle.start(AbstractBundle.java:282)at org.eclipse.osgi.framework.internal.core.AbstractBundle.start(AbstractBundle.java:276)at com.ibm.team.repository.provision.internal.ProvisionService.startFeatureBundles(ProvisionService.java:1376)at com.ibm.team.repository.provision.internal.ProvisionService.startFeatureBundles(ProvisionService.java:1331)at com.ibm.team.repository.provision.internal.ProvisionService.startFeatureBundles(ProvisionService.java:1331)at com.ibm.team.repository.provision.internal.ProvisionService.startFeatureBundles(ProvisionService.java:1331)at com.ibm.team.repository.provision.internal.ProvisionService.startFeatureBundles(ProvisionService.java:1331)at com.ibm.team.repository.provision.internal.ProvisionService.installFromProfiles(ProvisionService.java:1046)at com.ibm.team.repository.provision.internal.ProvisionService.installFromProfileNoWait(ProvisionService.java:1143)at com.ibm.team.repository.provision.internal.ProvisionService.installFromProfile(ProvisionService.java:1081)at com.ibm.team.repository.provision.internal.ProvisionService.installFromContextConfigurationUrl(ProvisionService.java:1160)at com.ibm.team.jfs.app.install.Activator\$1.run(Activator.java:172)at java.lang.Thread.run(Thread.java:761)Caused by: org.eclipse.osgi.framework.internal.core.AbstractBundle\$BundleStatusException... 14 more ``` We would try your solution but we do not understand which is the right solution for #2: #2 - you should only delete the folders under webapps that correspond to the WAR files deployed by the Jazz applications.  for ex., jts, ccm, qm. Don't delete the folder called ROOT.  Regarding #2, I deleted the files (war files) directly under the Tomcat webapps directory, but none of the sub folders or files under the sub folders.  That seemed to do the trick. Should we delete the /JazzTeamServer/server/tomcat/webapps/jazz folder or the /JazzTeamServer/server/tomcat/webapps/jazz.war file? Thanks in advance. Cheers. Comments Neal Schrauf commented May 07 '15, 11:26 a.m. Although I dealt with this issue a long time ago, I don't believe I deleted any of the folders, just the war files under (i.e. ,,,\server\tomcat\webapps) and restarted the jazz server.  I believe tomcat redeployed the war files and then the JAZZ server started up fine.  Hope this helps. SEC Servizi commented May 12 '15, 6:21 a.m. We deleted the /JazzTeamServer/server/tomcat/temp folder to resolve the issue.


text:
aliquote

< a quantity that can be divided into another a whole number of time />

Introspective Emacs. #emacs


text:
# Prime Numbers

Go back to  'Factors, Multiples, Primes'

## An easy way to understand Prime Numbers

If a number can be divided into equal groups, then it is not a Prime number. It is a Composite number.

If a number cannot be divided into equal groups, then it is a Prime number.
Keep in mind:

• groups cannot be of size one.
• groups cannot have fractional quantities like $$\frac{1}{2}$$ in each group.

Let’s take a few examples: 12 can be divided into:

• 2 groups of 6
• 3 groups of 4
• 4 groups of 3
• 6 groups of 2
• (1 group of 12 or 12 groups of 1 don't count in our current system)

Hence 12 is not a Prime number. 12 is a composite number.

Quick recall: 1, 2, 3, 4, 6 and 12 are called factors or 12

Another example: 7 cannot be divided into groups of 2 or 3 or 4 or any other whole number (remember 7 groups of 1 or 1 group of 7 doesn't count). So, 7 is a Prime number.

Quick recall: 1 and 7 are the only factors of 7.

### An Easy Definition of Prime Numbers

 Any whole number greater than 1 that has exactly two factors, 1 and itself, is defined to be a prime number.

Another equivalent definition is

 Any whole number greater than 1 that is divisible only by 1 and itself, is defined as a prime number.

Any number greater than 1 that is not a Prime number, is defined to be a composite number. Thus composite numbers will always have more than 2 factors.

1 is neither prime nor composite. See the reason for it below.

Any whole number greater than 1 is either a Prime or Composite.

For example, the number 15 has more than two factors: 1, 3, 5, and 15. Hence it is a composite number. On the other hand, 13 has just two factors: 1 and 13. Hence it is a prime number.

### Examples of Prime Numbers and Composite Numbers

Here are some examples of Prime and Composite numbers with the reasoning behind that label. Using this reasoning you can find out whether any given number is prime or composite.

1. Is 4 a Prime number?
• How many factors does 4 have?
• Factors of 4 are: 1, 2 and 4.
• Since 4 has more than two factors, it is not a prime number. It is a composite number.
1. Is 7 a Prime number?
• How many factors does 7 have?
• Factors of 7 are: 1 and 7.
• 7 has just two factors: 1 and itself. Hence it is a prime number.
1. Is 13 a Prime number?
• How many factors does 13 have?
• Factors of 13 are: 1 and 13.
• 13 has just two factors: 1 and itself. Hence it is a prime number.
1. Is 20 a Prime number?
• How many factors does 20 have?
• Factors of 20 are: 1, 2, 4, 5, 10 and 20.
• Since 20 has more than two factors, it is not a prime number. It is a composite number.

## How to find out if a Number is a Prime Number?

Math is best learnt through activities. Here’s a beautiful activity that helps you find out if a number is prime or composite. This activity works best for numbers between 2 to 100.

### Activity to find out all Prime Numbers till 100

Start with a grid with numbers listed from 1 to 100. We cross out 1 because Prime numbers are defined for numbers greater than 1.

• Step 1: We have crossed out 1. So we start with the next available number. That’s 2. Circle it. This is your first prime number.

• Step 2: Now cross out all multiples of 2. Continue until you reach the end of the grid. So almost half the number up to 100 will get crossed out. These cannot be prime as they have 2 as their factor apart from 1 and themselves. So all multiples are composite numbers.

• Step 3: Move to the next available number. That will be 3. Circe it. You’ve found another prime number!

• Step 4: Cross out all multiples of 3. Continue until you reach the end of the grid. If a number is already crossed out, e.g. 6 just move on. The last number you’ll cross out will be 99.

• Step 5: Move to the next available number. This will be 5. Circle it. You’ve found another prime number!

• Step 6: Cross out all multiples of 5.

• Step 7: Move to the next available number and circle it. Then cross out its multiples. Repeat step 7 till all numbers on the grid are either crossed or circled.

The numbers you circled are Prime Numbers. The numbers you crossed are composite numbers.

### Another quick way to find if a number is a prime number or not

Is 97 a prime number?

1. We use divisibility rules to quickly find out if 97 is divisible by 2, 3, 5.
• 97 is an odd number - so not divisible by 2.
• The digits of 97 add up to 16 which is not divisible by 3. So 97 is not divisible by 3.
• Since the units place digit is not 0 or 5, 97 is not divisible by 5.
• We can skip 4, 6, 8, 9, 10, 12 and 15 as any number divisible by these would have also been divisible by 2 or 3 or 5.
1. Now let’s check some of the subsequent numbers like 7 and 11. Using whichever technique including long division we see that 97 is not divisible by any of these numbers.
• Also, observe that 11 x 9 = 99, a number greater than 97. That means we’ve gone past the square root of 97. This gives us an important clue. If you’ve gone past the square root of the given number without finding any factors, then you need not continue. You won't find any new factors.
• The reason for this is for all subsequent numbers (larger than 11), the multiplicand will be smaller than 9. Since we’ve checked all numbers till 9 (in fact till 11 already), we know that no pair of factors is possible. Remember all factors pairs will have one number smaller than the square root and the other larger than the square root. (A factor pair is a set of two numbers which when multiplied give you the original number whose factors you’re identifying).
1. Thus 97 does not have any factors apart from 1 and 97. So, 97 is a prime number.

### Why are Prime Numbers Important?

Prime numbers can be considered the building blocks of numbers. Any whole number is either a prime number or can be built by multiplying a unique combination of prime numbers.

That means all composite numbers have a recipe to be built using prime numbers.

Let’s consider a few composite numbers:

• 4 is a composite number.
It can be built by multiplying the prime number 2 twice.
2 x 2 = 4
• 12 is a composite number.
It can be built by multiplying the prime numbers 2 and 3 is a certain combination.
2 x 2 x 3 = 12
• 105 is a composite number.
It can be built by multiplying the prime number 3, 5 and 7.
3 x 5 x 7 = 105

In this way, any whole number you pick will either be a prime number itself or can be built using a unique combination of prime numbers.

This is a very important idea in mathematics and is called the Fundamental Theorem of Arithmetic.

Hence prime numbers are sometimes called the building blocks of all numbers.

### Why is 1 not a Prime Number?

There are many ways to think about this. Prime numbers are most effective because using them we can create a unique combination for any composite number.
E.g. the composite number 147 is built using 3 x 7 x 7.

If we consider 1 to be a prime number, the combination of prime numbers that create 147 will no longer remain unique:

• Combination 1: 147 = 3 x 7 x 7 x 1
• Combination 2: 147 = 3 x 7 x 7 x 1 x 1

Etc.

This is not very practical. Hence the definition of prime numbers excludes 1.

### What are twin Primes?

A pair of Prime Numbers that differs by 2 are called twin primes.
E.g: 3 and 5 are twin primes. 5 and 7 are twin primes. 11 and 13 is another pair of twin primes.

### What are co Primes?

If a pair of numbers has no common factor apart from 1, then they are called co-prime numbers.
E.g. 8 and 21 are co-primes. That’s because factors of 8 are 1, 2, 4 and 8. Factors of 21 are 1, 3, 7 and 21. So 8 and 21 have no common factors apart from 1, making them co-prime.
List of all prime numbers between 1 to 100
There are 25 prime numbers till 100. They are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.

 Prime numbers between 1 to 10 2, 3, 5, 7 Prime numbers between 11 to 20 11, 13, 17, 19 Prime numbers between 21 to 30 23, 29 Prime numbers between 31 to 40 31, 37 Prime numbers between 41 to 50 41, 43, 47 Prime numbers between 51 to 100 53, 59, 61, 67, 71, 73, 79, 83, 89, 97

### All prime numbers between 1 and 200

There are 46 prime numbers till 200. They are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, and 199.

## Tips and Tricks to Identify Prime Numbers

• An even number greater than 2 will always be composite. 2 is the only even Prime number.
• If a number with 2 or more digits ends with 5 or 0 it cannot be a prime number. That’s because such numbers will always have 5 as a factor.

### Common mistakes or misconceptions related to Prime numbers

• Misconception: 1 is a Prime number
The definition of Prime numbers applies for numbers greater than 1. So 1 is not included. It is neither Prime nor Composite.
• Misconception: Many other numbers can also be neither prime, not composite
All whole numbers greater than 1 are either Prime or Composite.
• What about fractions, decimals, rational numbers, and integers? Can they also be Prime numbers?
The definition of Prime numbers is based on whole numbers greater than 1. So fractions, decimals, rational numbers, that are not whole numbers are not considered to be Prime or Composite. Negative integers get eliminated as they are not greater than 1.

1. Is 53 a Prime number?
____________________

2. Is 72 a Prime number?
____________________

3. How many factors does 15 have?
____________________

4. Is \begin{align} \frac{3}{4} \end{align} a Prime number?
____________________

5. How many Prime Numbers are there between 60 and 70?
____________________