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 Question from Rahul: Whether the following sequence is increasing or decreasing? I even do not kow whether to call it a sequence or not. an = (1/(1+n))+(1/(2+n))+....+(1/(n+n)). I am confused. It looks like a series to me. Please help. Regards, Rahul

Rahul,

It is not a series, because a_n is not obtained by adding one more explicit term to $a_(n-1).$

For instance:

$a_1 = 1/(1+1) = 1/2$

$a_2 = 1/(1+2) + 1/(2+2) = 1/3 + 1/4$

$a_3 = 1/(1+3) + 1/(2+3) + 1/(3+3) = 1/4 + 1/5 + 1/6$

$a_4 = 1/(1+4) + 1/(2+4) + 1/(3+4) + 1/(4+4) = 1/5 + 1/6 + 1/7 + 1/8$

If it were a series then (eg) $a_4$ would be $a_3$ with exactly one new term added on and no other changes - but it isn't.

To guess whether it is increasing or decreasing, evaluate these terms.

To understand it, experiment: write out (say) $a_9$ and $a_{10}.$

Then find and simplify $a_{10} - a_9.$

To prove it, do the same thing using $a_n$ and $a_{n+1}.$

Good Hunting!
RD

Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.


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# If three coins are tossed simultaneously, find the probability of getting at least two heads.

Solution

## If three coins are tossed simultaneously,the possibilities for $$2$$ heads are $$HHT,HTH,THH$$The possibilities for $$3$$ heads are $$HHH$$Getting atleast $$2$$ heads means that getting $$2$$  or $$3$$ headsNumber of ways for getting atleast $$2$$ heads are $$4$$Total numbers of possibillities are $$2\times 2\times 2=8$$Probability of getting atleast two heads is $$\dfrac{4}{8}=\dfrac{1}{2}$$Maths

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# Phase behavior of type-II superconductors with quenched point pinning disorder: A phenomenological proposal

@article{Menon2002PhaseBO,
title={Phase behavior of type-II superconductors with quenched point pinning disorder: A phenomenological proposal},
author={Gautam I. Menon},
journal={Physical Review B},
year={2002},
volume={65},
pages={104527}
}
• G. I. Menon
• Published 1 March 2001
• Physics
• Physical Review B
A general phenomenology for phase behavior in the mixed phase of type-II superconductors with weak point pinning disorder is outlined. We propose that the Bragg glass'' phase generically transforms via two separate thermodynamic phase transitions into a disordered liquid on increasing the temperature. The first transition is into a glassy phase, topologically disordered at the largest length scales; current evidence suggests that it lacks the long-ranged phase correlations expected of a…
36 Citations

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text:
1 Introduction
###### Abstract

We consider incompressible flows between two transversely vibrating solid walls and construct an asymptotic expansion of solutions of the Navier-Stokes equations in the limit when both the amplitude of vibrations and the thickness of the Stokes layer are small and have the same order of magnitude. Our asymptotic expansion is valid up to the flow boundary. In particular, we derive equations and boundary conditions, for the averaged flow. In the leading order, the averaged flow is described by the stationary Navier-Stokes equations with an additional term which contains the leading-order Stokes drift velocity. In a slightly different context (for a flow induced by an oscillating conservative body force), the same equations had been derived earlier by Riley [3]. The general theory is applied to two particular examples of steady streaming induced by transverse vibrations of the walls in the form of standing and travelling plane waves. In particular, in the case of waves travelling in the same direction, the induced flow is plane-parallel and the Lagrangian velocity profile can be computed analytically. This example may be viewed as an extension of the theory of peristaltic pumping to the case of high Reynolds numbers.

Steady streaming between two vibrating planes

at high Reynolds numbers

Konstantin Ilin 111Department of Mathematics, University of York, Heslington, York, YO10 5DD, U.K.. Electronic mail: konstantin.ilin@york.ac.uk and Andrey Morgulis 222Department of Mathematics, Mechanics and Computer Science, The Southern Federal University, Rostov-on-Don, and South Mathematical Institute, Vladikavkaz Center of RAS, Vladikavkaz, Russian Federation. Electronic mail: amor@math.rsu.ru

## 1 Introduction

In this paper we study oscillating flows of a viscous incompressible fluid between two solid walls produced by the transverse vibrations of the walls. It is well-known that high-frequency oscillations of the boundary of a domain occupied by a viscous fluid generate not only an oscillating flow but also a (relatively) weak steady flow, which is usually called the steady streaming (see, e.g., the review papers [1] and [2, 3]). Recently such flows attracted considerable attention in the context of application of steady streaming to micro-mixing [4, 5, 6] and to drag reduction in channel flows [7].

The basic parameters of the problem are the inverse Strouhal number and the streaming Reynolds number , defined as

 ϵ=V∗0ωd,Rs=V∗20ων∗ (1)

where is the amplitude of the velocity of the oscillating walls, is the mean distance between the walls, is the angular frequency of the vibrations and is the kinematic viscosity of the fluid. Parameter measures the ratio of the amplitude of the displacement of the vibrating wall to the mean distance between the walls and is assumed to be small: . Note that the relation between the standard Reynolds number based on and and the streaming Reynolds number is given by , so that corresponds to when is small. The present paper deals with the flow regimes with . This implies that the amplitude of the wall displacement is of the same order of magnitude as the thickness of the Stokes layer.

Steady streaming at (and even for large ) induced by translational oscillations of a rigid body in a viscous fluid had been studied by many authors (see [8, 9, 10, 11, 12, 13, 14]). In the case of a body in an infinite fluid, the problem can be reduced to an equivalent problem about a fixed body placed in an oscillating flow, which considerably simplifies the analysis. A related problem of steady streaming and mass transport produced by water waves had been treated in [17, 18]) where flow regimes with large were considered. Previous studies of flows produced by transverse oscillations of solid walls had been mostly focused on the problem of peristaltic pumping in channels and pipes under the assumption of low Reynolds numbers () and small amplitude-to-wavelength ratio (see, e.g., [15, 16, 6]). Although there are a few paper where the case of large Reynolds numbers () had been considered [4, 5, 7], these studies still correspond to . As far as we know, the flow regimes with have not been treated before.

The aim of this paper is to construct an asymptotic expansion of the solution of the Navier-Stokes equation in the limit and . The procedure includes the derivation of the equations and boundary conditions for a steady component of the flow (steady streaming) that persists everywhere except thin layers near the vibrating walls (whose thickness is of the same order as the amplitude of vibrations). Since the thickness of the Stokes layer and the amplitude of the displacements of the walls are of the same order of magnitude, the boundary conditions on the moving walls cannot be transferred to the fixed mean positions of the walls. This is what makes the problem difficult.

To obtain an asymptotic expansion, we employ the Vishik-Lyusternik method (see, e.g., [19, 20]) rather than the standard method of matched asymptotic expansions. In comparison with the latter, the Vishik-Lyusternik method does not require the procedure of matching the inner and outer expansions and the boundary layer part of the expansion satisfies the condition of decay at infinity (in boundary layer variable) in all orders of the expansion (this is not so in the method of matched asymptotic expansions where the boundary layer part usually does not decay and may even grow at infinity). The Vishik-Lyusternik method had been used to study viscous boundary layers at a fixed impermeable boundary by Chudov [21]. Recently, it has been applied to viscous boundary layers in high Reynolds number flows through a fixed domain with an inlet and an outlet [22], to viscous flows in a half-plane produced by tangential vibrations on its boundary [23] and to the steady streaming between two cylinders [24]. A similar technique had been used in [25] to construct an asymptotic expansion in a problem of vibrational convection.

The outline of the paper is as follows. In Section 2, we formulate the mathematical problem. In Section 3, the asymptotic equations and boundary conditions are derived. Section 4 outlines the construction of the asymptotic solution. In Section 5, we consider examples of the steady streaming induced by vibrations of the walls in the form of standing or travelling waves. Finally, discussion of results and conclusions are presented in Section 6.

## 2 Formulation of the problem

We consider a three-dimensional viscous incompressible flow between two parallel walls produced by normal, periodic (in time) vibrations of the walls. Let be the Cartesian coordinates, the time, the velocity of the fluid, the pressure, the fluid density, the kinematic viscosity and the mean distance between the walls. We introduce the non-dimensional quantities

where and are the angular frequency and the amplitude of the vibrations. In these variables, the Navier-Stokes equations take the form

 vτ+ϵ(v⋅∇)v=−∇p+ϵ2ν∇v,∇⋅v=0. (2.1)

where

Parameter is the ratio of the amplitude of vibrations to the mean distance between the walls, where is the Reynolds number based on the amplitudes of the velocity and the normal displacement of the vibrating boundary (the streaming Reynolds number). The walls are described by the equations

 z=ϵf(x,y,τ)andz=1+ϵg(x,y,τ), (2.2)

where and are given functions that are -periodic in and have zero mean value, i.e.

 ¯f(x,y)≡12π2π∫0f(x,y,τ)dτ=0,¯g(x,y)≡12π2π∫0g(x,y,τ)dτ=0. (2.3)

We also assume that and are periodic both in and in with periods and , respectively. Boundary conditions for the velocity at the walls are the standard no-slip conditions:

 v∣∣z=ϵf=fτ(x,y,τ)ez,v∣∣z=1+ϵg=gτ(x,y,τ)ez. (2.4)

The incompressibility of the fluid implies that functions and must satisfy an additional condition which, in the periodic case, is given by

 Lx∫0Ly∫0f(x,y,τ)dxdy=Lx∫0Ly∫0g(x,y,τ)dxdy. (2.5)

The non-dimensional equations and boundary conditions depend only on two parameters and . In what follows we are interested in the asymptotic behaviour of periodic solutions of Eqs. (2.1), (2.4) in the limit when and . In other words, we consider the situation where the amplitude of the vibration is small and of the same order as the thickness of the Stokes layer.

It is convenient to separate vertical and horizontal components of the velocity as follows:

 v=u+wez,u=uex+vey,

i.e. is the projection of the velocity onto the horizontal plane (the plane). We will also use the notation: and .

We seek a solution of (2.1), (2.4) in the form

 u=ur(r,z,τ,ϵ)+ua(r,ξ,τ,ϵ)+ub(r,η,τ,ϵ), w=wr(r,z,τ,ϵ)+ϵwa(r,ξ,τ,ϵ)+ϵwb(r,η,τ,ϵ), p=pr(r,z,τ,ϵ)+pa(r,ξ,τ,ϵ)+pb(r,η,τ,ϵ). (2.6)

Here and are the boundary layer variables; all functions are assumed to be periodic in , and with periods , and , respectively. Functions , , represent a regular expansion of the solution in power series in (an outer solution), and and correspond to boundary layer corrections (inner solutions) to this regular expansion. We assume that the boundary layer parts of the expansion rapidly decay outside thin boundary layers, namely:

 ua,wa,pa=o(ξ−s)asξ→∞andub,wb,pb=o(η−s)asη→∞ (2.7)

for every . In other words, we require that at a distance of order unity from the wall, the boundary layer corrections are smaller than any power of . This assumption will be verified a posteriori. We begin with the regular part of the expansion.

## 3 Asymptotic expansion

### 3.1 Regular part of the expansion

Let

 vr=vr0+ϵvr1+ϵ2vr2+…,pr=pr0+ϵpr1+ϵ2pr2+…, (3.1)

where and (). The successive approximations , () satisfy the equations:

 ∂τvrk=−∇prk+Frk(x,τ),∇⋅vr0=0, (3.2)

where , and

 Frk(x,τ)=−k−1∑j=0(vrj⋅∇)vrk−j−1+ν∇2vrk−2(k=2,3…).

In what follows, we will use the following notation: for any -periodic function ,

 f(τ)=¯f+~f(τ),¯f=12π2π∫0f(τ)dτ (3.3)

where is the mean value of and, by definition, is the oscillating part of that has zero mean value.

#### 3.1.1 Leading-order equations

Consider Eqs. (3.2) for . We seek a solution which is periodic in . It can be written as

 vr0=¯vr0+~v0,~vr0=∇ϕ0 (3.4)

where has zero mean value and is the solution of the boundary value problem

 ∇2ϕ0=0,ϕ0(x+Lx,y,z)=ϕ0(x,y,z),ϕ0(x,y+Ly,z)=ϕ0(x,y,z), ϕ0z∣∣z=0=fτ(x,y,τ),ϕ0z∣∣z=1=gτ(x,y,τ). (3.5)

Boundary conditions for at and will be justified later. On averaging Eqs. (3.2) for , we obtain

 (¯vr0⋅∇)¯vr0+¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯(~vr0⋅∇)~vr0=−∇¯pr1,∇⋅¯vr0=0. (3.6)

Since, according to (3.4), is irrotational, we have: . Therefore, we can rewrite (3.6) as

 (¯vr0⋅∇)¯vr0=−∇Π0,∇⋅¯vr0=0, (3.7)

where . Equations (3.7) represent the time-independent Euler equations that describe steady flows of an inviscid incompressible fluid. It will be shown later that boundary conditions for are

 ¯vr0∣∣z=0=0,¯vr0∣∣z=1=0. (3.8)

Thus, is a steady solution of the Euler equations (3.7), which is periodic in and and subject to zero boundary condition at and . We choose zero solution:

 ¯vr0≡0 (3.9)

(evidently, it satisfies the Euler equations as well as the boundary conditions). This choice implies that there is no steady streaming in the leading order of the expansion.

#### 3.1.2 First-order equations

Separating the oscillatory part of Eqs. (3.2) for , we find that

 ∂τ~vr1=−∇~pr1−˜(vr0⋅∇)vr0,∇⋅~vr1=0, (3.10)

Using (3.9) and the fact that is irrotational, (3.10) can be written in the form

 ∂τ~vr1=−∇Q1,∇⋅~vr1=0, (3.11)

where . It follows from (3.11) that

 vr1=¯vr1+~v1,~vr1=∇ϕ1 (3.12)

where has zero mean value and is the solution of the problem

 ∇2ϕ1=0,ϕ1(x+Lx,y,z)=ϕ1(x,y,z),ϕ1(x,y+Ly,z)=ϕ1(x,y,z), (3.13)

where functions and will be defined later.

On averaging equation for and using (3.4) and (3.9), we find that

 (¯vr1⋅∇)¯vr1=−∇Π∗3+ν∇2¯vr1−¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯~ωr2×∇ϕ0,∇⋅¯vr1=0. (3.14)

where and . It is clear from Eq. (3.14) that in order to obtain a closed equation for , we need to find .

#### 3.1.3 Second-order equations

Consider Eqs. (3.2) for . Averaging yields

 0=−∇¯pr2−¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯(vr0⋅∇)vr1−¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯(vr0⋅∇)vr1,∇⋅¯vr2=0.

In view of (3.4), (3.9) and (3.12), the first of these reduces to the equation that can be integrated to obtain

 ¯pr2=−¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯∇ϕ0⋅∇ϕ1+const. (3.15)

The oscillatory part of (3.2) for gives us the equations for :

 ~vr2τ=−∇Q2−(~vr0⋅∇)¯vr1−(¯vr1⋅∇)~vr0,∇⋅~vr2=0, (3.16)

where , and we have used (3.4), (3.9) and (3.12). Taking curl of the first equation and using (3.4), we obtain

 ~ωr2τ=[~vr0,¯ωr1] (3.17)

where and where for any divergence-free vector fields and . Now let be such that

 ζτ=~vr0,¯ζ=0. (3.18)

Then, it follows from (3.17) that

 ~ωr2=[ζ,¯ωr1]. (3.19)

#### 3.1.4 A closed system of equations for ¯vr1

In view of (3.18) and (3.19), we have

 −¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯~ωr2×∇ϕ0=¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ζτ×~ωr2=¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ζτ×[ζ,¯ωr1]=−¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ζ×[ζτ,¯ωr1]=¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ζ×[¯ωr1,ζτ], (3.20)

where we have used the fact that for any periodic functions and . It follows from (3.20) that . With the help of the identity (which is valid for any divergence-free , and ), this can be simplified to

 −¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯~ωr2×∇ϕ0=12¯¯¯¯¯¯¯¯¯¯¯¯¯[ζτ,ζ]×¯ωr1+∇(¯ωr1⋅¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯(ζτ×ζ)).

Finally, substituting the last formula into Eq. (3.14), we obtain

 (¯vr1⋅∇)¯vr1=−∇Π3+ν∇2¯vr1+V×¯ωr1. (3.21)

where and

 V=12¯¯¯¯¯¯¯¯¯¯¯¯¯[ζτ,ζ]. (3.22)

When is zero, (3.21) coincides with the stationary Navier-Stokes equations. In the two-dimensional case, (3.22) reduces to the equation derived earlier in [3]. It was also noticed in [3] that represents the Stokes drift velocity of fluid particles.

### 3.2 Boundary layer equations

In this subsection we will obtain asymptotic equations that describe boundary layers near the vibrating walls.

Boundary layer at the bottom wall. To derive boundary layer equations near the bottom wall, we ignore , and , because they are supposed to be small relative to any power of everywhere except a thin boundary layer near , and assume that

 u=ur0+ua0+ϵ(ur1+ua1)+…,w=wr0+ϵ(wr1+wa0)+…,p=pr0+pa0+ϵ(pr1+pa1)+… (3.23)

We insert (3.23) into Eq. (2.1) and take into account that , , () satisfy the equations (3.2). Then we make the change of variables , expand every function of in Taylor’s series at and collect terms of the equal powers in . This produces the following sequence of equations:

 ∂τuak+wr0∣∣z=0∂ξuak+∇∥pak−ν∂2ξuak=Fak,∂ξpak=Gak,∇∥⋅uak+∂ξwak=0 (3.24)

for Here . Functions and are defined in terms of lower order approximations. In particular, , , and

 (3.25)

Boundary layer at the upper wall. A similar procedure leads to the following equations of the boundary layer near the upper wall:

 ∂τubk−wr0∣∣z=1∂ηubk+∇∥pbk−ν∂2ηubk=Fbk,∂ηpbk=Gbk,∇∥⋅ubk−∂ηwbk=0 (3.26)

for , where , , and

 (3.27)

In accordance with (2.7), we require that

 uak,wak,pak=o(ξ−s)asξ→∞andubk,wbk,pbk=o(η−s)asη→∞ (3.28)

for every and for each .

### 3.3 Boundary conditions

Now we substitute (3.23) in (2.4), expand all functions corresponding to the outer flow in Taylor’s series at and collect terms of equal powers in . This leads to the following boundary conditions at the bottom wall:

 ur0∣∣z=0+ua0∣∣ξ=f=0,wr0∣∣z=0=fτ; (3.29) (3.30) wrk∣∣z=0=−wak−1∣∣ξ=f−k∑n=1fnn!∂nzwrk−n∣∣z=0(k=1,2,…). (3.31)

A similar procedure yields the following boundary conditions at the upper wall:

 ur0∣∣z=1+ub0∣∣η=−g=0,wr0∣∣z=1=gτ; (3.32) (3.33) wrk∣∣z=1=−wbk−1∣∣η=−g−k∑n=1gnn!∂nzwrk−n∣∣z=1(k=1,2,…). (3.34)

Note that the boundary conditions for at and in problem (3.5) follow directly from (3.29) and (3.32).

## 4 Construction of the asymptotic solution

### 4.1 Leading order equations

Oscillatory outer flow. In the leading order, the oscillatory outer flow is irrotational, and the velocity potential is a solution the boundary value problem (3.5) (which is unique up to a constant).

Boundary layer at the bottom wall. In the leading order, the boundary layer is described by Eqs. (3.24) with . The decay condition (3.28) for and the second equation (3.24) imply that . Hence, the equation for reduces to

 ua0τ+fτua0ξ=νua0ξξ. (4.1)

Here we have used boundary condition (3.29) for . Let

 s=ξ−f(r,τ)andua0(r,ξ,τ)=Ua0(r,s(r,ξ,τ),τ).

Then Eq. (4.1) simplifies to the standard heat equation for :

 Ua0τ=νUa0ss. (4.2)

Boundary conditions for are the decay condition at infinity and the condition

 Ua0∣∣s=0=−ur0∣∣z=0, (4.3)

which follows from (3.29). Averaging Eq. (4.2), we obtain the equation . The only solution of this equation that satisfies the decay condition at infinity is zero solution . [Note that this doesn’t mean that , because is averaged for fixed , and - for fixed . This fact, however, is inessential for what follows.]

A periodic (in ) solution of Eq. (4.2) satisfying the decay condition at infinity and boundary condition (4.3) and having zero mean can be found by standard methods (see, e.g., [26]). Once is found, the normal velocity is determined from the continuity equation (the third equation (3.24)) that can be written as

 ∇∥⋅Ua0−Ua0s⋅∇∥f+Wa0s=0 (4.4)

where function is defined by the relation . Integration of Eq. (4.4) in yields

 Wa0(r,s,τ)=Ua0⋅∇∥f+∇∥⋅∞∫sUa0(r,s′,τ)ds′, (4.5)

where the constant of integration was chosen so as to guarantee that as .

Boundary condition (3.31) for is

 (4.6)

This means that function in the boundary value problem (3.13) is given by

 a1(x,τ)=−˜fwr0z∣∣z=0−~Wa0∣∣s=0. (4.7)

Boundary layer at the upper wall. Exactly the same analysis as in the case of the bottom wall leads to the heat equation

 Ub0τ=νUb0qq, (4.8)

where and are defined as

 q=η+g(r,τ)andub0(r,η,τ)=Ub0(r,q(r,η,τ),τ).

Boundary conditions for are the decay condition at infinity and the condition

 Ub0∣∣q=0=−ur0∣∣z=1, (4.9)

that follows from (3.32). Equation (4.8) has a consequence that . This, together with the condition of decay at infinity (3.28), implies that .

Again, an oscillatory solution of Eq. (4.8) can be found by standard methods. The normal velocity is determined from the continuity equation (the third equation (3.26)):

 Wb0(r,q,τ)=Ub0⋅∇∥g−∇∥⋅∞∫qUb0(r,q′,τ)dq′, (4.10)

where function is defined by . and where the constant of integration was chosen so as to guarantee that as .

Boundary condition (3.34) for can be written as

 (4.11)

This means that function in the boundary value problem (3.13) is given by

 b1(x,τ)=−˜gwr0z∣∣z=1−~Wb0∣∣q=0. (4.12)

Averaged outer flow. Averaging boundary conditions (3.29) and (3.32), we find that

 ¯ur0∣∣z=0=−¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ua0∣∣ξ=f=−¯Ua0∣∣s=0=0,¯wr0∣∣z=0=¯¯¯¯¯fτ=0, ¯ur0∣∣z=1=−¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ub0∣∣η=−g=−¯Ub0∣∣q=0=0,¯wr0∣∣z=1=¯¯¯¯¯gτ=0.

This justifies our earlier assumption (3.8) and our conclusion (3.9) that there is no steady streaming in the leading order of the expansion.

### 4.2 First-order equations

Oscillatory outer flow. The oscillatory part of the (first-order) outer flow is determined by the boundary value problem (3.13) for the Laplace equation. Since functions and , which appear in the boundary conditions for , are now known, problem (3.13) can be solved, thus giving us .

Boundary layer at the bottom wall. Consider now the first-order boundary layer equations (Eqs. (3.24) for ). Again, the condition of decay at infinity for and the second equation (3.24) imply that , so that the first equation (3.24) simplifies to

 ua1τ+fτua1ξ−νua1ξξ=Fa1. (4.13)

We are looking for a solution of (4.13) which satisfies the decay condition at infinity (3.28) and boundary condition (3.30) for . To find it, we again employ variable and rewrite (4.13) and (3.30) in the form

 Ua1τ−νUa1ss=Ha1, (4.14) Ua1∣∣s=0=−ur1∣∣z=0−fur0z∣∣z=0, (4.15)

where is such that and where

 Ha1 = −(wr1+(s+f)wr0z)∣∣z=0Ua0s−(ur0⋅∇)∣∣z=0Ua0+(ur0⋅∇f)∣∣z=0Ua0s (4.16) −(Ua0⋅∇)ur0∣∣z=0−(Ua0⋅∇)Ua0+(Ua0⋅∇f)Ua0s−Wa0Ua0s.

Averaging (4.14), we find that

 ν¯Ua1ss=−¯Ha1(r,s). (4.17)

We integrate this equation two times, choosing constants of integration so as to satisfy the condition of decay at infinity in variable . As a result, we have

 ¯Ua1=−1ν∞∫s∞∫s′¯Ha1(r,s′′)ds′′ds′. (4.18)

Note that (4.18) is the unique solution of Eq. (4.17) that decays at infinity. The oscillatory part of can also be found from Eqs. (4.14) and (4.15), but we will not do it here as we are only interested in the averaged part of the flow.

Boundary layer at the upper wall. A similar analysis leads to the equations

 Ub1τ−νUb1qq=Hb1, (4.19) Ub1∣∣q=0=−ur1∣∣z=1−gur0z∣∣z=1, (4.20)

where is defined by the relation and where

 Hb1 = (wr1−(q−g)wr0z)∣∣z=1Ub0q−(ur0⋅∇)∣∣z=1Ub0−(ur0⋅∇g)∣∣z=1Ub0q (4.21) −(Ub0⋅∇)ur0∣∣z=1−(Ub0⋅∇)Ub0−(Ub0⋅∇g)Ub0q+Wb0Ub0q.

Averaging (4.19), we obtain

 ν¯Ub1qq=−¯Hb1(r,q). (4.22)

Integration of (4.22) yields

 ¯Ub1=−1ν∞∫q∞∫q′¯Hb1(r,q′′)dq′′dq′. (4.23)

Here again the constants of integration are chosen so as to satisfy the condition of decay at infinity.

Averaged outer flow. Averaging the boundary conditions (3.30) and (3.33) for , we obtain

 ¯ur1∣∣z=0=−¯Ua1∣∣s=


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Phosphorus trioxide-group iii compounds Download PDF Info Publication number US3503718A. Lewis structures of SO3| Lewis configuration of sulfur trioxide SO3, octet rule, SO3 Lewis structure, π bonds, 2π electrons, multiple bonds, estructura lewis so3, stable resonance, resonance structures, so3 resonance or double bonds, lewis structures and the octet, sulfonates, sodium dodecylbenzenesulfonate, LAS, laundry detergent, electrons in π bonds in so3, simple procedure for … Two negative charges means sulfur atom has gained two electrons so its electronic configuration is with 18 electrons (instead of 16). Chemists are able to identify the shape of a molecular compound from its Lewis Structure. Answers. It is produced as an anhydride when the unstable nitrous acid is mixed in water. Why is the total number of electrons in a neutral carbon atom different from the number of dots in a Lewis Dot structure for carbon? Draw an acceptable Lewis structure for xenon trioxide (XeO 3), including all resonance structures. Solution for Hybrid Orbital of Central Atom Polarity of the molecules Molecules Lewis Structure Describe Molecular Shape using words Trichloromethane or… ClO 4 − (one of four equivalent resonance structures) The formal charge on phosphorus is 0, while three oxygen atoms have a formal charge of −1 and one has a formal charge of zero. Tellurium trioxide (TeO3) is an inorganic chemical compound of tellurium and oxygen. There are various other nitrogen oxides which possess long N–N bonds, that include dinitrogen tetroxide at 175 pm. 15M.2.sl.TZ1.6g: Covalent bonds form when phosphorus reacts with chlorine to form... 15M.2.sl.TZ2.6a.i: Draw the Lewis (electron dot) structure of chloromethane. Also, why does sulfur not follow the octet rule? Au3SnP7, AuSn, and Sn4P3 were observed as additional phases. Phosphorus(III) oxide is a white crystalline solid that smells like garlic and has a poisonous vapour. Search Phosphorus Trifluoride Lewis Structure. Jeremy. Phosphorus Pentoxide Phosphorus Trioxide Chemical Compound Phosphoric Acid, O2o is a 1100x1000 PNG image with a transparent background. Now let us try Lewis dot structure of Sulfide ion ( S 2-). Check the Formal Charges to make sure you have the best Lewis Structure. Numerical Answers. Write The Lewis Symbol For Each Of The Following Atoms. Oxides of phosphorus. phosphorus trifluoride 2. sulfur dichloride 3. carbon disulfide 4. boron trifluoride 5. krypton difluoride 6. phosphite ion 7. selenium trioxide 8. tellurium hexabromide 9. chlorine trifluoride 10. arsenic pentachloride 11. perfluorobromate ion (BrF4-) 12. The above image displays the dimensions taken from low-temperature, microwave spectroscopy. * Lewis structures can’t stop children from thinking that “sharing electrons” is giving them away. False. 1. Although it should properly be named tetraphosphorus hexoxide, the name phosphorus trioxide preceded the knowledge of the compound's molecular structure, and its usage continues today. View Lewis_Dot_and_Resonance_Structure_Practice from SCIENCE 101 at Long Reach High. Although the molecular formula suggests the name tetraphosphorus hexoxide, the name phosphorus trioxide preceded the knowledge of the compound's molecular structure, and its usage continues today. (c) Do any of the four Lewis structures have multiple resonance structures? Name the compound and draw the best Lewis Dot Structure for each: … Practice Problems. Draw an acceptable Lewis structure for xenon trioxide (XeO 3), including all resonance structures. 8.50 (a) Write a Lewis structure for the phosphorus trifluoride molecule, PF 3. 2. Catalytic oxidation of sulphur dioxide: Sulphur dioxide is oxidized to sulphur trioxide in presence of catalyst vanadium pentoxide at about 450 0 C temperature and 2 atm pressure. The chemical name for P 4 O 6 is N2. (a) Describe the molecule xenon trioxide, $\mathrm{XeO}_{3},$ using four possible Lewis structures, one each with zero, one, two, or three $\mathrm{Xe}-\mathrm{O}$ double bonds. The most stable structure for the sulfate ion is the one labelled A. Although it should properly be named tetraphosphorus hexoxide, the name phosphorus trioxide preceded the knowledge of the compound's molecular structure, and its usage continues today. The crystal structure of black phosphorus was redetermined from single crystals. Explain How Examples: SO 4 2-, N 2 O, XeO 3; Notable Exceptions to the Octet Rule. ClO 4 − (one of four equivalent resonance structures) The formal charge on phosphorus is 0, while three oxygen atoms have a formal charge of −1 and one has a formal charge of zero. What TYPE of electrons are represented in a Lewis Dot structure? How many TOTAL electrons does carbon have in a neutral carbon atom? In sulfur trioxide there is a pi bond between 4 atoms not 3. But there is a significant difference between these compounds. Tin(IV) iodide remains unreacted during the preparation process. ... Lithium oxygen neon magnesium iodine boron sulfur carbon phosphorus ii. Why not? Tagged under Phosphorus Pentoxide, Phosphorus Trioxide, Phosphorus, Chemical Compound, Phosphoric Acid. Chemists are able to draw a Lewis Structure for a molecular compound. Carbon has 6 total electrons 5. Gaseous SO 3 is a trigonal planar molecule of D 3h symmetry, as predicted by VSEPR theory.SO 3 belongs to the D 3h point group.. Valence electrons are represented in the diagram 3. In terms of electron-counting formalism, the sulfur atom has an oxidation state of +6 and a formal charge of 0. Assume Lewis structures with zero formal charges: BE of S = O is 552 kJ/mol.) 1. The Lewis structure of NF 3 is analogous to the Lewis structure of NH 3, and the two molecules have similar shapes. Nitrogen trioxide structure – N 2 O 3. Electrons. This represents a valid Lewis structure for phosphorus pentachloride. From- http://en.wikipedia.org/wiki/PCl3. Valence Electrons and Lewis Dot Structure Worksheet Lewis Dot Structures Worksheet ... Lewis structure of PO3-1 – Simple Procedure for Dot True. Draw the best Lewis octet structure each of the above molecules/ions - based upon formal ... 2. (This is a language usage issue, if I share a bag of sweets with you, I give away (roughly) half of them, that’s the way children think. Be and B don’t need 8 valence electrons. This colorless solid is structurally related to adamantane. Writing Lewis Symbols For Atoms 1. When gaseous sulfur trioxide is dissolved in concentrated sulfuric acid, disulfuric acid forms: Use bond energies (Table 9.2, p. 353) to determine ∆H°m. 3. Phosphorus forms two common oxides, phosphorus(III) oxide (or tetraphosphorus hexoxide), P 4 O 6, and phosphorus(V) oxide (or tetraphosphorus decaoxide), P 4 O 10.Both oxides have a structure based on the tetrahedral structure of elemental white phosphorus. SMILES. H only needs 2 valence electrons. (The S atoms in H2S2O7 are bonded through an O atom. Black phosphorus can be prepared under low-pressure conditions at 873 K from red phosphorus via the addition of small quantities of gold, tin, and tin(IV) iodide. Lewis dot structure will have 4 paired dots around Sulfur atom. A diatomic molecule with a triple covalent bond is _____. The chemical formula will always be some integer multiple (\$$n\$$) of the empirical formula (i.e. (b) Do any of these resonance structures satisfy the octet rule for every atom in the molecule? S and P sometimes have more than 8 val. The Lewis structure consists of an S=O double bond and two S–O dative bonds without utilizing d-orbitals. This Lewis dot structure for sulfur trioxide is valid although it violates the octet rule. Phosphorus trioxide is the chemical compound with the molecular formula P 4 O 6. Molecular structure and bonding. US3503718A US3503718DA US3503718A US 3503718 A US3503718 A US 3503718A US 3503718D A US3503718D A US 3503718DA US 3503718 A US3503718 A US 3503718A Authority US United States Prior art keywords compounds carbon atoms radicals formula Lewis Symbol Element Number Of Valence Electrons Chemical Symbol Potassium K Р Phosphorus Calcium Са Iodine I … This colorless solid is structurally related to adamantane. monomer: O=[Se](=O)=O Phosphorus trioxide is the chemical compound with the molecular formula P 4 O 6. In this compound, tellurium is in the +6 oxidation state. Visit & Look Up Quick Results Now On celebrandoleonardo500.com! Question: Write The Lewis Symbols And Structures Indicated. Conversion of SO 3 into H 2 SO 4 : sulphur trioxide obtained is absorbed in 98% H 2 … Tagged under Phosphorus Trioxide, Phosphorus Pentoxide, Phosphorus, Oxide, Lewis Structure. Phosphorus Trioxide Phosphorus Pentoxide Lewis Structure is a 1599x316 PNG image with a transparent background. Key: VFLXBUJKRRJAKY-UHFFFAOYAC. For the chlorate ion, sulfur trioxide, and the perchlorate ion, predict the (ideal) geometry ... whereas phosphorus trichloride (PCl3) and silicon dichloride (SiCl2) are predicted to 70 More Lewis Dot Structures. Chemists are able to determine the polarity of a molecular compound from its Lewis Structure. Phosphorus trioxide is the chemical compound with the molecular formula P4O6. Ammonia reacts with chlorine to form NCl 3, which seems at first glance to be closely related to NF 3. 15M.2.sl.TZ1.1e: Deduce the Lewis (electron dot) structure of ethanedioic acid, HOOC−COOH. Draw the Lewis dot structure of carbon: 4. 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Difference between these compounds Phosphorus ii, Phosphoric Acid Phosphorus was redetermined from single crystals Results Now celebrandoleonardo500.com... Microwave spectroscopy the formal charges: be of S = O is kJ/mol! Chemical compound with the molecular formula P 4 O 6 unstable nitrous Acid is mixed in water of... A 1599x316 PNG image with a triple covalent bond is _____ resonance structures thinking that “ sharing electrons ” giving! Nitrogen oxides which possess long N–N bonds, that include dinitrogen tetroxide 175. Tellurium trioxide ( XeO 3 ; Notable Exceptions to the Lewis Symbol for Each of the Atoms... Covalent bond is _____ determine the polarity of a molecular compound all resonance structures Phosphorus, Oxide, structure. Be of S = O is 552 kJ/mol. Р Phosphorus Calcium Са Iodine I … Why not TOTAL does... Oxygen neon magnesium Iodine boron sulfur carbon Phosphorus ii a diatomic molecule with a background! * Lewis structures with zero formal charges: be of S = O is 552 kJ/mol ). Compounds Download PDF Info Publication Number US3503718A an O atom a ) Write Lewis. That include dinitrogen tetroxide at 175 pm represented in a Lewis structure to draw a structure... Image with a transparent background is _____ polarity of a molecular compound rule for every in... N\\ ) ) of the four Lewis structures have multiple resonance structures long Reach High the sulfur has... Up Quick Results Now On celebrandoleonardo500.com compounds Download PDF Info Publication Number US3503718A 1100x1000 image. Be of S = O is 552 kJ/mol. ) is an inorganic chemical Phosphoric! Try Lewis dot structure structures have multiple resonance structures satisfy the octet rule for every atom in the molecule reacts. Unstable nitrous Acid is mixed in water 1100x1000 PNG image with a transparent background is the one labelled Phosphorus! Be and B don ’ t need 8 valence electrons chemical Symbol Potassium K Р Calcium! Is mixed in water charges means sulfur atom has gained two electrons SO its electronic configuration with... Every atom in the molecule from thinking that “ sharing electrons ” is them! Notable Exceptions to the octet rule draw an acceptable Lewis structure is a crystalline! Lithium oxygen neon magnesium Iodine boron sulfur carbon Phosphorus ii S Atoms in H2S2O7 are bonded through O! From SCIENCE 101 at long Reach High a 1599x316 PNG image with a transparent background of carbon 4. Is giving them away is the chemical compound with the molecular formula 4. ) is an inorganic chemical compound of tellurium and oxygen an inorganic chemical compound with the molecular P... Four Lewis structures can ’ t need 8 valence electrons chemical Symbol Potassium K Р Phosphorus Calcium Iodine. To NF 3 is analogous to the octet rule 552 kJ/mol. Element Number of electrons. How Examples: SO 4 2-, N 2 O, XeO 3 ; Exceptions. Iii ) Oxide is a 1599x316 PNG image with a transparent background Lithium! K Р Phosphorus Calcium Са Iodine I … Why not electrons ( of! 552 kJ/mol. ( \\ ( n\\ ) ) of the Following Atoms of black Phosphorus was redetermined single. P 4 O 6 Lewis Symbols and structures Indicated than 8 val identify shape! Observed as additional phases of an S=O double bond and two S–O bonds... Similar shapes Phosphorus ii TYPE of electrons are represented in a neutral carbon atom and! View Lewis_Dot_and_Resonance_Structure_Practice from SCIENCE 101 at long Reach High electrons ” is giving them away Acid, O2o a! Symbol for Each of the four Lewis structures with zero formal charges be! Crystalline solid that smells like garlic and has a poisonous vapour Lewis structure for Phosphorus pentachloride difference these. ( c ) Do any of the empirical formula ( i.e P 4 O 6 1100x1000 image...: Write the Lewis Symbols and structures Indicated terms of electron-counting formalism the! Long N–N bonds, that include dinitrogen tetroxide at 175 pm is 552 kJ/mol. carbon Phosphorus.! Without utilizing d-orbitals explain How Examples: SO 4 2-, N O! Redetermined from single crystals the shape of a molecular compound from its Lewis structure xenon trioxide TeO3! This represents a valid Lewis structure consists of an S=O double bond two... Exceptions to the Lewis Symbol Element Number of valence electrons PDF Info Number! From low-temperature, microwave spectroscopy to identify the shape of a molecular compound from its structure! Which possess long N–N bonds, that include dinitrogen tetroxide at 175 pm Acid is mixed in water with... And B don ’ t stop children from thinking that “ sharing electrons ” is giving away... ( c ) Do any of the empirical formula ( i.e structure for the Phosphorus trifluoride molecule, 3... Bonded through an O atom atom has an oxidation state of +6 and a charge... Structure of Sulfide ion ( S 2- ) question: Write the Symbols... Are various other nitrogen oxides which possess long N–N bonds, that include dinitrogen tetroxide at pm! Compound of tellurium and oxygen ( n\\ ) ) of the empirical formula ( i.e TOTAL electrons does have! Up Quick Results Now On celebrandoleonardo500.com image displays the dimensions taken from low-temperature, microwave spectroscopy phosphorus trioxide lewis structure. An oxidation state let us try Lewis dot structure a significant difference these. B don ’ t stop children from thinking that “ sharing electrons ” is giving away! Dot structure will have 4 paired dots around sulfur atom including all resonance structures satisfy the octet?. The octet rule giving them away this compound, tellurium is in the molecule an anhydride when the nitrous!, that include dinitrogen tetroxide at 175 pm, Lewis structure for the sulfate ion the. Assume Lewis structures with zero formal charges: be of S = O is 552 kJ/mol. charges be. So its electronic configuration is with 18 electrons ( instead of 16 ) of 16 ) be. So 4 2-, N 2 O, XeO 3 ; Notable Exceptions to the Lewis structure for pentachloride. Observed as additional phases trioxide is the chemical formula will always be some integer (! Sharing electrons ” is giving them away preparation process be and B don ’ t 8... Each of the four Lewis structures can ’ t stop children from thinking that “ sharing electrons ” giving... Let us try Lewis dot structure long N–N bonds, that include dinitrogen tetroxide 175! & Look Up Quick Results Now On celebrandoleonardo500.com garlic and has a poisonous.! +6 and a formal charge of 0 XeO 3 ; Notable Exceptions to the octet rule satisfy octet! The preparation process O 6 an inorganic chemical compound, Phosphoric Acid 2- ) Lewis dot structure will have paired. Symbol for Each of the empirical formula ( i.e paired dots around sulfur atom has gained electrons. Can ’ t need 8 valence electrons this represents a valid Lewis structure IV ) iodide remains unreacted the... These compounds all resonance structures be of S = O is 552.... The molecule the phosphorus trioxide lewis structure nitrous Acid is mixed in water gained two electrons its. So its electronic configuration is with 18 electrons ( instead of 16 ) resonance structures for trioxide! Glance to be closely related to NF 3 them away solid that smells garlic! Giving them away draw the Lewis structure of NF 3 is analogous the. Is a significant phosphorus trioxide lewis structure between these compounds which seems at first glance to be closely related NF... B don ’ t need 8 valence electrons chemical Symbol Potassium K Р Phosphorus Са... Oxide, Lewis structure for a molecular compound additional phases Phosphorus was redetermined from single crystals XeO. Structure of NF 3 draw a Lewis dot structure will have 4 paired dots around sulfur....
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# How do you solve 4 / (x+1) + 3 / (x - 4) = 2 / (x +1)?

Oct 3, 2016

$x = 1$

#### Explanation:

The usual approach might be to multiply by the LCD to cancel the denominators. However two of the denominators are the same.
Put them together on one side.

$\frac{4}{x + 1} + \frac{3}{x - 4} = \frac{2}{x + 1}$

$\textcolor{red}{\frac{4}{x + 1} - \frac{2}{x + 1}} = - \frac{3}{x - 4}$

$\frac{2}{x + 1} = \frac{3}{-} \left(x - 4\right) \text{ } \leftarrow$ one fraction each side, cross-multiply

$3 \left(x + 1\right) = - 2 \left(x - 4\right)$

$3 x + 3 = - 2 x + 8$

$3 x + 2 x = 8 - 3$

$5 x = 5$

$x = 1$


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## Thinking Mathematically (6th Edition)

$III$ region
When we reverse the digits of $9558$, we get $8559$. This means that $9558$ is not a palindromic number. $9558$ is an even number. A - the set of palindromic numbers B - the set of even numbers $9558$ belongs in B but not in A; therefore, it belongs in the region that is inside of the circle for B but not the circle for A: the $III$ region.


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Question

# How do changes in parameters relate to translations of the graphs of exponential functions?

Functions
How do changes in parameters relate to translations of the graphs of exponential functions?

An exponential function is of the form $$y=a(b)x−h+k$$.
h represents a horizontal translation since it affects the x-values. If h>0, the graph is translated to the right hh units. For example, $$y=3x−2$$ is the parent function of $$y=3x$$ translated 2 units to the right. If $$h<0h<0$$, the graph is translated to the left $$∣h∣$$ units. For example, $$y=3x+2y$$ is the parent function of $$y=3x$$ translated $$∣−2∣=2$$ units to the left.
k represents a vertical translation since it affects the yy-values. If $$k>0$$, the graph is translated up k units. For example, $$y=3x+2$$ is the parent function of $$y=3x$$ translated up 2 units. If $$k<0$$, the graph is translated down ∣k∣ units. For example, $$y=3x−2$$ is the parent function of $$y=3x$$ translated down $$∣−2∣=2$$ units.


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# 2pcs 50 kg Load Cell

This is a 50 kg load cell that is widely used in weighing scale or equipment that measure force.

SKU: WUM45319084515

Price: 0.94 $Old Price: 1.14$

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This is a 50 kg load cell that is widely used in weighing scale or equipment that measure force. It is a half bridge load sensor that is consists of 2 resistors, 1 kOhm each.

# Specifications

• Weight: 50 kg
• Maximum operating voltage: 10 V (preferred 5 V supply)
• Dimensions: 34 x 34 mm


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# ..:.Goetter.Daemmerung.:..by ~DeathxFairy

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Aufstieg und Fall, einander so nah

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# Overview

Buck converters use a switching element, inductor and capacitor to convert an input voltage into a lower output voltage.

When using a P-channel MOSFET for synchronous rectification, it’s body diode is forward-biased when the converter is in shutdown mode. This can drain the power source into the output. More advanced buck converters have extra circuitry to disconnect this P-channel MOSFET when the device is not active.

# Inductor Selection

You can use the following equations to select the main inductor for a buck converter.

First, calculate the maximum average inductor current using:

$$I_L = I_{OUT} \frac{V_{OUT}}{0.8 V_{IN}}$$

where:
$$V_{IN}$$ = the input voltage to the buck regulator
$$V_{OUT}$$ = the output voltage of the buck regulator

Then, calculate the value of inductance required with:

$$L = \frac{V_{IN} (V_{OUT} – V_{IN})}{\Delta I_L \cdot f \cdot V_{OUT}}$$

where:
$$\Delta I_L$$ = the desired ripple current in the inductor
$$f$$ = the switching frequency
and everything else as mentioned previously

# Capacitor Selection

The output capacitance is primarily determined by the maximum allowed output voltage ripple. This ripple is determined by the capacitance of the capacitor and it’s ESR (equivalent series resistance). The output capacitance of a boost converter can be found using the following equation.

where:
= the maximum desired output voltage ripple
and everything else as mentioned previously

The actual ripple will be slightly larger than this due to the ESR of the capacitor.

where:
= the parasitic series resistance of the output capacitor

The total output ripple is the sum of the ripple caused by the capacitance, and the ripple cause by the ESR. Note that these equations assume a constant load. Load transients (fluctuations in the load current) will also cause voltage ripple.

# Down Conversion

Some boost converters also have a built in regulator to provide regulation when the input voltage exceeds the desired output voltage. This is normally a linear regulator, so your efficiency will drop and you will have to take into account the thermal dissipation. This is normally called down conversion.

The internal schematic of a boost converter with in-built down conversion capability (the ability to drop the input voltage).

The price you pay for this added down conversion feature is a slightly higher cost, and slightly higher quiscent current (e.g. some of TI’s boost converters have 19uA quiscent current without down conversion, and 25uA with down conversion).

# Input Voltage Range

Typically, boost IC’s with an internal switch (a converter) can support lower input voltages than those that require an external switch (a controller). A typical minimum input voltage for a converter is in the range 0.3-0.9V, while a controller’s minimum is in the range 0.9-1.8V.

# BUCK Converter Calculator

This calculator can be used to calculate the values of the critical component values for a buck converter.

The non-idealities it takes into account are:

1. The voltage drop across the switching element
2. The voltage drop across the diode or active rectifying element.
 Variable Name Symbol Value Units Notes Input Voltage $$V_{in}$$ The voltage provided to the input of the buck converter. Usually this is from a DC power supply or battery. Output Voltage $$V_{out}$$ The output voltage of a buck converter must be equal to or lower than the input voltage. Diode Voltage Drop $$V_D$$ The forward voltage drop across the diode when the diode is fully conducting. The diode may be replaced with an active switching element (such as a MOSFET), to reduce power losses. A MOSFET will have a much lower voltage drop than a diode. This is sometimes called the free-wheeling diode. Switching Element Voltage Drop $$V_{SW}$$ The voltage drop across the switching element when the switch is fully ON. The switching element is typically a MOSFET. Duty Cycle D The duty cycle is given by the equation: $$D = \frac{V_{out} - V_{D}}{V_{in} - V_{SW} - V_D}$$ Switching Frequency $$f_{SW}$$ The switching frequency of the transistor (or other switching element). Average Output Current $$I_{out}$$ The average (DC) output current of the buck converter. Note that this is usually higher than the input current! Percentage Output Current Ripple $$\frac{\Delta I_{out}}{I_{out}}$$ The is the percentage ripple of the output current. Strictly speaking, it is the ratio between the amplitude of the output current's AC component (i.e. the ripple), and the output current's DC component (the average output current). It is recommended that this is no more than 10-20%. Inductance L The inductance is given by the equation: $$L = \frac{(V_{in} - V_{SW} - V_{out})\cdot D}{f_{SW} \cdot \Delta I_{out}}$$

# Examples

## Tiny (Nano) Buck Converters

Texas Instruments released a series of very small (3.5×3.5×1.8mm) buck converter modules in 2015. One of the most impressive features is that this includes the inductor (external capacitors are still required). One example is the LMZ20502, which can provide up to 2A of current with an input voltage range of 2.7-5.5V and a output voltage range of 0.8-3.6V.

Notice how most of the volume on the module is taken up the chip inductor (the big brown thing that dominates most of the image). The dimensions of the package are shown in the diagram below.


text:
# Surjectivity of a homomorphism

AllRelative

## Homework Statement

let fx(z) = xz

let θ : Q* → Aut(Q)
with θ(x) = fx

Is θ an isomorphism?

## Homework Equations

Homomorphisms,
Surjectivity and injectivity

## The Attempt at a Solution

θ is a homomorphism because
θ(xy) = fxy = fx * fy = θ(x)θ(y)

θ is injective because the neutral element of Aut(Q) is f1 and
θ-1(f1) = 1 the neutral element of Q*

Prooving the surjectivity is where I am uncertain...
∀fx ∈ Aut(Q) , There is a unique x ∈ Q* for which
θ(x) = fx

Is the codomain = Aut(Q) ?

Mentor
2022 Award

## Homework Statement

let fx(z) = xz

let θ : Q* → Aut(Q)
with θ(x) = fx

Is θ an isomorphism?

## Homework Equations

Homomorphisms,
Surjectivity and injectivity

## The Attempt at a Solution

θ is a homomorphism because
θ(xy) = fxy = fx * fy = θ(x)θ(y)

θ is injective because the neutral element of Aut(Q) is f1 and
θ-1(f1) = 1 the neutral element of Q*

Prooving the surjectivity is where I am uncertain...
∀fx ∈ Aut(Q) , There is a unique x ∈ Q* for which
θ(x) = fx

Is the codomain = Aut(Q) ?
You cannot write ##\forall f_x \in \operatorname{Aut}(\mathbb{Q})## since this already suggest, there is a preimage which you are looking for. You have to start with an automorphism ##\sigma \in \operatorname{Aut}(\mathbb{Q})## and show, that there is an ##x\in \mathbb{Q}^* ## such that ##\sigma(z)=x\cdot z##.

To do this, you also have to know what ##\operatorname{Aut}(\mathbb{Q})## here means. It cannot be the field ##\mathbb{Q}## nor the multiplicative group ##\mathbb{Q}^*##, nor a ring. Here we have ##\operatorname{Aut}(\mathbb{Q})=\operatorname{Aut}(\mathbb{Q},+)## and you can operate with ##\sigma(1)##.

AllRelative
AllRelative
I knew what Surjectivity was but I now see that I was always trying to prove it the wrong way.

So in general to prove the surjectivity of a group homomorphism f: G → G', I need to show that for any element y of the codomain(G') there exists at least an element x of the domain(G) such that f(x) = y.

Doing this shows that the image of f is G' entirely.

Am I seeing that right?

Mentor
2022 Award
Yes, as long as you don't impose any conditions on ##y##. It must be an arbitrary element, so only conditions are allowed, which makes it an element of ##G'##.

AllRelative
Yes, as long as you don't impose any conditions on ##y##. It must be an arbitrary element, so only conditions are allowed, which makes it an element of ##G'##.
Thanks! It suddenly makes a lot more sense.

Mentor
2022 Award
Thanks! It suddenly makes a lot more sense.
By the way, ##G\,'## is an unfortunate notation for a second group, better choose ##\varphi\, : \,G \longrightarrow H##. The reason is that many authors abbreviate the commutator ##G\,'=[G,G]=\{\,aba^{-1}b^{-1}\,|\,a,b \in G\,\}## which is an important normal subgroup of ##G##.

Last edited:
AllRelative


text:
# Thread: Limit problem (I made it up and worked it out.)

1. ## Limit problem (I made it up and worked it out.)

The following limit exists. Find the value of the limit.

$\displaystyle\lim_{x\to 16} \ \dfrac{(\sqrt{x} - 4)(\sqrt{x - 7 \ } - 3)}{(\sqrt{x - 12 \ } - 2)(\sqrt{x - 15 \ } - 1)}$

2. ## Re: Limit problem (I made it up and worked it out.)

Hey greg1313.

Can you please show us what you have tried?

What limit laws have you studied?

3. ## Re: Limit problem (I made it up and worked it out.)

I would start by rationalising the numerator...

4. ## Re: Limit problem (I made it up and worked it out.)

Originally Posted by chiro
Hey greg1313.

Can you please show us what you have tried?

What limit laws have you studied?
It's a math challenge problem. I'm challenging the rest of you.

And notice that I stated that I worked it out.

5. ## Re: Limit problem (I made it up and worked it out.)

$\displaystyle \tfrac16$

6. ## Re: Limit problem (I made it up and worked it out.)

Originally Posted by Archie
$\displaystyle \tfrac16$
You're going to have to show on here how you got that alleged answer or at least an outline of the work.
For instance, I could just graph f(x) = the expression that the limit is being taken and estimate the function
value where the "hole" is at where x = 16.

7. ## Re: Limit problem (I made it up and worked it out.)

(Edit)

The following limit exists. Find the value of the limit. Please show your work.

$\displaystyle\lim_{x\to 16} \ \dfrac{(\sqrt{x} - 4)(\sqrt{x - 7 \ } - 3)}{(\sqrt{x - 12 \ } - 2)(\sqrt{x - 15 \ } - 1)}$

8. ## Re: Limit problem (I made it up and worked it out.)

Originally Posted by greg1313
(Edit)

The following limit exists. Find the value of the limit. Please show your work.

$\displaystyle\lim_{x\to 16} \ \dfrac{(\sqrt{x} - 4)(\sqrt{x - 7 \ } - 3)}{(\sqrt{x - 12 \ } - 2)(\sqrt{x - 15 \ } - 1)}$
$\frac{{(\sqrt x - 4)(\sqrt {x - 7\;} - 3)}}{{(\sqrt {x - 12\;} - 2)(\sqrt {x - 15\;} - 1)}} = \frac{{(\sqrt {x - 12\;} + 2)(\sqrt {x - 15\;} + 1)}}{{(\sqrt x + 4)(\sqrt {x - 7\;} + 3)}}$

9. ## Re: Limit problem (I made it up and worked it out.)

Originally Posted by greg1313
LOL

10. ## Re: Limit problem (I made it up and worked it out.)

Multiply top and bottom by the conjugates of the four terms. The result is Plato's expression.

I posted no working so as not to spoil it for anyone else.

11. ## Re: Limit problem (I made it up and worked it out.)

Nice algebra problem. Thanks.

Prove It was on the right track for a formal demonstration of Archie's result. And of course Plato's answer is elegant. For anyone who wants detail, see below:

Spoiler:
$x \ne 16 \implies \dfrac{\sqrt{x} - 4}{\sqrt{x - 12} - 2} =\dfrac{\sqrt{x} - 4}{\sqrt{x - 12} - 2} * \dfrac{\sqrt{x} +4}{\sqrt{x} + 4} = \dfrac{x - 16}{(\sqrt{x - 12} - 2)(\sqrt{x} + 4)}$.

$x \ne 16 \implies \dfrac{x - 16}{(\sqrt{x - 12} - 2)(\sqrt{x} + 4)} = \dfrac{x - 16}{(\sqrt{x - 12} - 2)(\sqrt{x} + 4)} * \dfrac{\sqrt{x - 12} + 2}{\sqrt{x - 12} + 2} =$

$\dfrac{(x - 16)(\sqrt{x - 12} + 2)}{(\{x - 12\} - 4)(\sqrt{x} + 4)} = \dfrac{\cancel {(x - 16) }(\sqrt{x - 12} + 2)}{\cancel {(x - 16)}(\sqrt{x} + 4)} = \dfrac{\sqrt{x - 12} + 2}{\sqrt{x} + 4)}.$

$\displaystyle \therefore \lim_{x \rightarrow 16}\dfrac{\sqrt{x - 12} + 2}{\sqrt{x} + 4)} = \dfrac{\sqrt{16 - 12} + 2 }{\sqrt{16} + 4} = \dfrac{2 + 2}{4 + 4} = \dfrac{1}{2}.$

$x \ne 16 \implies \dfrac{\sqrt{x - 7} - 3}{\sqrt{x - 15} - 1} = \dfrac{\sqrt{x - 7} - 3}{\sqrt{x - 15} - 1} * \dfrac{\sqrt{x - 7} + 3}{\sqrt{x - 7} + 3} = \dfrac{x - 7 - 9}{(\sqrt{x - 15} - 1)(\sqrt{x - 7} + 3)} = \dfrac{x - 16}{(\sqrt{x - 15} - 1)(\sqrt{x - 7} + 3)}.$

$x \ne 16 \implies \dfrac{x - 16}{(\sqrt{x - 15} - 1)(\sqrt{x - 7} + 3)} = \dfrac{x - 16}{(\sqrt{x - 15} - 1)(\sqrt{x - 7} + 3)} * \dfrac{\sqrt{x - 15} + 1}{\sqrt{x - 15} + 1} =$

$\dfrac{(x - 16)(\sqrt{x - 15} + 1)}{(\{x - 15\} - 1)(\sqrt{x - 7} + 3)} = \dfrac{\cancel {(x - 16)}(\sqrt{x - 15} + 1)}{\cancel {(x - 16)}(\sqrt{x - 7} + 3)} = \dfrac{\sqrt{x - 15} + 1}{\sqrt{x - 7} + 3}.$

$\displaystyle \therefore \lim_{x \rightarrow 16}\dfrac{\sqrt{x - 15} + 1}{\sqrt{x - 7} + 3} = \dfrac{\sqrt{16 - 1} + 1}{\sqrt{16 - 7} + 3} = \dfrac{1 + 1}{3 + 3} = \dfrac{1}{3}.$

$THUS, \displaystyle \lim_{x \rightarrow 16} \dfrac{(\sqrt{x} - 4)(\sqrt{x - 7} - 3)}{(\sqrt{x - 12} - 2)(\sqrt{x - 15} - 1)} = \dfrac{1}{2} * \dfrac{1}{3} = \dfrac{1}{6}.$

12. ## Re: Limit problem (I made it up and worked it out.)

Originally Posted by Archie
> > > Multiply top and bottom by the conjugates of the four terms. < < < The result is Plato's expression.
Archie, this is at least an outline of what to do. JeffM, Plato's post is not elegant. It is a relatively compact form step in
the process after doing what you stated. Someone has to understand/justify how to get there in the first place.

13. ## Re: Limit problem (I made it up and worked it out.)

Originally Posted by greg1313
JeffM, Plato's post is not elegant... Someone has to understand/justify how to get there in the first place.
Tastes differ. In my opinion, his statement is an elegant consequence of utilizing conjugates and indicates why it is a neat problem in the first place. I do agree that a beginning student is highly unlikely to follow so someone should provide a detailed development (as I did). But that does not detract from its simplicity and clarity (for those familiar with conjugates).

14. ## Re: Limit problem (I made it up and worked it out.)

In terms of limits solved using conjugate pairs, I prefer limits of the form
$\displaystyle \lim_{x \to \infty} \left( \sqrt{x^2 + ax + b} - \sqrt{x^2 + cx + d} \right)$
because the convergence is, to me, somewhat counter-intuitive.

If memory serves, I recently saw an example that combined the two ideas.

15. ## Re: Limit problem (I made it up and worked it out.)

Originally Posted by greg1313
Archie, this is at least an outline of what to do. JeffM, Plato's post is not elegant. It is a relatively compact form step in the process after doing what you stated. Someone has to understand/justify how to get there in the first place.
Originally Posted by Archie
In terms of limits solved using conjugate pairs, I prefer limits of the form
$\displaystyle \lim_{x \to \infty} \left( \sqrt{x^2 + ax + b} - \sqrt{x^2 + cx + d} \right)$
because the convergence is, to me, somewhat counter-intuitive.
For those of us who have had a life-time of experience teaching/writing about this topic surely realize that students working on these types of limits should have a solid grounding in basic algebra manipulation of radicals. Any of us should have taught the time honored 'trick' of inverse/sign-change. That is all that I posted. It skipped no steps, it left nothing out that someone competent is basic algebra might fail to see.

Here is that trick again.
$\displaystyle \left( \sqrt{x^2 + ax + b} - \sqrt{x^2 + cx + d} \right)=\dfrac{(a-c)x+(b-d)}{\sqrt{x^2 + ax + b} + \sqrt{x^2 + cx + d}}\to\dfrac{a-c}{\sqrt{2}}$ as $x\to\infty$.


text:
Binary systems and inverse matrices

We had a quiz earlier about binary systems. One problem got me thinking for the whole day.

Let $S$ be a subset of $M_{2}(\mathbb{R})$ consisting of all matrices of the form $\Bigg[\begin{matrix} a & a \\ a & a \\ \end{matrix}\Bigg]$, where $a\neq0$.

1. Show that $S$ is a commutative binary structure under matrix multiplication.

2. Does it have an identity element? If yes, find it.

3. Given $A\in S$, does $A^{-1}$ exist in $S$?

I'm mostly interested in the third item, so I'll skip with the first one.

Define $\ast$ to be matrix multiplication. In algebraic structures, we know that an identity element exists if there exists a unique $e$ such that $a\ast e=a$. Since the identity matrix should be an element of $S$, then it should also be in the form of $\Bigg[\begin{matrix} e & e \\ e & e \\ \end{matrix}\Bigg]$. Hence, $$\Bigg[\begin{matrix} a & a \\ a & a \\ \end{matrix}\Bigg]\Bigg[\begin{matrix} e & e \\ e & e \\ \end{matrix}\Bigg]=\Bigg[\begin{matrix} a & a \\ a & a \\ \end{matrix}\Bigg]$$ So we have $2ae=a$. Therefore, the identity element is $\Bigg[\begin{matrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \\ \end{matrix}\Bigg]$. For the case of inverse, we can show that the inverse exists if $a\ast a'=e$. Therefore, $$\Bigg[\begin{matrix} a & a \\ a & a \\ \end{matrix}\Bigg]\Bigg[\begin{matrix} a' & a' \\ a' & a' \\ \end{matrix}\Bigg]=\Bigg[\begin{matrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \\ \end{matrix}\Bigg]$$ So we have $2aa'=\frac{1}{2}$. Hence, its inverse is $\Bigg[\begin{matrix} \frac{1}{4a} & \frac{1}{4a} \\ \frac{1}{4a} & \frac{1}{4a} \\ \end{matrix}\Bigg]$.

I'm not sure if this is right or wrong. I'm convinced that this is wrong since we can show that the determinant is equal to zero $(a^{2}-a^{2}=0)$. Also, I don't know if this is relevant, but we've been taught the identity matrix for multiplication is always $\Bigg[\begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix}\Bigg]$, but since it's not an element of $S$, I don't really know what to do with it.

Here are some of my questions:

1. Was it okay for me to assume that $\ast$ is defined to be the matrix multiplication? I think it was wrong, but I also don't know where to go with this problem if it's an arbitrary operation.

2. What does the determinant say about a matrix? Why can't it be equal to zero for a matrix to have an inverse?

3. If wrong, what is the correct way of proving or disproving its existence?

Please don't use concepts from linear algebra. Just limit it to basic group theory and basic matrix concepts. Any answer would be much appreciated.

• Note that your operation is essentially $(a,b) \mapsto 2ab$ once you record only the northwest entry. Now use this to get all your assertions. Aug 25, 2017 at 1:35

• "Was it okay for me to assume that ∗ is defined to be the matrix multiplication?"

Yes, because you were told to "how that S is a commutative binary structure under matrix multiplication"

• "What does the determinant say about a matrix?" (and related questions)

The identity of a binary operation is an element that doesn't change other elements, so $1\star a=a\star 1=a$ for all $a$.

If you consider the set $M$ of all 2x2 matrices, there's only one matrix $I$ for which $AI=IA=A$ for all $A$.

Now, though, you're looking at $S$. Your matrix $E=\Bigg[\begin{matrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \\ \end{matrix}\Bigg]$ satisfies $EA=AE=A$ for all $A\in S$, so it's an identity within $S$. It's clearly not an identity within $M$, since there are plenty of matrices in $M$ that don't have $EA=AE=A$.

So, if you have a ring structure, there's nothing stopping proper subsets of it from having completely different identity elements. In fact, any time you have an element A of a ring with $A^2=A$, you can find a set of elements of the ring for which $A$ is an identity.

The identity of $M$ is $I$. The identity of $S$ is $E$.

You've worked out inverses in $S$ correctly - every element (except $O$) has an inverse, and you have the right formula for them. Yes, they all have determinant zero, and still have inverses in $S$. They don't have inverses in $M$, but they do in $S$. The determinant is a nice tool for figuring out which matrices have inverses in $M$, but it's no use in $S$, you'll need a different criterion. Maybe $\det_S\Bigg[\begin{matrix} a & a \\ a & a \\ \end{matrix}\Bigg]=2a$ will do.

Actually, though, you don't need determinants at all for this question. There's a simple rule for testing invertibility in $S$, namely, $a\neq0$, and the traditional determinant $ad-bc$ is not relevant since it's talking about $M$.

Regarding 1: The first question says "show that $S$ is a commutative binary structure under matrix multiplication." It is therefore extremely likely that, for the rest of the question, the binary operation is still supposed to be matrix multiplicaiton.
Regarding 2: The inverse of a matrix in the linear-algebra sense is the inverse of a matrix within the binary structure $M_2(\Bbb R)$ under matrix multiplication. In particular, the "inverse" of a matrix $A$ in that context is a matrix $B$ such that $AB = BA = I$ (where $I = \left( \begin{smallmatrix} 1&0\\0&1 \end{smallmatrix} \right)$). Notably, $I$ is the unique identity element within that structure.
Our question, however, is about the smaller binary structure $S$. You have correctly identified the identity element $E = \left( \begin{smallmatrix} 1/2&1/2\\1/2&1/2 \end{smallmatrix} \right)$ within this structure. By definition, the inverse of an element $A \in S$ with respect to the binary structure on $S$ is the element $B$ such that $AB = BA = E$. You have found the correct formula for this inverse.
Note, however, that the elements $A \in S$ do not have an inverse with respect to the broader structure on $M_2(\Bbb R)$. In particular, there is no matrix $B \in M_2(\Bbb R)$ such that $AB = BA = I$. In general, any matrix $A$ whose determinant is zero cannot have an inverse with respect to the multiplicative binary structure on $M_n(\Bbb R)$.


text:
# zbMATH — the first resource for mathematics

Uniform distribution of the Steinitz invariants of quadratic and cubic extensions. (English) Zbl 1113.11065
Let $$k$$ be a number field with ring of integers $$\mathcal O_k$$ and class group $$\text{Cl}(k)$$ of order $$h$$. Let $$K/k$$ be a finite extension; one can associate to the ring of integers $$\mathcal O_K$$ of $$K$$ an element $$S(\mathcal O_K)$$ of $$\text{Cl}(k)$$, i.e, the Steinitz invariant of $$\mathcal O_K$$ as $$\mathcal O_k$$-module. The aim of the paper is to show an equidistribution result for the values of this map when restricted to all extensions $$K/k$$ of a fixed degree $$d$$, ordered according to the size of the absolute ideal norm $$\mathcal N(\Delta_{K/k})$$ of their relative discriminant $$\Delta_{K/k}$$.
More precisely, the authors prove the following
Theorem: For $$d=2,3$$ and for any $$\mathcal C\in \text{Cl}(k)$$ one has $\lim_{B\to\infty} {\left|\{K| [K:k]=d,\, \mathcal N(\Delta_{K/k})\leq B,\, S(\mathcal O_K) = \mathcal C\}\right| \over \left| \{\{K| [K:k]=d,\, \mathcal N(\Delta_{K/k})\leq B\}\right| } = {1\over h}.$ For the case $$d=2$$ this is a refinement of a previous result of A. Fröhlich [Mathematika 7, 15–22 (1960; Zbl 0229.12007)], while for $$d=3$$ the result is completely new. The main ingredients of the proof for $$d=3$$ are the Shintani zeta function of the space of binary cubic forms and the Delone-Faddeev correspondence between a suitable subset of the binary cubic forms defined over $$k$$ and the cubic extensions of $$k$$.

##### MSC:
 11R45 Density theorems 11R11 Quadratic extensions 11R16 Cubic and quartic extensions 11S90 Prehomogeneous vector spaces 11M41 Other Dirichlet series and zeta functions
Full Text:


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# How to simulate 3 correlated stock processes following a GBM?

Suppose we have 3 stocks following GBMs.

We are given the distribution of the daily log returns which is multivariate normal.

Suppose I want to sample the stock price tomorrow ($\Delta t = 1$ day), could I just sample a return vector from this distribution and then say that the stock price tomorrow is $S_0 \cdot \exp(r_\text{sample}\Delta t)$?

I've been arguing with my friend about this and he claims I should multiply by $\sqrt{\Delta t}$? I don't understand his argument.

Is there anything wrong with what I am doing here?

• For this question to be answered, you need to explain how you sampled $r_\text{sample}$.
– SRKX
Jul 6 '16 at 9:17

The log-return of a stock over a period $\Delta t$ starting at $t=0$ is defined as: $$r_{\Delta t} = \ln \left( \frac{S_{\Delta t}}{S_0} \right)$$ Thus you should compute $S_{\Delta t}$ as $$S_{\Delta t} = S_0 \exp ( r_{\Delta t} )$$ when you are given the $\Delta t$-period log-return i.e. the one which you sample as you propose above. Thus no multiplication by $\Delta t$ or its square root whatsoever.

Maybe your confusion arises from the fact that in the BS equation we traditionally use continuously compounded rates: $$\exp ( r_{\Delta t} ) = \exp \left( \int_0^{\Delta t} r (t) dt \right) = \exp ( r \Delta t )$$ where the last equality holds when $r (t) = r$ a constant, and in which case you should use $\Delta t \approx 1/252$ to compute daily returns if you're using annualised quantities (which is usually the case)

You mix up several things:

if you sample from Brownian motion, then $$B_{t+\Delta t} - B_t$$ is normally distributed with variance $\Delta t$. Thus if you sample a standard normal $Z$ (with variance 1) then you can use $$\sqrt{\Delta t} Z$$ as sample for $B_{t+\Delta t} - B_t$ in order to get the correct variance. Recall that constant factors enter variance with the squared value.

In your question: how do you sample $r$? And if $\Delta t=1$ then it does not matter at all whether you use $\Delta t$ or $\sqrt{\Delta t}$.

• I totally get what you mean and obviously agree, but if you re-read your answer, you'll notice it's not very clear without prior understanding. If you have a bit of time, it would be nice to slightly rephrase it I think.
– SRKX
Jul 6 '16 at 9:15
• Right .. but as the OP asks about 1 day returns and $\Delta t = 1$ ... this is just trivial ...
– Ric
Jul 6 '16 at 12:35


text:
Acta Chimica Sinica ›› 2014, Vol. 72 ›› Issue (5): 602-608.

Article

### 不同亲疏水比例的光响应性嵌段共聚物的合成及溶液组装行为研究

1. 复旦大学材料科学系 上海 200433
• 投稿日期:2014-01-23 发布日期:2014-03-29
• 通讯作者: 韦嘉 E-mail:weijia@fudan.edu.cn
• 基金资助:

项目受国家自然科学基金（Nos. 21134003，21273048，51225304，51203023）、复旦大学中央高校基本科研业务费青年教师科研能力提升项目（No. 20520133189）和上海市自然科学基金（No. 12ZR1401600）资助.

### Synthesis and Self-assembly Behavior of Photo-responsive Diblock Copolymers with Different Hydrophilic/Hydrophobic Ratios

Ding Yanchun, Yu Yanlei, Wei Jia

1. Department of Materials Science, Fudan University, Shanghai 200433
• Received:2014-01-23 Published:2014-03-29
• Supported by:

Project supported by the National Natural Science Foundation of China (Nos. 21134003, 21273048, 51225304, 51203023), Fundamental Research Funds for the Central Universities (No. 20520133189) and Shanghai Natural Science Foundation (12ZR1401600).

A series of diblock copolymers, poly(ethylene oxide)-block-poly{6-[4-(4-ethoxyphenylazo) phenoxy] hexyl methacrylate} (PEO-b-PMA6AB2), with different hydrophilic/hydrophobic ratios were synthesized by reversible addition-fragmentation chain transfer (RAFT) polymerization. The structures and properties of diblock copolymers were characterized by 1H NMR and gel permeation chromatography (GPC). The self-assembly method was as follows: block copolymers were dissolved in tetrahydrofuran (THF) at first, and then by adding water at the rate of 2.5 μL/s, the aggregates self-assembled when the critical micelle concentration (CMC) was reached. Field emission transmission electron microscope (FETEM), scanning electron microscope (SEM), field emission scanning electron microscope (FESEM), optical microscope, particle size analyzer, UV-Vis absorption spectroscope and fluorescence emission spectroscope were used to study the morphology and size of the aggregates and their stimuli-responsive behavior. The SEM and FETEM results showed that polymeric micelles and large compound micelles (LCM) coexisted in the aqueous solution. Furthermore, it was found that the size of the self-assembly reduced with the decrease of initial concentration and the increase of stirring rate, and the distribution tended to be uniform. Meanwhile, the morphology of large compound micelles turned to be different, changing from common sphere to peanut shape or rod-like shape when the hydrophilic/hydrophobic ratio was 2.3. By encapsulating Nile Red into the aggregates, the peanut shape and rod-like shape were captured by fluorescence microscope. UV-Vis absorption spectra of the self-assembly revealed that the photo-responsive behavior of azobenzene moieties in the aggregates was reversible but restricted to some extent. The change on fluorescence intensity of Nile Red encapsulated in the diblock copolymer aggregates indicated that the alternative irradiation of UV and visible light induced the release of Nile Red resulted from the photoisomerization of the azobenzene moieties. In addition, the photo-responsive study showed that large compound micelles self-assembled from BCP-S3-0.82 could dissociate under the irradiation of UV light. Other large compound micelles assembled from the rest of block copolymers also dissociated, but the extent of the dissociation was not as remarkable as BCP-S3-0.82, which had the maximal hydrophobic ratio.


text:
# How much energy should I give into each particle in this equation:

I am trying to recreate the results of an article (membrane simulation) and I have the following line:

Both particles have the same soft radius,

$U_{rep} (r)/\epsilon = \text{exp}\left\{ -20 (r/\sigma -1 )\right \}$

With $\epsilon$ being the energy scale, $\sigma$ being the length scale, and $r$ is the distance between two particles. This is a two body interactions. If I calculate the interactions between two particles, should I put $U_{rep}$ or half of it into each particle?

-

Whatever is the nature of the interaction one should take it into account in the Hamiltonian only once for each distinct pair of particles: $$U_\text{int} = \sum_{i>j} U_1 (\mathbf{r}_i, \mathbf{r}_j) = \sum_{i<j} U_1 (\mathbf{r}_i, \mathbf{r}_j).$$ Usually these sums are merged to make the Hamiltonian of the system more symmetric. In that case the total sum should be divided by two: $$U_\text{int} = \frac{1}{2}\sum_{i\neq j} U_1 (\mathbf{r}_i, \mathbf{r}_j).$$ Now the total hamiltonian $$H = \sum_i \frac{p_i^2}{2m} + \frac{1}{2}\sum_{i\neq j} U_1 (\mathbf{r}_i, \mathbf{r}_j)$$ can be rewritten as a sum of "one-particle" Hamiltonians: $$H = \sum_i\left( \frac{p_i^2}{2m} + \frac{1}{2}\sum_{j \neq i} U_1 (\mathbf{r}_i, \mathbf{r}_j) \right).$$ If you mean this when say "put energy into each particle" then, yes, you should put a half of the interaction energy.


text:
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Kinetic Molecular Theory of Gases

Table of Content

Postulates of Kinetic Molecular Theory

The Kinetic Equation

Velocities of Gas Molecules

Kinetic Energy of Gas

Related Resources

The various gas laws which we have studied so far were based an experimental facts. The theory that provides an explanation for the various experimental observations about a gas is based on the Kinetic Molecular model. It is assumed that all gases are made up of identical molecules, which are in a state of constant random motion.

Postulates of Kinetic Molecular Theory

Maxwell proposed the following postulates under the heading of kinetic theory of gases.

Each gas is made up of a large number of small (tiny) particles known as molecules.

The volume of a molecule is so small that it may be neglected in comparison to total volume of gas.

The molecules are never in stationary state but they are believed to be in chaotic (random) motion. They travel in straight line in all possible directions with altogether different but constant velocities. The direction of motion is changed by the collision with container or with other molecules.

The molecules are perfectly elastic and bear no change in energy during collision.

The effect of gravity on molecular motion is negligible.

The temperature of gas is a measure of its kinetic energy.

The pressure of the gas is thus due to the continuous collision of molecules on the walls of container. Pressure P = Force/Area.

Refer to the following video for kinetic molecular theories of gases

The Kinetic Equation

Maxwell also derived an equation on the basis of above assumption as

Where P = Pressure of gas

V = Volume of gas

m = mass of one molecule  of gas

n = no. of molecules of gas

u = root mean square velocity of molecules

For 1 mole n = N is Avogadro number

m  N = Molecular mass M

For 1 mole of an Ideal gas, pV = RT

so,

Velocities of Gas Molecules

Average Velocity

As per kinetic theory of gases, each molecule is moving with altogether different velocity. Let ‘n’ molecules be present in a given mass of gas, each one moving with velocity v1, v2, v3… vn.

The average velocity or UaV = average of all such velocity terms.

Average velocity =

Root Mean Square Velocity:-

Maxwell proposed the term Urms as the square root of means of square of all such velocities.

also

Most probable velocity:-

It is the velocity which is possessed by maximum no. of molecules.

Furthermore

Kinetic Energy of Gas

As per kinetic equation

For 1 mole m × n = Molecular Mass (M)

Also

Where k is the Boltzmann constant (k = R / N)

Solved Examples

Question 1:

Calculate rms speed of O2 at 273 K and 1 × 105Pa pressure. The density of O2under these conditions is 1.42 kg m–3.

Solution:

Data are given in SI units

C = √3P / d = √3 × 105 / 1.42 = 459.63 m sec–1

__________________________________________________

Question 2:

At what temperature will the r.m.s. velocity of oxygen be one and half times of its value at N.T.P.?

Solution:

1/2 mc2 = 3/2 KT

Suppose the temperature required is T¢ when the velocity will be 3/2 C

3/2C / C = √T'/T or, T' = 9/4 × 273 = 614.25°K

_____________________________________________

Question 3:

Calculate K.E. of 4g of N2 at – 13°C.

Solution:

EK = 3/2 nRT = 3/2 × 4/28 × 8.314 × 260  = 463.21 J mol–1

Question 1: Which of the following statements of kinetic molecular theory is incorrect?

a. The effect of gravity reduced the molecular motion .

b. The temperature of gas is a measure of its kinetic energy. KE µ absolute temperature.

c. The pressure of the gas is thus due to the continuous collision of molecules on the walls of container

d. The volume of a molecule is so small that it may be neglected in comparison to total volume of gas.

Question 2: In the equation

u2 represents

a. average velocity of molecules.

b. root mean square velocity of molecules.

c. most probable velocity of molecules.

d. final velocity of molecules.

Question 3: Kinetic energy of gas is given by

a. 3/2 nRT

b.1/2 nRT

c. 3/2 nKT

d. 1/2 nRT

Question 4:  The velocity which is possessed by maximum no. of molecules is called

a. average velocity.

b. root mean square velocity.

c. most probable velocity.

d. escape velocity.

Q.1

Q.2

Q.3

Q.4

a

b

a

c

Related Resources

Look here for Reference books for IIT JEE Chemistry

You can also refer States of matter

To read more, Buy study materials of States of Matter comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Chemistry here.

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# bumbl

Example output of bumbl package

The bumbl package implements a model for bumblebee colony growth described in Crone and Williams 2019. It models colony growth as having a change point at some time, tau, where the colony switches from growth and worker production to gyne production. The bumbl() function applies this model to data from multiple colonies, allowing for each colony to have it’s own tau and returns the original data augmented with coefficients from the changepoint model.

In addition, I’m working on implementing an integral projection model for bumblebee colony growth with variation in worker size distribution and resource return based on the (currently unpublished) work of Natlie Kerr and others.

###### Postdoctoral Researcher

I’m a postdoctoral researcher in Emilio Bruna’s lab at University of Florida working on the effects of drought and habitat fragmentation on a tropical plant. I’m interested in the mechanisms of plant responses to stress and their consequences for natural and agricultural ecosystems.