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Advanced Interactive Media/Design Team/Goals and Description

Goals and Description
The role of the Design Team in the Multimedia Recruiting Project is one of the most essential and vital roles of all the teams. They are in charge of designing the look and style of all the elements of the project. They are developing design strategies and screen templates for the video and interactive teams as well as shooting VR's. They design the marketing materials and are striving to hit on alternative advertising. They will create innovative and creative electronic and print materials. They will form a focus group of high school students from the Tulsa area in order to understand the target market more in depth.

The team's goal and objective is to make the project appealing to the target market. Our vision is for not only the design be accepted and enjoyed by the target market, but also create a memorable experience that will make an impression on potential students. They understand the position of influence they have with their audience. They see beyond the design into the possibility that they have to help potential students realize a purpose for their life.


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Korean/Lesson II8

Vocabulary

 * hospital  병원
 * emergency 응급, 긴급
 * ambulance 구급차
 * doctor    의사
 * nurse     간호사
 * fever     열
 * headache 두통
 * stomachache 복통
 * cough 기침
 * throat  목, 목구멍
 * pain    고통, 아픔
 * medicine 약
 * prescription 처방전
 * flu       독감
 * catch a cold 감기에 걸리다
 * illness,disease   병
 * surgery 수술

Dialogue
D: 어디가 아프세요? Where does it hurt? P: 두통이 있어요. I have a headache. 열도 나고, 기침도 나요. I have a fever and a cough. D: 감기에 걸리신 것 같아요. You seem to have caught a cold. 처방전을 들고 약국에 가세요. Visit pharmacy with this prescriptrion. P: 감사합니다. Thank you.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: Korean/Lesson II8


text:
Korean/Lesson II7

Vocabulary

 * sheep  양(羊)
 * field  밭
 * cow    소
 * rice paddy 논
 * farmer 농부(農夫)

title: Korean/Lesson II7


text:
Korean/Lesson II3

Vocabulary

 * Movie theater     영화관/극장
 * Karaoke Room     노래방
 * DVD room          DVD 방
 * PC room           PC 방
 * small restaurant  식당
 * Station             역
 * Subway            지하철
 * store                가게
 * Department store 백화점(←every[百]+goods[貨]+store[店])
 * Convenient store 편의점
 * book store        서점
 * bank               은행
 * police office     경찰서
 * fire station        소방서
 * hospital           병원
 * parking lot       주차장
 * University  대학교

Conversation

 * PC방이 어디에 있어요?  Where is the PC room?
 * 영화가 언제 시작해요?  When does the movie start?
 * 동대문역 어디에 있십니까? Where is Dongdaemun station?

title: Korean/Lesson II3


text:
This Quantum World/Feynman route/From quantum to classical

< PREVIOUS

Action
Let's go back to the propagator

\langle B|A\rangle=\int\!\mathcal{DC}\,Z[\mathcal{C}:A\rightarrow B]. $$ For a free and stable particle we found that

Z[\mathcal{C}]=e^{-(i/\hbar)\,m\,c^2\,s[\mathcal{C}]},\qquad s[\mathcal{C}]= \int_\mathcal{C}ds, $$ where $$ds=\sqrt{dt^2-(dx^2+dy^2+dz^2)/c^2}$$ is the proper-time interval associated with the path element $$d\mathcal{C}$$. For the general case we found that the amplitude $$Z(d\mathcal{C})$$ is a function of $$t,x,y,z$$ and $$dt,dx,dy,dz$$ or, equivalently, of the coordinates $$t,x,y,z$$, the components $$c\,dt/ds, dx/ds, dy/ds, dz/ds$$ of the 4-velocity, as well as $$ds$$. For a particle that is stable but not free, we obtain, by the same argument that led to the above amplitude,

Z[\mathcal{C}]=e^{(i/\hbar)\,S[\mathcal{C}]}, $$ where we have introduced the functional $$S[\mathcal{C}]=\int_\mathcal{C}dS$$, which goes by the name action.

For a free and stable particle, $$S[\mathcal{C}]$$ is the proper time (or proper duration) $$s[\mathcal{C}]=\int_\mathcal{C}ds$$ multiplied by $$-mc^2$$, and the infinitesimal action $$dS[d\mathcal{C}]$$ is proportional to $$ds$$:

S[\mathcal{C}]=-m\,c^2\,s[\mathcal{C}],\qquad dS[d\mathcal{C}]=-m\,c^2\,ds. $$

Let's recap. We know all about the motion of a stable particle if we know how to calculate the probability $$p(B|A)$$ (in all circumstances). We know this if we know the amplitude $$\langle B|A\rangle$$. We know the latter if we know the functional $$Z[\mathcal{C}]$$. And we know this functional if we know the infinitesimal action $$dS(t,x,y,z,dt,dx,dy,dz)$$ or $$dS(t,\mathbf{r},dt,d\mathbf{r})$$ (in all circumstances).

What do we know about $$dS$$?

The multiplicativity of successive propagators implies the additivity of actions associated with neighboring infinitesimal path segments $$d\mathcal{C}_1$$ and $$d\mathcal{C}_2$$. In other words,

e^{(i/\hbar)\,dS(d\mathcal{C}_1+d\mathcal{C}_2)}= e^{(i/\hbar)\,dS(d\mathcal{C}_2)}\, e^{(i/\hbar)\,dS(d\mathcal{C}_1)} $$ implies

dS(d\mathcal{C}_1+d\mathcal{C}_2)= dS(d\mathcal{C}_1)+ dS(d\mathcal{C}_2). $$ It follows that the differential $$dS$$ is homogeneous (of degree 1) in the differentials $$dt,d\mathbf{r}$$:

dS(t,\mathbf{r},\lambda\,dt,\lambda\,d\mathbf{r})=\lambda\,dS(t,\mathbf{r},dt,d\mathbf{r}). $$ This property of $$dS$$ makes it possible to think of the action $$S[\mathcal{C}]$$ as a (particle-specific) length associated with $$\mathcal{C}$$, and of $$dS$$ as defining a (particle-specific) spacetime geometry. By substituting $$1/dt$$ for $$\lambda$$ we get:

dS(t,\mathbf{r},\mathbf{v})=\frac{dS}{dt}. $$ Something is wrong, isn't it? Since the right-hand side is now a finite quantity, we shouldn't use the symbol $$dS$$ for the left-hand side. What we have actually found is that there is a function $$L(t,\mathbf{r},\mathbf{v})$$, which goes by the name Lagrange function, such that $$dS=L\,dt$$.

Geodesic equations
Consider a spacetime path $$\mathcal{C}$$ from $$A$$ to $$B.$$ Let's change ("vary") it in such a way that every point $$(t,\mathbf{r})$$ of $$\mathcal{C}$$ gets shifted by an infinitesimal amount to a corresponding point $$(t+\delta t,\mathbf{r}+\delta\mathbf{r}),$$ except the end points, which are held fixed: $$\delta t=0$$ and $$\delta\mathbf{r}=0$$ at both $$A$$ and $$B.$$

If $$t\rightarrow t+\delta t,$$ then $$dt=t_2-t_1\longrightarrow t_2+\delta t_2-(t_1+\delta t_1)= (t_2-t_1)+(\delta t_2-\delta t_1)=dt+d\delta t.$$

By the same token, $$d\mathbf{r}\rightarrow d\mathbf{r} + d\delta\mathbf{r}.$$

In general, the change $$\mathcal{C}\rightarrow\mathcal{C}'$$ will cause a corresponding change in the action: $$S[\mathcal{C}]\rightarrow S[\mathcal{C}']\neq S[\mathcal{C}].$$ If the action does not change (that is, if it is stationary at $$\mathcal{C}$$ ),

\delta S=\int_{\mathcal{C}'} dS-\int_\mathcal{C} dS=0, $$ then $$\mathcal{C}$$ is a geodesic of the geometry defined by $$dS.$$ (A function $$f(x)$$ is stationary at those values of $$x$$ at which its value does not change if $$x$$ changes infinitesimally. By the same token we call a functional $$S[\mathcal{C}]$$ stationary if its value does not change if $$\mathcal{C}$$ changes infinitesimally.)

To obtain a handier way to characterize geodesics, we begin by expanding

dS(\mathcal{C}')=dS(t+\delta t,\mathbf{r}+\delta\mathbf{r},dt+d\delta t,d\mathbf{r}+d\delta\mathbf{r}) $$

=dS(t,\mathbf{r},dt,d\mathbf{r})+\frac{\partial dS}{\partial t}\,\delta t+\frac{\partial dS}{\partial\mathbf{r}}\cdot\delta\mathbf{r}+ \frac{\partial dS}{\partial dt}\,d\delta t+\frac{\partial dS}{\partial d\mathbf{r}}\cdot d\delta\mathbf{r}. $$

This gives us

(^*)\quad\int_{\mathcal{C}'} dS-\int_\mathcal{C} dS=\int_\mathcal{C}\left[{\partial dS\over\partial t}\delta t+ {\partial dS\over\partial\mathbf{r}}\cdot\delta\mathbf{r}+{\partial dS\over\partial dt}d\,\delta t+ {\partial dS\over\partial d\mathbf{r}}\cdot d\,\delta\mathbf{r}\right]. $$

Next we use the product rule for derivatives,

d\left({\partial dS\over\partial dt}\delta t\right)= \left(d{\partial dS\over\partial dt}\right)\delta t+{\partial dS\over\partial dt}d\,\delta t, $$

d\left({\partial dS\over\partial d\mathbf{r}}\cdot\delta\mathbf{r}\right)= \left(d{\partial dS\over\partial d\mathbf{r}}\right)\cdot\delta\mathbf{r}+ {\partial dS\over\partial d\mathbf{r}}\cdot d\,\delta\mathbf{r}, $$

to replace the last two terms of (*), which takes us to

\delta S=\int\left[\left({\partial dS\over\partial t}-d{\partial dS\over\partial dt}\right)\delta t+\left({\partial dS\over\partial\mathbf{r}}-d{\partial dS\over\partial d\mathbf{r}}\right)\cdot\delta\mathbf{r}\right]+\int d\left({\partial dS\over\partial dt}\delta t+{\partial dS\over\partial d\mathbf{r}}\cdot \delta\mathbf{r}\right). $$

The second integral vanishes because it is equal to the difference between the values of the expression in brackets at the end points $$A$$ and $$B,$$ where $$\delta t=0$$ and $$\delta\mathbf{r}=0.$$ If $$\mathcal{C}$$ is a geodesic, then the first integral vanishes, too. In fact, in this case $$\delta S=0$$ must hold for all possible (infinitesimal) variations $$\delta t$$ and $$\delta\mathbf{r},$$ whence it follows that the integrand of the first integral vanishes. The bottom line is that the geodesics defined by $$dS$$ satisfy the geodesic equations

Principle of least action
If an object travels from $$A$$ to $$B,$$ it travels along all paths from $$A$$ to $$B,$$ in the same sense in which an electron goes through both slits. Then how is it that a big thing (such as a planet, a tennis ball, or a mosquito) appears to move along a single well-defined path?

There are at least two reasons. One of them is that the bigger an object is, the harder it is to satisfy the conditions stipulated by Rule $$B.$$ Another reason is that even if these conditions are satisfied, the likelihood of finding an object of mass $$m$$ where according to the laws of classical physics it should not be, decreases as $$m$$ increases.

To see this, we need to take account of the fact that it is strictly impossible to check whether an object that has travelled from $$A$$ to $$B,$$ has done so along a mathematically precise path $$\mathcal{C}.$$ Let us make the half realistic assumption that what we can check is whether an object has travelled from $$A$$ to $$B$$ within a a narrow bundle of paths — the paths contained in a narrow tube $$\mathcal{T}.$$ The probability of finding that it has, is the absolute square of the path integral $$I(\mathcal{T})=\int_\mathcal{T}\mathcal{DC} e^{(i/\hbar)S[\mathcal{C}]},$$ which sums over the paths contained in $$\mathcal{T}.$$

Let us assume that there is exactly one path from $$A$$ to $$B$$ for which $$S[\mathcal{C}]$$ is stationary: its length does not change if we vary the path ever so slightly, no matter how. In other words, we assume that there is exactly one geodesic. Let's call it $$\mathcal{G},$$ and let's assume it lies in $$\mathcal{T}.$$

No matter how rapidly the phase $$S[\mathcal{C}]/\hbar$$ changes under variation of a generic path $$\mathcal{C},$$ it will be stationary at $$\mathcal{G}.$$ This means, loosely speaking, that a large number of paths near $$\mathcal{G}$$ contribute to $$I(\mathcal{T})$$ with almost equal phases. As a consequence, the magnitude of the sum of the corresponding phase factors $$e^{(i/\hbar)S[\mathcal{C}]}$$ is large.

If $$S[\mathcal{C}]/\hbar$$ is not stationary at $$\mathcal{C},$$ all depends on how rapidly it changes under variation of $$\mathcal{C}.$$ If it changes sufficiently rapidly, the phases associated with paths near $$\mathcal{C}$$ are more or less equally distributed over the interval $$[0,2\pi],$$ so that the corresponding phase factors add up to a complex number of comparatively small magnitude. In the limit $$S[\mathcal{C}]/\hbar\rightarrow\infty,$$ the only significant contributions to $$I(\mathcal{T})$$ come from paths in the infinitesimal neighborhood of $$\mathcal{G}.$$

We have assumed that $$\mathcal{G}$$ lies in $$\mathcal{T}.$$ If it does not, and if $$S[\mathcal{C}]/\hbar$$ changes sufficiently rapidly, the phases associated with paths near any path in $$\mathcal{T}$$ are more or less equally distributed over the interval $$[0,2\pi],$$ so that in the limit $$S[\mathcal{C}]/\hbar\rightarrow\infty$$ there are no significant contributions to $$I(\mathcal{T}).$$

For a free particle, as you will remember, $$S[\mathcal{C}]=-m\,c^2\,s[\mathcal{C}].$$ From this we gather that the likelihood of finding a freely moving object where according to the laws of classical physics it should not be, decreases as its mass increases. Since for sufficiently massive objects the contributions to the action due to influences on their motion are small compared to $$|-m\,c^2\,s[\mathcal{C}]|,$$ this is equally true of objects that are not moving freely.

What, then, are the laws of classical physics?

They are what the laws of quantum physics degenerate into in the limit $$\hbar\rightarrow0.$$ In this limit, as you will gather from the above, the probability of finding that a particle has traveled within a tube (however narrow) containing a geodesic, is 1, and the probability of finding that a particle has traveled within a tube (however wide) not containing a geodesic, is 0. Thus we may state the laws of classical physics (for a single "point mass", to begin with) by saying that it follows a geodesic of the geometry defined by $$dS.$$

This is readily generalized. The propagator for a system with $$n$$ degrees of freedom — such as an $$m$$-particle system with $$n=3m$$ degrees of freedom — is

\langle \mathcal{P}_f,t_f|\mathcal{P}_i,t_i\rangle=\int\!\mathcal{DC}\,e^{(i/\hbar)S[\mathcal{C}]}, $$ where $$\mathcal{P}_i$$ and $$\mathcal{P}_f$$ are the system's respective configurations at the initial time $$t_i$$ and the final time $$t_f,$$ and the integral sums over all paths in the system's $$n{+}1$$-dimensional configuration spacetime leading from $$(\mathcal{P}_i,t_i)$$ to $$(\mathcal{P}_f,t_f).$$ In this case, too, the corresponding classical system follows a geodesic of the geometry defined by the action differential $$dS,$$ which now depends on $$n$$ spatial coordinates, one time coordinate, and the corresponding $$n{+}1$$ differentials.

The statement that a classical system follows a geodesic of the geometry defined by its action, is often referred to as the principle of least action. A more appropriate name is principle of stationary action.

Energy and momentum
Observe that if $$dS$$ does not depend on $$t$$ (that is, $$\partial dS/\partial t=0$$ ) then

E=-{\partial dS\over\partial dt} $$ is constant along geodesics. (We'll discover the reason for the negative sign in a moment.)

Likewise, if $$dS$$ does not depend on $$\mathbf{r}$$ (that is, $$\partial dS/\partial\mathbf{r}=0$$ ) then

\mathbf{p}={\partial dS\over\partial d\mathbf{r}} $$ is constant along geodesics.

$$E$$ tells us how much the projection $$dt$$ of a segment $$d\mathcal{C}$$ of a path $$\mathcal{C}$$ onto the time axis contributes to the action of $$\mathcal{C}.$$ $$\mathbf{p}$$ tells us how much the projection $$d\mathbf{r}$$ of $$d\mathcal{C}$$ onto space contributes to $$S[\mathcal{C}].$$ If $$dS$$ has no explicit time dependence, then equal intervals of the time axis make equal contributions to $$S[\mathcal{C}],$$ and if $$dS$$ has no explicit space dependence, then equal intervals of any spatial axis make equal contributions to $$S[\mathcal{C}].$$ In the former case, equal time intervals are physically equivalent: they represent equal durations. In the latter case, equal space intervals are physically equivalent: they represent equal distances.

If equal intervals of the time coordinate or equal intervals of a space coordinate are not physically equivalent, this is so for either of two reasons. The first is that non-inertial coordinates are used. For if inertial coordinates are used, then every freely moving point mass moves by equal intervals of the space coordinates in equal intervals of the time coordinate, which means that equal coordinate intervals are physically equivalent. The second is that whatever it is that is moving is not moving freely: something, no matter what, influences its motion, no matter how. This is because one way of incorporating effects on the motion of an object into the mathematical formalism of quantum physics, is to make inertial coordinate intervals physically inequivalent, by letting $$dS$$ depend on $$t$$ and/or $$\mathbf{r}.$$

Thus for a freely moving classical object, both $$E$$ and $$\mathbf{p}$$ are constant. Since the constancy of $$E$$ follows from the physical equivalence of equal intervals of coordinate time (a.k.a. the "homogeneity" of time), and since (classically) energy is defined as the quantity whose constancy is implied by the homogeneity of time, $$E$$ is the object's energy.

By the same token, since the constancy of $$\mathbf{p}$$ follows from the physical equivalence of equal intervals of any spatial coordinate axis (a.k.a. the "homogeneity" of space), and since (classically) momentum is defined as the quantity whose constancy is implied by the homogeneity of space, $$\mathbf{p}$$ is the object's momentum.

Let us differentiate a former result,

dS(t,\mathbf{r},\lambda\,dt,\lambda\,d\mathbf{r})=\lambda\,dS(t,\mathbf{r},dt,d\mathbf{r}), $$ with respect to $$\lambda.$$ The left-hand side becomes

{d(dS)\over d\lambda}={\partial dS\over\partial(\lambda dt)}{\partial(\lambda dt)\over\partial\lambda}+ {\partial dS\over\partial(\lambda d\mathbf{r})}\cdot{\partial(\lambda d\mathbf{r})\over\partial\lambda}= {\partial dS\over\partial(\lambda dt)}dt+{\partial dS\over\partial(\lambda d\mathbf{r})}\cdot d\mathbf{r}, $$ while the right-hand side becomes just $$dS.$$ Setting $$\lambda=1$$ and using the above definitions of $$E$$ and $$\mathbf{p},$$ we obtain

$$dS=-m\,c^2\,ds$$ is a 4-scalar. Since $$(c\,dt,d\mathbf{r})$$ are the components of a 4-vector, the left-hand side, $$-E\,dt+\mathbf{p}\cdot d\mathbf{r},$$ is a 4-scalar if and only if $$(E/c,\mathbf{p})$$ are the components of another 4-vector.

(If we had defined $$E$$ without the minus, this 4-vector would have the components $$(-E/c,\mathbf{p}).$$)

In the rest frame $$\mathcal{F}'$$ of a free point mass, $$dt'=ds$$ and $$dS=-m\,c^2\,dt'.$$ Using the Lorentz transformations, we find that this equals

dS=-mc^2{dt-v\,dx/c^2\over\sqrt{1-v^2/c^2}}=-{mc^2\over\sqrt{1-v^2/c^2}}\,dt+ {m\mathbf{v}\over\sqrt{1-v^2/c^2}}\cdot d\mathbf{r}, $$ where $$\mathbf{v}=(v,0,0)$$ is the velocity of the point mass in $$\mathcal{F}.$$ Compare with the above framed equation to find that for a free point mass,

E={mc^2\over\sqrt{1-v^2/c^2}}\qquad\mathbf{p}={m\mathbf{v}\over\sqrt{1-v^2/c^2}}\;. $$

Lorentz force law
To incorporate effects on the motion of a particle (regardless of their causes), we must modify the action differential $$dS=-mc^2\,dt\sqrt{1-v^2/c^2}$$ that a free particle associates with a path segment $$d\mathcal{C}.$$ In doing so we must take care that the modified $$dS$$ (i) remains homogeneous in the differentials and (ii) remains a 4-scalar. The most straightforward way to do this is to add a term that is not just homogeneous but linear in the coordinate differentials:

(^*)\quad dS=-mc^2\,dt\sqrt{1-v^2/c^2}-qV(t,\mathbf{r})\,dt+ (q/c)\mathbf{A}(t,\mathbf{r})\cdot d\mathbf{r}. $$ Believe it or not, all classical electromagnetic effects (as against their causes) are accounted for by this expression. $$V(t,\mathbf{r})$$ is a scalar field (that is, a function of time and space coordinates that is invariant under rotations of the space coordinates), $$\mathbf{A}(t,\mathbf{r})$$ is a 3-vector field, and $$(V,\mathbf{A})$$ is a 4-vector field. We call $$V$$ and $$\mathbf{A}$$ the scalar potential and the vector potential, respectively. The particle-specific constant $$q$$ is the electric charge, which determines how strongly a particle of a given species is affected by influences of the electromagnetic kind.

If a point mass is not free, the expressions at the end of the previous section give its kinetic energy $$E_k$$ and its kinetic momentum $$\mathbf{p}_k.$$ Casting (*) into the form

dS=-(E_k+qV)\,dt+[\mathbf{p}_k+(q/c)\mathbf{A}]\cdot d\mathbf{r} $$ and plugging it into the definitions

(^*{}^*)\quad E=-{\partial dS\over\partial dt},\qquad \mathbf{p}={\partial dS\over\partial d\mathbf{r}}, $$ we obtain

E=E_k+qV,\qquad \mathbf{p}=\mathbf{p}_k+(q/c)\mathbf{A}. $$ $$qV$$ and $$(q/c)\mathbf{A}$$ are the particle's potential energy and potential momentum, respectively.

Now we plug (**) into the geodesic equation

{\partial dS\over\partial\mathbf{r}}=d\,{\partial dS\over\partial d\mathbf{r}}. $$ For the right-hand side we obtain

d\mathbf{p}_k+{q\over c}d\mathbf{A}=d\mathbf{p}_k+{q\over c}\left[dt{\partial\mathbf{A}\over\partial t}+\left(d\mathbf{r}\cdot{\partial\over\partial\mathbf{r}}\right)\mathbf{A}\right], $$ while the left-hand side works out at

-q{\partial V\over\partial\mathbf{r}}dt+{q\over c}{\partial(\mathbf{A}\cdot d\mathbf{r})\over\partial\mathbf{r}}= -q{\partial V\over\partial\mathbf{r}}dt+{q\over c}\left[\left(d\mathbf{r}\cdot{\partial\over\partial\mathbf{r}}\right)\mathbf{A}+ d\mathbf{r}\times\left({\partial\over\partial\mathbf{r}}\times\mathbf{A}\right)\right]. $$ Two terms cancel out, and the final result is

d\mathbf{p}_k=q\underbrace{\left(-{\partial V\over\partial\mathbf{r}}-{1\over c}{\partial\mathbf{A}\over\partial t}\right)}_{\displaystyle\equiv\mathbf{E}}dt+ d\mathbf{r}\times {q\over c}\underbrace{\left({\partial\over\partial\mathbf{r}}\times \mathbf{A}\right)}_{\displaystyle\equiv\mathbf{B}}= q\,\mathbf{E}\,dt+ d\mathbf{r}\times {q\over c}\,\mathbf{B}. $$ As a classical object travels along the segment $$d\mathcal{G}$$ of a geodesic, its kinetic momentum changes by the sum of two terms, one linear in the temporal component $$dt$$ of $$d\mathcal{G}$$ and one linear in the spatial component $$d\mathbf{r}.$$ How much $$dt$$ contributes to the change of $$\mathbf{p}_k$$ depends on the electric field $$\mathbf{E},$$ and how much $$d\mathbf{r}$$ contributes depends on the magnetic field $$\mathbf{B}.$$ The last equation is usually written in the form

{d\mathbf{p}_k\over dt}=q\,\mathbf{E}+{q\over c}\,\mathbf{v}\times\mathbf{B}, $$ called the Lorentz force law, and accompanied by the following story: there is a physical entity known as the electromagnetic field, which is present everywhere, and which exerts on a charge $$q$$ an electric force $$q\mathbf{E}$$ and a magnetic force $$(q/c)\,\mathbf{v}\times\mathbf{B}.$$

(Note: This form of the Lorentz force law holds in the Gaussian system of units. In the MKSA system of units the $$c$$ is missing.)

Whence the classical story?
Imagine a small rectangle in spacetime with corners

 * $$A=(0,0,0,0),\;B=(dt,0,0,0),\;C=(0,dx,0,0),\;D=(dt,dx,0,0).$$

Let's calculate the electromagnetic contribution to the action of the path from $$A$$ to $$D$$ via $$B$$ for a unit charge ($$q=1$$) in natural units ( $$c=1$$ ):

S_{ABD}=-V(dt/2,0,0,0)\,dt+A_x(dt,dx/2,0,0)\,dx $$

\quad=-V(dt/2,0,0,0)\,dt+\left[A_x(0,dx/2,0,0)+{\partial A_x\over\partial t}dt\right]dx. $$ Next, the contribution to the action of the path from $$A$$ to $$D$$ via $$C$$:

S_{ACD}=A_x(0,dx/2,0,0)\,dx-V(dt/2,dx,0,0)\,dt $$

=A_x(0,dx/2,0,0)\,dx-\left[V(dt/2,0,0,0)+{\partial V\over\partial x}dx\right]dt. $$ Look at the difference:

\Delta S=S_{ACD}-S_{ABD}=\left(-{\partial V\over\partial x}-{\partial A_x\over\partial t}\right)dt\,dx =E_x\,dt\,dx. $$ Alternatively, you may think of $$\Delta S$$ as the electromagnetic contribution to the action of the loop $$A\rightarrow B \rightarrow D\rightarrow C\rightarrow A.$$

Let's repeat the calculation for a small rectangle with corners

 * $$A=(0,0,0,0),\;B=(0,0,dy,0),\;C=(0,0,0,dz),\;D=(0,0,dy,dz).$$

S_{ABD}=A_z(0,0,0,dz/2)\,dz+A_y(0,0,dy/2,dz)\,dy$$

=A_z(0,0,0,dz/2)\,dz+\left[A_y(0,0,dy/2,0)+{\partial A_y\over\partial z}dz\right]dy, $$

S_{ACD}=A_y(0,0,dy/2,0)\,dy+A_z(0,0,dy,dz/2)\,dz $$

=A_y(0,0,dy/2,0)\,dy+\left[A_z(0,0,0,dz/2)+{\partial A_z\over\partial y}dy\right]dz, $$

\Delta S=S_{ACD}-S_{ABD}= \left({\partial A_z\over\partial y}-{\partial A_y\over\partial z}\right)dy\,dz=B_x\,dy\,dz. $$ Thus the electromagnetic contribution to the action of this loop equals the flux of $$\mathbf{B}$$ through the loop. Remembering (i) Stokes' theorem and (ii) the definition of $$\mathbf{B}$$ in terms of $$\mathbf{A},$$ we find that

\oint_{\partial\Sigma}\mathbf{A}\cdot d\mathbf{r}=\int_\Sigma\hbox{curl}\,\mathbf{A}\cdot d\mathbf{\Sigma}=\int_\Sigma\mathbf{B}\cdot d\mathbf{\Sigma}. $$ In (other) words, the magnetic flux through a loop $$\partial\Sigma$$ (or through any surface $$\Sigma$$ bounded by $$\partial\Sigma$$ ) equals the circulation of $$\mathbf{A}$$ around the loop (or around any surface bounded by the loop).

The effect of a circulation $$\oint_{\partial\Sigma}\mathbf{A}\cdot d\mathbf{r}$$ around the finite rectangle $$A\rightarrow B\rightarrow D\rightarrow C\rightarrow A$$ is to increase (or decrease) the action associated with the segment $$A\rightarrow B\rightarrow D$$ relative to the action associated with the segment $$A\rightarrow C\rightarrow D.$$ If the actions of the two segments are equal, then we can expect the path of least action from $$A$$ to $$D$$ to be a straight line. If one segment has a greater action than the other, then we can expect the path of least action from $$A$$ to $$D$$ to curve away from the segment with the larger action.

Compare this with the classical story, which explains the curvature of the path of a charged particle in a magnetic field by invoking a force that acts at right angles to both the magnetic field and the particle's direction of motion. The quantum-mechanical treatment of the same effect offers no such explanation. Quantum mechanics invokes no mechanism of any kind. It simply tells us that for a sufficiently massive charge traveling from $$A$$ to $$D,$$ the probability of finding that it has done so within any bundle of paths not containing the action-geodesic connecting $$A$$ with $$D,$$ is virtually 0.

Much the same goes for the classical story according to which the curvature of the path of a charged particle in a spacetime plane is due to a force that acts in the direction of the electric field. (Observe that curvature in a spacetime plane is equivalent to acceleration or deceleration. In particular, curvature in a spacetime plane containing the $$x$$ axis is equivalent to acceleration in a direction parallel to the $$x$$ axis.) In this case the corresponding circulation is that of the 4-vector potential $$(cV,\mathbf{A})$$ around a spacetime loop.

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text:
Korean/Lesson III4

Vocabulary

 * the Universe   우주
 * astronaut      우주인
 * star           별
 * constellation  별자리
 * sun            태양
 * solar system   태양계
 * Mercury        수성
 * Venus          금성
 * Earth          지구
 * Mars           화성
 * Jupiter        목성
 * Saturn         토성
 * Uranus         천왕성
 * Neptune        해왕성
 * Pluto          명왕성
 * Comet          혜성

title: Korean/Lesson III4


text:
Korean/Lesson I7

Vocabulary

 * house       집
 * home        가정
 * apartment   아파트
 * bathroom    화장실 or 욕실
 * kitchen     부엌 or 주방
 * bedroom     침실 or 방
 * living room 거실
 * garden      정원
 * garage      차고
 * table       탁자
 * dining table 식탁
 * chair       의자
 * toilet      변기
 * sink        싱크대
 * window      창문
 * door        문

title: Korean/Lesson I7


text:
Korean/Lesson I5

Vocabulary

 * Taxi 택시
 * Hour 시간
 * Minute 분
 * Won (Korean currency) 원
 * Around, About ~ 정도

Conversation
Joie: 강남역까지 가고 싶습니다. / I want to go to Kang-nam station

Taxi driver: 네 강남역까지 가겠습니다. / Yes I will drive to Kang-nam station

Joie: (시간이) 얼마나 걸립니까? / How long does it take?

Taxi driver: 20분 정도 걸립니다. / It takes around 20 minutes

Joie: 얼마입니까? / How much is it?

Taxi driver: 5000원 입니다. / It is 5000 wons

Joie: 감사합니다. / Thank you

 * How far is it there? (거리가) 얼마나 멉니까?
 * How long does it take? (거리가) 얼마나 걸립니까?
 * Turn left  왼쪽으로 돌으십시오.
 * Turn right 오른쪽으로 돌으십시오.
 * Go straight 직진 하십시오.

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title: Korean/Lesson I5


text:
HSC Mathematics Advanced, Extension 1, and Extension 2

This book covers the New South Wales mathematics syllabi for Advanced, Extension 1 and Extension 2 levels. This aims to be a comprehensive guide to the course and heavily encourages contributions to improve its usefulness and keep it current in the face of more syllabus changes.

Preliminary

 * Basic arithmetic and algebra
 * /Real functions/
 * Trigonometric ratios
 * /Linear functions/
 * The quadratic polynomial and the parabola
 * Plane geometry — geometrical properties
 * /Tangent to a curve and derivative of a function/

HSC

 * Coordinate methods in geometry
 * Applications of geometrical properties
 * Geometrical applications of differentiation
 * /Integration/
 * /Trigonometric functions/ (including applications of trigonometric ratios)
 * /Exponential and logarithmic functions/
 * Applications of calculus to the physical world
 * Probability
 * /Series and series applications/

Extension 1 course

 * Graphs
 * Complex numbers
 * Conics
 * Integration
 * Volumes
 * Mechanics
 * Polynomials
 * Harder 3-Unit Topics
 * /4-Unit/Mathematical inductions

Preliminary

 * Other inequalities
 * Circle geometry
 * Further trigonometry (sums and differences, t-formulae, identities and equations)
 * Lines and linear functions
 * Internal and external division of lines into given ratios
 * Parametric representation
 * Permutations and combinations
 * Polynomials
 * Harder applications of the Preliminary 2-Unit course

HSC

 * Methods of integration
 * Primitive of sin²α and cos²α
 * Equation dN / dt = k(N - P)
 * Projectile motion
 * Further probability
 * Iterative methods for numerical estimation of the roots of a polynomial equation
 * Harder applications of HSC 2-Unit topics


text:
HSC Extension 1 and 2 Mathematics/3-Unit/Preliminary/Parametrics

The parametric form of a curve is an algebraic representation which expresses the co-ordinates of each point on the curve as a function of an introduced parameter, most frequently $$t$$. This contrasts with Cartesian form in that parametric equations do not describe an explicit relation between $$x$$ and $$y$$. This relation must be derived in order to convert from parametric to Cartesian form.

In 3-unit, parametrics focuses on a parametric representation of the quadratic, moving from parametric to Cartesian form and vice-versa, and manipulating geometrical aspects of the quadratic with parametrics. Recognition of other parametric forms is also useful, and more forms are introduced and dealt with in the 4-unit topic, conics.

Reasons for using the parametric form
In specific situations (in the school syllabus, primarily Conic sections), the parametric representation can be useful because:
 * points on the curve are represented by a single number, not two, simplifying algebra;
 * some elegant results are possible; for instance, in the standard paramaterisation of the quadratic, the gradient is equal to the parameter, $$t$$;
 * some curves, which cannot be expressed in functional form (for example, the circle, which is neither a function of $$x$$ nor of $$y$$) can be conveniently expressed in parametric form;
 * intuitively, it allows for an easier way to find points on the graph: you can sub in any value for the parameter and instantly find a point, whereas a relational form is not deterministic in the same way.

Furthermore, the parametric form occurs in certain natural phenomena. For instance, using the equations of motion, a thrown ball's location at any time can be calculated using the laws of projectile motion. This is implicitly paramaterising the ball's path by time; to find the shape of the ball's path, (which we know is a parabola) we must use parametrics, to eliminate time, $$t$$, from the equations.

Converting parametric to Cartesian forms
One of the simplest parametric forms is the line:
 * $$\begin{align}

x & = t \\ y & = t\end{align}$$ By inspection, it is obvious that this describes the line $$y = x$$. However, what is the formalised approach for doing this?

We are looking for some relation between $$x$$ and $$y$$ which has no $$t$$ in it. In other words, we want to eliminate $$t$$ from the equations. In the above example, we did this by equating the first and second equations, eliminating $$t$$.

Another example:

 * $$\begin{align}x & = 3t + 1 \\

y & = 9t^2 - 1\end{align}$$ We solve equation 1 for t, and substitute into equation 2:

\begin{align} x & = 3t + 1 \\ t & = \frac{x - 1}{3} \end{align} $$ Sub into equation 2:

\begin{align} y & = 9t^2 - 1 \\ & = 9\left(\frac{x - 1}{3}\right)^2 - 1 \\ & = x^2 - 2x + 1 - 1 \\ & = x^2 - 2x \end{align} $$ Here we have a Cartesian form, as required.

Three standard parametric forms
These parametric forms occur frequently, and should be recognised by 3- and 4-unit students.

Parabola
The standard paramaterisation of the parabola describes one with focal length $$a\,$$ and with its vertex at the origin. In Cartesian form, this is given as
 * $$x^2 = 4ay\,$$

In parametric form, this is given as
 * $$\begin{align}x & = 2at \\

y & = at^2\end{align}$$

Eliminating $$t\,$$ allows us to verify that this is equivalent to the Cartesian form:

\begin{align} x  & = 2at \\ x^2 & = 4a^2t^2 \\ & = 4a (at^2) \\ & = 4ay \end{align} $$ as required.

Circle
The circle with radius $$r\,$$ and center at the origin can be written in Cartesian form as
 * $$x^2 + y^2 = r^2\,$$

Introducing the parameter, $$\theta\,$$, this is:
 * $$\begin{align}x &= r\cos\theta \\

y &= r\sin\theta\end{align}$$ We convert to Cartesian as follows:

\begin{align} x^2      & = r^2\cos^2\theta \\ y^2      & = r^2\sin^2\theta \\ x^2 + y^2 & = r^2\cos^2\theta + r^2\sin^2\theta \\ & = r^2(\cos^2\theta + \sin^2\theta) \end{align} $$ Recalling the trig identity,
 * $$\sin^2\theta + \cos^2\theta = 1\,$$

we conclude
 * $$x^2 + y^2 = r^2\,$$

as required.

Geometrical interpretation
Unlike most parametrics, the conic sections are paramaterised by $$\theta\,$$ instead of $$t\,$$. For the circle, this implies a geometric representation: $$\theta\,$$ represents the angle the point makes with the $$x$$ axis.

Ellipse
Conceptually, an ellipse is just a 'squished' circle. The parametric form makes this clear:
 * $$\begin{align}x &= a\cos\theta\\

y &= b\sin\theta\end{align}$$ The cos and sin are still there, but they are now multiplied by different constants so that the $$x$$ and $$y$$ components are stretched differently. We can turn this into the parametric form in a similar fashion to the circle:

\begin{align} x^2                              & = a^2\cos^2\theta \\ y^2                              & = b^2\sin^2\theta \\ \frac{x^2}{a^2} + \frac{y^2}{b^2} & = \cos^2\theta + \sin^2\theta \\ & = 1 \end{align} $$ which is the standard form of an ellipse, with $$x$$ and $$y$$ intercepts at $$\pm a\,$$ and $$\pm b\,$$, respectively.

Properties of the parabola
3-Unit students are expected to remember the parametric description of a parabola ($$x = 2at, y = at^2$$). They are also expected to know (and/or be able to quickly derive) the equations of the tangent and the normal to the parabola at the point $$t$$ and at the point $$(x_1,y_1)$$.

Derivation of equations of the tangent and the normal
Differentiating each parametric equation with respect to $$t\,$$,
 * $$\begin{align}

\frac{dy}{dt} & = 2at \\ \frac{dx}{dt} & = 2a \end{align}$$ Then the gradient $$\tfrac{dy}{dx}$$ can be obtained by dividing $$\tfrac{dy}{dt}$$ by $$\tfrac{dx}{dt}$$ (this is the chain rule):
 * $$\begin{align}

\frac{dy}{dx} & = \frac{dy}{dt} \frac{dt}{dx} \\ & = \frac{2at}{1} \frac{1}{2a} \\ & = t \end{align}$$

Note that this result could have been derived without the chain rule, by taking the derivative of the Cartesian form (with respect to $$x\,$$) and solving for $$t\,$$. However, the above derivation is faster and more elegant.

The equation of the tangent at P
The gradient of the tangent at a point $$P(2at,at^2)$$ then is $$t$$. The equation of the tangent at $$P$$ is:
 * Using the point-gradient formula $$y-y_1=m(x-x_1)\,$$, the equation is:
 * $$\begin{align}

y-at^2 & = t(x-2at) \\ & = tx - 2at^2 \\ y+at^2 & = tx \\ y-tx+at^2 & = 0 & \mbox{ or } y = xt - at^2 \end{align}$$

The equation of the normal at P
The gradient of the normal at the point $$P$$ is $$\tfrac{-1}{t}$$ (since two perpendicular lines with gradients $$m_1\,$$ and $$m_2\,$$ must have $$m_1 m_2 = -1\,$$). Similarly to the derivation of the equation of the tangent:
 * $$\begin{align}

y - at^2 & = \tfrac{-1}{t}(x - 2at) \\ & = \tfrac{-x}{t} + \tfrac{2at}{t} \\ yt - at^3 & = -x + 2at \\ yt + x - at(t^2 + 2) & = 0 & \mbox{or } x + ty = 2at + at^3 \end{align}$$

Chords of a parabola

Suppose that $$P(2ap,ap^2)$$ and $$Q(2aq,aq^2)$$ are two distinct points on the parabola $$x^2=4ay$$. We can derive the equation for the line $$PQ$$ by finding the gradient and using the point-gradient formula:
 * $$\begin{align}

m &= \frac{ap^2-aq^2}{2ap-2aq} \\ &= \frac{a(p-q)(p+q)}{2a(p-q)} \\ &= \tfrac{1}{2}(p+q) \end{align}$$ so the chord is
 * $$\begin{align}

y - ap^2 &= \tfrac{1}{2}(p+q)(x-2ap) \\ &= \tfrac{1}{2}(p+q)x - ap^2 - apq \\ y &= \tfrac{1}{2}(p+q)x - apq \end{align}$$

Intersection of tangents
The points $$P(2ap,ap^2)$$ and $$Q(2aq,aq^2)$$ lie on the parabola $$x^2=4ay$$. The intersection of the tangents at $$P$$ and $$Q$$ can be found in terms of $$p$$ and $$q$$:
 * The tangents are:
 * $$\begin{align}

y & = px - ap^2 \\ y & = qx - aq^2 \end{align}$$
 * Subtracting these,
 * $$\begin{align}

y - y & = px - ap^2 - (qx - aq^2) \\ 0 & = px - ap^2 - qx + aq^2 \\ 0 & = px - qx + aq^2 - ap^2 \\ x(p-q) & = a(p^2 - q^2) \\ x(p-q) & = a(p+q)(p-q) \\ x & = a(p+q) \end{align}$$
 * Substituting into the original tangent formula,
 * $$\begin{align}

y & = p(a(p+q)) - ap^2 \\ & = ap(p+q) - ap^2 \\ & = ap^2 + apq - ap^2 \\ & = apq \end{align}$$ So the point of intersection is described by $$T(a(p+q),apq)\,$$.


text:
HSC Extension 1 and 2 Mathematics/3-Unit/Preliminary/Polynomials

A polynomial is any curve that can be described by the set of numbers $$\{a_0,a_1,...,a_n\}$$ such that $$f(x) = a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0$$, or alternatively
 * $$f(x) = \sum_{k=0}^n a_kx^k$$


text:
User:Rotter 00750682

Fenugreek
Fenugreek, its formal name being Trigonella foenum-graecum, which translates to three angled Greek hay. Fenugreek is native to the Europe and Asia areas and was commonly used by the Greeks, Romans, and Egyptians. It is now commonly cultivated in India, Africa, Egypt, Morocco, and a little in England. This seed could easily be considered a cure-all due to the multitude of ills that it can prevent or cure whether externally or internally. Externally the paste of fenugreek seed, which is made by mixing 1oz of fenugreek with 1 pint of water, can be used to treat abscesses, boils, carbuncles, and other skin issues of that ilk. Internally fenugreek seed has been known to treat bronchitis, fevers, sore throats, swollen glands, diabetes, ulcers, and even cancer. One will see fenugreek more commonly used as a digestive aid, as a way to promote lactation, and as an aphrodisiac. It has also been found to lower blood glucose levels in lab tests done with animals. Fenugreek can easily be taken through out the day due to the fact that the crushed seed has a maple-like flavor that can be slipped into one’s favorite cookie recipe or the jelly of their peanut butter and jelly sandwich. But be careful how much is consumed. Fenugreek is used as a laxative in some cases and shouldn’t be taken in too large of amounts right away. So, when using fenugreek seed one should start with a small amount and logically progress from there until they find the amount that is right for their body’s comfort level.

Fennel
Fennel, its formal name is Foeniculum Vulgare. Fennel is native to southern Europe, namely the Mediterranean. Fennel is used for infants and small children in order to cure colic, croup or gas. It is also used for congestion, bronchitis, and asthma. It can be commonly found as an ingredient in cold medicines. For a woman who is nursing it can be used to increase the flow of breast milk, for woman in general it can be used to increase breast size. Also, commonly used for better breath. Much cheaper than breath mints and has many more health benefits.

Omega-3 Fatty Acids
Omega-3 Fatty Acids are a natural supplement commonly found in fish. People are supposed to have anywhere from 0.5 to 1.8 grams per day either from fish or from a supplement, although it is better for one’s body to take Omega-3 through food. Omega-3’s can prevent sudden cardiovascular issues such as non-fatal heart attacks, non-fatal strokes, and death. It can also prevent health problems such as atherosclerosis, which is a hardening of arteries that can lead to serious health problems. Omega-3’s can slightly lower blood pressure, but they just generally prevent cardiovascular disease.

title: User:Rotter 00750682


text:
HSC Mathematics Advanced, Extension 1, and Extension 2/3-Unit/HSC/Primitive of sin and cos squared

The integration of these functions can be performed by the substitution of $$\sin^2\alpha$$ or $$\cos^2\alpha$$ by a function in terms of $$\cos2\alpha$$ respectively.

Deriving the substituted functions
By the double-angle formula, $$\left.\cos 2\alpha = 2\cos^2 \alpha - 1\right.$$. Rearranging to make $$\cos^2\alpha$$ the subject, $$\cos^2\alpha = \tfrac{1}{2} + \tfrac{1}{2}\cos2\alpha$$.

We know by the Pythagorean identity $$\left.\sin^2\theta + \cos^2\theta = 1\right.$$ that $$\left.\cos^2\alpha = 1 - \sin^2\alpha\right.$$, so $$1 - \sin^2\alpha = \tfrac{1}{2} + \tfrac{1}{2}\cos2\alpha$$. Rearranging, $$\sin^2\alpha = \tfrac{1}{2} - \tfrac{1}{2}\cos2\alpha$$.

Finding the primitive
By substituting $$\sin^2\alpha\,$$ for $$\tfrac{1}{2} - \tfrac{1}{2}\cos2\alpha$$, $$\int \sin^2 \alpha d\alpha = \int \tfrac{1}{2} - \tfrac{1}{2}\cos2\alpha d\alpha$$. Similarly for $$\cos^2\alpha \,$$,

$$\int \cos^2 \alpha \, d\alpha = \int \tfrac{1}{2} + \tfrac{1}{2}\cos2\alpha \, d\alpha$$

These can be integrated by using the primitives of cosine and sine

$$ \int \cos \alpha d\alpha = \sin \alpha + c$$

$$ \int \sin \alpha d\alpha = - \cos \alpha + c$$

For cos2 α

$$\int \cos^2 \alpha \, d\alpha = \int \tfrac{1}{2} + \tfrac{1}{2}\cos2\alpha \, d\alpha$$

$$\int \tfrac{1}{2} + \tfrac{1}{2}\cos2\alpha \, d\alpha = \tfrac{1}{2} \alpha + \tfrac{1}{4} \sin2\alpha + c $$

$$\int \cos^2 \alpha \, d\alpha = \tfrac{1}{2} \alpha + \tfrac{1}{4} \sin2\alpha + c $$

For sin2 α

$$\int \sin^2 \alpha \, d\alpha = \int \tfrac{1}{2} - \tfrac{1}{2}\cos2\alpha \, d\alpha$$

$$\int \tfrac{1}{2} - \tfrac{1}{2}\cos2\alpha \, d\alpha = \tfrac{1}{2} \alpha - \tfrac{1}{4} \sin2\alpha + c $$

$$\int \sin^2 \alpha \, d\alpha = \tfrac{1}{2} \alpha - \tfrac{1}{4} \sin2\alpha + c $$


text:
HSC Mathematics Advanced, Extension 1, and Extension 2/3-Unit/HSC/Induction

Induction is a form of proof useful for proving equations involving non-closed expressions (i.e., expressions with $$n$$ terms; sequences).

Explanation
Induction involves first proving that the equation is true for $$n = 1$$, then proving true for $$n = k + 1$$ (assuming for the purpose of the proof that the equation holds true for $$n = k$$). Since it is true for $$n = k$$ and true for $$n = k+1$$, and also true for $$n = 1$$, it is true for $$n = 2$$. It follows that it is true for all positive integers $$n$$.

Proving the formula for the sum of a series
Q: Prove by mathematical induction that for all integers $$n \ge 1$$,
 * $$1^3 + 2^3 + 3^3 + 4^3 + \cdots + n^3 = (1+2+3+....n)^2$$

A:  When $$n = 1$$, $$1^3 = 1 = \tfrac{1}{4} (1)^2((1)+1)^2 = \tfrac{1}{4}(4) = 1$$, so it is true for $$n = 1$$ Suppose that the statement is true for $$k, k \in \mathbb{N}$$. That is, suppose that $$1^3 + 2^3 + 3^3 + 4^3 + \cdots + k^3 = \tfrac{1}{4} k^2 (k+1)^2$$. This is sometimes called the induction hypothesis. Then prove the statement for $$n = k+1$$ (that is, prove that $$1^3 + 2^3 + 3^3 + \cdots + (k+1)^3 = \tfrac{1}{4} (k+1)^2(k+2)^2$$:

\begin{align} \mbox{LHS} & = 1^3 + 2^3 + 3^3 + 4+3 + \cdots + k^3 + (k+1)^3 \\ & = \tfrac{1}{4} k^2 (k+1)^2 + (k+1)^3               & \mbox{ (by the induction hypothesis)} \\ & = \tfrac{1}{4}(k+1)^2(k^2 + 4(k+1)) \\ & = \tfrac{1}{4}(k+1)^2(k^2 + 4k + 4) \\ & = \tfrac{1}{4}(k+1)^2(k+2)^2 \\ & = \mbox{RHS} \end{align} $$  It follows from parts 1 and 2 by mathematical induction that the statement is true for all positive integers $$n$$.


text:
Korean/Comparatives and superlatives

Introduction
This chapter talks about how to do comparisons in the Korean language and express superlatives. For those unfamiliar with the grammar terms, a superlative term basically expresses whether something is the best, the biggest, the tallest, etc.

Superlatives
To express a superlative in Korean, a speaker can use either 가장 or 제일 followed by the verb. One example is telling someone what your favorite movie is. Thus, to say that your favorite movie is the Matrix, you can say 나는 Matrix를 제일 좋아해요. This would literally translate to 'I like Matrix the most.'

Comparatives
To do a comparison in Korean, a speaker can use '보다 (더)' followed by a verb. Here are some examples.
 * 나는 차보다 커피를 즐겨요. I like coffee better than tea.
 * 나는 수학을 역사보다 좋아해요. I like math better than history.

If there are 2 objects being compared, the object that that is being compared against is placed right before the 보다. So in the examples above, the Korean word for tea, 차, is placed right before the 보다 and the Korean word for history, 역사, is put right before the 보다.

Degree of adjectives
To form varying degrees of adjectives, prefix an adjective with the adverbs 매우 or 아주:
 * 비싸다: (to be) expensive
 * 매우 비싸다: (to be) very expensive
 * 아주 비싸다: (to be) extremely expensive

title: Korean/Comparatives and superlatives


text:
Korean/Lesson II1

Vocabulary

 * 축구 (chukgu)            soccer
 * 농구 (nonggu)            basketball
 * 배구 (baegu)            volleyball
 * 야구 (yagu)            baseball
 * 테니스 (teniseu)         tennis
 * 수영 (suyeong)            swimming
 * 골프 (golpeu)            golf
 * 스키 (seuki)            skiing

  Using Common form - 요

Here we will be using the common level of politeness used in Korean which includes - 요 This is one of the most common levels of politeness used in Korea which is quite universal if you are speaking to people who you are meeting for the first time (NOT if they are older than you or in a higher position) and so you should become familiar with it before learning the casual form - 야 and the honorific form - 습니다, 습니까.

The common level is easier to conjugate into because you use the same endings if you were to form a question. All you have to do to let the person/persons know that you are asking a question is raise your voice! This only works in the common and casual form of conjugation and this is why it is so commonly used in Korean everyday life. However, do not make the mistake of using this level of politeness to older people or people with higher authority than you because you could get into a lot of trouble!

  Conversation   You can see the vocabulary included above about different types of sports! It just so happens that 가영 and 수연 are having a conversation about which sports they like and which they don't!

(가영)=            안녕하세요, 수연씨!

(수연)=            '''네, 안녕하세요. 가영씨, 내일 스키를 갈거예요?'''

(가영)=            '''아니요. 저는 스키를 싫어해요. '''

(수연)= '''아, 그래요? 저도요.'''

(가영)= 하지만 저는 농구를 좋아해요.

(수연)= '''저도요. 저는 학교에서 농구를 하는 것을 좋아해요.'''

(가영)= '''맞아요. 저도 그래요.'''

  Explanation

''' 안녕하세요, 수연씨! '''

There's nothing new here if you have been through all of the other beginner lessons.

안녕하세요

Hello

We know that this phrase means hello and there isn't so much of a difference here between the honorific form of saying it -안녕하십니까 Later on, you should learn about verb stems(VS) if you haven't already and these verb stems are present throughout all of the conjugations in Korean (unless they're irregular)

Verb Stems

The verb stem is taken from the verb that you see in the dictionary. In the dictionary, all verbs are written with a -다 at the end. This -다 indicates that the verb is in its dictionary form and in order for us to use this verb, we have to conjugate it!

I will not go over the basics of conjugation, however, the verb always goes at the end of the sentence unless you are making quite a complex sentence that might include more than one verb.

수연 replies and says hello back. Then she asks a question.

''' 가영씨,     내일       스키를          갈거예요? ''' (Ga Young)    Tomorrow,    Skiing(Object)    going to go?(Future Tense)

이에요/예요

You can see here that the verb ending didn't have to change in order to phrase this sentence as a question! However, since this would be your first time in studying in the common level of politeness (CLP), this - 예요 that has been added at the end of the verb (go) will be new. Just like most Korean rules, you have to add either - 예요 or - 이예요 to the end of the verb. Remember 을/를? It's exactly the same!

 * 1) You add - 예요 if the verb ends in a vowel.

 * 1) You add - 이예요 if the verb ends in a consonant!

Even though this is the rule, in normal speed conversation, this will be said fast and so the two will virtually sound the same. Don't worry though, they both have the same meaning but it's just to aide to a smooth flow of conversation.

''' 아니요. 저는 스키를 싫어해요. '''

Now we can see our very first negative sentence! Here 가영 replies to 수연.

No. I(Topic)   skiing(Object)    dislike.

This is quite self-explanatory with regards to the correct order. This sentence means "No, I don't like skiing." However, Koreans use the word 'no' differently to that in English, this is why I have used this form of 'no' and not the other form of it which is '아니예요.'The conjugated form of "to dislike" into the CLP is what we can see here. So to say you dislike something, all you have to do is add - 싫어해요 onto the end of your noun when you are speaking! It's that easy. Remember, this is only the conjugation for the common level so you can NOT use this when speaking to your elders or those in higher positions than you unless stated.

수연 replies and there's nothing new until we see - 저도요. This literally means "Me too". Great, right?

가영 wants to lighten the conversation and so she starts to talk about something that she likes.

하지만 저는 농구를 좋아해요.

But,   I(Topic)    Basketball(Object)    Like!

Think you can translate this? Well done if you got it right! It means "But, I like basketball!" Now we can see how to say that we 'like' something in the CLP form! - 좋아해요

Now, all you have to do when you want to say that you like something is to add - 좋아해요 onto the end of your noun!

저도요. 저는 학교에서 농구를 하는 것을 좋아해요.

Here we can see quite a complex sentence and you might have to think hard to figure what the meaning of this is!

Yes, me too. I(Topic)   School(at)    Basketball-Doing(Object)    like.

This means, "Me too. I like to play basketball at school." Good. Plenty of new, helpful vocabulary for us to learn! Remember that - 에서 means 'at' here.

Also, 수연 says '농구를(Object) 하는것(Verb->Noun)' to mean 'to do(play) basketball'. Keep in mind. To make a verb a noun, paste '-는 것' (for present tense), '-ㄴor은 것' (for past tense), '-ㄹor을 것' (for future tense) after the verb stem ('하-').

Let's practice this: How we can speak "Are you going to eat an apple ?" The answer is " 사과를 먹을 것 이에요?" or " 사과 먹을 거 예요?"(를/을 after noun is often skipped). 먹다 is to eat, and 사과 is an apple.

Isn't it easy? Sorry. It'll be pretty difficult. This is 'quite a complex sentence' that might include more than one verb.

Now 가영 replies with 맞아요. 저도 그래요.

This is something that you will hear quite frequently in Korean and so you should learn it here. This means "Right. Me, too."

안녕! 공부하세요!

title: Korean/Lesson II1


text:
Polish/Basic sentences 2

< Polish pronunciation < ^ Polish ^

Let's try to make some simple sentences.

Two things you should know are that:
 * 1) verbs take different forms depending on person, number and tense
 * 2) nouns take different forms depending on number and function in the sentence, or to be more exact, grammatical case

 * "Basia je jabłko" - Basia eats/is eating an apple.
 * "Basia je ciastko" - Basia eats/is eating a cookie.

"Basia" (English Barb) is the short form of the name Barbara. "Jabłko" means an apple. "Ciastko" means cookie. "je" is the appropriate form of verb that means "to eat".

Note: Polish does not have a word for "the" or "a", therefore when translating from Polish you have to add these in where they would be appropriate in an English sentence, using the context or meaning of the sentence as a guide.

There are 7 cases in Polish, with the two most important being nominative (Polish "mianownik") - for subjects, and accusative (Polish "biernik") - for direct objects. Subject in Polish can adopt two forms - logical and gramatical; grammatical subject is a noun in nominative and logical - a noun in accusative. Logical subject is used when we want to express lack or excess of something. Grammatical subject can be represented by any part of speech. For example:

 * noun (Pol. rzeczownik) – Dom stoi na wzgórzu.
 * adjective (Pol. przymiotnik) – Chory wziął lekarstwo.
 * numeral (pol. liczebnik) – Wybiła dziesiąta.
 * verb (Pol. czasownik) - Przyjemnie jest Tańczyć.
 * pronoun (Pol. zaimek) - On czyta.
 * adverb (Pol. przysłówek) – Jutro należy do nas.
 * conjunctive (Pol. łącznik) – "I" jest spójnikiem.
 * preposition (Pol. przyimek) – "W" jest przyimkiem.
 * particle (Pol. partykuła) - "Li", "że" są partykułami.
 * interjection (Pol. wykrzyknik) – "Och" rozległo się nagle.

While word order in Polish can be changed, the most typical is:
 * Subject (in nominative) - Verb - Object (in accusative).

Declension of Polish nouns is rather regular. Nouns that end in "-o" have the same nominative and accusative (these are neutral), nouns that end in "-a" in the nominative have "-ę" in the accusative (these are feminine).

Knowing that, we can create more sentences.

For example:
 * "Dziecko je ciastko" - a child eats a cookie.
 * "Dziecko je czekoladę" - a child eats a chocolate.

 * "dziecko" - a child
 * "czekolada" - a chocolate

To practise a little more, let's use "widzi", which is the appropriate form of the verb "to see".

 * "Piotr widzi Basię" - Piotr sees Basia
 * "Basia widzi Piotra" - Basia sees Piotr
 * "Basia widzi dziecko" - Basia sees a child

< Some useful expressions < ^ Polish ^ > Polish pronunciation >

title: Polish/Basic sentences 2


text:
User talk:72.23.40.23

Regarding edits made during March 12 2007
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title: User talk:72.23.40.23


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User talk:203.199.255.131

Regarding edits made during March 12 2007
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title: User talk:203.199.255.131


text:
Korean/Lesson II2

Vocabulary

 * 경찰관 (gyeongchalgwan)    police officer
 * 소방관 (sobanggwan)   fire fighter
 * 교사/선생님 (gyosa/seonsaengnim)     teacher
 * 학생 (haksaeng)      student
 * 예술가 (yesulga)   artist
 * 의사 (uisa)     doctor
 * 간호사 (ganhosa)   nurse
 * 정치인 (jeongchiin)   politician
 * 대통령 (daetongryeong)   the president
 * 국회의원 (gukhoe-uiwon)   member of the National Assembly
 * 이발사 (ibalsa)   barber
 * 회사원 (hwoesawon)   office worker
 * 운동 선수 (undong seonsu) athlete
 * 육상 선수 (yuksang seonsu) (track-and-field) athlete
 * 야구 선수 (yagu seonsu) baseball player
 * 축구 선수 (chukgu seonsu) soccer player
 * 농구 선수 (nong-gu seonsu) basketball player
 * 수영 선수 (suyeong seonsu) swimmer
 * 과학자 (gwahakja)    scientist
 * 가수 (gasu)      singer
 * 연예인 (yeonyein)    entertainer
 * 배우/여배우 (bae-woo/yeobae-woo)      acter/actress
 * 군인 (goon-in)     soldier
 * 조종사 (jojongsa)   pilot
 * 기사 (gisa)     engineer
 * 기술자 (gisulja)   mechanic
 * 음악가 (eumakga)   musician

title: Korean/Lesson II2


text:
User talk:195.8.171.130

Regarding edits made during March 12 2007
Thanks for experimenting with Wikibooks. Your test worked, and has now been removed. Please use Sandbox for any other tests you want to do, since testing material in modules will normally be removed quickly. Please see the welcome page if you would like to learn more about contributing to our project. Thanks. If this is an IP address, and it is shared by multiple users, ignore this warning if you did not make any unconstructive edits. Az1568(Talk) 14:46, 12 March 2007 (UTC)

title: User talk:195.8.171.130


text:
Talk:Esperanto/Quick and dirty guide

Untitled
I am not sure if this is the right place but the links for the first two practice exercises in the book are no longer being hosted at the site they were. They should probably either be redirected to another suitable location or taken off.

-- http://de.wikipedia.org/wiki/Klempen eble estus utila ligilo por vi, kiu helpas eviti la "klempadon", cxefe do forlasi la spacon antaux "(".. Vere, mi vidis tiun fenomenon cxe cxinaj auxtoroj, sed opinias, ke oni ja enmetu la spacon por pli bona kaj kutima legebleco. Plian sukceson, vi sxajne estas tre okupita kaj ambicia.. -- brilanto (momente malfacilas ensaluti..), 30.10.07

Ĉu vi parolas la hispanan? is ambiguous?
Ĉu vi parolas la hispanan = do you speak spanish Ĉu vi parolanta la hispanan = are you speaking spanish

It's not ambiguous as far as I know. Esperanto does have a present continuous tense, this statement implies it doesn't.

-lackinglatin (not logged in at the moment, just in and out while googling something else)

"Ĉu vi parolanta la hispanan" means something like "Speaking you Spanish (adj)?" I believe you mean "Ĉu vi estas parolanta/parolantas hispanon". Interchangeable (discuss • contribs) 16:18, 19 November 2011 (UTC)

title: Talk:Esperanto/Quick and dirty guide


text:
Animal Behavior/Motor Systems

Neural Control of Motor Systems
I am coming more and more to the conviction that the rudiments of every human behavioral mechanism will be found far down in the evolutionary scale and also represented even in primitive activities of the nervous system—Karl Lashley, 1951

The neural control of behavior will be most obvious in cases where the brain performs only a small amount of processing. Alternatively we can explore neural systems where functions are limited to a primary goal and which are highly optimized to achieve it. The latter is particularly true for sensory systems which encode a single stimulus parameter, or motor systems in escape behaviors. We can ask how a particular behavior is produced, with characterizing muscle activities in the behavior, exploring the neural motor centers that control them, working our way upstream in the flow of information processing, until we reach the mechanisms responsible for sensory input.

Executive control
A set of cognitive processes serve as a supervisory attentional system in behavior. These system impact behavior via control of attention, inhibition, working memory, cognitive flexibility, reasoning, problem solving, and planning. In humans, the prefrontal cortex, the caudate nucleus and the subthalamic nucleus are thought to play central roles in the inhibitory control of behavior. Addiction and attention deficit hyperactivity disorder may result from an impairment in executive control.

Motor Patterns
Complex and coordinated motor patterns can often be centrally generated without requiring sensory input. Such fictive motor patterns may often even be elicited in reduced preparations and studied in vitro. Such fictive motor patterns arise from bursts of activity in isolation, or as a consequence of circuit interactions.

Command Neurons
Command Neurons are cells, when activated, are capable of producing complex behavior patterns in the absence of any meaningful external stimuli. Activation of these neurons is both necessary and sufficient. Examples include the production centers for mating song in crickets, the giant fibers in crayfish escape circuitry, or the Mauthner neurons in goldfish. The concept in which single spikes are viewed as sufficient for behavior may be somewhat dated. There is little evidence that any single neuron is able to exert control over a complex behavior and its motor programs. Neurons cannot be viewed as single identities, they are always part of larger circuits, they control as well as receive input, and their activity thus only makes sense within the context into which they are embedded.

Central Pattern Generator
Central Pattern Generators (CPGs) are circuits that organize repetitive motor patterns, such as those underlying feeding, locomotion or breathing. The production of rhythmic motor patterns and its control by higher-order command and modulatory interneurons can be explained by the intrinsic membrane properties and connections of these neurons. Studies concern the specific timing of activation of the component neurons, how sensory neurons alter or gate CPG output, or how motor patterns are activated, inhibited, or modified by modulatory drive (Marder et al., 2005). Reciprocal Inhibition of functional antagonists can produce alternating discharges in neurons and its associated motor output. Complex, multidimensional oscillations can emerge from multiple, overlapping mechanisms.

Stomatogastric System of Crustaceans
The crustacean stomatogastric nervous system (STNS) generates rhythmic motor patterns in the stomach and other regions of the foregut. The patterns, controlling more than 40 pairs of striated muscles, serve to grind and filter food using multiple and variable cycles. Over the past 40 years the small section of the nervous system that controls it, has been used to gain an understanding how neural circuit dynamics arise from the properties of its individual neurons and their connections. The STNS consists of a group of four linked ganglia, the paired commissural ganglia (CoGs), the unpaired esophageal ganglion (OG), and the STG. The STG consists of ∼30 motor neurons that move the muscles of the gastric mill and pyloric regions of the stomach. The CoGs and OG control STG activity via descending modulatory input. Accessible to simultaneous recordings of all relevant circuit neurons, the electrical, chemical and molecular properties of individual nerve cells and the direct and modulatory connections between them, can be studied along with the various emerging network properties.

Crayfish Escape
A tailflip consists of a rapid, ventral flexion of the abdomen. It is produced in response to stimulation from a potential predator and results in a backward or upward propulsion that causes the animal to escape from it. The control system for this behavior is formed by 2 pairs of giant neurons - the lateral giant neurons (LGN) and the medial giant neurons (MGN). The concept of command neurons was used to explain the causation of that behavior. The actual components of the system are:
 * Bipolar sensory neurons: tuned to touch or high frequency water movements (>80 Hz) - sensory filtering. Sensory neurons produce chemical synapses onto non-giant interneurons, and electrical synapses to giant interneurons subthreshold stimulation triggers non-giant escape swimming, produced by a pattern generator consisting of interneuron networks
 * Medial Giant Interneurons: brain to last abdominal ganglion, contraction of fast flexor muscles in all abdominal segments, direction of escape: straight back
 * Lateral Giant Interneurons: segmentally repeated; input in abdominal segments, flexor contraction in anterior segments of abdomen, direction of escape: upwards giant motor neurons and fast flexor motor neurons for flexion of abdomen via giant fibers. Habituation only occurs if stimulation of sensory neurons is sufficiently intense to drive an action potential in the giant fiber system
 * Non-Giant Interneurons: escape swimming
 * Re-Extension of abdomen follows lateral giant tailflips within 100ms of stimulus, reflex of hair receptors and muscle receptor organs
 * Excitation of extensor systems via excitatory interneurons with onset of tailflip to prepare for re-extension
 * Inhibition of extensor systems via inhibitory interneurons while tailflip is happening then Inhibition of further tailflips while re-extension is happening: inhibition of sensory neurons driving giant fibers, inhibition of flexor muscles

title: Animal Behavior/Motor Systems


text:
User talk:Dablur

Kapampangan wikibook
Hi Dablur, I have seen your edit, thanks a lot for helping out with the Kapampangan wikibook.

As you have probably seen, there isn't much activity at the moment, as I have been quite busy all winter. If you speak Kapampangan and have any other ideas on how to improve this book, please go ahead an edit, I would certainly welcome some help.

I hope to see you editing some more. --Lgriot 15:55, 12 March 2007 (UTC)

title: User talk:Dablur


text:
Animal Behavior/Hormones in Behavior

The Glandular Systems
The body has two types of glandular systems, the endocrine, which generally secrete hormones through the bloodstream, and the exocrine which secrete fluids to the outer surfaces of the body, such as sweating.

Endocrine glands
The Endocrine System combines neural and glandular mechanisms which control physiological functions/behavior via the secretion of hormones. Hormones are chemical signaling molecules which play an integral role during development (organizational effects) and day-to-day functioning (activational effects) of target tissues at critical times. Secretory cells of a particular type are often clumped together into a well defined gland (e.g. pituitary, thyroid, adrenal, testes, ovaries). Secreted at that site they distribute throughout the body via the blood stream, and cause physiological changes at any other sites.

 * Steroids: derived from cholesterol
 * Amines: derived from amino acids
 * Peptides: short chains of amino acids via protein synthesis

Sex hormones, largely steroids, are secreted from gonads and adrenal cortex. Androgens (e.g., testosterone) are usually higher in male mammals while levels of estrogens (e.g., estradiol) in female mammals exceed those in males. Circulating levels of sex hormones then provide the basic organization for gender phenotypes.

Neurohormone refers to a compound that is released into the bloodstream at specialized neurohemal release sites. It binds to receptors anywhere in the body and thereby coordinates disparate biochemical responses.They are released from glands, transported via the circulatory system and influence the activity of target organs.

Functionally hormones are categorized as Effector hormones (e.g. Vasopressin, Oxytocin) or Tropic hormones, releasing factors (e.g. Gonadotropin Releasing Hormone - GnRH, Growth Hormone Releasing Hormone - GHRH). Target Organs receive hormones via blood stream, respond directly or release their own hormones in response (steroid hormones), and these hormones circulate back to turn off hormonal secretion: endocrine feed back loops. Compensatory hypertrophy results from feedback loops control levels of activity where systems are upregulated until they achieve sufficient functional effects (e.g. thyroxine and goiter). Activational Effects and Organizational Effects

Pituitary gland
The Pituitary gland is a small endocrine organ located within the mammalian brain, dorsal to the hypothalamus. The gland is composed of anterior and posterior sections, as well as an intermediate lobe. These lobes are largely responsible for the storage of many hormones that are associated with many different facets of bodily functioning. The pituitary gland is largely responsible for the release of a variety of important hormones that regulate thyroid activity. The Pituitary additionally releases the lactotrophin’s oxytocin and vasopressin, the gonadotropin hormones follicle stimulating hormone (FSH) and Luteinizing hormone (LH), and hormones responsible for development, growth, and the reproductive process/ovulation. It has been shown to regulate water and osmolarity in the body, as well as the ability to maintain homeostatic temperature. Malfunctioning of the pituitary can lead to a number of diseases, including hyper/hypothyroidism, gigantism, central diabetes insipidus, and pituitary tumors. Due its role in cortisol regulation, mental disorders can be affected by pituitary functioning. Due to its critical involvement in the regulation of hormonal homeostasis, medical problems such as pituitary tumors can have significant effects on mood, emotions, sexual functioning, and body weight. It is likely that someone with a malfunctioning pituitary would experience mental disorders such as depression, anxiety, and personality changes.

Oxytocin
Oxytocin is an important regulatory hormone that acts as a neurotransmitter in the brain, and is known to be involved with a number of critical bodily functions in humans. Secreted from the pituitary gland, oxytocin stimulates uterine contraction when giving birth, and lactation in nursing mothers. It has been found to modulate inflammation at wound sites, allowing the wound to heal more quickly. Social bonding behaviors are closely associated with oxytocin, such as facilitating inter-personal trust while decreasing feelings of fear and anxiety. It can also play a role in enhancement of social memories, leading to more positive feelings when thinking about the memory in the future. Due to its release from hugging and cuddling behaviors, it is important in the positive feelings people experience during periods of close contact.

Sex Determination
Hormonal influences during critical periods produce fairly permanent changes in nervous and endocrine systems. Sex-determination may be chromosomal or environmental:
 * Mammals: hormonal cascade of events takes an individual down one of two paths (default is female). testes determining factor (tdf) in mammals or Sexchromosomal Abnormalities: Turner syndrome (XO monosomy), Klinefelter syndrome (XXY trisomy), XYY syndrome ("supermale"), Multi-X syndrome ("superfemale"), XX Male syndrome (SRY gene transference) XY Female syndrome (SRY gene missing)
 * Birds: Males are homogametic (ZZ), females are heterogametic (ZW)
 * Turtles: temperature
 * Fish: social stimuli

internal sex organs: Precursors for both internal sex organs are present in the embryo: the Mullerian system for the female sex; the Wolffian system for males. Controlled by the levels of circulating hormones, only one set of organs develops while the other shrinks. Secreted by the testes, Müllerian inhibiting hormone (i.e., a peptide hormone) has defeminizing effects while androgens (i.e., steroid hormones) exert a masculinizing effect. Gonads: testes or ovaries, first to develop. External genitalia: Androgens are essential for the development of primary sexual characteristics.

Androgen insensitivity syndrome: A genetic male lacking functional androgen receptors develops testes, secretes Müllerian inhibiting substance and androgens. Internal sex organs progress onto the intended male path, but female external genitalia result from the inability to respond to androgens correctly. Turner’s syndrome, with genotype XO (i.e., a single X chromosome and no Y), lacks testes and ovaries.

Maternal effects explain influenced by effects of the mother that are not due to direct genetic inheritance. Positional effects in the mammalian uterus

Homeostasis
Hormonal secretion or its inhibition may bring about some rapid physiological changes in a variety of target tissues. Vasopressin.

Metabolic Balance
Hormones influence sensory perception (human visual, rat odor, preference in castrated vs. intact). Hypothalamus neurons monitor the internal state (thirst, hunger) and send neurosecretory cell axons to the posterior pituitary.

Reproductive Behavior
In females, Oxytocin, released from the posterior pituitary, triggers milk let-down in mammary glands or contraction in the uterus during child-birth. It is also associated with powerful emotional effects (affective states) in parental behavior and maternal competence. Moreover, the hypothalamus secretes releasing factors to stimulate or inhibit manufacture and secretion of hormones in anterior pituitary portal system: Prolactin, Follicle stimulating hormone (FSH), Luteinizing hormone (LH); receives, stores and releases neurohormones from hypothalamus. Sex behaviors: gonadotropin releasing hormone (GnRH) -> gonadotropin -> gonad maturation and gonadal steroid production -> feed back to brain. Sex steroids initiate the basic female and male anatomies (i.e., organizing effects in sexual differentiation cascade) but also allow animals to later respond to activational effects. Expression of sex differences in behavior also requires that steroids activate many aspects of the phenotype during maturation. Reproductive Neuroendocrinology of Ring Doves (Streptopelia risoria): monogamous, sexually monomorphic, parental care from both male and female (incubate eggs, crop "milk"); Lordosis in female hamsters: female soliciting behavior triggered by a touch on the side of the haunches.

McClintock Effect
synchronisation of menstrual cycles in female mammals housed together

Lee-Boot Effect
Oestrous cycles slow and stop in female mice housed together in the absence of any males in the surrounding environment

Whitten Effect
individuals start cycling again in synchrony if exposed to male odor

Vandenbergh Effect
acceleration of onset of puberty in female rat when exposed to odor of male

Bruce Effect
failure of pregnancy if female rat is housed with a male who is not the father

Example: Arthropod Molting
Hormonal Control of Arthropod Molting: Brain: releases ecdysiotropin which stimulates release of molting hormone (MH) from Prothoracic Gland; Corpora allata: Juvenile Hormone (JH) suppresses metamorphosis; Prothoracic Gland; Corpora allata: Juvenile Hormone (JH) suppresses metamorphosis

title: Animal Behavior/Hormones in Behavior