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Now $E=u^2/2-k/R<0$, so the orbit is bound. Since $L=ub$, the initial energy isAt the farthest central-force radial turning point $q$, the radial speed vanishes and the transverse speed is $L/q$, givingEquating these expressions and multiplying by $2q^2$ yieldsThus $A=L^2/b^2-2k/R$, $B=2k$, and $C=-L^2$.

The central force conserves angular momentum. The initial impact parameter gives $L=ub$, while at closest approach the velocity is transverse and $L=wp$. Thus $u=wp/b$. Conservation of energy givesUnder $u^2\gg 2k/R$, neglect the initial potential term and substitute $u=wp/b$:Consequently

Green theorem states that for a positively oriented planar region $D$,Here $P=x{y}^2-\lambda y$ and $Q=x^{2}y$, soThe stated orientations make $C_1\cup C_2$ the positive boundary of the region inside the radius-$a$ circle and outside the radius-$b$ circle. Its area is $\pi (a^2-b^2)$ regardless of the inner circle's center. Therefore

Set $F(x)=|x^3-x|$. The relation is exactlyIt is therefore the equivalence relation induced by a function: equality of the two function values is reflexive, symmetric, and transitive. Hence this is an equivalence relation.

This relation is reflexive because choosing $n=1$ gives $|x/x|=1$. It need not be symmetric: take positive real numbers $y=e$ and $x=e^2$. Then $|x/y^2|=1$ is rational, so $x\sim y$. Butis transcendental, hence irrational, for every integer $m\ge 1$. Thus $y\nsim x$, and the relation is not an equivalence relation.

By the equality condition in the complex triangle inequality,for nonzero $x,y$ exactly when $x/y$ is a positive real number. Having the same complex argument is reflexive, symmetric, and transitive, so this is an equivalence relation. Its classes are the open rays from the origin.

An equivalence relation is reflexive, symmetric, and transitive. HereThe relation is reflexive and symmetric, but it is not transitive. For example, takeThe real parts of $x\overline{y}$ and $y\overline{z}$ are positive, whereas that of $x\overline{z}$ is negative. Hence this is not an equivalence relation.

The product of two transpositions has three possible forms. Equal transpositions produce the identity, transpositions meeting in one letter produce a three-cycle, and disjoint transpositions produce two disjoint two-cycles. There are two three-cycles on each chosen triple and three pairings of each chosen four-set. Therefore the number of distinct permutations representable this way is

Choose the support of the four-cycle and then cyclically order it. A chosen set of four letters has $(4-1)!=6$ distinct cycles, so the answer is

There are $\lfloor n/2\rfloor$ even letters. The stated condition and injectivity force the even letters to permute among themselves, which can be done in $\lfloor n/2\rfloor !$ ways. The remaining odd letters must then permute among themselves. Thus the number is

A transposition exchanges exactly two letters and fixes the rest. A permutation cycle $(a_1{a}_2\dots ,a_k)$ sends each displayed letter to the next, sends $a_k$ to $a_1$, and fixes all other letters.

After imposing $\sigma (1)=2$, the remaining $n-1$ domain elements may be sent bijectively to the remaining $n-1$ values in arbitrary order. Hence the number is

In the parent's rest frame, momentum conservation makes the massive daughter and photon momenta equal and opposite, of magnitude $p$. The relativistic two-body decay formulas giveSince $v=p{c}^2/E_R$ and $r=m/M$,The daughter's Lorentz factor isIf it decays after proper time $\tau$, time dilation gives $t=\gamma \tau$, while uniform motion gives $x=v\gamma \tau$. Therefore the event coordinates in the parent frame are

Let the photon energies be $E_1,E_2$ and their momentum angle be $\theta$. Conservation of four-momentum and the null photon momenta giveUsing $1-\cos \theta =2{\sin }^2(\theta /2)$,Thus the opening angle depends only on the parent rest energy and the two photon energies.

Suppose $u_1,u_2$ are solutions and put $w=u_1-u_2$. Then $\Delta w=0$ and $w$ satisfies the homogeneous Robin boundary conditionFor $0<\lambda <1$, Green's first identity givesThe left side is nonnegative, so both sides vanish and $w=0$. For $\lambda =0$, the Dirichlet boundary condition $w=0$ on $S$ and the same identity again imply $w=0$. The solution is therefore unique for every $0\le \lambda <1$.

For a regular spherically symmetric solution, the Laplacian equation isRegularity at the origin gives $u^{\prime }=2r$, hence $u=r^2+C$. Applying the boundary condition at $r=b$ yieldsAt $\lambda =1$ the Neumann boundary condition would require $u^{\prime }(b)=2b=0$, which is impossible for $b>0$. Equivalently, its compatibility condition would require ${\int }_{V}6dV=0$.

Writing $a_k{x}_k=\mathbf{a}\cdot \mathbf{x}$ and differentiating givesIts symmetric second-rank tensor part and antisymmetric second-rank tensor part areTaking the trace makes the two symmetric terms cancel, so the divergence isContracting the antisymmetric part with the Levi-Civita symbol gives the curl

In suffix notation,where ${\delta }_{ij}$ is the Kronecker delta. Since $r=(x_k{x}_k{)}^{1/2}$, the chain rule gives