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For the proposed circular motion,The separation is ${\mathbf{x}}_1-{\mathbf{x}}_2=2a(\cos \omega t,\sin \omega t,0)$, so the inverse-square force on particle 1 isThe equation $m{\ddot{\mathbf{x}}}_1={\mathbf{F}}_{12}$ holds exactly whenThe equation for particle 2 follows by symmetry.

The center of mass isBy Newton's third law, ${\mathbf{F}}_{21}=-{\mathbf{F}}_{12}$, soThus $\dot{\mathbf{R}}$ is constant. For $\mathbf{r}={\mathbf{x}}_1-{\mathbf{x}}_2$,Thereforewhere $\mu$ is the reduced mass.

Suppose first that $|A_n|=1$ for every $n>N$. Then an element of $B$ is determined by its first $N$ coordinates, so $B$ is in bijection withA finite Cartesian product of countable sets is countable by repeated application of the diagonal enumeration in part (b)(i). Hence $B$ is countable.

Conversely, suppose infinitely many factors contain at least two elements. Choose increasing indices $n_1<n_2<\cdots$ and distinct elementsFix one element in every remaining factor. Each binary sequence $\epsilon =({\epsilon }_1,{\epsilon }_2,\dots )$ then defines an element of $B$ by placing $a_j^{{\epsilon }_j}$ in coordinate $n_j$ and the fixed element elsewhere. This map is injective.

The set $\{0,1{\}}^{\mathbb{N}}$ is uncountable by Cantor's diagonal argument: from any proposed list of binary sequences, form a new sequence whose $j$th digit differs from the $j$th digit of the $j$th listed sequence. It is absent from the list. Thus $B$ contains an uncountable subset and cannot be countable.

Therefore

A set is countable when it is finite or its elements can be put in one-to-one correspondence with a subset of $\mathbb{N}$. Equivalently, there is an injection from the set into $\mathbb{N}$.

DefineThen $\mathbf{x}=\mathbf{y}+\mathbf{z}$, the support of $\mathbf{y}$ lies in $P(\mathbf{s})$, and the support of $\mathbf{z}$ is disjoint from $P(\mathbf{s})$. Thus $\mathbf{y}⪯\mathbf{s}$ and $\mathbf{z}\perp \mathbf{s}$.

For uniqueness, if $s_i>0$, orthogonality forces $z_i=0$, hence $y_i=x_i$. If $s_i=0$, the condition $\mathbf{y}⪯\mathbf{s}$ forces $y_i=0$, hence $z_i=x_i$. Every coordinate is therefore forced, proving uniqueness.

The definition saysEquality of sets is reflexive, symmetric, and transitive, so $\sim$ is an equivalence relation. Its classes are indexed by all subsets of $\{1,\dots ,n\}$, and every subset occurs as the support of its zero-one indicator vector. The number of classes is therefore

Suppose $\mathbf{x}⪯\mathbf{y}$. Define the nonnegative tuple $\mathbf{z}$ coordinatewise byWhen $y_i=0$, support inclusion forces $x_i=0$, so $x_i=y_i{z}_i$ in every coordinate.

Conversely, if $x_i=y_i{z}_i$ with $y_i,z_i\ge 0$, then $x_i>0$ implies $y_i>0$. Hence $P(\mathbf{x})\subseteq P(\mathbf{y})$ and $\mathbf{x}⪯\mathbf{y}$. This proves the equivalence.

Because $P(\mathbf{x})\subseteq P(\mathbf{x})$, the relation is reflexive. If $\mathbf{x}⪯\mathbf{y}$ and $\mathbf{y}⪯\mathbf{z}$, thenso it is transitive. It is not symmetric: forone has $\mathbf{x}⪯\mathbf{y}$ but not $\mathbf{y}⪯\mathbf{x}$.

PutThe initial values are $S_1=S_2=1$. Using Pascal's identity and the convention that an out-of-range binomial coefficient is zero,Thus $(S_n)$ has the same initial values and linear recurrence relation as the Fibonacci numbers. Mathematical induction gives

The special binomial theoremfollows by mathematical induction. Multiplication of the formula for $n$ by $1+t$ and collection of the coefficient of $t^k$ uses Pascal's identity from part (a).

Termwise integration from $0$ to $1$ givesReplacing $t$ by $-t$ gives

Finally,Writing $y=1-x$ and using the finite geometric series,turns this into

For $0\le k\le n$, the binomial coefficient isWhen $1\le k\le n-1$,This is Pascal's identity.

For the special-relativistic matrix,Thus the eigenvalues and corresponding eigenvectors areThe eigenvector equations are $x=ct$ and $x=-ct$. They are the two null directions of the light cone, representing right-moving and left-moving light rays. A Lorentz boost preserves each light-ray direction while rescaling its null coordinate.

Let $\mathbf{E}=E\hat{\mathbf{x}}$ and $\mathbf{B}=B\hat{\mathbf{y}}$, and choose $\hat{\mathbf{z}}=\hat{\mathbf{x}}\times \hat{\mathbf{y}}$. Part (a) gives $y=0$. With the origin at ${\mathbf{x}}_0$, the remaining Cartesian components areThe initial conditions are $x=z=\dot{x}=\dot{z}=0$. Integrating the second equation gives $\dot{z}=\omega x$, soSolving this forced harmonic oscillator and then integrating $\dot{z}=\omega x$ yieldsThus the position vector is ${\mathbf{x}}_0+x(t)\hat{\mathbf{x}}+z(t)\hat{\mathbf{z}}$. The oscillation occurs at the signed cyclotron frequency, superposed on the usual E-cross-B drift.