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Write $E=|\mathbf{E}|$ and $\hat{\mathbf{E}}=\mathbf{E}/E$. The constant electric force gives momentumCombining $\mathbf{p}=\gamma m\dot{\mathbf{x}}$ with the relativistic energy-momentum relation givesThe dimensional analysis from part (a) confirms that $qEt$ has the dimensions of momentum, so $qEt/(mc)$ is dimensionless and the expression has the dimensions of velocity.

The speed increases monotonically from zero and approaches the speed of light asymptotically:Integrating in the direction of the field givesAt small times, the Taylor expansion of the square root gives the nonrelativistic limitwhich is motion from rest with the constant acceleration $q\mathbf{E}/m$.

The Lorentz force and Newton's second law giveWriting $\Omega =qB/m$, the component equations areThe initial velocity therefore givesand integration with the initial position givesThis is helical motion in a uniform magnetic field: a helix of radius $v/\Omega$ around the line $x=1+v/\Omega$, $y=0$, with its axis parallel to the magnetic field.

A three-cycle is an even permutation, as is a product of two disjoint three-cycles. Thus every element counted in part (c) lies in $A_7$, and every order-three element of $A_7$ was already counted there. The answer remains $\boxed{350}$.

An element of order three in $S_7$ has cycle type $31^4$ or $3^{2}1$. There aresingle three-cycles. For two disjoint three-cycles, choose the two unordered three-element supports and an orientation on each:Hence $S_7$ contains $\boxed{350}$ elements of order three.

Write the dihedral group as $D_{2n}=⟨r,s:r^n=s^2=e,srs=r^{-1}⟩$. Every reflection has order two, so an element of order three must be a rotation. Such rotations exist exactly when $3\mid n$, and then they are $r^{n/3}$ and $r^{2n/3}$. The answer is therefore

Lagrange's theorem for finite groups states that if $H\le G$ and $G$ is finite, thenso $|H|$ divides $|G|$. Indeed, the left cosets of $H$ partition $G$, and multiplication by a coset representative is a bijection from $H$ to each coset.

Applying the theorem to the cyclic subgroup $⟨g⟩$ gives

In $C_3\times C_9$, an element has order dividing three exactly when its first coordinate is arbitrary and its second belongs to the unique subgroup of order three. There are $3\cdot 3=9$ such elements, and after removing the identity the answer is $\boxed{8}$.

Cauchy's theorem for finite groups states that if a prime number $p$ divides the order of a finite group $G$, then $G$ contains an element of order $p$.

ConsiderThe first $p-1$ entries determine the last, so $|X|=|G{|}^{p-1}$ is divisible by $p$. The cyclic group $C_p$ acts on $X$ by cyclically shifting the coordinates; the shifted tuple remains in $X$ because $x_2\cdots x_p{x}_1=x_1^{-1}(x_1\cdots x_p)x_1=e$. Every orbit has size one or $p$. The fixed points are exactly $(x,\dots ,x)$ with $x^p=e$. Their number is therefore divisible by $p$. Since the identity is one fixed point, another exists; its order is $p$.

Now let the abelian group $G$ have order $pq$ for distinct primes $p,q$. Cauchy's theorem supplies $x,y$ of orders $p,q$. Their cyclic subgroups intersect trivially, and commutativity shows that $xy$ has order $pq$. Thus $G$ is cyclic andThe corresponding assertion for order $p^2$ is false: $C_{p^2}$ is not isomorphic to $C_p\times C_p$, since the former has an element of order $p^2$ while every nonidentity element of the latter has order $p$.

Suppose first that $H$ and $K$ are normal subgroups. For $h\in H$ and $k\in K$, the commutatorlies in $H$, because $k{h}^{-1}{k}^{-1}\in H$, and in $K$, because $hk{h}^{-1}\in K$. Since $H\cap K=\{e\}$, $[h,k]=e$, so $h$ and $k$ commute.

Conversely, suppose every element of $H$ commutes with every element of $K$. Write an arbitrary $g\in G$ as $g=h_0{k}_0$. For $h\in H$,so $H⊴G$. The symmetric calculation proves $K⊴G$.

For a spherically symmetric function $\Phi (r)$, the Laplacian in spherical coordinates isIntegrating from zero to $r$ and using regularity at the origin gives

For the stated density, integration and the boundary condition $\Phi (a)=0$ giveThe two pieces and their first derivatives agree at $r=b$. If another solution existed, its difference from $\Phi$ would be harmonic with zero boundary data, so part (a) proves Uniqueness of the Dirichlet problem.

On the shell $U(b,a)$, normalize the outer radial part to obtain the harmonic functionBy the Dirichlet principle, this function minimizes the energy among all $w$ with the same boundary values. Sinceits energy isTherefore

Green's first identity and the boundary condition giveThus $\nabla u=0$, so $u$ is constant on each connected component of $V$; its zero boundary value forces $\boxed{u=0}$.

Since $w-v=0$ on $S$ and ${\nabla }^{2}v=0$, another application of Green's first identity givesConsequentlyExpanding $\nabla w=\nabla v+\nabla (w-v)$ and using the vanishing cross term yields the Dirichlet principle

The unit sphere and the kernel $e^{-c|y-x{|}^2}$ are invariant under every rotation in three dimensions fixing $y$. Therefore $T(y)$ has the same axisymmetry, so part (a) gives

Choose spherical polar coordinates with polar axis $y$ and put $u=\cos \theta =x\cdot y$. Since $|x|=|y|=1$,The tensor contraction giving the trace isSimilarly,At $c=0$ these expressions have the continuous limiting values $4\pi$ and $4\pi /3$. Writing them as $A=T_{kk}=3\alpha +\beta$ and $B=T_{ij}{y}_i{y}_j=\alpha +\beta$ gives

Under an orthogonal change of coordinates represented by $Q$, the component matrix changes byThus $T^{\prime }$ is similar to $T$, so its characteristic polynomial, eigenvalues, and their algebraic multiplicities are unchanged.

If $T$ is symmetric, the spectral theorem for real symmetric matrices makes it diagonalizable, so algebraic and geometric multiplicities agree. Consequently the dimensions of its eigenspaces are also independent of the coordinate frame.

For the stated tensor, choose a unit eigenvector $n$ belonging to the simple eigenvalue $\mu$. Its perpendicular plane is the eigenspace of $\lambda$, soHence

Urn $i$ has $2^i$ balls, so a draw from it is white with probability $2^{-i}$. Given that the first ball is white, Bayes' theorem gives the posterior probabilityAfter replacement, the second draw is conditionally independent of the first given the urn. HenceConsequently $\boxed{p(n)\to 1/3}$.