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Put $z=2\cos (\pi /n)$. The recurrencedefines monic integer polynomials and gives $S_k(z)=2\cos (k\pi /n)$. Hence $z$ is an algebraic integer. If $\cos (\pi /n)$ is rational, then the rational algebraic integer $z$ is an integer. Since $-2\le z<2$, this leaves the values corresponding toThis is the rational cosine of an integral submultiple of pi result. For every $n\ge 4$, the value lies strictly between $1/2$ and $1$, so it is irrational.

The recurrence shows that $T_n$ is a polynomial with integer coefficients. Part (a) givesThus $\cos (\pi /n)$ is a root of a nonzero integer polynomial and is algebraic.

The recurrence and the addition formula give the result by induction. It is true for $n=0,1$, and if true for $n,n-1$, then

By part (a), there are $A_{n+1}$ permutations of each alternating type. In an up-down permutation the maximum $n+1$ can occur only at a peak, while in a down-up permutation it can occur only at the complementary positions. Thus, after combining the two types, each possible number $k$ of entries to the left of the maximum occurs once.

Choose those $k$ labels in $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ ways. The entries on the two sides must independently alternate, and after order-preserving relabelling can be chosen in $A_k$ and $A_{n-k}$ ways. This alternating-permutation convolution is

DefineThis is an involution on permutations. Every inequality is reversed, so it maps up-down permutations bijectively to down-up permutations.

Take $\theta =0$ initially. The velocity components areso $l=2\sqrt{3}$. Since $f(1/u)=3u^2+9u^3$, the orbit equation becomesThusThe data $u(0)=1$ and $\dot{r}=-l{u}^{\prime }$ give $A=0$, $B=2/\sqrt{3}$, henceAt $\theta =2\pi$, $u=1$ again, so the particle returns to its initial position after one revolution; now $u^{\prime }=-1/\sqrt{3}$, so it is moving outward. Subsequently $u$ first reaches zero at $\theta =8\pi /3$, where $r=1/u\to \infty$. It therefore flies off to infinity.

The transverse equation of motion is $r^{-1}d(r^2\dot{\theta })/dt=0$, sois constant. With $u=1/r$,Differentiating once more and substituting into the radial equation $\ddot{r}-r{\dot{\theta }}^2=-f(r)$ gives the Binet equation

Momentum conservation makes the sum of the two photon momenta parallel to $u$, so all three vectors are coplanar. Put the photons on opposite sides of the incident direction. Transverse and longitudinal momentum and energy conservation giveEliminating $E_1,E_2$ yieldsUsing the half-angle identities on the left gives exactly

With signature $(+,-,-,-)$, the four-momenta areIn the rest frame of a future timelike $P_1$, one has $P_1=(Mc,0)$ and $P_2^0>0$, so $P_1\cdot P_2=Mc{P}_2^0>0$; Lorentz invariance proves the assertion in every frame.

The impossibility of photon decay into two massive particles follows because a photon has squared four-momentum zero. If it decayed into an electron and positron, conservation would givea contradiction.

The sign map has abelian quotient, so $S_5^{\prime }\le A_5$. On the other hand $A_5^{\prime }=A_5$ and $A_5\le S_5$ imply $A_5\le S_5^{\prime }$. Hence

The subgroup $A_5^{\prime }$ is normal in $A_5$. It is nontrivial because $A_5$ is nonabelian; simplicity of $A_5$ therefore gives

Writing $D_8=⟨r,s:r^4=s^2=e,srs=r^{-1}⟩$, one has $[r,s]=r^{-2}=r^2$. The quotient by $⟨r^2⟩$ is $C_2\times C_2$, so

No. In any group choose a nonidentity $x$ and put $y=x^{-1}$. They commute, but $xy=e$ has order one rather than $o(x)o(y)$.

No. In $S_3$, take $x=(12)$ and $y=(123)$. Their orders are two and three, but $xy$ is a transposition and has order two.

False. Let $G=C_4$ and $H=C_2\times C_2$, and in each choose a subgroup of order two. Then the chosen normal subgroups are isomorphic and both quotients are $C_2$, while $G$ and $H$ are not isomorphic.

True. The group has order six. A Sylow $3$-subgroup $P$ has order three, while the given normal subgroup $N$ has order two. Their intersection is trivial and $NP=G$. Conjugation by $P$ on $N$ is trivial because $\mathrm{Aut}(C_2)$ is trivial, so $G=N\times P\cong C_2\times C_3\cong C_6$.

False. Take $G=C_2\times C_2$ and let $N$ be either factor. Then $N\cong C_2$ and $G/N\cong C_2$, but $G$ is not cyclic and hence is not $C_4$.

The measure, Gaussian factor, $r^{-1}$, coordinates, and derivatives all transform covariantly under rotations, so $T_{ijkl}$ is an isotropic tensor integral. Symmetry gives $\beta =\gamma$.

Contracting $i=j$ and using ${\nabla }^2(1/r)=0$ away from the origin givesNext contract $i=k$ and $j=l$. Sinceone obtainsSolving yields

Let the new orthonormal basis satisfy $e_i^{\prime }=R_{ij}{e}_j$. Then an $n$th-rank tensor transforms bySince $\partial /\partial x_i^{\prime }=R_{ij}\partial /\partial x_j$, the Hessian of a scalar obeys this law with two factors of $R$.

An isotropic tensor is unchanged under every rotation. In three dimensions the most general isotropic tensors of ranks zero through three are respectively