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A finite field ${\mathbb{F}}_q$ contains a primitive tenth root exactly when $10\mid (q-1)$. Its multiplicative group is cyclic: the exponent of a finite abelian subgroup of a field is attained by an element, by combining its prime-power components; since every element is a root of $X^e-1$, the polynomial root bound forces the group order to be at most that exponent, hence equal to it. A cyclic group contains an element of order ten exactly when ten divides its order.

Over ${\mathbb{F}}_q$, the conjugates of $\xi$ are its Frobenius automorphism orbit $\xi ,{\xi }^q,{\xi }^{q^2},\dots$. The extension degree is the least $d>0$ with $q^d\equiv 1(\mathrm{mod}10)$. For $q=3$ this order is four, for $q=11$ it is one, and for $q=19$ it is two. Therefore

A primitive tenth root $\xi$ has ${\xi }^5=-1$ and $\xi \ne -1$, so division of $X^5+1$ by $X+1$ givesIt is irreducible over $\mathbb{Q}$: replacing $X$ by $-X$ gives ${\Phi }_5(X)$, and then replacing $X$ by $X+1$ gives $X^4+5X^3+10X^2+10X+5$. Any nontrivial monic integer factorization would reduce modulo five to factors that are powers of $X$, forcing both constant terms divisible by five and their product divisible by $25$, contrary to the constant term five. Rational factorization reduces to integer factorization by Gauss lemma: products of primitive integer polynomials are primitive, since reduction modulo any prime preserves a nonzero product. ThusLet $t=\xi +{\xi }^{-1}$. Dividing ${\Phi }_{10}(\xi )=0$ by ${\xi }^2$ gives $t^2-t-1=0$. Hence $(2t-1{)}^2=5$, so $\boxed{\sqrt{5}\in \mathbb{Q}(\xi )}$. Either sign of this square root suffices; for $\xi =e^{\pi i/5}$ the positive one is $2t-1$.

A primitive root of unity ${\xi }_n$ satisfies ${\xi }_n^n=1$ and ${\xi }_n^j\ne 1$ for $1\le j<n$, equivalently it has multiplicative order $n$. A positive characteristic of a field must be prime: if its least positive value $p$ factored as $ab$ with $1<a,b<p$, the two nonzero field elements $a\cdot 1$ and $b\cdot 1$ would have product zero. If this prime $p$ divides $n$, put $m=n/p$. In characteristic $p$, the binomial coefficients $\left(\genfrac{}{}{0ex}{}{p}{j}\right)$ for $0<j<p$ vanish because their integer numerators contain $p$ and their denominators do not. Thus $(X-1{)}^p=X^p-1$. It follows that $({\xi }_n^m-1{)}^p={\xi }_n^n-1=0$. A field has no nonzero nilpotent element, so ${\xi }_n^m=1$, contradicting primitivity. Consequently $\boxed{\mathrm{char}K\nmid n}$.

Represent the Möbius transformation by $A=\left(\begin{array}{cc}3& 2\\ 7& 5\end{array}\right)$, which has determinant one and trace eight. Its distinct eigenvalues are ${\lambda }_{\pm }=4\pm \sqrt{15}$ and their product is one. A change of basis to eigenvectors conjugates its projective action to a dilation by ${\lambda }_{+}/{\lambda }_{-}=(4+\sqrt{15}{)}^2$ or its reciprocal. Translation length is invariant under conjugacy, so part (i)'s dilation calculation givesThe fixed boundary points solve $7z^2+2z-2=0$, namely $(-1\pm \sqrt{15})/7$. The axis of a hyperbolic-space isometry is the semicircular geodesic with those endpoints in the vertical real plane.

The physical particle horizon is the greatest physical distance at time $t$ from which a light signal could have reached the observer since the beginning of the cosmological model. If that beginning is $t=0$, radial null propagation has comoving distance ${\int }_0^{t}cd{t}^{\prime }/a(t^{\prime })$, soFor $a(t)=t^{\alpha }$ and $0<\alpha <1$, the integral converges at zero and equals $t^{1-\alpha }/(1-\alpha )$. ConsequentlyThe horizon problem is that widely separated regions of the observed cosmic microwave background have nearly equal temperatures although their past causal regions did not overlap before recombination in the usual decelerating hot-big-bang history. That history offers no causal equilibration mechanism for their observed uniformity. The particle horizon used here is distinct from the Hubble radius $c/H$.

Take velocities relative to a comoving observer at $O$. Spatial homogeneity means that changing the observer to $A$ gives the same relative velocity law, soWith $O$ as origin this implies $v(0)=0$ and $v(a+b)=v(a)+v(b)$. A physical velocity field is continuous; additivity first gives rational homogeneity and then, by continuity, real homogeneity. Hencewith coefficients independent of position. Without a regularity assumption, the additive functional equation alone admits non-linear pathological solutions; continuity is the usual implicit velocity-field assumption.

Isotropy requires $v(Rr)=Rv(r)$ for every spatial rotation, or $HR=RH$. Rotations about $e_i$ force $H{e}_i$ to be parallel to $e_i$, and rotations exchanging coordinate directions force those diagonal coefficients to agree. Therefore $H_{ij}=H{\delta }_{ij}$ and $\boxed{v=Hr}$, the Hubble law.

Allow $H=H(t)$. Solving $\dot{r}=H(t)r$ gives $r(t)=a(t)x$ for a constant comoving coordinate $x$, where $a(t)=a(t_0)\exp ({\int }_{t_0}^{t}H(s)ds)$. Thus $\boxed{H=\dot{a}/a}$, with $a$ the scale factor.

Set $I=I_1=I_2$. The third of the Euler equations for a torque-free rigid body gives $I_3{\dot{\omega }}_3=0$, so ${\omega }_3$ is constant. Define $\Omega =(I_3-I){\omega }_3/I$. The first two Euler equations for a torque-free rigid body becomeso ${\omega }_1+i{\omega }_2=C{e}^{i\Omega t}$. Since the body-frame angular momentum has components $(I{\omega }_1,I{\omega }_2,I_3{\omega }_3)$, its transverse components rotate at $\Omega$ while its axial component is constant. Substitution of the principal moments of inertia givesThis is precession relative to the body axes; the angular momentum is constant in an inertial frame. If $h^2=3a^2$, the rate is zero even when ${\omega }_3\ne 0$. If the transverse components vanish, the precession cone degenerates to its axis.

Put the cylinder's centre at the origin, with $-h/2\le z\le h/2$, $0\le r\le a$, and uniform density $\rho =M/(\pi a^{2}h)$. Symmetry makes the inertia tensor diagonal and gives $I_1=I_2$. Direct integration yieldsFor a transverse principal moment of inertia, the two contributions areTherefore $\boxed{I_1=I_2=M(a^2/4+h^2/12),I_3=M{a}^2/2}$ are the three principal moments of inertia.

In the classical limit the fugacity is small and ${\mathrm{Li}}_s(z)=z+O(z^2)$. The Bose-Einstein distribution approaches the Maxwell-Boltzmann distribution, givingTaking the ratio eliminates both $z$ and $B$, yielding $\boxed{E\simeq \frac{9}{2}N{k}_{B}T}$. The coefficient $9/2$ follows from this density of states, rather than the $3/2$ of a free three-dimensional gas.

Below $T_c$ the fugacity is $z=1$ in the continuum ideal-gas approximation. The zero-energy condensate contributes no internal energy, so the same Bose integral with one additional factor of energy givesEliminate $B$ using $N=B\Gamma (9/2)\zeta (9/2)(k_B{T}_c{)}^{9/2}$ and $\Gamma (11/2)=(9/2)\Gamma (9/2)$:The excited fraction is $(T/T_c{)}^{9/2}$ and the remaining particles occupy the Bose-Einstein condensate. For fixed $B$, the condensed-state energy at fixed $T$ is independent of how many additional particles are placed in the zero-energy ground state.

Measure energy from the single-particle ground state and write $z=e^{\mu /(k_{B}T)}\le 1$. The Bose-Einstein distribution gives the number in excited levels asHere ${\mathrm{Li}}_s(z)={\sum }_{j\ge 1}{z}^j/j^s$, obtained directly from the supplied integral identity. At the Bose-Einstein condensation threshold, $z\uparrow 1$ and the excited states are just able to hold all $N$ particles. ThusThe ground-state population is treated separately from the continuum density of states.

The supplied commutators say that the three operators $X_q$ transform as an angular-momentum-one tensor. Thus the span of $X_q|n\ell m⟩$ transforms inside the product representation $1\otimes \ell$. The angular-momentum addition rule requires $\boxed{|\ell -1|\le {\ell }^{\prime }\le \ell +1}$; specifically $\ell ={\ell }^{\prime }=0$ is forbidden. Alsoso a nonzero amplitude requires $m^{\prime }=m+q$. Since ${\hat{x}}_3=X_0$, its rule is $\Delta m=0$. The components ${\hat{x}}_1,{\hat{x}}_2$ are linear combinations of $X_{+1},X_{-1}$, so each requires $\Delta m=\pm 1$. Projection bounds $|m|\le \ell$, $|m^{\prime }|\le {\ell }^{\prime }$ remain necessary.

Hydrogen orbital states have parity $(-1{)}^{\ell }$, while the position operator is odd. Inserting parity on both sides of a position matrix element multiplies it by $(-1{)}^{{\ell }^{\prime }+\ell +1}$; a nonzero element therefore requires this factor to equal one. Combining the resulting odd ${\ell }^{\prime }+\ell$ with the angular-momentum condition leaves the electric dipole selection ruleThese are necessary conditions, not guarantees of a nonzero radial integral. Angular momentum supplies no additional selection rule on $n,n^{\prime }$ apart from the allowed orbital labels.

Addition of angular momenta one and $j$ gives $\boxed{J=|j-1|,|j-1|+1,\dots ,j+1}$, with projection condition $\boxed{M=q+m}$. The usual bounds $q\in \{-1,0,1\}$, $-j\le m\le j$, $-J\le M\le J$ must also hold. For $j=0$ only $J=1$ occurs; for $j=1/2$ the values are $3/2,1/2$.

For $j=1/2$, abbreviate the product state $|1,q⟩|1/2,m⟩$ by $|q,\uparrow ⟩$ or $|q,\downarrow ⟩$. The highest-weight state is $|3/2,3/2⟩=|1,\uparrow ⟩$. Applying the total lowering operator, the sum of the two individual lowering operators, and dividing by the required ladder coefficient gives the whole quartet:The normalized vector orthogonal to the $M=1/2$ quartet state is annihilated by total raising, hence is the highest state of the remaining spin-$1/2$ representation. Lowering it givesThese six orthonormal states span the six-dimensional product space, and are therefore all the states. Their relative signs are consistent with the lowering convention; a common phase for either multiplet is immaterial. The coefficients are the corresponding Clebsch-Gordan coefficients.

Assume periodic boundary conditions or sufficient decay so that integrations by parts have no boundary terms. Use the Hamiltonian operator $\mathcal{J}={\partial }_x$ and the functionalFor a variation $u+\epsilon v$, integrate the second term by parts $k$ times. The two factors $(-1{)}^k$ cancel, givingThus $\boxed{u_t=\mathcal{J}\delta H/\delta u}$ is exactly the equation, including $k=0$ and $n=0$. The bracket $\{F,G\}=\int (\delta F/\delta u){\partial }_x(\delta G/\delta u)dx$ is antisymmetric by integration by parts. Since this differential operator has coefficients independent of $u$, its associated constant Poisson bracket satisfies the Jacobi identity. This supplies a genuine Hamiltonian field equation, not just a conserved scalar functional.

For $k,n>0$, substitute $x^{\prime }=\alpha x$, $t^{\prime }=\beta t$, $u^{\prime }=\gamma u$ and compare the two distinct right-hand terms. Invariance requires $\beta ={\alpha }^{2k+1}$ and ${\gamma }^n={\alpha }^{-2k}$. Taking $\alpha =e^s$ gives the Lie point symmetryTranslations in $x$ and $t$ are also Lie point symmetries, generated by $X={\partial }_x$ and $T={\partial }_t$. Direct differentiation givesso their span is a three-dimensional Lie algebra.

The printed nonnegative-integer range also includes cases where the generic scaling would divide by zero. If $k>0,n=0$, the equation is linear, $u_t=u_x+{\partial }_x^{2k+1}u$. Amplitude scaling $(x,t,u)\mapsto (x,t,e^{s}u)$ is a symmetry, with $D=u{\partial }_u$. Along with $X,T$ all brackets are zero, giving an abelian three-dimensional Lie algebra. A nontrivial pure rescaling of space and time cannot preserve both derivative orders in this case.

If $k=0$, the equation becomes $u_t=(u^n+1)u_x$, including $u_t=2u_x$ when $n=0$. In either case choose $(x,t,u)\mapsto (e^{s}x,e^{s}t,u)$, with $D=x{\partial }_x+t{\partial }_t$. Then $[D,X]=-X$, $[D,T]=-T$, and $[X,T]=0$. This covers every allowed pair $(k,n)$; when both vanish, there is also an independent amplitude scaling.

The real Lax-Milgram theorem states that if $V$ is a Hilbert space, $B:V\times V\to \mathbb{R}$ is a continuous bilinear form with $B(v,v)\ge \alpha \Vert v{\Vert }_V^2$ for some $\alpha >0$, and $\ell \in V^{\ast }$ is continuous, then there is a unique $u\in V$ such that $B(u,v)=\ell (v)$ for all $v\in V$. Moreover $\Vert u{\Vert }_V\le \Vert \ell \Vert /\alpha$.

Take the Sobolev space $V=H_0^1(\Omega )$ with norm $\Vert v{\Vert }_V=\Vert \nabla v{\Vert }_2$. Since $\Omega$ is bounded, the Poincaré inequality gives $\Vert v{\Vert }_2\le C_P\Vert \nabla v{\Vert }_2$, making this norm equivalent to the full Sobolev norm. For the given bounded measurable coefficient defineThen $|B(u,v)|\le (1+\bar{a}{C}_P^2)\Vert u{\Vert }_V\Vert v{\Vert }_V$, $B(v,v)\ge \Vert v{\Vert }_V^2$ since $a\ge 0$, and $|\ell (v)|\le C_P\Vert f{\Vert }_2\Vert v{\Vert }_V$. The Lax-Milgram theorem therefore gives a unique $u\in H_0^1(\Omega )$ satisfying this identity, which is exactly the required weak solution. Strict positivity of the lower coefficient bound is unnecessary: the zero boundary condition and boundedness supply coercivity through the gradient term.

Taking the Fourier transform gives $(1+|\xi {|}^2)\hat{u}(\xi )=\hat{\varphi }(\xi )$. Thus the canonical solution isA smooth compactly supported function is a Schwartz function, as is its Fourier transform. The multiplier $(1+|\xi {|}^2{)}^{-1}$ and all its derivatives have polynomial bounds, so their product with $\hat{\varphi }$ is again a Schwartz function. The inverse integral can therefore be differentiated to verify $-\Delta u+u=\varphi$ classically. By part (b), its Fourier-space integrand factor $\hat{\varphi }/(1+|\xi {|}^2)$ is radial. Changing variables by any orthogonal transformation in the inverse integral gives $u(Qx)=u(x)$, proving radiality.

This is also the unique tempered-distribution solution: the nonvanishing multiplier $1+|\xi {|}^2$ can be inverted on that space. Unrestricted classical solutions need not be unique, since one may add, for example, $e^{x_1}$; the Fourier construction selects the rapidly decaying solution.

Use the Fourier transform convention $\hat{\varphi }(\xi )={\int }_{{\mathbb{R}}^n}{e}^{-ix\cdot \xi }\varphi (x)dx$. For every orthogonal matrix $Q$, changing variables $x=Qy$ givessince the Jacobian has absolute determinant one and a radial function satisfies $\varphi (Qy)=\varphi (y)$. Any two vectors of the same length are related by an orthogonal transformation, so $\boxed{\hat{\varphi }\text{ is radial}}$.

There is a missing hypothesis in the printed uniqueness claim: on an unbounded domain it is false without a condition at infinity. For example, on $\Omega =\{x_d>0\}$ take $f=0$ and $u_D=0$. Both $u=0$ and $u=x_d$ are classical solutions, although $f_u=0$.

For the intended bounded-domain statement, let $u,v\in C^2(\Omega )\cap C(\overline{\Omega })$ be solutions and put $w=u-v$. The derivative assumption makes $f$ nondecreasing in its first argument. On the open set $D=\{x\in \Omega :w(x)>0\}$,The difference $w$ vanishes on $\partial \Omega$, and continuity makes it zero on every interior boundary point of $D$ as well. Since $D$ is bounded, the weak maximum principle for subharmonic functions gives ${\sup }_{\overline{D}}w\le {\sup }_{\partial D}w=0$, contradicting $w>0$ there unless $D$ is empty. Thus $u\le v$. Interchanging $u,v$ gives the reverse inequality, proving $\boxed{u=v}$. An appropriate bound or decay at infinity can replace boundedness, but the printed hypotheses alone do not include it.

Write $M={\sup }_{(0,1)}|f|<\infty$ and use the same constant extensions $f_n$ as in part (a). They satisfy $\Vert f_n{\Vert }_{\infty }\le M$. Choose polynomials $Q_n$ with $\Vert Q_n-f_n{\Vert }_{\infty }<1/n$ for $n\ge 3$, and set the first two polynomials equal to zero. Then $\boxed{\underset{n}{\sup }\underset{0<x<1}{\sup }|Q_n(x)|\le M+1}$, while the argument of part (a) proves uniform convergence to $f$ on every compact subset of $(0,1)$. The bound concerns the entire interval, rather than just each separate compact subset.