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The polar form of a complex number and the roots of unity givesince $512=2^9$. Using the complex conjugate to compute the squared complex modulus,The residue $3j-k$ runs through every class modulo $9$, already with $j=0$. The cosine is even, so the nine angles give five distinct values, represented by $r=0,1,2,3,4$. They are distinct because cosine is strictly decreasing on $[0,\pi ]$ and these representative angles lie in that interval. HenceEvery listed value is attained by choosing $z_1=1$ and $z_2=2e^{2\pi ir/9}$.

Away from zero and the zeros of $\cos (\pi /x)$, the sign function is locally constant and the absolute value function is differentiable at the value of the cosine. The product rule and chain rule giveThus the triangle inequality and the bounds on sine and cosine imply $|f^{\prime }(x)|\le 2|x|+\pi$. Part (c) rules out differentiability at every nonzero cosine zero, and part (b) gives $f^{\prime }(0)=0$, so this estimate covers every point where the derivative exists.

For a finite interval $I$, choose $L<\infty$ such that $|x|\le L$ throughout $I$. ThenThe apparent $x^{-2}$ factor in the derivative of the oscillation cancels against the multiplying $x^2$, which is what makes a uniform bound possible even near zero.

The zeros of $\cos (\pi /x)$ occur atLet $g(x)=\cos (\pi /x)$. At these nonzero points,The definition of the derivative gives $g(x_k+h)=g^{\prime }(x_k)h+o(|h|)$, henceThe sign function is constant sufficiently near the nonzero point $x_k$, and $(x_k+h{)}^2=x_k^2+O(h)$. As $f(x_k)=0$, multiplication yieldsIts right and left one-sided derivatives are therefore $\pi \mathrm{sign}(x_k)$ and $-\pi \mathrm{sign}(x_k)$ respectively. They differ, so $f$ is not differentiable at any $x_k$. In particular, $x_k\to 0$ as $k\to +\infty$, provingThe absolute value at a simple zero mechanism creates unequal one-sided derivatives. It also shows why the existence of $f^{\prime }(0)$ alone does not imply higher-order differentiability at a point of order two.

Since $f(0)=0$, its difference quotient isThe absolute value of cosine is at most $1$, so this quotient lies between $0$ and $|h|$. The squeeze theorem proves

For $n=1$, differentiability at $x_0$ means existence of the finite two-sided limit of a function ${\lim }_{h\to 0}(f(x_0+h)-f(x_0))/h$. More generally, higher-order differentiability at a point is defined recursively: for $n\ge 2$, the derivatives $f^{\prime },\dots ,f^{(n-1)}$ must exist throughout some open interval containing $x_0$, and $f^{(n-1)}$ must be differentiable at $x_0$. With $f^{(0)}=f$, the final condition isThe neighbourhood requirement makes the numerator meaningful for every sufficiently small nonzero $h$. It does not require $f^{(n)}$ to exist at other points of that neighbourhood, or to be continuous at $x_0$.

The assumed differentiability makes $e$ continuous. The identity $e^{\prime }=e$ then implies inductively that $e$ has continuous derivatives of every order andFix $x\ne 0$. By the extreme value theorem, the continuous function $|e|$ is bounded on the closed interval between $0$ and $x$; let $M_x$ be its maximum. Applying the Taylor theorem with Lagrange remainder about $0$ gives some ${\xi }_n$ between $0$ and $x$ such thatFor the last factor, successive ratios are $|x|/(n+2)$, which are at most $1/2$ once $n$ is sufficiently large. A geometric sequence bound therefore proves that $|x{|}^{n+1}/(n+1)!\to 0$. The partial sums converge to $e(x)$, soAt $x=0$ the equality follows directly from $e(0)=1$. The proof obtains this Taylor series solely from the differential identity, continuity on a fixed finite interval, and the Taylor remainder estimate.

One precise version of the Taylor theorem with Lagrange remainder is as follows. Let $n\ge 0$, let $f$ and its derivatives through order $n$ be continuous on the closed interval between distinct points $a$ and $x$, and suppose $f^{(n+1)}$ exists throughout its interior. Then some $\xi$ strictly between $a$ and $x$ satisfiesHere the endpoint derivatives are understood as restrictions of derivatives defined near the interval, or as their continuous one-sided extensions. No continuity of $f^{(n+1)}$ is required. For $x=a$ the equality is immediate with zero remainder.

To prove the formula, write the Taylor polynomial as $P_n(t)={\sum }_{k=0}^n{f}^{(k)}(a)(t-a{)}^k/k!$ and defineThen $h(a)=h(x)=0$, and $h^{(j)}(a)=0$ for $j=1,\dots ,n$. The Rolle theorem gives a zero ${\xi }_1$ of $h^{\prime }$ strictly between $a$ and $x$. If $n\ge 1$, apply the Rolle theorem to $h^{\prime }$ between $a$ and ${\xi }_1$, using $h^{\prime }(a)=0$, to get a zero ${\xi }_2$ of $h^{\prime \prime }$. Repeating produces a point ${\xi }_{n+1}$ strictly between $a$ and $x$ with $h^{(n+1)}({\xi }_{n+1})=0$. Each application is justified by continuity on the relevant closed interval and differentiability in its interior; the same construction works when $x<a$.

Since $P_n$ has degree at most $n$,Substituting this value of $K$ into its definition proves the displayed Taylor theorem with Lagrange remainder.

Write $H=F(P,Q)$ with $P=\sum p_k^2$ and $Q=\sum q_k^2$. The angular momentum component $L_i={ϵ}_{ijk}{q}_j{p}_k$ has ${\partial }_{p_{\ell }}{L}_i={ϵ}_{ij\ell }{q}_j$ and ${\partial }_{q_{\ell }}{L}_i={ϵ}_{i\ell k}{p}_k$. Thereforesince a symmetric product contracts to zero with the antisymmetric Levi-Civita symbol. All three components of angular momentum are conserved by this rotationally invariant Hamiltonian.

Use the canonical Poisson bracket convention $\{H,G\}={\sum }_k(H_{q_k}{G}_{p_k}-H_{p_k}{G}_{q_k})$. The Hamiltonian flow generated by $G$ has ${\dot{q}}_k=G_{p_k}$ and ${\dot{p}}_k=-G_{q_k}$. The chain rule therefore givesThus $H$ is invariant under every such trajectory exactly when $\boxed{\{H,G\}=0}$ identically. Antisymmetry then gives $dG/dt=\{G,H\}=0$ along the flow of $H$, so $G$ is a constant of motion.

For $G=\alpha {\sum }_k{p}_k$, the generated equations are ${\dot{q}}_k=\alpha$, ${\dot{p}}_k=0$, hence $q_k(t)=q_k(0)+\alpha t$, $p_k(t)=p_k(0)$. Invariance under this simultaneous translation implies conservation of $\boxed{\sum _k{p}_k}$ when $\alpha \ne 0$; the zero generator alone conveys no additional conservation law.

True. The length-one and length-two registers in part (ii), with the displayed nonzero fills, provide an explicit example. Register length need not equal the minimal recurrence length of every sequence it can generate.

True. A length-$d$ linear-feedback shift register has $2^d-1$ nonzero fills, and its first $d$ output bits recover its fill, so these fills give distinct output sequences. If lengths satisfy $r<s$, the longer register has strictly more nonzero outputs. At least one is absent from the shorter register's output set; choosing any nonzero shorter fill then gives unequal outputs. This is the counting property of output sets of different-length linear-feedback registers.

False. Take $s_{n+1}=s_n$ and $t_{n+2}=t_n$. Fills $1$ and $(1,1)$ both produce the constant-one sequence, despite different lengths. Both state updates are invertible.

False. Take $s_{n+1}=s_n$ and $t_{n+2}=t_{n+1}+t_n$. The only nonzero length-one fill produces constant ones. Constant ones cannot satisfy the second linear recurrence relation, since $1+1=0$ in ${\mathbb{F}}_2$. Thus these linear-feedback shift registers have no identical outputs from nonzero fills.

For $s=2$, part (a) gives $\rho (w)=\frac{3}{2}{w}^2-2w+\frac{1}{2}$ and the right-side polynomial is $w^2$. On the scalar test equation $y^{\prime }=\lambda y$, put $z=h\lambda$. Its characteristic equation isThe linear stability domain contains every $z$ for which both roots have modulus strictly less than one. For real $-1/2\le z<0$, the roots are positive; their larger numerator is less than three while their denominator exceeds three, so both are less than one. At $z=-1/2$ the repeated root is $1/2$, also strictly inside the unit circle. For $z<-1/2$, the roots are complex conjugates and their product isThus in every case both roots lie strictly inside the unit disk. Repeated roots there are allowed, since polynomial factors times their powers still tend to zero. Therefore

Interpolate through the $s+1$ values at $t_n,\dots ,t_{n+s}$, and differentiate the interpolation polynomial at its newest node. In the backward coordinate $\tau =(t-t_{n+s})/h$, the Newton form iswith the $j=0$ factor interpreted as one and $\nabla y_m=y_m-y_{m-1}$. For $j\ge 1$, the derivative of the factor at zero is $(j-1)!/j!=1/j$. Thus the backward differentiation formula isIf $w$ denotes the forward-shift variable, ${\nabla }^j{y}_{n+s}$ corresponds to $w^{s-j}(w-1{)}^j$. Consequently the requested polynomial isThe first form is a polynomial valid also at $w=0$. The second form is convenient for relating it to the truncated series of $\log w$. Differentiation of the degree-$s$ interpolant has local derivative error $O(h^s)$, giving order $s$ for smooth solutions.

The Maxwell-Boltzmann velocity distribution is unaffected by the position-dependent potential, since the canonical density factors. Its speed density is proportional to $v^2{e}^{-m{v}^2/(2k_{B}T)}$ for $v\ge 0$. For $v>0$, differentiating its logarithm gives $2/v-mv/(k_{B}T)=0$, soThe density vanishes at zero and at infinity and has only this positive stationary point, so it is the maximum. It is the same at every height.

The Boltzmann distribution factors between momentum and position. Normalizing the height factor givesIn the tall-box limit this becomes $\boxed{\beta mg{e}^{-\beta mgz}dz}$ on $z\ge 0$, an exponential distribution with scale height $k_{B}T/(mg)$.

The equipartition theorem assigns $k_{B}T/2$ to each of the three quadratic momentum terms, giving kinetic energy $3N{k}_{B}T/2$. The gravitational term is linear in $z$, so it must not be counted as a quadratic degree of freedom. In the tall-box limit its distribution is proportional to $e^{-\beta mgz}$, with mean height $1/(\beta mg)$ and mean potential $k_{B}T$ per atom. Equivalently the generalized equipartition identity $⟨z\partial H/\partial z⟩=k_{B}T$ follows by integration by parts on $[0,\infty )$, where the boundary term vanishes. Three quadratic kinetic contributions plus one linear potential contribution give $5k_{B}T/2$ per atom. At finite $L$, the integration-by-parts boundary term is precisely the correction in part (i).

Take the floor at $z=0$, gas-particle mass $m$, $N$ atoms and box side $L$. The one-particle Hamiltonian is $p^2/(2m)+mgz$. With $\beta =1/(k_{B}T)$ and horizontal area $A=L^2$, the partition function isThis follows by independent Gaussian momentum integrals and the elementary height integral. Taking $E=-{\partial }_{\beta }\log Z_N$ givesFor an arbitrarily tall box compared with the gravitational scale height, $L\gg k_{B}T/(mg)$, the upper-wall correction is negligible andThe energy includes gravitational potential relative to the floor; the kinetic part alone is $3N{k}_{B}T/2$. Retaining the finite-height correction also gives the ordinary ideal-gas limit as $g\to 0$.

The fixed terminal height makes $-y(T)+T=T-1$, so this is a minimum-time problem. An optimal path can be taken entirely in $y\ge 1/2$. To justify the positive-$y$ calculation despite the absolute value in the dynamics, the time of a spatial path is its length in the metric $ds/\sqrt{|y|}$. Reflection to $|y|$ preserves this time; clipping the reflected height below $1/2$ to $1/2$ leaves the endpoints unchanged and cannot increase time, since it decreases vertical length and increases the speed available. Any excursion below $1/2$ strictly wastes time. Thus an optimum does not meet $y=0$.

Use $H=\sqrt{y}(pu+qv)-1$. Its maximum is $\sqrt{y}\sqrt{p^2+q^2}-1$, attained atThe costate equations and free-terminal-time transversality condition giveThe last identity holds throughout because the maximized autonomous Hamiltonian is constant. An abnormal extremal would instead have maximized Hamiltonian $\sqrt{y}\sqrt{p^2+q^2}=0$, forcing both costates to vanish and contradicting their nontriviality. Thus the normal choice is valid. Also $p\ne 0$, since otherwise $x$ cannot reach one; in fact $p>0$ because $\dot{x}$ has its constant sign.

Now express both requested derivatives using the costates:Consequently the optimal trajectory satisfiesThe maximizing controls above and the fixed endpoints determine the relevant extremal; the displayed equation is the requested necessary trajectory equation.