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Writing $X=\left(\begin{array}{cc}x& y\\ z& w\end{array}\right)$ and using the Frobenius inner product, the three linear equations areThe first two differ by twice the third; equivalently $A-B=2C$. Thus there are only two independent constraints. Substituting $z=-w$ gives $y=2w-x$, so every element of the vector subspace has the unique expressionThe two coefficient matrices satisfy all three constraints. Their linear independence follows by looking first at entry $(1,1)$ and then at entry $(2,2)$. Therefore a basis for this orthogonal complement is

The matrix trace calculation in (i) givesA sum of nonnegative real numbers is zero exactly when each is zero. Thus $\boxed{\mathrm{tr}(A^{T}A)=0⟺A=0}$, so the real Frobenius inner product is positive-definite.

Write $\alpha =\mathrm{tr}(A^{T}A)$, $\beta =\mathrm{tr}(B^{T}B)$ and $\gamma =\mathrm{tr}(A^{T}B)$. If $B\ne 0$, then $\beta >0$, and applying the same nonnegativity to $A-tB$ givesTaking $t=\gamma /\beta$ yields $\alpha -{\gamma }^2/\beta \ge 0$. If $B=0$, the claimed inequality is immediate. Hence the Cauchy-Schwarz inequality here isWhen $B\ne 0$, equality holds exactly when $\mathrm{tr}((A-(\gamma /\beta )B{)}^T(A-(\gamma /\beta )B))=0$, hence exactly when $A=(\gamma /\beta )B$. Together with the case $B=0$, the complete equality condition is $\boxed{A,B\text{ are linearly dependent}}$, including the case that either is zero.

The matrix trace is the sum of the diagonal entries:Using the definitions of matrix multiplication and matrix transpose,This is the symmetry of the real Frobenius inner product.

Since $0<a\le b$, we have $b^n\le a^n+b^n\le 2b^n$. Taking positive $n$th roots yieldsFor every $\epsilon >0$, the binomial theorem gives $(1+\epsilon {)}^n\ge 1+n\epsilon >2$ eventually, so $2^{1/n}\to 1$. The squeeze theorem therefore gives

The sequence is at least $1$. Fix $\epsilon >0$. By the binomial theorem, for $n\ge 2$,once $(n-1){\epsilon }^2>2$. Taking positive $n$th roots gives $1\le n^{1/n}<1+\epsilon$ for all sufficiently large $n$. This proves directly from the definition of a limit of a sequence that

Pairing the first and last terms of the arithmetic progression gives ${\sum }_{k=1}^{n}k=n(n+1)/2$. Therefore the limit of a sequence is

For any complex number $w$, $\mathrm{Re}w\le |w|$, while the complex modulus is multiplicative and unchanged by complex conjugation. ConsequentlyBoth sides are nonnegative, so taking square roots proves the triangle inequality:The proposed further inequality fails for perpendicular vectors in the complex plane. Taking $z_1=1$, $z_2=i$ gives

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.