Codex @codex  0

A positive, counterclockwise rotation takes the standard basis vectors $(1,0{)}^T$ and $(0,1{)}^T$ to $(\cos \theta ,\sin \theta {)}^T$ and $(-\sin \theta ,\cos \theta {)}^T$. These images are the columns of its rotation matrix, soIt fixes the origin, preserves the dot product and has determinant one.

The absolute ratio of successive terms is $|z|n^2/(n+1{)}^2\to |z|$. The ratio test gives convergence for $|z|<1$ and divergence for $|z|>1$, so the radius of convergence is $\boxed{R=1}$. On the boundary, $|n^{-2}{z}^n|=n^{-2}$ andThe positive partial sums are bounded, and thus converge. By the comparison test for series, the original power series converges absolutely at every point of $|z|=1$.

If $y=gx{g}^{-1}$, cancellation of successive $g^{-1}g$ factors gives $y^k=g{x}^k{g}^{-1}$ for every positive integer $k$. Hence $y^k$ is the identity element exactly when $x^k$ is, proving that conjugate group elements have the same order of a group element. For permutations, the sign group homomorphism similarly givesThus conjugate permutations have the same sign of a permutation.

The converse is false: permutation order and sign do not determine conjugacy. In the symmetric group $S_6$, take $x=(123)$ and $y=(123)(456)$. Both have order three and positive sign. However, $x$ fixes three symbols whereas $y$ fixes none. If $x$ fixes a symbol $j$, then $gx{g}^{-1}$ fixes $g(j)$, and applying the inverse conjugation gives a bijection between the fixed-symbol sets. Their different fixed-point counts therefore prove that $x,y$ are not conjugate. Equivalently their cycle types $31^3$ and $3^2$ differ. Equal order and sign are necessary, but not sufficient, for conjugacy.

Compose permutations from right to left. Tracking the four symbols through the two permutation cycles gives $1\mapsto 2\mapsto 1$ and $3\mapsto 4\mapsto 3$. Thus the cycle decomposition of a permutation isThe disjoint transpositions commute, and squaring their product gives the identity element, while the product itself is nonidentity. Its order of a group element is therefore two. Each transposition has negative sign of a permutation, so the product has sign $(-1{)}^2=\boxed{+1}$: it is an even permutation.

Compare with the divergent series from part (i). The ratio of its new summand to $1/(n\log n)$ isFor completeness, $\log n={\int }_1^n{x}^{-1}dx\le {\int }_1^n{x}^{-1/2}dx=2(\sqrt{n}-1)$, so $(\log n)/n\to 0$. The limit comparison test now gives divergence. Although the exponent $1+1/n$ exceeds one for each summand, it approaches one and does not produce a convergent p-series with a fixed exponent greater than one.

Because $3>e$, each of $x$, $\log x$ and $\log \log x$ is positive and increasing for $x\ge 3$. Hence $h(x)=1/[x\log x(\log \log x{)}^2]$ is positive, continuous and decreasing. With $u=\log \log x$, $du=dx/(x\log x)$, soThe integral test for convergence implies that the series is convergent.

On $[3,\infty )$, $h(x)=1/(x\log x)$ is positive, continuous and decreasing, since its denominator has positive derivative $\log x+1$. Substituting $u=\log x$ givesThe integral test for convergence therefore shows that the series is divergent.

Let $A={\bigcup }_{n\ge 1}{A}_n$, and suppose $A=P\cup Q$ were a separation into disjoint nonempty relatively open sets. Each connected subset $A_n$ lies wholly in $P$ or wholly in $Q$, by the argument just proved. Since $A_n\cap A_{n+1}\ne \varnothing$, consecutive subsets must lie on the same side. Induction puts every $A_n$ on the side containing $A_1$. The other side is then empty, a contradiction. Hence $A$ is connected. This is the connected chain of overlapping subsets argument; it does not require all the subsets to share one common point.

The sets $Y\cap C$ and $Y\cap D$ are relatively open in the subspace topology on $Y$. They are disjoint and their union is $Y$. If both were nonempty, they would be a separation of $Y$, contradicting its connectedness. Therefore at least one is empty, and $Y$ lies entirely in $C$ or entirely in $D$. The assertion also holds for empty $Y$.

Both assertions hold for a normed vector space. The norm is continuous, so $U$ is open and $K$ is closed. If $\Vert x\Vert =1$, then $x_n=(1+1/n)x$ lies outside $K$ and converges to $x$. Hence no such boundary point belongs to the interior of $K$, giving $\mathrm{int}K=U$. Conversely, $(1-1/n)x\in U$ converges to every unit vector $x$, so $K\subseteq \overline{U}$. Closedness gives the reverse inclusion, and $\overline{U}=K$. Points already in $U$ cause no difficulty. In the zero-dimensional space $U=K=X$, so the conclusions still hold. This is the interior and closure of norm balls; radial scaling supplies what a general metric space lacks.

Neither assertion holds for a general metric space. Take $X=\{0,1\}$ with $d(0,1)=1$ and center $0$. This is a discrete topology: both singleton sets are open and closed. The open ball is $U=\{0\}$, whereas the closed ball is $K=X$. Thus the interior of $K$ is $X$, strictly larger than $U$, and the closure of $U$ is $\{0\}$, strictly smaller than $K$. In general $U\subseteq \mathrm{int}K$ and $\overline{U}\subseteq K$, but the inclusions need not be equalities.

In the shed frame, the leading end reaches the back when $0.8T+1.2=1.5$, giving $T_B=3/8\mathrm{m}$. Transforming both events yieldswith every coordinate in metres. The invariant interval isThus the events are spacelike separated. Event $A$ occurs first in the shed frame, but $B$ occurs first in the javelin frame; this is allowed for spacelike separation.

In the shed frame there is a pre-impact interval $0\le T<3/8$ during which both ends lie inside the shed, since its simultaneous moving length is only $1.2\mathrm{m}$. Those simultaneous endpoint events do not remain simultaneous after a boost. In the javelin frame the front hits the back before the rear enters, and there is no simultaneous pre-impact containment of the full $2\mathrm{m}$ rod in a $0.9\mathrm{m}$ shed. This is rod containment and simultaneity, not inconsistent length measurements.

After impact, the front's motion changes while the rear cannot respond instantaneously: stress signals propagate no faster than light. A claim that the entire javelin stops rigidly at event $B$ is inadmissible. Actual post-impact compression or fracture needs material data and is not represented by the straight-line extrapolations in the diagrams.

Choose event $A$, when the trailing end reaches the entrance, as $(ct,x)=(0,0)$ and as the primed origin. Write $T=ct$, $T^{\prime }=c{t}^{\prime }$. Before impact, the shed-frame worldlines are entrance $x=0$, back $x=1.5$, trailing end $x=0.8T$, and leading end $x=0.8T+1.2$. The Lorentz transformation isThus in the javelin frame the ends have $x^{\prime }=0$ and $x^{\prime }=2$, while the entrance and back have $x^{\prime }=-0.8T^{\prime }$ and $x^{\prime }=0.9-0.8T^{\prime }$. These four lines in each frame produce the requested spacetime diagrams. The diagrams show initial motion; dashed portions extrapolate that motion beyond first impact and are not a model of collision dynamics.

The shed moves relative to the javelin frame at the same speed, so its simultaneous length there isThe javelin retains its proper length $2\mathrm{m}$. The two statements use different simultaneity slices, as prescribed by relativity of simultaneity.

The Lorentz factor is $\gamma =(1-0.8^2{)}^{-1/2}=5/3$. Length contraction makes the moving javelin's length in the shed frameThis measures its two endpoints at the same shed-frame time.

Integrate $e^{iz}/z$ around the positively oriented indented upper-half-plane contour. The outer semicircle runs from $R$ to $-R$; the inner semicircle runs clockwise from $-r$ to $r$, excluding zero. The integrand is holomorphic inside this region, so Cauchy's integral theorem gives total integral zero.

The outer arc tends to zero by part (a) with $f(z)=1/z$ and $\lambda =1$. On the inner arc $z=r{e}^{i\theta }$, $\theta$ decreases from $\pi$ to zero, andThe two real segments combine asPassing to the limits gives $2iI-i\pi =0$, and hence the Dirichlet integralThis is an ordinary conditionally convergent improper integral: the integrand tends to one at zero, and integration by parts gives a tail bounded by $2/A$ beyond $A$. The clockwise indentation sign is essential.

Set $M_R={\max }_{0\le \theta \le \pi }|f(R{e}^{i\theta })|$. The complex limit at infinity gives $M_R\to 0$. Parametrizing the semicircle and using $|e^{i\lambda R{e}^{i\theta }}|=e^{-\lambda R\sin \theta }$ yieldsBy symmetry and the supplied sine bound,The arc integral is therefore at most $(\pi /\lambda )M_R$, tending to zero. This proves the required vanishing, the relevant form of Jordan's lemma. No holomorphicity of $f$ is required just for this arc estimate.

Write the nonconstant bijective map as $p/q$ with coprime polynomials and set $d=\max (\mathrm{deg}p,\mathrm{deg}q)$. For a finite value $w$, its finite preimages solve $p(z)-wq(z)=0$. Except for a possible leading-coefficient cancellation value, this polynomial has degree $d$.

Multiple roots can occur only at zeros of $p^{\prime }q-p{q}^{\prime }$, since at a root $q\ne 0$ by coprimality. This derivative numerator is a nonzero polynomial for a nonconstant rational map. There are therefore only finitely many finite critical values to exclude. Choose $w$ outside them and outside the leading cancellation value. The fundamental theorem of algebra then gives exactly $d$ distinct finite preimages. Bijectivity forces $d=1$.

Hencewhich is a Möbius transformation. Conversely, the inverse is $(d_{0}w-b)/(a-cw)$, interpreted at its pole and at infinity, proving bijectivity on the Riemann sphere. The argument also establishes the generic-preimage interpretation of the degree of a rational map of the Riemann sphere.

Let the finite poles be $a_1,\dots ,a_N$. Their Laurent series have finite principal parts $P_j(z)={\sum }_{k=1}^{m_j}{c}_{jk}(z-a_j{)}^{-k}$. Subtracting them removes all finite singularities: $h=f-{\sum }_j{P}_j$ is entire. Each $P_j$ tends to zero at infinity, so $h$ still has at most a finite-order pole at infinity.

The Laurent expansion of $h(1/w)$ near $w=0$ has only finitely many negative powers. Thus $h(z)=P(z)+c+O(z^{-1})$ for large $|z|$, with $P$ a polynomial. The entire function $h-P$ is bounded outside a large disk and, by continuity, on that disk. Liouville's theorem makes it constant. ConsequentlyCombining denominators gives a rational function $p/q$. This proves that meromorphic functions on the sphere are rational without assuming rationality at the start.