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Write $F=\mathcal{L}f$. Using the Laplace transform of a derivative and the initial data givesThe transformed equation is $(s^4+2s^3+s^2)F=1$, so $F=1/[s^2(s+1{)}^2]$. The preceding inverse Laplace transform therefore givesFor a direct check, $f^{\prime }=1-(t+1)e^{-t}$, $f^{\prime \prime }=t{e}^{-t}$, $f^{(3)}=(1-t)e^{-t}$, and $f^{(4)}=(t-2)e^{-t}$. These satisfy both the ordinary differential equation and all four initial conditions.

Part (i) gives $\mathcal{L}(t)=1/s^2$ and $\mathcal{L}(t{e}^{-t})=1/(s+1{)}^2$. The convolution theorem for Laplace transforms therefore givesRepeated integration by parts yieldsSubstitution gives $f(t)=t-2+(t+2)e^{-t}$, agreeing with the contour integration result.

By the Bromwich inversion formula, for $t>0$ and $\gamma >0$,Close the upward vertical segment by a left semicircle of radius $R$ centred at $\gamma$. On that arc $\mathrm{Re}s\le \gamma$, so $|e^{st}|\le e^{\gamma t}$, while $|s^2(s+1{)}^2|$ is bounded below by a constant times $R^4$ for large $R$. Its length is $\pi R$, hence the arc integral tends to zero as $O(R^{-3})$. The contour is positively oriented and encloses the double poles at $0$ and $-1$.

Their residues areandThe residue theorem therefore yieldsThe value at zero follows by continuity and is $f(0)=0$.

For an integer $n\ge 0$ and $\mathrm{Re}s>0$, the definition of the Laplace transform gives $F_0(s)={\int }_0^{\infty }{e}^{-st}dt=1/s$. For $n\ge 1$, integration by parts givesThe endpoint term vanishes because exponential decay dominates every fixed power of $t$. Induction therefore gives $F_n(s)=n!/s^{n+1}$. Multiplication by $e^{at}$ replaces $e^{-st}$ by $e^{-(s-a)t}$ in the defining integral, soThe second formula holds for $\mathrm{Re}(s-a)>0$ and is the Laplace transform shift theorem.

For each integer $n\ge 1$, the open balls $\{B(x,1/n):x\in M\}$ cover the compact metric space $M$. Compactness supplies a finite subcover with centres forming a finite set $S_n$. Define $S={\bigcup }_{n\ge 1}{S}_n$, a countable union of finite sets and therefore countable.

For any $x\in M$ and $\epsilon >0$, choose $n$ with $1/n<\epsilon$. One of the covering balls contains $x$, so there is $s\in S_n\subseteq S$ with $d(x,s)<1/n<\epsilon$. Every open neighbourhood of $x$ therefore meets $S$, giving $x\in \mathrm{cl}(S)$. Thus $\mathrm{cl}(S)=M$, and $S$ is a countable dense subset. If $M$ is empty, simply take $S=\varnothing$.

The closure $\mathrm{cl}(A)$ is the intersection of all closed subsets containing $A$; equivalently it consists of the points whose every open neighbourhood meets $A$.

Suppose $f$ is continuous and $x\in \mathrm{cl}(A)$. For any open neighbourhood $V$ of $f(x)$, continuity makes $f^{-1}(V)$ an open neighbourhood of $x$, so it meets $A$. Hence $V$ meets $f(A)$, showing $f(x)\in \mathrm{cl}(f(A))$. This proves $f(\mathrm{cl}(A))\subseteq \mathrm{cl}(f(A))$.

Conversely suppose this inclusion holds for every subset $A$. For a closed subset $C\subseteq Y$, let $A=f^{-1}(C)$. Then $f(A)\subseteq C$ and therefore $\mathrm{cl}(f(A))\subseteq C$. By the assumed inclusion, $f(\mathrm{cl}(A))\subseteq C$, so $\mathrm{cl}(A)\subseteq f^{-1}(C)=A$. Since $A\subseteq \mathrm{cl}(A)$ always, $A$ is closed. Thus every closed set has closed preimage, equivalently every open set has open preimage. This is continuity, proving the claimed equivalence.

The face action gives $|G|=60$, as calculated above. Every normal subgroup contains the identity and is a union of whole conjugacy classes. Thus its possible order is $1$ plus a subset sum of $15,20,12,12$. The distinct resulting numbers areBy Lagrange's theorem for finite groups, a subgroup order divides $60$. Only $1$ and $60$ in this list do so. Consequently the normal subgroups are $\{1\}$ and $G$, and the rotational symmetry group of a regular icosahedron is a simple group.

First, the orbit-stabilizer theorem applied to a face gives $|G|=20\cdot 3=60$: its stabilizer subgroup consists of the identity and the two rotations through $120^{\circ }$ about the axis through the centres of that face and its opposite. A rotation preserving a face must preserve its outward normal and cyclically permute its three vertices, so there are no further stabilizing rotations.

The conjugacy classes are as follows:

There are $30/2=15$ edge axes, $20/2=10$ face axes, and $12/2=6$ vertex axes. They give respectively $15$, $2\cdot 10=20$, and $4\cdot 6=24$ distinct nonidentity rotations, already accounting for all $59$ nonidentity elements.

To justify the class assertions, conjugation by a rotation carries the rotation axis to its image and preserves the rotation angle about the transported oriented axis. The transitive group action on edges puts all half-turns in one conjugacy class. The transitive group action on faces puts the $120^{\circ }$ rotations defined using outward face normals in one conjugacy class; opposite faces encode opposite signs about the same unoriented axis. The same reasoning with outward vertex directions gives a class of twelve $72^{\circ }$ rotations and a class of twelve $144^{\circ }$ rotations, including their inverses. These last classes cannot merge: the matrix trace of a three-dimensional rotation is $1+2\cos \theta$, which differs for $\theta =2\pi /5$ and $4\pi /5$. Thus the list is complete.

For a $2\times 2$ matrix, the characteristic polynomial is $t^2-\mathrm{tr}(A)t+\det A$. The given matrix trace and determinant therefore make it $t^2+1$. By the Cayley-Hamilton theorem, $A^2=-I$, and hence $A^4=I$. Since $A^2\ne I$, the order of a group element $A$ cannot be $1$ or $2$. If $A^3=I$, then $A=A^4(A^3{)}^{-1}=I$, also impossible. Thus the order is $4$.

The Cayley-Hamilton theorem states that every square complex matrix satisfies its own characteristic polynomial: if ${\chi }_A(t)=\det (tI-A)=t^n+c_{n-1}{t}^{n-1}+\cdots +c_0$, thenFor a proof, expand the adjugate matrix as $\mathrm{adj}(tI-A)={\sum }_{k=0}^{n-1}{B}_k{t}^k$. The adjugate identity givesComparing coefficients yields $-A{B}_0=c_{0}I$, $B_{k-1}-A{B}_k=c_{k}I$ for $1\le k\le n-1$, and $B_{n-1}=I$. Multiply the middle identities on the left by $A^k$, include the first identity, and add $A^n{B}_{n-1}=A^n$. The terms involving the $B_k$ cancel consecutively:Their right sides sum to $c_{0}I+{\sum }_{k=1}^{n-1}{c}_k{A}^k+A^n$. This proves the theorem without any substitution of noncommuting matrices into a scalar polynomial identity.

Every nonempty open set in the given topology contains $1$. Suppose a continuous map $f$ took distinct values $f(m)$ and $f(n)$. Choose disjoint open intervals $U,V\subset \mathbb{R}$ containing these respective values. By continuity, $f^{-1}(U)$ and $f^{-1}(V)$ are nonempty open sets in $\mathbb{N}$, and therefore both contain $1$. This is impossible because the inverse images of disjoint sets are disjoint. Thus every such continuous map is constant. Conversely, constant maps are continuous. This is the continuous real maps from the initial-segment topology argument, using the Hausdorff separation of distinct real points.

Every point has arbitrarily small path-connected open neighbourhoods. Indeed, in a coordinate chart homeomorphic to ${\mathbb{R}}^n$, take the inverse image of a sufficiently small open Euclidean ball. The ball is convex, hence path-connected, and its inverse image is path-connected by the homeomorphism. If the supplied neighbourhood is not itself open, first take an open neighbourhood within it and a ball contained in its chart image. This proves that $X$ is a locally path-connected space.

Fix $x_0\in X$ and let $A$ consist of the points joined to $x_0$ by a path. For $x\in A$, a path-connected open neighbourhood $U_x$ is contained in $A$: concatenate a path from $x_0$ to $x$ with one from $x$ to any point of $U_x$. Hence $A$ is open. For $x\notin A$, the same kind of neighbourhood cannot meet $A$, since a meeting point would provide a path to $x$. Hence $X\setminus A$ is also open. Since $X$ is a connected space and $A$ is nonempty, $A=X$. This proves $X$ is path-connected, including the general principle that connected locally path-connected spaces are path-connected. The empty-space case, if allowed, is vacuous.

The derivative is $f^{\prime }(x)=-4x^3{e}^{-x^4}$. Its absolute value is continuous and tends to zero as $x\to \pm \infty$, so it is bounded. More explicitly,obtained by differentiating $4t^3{e}^{-t^4}$ for $t\ge 0$. The mean value theorem gives $|f(x)-f(y)|\le M|x-y|$ for every $x,y$. Thus $f$ is Lipschitz continuous, and choosing $\delta =\epsilon /M$ proves $e^{-x^4}$ is uniformly continuous on $\mathbb{R}$.

Take $x_n=2\pi n$ and $y_n=2\pi n+1/n$. Then $|y_n-x_n|=1/n\to 0$, whileThe sequential criterion for uniform continuity would force $f(y_n)-f(x_n)\to 0$. These pairs contradict it: for example, for all sufficiently large $n$ their output separation is greater than $\pi$, despite arbitrarily small input separation. Thus $x\sin x$ is not uniformly continuous on $\mathbb{R}$.

The symmetric group $S_5$ has cardinality $5!=120=3\cdot 40$, so its Sylow subgroups for $p=3$ have cardinality $3$. One example isEvery nonidentity permutation of order of a group element $3$ in $S_5$ is a single permutation cycle of length $3$: two disjoint length-three permutation cycles would require six letters. Choose the three moved letters in $\left(\genfrac{}{}{0ex}{}{5}{3}\right)$ ways, then one of their two length-three permutation cycles, giving $20$ such elements. Each subgroup of cardinality $3$ has exactly two nonidentity elements, and distinct such subgroups have no nonidentity element in common. ConsequentlyThis also satisfies the numerical Sylow theorem restrictions $10\mid 40$ and $10\equiv 1(\mathrm{mod}3)$.

Write the cardinality of the finite group as $|G|=p^{a}m$, where $p$ is prime and $p\nmid m$. The Sylow theorems say that a Sylow subgroup of order $p^a$ exists; every p-group is contained in a Sylow subgroup; and all Sylow subgroups are conjugate subgroups. If $n_p$ is their number, thenEquivalently, for a Sylow subgroup $P$, conjugation and the orbit-stabilizer theorem give $n_p=[G:N_G(P)]$, where $N_G(P)$ is the normalizer. The containment statement is often included in the conjugacy form: every p-group lies in a conjugate of any fixed Sylow subgroup.

Regard $B$ as a linear map ${\mathbb{R}}^n\to {\mathbb{R}}^p$ and $A$ as a linear map ${\mathbb{R}}^p\to {\mathbb{R}}^m$ (the same argument holds over any field). ThenIts dimension cannot exceed either $\dim \mathrm{im}A$ or $\dim \mathrm{im}B$, since applying a linear map cannot increase dimension. The rank bound for a matrix product is thereforeThe bound $p$ is attained whenTheir product has an $I_p$ block and zeros elsewhere, so its rank of a matrix is $p$. Thus the bound depending only on the given matrix sizes is sharp.

The row rank of a matrix is the dimension of the linear span of its row vectors. Its column rank is the dimension of the linear span of its column vectors, equivalently the dimension of the image of the corresponding linear map. The equality of row rank and column rank states that these two numbers are equal for every matrix. Their common value is the rank of a matrix.

False. The real line with its usual metric is complete and is homeomorphic to $(0,1)$ with its usual metric, throughIts inverse is $h^{-1}(u)=\tan (\pi (u-1/2))$, so both maps are continuous. However, $(1/n{)}_{n\ge 2}$ is a Cauchy sequence in $(0,1)$ and has no limit there: its only possible real limit is zero. Hence $(0,1)$ is not a complete metric space. Completeness is a property of the metric, not just of its topology.

False. Give $X=\mathbb{R}$ its usual topology, let $Y$ be a one-point space, and take the constant continuous map $f:X\to Y$. The set $K=Y$ is compact, but $f^{-1}(K)=\mathbb{R}$ is not compact. For example, the open cover $\{(-n,n):n\ge 1\}$ has no finite subcover. Thus continuity alone does not preserve compactness under inverse images.