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The Fourier sine series coefficients of $x$ on $(0,\pi )$ areby integration by parts. The odd periodic extension is piecewise smooth, so the Fourier series converges to $x$ at every interior point. HenceTo justify integration from zero, the Fourier sine series partial sums converge to $x$ in L2 norm. By the Cauchy-Schwarz inequality, integration from zero to any $x\in [0,\pi ]$ has error at most $\sqrt{\pi }$ times that L2 norm error, uniformly in $x$. Integrating the partial sums and taking the limit therefore givesThe integrated series is absolutely and uniformly convergent, since its terms are bounded by $4/n^2$. Thus the resulting Fourier cosine series hasIndependently, the constant coefficient isComparison yields the requested exact sum

A Hausdorff space is a topological space in which any two distinct points have disjoint open neighborhoods. Let $K$ be a compact subset of a Hausdorff space and $x\notin K$. For each $y\in K$, choose disjoint open neighborhoods $U_y$ of $y$ and $W_y$ of $x$. The $U_y$ cover $K$, so compactness gives a finite subcover $U_{y_1},\dots ,U_{y_m}$. Their paired intersection $W={\bigcap }_{r=1}^m{W}_{y_r}$ is an open neighborhood of $x$ disjoint from $K$. Thus every point of its complement has an open neighborhood in the complement, proving every compact subset of a Hausdorff space is closed. The empty set case is immediate.

In the cocountable topology on an uncountable set, any two nonempty open sets intersect: their combined complements are countable and cannot exhaust $X$. Thus the space is not Hausdorff.

Nevertheless every compact subset is closed here. In fact only finite subsets are compact. To prove this, suppose $K$ is infinite and choose distinct points $d_1,d_2,\dots \in K$. The open setsform an increasing open cover of $X$, and hence of $K$. Any finite selection is contained in $U_N$ for its largest index $N$, which misses $d_N\in K$. So $K$ is not compact. Finite sets are compact in any topological space, and in this cocountable topology they are closed because their complements are open. This proves the claimed behavior of compact subsets of a cocountable space.

The contraction mapping theorem states that a contraction mapping $T:X\to X$ on a nonempty complete metric space has a unique fixed point. Here contraction means that some $0\le q<1$ satisfies $d(Tx,Ty)\le qd(x,y)$ for all $x,y$.

For a proof, choose $x_0$ and iterate $x_{r+1}=T{x}_r$. Induction gives $d(x_{r+1},x_r)\le q^{r}d(x_1,x_0)$. For $m>n$, the triangle inequality and a geometric series giveThus the iterates form a Cauchy sequence, and completeness supplies a limit $a\in X$. A contraction mapping is continuous, so $Ta=\lim T{x}_n=\lim x_{n+1}=a$. If $b$ is another fixed point, $d(a,b)=d(Ta,Tb)\le qd(a,b)$, forcing $d(a,b)=0$. This proves existence and uniqueness.

For the exponential problem, use $h=f\circ f$ on the complete metric space $\mathbb{R}$. Its derivative isPutting $t=e^{-x}>0$, the function $t{e}^{-t}$ has maximum $1/e$ at $t=1$, as its derivative is $(1-t)e^{-t}$. By the mean value theorem, $|h(x)-h(y)|\le e^{-1}|x-y|$, so $h$ is a contraction mapping on all of $\mathbb{R}$. Let $a$ be its unique fixed point. The compositions commute, so $h(f(a))=f(h(a))=f(a)$; uniqueness of the fixed point of $h$ forces $f(a)=a$. Conversely every fixed point of $f$ is fixed by $h$. Hence there is exactly one real solution of $x=e^{-x}$. It lies in $(0,1)$ because $a=e^{-a}>0$ and then $e^{-a}<1$. This uses the iterated contraction argument without claiming that $f$ itself contracts on all of $\mathbb{R}$.

For a finite group of order $p^{a}m$ with $p$ prime and $p\nmid m$, the Sylow theorems assert that a subgroup of order $p^a$ exists; every subgroup whose order is a power of $p$ is contained in one such Sylow subgroup; all Sylow subgroups for that prime are conjugate; and their number $n_p$ divides $m$ and is congruent to one modulo $p$. A unique Sylow subgroup is a normal subgroup, since conjugation cannot move it to another one.

Here $56=7\cdot 8$, so $n_7$ divides eight and is congruent to one modulo seven. Hence $n_7=1$ or $8$. In the first case the Sylow subgroup of order seven is normal. In the second case, any two distinct order-seven subgroups intersect only at the identity, because a nontrivial intersection would generate both prime-order subgroups. Their nonidentity elements therefore account for $8\cdot 6=48$ elements.

Exactly eight elements of the group remain. Every Sylow subgroup of order eight consists entirely of these remaining elements: a nonidentity element in it has order a power of two and cannot be one of the order-seven elements already counted. Each such Sylow subgroup must consequently equal the remaining eight-element set. It is unique and hence normal. Thus a group of order 56 has a normal subgroup of order seven or of order eight.

The monomials $x^i{y}^j$ with $i,j\ge 0$ and $i+j\le n$ form a basis of the given vector space. There are $k+1$ of total degree $k$, soOn that basis the linear operator is diagonal:Both summands of the eigenvalue are nonnegative integers. Their sum is zero precisely when $i,j\in \{0,1\}$. Thus the kernel is the span of those members of $1,x,y,xy$ whose total degree is at most $n$. In particular it is $\mathrm{span}\{1,x,y\}$ for $n=1$ and $\mathrm{span}\{1,x,y,xy\}$ for $n\ge 2$; for $n=0$ it is the constants. The image of a linear map is the span of all the other permitted monomials, since each has a nonzero diagonal eigenvalue and hence is the image of its reciprocal multiple. No excluded monomial can occur in an image because its diagonal coefficient is zero.

The trace is the sum of these eigenvalues. At degrees zero and one all contributions vanish. At degree two the contributions are $2,0,2$, summing to four; at degree three they are $6,2,2,6$, summing to sixteen; at degree four they are $12,6,4,6,12$, summing to forty. ThereforeThis is the diagonal monomial action of a second-order Euler operator.

A random vector $X$ has multivariate normal distribution $N_d(\mu ,\Sigma )$ if every real linear combination $t^{T}X$ is normally distributed, with mean $t^T\mu$ and variance $t^T\Sigma t$. The covariance matrix is symmetric positive semidefinite; zero variances permit degenerate normals. Its moment-generating function is $M_X(t)=\exp (t^T\mu +t^T\Sigma t/2)$, including the degenerate cases.

For the given $X$, let $t_j\in {\mathbb{R}}^{d_j}$. The joint moment-generating function of the $Y_j=A_{j}X$ isThe last equality uses the assumed orthogonality of the row spaces. Its factors are exactly the individual moment-generating functions, so the hint implies the random vectors $Y_j$ are independent. Also every linear combination of the stacked vector $Y$ is a linear combination of $X$, hence normally distributed. ConsequentlyThis establishes both claims without excluding singular block covariances.

For the sample mean and sample variance result, assume $n\ge 2$ and set $W=(Z_1-\mu ,\dots ,Z_n-\mu {)}^T$. Independence of the observations gives $W\sim N_n(0,{\sigma }^{2}I)$. Let $u=(1,\dots ,1{)}^T$, $A_1=u^T/n$, and $A_2=I-u{u}^T/n$. Then $A_1{A}_2^T=0$. The first transformed vector is the scalar $\overline{Z}-\mu$, and the second is the residual vector $(Z_i-\overline{Z}{)}_i$. They are independent by the just-proved result. Applying the measurable function $v\mapsto \Vert v{\Vert }^2/(n-1)$ to the residual vector preserves independence from the sample mean. Thus

Choose the Givens rotation convention in which ${\Omega }^{[p,q]}$ equals the identity outside rows and columns $p,q$, and has blockin those coordinates. It is an orthogonal matrix. For any chosen column $j$, write $u=A_{pj}$, $v=A_{qj}$ and $r=\sqrt{u^2+v^2}$. If $r>0$, take $\cos \theta =u/r$, $\sin \theta =v/r$. Then the new entries are $r$ and $-vu/r+uv/r=0$. If both entries vanish, take $\theta =0$. This proves the requested elimination for each chosen $j$; the angle is allowed to depend on $j$, and one angle need not eliminate an entire arbitrary row.

For the given matrix, use $\cos \theta =\sin \theta =1/\sqrt{2}$ first for ${\Omega }^{[1,2]}$, obtainingUse the same cosine and sine for ${\Omega }^{[1,3]}$. It makesThis is already upper triangular, including the zero $(3,2)$ entry, and every leading nonzero row entry is positive. Thus the requested QR decomposition isThe orthogonal matrix property follows from the two rotations, and direct multiplication verifies $A=QR$.

The vacuum Maxwell equations are $\nabla \cdot E=\nabla \cdot B=0$, $\nabla \times E=-B_t$ and $\nabla \times B=E_t/c^2$. Taking the curl of Faraday's equation and using $\nabla \times (\nabla \times E)=\nabla (\nabla \cdot E)-{\nabla }^{2}E$ gives $-{\nabla }^{2}E=-E_{tt}/c^2$. The same calculation for Ampère's equation gives ${\nabla }^{2}B=B_{tt}/c^2$. Hence each Cartesian component satisfies the electromagnetic wave equation.

Model the walls as a perfect conductor, with no oscillating fields in its interior. The tangential electric field is zero at the surface, otherwise it would drive unbounded conductor currents. The normal magnetic field is zero, since the divergence-free field has a continuous normal component and the interior oscillating field is zero. Thus $n\times E=0$ and $n\cdot B=0$. Normal $E$ and tangential $B$ may be nonzero because surface charge and surface current support their jumps.

The PDF writes the full magnetic field as $B=B_0(x,y)\hat{z}{e}^{i(kz-\omega t)}$. Taken literally, this cannot be a nonzero travelling wave: $\nabla \cdot B=ik{B}_z$, so for $k\ne 0$ it forces $B_z=0$. The subsequent requests require transverse components. The consistent intended reading is $B_z=B_0(x,y)e^{i(kz-\omega t)}$, with transverse fields determined by the Maxwell equations. Use that reading below.

All fields have the same $z,t$ factor. Put $\Delta =(\omega /c{)}^2-k^2$. The harmonic Maxwell equations are $\nabla \times E=i\omega B$ and $\nabla \times B=-i\omega E/c^2$. Their $x$ and $y$ components givetogether with ${\partial }_y{E}_z-ik{E}_y=i\omega B_x$ and $ik{E}_x-{\partial }_x{E}_z=i\omega B_y$. Substitution and solving, for $\Delta \ne 0$, givesThe corresponding electric components areFor a transverse electric mode of a rectangular waveguide, $E_z=0$. The electromagnetic wave equation becomesAt $x=0,a$, the tangential component $E_y=-i\omega {\partial }_x{B}_z/\Delta$ vanishes, so ${\partial }_x{B}_0=0$; at $y=0,b$, the corresponding condition $E_x=0$ gives ${\partial }_y{B}_0=0$. These are Neumann boundary conditions, and the normal-$B$ conditions give the same conclusions through the displayed transverse magnetic components. Separation giveswhere $m,n\ge 0$ are integers and $(m,n)\ne (0,0)$. The constant candidate is not a physical oscillating mode: its transverse derivatives vanish, so it has $E=0$ and Faraday's equation forces $B=0$ for $\omega \ne 0$. More generally a hollow single-conductor guide has no transverse-electromagnetic zero-cutoff mode: its transverse electrostatic potential is harmonic with the same constant boundary value, hence constant, leaving zero transverse field.

The dispersion relation is ${\omega }^2=c^2(k^2+{\kappa }_{mn}^2)$. The waveguide cutoff frequency for this mode is ${\omega }_{mn}=c{\kappa }_{mn}$; below it $k$ is imaginary and the field is evanescent. Since $a\ge b$, the lowest cutoff occurs at $(m,n)=(1,0)$:For a mode propagating in the positive $z$ direction ($k>0$), above its own cutoff ${\omega }_{mn}$, the phase velocity and group velocity areThese waveguide phase and group velocities have product $c^2$. The superluminal phase speed describes motion of a constant phase, while a narrow wave packet travels at the subluminal group velocity; it does not transmit information faster than light. At cutoff $k=0$ there is no travelling mode with nonzero longitudinal flux.

For a one-dimensional wavefunction, the probability density and probability current areThe Time-dependent Schrödinger equation with a real potential is $i\hslash {\psi }_t=-{\hslash }^2{\psi }_{xx}/(2m)+V\psi$. Multiply it and its complex conjugate by ${\psi }^{\ast }$ and $\psi$, respectively. Subtracting cancels the real-potential terms and givesThis derives the continuity equation ${\rho }_t+j_x=0$.

For $k>0$, the specified free-region wavefunction combines a right-moving incoming wave of unit amplitude and a left-moving reflected wave with complex reflection amplitude $R$. Applying the free Schrödinger operator to either plane wave gives $E={\hslash }^2{k}^2/(2m)$. Its time factor has modulus one, soThe density includes the stationary interference pattern; $|R{|}^2$, rather than $R$, is the reflection probability.

For the finite square well, let $k=\sqrt{2mE}/\hslash$ outside and $q=\sqrt{2m(E+V_0)}/\hslash$ inside. Write the stationary parts as $e^{ikx}+R{e}^{-ikx}$ to the left, $A{e}^{iqx}+B{e}^{-iqx}$ inside, and $T{e}^{ikx}$ to the right. At both finite potential steps, the wavefunction and its first spatial derivative are continuous. Integrating the stationary Schrödinger equation across a shrinking step proves derivative continuity; a jump of the wavefunction would create an unmatched distributional derivative and is excluded.

Propagate the boundary values at $a$ back through the well. The values at zero areMatching them to $1+R$ and $ik(1-R)$, and adding the two equations after division of the second by $ik$, givesSince the exterior wave numbers agree, the transmission probability is the current ratio $|T{|}^2$, soWhen $V_0=3E$, $q=2k=\sqrt{8mE}/\hslash$, yieldingClassically an incoming particle of positive energy crosses the attractive well and emerges with its original energy, so transmission is one. The quantum result agrees exactly when $qa$ is an integer multiple of $\pi$:These are widths with resonant transmission through a square well; the degenerate width zero also has unit transmission.

For small transverse displacement, the wave equation on a string iswhere $T$ is the tension and $\rho$ is mass per unit length. Using separation of variables, put $y=X(x)F(t)$, the ratios $X^{\prime \prime }/X=F^{\prime \prime }/(c^{2}F)$ equal a constant. The fixed-end conditions $X(0)=X(L)=0$ exclude nontrivial solutions for nonnegative separation constants; for $-{\kappa }^2$ they give $X=\sin (\kappa x)$ and $\kappa =n\pi /L$, $n\ge 1$. The associated angular frequencies are ${\omega }_n=n\pi c/L$. Thus the general superposition of separated normal modes isThe string is initially straight and at rest before the impulse, so $A_n=0$. Assume the hammer interval lies inside $[0,L]$. The imposed velocity $g(x)=y_t(x,0)$ has Fourier sine series coefficientTherefore $B_n=g_n/{\omega }_n$, giving the amplitude of every excited normal mode.

The string energy is $\frac{\rho }{2}{\int }_0^L(y_t^2+c^2{y}_x^2)dx$. At the impulse it is $E_{\mathrm{total}}=\rho a{v}^2/2$. Orthogonality of the sine modes gives $E_n=\rho L{g}_n^2/4$ initially; each mode subsequently exchanges kinetic and elastic energy while retaining that total. Thus the modal energy of a localized velocity impulse on a string has fractionsFor $l=L/3$ and $a=L/10$, a mode is absent exactly when $\sin (n\pi /3)=0$ or $\sin (n\pi /20)=0$. Hence the unexcited modes are precisely

A subset $U\subseteq {\mathbb{R}}^n$ is an open set when every $x\in U$ has some radius $r>0$ for which $B(x,r)\subseteq U$. If a continuous function takes only the values zero and one, continuity at $x$ with tolerance $1/2$ gives a neighborhood on which $|f(y)-f(x)|<1/2$. The two possible values differ by one, so $f(y)=f(x)$ throughout that neighborhood. Thus $f$ is locally constant, and the zero linear map is its derivative, since the difference $f(x+h)-f(x)$ is exactly zero for all sufficiently small $h$. Hence $D{f}_x=0$ everywhere.

For the bound on $g$, parametrize the segment by $\gamma (t)=b+t(a-b)$, $0\le t\le 1$. The chain rule givesby the definition of operator norm. The mean value theorem applied to the real function $g\circ \gamma$ provesIf the endpoints agree the estimate is immediate. No continuity of the derivative is required.

For the finite-line complement, each removed line ${\ell }_i$ is closed, so $V$ is open. Because $a,b\notin {\ell }_i$, there is a unique affine plane $P_i$ containing $a$ and ${\ell }_i$, and a unique affine plane $Q_i$ containing $b$ and ${\ell }_i$. Choose $c$ outside the union of these finitely many planes, using the permitted fact about finite plane covers. In particular $c\in V$. If $[a,c]$ met ${\ell }_i$ at a point $d$, then $d\ne a$, and the line through $a,d$ lies in $P_i$. Since $c$ is on that same line, it would belong to $P_i$, a contradiction. The same argument using $Q_i$ applies to $[c,b]$. Thus $a$ and $b$ can be joined in $V$ by these two segments, proving polygonal connectivity of a finite-line complement.

Finally a continuous $f:V\to \{0,1\}$ has zero derivative by the first argument. Apply the proved segment estimate with $M=0$ separately on $[a,c]$ and $[c,b]$. It gives $f(a)=f(c)=f(b)$. Since $a,b$ were arbitrary, $f$ is constant on $V$.

The centre of a group is $Z(G)=\{z\in G:zg=gz\text{ for every }g\in G\}$. It is a normal subgroup. For $|G|=p^r$ with $r\ge 1$, let $G$ act on itself by conjugation. The orbit-stabilizer theorem makes the size of the conjugacy class of $g$ equal to $[G:C_G(g)]$, a power of $p$. A class has size one exactly when $g\in Z(G)$; every other class size is divisible by $p$. The class equation therefore givesso $p$ divides $|Z(G)|$. Since the identity belongs to the centre of a group, its size is a positive multiple of $p$. This proves the nontrivial center of a finite p-group.

If $G/Z(G)$ is cyclic, choose a representative $a$ of a generator. Every element is $a^{m}z$ for an integer $m$ and $z\in Z(G)$. Any two such elements commute, since their central factors commute with everything and their remaining factors are powers of the same element. Thus $G$ is an abelian group, $Z(G)=G$, and the cyclic quotient by the centre of a group is trivial.

For a non-abelian group of order $p^3$, the centre of a group has order $p$, $p^2$ or $p^3$ by the proved result and Lagrange's theorem. The last would make the group abelian. The middle would make $G/Z(G)$ have prime order and hence cyclic, also making $G$ abelian by the preceding argument. Therefore its centre has order $p$.

Finally write the given upper-unitriangular matrix as $M(a,b,c)$. Direct matrix multiplication givesIt commutes with every $M(a^{\prime },b^{\prime },c^{\prime })$ exactly when $a{c}^{\prime }=a^{\prime }c$ for all $a^{\prime },c^{\prime }\in F$. Taking $(a^{\prime },c^{\prime })=(0,1)$ forces $a=0$, and taking $(1,0)$ forces $c=0$; those conditions also suffice. Thus the centre of a group in this Heisenberg group over a prime field is