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The closed bounded set $A\subset {\mathbb{R}}^m$ is compact by the Heine-Borel theorem. Choose a convergent subsequence $x_{n_j}\to x_0$. For fixed $N$ and sufficiently large $j$, monotonicity givesContinuity and passage to the limit give $g_N(x_0)\ge \delta$ for every $N$, hence ${\lim }_N{g}_N(x_0)\ge \delta$.

For the deduction, set $g_n=f_n-f$. These functions are continuous, decrease pointwise, and tend to zero. If convergence were not uniform, some $\delta >0$ would admit $x_n$ with $g_n(x_n)\ge \delta$. The result just proved would produce $x_0$ with $\lim g_n(x_0)\ge \delta$, contradicting pointwise convergence. Therefore $f_n\to f$ uniformly. This is Dini's theorem.

The sequence $(f_n)$ converges uniformly to $f$ on $A$ when, for every $\epsilon >0$, there is $N$ such thatfor every $x\in A$.

The function $f$ is uniformly continuous when, for every $\epsilon >0$, there is $\delta >0$ such that $x,y\in A$ and $\Vert x-y\Vert <\delta$ imply $|f(x)-f(y)|<\epsilon$. The same $\delta$ works throughout the domain.

Translate the variable by one:This satisfies Eisenstein criterion with $p=5$: every nonleading coefficient is divisible by $5$, while $25\nmid 30$. Translation $X\mapsto X+1$ is an automorphism of a polynomial ring, so it preserves irreducibility. Hence $f$ is irreducible over $\mathbb{Q}$.

Eisenstein criterion states that ifhas a prime number $p$ such that $p\nmid a_n$, $p\mid a_i$ for every $i<n$, and $p^2\nmid a_0$, then $f$ is irreducible over $\mathbb{Q}$.

After dividing out the polynomial content, suppose for contradiction that the resulting primitive polynomial factors as $f=gh$ with positive-degree $g,h\in \mathbb{Z}[X]$, using Gauss lemma for polynomials. Reduction modulo $p$ gives $\bar{f}=a_n{X}^n$, so in the polynomial ring ${\mathbb{F}}_p[X]$ both $\bar{g}$ and $\bar{h}$ are monomials. Because both factors have positive degree, their constant coefficients are divisible by $p$. Their product $a_0$ is then divisible by $p^2$, a contradiction.

Gauss lemma for polynomials says that the product of two primitive polynomials in $\mathbb{Z}[X]$ is primitive. Equivalently, a primitive polynomial in $\mathbb{Z}[X]$ is reducible over $\mathbb{Q}$ exactly when it is reducible over $\mathbb{Z}$.

Let $J$ be an ideal of $R^{\prime }=S^{-1}R$ and contract it toThis is an ideal of $R$, so $I=(d)$ because $R$ is a principal ideal domain. Every $x^n/1$ is a unit in a ring $R^{\prime }$. Consequently, if $r/x^n\in J$, then $r/1=(x^n/1)(r/x^n)\in J$, so $r\in (d)$ and $r/x^n\in (d/1)$. Conversely $d/1\in J$, hence every multiple of it belongs to $J$. Thusso every ideal of $R^{\prime }$ is principal. The localization remains an integral domain; hence $R^{\prime }$ is a principal ideal domain.

Let $S=\{1,x,x^2,\dots \}$. Since a principal ideal domain is an integral domain and $x\ne 0$, $S$ is a multiplicative subset avoiding zero. The stated equivalence relation and operations are precisely those of the localization of a ring $S^{-1}R$.

For completeness, reflexivity and symmetry are immediate, while transitivity follows by multiplying the two cross-multiplication identities and cancelling a nonzero power of $x$. If $(r,x^n)\sim (s,x^m)$ and $(r^{\prime },x^{n^{\prime }})\sim (s^{\prime },x^{m^{\prime }})$, cross multiplication shows that the proposed sums and products are equivalent. Thus the operations do not depend on representatives. The usual fraction calculations then prove associativity, commutativity and distributivity, with identities $[(0,1)]$ and $[(1,1)]$. Therefore $R^{\prime }$ is a well-defined commutative ring.

Measure vertices by their Hamming distance from the target and let $h_k$ be the expected hitting time when that distance is $k$. We already know $h_1=2^n-1$. From distance one, one of the $n$ coordinate flips reaches the target and the other $n-1$ flips move to distance two. First-step analysis therefore givesFor $n\ge 2$,

Let $h$ be the expected hitting time from any neighbour of the origin back to the origin. After the first step, the return-time calculation from part (i) givesso $h=2^n-1$. Translation by the target vertex is a graph automorphism of the hypercube that exchanges the origin and that adjacent vertex, so the reverse expected hitting time is the same. Hence

For a finite irreducible Markov chain with invariant distribution $\pi$, the mean recurrence time of state $i$ isThe simple random walk on the hypercube has the uniform invariant distribution because the hypercube graph is regular, so every vertex has mass $2^{-n}$. Therefore the expected first positive return time to the initial vertex is

The homogeneous solutions $(C_1+C_{2}x)e^{-x}$ are not $2\pi$-periodic unless $C_1=C_2=0$, so there is a unique periodic solution. For a forcing term $a_n\cos (nx)$, a trial solution $A_n\cos (nx)+B_n\sin (nx)$ givesUsing the Fourier series from part (a), for which only odd $n$ occur and $a_n=-4/(\pi n^2)$, yieldsThe coefficients decay fast enough to differentiate the series twice term by term, verifying the equation and periodicity.

The function is an even function, so every sine Fourier coefficient vanishes. Direct integration givesThusfor the $2\pi$-periodic extension.

Take the parabolic element of PSL2(R)It fixes no point in $\mathbb{H}$. A Geodesic in the Poincare half-plane model is determined by its unordered pair of ideal endpoints: either two real numbers or one real number together with $\infty$. Translation adds one to every finite endpoint, so no such pair is invariant. Therefore this $A$ preserves neither a point nor a hyperbolic line in $\mathbb{H}$.

A fixed point of $A$ satisfiesIts discriminant iswhere $ad-bc=1$ was used. Thus the two roots are nonreal complex conjugates, and exactly one lies in the upper half-plane. Consequently every elliptic element of PSL2(R) fixes a point of $\mathbb{H}$.

Write $B=\mathrm{diag}(\lambda ,{\lambda }^{-1})$ and $q={\lambda }^2>0$, so its Möbius transformation is $z\mapsto qz$, with $q\ne 1$. For $z=u+iv$, the Hyperbolic distance in the Poincare half-plane satisfiesAt $z=i$, the final parenthesis is $1$. Since $\cosh$ is strictly increasing on nonnegative arguments, every $z$ off the imaginary axis has

The characteristic polynomial of $A$ isWhen $|\mathrm{tr}A|>2$, its discriminant is positive, so $A$ has two distinct real eigenvalues $\lambda ,{\lambda }^{-1}$ and two independent real eigenvectors. Let $P$ have those eigenvectors as columns. Rescaling one column makes $\det P=1$ without changing the diagonalization. Hence $P\in S{L}_2(\mathbb{R})$ andThis is the normal form of a hyperbolic element of PSL2(R).

Write the function from the preceding part as $F_m$. Its expansion at zero isHence the residue theorem givesChoose $R=R_m=(m+\frac{1}{2})\pi$. This circle stays a distance at least $\pi /2$ from every pole of $f$, so $f$ is bounded uniformly on it. MoreoverThus ${\max }_{C_{R_m}}|F_m|=O(\log m)$, and the ML inequality yieldsLetting $m\to \infty$ solves the Basel problem:

The function $f$ has a simple pole of residue $1/2$ at every $z=k\pi i$, because $(e^{2z}-1{)}^{\prime }=2$ there. At $z=0$, the term $1/(2z)$ in $g$ cancels the principal part of $f$. For $1\le k\le m$, the summandhas residue $1/2$ at each of $z=\pm k\pi i$, so it cancels the corresponding pole of $f$. Therefore all possible singularities of $F$ in $|z|<(m+1)\pi$ are removable singularities: they are $0$ and $\pm k\pi i$ for $1\le k\le m$. After filling them in, $F$ is analytic throughout that disc.