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Let $\epsilon >0$. By the definition of Cauchy sequence, choose an integer $K\ge 1$ such that $|x_j-x_k|<\epsilon /2$ whenever $j,k\ge K$. Apply the given inequality with $N=K+1$. It supplies $n>K+1$ for whichBoth $n$ and $K$ are in the Cauchy sequence estimate, soFor every $m\ge K$ the triangle inequality givesThis is the definition of convergence to zero: $\boxed{x_m\to 0}$. The proof uses only the Cauchy sequence estimate, positivity, and the stipulated inequality; it does not assume the general completeness of the real numbers theorem for Cauchy sequences. Taking $N=K+1\ge 2$ also avoids the empty initial set in a literal reading of the condition for very small $N$.

For $n=1$, choose $x=0$, $y=1$. For $n\ge 2$, define the continuous functionIts values at the grid points have the telescoping sumIf one summand is zero, we already have a zero of $g$. Otherwise the summands cannot all have the same sign, so two of these values have opposite signs. The intermediate value theorem between their arguments gives $t\in [0,1-1/n]$ with $g(t)=0$. Set $x=t$, $y=t+1/n$. ThenThis is the horizontal chord of reciprocal-integer length property; the telescoping sum explains why reciprocal integers appear.

The greatest common divisor of $54$ and $21$ is $3$, which divides $906$. Divide the linear Diophantine equation by $3$ to obtainReducing modulo $7$ gives $4x\equiv 1(\mathrm{mod}7)$, so $x\equiv 2(\mathrm{mod}7)$. Substitution gives the complete family of integer solutionsEvery such pair satisfies the equation, and every solution must have this residue for $x$, so none are omitted. Positivity requires $2+7t>0$ and $38-18t>0$. For integer $t$ these are exactly $0\le t\le 2$. Thus the pairs with both entries positive are

Let $d$ be any positive integer. If $d$ divides $a$ and $b$, it divides the integer linear combination $a-qb=r$. Conversely, if $d$ divides $b$ and $r$, it divides $qb+r=a$. Thus $(a,b)$ and $(b,r)$ have exactly the same positive common divisors. Their greatest elements coincide:This greatest common divisor invariance is the step behind the Euclidean algorithm; it does not require $r<b$, although that additional bound ensures the algorithm terminates. If the convention $0\in \mathbb{N}$ permits $r=0$, the same argument works and $\gcd (b,0)=b$ for $b>0$.

Back-substitute in the Euclidean algorithm to obtain an integer linear combination:Multiplying by $3$ gives the particular solution $(x_0,y_0)=(-45,21)$ of the linear Diophantine equation. Subtracting it from any other solution and dividing by $9$ givesThe integers $34$ and $73$ are coprime, so $73$ divides $x+45$. Writing $x+45=73t$ forces $y-21=-34t$. Conversely these expressions satisfy the original equation for every integer $t$. Thus all solutions of a linear Diophantine equation here are

Treat the boundary data as $2\pi$-periodic in the angular variable. Denote the outer data by $g(\theta )$, and set $h(\theta )=f(a,\theta )$ for the inner data. Define their mean and nonconstant Fourier coefficients byThe outer data have reflection symmetry, so all their sine Fourier coefficients vanish. Their mean is $1/2$, and their cosine Fourier coefficients are

The angular Fourier harmonics for a single-valued harmonic function in an annulus are $\cos (n\theta )$ and $\sin (n\theta )$, with integer $n\ge 1$. Their radial factors are independent combinations of $r^n$ and $r^{-n}$; the zero angular mode has radial factor $A_0+B_0\log r$. For arbitrary inner data, the sine and cosine modes need independent radial coefficients: forcing them to share one radial factor, as in the product form supplied in the paper, would impose an extra restriction. For either sine or cosine mode, if its outer and inner coefficients are $v$ and $w$, the radial coefficients $\alpha ,\beta$ obeySolving this linear system givesIt is convenient to collect the result into the annular harmonic interpolation factorsThey satisfy $H_n(1)=1$, $H_n(a)=0$, $K_n(1)=0$ and $K_n(a)=1$. The zero mode satisfies $A_0=1/2$ and $B_0=(p_0-1/2)/\log a$. Therefore the required separation of variables solution isAll dependence on $f(a,\theta )$ is given by the displayed integrals. Substituting $r=a$ leaves its Fourier series, and substituting $r=1$ leavesthe Fourier series of the specified outer step. The radial factors satisfy $r^2{R}^{\prime \prime }+r{R}^{\prime }-n^{2}R=0$, so each nonconstant term satisfies Laplace's equation; $1$ and $\log r$ do also. For integrable boundary data the radial factors give exponential decay on every closed subannulus strictly inside $a<r<1$, justifying the necessary termwise partial derivatives there.

There is a qualification at the two outer jump discontinuities. By pointwise convergence of a piecewise smooth Fourier series, the outer series gives $1/2$ at $\theta =\pm \pi /2$, whereas the printed data assign $1$ at those two points. Thus the displayed solution attains the outer data at their continuity points, and has radial limit $1/2$ at the jumps. One may separately assign the boundary value $1$ at the two exceptional points without changing the interior harmonic function, but no continuous extension can match the discontinuous data there. For piecewise smooth inner data, the analogous boundary trace agrees with $h$ at continuity points and with its midpoint at jumps. Some regularity of $h$ is necessary for a classical interpretation; the problem imposes none explicitly.

Set $u=X(x)Y(y)$ in Laplace's equation. Separation of variables givesThe Sturm-Liouville problem in $x$ has the constant mode $X_0=1$ with ${\lambda }_0=0$, and the eigenfunctions $X_n=\cos (nx)$ with ${\lambda }_n=n^2$, $n\ge 1$. Indeed, integration by parts makes $\lambda$ nonnegative; for $\lambda >0$ the first Neumann boundary condition removes the sine term and the second requires $\sqrt{\lambda }$ to be an integer. For $\lambda =0$, both derivative conditions make $X$ constant.

Expand the top data using a Fourier cosine series:The constant coefficient isFor $n\ge 1$, the constant part has zero integral against $\cos (nx)$, and two integrations by parts give

Since the bottom value is zero, a convenient factor depending on $y$ for the nonconstant modes is $\sinh (n(y+1))$, which vanishes at $y=-1$. Its value at $y=1$ is $\sinh (2n)$. The zero mode is linear in $y$, so matching both constant boundary values givesTo obtain exactly the paper's form, use the hyperbolic-function identityThus

Each product of hyperbolic sine or hyperbolic cosine with $\cos (nx)$ has $u_{xx}+u_{yy}=0$. The zero mode is also a harmonic function. On $-1\le y\le 1$, the factor $\sinh (n(y+1))/\sinh (2n)$ lies between zero and one. Since $\sum n^{-2}$ converges, the Weierstrass M-test gives uniform convergence on the closed rectangle. The series therefore gives zero at the bottom and the continuous Fourier cosine series of ${\pi }^2-x^2$ at the top, including its endpoints. On compact subsets with $y<1$, the factors decay exponentially with $n$, justifying termwise partial derivatives. In particular $u_x=0$ on both open vertical edges because every differentiated cosine factor vanishes there. The solution has finite Dirichlet energy: Fourier orthogonality gives the $n$th nonconstant mode's energy as $(\pi /2)n{g}_n^2\coth (2n)=O(n^{-3})$, whose sum is finite. Here $\coth$ is the hyperbolic cotangent. Thus a proves uniqueness in that class.

There is a corner compatibility condition for mixed boundary data in the printed wording. The top data have $g^{\prime }(\pi )=-2\pi$, but the paper writes $u_x(\pi ,y)=0$ even at $y=1$. The series solves the interior equation and all the open-edge boundary conditions; the literal derivative condition at $(\pi ,1)$ is incompatible with the top data. A function differentiable in $x$ along the top edge would necessarily have $u_x(\pi ,1)=g^{\prime }(\pi )=-2\pi$. Hence it cannot simultaneously satisfy the printed zero derivative at that corner, and there is no solution with all those literal corner requirements. The standard interpretation of this mixed boundary condition imposes the Neumann boundary condition on the open sides and the Dirichlet boundary condition continuously on the horizontal edges. Differentiating the series term by term on the top edge would be invalid at this corner; the uniform convergence of the undifferentiated series does not license that operation.

Write $A=I_3+N$, whereThe binomial theorem for commuting matrices applies because $I_{3}N=N{I}_3$. All terms of degree at least three vanish, so for every positive integer $n$,The same expression directly gives $A$ when $n=1$, so no small exponent is excluded.

Neither $1$ nor $-1$ is divisible by a prime number, so both lie outside $U$. On the other hand, $0\in 2\mathbb{Z}\subseteq U$. If an integer $z$ has $|z|>1$, part (a)(i) supplies a prime number $p$ dividing the positive integer $|z|$; the same $p$ divides $z$, so $z\in p\mathbb{Z}\subseteq U$. These cases exhaust $\mathbb{Z}$, proving

A limit point $p$ of $E\subseteq \mathbb{R}$ is a point such that, for every $\epsilon >0$, the punctured neighbourhood $(p-\epsilon ,p+\epsilon )\setminus \{p\}$ meets $E$. A closed set contains all of its limit points.

Every nonempty open interval contains a rational number; applying this to $(p,p+\epsilon )$ shows that every $p\in \mathbb{R}$ is a limit point of $\mathbb{Q}$. Thus the limit points of $\mathbb{Q}$ form $\mathbb{R}$, and $\mathbb{Q}$ is not closed.

If $p\in \mathbb{Z}$, radius $1/2$ excludes all other integers. If $p\notin \mathbb{Z}$, its distances to the two neighbouring integers have a positive minimum, and a smaller radius excludes every integer. Thus $\mathbb{Z}$ has no limit points in $\mathbb{R}$ and is closed.

The integers form a countable set, and the events $\{X=k\}$ are disjoint. By countable additivity and part ii,Countability is essential: one cannot use the same sum argument over an arbitrary uncountable collection of zero-probability events.

For an integer $k\ge 1$, $\lfloor X\rfloor \ge k$ is the same event as $X\ge k$. Integration for the Pareto distribution givesThe second identity subtracts two successive tail probabilities, equivalently integrating over $[k,k+1)$. If $\mathbb{N}$ includes zero, the extra case is $\mathbb{P}(\lfloor X\rfloor \ge 0)=1$ and $\mathbb{P}(\lfloor X\rfloor =0)=0$, since $X\ge 1$ almost surely.

By part iii, $X$ is an integer with probability zero. Thus $\lceil X\rceil =\lfloor X\rfloor +1$ almost surely, andThe pointwise inequalities $\lfloor X\rfloor \le X\le \lceil X\rceil$ implyPutting $s=t-1>1$ and rearranging yieldsThis extracts bounds for an infinite series from simple rounding inequalities.

For a nonzero separated solution $U=X(x)T(t)$, separation of variables gives $X^{\prime \prime }/X=T^{\prime }/T=-\lambda$. The zero Dirichlet boundary conditions force $X=\sin nx$, $\lambda =n^2$, with integers $n\ge 1$; the time factors are $e^{-n^{2}t}$. Expanding the initial profile in this Fourier sine series requires only the two already displayed spatial modes. ThusBoth terms individually solve the heat equation, vanish at the endpoints and give the required initial profile at $t=0$. The initial function is also $4{\sin }^{3}x\ge 0$ on the bar; the later solution stays nonnegative and tends to zero as heat leaves through the fixed-temperature ends.

Normalization of a probability distribution and the geometric series give $1=A/(1-\rho )$, so $A=1-\rho$. For a nonnegative integer $n$, the strict tail starts at $n+1$:For nonnegative integers $n>m$, the event $\{N>n\}$ is contained in $\{N>m\}$. ThereforeThis is the memorylessness of the geometric distribution with the zero-based counting convention. Keeping strict versus non-strict inequalities explicit is essential: $N-m-1$, conditional on $N>m$, has the original distribution of $N$.

The values of the complex logarithm at $-1$ are $i(2m+1)\pi$, $m\in \mathbb{Z}$. For any such value and any integer $n$,This is an integer multiple of $\pi$, so its sine is zero. In particular the result holds with the principal value $\log (-1)=i\pi$. The argument does not require treating the multivalued logarithm as a single globally defined function.

For $p>0$, integration by parts gives the Gamma function recurrence:The endpoint terms vanish because $x^p\to 0$ at zero and the exponential dominates the power at infinity. Also $\Gamma (1)={\int }_0^{\infty }{e}^{-x}dx=1$. Iterating for nonnegative integers $n$ therefore yieldsNegative integers are outside this integral definition and factorial conclusion.

For $\lambda \ge 0$, the exponential power series gives the normalization of the Poisson distribution:For $\lambda =0$, the limiting law assigns probability one to $k=0$ and zero to every positive integer $k$. Negative $\lambda$ would not define the stated probability distribution.

For a vacuum electric potential, put $\Phi =R(r)Y(\theta )$ in Laplace's equation in spherical coordinates. Multiplication by $r^2/(RY)$ separates the radial and angular terms. With separation constant $\ell (\ell +1)$ and $x=\cos \theta$, the angular Legendre differential equation isRegularity at both polar axes selects nonnegative integers $\ell$ and the Legendre polynomials. The radial equation is $(r^2{R}^{\prime }{)}^{\prime }-\ell (\ell +1)R=0$, whose independent solutions are $r^{\ell }$ and $r^{-\ell -1}$. Hence the angularly regular axisymmetric vacuum solution isThe radial domain and its boundary conditions determine which terms are admissible.

For the given charge density, only $P_1(\cos \theta )=\cos \theta$ is forced. Write $\Phi =f(r)\cos \theta$. The interior radial Poisson equation is $f^{\prime \prime }+2f^{\prime }/r-2f/r^2=-4$. The identity for the Poisson equation with a spherical harmonic source gives the particular solution $f_p=-r^2$, since the radial operator takes $r^2$ to four. Finiteness at zero and decay at infinity therefore giveContinuity of the electric potential and its radial derivative at $a$ requires $Aa-a^2=B/a^2$ and $A-2a=-2B/a^3$. Solving gives $A=4a/3$, $B=a^4/3$, soAll other angular modes have zero coefficients after imposing regularity, matching and decay. At the origin the interior expression is $(4a/3)z-rz$, which extends with a continuous gradient and value zero. The angular charge density has no unique value exactly at that point; Poisson's equation holds classically for $r>0$ and as a weak solution through the origin. There is no additional point charge.

The Laplace transform is $\overline{f}(p)={\int }_0^{\infty }{e}^{-pt}f(t)dt$ wherever it converges. For sufficiently large $\mathrm{Re}p$, suitable exponential bounds justify integration by parts and differentiation under the integral sign, givingFor a nonnegative integer $n$ and real $p>0$, substitute $u=pt$ in the gamma integral. The result extends holomorphically to $\mathrm{Re}p>0$, givingThis also follows recursively from the second identity, starting with $\mathcal{L}[1]=1/p$.

Put $G=\overline{g}$, $g_0=g(0^{+})$ and $g_1=g^{\prime }(0^{+})$. For a solution with finite initial values and the required transform bounds, the transformed equation isThe initial-value terms cancel, leaving $(p-p^2)G^{\prime }+(3-p)G=0$. On the real interval $p>1$, a nonzero solution satisfiesIntegration gives $G=C(p-1{)}^2/p^3=C(1/p-2/p^2+1/p^3)$. Inverting term by term yieldsTaking $C=1$ gives a nonzero solution, verified by direct substitution. It is $L_2(t)$, and the rational transform agrees with the general Laplace transform of a Laguerre polynomial. No claim about solutions with singular initial behavior is needed to produce this regular solution.