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A metric space is totally bounded if, for every $ϵ>0$, finitely many open balls of radius $ϵ$ cover it. The Bolzano-Weierstrass property here means that every sequence has a subsequence converging to a point of the space, namely sequential compactness.

First suppose that property holds. A Cauchy sequence has a convergent subsequence with limit $x$ in the space. Given $ϵ>0$, choose an index after which all sequence terms lie within $ϵ/2$ of each other, and a subsequence term beyond that index within $ϵ/2$ of $x$. The triangle inequality then puts every subsequent term within $ϵ$ of $x$. Thus the whole Cauchy sequence converges, proving completeness.

If total boundedness failed, some $ϵ>0$ would admit no finite ball cover. Choose points inductively outside the union of the $ϵ$-balls about previously chosen points. Distinct sequence terms then have distance at least $ϵ$. No subsequence can be Cauchy, and therefore none can converge, a contradiction. This proves total boundedness.

Conversely, suppose the space is complete and totally bounded. For any sequence, cover the space by finitely many radius-$2^{-1}$ balls and retain the infinitely many indices in one ball. Within those indices, a finite cover by radius-$2^{-2}$ balls retains an infinite subset in one smaller ball. Repeat with radii $2^{-k}$ to obtain nested infinite index sets $I_k$. Choose increasing indices $n_k\in I_k$. For $l,j\ge k$, both terms lie in the ball selected at step $k$, soThe subsequence is Cauchy and completeness makes it converge within the space. Hence the Bolzano-Weierstrass property is equivalent to completeness plus total boundedness.

For an incident particle from the left, take $E>0$ and $E>V_0$, and defineThe time-independent Schrödinger equation gives an incident plus reflected plane wave on the left and an outgoing transmitted plane wave on the right:Continuity of the wavefunction and its derivative at a finite potential jump yields $1+r=t$ and $k(1-r)=qt$. Hence $r=(k-q)/(k+q)$ and $t=2k/(k+q)$. The incident and reflected probability currents have equal wave-number magnitude, soAs a check, the transmitted current fraction is $(q/k)|t{|}^2=4kq/(k+q{)}^2$, and the two probabilities sum to one.

For a positive step, $E\downarrow V_0$ makes $q\downarrow 0$, so $P\to 1$. For a negative step with fixed positive incident energy, $V_0\to -\infty$ makes $q\to \infty$, so again $P\to 1$. The latter is quantum reflection from an increasingly large wave-number mismatch, even though the step is downward.

The discriminant of a binary quadratic form is $d=b^2-4ac$. If the form is positive-definite, evaluating it at $(1,0)$ gives $a>0$. For $a\ne 0$, completing the square givesEvaluating along $x=-by/(2a)$ shows that positivity also requires $d<0$. Conversely these two inequalities make both displayed coefficients positive, provingA reduced positive definite binary quadratic form satisfies $|b|\le a\le c$, with $b\ge 0$ if $|b|=a$ or $a=c$. The last convention avoids counting both boundary representatives. For fixed negative discriminant, these inequalities implyThere are finitely many such integers $a$, finitely many integers $b$ with $|b|\le a$, and then $c=(b^2-d)/(4a)$ is determined. Hence the number of reduced forms is finite. If no admissible integer coefficients have the given discriminant, this number is zero.

For $d=-39$, the bound gives $a=1,2,3$. The integrality of $c=(b^2+39)/(4a)$ and the boundary sign convention give respectivelyThese are all the reduced positive definite binary quadratic forms of discriminant minus thirty-nine. Thusrepresented by $x^2+xy+10y^2$, $2x^2+xy+5y^2$, $2x^2-xy+5y^2$, and $3x^2+3xy+4y^2$.

The Cauchy integral formula states that if $f$ is holomorphic on a neighbourhood of a simple closed positively oriented contour and its interior, then, for an interior point $z$,Fix $0<r<1$ and apply it on $|\xi |=r$. For $|z|<r$, the geometric seriesconverges uniformly in $\xi$ and locally uniformly in $z$. Indeed, for $|z|\le s<r$ its absolute terms are bounded by $s^n/r^{n+1}$. Since $f$ is bounded on the contour, termwise contour integration is valid, givingTermwise differentiation of this locally convergent power series gives $f^{(n)}(0)=n!c_n$. ThereforeThis also derives the needed Taylor series representation rather than assuming an unjustified interchange at the boundary of the analytic disk.

Take $z$ positive downwards, and let the surface pressure be $p_0$. Static force balance gives the hydrostatic pressure equation $dp/dz=\rho g$. HenceOn a submerged body's surface, pressure traction is $-p\mathbf{n}$, where $\mathbf{n}$ points out of the body. The divergence theorem gives the resultantThus the buoyancy is upward with magnitude $\rho gV$, proving Archimedes' principle. The constant surface pressure makes no contribution because ${\int }_{\partial V}\mathbf{n}dS=0$.

For a floating iceberg of total volume $V$, the displaced seawater volume is $V-V_I$. Neglecting air buoyancy, equality of weight and buoyancy gives ${\rho }_{I}gV={\rho }_{w}g(V-V_I)$. Consequentlywith $0<{\rho }_I<{\rho }_w$, as required for partial flotation.

Expansion of the determinant along the first row givesThus the matrix is invertible exactly when $x\ne 1$. The minor formed from the first two rows and the last two columns is $-1$, independently of $x$, so the matrix rank is always at least two. ThereforeTaking cofactors and transposing them gives the adjugate matrixThe adjugate identity $M\mathrm{adj}M=(\det M)I$ then givesIn particular, the singular case has rank two, rather than a further exceptional rank-one value.

The curvature-minus-one Poincaré disk model has Riemannian metricThe geodesics through zero are Euclidean diameters. Indeed, for a path written in polar coordinates, its length is at least $\int 2|\dot{r}|/(1-r^2)dt$, and hence at least $2\mathrm{artanh}r$ between zero and a point of modulus $r$. The radial segment attains this lower bound. By uniqueness of a geodesic with given initial tangent, these radial paths describe all geodesics through the origin. The radial hyperbolic distance isThus a hyperbolic circle in the Poincare disc is the indicated Euclidean circle.

An isometry from the Poincaré half-plane model to the Poincaré disk model is the Cayley transform between the half-plane and disk,Writing $z=x+iy$, one has $1-|w{|}^2=4y/|z+i{|}^2$ and $|dw{|}^2=4|dz{|}^2/|z+i{|}^4$. Substitution into the disk Riemannian metric gives $d{s}^2=|dz{|}^2/y^2$, the half-plane metric, and $i$ maps to zero.

Put $r=\tanh (\rho /2)$. The inverse image of $|w|=r$ obeysSince $(1+r^2)/(1-r^2)=\cosh \rho$, completing the square yieldsThis proves that the hyperbolic circle in the upper half-plane has Euclidean centre $i\cosh \rho$ and radius $\sinh \rho$.

Put $S_1={\sum }_{i=1}^n{X}_i$, $S_2={\sum }_{i=1}^n{X}_i^2$. The joint normal distribution density can be written asThe Fisher-Neyman factorization theorem says that, for a family dominated by a common measure, a statistic $T$ is a sufficient statistic when its densities have the form $g_{\mu }(T(\mathbf{x}))h(\mathbf{x})$ with $h$ independent of the parameter. The displayed factorization therefore shows that $(S_1,S_2)$ is a two-dimensional sufficient statistic.

For maximum likelihood estimation, discard parameter-independent terms from the log-likelihood:If $S_2>0$, the quadratic equation has exactly one positive root. The numerator is positive before that root and negative after it, while the likelihood tends to zero at both ends of $\mu >0$. Hence the normal likelihood with variance equal to squared mean gives the global maximizerThis holds almost surely. The exceptional all-zero sample has $S_2=0$ and likelihood proportional to ${\mu }^{-n}$, so its supremum is approached as $\mu \downarrow 0$ and there is no positive maximizer.

For the real even function, let $s_N=a_0/2+{\sum }_{n=1}^N{a}_n\cos nx$. Fourier orthogonality gives ${\int }_{-\pi }^{\pi }\cos nxdx=0$ for $n\ge 1$ and ${\int }_{-\pi }^{\pi }\cos nx\cos mxdx=\pi {\delta }_{nm}$. ConsequentlyFor an even square-integrable function, the complete Fourier cosine series converges in the L2 norm; hence its norms converge. Taking $N\to \infty$ proves Parseval's identity in the required normalization:This limit argument avoids multiplying an infinite series without a justification of L2 norm convergence.

For the quadratic function, the Fourier coefficients areThe last equality follows from two integrations by parts, using $\sin n\pi =0$. ThusThis Fourier series is absolutely convergent; the periodically extended quadratic is continuous, so it represents its values also at the endpoints. Substituting its coefficients into Parseval's identity gives

First work over a field of characteristic different from two. If a nonzero symmetric bilinear form had $\varphi (v,v)=0$ for every vector, polarization would givea contradiction. Hence choose $e$ with $\varphi (e,e)\ne 0$. Every $v$ decomposes uniquely aswith the second summand orthogonal to $e$. Induct on the orthogonal complement to obtain a diagonal basis. If the restricted form becomes zero, any basis of that remaining space completes the diagonalization. In this basis the rank of a quadratic form is exactly the number of nonzero diagonal coefficients; the remaining basis vectors span its radical of a bilinear form.

Over the reals, rescale the nonzero basis vectors so the diagonal entries are $p$ copies of $+1$, $q$ copies of $-1$ and $z=n-p-q$ zeros. Define the signature of a quadratic form by $\sigma =p-q$, so $r=p+q$. To prove it is basis independent, note that $p$ is the largest dimension of a linear subspace on which the form is positive-definite. The positive coordinate linear subspace attains dimension $p$; on any other positive linear subspace, projection to those positive coordinates is injective, since its kernel has nonpositive quadratic value. Thus its dimension is at most $p$. The same argument for the negative form characterizes $q$. This proves Sylvester's law of inertia and that the signature is well-defined.

Let $e_i$ and $f_i$ be positive and negative normalized basis vectors, and let $Z$ be the radical of a bilinear form. The span of $Z$ and $e_i+f_i$ for $1\le i\le \min (p,q)$ is a totally isotropic subspace: all mutual bilinear pairings vanish. Its dimension is $z+\min (p,q)$. Conversely, project any totally isotropic subspace $U$ to the nondegenerate positive-plus-negative quotient. The projected linear subspace is still null, and its projections to both the positive and negative coordinate spaces are injective: a vector with one component zero would have strictly signed quadratic value unless it vanished. Its dimension is therefore at most $\min (p,q)$, while the kernel of the quotient map on $U$ has dimension at most $z$. HenceThis proves both construction and sharpness of the maximum dimension of a totally isotropic subspace.

For the five-variable example, the symmetric coefficient matrix isIts cyclic structure makes the vectors $(1,{\zeta }^k,{\zeta }^{2k},{\zeta }^{3k},{\zeta }^{4k})$, $\zeta =e^{2\pi i/5}$, eigenvectors with eigenvalues ${\zeta }^k+{\zeta }^{-k}=2\cos (2\pi k/5)$. Thus the eigenvalues areTheir product is $2$, and three are positive and two negative. ThereforeThe real and imaginary parts of each conjugate pair of eigenvectors give the corresponding real two-dimensional eigenspaces, consistent with the real inertia calculation.

Use the Fourier transform convention $\hat{F}(k)={\int }_{\mathbb{R}}F(x)e^{-ikx}dx$. First compute the Fourier coefficients of the periodized sum. Absolute integrability givesThus Fubini's theorem justifies termwise integration. For integer $m$, the change of variable $x=2\pi j+\tau$ and $e^{2\pi imj}=1$ giveReindexing a convergent periodization gives periodicity; the integral argument always gives its periodic representative almost everywhere. This proves the Fourier-coefficient content of the Poisson summation formula.

There is a genuine qualification to the printed pointwise claim: the periodization of an integrable function need not agree pointwise with its Fourier series under the stated hypotheses alone. Set $F(0)=1$ and $F(x)=0$ for $x\ne 0$. This is absolutely integrable, and on one closed period the periodization has only two possibly nonzero summands, so the series is uniformly convergent. Yet $f(0)=1$, while $\hat{F}(m)=0$ for every $m$. The claimed pointwise equality at zero is therefore false. This counterexample also works for Riemann integrability, not only for an almost-everywhere convention.

The coefficient identity gives a rigorous general replacement: the weighted Fejér sumsconverge to the periodization in the integral norm, and at every continuity point. Indeed they are its convolution with the normalized Fejér kernel; that kernel is nonnegative, has integral one, and has integral tending to zero outside every fixed neighbourhood of zero. For convergence in the L1 norm, split the convolution error into small translations, controlled by continuity of translation in the integral norm, and the remaining tail. The same split gives pointwise convergence at a continuity point. If the periodization is piecewise continuously differentiable, the ordinary Fourier series has the usual midpoint limit; if it is continuous and its coefficients are absolutely summable, the displayed unweighted formula holds pointwise. These are sufficient interpretations of the intended identity.

For the specified exponential example all the needed Fourier series convergence properties do hold. Split its Fourier transform at zero to obtainIts periodization is continuous, with uniformly geometrically bounded tails, and the sampled coefficients are absolutely summable. The absolutely summable coefficients define a uniformly convergent Fourier series with the computed coefficients. Its continuous sum equals the continuous periodization: their difference has every Fourier coefficient zero, hence is zero in the L2 norm by Fourier completeness, and continuity upgrades that to equality everywhere. Therefore its ordinary Poisson summation formula is valid. Evaluate it at zero:Thus the fully justified requested sum is

Put $\theta =x-t$. Taking the gradient of the velocity potential givesThe motion is not generally incompressible; a velocity potential ensures irrotationality, not zero divergence. Both components have zero time average at each fixed point because their sine and cosine averages vanish.

The fluid acceleration is the material derivative $D\mathbf{u}/Dt={\partial }_t\mathbf{u}+(\mathbf{u}\cdot \nabla )\mathbf{u}$. Direct differentiation givesAveraging over one period at fixed $(x,y)$ therefore yieldsThis need not be the derivative of the mean velocity: the nonlinear advective term remains after averaging.

The dyed particle obeys the Lagrangian trajectory equationsExpand $x=ϵx_1+{ϵ}^2{x}_2+O({ϵ}^3)$ and $y=ϵy_1+{ϵ}^2{y}_2+O({ϵ}^3)$, for bounded times. The first-order equations give $x_1=0$, ${\dot{y}}_1=\cos t$, so $y_1=\sin t$. At second order,Integrating with zero initial values verifiesThe velocity averaged along this particle over $0\le t\le 2\pi$ is its displacement divided by $2\pi$. HenceThis Stokes drift is a Lagrangian mean; it does not contradict the zero Eulerian mean velocity at a fixed point. The secular term is a small-parameter expansion, not an assertion of uniform accuracy as $t\to \infty$.

Project from the north pole $N=(0,0,1)$ onto the equatorial plane, identifying that plane with the complex plane. The line from $N$ to $P=(X,Y,Z)$ meets the plane inThis is stereographic projection. Its inverse isRotation through $\theta$ about the $z$-axis changes $X+iY$ to $e^{i\theta }(X+iY)$ and preserves $Z$, so it becomes the Möbius transformation $u\mapsto e^{i\theta }u$.

Let $B$ be the rotation matrix supplied in the question. It sends the $z$-axis to the $x$-axis, and therefore $R_x(\theta )=B{R}_z(\theta )B^{-1}$. If $b=\varphi B{\varphi }^{-1}$ is its given Möbius transformation, thenwhich is again a Möbius transformation. Such $z$- and $x$-axis rotations can send any chosen point of the sphere to the south pole: first rotate its horizontal projection onto the $y$-axis, then rotate in the $yz$-plane. This supplies the rotations needed below.

The antipodal stereographic coordinate relation is $\varphi (-P)=-1/\overline{\varphi (P)}$, as follows immediately from the inverse formula. Fix the cross-ratio conventionWe claim the required ordering isFor the antipodal cross-ratio and spherical distance identity, rotate $P$ to the south pole. Its coordinate becomes zero and its antipode becomes infinity. A point at angular spherical distance $d$ from the south pole has vertical coordinate $-\cos d$ and horizontal magnitude $\sin d$, so its new projected coordinate $v^{\prime }$ has modulus $\sin d/(1+\cos d)=\tan (d/2)$. By Möbius invariance of the cross-ratio, the displayed expression equalsFor finite nonsingular coordinates the same ratio is $-|u-v{|}^2/|1+u\bar{v}{|}^2$. Zero or infinite antipodal coordinates are interpreted by limits. When $Q=-P$, $d=\pi$ and the ratio has the extended value infinity; the finite real formula applies for $0<d<\pi$.