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If the bound is to hold with finite $c$, the error functional must vanish on every polynomial whose fourth derivative is zero. Thus the scheme must be exact for degrees zero through three. Applying it to powers of $t=x+1$ givesSolving,This is the four-point one-sided second-derivative formula.

The Peano kernel theorem says that if a linear functional $L$ annihilates all polynomials of degree below $r$, then for $f\in C^r[a,b]$,

Here $r=4$ and $L=e$. Under the stated nonnegativity assumption,The integral equals $e(q)$ for $q(x)=(x+1{)}^4/24$, since $q^{(4)}=1$. Now $q^{\prime \prime }(-1)=0$ andThereforeEquality is attained by $q$, so the sharp Peano-kernel constant for the four-point endpoint second derivative is

Take the sphere to move in the positive $z$-direction and use spherical coordinates centred on it. An axisymmetric harmonic potential that decays at infinity has the dipole form $A\cos \theta /r^2$. The no-penetration condition in the laboratory frame isIt fixes $A=-U{a}^3/2$, so the potential flow around a translating sphere isThe laboratory-frame velocity components are

The fluid kinetic energy iswhere $V=4\pi a^3/3$. Thus the added mass of a sphere is $\rho V/2$.

When the sphere falls, replacing fluid of density $\rho$ by material of density ${\rho }_s$ lowers the gravitational potential energy at rateThe total kinetic energy of sphere and fluid isConservation of total energy givesFor $U\ne 0$, the acceleration of a freely falling sphere with added mass isThe same formula holds at release by continuity.

In electrostatic equilibrium, a nonzero electric field inside a conductor would move its free charges. Hence $E=0$ in the conducting material and the potential is constant throughout each connected conductor. Immediately outside its surface the tangential electric field is zero. A Gaussian pillbox across the surface givesSince $E_{\mathrm{in}}=0$, the electrostatic boundary conditions at a conductor give

For the widely separated shells, let their charges be $Q_1,Q_2$. Their common potential requiresThus the charge sharing between distant connected spheres is

For the charge on connected concentric spherical shells, the potentials at the two radii areEquality forces $Q_1=0$. Henceso all charge lies on the exterior of the outer shell.

For the neutral sphere in the uniform field, the far-field condition gives $\alpha =-E$. Constancy of the potential on $r=R$ gives $\beta =E{R}^3$. ThereforeThe outward normal field at the surface isThe induced charge on a conducting sphere in a uniform electric field is consequentlyFinally,confirming neutrality.

Separation $y=T(t)X(x)$ and the fixed-end conditions givePutThe initial displacement and zero initial velocity then give the separated solution of the damped string equation

For the triangular initial displacement,Thus the even coefficients vanish and the odd ones alternate in sign.

LetThenOrthogonality of the sine and cosine modes, equivalently the Parseval identity, gives

The sign of its derivative is clearest directly from the equation. Integration by parts, using $y_t=0$ at the fixed endpoints, yieldsThis is the energy dissipation identity for a linearly damped string.

For a closed piecewise smooth curve $\gamma$ avoiding $w$, its winding number about $w$ is

The argument principle says that if a positively oriented closed curve $\gamma$ bounds a domain $U$, and a meromorphic function $F$ has no zeros or poles on $\gamma$, thenwhere zeros and poles are counted with multiplicity.

Suppose $F$ and $G$ are holomorphic on a neighbourhood of $\overline{U}$ andFor $0\le t\le 1$, $F+tG$ has no boundary zero, because a zero would imply $|F|=t|G|<|F|$. Its argument-principle count is integer-valued and continuous in $t$, hence constant. Thus Rouché's theorem states that

Now let $w_0=f(z_0)$. Since $|w_0|<1\le |f|$ on the unit circle, Rouché's theorem shows that $f$ and $f-w_0$ have the same number of zeros in the unit disc. The latter has the zero $z_0$, so $f$ has at least one zero there. For any $w$ with $|w|<1$, the same boundary inequality shows that $f-w$ has the same positive number of zeros as $f$. ThereforeThis is the unit-disc image from a boundary modulus lower bound.

Finally, take a sufficiently small positively oriented circle $C$ around zero. The residueis the winding number of $g(C)$ around zero, so $k\in \mathbb{Z}$. The pole is simple, hence its residue is nonzero and $k\ne 0$. Forwe haveThe second term cancels the complete principal part at zero, so the integer residue of a logarithmic derivative proves that $h^{\prime }/h$ has a removable singularity there.

In polar coordinates the metric isAlong a radius, the hyperbolic distance from the origin to Euclidean radius $r$ is thereforeso $r=\tanh (R/2)$. The Riemannian area element isConsequently a hyperbolic disc of radius $R$ has areaThis proves the area of a hyperbolic disc formula from the metric.

For area $\pi /2$ we get $\cosh R=5/4$. Hence $\sinh R=3/4$ and $e^R=2$, so $R=\log 2$. Tangent equal discs have centers at distance $2R$. Since the centers lie successively on the same radial geodesic,If $r_n$ is the Euclidean coordinate of $c_n$, the radial distance formula givesTherefore the radial chain of equal hyperbolic discs has

No such isometry to the stated upper-half-plane configuration exists for $n\ge 3$. The centers $c_0,\dots ,c_n$ lie on one hyperbolic geodesic, so their images under an isometry would also lie on one geodesic. In the upper-half-plane model, geodesics are vertical lines or semicircles orthogonal to the real axis. Neither type can contain three distinct points of the horizontal line $y=1$, while the centers of the distinct discs $D_0^{\prime },\dots ,D_n^{\prime }$ all lie on that line. This is the geodesic obstruction for a horizontal chain of hyperbolic discs.