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A catenary is the static shape of an ideal flexible, inextensible chain hanging under its own gravitational weight in a uniform gravitational field, with uniform linear mass density along its arc length. With the lowest point above $x=0$, one convenient vertical origin givesHere $a$ has units of length. The supports and the chain's arc length determine the scale and position; horizontal and vertical shifts give the general catenary. ^[ref]

For $a=1$, the three points at $x=-1,0,1$ have heights $\cosh 1\approx 1.543081$, $1$, and $\cosh 1\approx 1.543081$. If heights should instead be measured from the lowest point, subtract $a$:The same three heights are then approximately $0.543081$, $0$, and $0.543081$. The shift changes the vertical origin and preserves the catenary shape.

The scale also measures tension: if $H>0$ is its constant horizontal component, $\mu$ is the linear mass density, and $g$ is the magnitude of gravitational acceleration, then $a=H/(\mu g)$. A short force balance explains this. For the portion from the lowest point to $x\ge 0$, with arc length $s(x)$, vertical tension balances its gravitational weight and horizontal tension givesApplying differentiation yields $a{y}^{\prime \prime }=\sqrt{1+y^{\prime 2}}$. The proposed hyperbolic cosine solves this differential equation, because $y^{\prime }=\sinh (x/a)$ and $a{y}^{\prime \prime }=\cosh (x/a)=\sqrt{1+{\sinh }^2(x/a)}$. This is the mechanism behind the shape, under the stated ideal chain assumptions.

The same identity simplifies the arc length between symmetric supports at $x=\pm b$:For $a=b=1$, $L\approx 2.350402$, exceeding the horizontal span of two. A larger $a$ at fixed $b$ gives a shallower sag: the Taylor series gives $a(\cosh (x/a)-1)=x^2/(2a)+x^4/(24a^3)+\cdots$. Thus a parabola is a local approximation, while the exact ideal chain shape is a catenary. The useful connection is that the same hyperbolic Pythagorean identity controls its coordinates, force balance, and arc length.

An inextensible uniform chain of length L and linear mass density $\rho$ links a mass m sliding on a frictionless table to a hanging mass M. If the initial hanging chain length is zero and x is its subsequent hanging length, all moving material has speed $\dot{x}$. The kinetic energy is $(M+m+\rho L){\dot{x}}^2/2$, while the gravitational potential energy is $-Mgx-\rho g{x}^2/2$. The Euler-Lagrange equation gives the displayed equation until a mass hits the floor or reaches the edge. Increasing hanging chain weight produces a hyperbolic cosine displacement from rest, rather than uniform acceleration.

The extension of the cosine to the complex plane is $\cos z=(e^{iz}+e^{-iz})/2$. Splitting the complex exponential function into its real part and imaginary part gives the displayed formula in terms of hyperbolic cosine and hyperbolic sine.

A finite rod generally contains both exponential branches. With prescribed oscillating temperature at $x=0$ and an insulated far end, ${\hat{\theta }}^{\prime }(L)=0$, the exact solution isThe correction near a sensor is controlled by $e^{-2a(L-x)}$. The important separation is between the farther sensor and the end, not merely between the heater and the end. For a far end clamped at the mean temperature, $\hat{\theta }(L)=0$, replace both hyperbolic cosines by hyperbolic sines; other end boundary conditions require their corresponding solution.

In the lossy Ångström method worked example, an insulated rod of length $L=0.080\mathrm{m}$ gives $D=0.354866$ and $\Delta \varphi =0.615651\mathrm{rad}$. Incorrectly applying the semi-infinite formula returns $1.29418\times 10^{-4}{\mathrm{m}}^2{\mathrm{s}}^{-1}$, about $29.4\%$ high. At $L=0.300\mathrm{m}$, it returns $9.99877\times 10^{-5}{\mathrm{m}}^2{\mathrm{s}}^{-1}$, about $0.0123\%$ low. Side loss remains real in both cases; the error arises because the measured spatial response no longer has one exponential complex wavenumber.

Choose the period and sensor spacing so that the lag and attenuation are resolvable while the downstream signal remains larger than sensor noise. Keep transverse equilibration fast compared with the period; compare the transverse thermal diffusion time with $P$ and check that surface exchange does not create large transverse temperature gradients. A short period reduces far-end influence but can expose sensor response times; a long period increases penetration and finite-end bias. Calibrate both sensors together, exchange their positions to check channel bias, compare successive cycles, and repeat at several spacings and periods. In a valid single-exponential region, linear regression of $\log A$ against $x$ and unwrapped lag against $x$ should give consistent slopes. These are proposed diagnostic controls. If they fail, fit the finite-rod or coupled-sensor model rather than interpreting every delay as intrinsic thermal diffusivity.

To isolate finite size from quantum vortex penetration, keep the same weak-field London equations but replace the half-space by an infinite slab $-d/2\le z\le d/2$. Put equal tangential magnetic fields $B_0\hat{\mathbf{x}}$ at its two faces. With $z=0$ now at the slab center, the even solution and its signed current density areThe hyperbolic cosine profile has equal boundary values, while the electric currents on the two sides have opposite signs. In particular,For the stipulated thickness $d=2{\lambda }_L$, the center retains about $0.648B_0$ and the mean is about $0.762B_0$, even though this is a Meissner state. For $d\gg {\lambda }_L$, the center magnetic field is exponentially small and the mean is about $2{\lambda }_L{B}_0/d$, reflecting the two surface layers. For $d\ll {\lambda }_L$, the mean tends to $B_0$: little screening is possible across such a short distance. These are analytic model examples, not material measurements. Real finite objects also need their demagnetizing field and edge geometry accounted for; the slab calculation only isolates the effect of thickness.

For $\beta >0$, periodically extend $\cosh (\beta x)$ and $\sinh (\beta x)$ from $-1<x\le 1$ with period $2$. Their Fourier series areand, away from odd integers,The first identity follows by integration of $\cosh (\beta x)\cos (n\pi x)$ on $[-1,1]$, using the antiderivativeFor the second identity, integration by parts and $d\cosh (\beta x)/dx=\beta \sinh (\beta x)$ show that its sine coefficient is $-n\pi /\beta$ times the first series' cosine coefficient. This avoids assuming that a Fourier series can always be differentiated term by term. The periodic hyperbolic cosine is continuous, so its series converges to it everywhere. The periodic hyperbolic sine has a jump discontinuity at each odd integer: the Dirichlet-Jordan convergence theorem gives the average of its two limiting values, namely zero, there. Both identities hold in the sense of mean-square convergence, which suffices for Parseval's identity.

Nonzero bending modes have the sine, cosine, hyperbolic sine and hyperbolic cosine form. Free conditions at the first end identify the cosine and hyperbolic cosine coefficients, and likewise the sine coefficients. The remaining boundary determinant is $2(1-\cos q\cosh q)$. Its first positive roots are about $4.73004,7.85320,10.99561$. The complete basis additionally includes two zero-energy filament modes, constant translation and affine tilt; these are absent from the finite-coefficient formula at $k=0$.

Hartmann flow is a fully developed viscous conducting-fluid flow driven along a planar channel while a uniform magnetic field crosses the walls. Induction and the Lorentz force density couple its velocity and tangential induced field. For no-slip, normal-field walls separated by $2L$, the velocity is a hyperbolic cosine profile that approaches plane Poiseuille flow at zero imposed field and develops a uniform core with Hartmann layers at large Hartmann number.

The hyperbolic cosine is the even functionIt has derivative $\sinh x$ and second derivative $\cosh x>0$, so it is a strictly convex function. Its unique global minimum is $\cosh 0=1$, and its image is $[1,\infty )$. For example, $\cosh (-1)=\cosh 1\approx 1.543081$. On the unit hyperbola, the hyperbolic cosine is the horizontal coordinate; as a graph against $x$, it is a symmetric bowl.

Its Taylor series beginsThe positive fourth-order term explains why the bowl resembles a parabola near its minimum yet differs from one away from it. Restricting to $x\ge 0$ gives the inverse hyperbolic cosine.

Hyperbolic functions combine the exponential function to describe a hyperbola much as trigonometric functions describe a circle. Their two basic members are the hyperbolic cosine and hyperbolic sine, defined for every real number $t$ byTheir derivatives exchange the two functions:These formulas follow by applying differentiation to the exponential function definitions. ^[ref]

Adding and subtracting gives $\cosh t+\sinh t=e^t$ and $\cosh t-\sinh t=e^{-t}$. Multiplying these equalities proves the hyperbolic Pythagorean identity:Consequently the parametric equation $(x,y)=(\cosh t,\sinh t)$ traverses the entire right branch of the unit hyperbola $x^2-y^2=1$, where $x\ge 1$. For example, $t=0$ gives $(1,0)$, while $t=\pm 1$ gives approximately $(1.543,\pm 1.175)$. Compare $(\cos \theta ,\sin \theta )$ on the unit circle, where the squared coordinates add to one. The minus sign changes the geometry from a closed circle to an unbounded hyperbola.

The other members are ratios and reciprocals: hyperbolic tangent is $\tanh t=\sinh t/\cosh t$, hyperbolic cotangent is $\coth t=\cosh t/\sinh t$, and hyperbolic secant is $\mathrm{sech}t=1/\cosh t$; denominators must be nonzero. The hyperbolic sine and hyperbolic cosine both solve the second-order linear differential equation $f^{\prime \prime }=f$. In fact $f(t)=A\cosh t+B\sinh t$ has $f(0)=A$ and $f^{\prime }(0)=B$, making this basis convenient for an initial value problem. Their geometry also makes hyperbolic substitution useful for integrals containing $\sqrt{1+u^2}$, and the hyperbolic cosine gives the physical catenary of a hanging chain.

In the Known-variance Gaussian mixture EM example, initialize both means at the same value $m$. The mixture responsibilities are then $1/2$ for every observation. Each M-step sets both means to the sample mean, zero here; the next E-step again gives $1/2$. Thus $(0,0)$ is a fixed point with log-likelihood $-14.593631199$, despite the better fit found from $(-1,1)$.

One can check the failure without trusting an iteration plot. Separate the means as $(-a,a)$ and use the hyperbolic cosine identity:The Taylor expansion uses ${\sum }_i{y}_i^2=18.16$. Small nonzero separation therefore improves the log-likelihood, whereas translating both means to $m$ changes it by $-3m^2$. This stationary point has both improving and worsening directions. Identical initialization cannot discover the separation because every observation keeps identical label probabilities. Trying several distinct initial fits and comparing their final log-likelihoods is useful, but does not prove global optimality.

Keeping the variance fixed also removes a different pathology. With known $\sigma =1$, every component probability density function is at most $(2\pi {)}^{-1/2}$, so $\ell \le -3\log (2\pi )$. If separate component variances are allowed to shrink without restriction, one mixture component can concentrate at an observation while another retains positive density at the other observations, making the likelihood function unbounded. The fixed positive variance in this example prevents that collapse; it does not remove initialization traps.

For $x\ge 1$, the inverse hyperbolic cosine on its nonnegative real branch is $\mathrm{arcosh}x=\log (x+\sqrt{x^2-1})$. It is the real cosine member of the inverse hyperbolic functions.

Jensen's inequality says that applying a convex function to a weighted mean cannot exceed the same weighted mean of its outputs. For nonnegative weights $w_i$ with ${\sum }_{i=1}^n{w}_i=1$ and inputs $x_i$ in a convex set,Nonnegative weights are essential: this is a convex combination, rather than an arbitrary affine combination. For two inputs it is precisely the defining inequality for a convex function; repeatedly grouping inputs gives the finite weighted version.

Take $f(x)=x^2$, weights $1/4$ and $3/4$, and inputs $1$ and $3$. Their weighted mean is $2.5$, so the two sides areThe Jensen gap is $0.75$. The inequality measures the effect of replacing distinct inputs by one representative input before applying a nonlinear function.

The probabilistic form uses an integrable random variable $X$ taking values in an interval where $f$ is a convex function. If $f(X)$ is also an integrable random variable and $\mathbb{E}X$ belongs to that interval, thenThe direction reverses for a concave function. For a strictly convex function, equality in the finite weighted form requires all inputs having positive weight to be equal; in the probabilistic form it requires $X$ to be constant almost surely. An affine function gives equality for any admissible distribution. ^[ref].

For the quadratic function $f(x)=x^2$ and a random variable with a finite second moment, the difference between the two sides is exactly its variance:Consider two equally likely values $2-d$ and $2+d$. Their expected value stays at $2$ as $d$ increases, but their variance is $d^2$. Thus the expected value of the square grows from $4$ to $4+d^2$. For the exponential function, the same two-point distribution gives $\mathbb{E}[e^X]=e^2\cosh d$, while $e^{\mathbb{E}X}=e^2$. Both are examples of Jensen's inequality, but their Jensen gaps depend differently on the spread of the inputs.

Another useful choice is the concave function $\log x$ on positive inputs. Its version of Jensen's inequality gives ${\sum }_i{w}_i\log x_i\le \log ({\sum }_i{w}_i{x}_i)$. Applying the increasing exponential function shows that the weighted geometric mean cannot exceed the weighted arithmetic mean. For the equally weighted inputs $1$ and $9$, these are $3$ and $5$. This recovers the arithmetic-geometric mean inequality and illustrates how choosing one function can generate an entire family of bounds.

A short proof idea explains the probabilistic form. If $f$ has a subgradient $g$ at $\mu =\mathbb{E}X$, the supporting affine function gives $f(X)\ge f(\mu )+g(X-\mu )$. Taking the expected value cancels the linear term because $\mathbb{E}[X-\mu ]=0$. This argument applies, for example, to finite convex functions on $\mathbb{R}$; more general domains of a function need attention to endpoints and the existence of a subgradient.

Johan Jensen's 1906 paper “Sur les fonctions convexes et les inégalités entre les valeurs moyennes” is an early source for the connection between convex functions and inequalities for means. ^[ref].

Scale the catenoid by its waist radius $a>0$ and choose a unit normal with outward radial component:The hyperbolic cosine controls the profile $r(z)=a\cosh (z/a)$. Direct differentiation yields the first fundamental form and second fundamental formThe coordinate tangent vectors are orthogonal, so the meridional and circumferential principal curvatures areThe opposite principal curvatures cancel at every point, not just at the waist. This is precisely why the catenoid is a minimal surface, while its nonzero second fundamental form and negative Gaussian curvature show that it is not a plane.

Put ${ϵ}_k={\hslash }^2{k}^2/(2m)>0$ for $\mathbf{k}\ne 0$ and $g=u\rho$, using $\rho =N/L^3$. Since the zero mode has zero kinetic energy, the exact kinetic energy in this trial state isTogether with part (d), the fixed-mean-number Bogoliubov variational energy isFor repulsive contact interaction, $u>0$, each ${ϵ}_k+g>g>0$. Using the double-angle hyperbolic-function identities,soThe stationary point therefore satisfiesThis is a minimum, not just a vanishing derivative. Define the positive Bogoliubov quasiparticle dispersionAt the displayed solution ${\theta }_k^{\ast }$, $\cosh 2{\theta }_k^{\ast }=({ϵ}_k+g)/E_k$ and $\sinh 2{\theta }_k^{\ast }=g/E_k$. HenceThe hyperbolic cosine has its unique minimum at zero, proving global minimization of each retained term. Treating equal-magnitude wavevectors together simply multiplies the same function by their multiplicity. At a fixed ultraviolet cutoff, the resulting energy change from $\theta =0$ isIndeed $E_k<{ϵ}_k+g$. More directly, for small positive $\theta$,The pair vacuum expectation value lowers the interaction energy at first order in $\theta$, whereas the nonzero occupation number and its kinetic energy cost start at second order. The two-mode squeezed vacuum admixes opposite-momentum pairs into the condensate, introducing spatial correlations that reduce the chance of two bosons coinciding. These correlations lower repulsive contact interaction enough to outweigh the added kinetic energy. A small quantum depletion of a Bose condensate can therefore lower the energy below the uncorrelated condensate value.

Two qualifications delimit this result. First, the paper does not state the sign of $u$. More generally a finite stable minimum exists only when ${ϵ}_k+g>|g|$ for every retained nonzero mode; repulsion ensures this. For example, with $g=-{ϵ}_k$ in some mode, $f_k(\theta )={ϵ}_k\sinh (2\theta )/2$ tends to $-\infty$ as $\theta \to -\infty$, so an unconditional minimum claim for arbitrary $u$ is false. Also the formula minimizes the truncated Bogoliubov theory energy, not the exact quartic Hamiltonian including its discarded terms, and it must remain in the regime of small quantum depletion of a Bose condensate.

Second, a three-dimensional unregulated contact interaction cannot give a finite absolute beyond-leading-order ground-state energy by this bare sum. For large $k$, $E_k-{ϵ}_k-g=-g^2/(2{ϵ}_k)+O({ϵ}_k^{-2})$, whose three-dimensional sum has an ultraviolet divergence. A finite ultraviolet cutoff, or renormalization of a three-dimensional dilute-gas contact coupling matched to a physical scattering length, is needed. This does not alter the modewise minimization or the finite-cutoff lowering mechanism just derived.

The hyperbolic cosine is entire and $\cosh 0=1$. Continuity makes it nonzero in a neighborhood of zero, so its reciprocal is holomorphic there and has a convergent Taylor series. Since $\cosh$ and its reciprocal are even, all odd coefficients vanish.

Multiply the proposed expansion by $\cosh z$:Coefficient comparison gives $c_2+1/2=0$ and $c_4+c_2/2+1/24=0$, henceThe zeros of $\cosh z$ solve $e^{2z}=-1$, so they are $z=i(\pi /2+k\pi )$, $k\in \mathbb{Z}$. The nearest ones are $\pm i\pi /2$, and they are simple because $\sinh$ is nonzero there. The hyperbolic secant has genuine poles at these points and is holomorphic in the disk of radius $\pi /2$. The radius of convergence of its Taylor series is therefore

Set $f(z)=\cosh (\pi z/a)$. The hyperbolic cosine addition formula givesFor $x>0$ and $0<y<a$, both factors in the imaginary part are positive. Thus the image lies in the upper half-plane. The three walls map onto the entire real axis:Also $f^{\prime }(z)=(\pi /a)\sinh (\pi z/a)\ne 0$ in the interior.

To verify that this is a one-to-one conformal map onto the whole upper half-plane, factor it through $s=e^{\pi z/a}$. This first map is one-to-one onto $|s|>1$, $\mathrm{Im}s>0$, and $f(z)=(s+s^{-1})/2$. Given any $\mathrm{Im}\zeta >0$, the roots of $s^2-2\zeta s+1=0$ have product one. Neither lies on $|s|=1$, since that would make $\zeta$ real. Exactly one root therefore has $|s|>1$, andforces that root into the upper half of the exterior disk. It has a unique preimage $z=(a/\pi )(\log |s|+i\arg s)$ with $0<\arg s<\pi$. This proves both surjectivity and injectivity. The construction is the hyperbolic-cosine map of a half-strip.

The source location $c$ is complex, with $\mathrm{Re}c>0$ and $0<\mathrm{Im}c<a$. Put $C=f(c)$, so $\mathrm{Im}C>0$. The method of images places an equal line source at $\bar{C}$ below the real axis:For real $\zeta$, the product of the two logarithm arguments is $|\zeta -C{|}^2>0$. Hence the streamfunction is constant along the wall, on locally consistent branches of the complex logarithm, and there is no normal velocity. Pulling back gives the requested complex potential of a two-dimensional potential flow:This is the line source in a closed-end semi-infinite strip, with no extra imposed flow. Since $f^{\prime }(c)\ne 0$, near the physical source $f(z)-f(c)=f^{\prime }(c)(z-c)+O((z-c{)}^2)$. Thus $w=(Q/(2\pi ))\log (z-c)+$ a regular term, verifying that the source strength remains $Q$ under the conformal map. The image source has no preimage inside the physical strip. Finally, as $x\to \infty$, $w(z)\sim Qz/a+$ a constant, so the far-field velocity is $(Q/a,0)$ and the channel carries exactly the injected volume flux $Q$.

Take eastward $x$, upward $z$, positive wavenumbers $k,K$, and a real instantaneous amplitude $A$. A common growing factor does not affect the phase geometry. Put $C=\cosh (Kz)$ and $S=\sinh (Kz)$, using the hyperbolic cosine and hyperbolic sine. The prescribed streamfunction iswhereFollowing a fixed phase givesTherefore the streamfunction pattern tilts westward with height. The lower boundary leads the upper boundary eastward.

In the conventional quasi-geostrophic streamfunction normalization, hydrostatic balance gives $b^{\prime }=f_0{\psi }_z^{\prime }$. ThereforeFor the phase calculation, multiplication by the constant $f_0$ is immaterial. Define a continuous phase ${\delta }_b=atan2(\alpha C,S)$; this avoids a spurious discontinuity at $z=0$. ThenThe buoyancy perturbation pattern tilts eastward with height, opposite to the streamfunction pattern. At $z=0$, ${\psi }^{\prime }=A\cos (kx)$ while $b^{\prime }=-f_{0}AK\alpha \sin (kx)$, so the two fields are a quarter wavelength apart there.

Both perturbation amplitudes are largest at the boundaries. Indeed,which increase with $|z|$. The physical reason is particularly clear through stratified quasi-geostrophic inversion: the Eady model has no interior background potential-vorticity gradient, and a growing normal mode has zero interior potential vorticity anomaly. The independent thermal anomalies reside at the rigid upper and lower boundaries, where meridional motion advects the basic horizontal buoyancy gradient. Each Eady edge wave induces an interior balanced flow that decays away from its boundary on scale $K^{-1}$. Their interacting, phase-locked combination produces the growing baroclinic instability. Boundary confinement of the thermal sources, rather than a rigid-wall requirement ${\psi }^{\prime }=0$, explains the amplitude pattern.

For a sketch with a quarter-wavelength difference between the boundary streamfunction patterns, chooseAn upper positive streamfunction crest is then ${\lambda }_x/4$ west of its lower counterpart, where ${\lambda }_x=2\pi /k$. The buoyancy perturbation crests slope the other way. The quarter-wavelength boundary assumption is used for the streamfunction; differentiating in $z$ fixes the buoyancy perturbation and does not generally give it the same boundary phase separation.

The connection to poleward heat transport can be checked without relying on the sketch. The meridional geostrophic flow and buoyancy perturbation areAverage over one zonal wavelength, using equal averages $1/2$ for squared sine and cosine and zero for their product:In the Northern Hemisphere this is positive: warm anomalies preferentially move northward and cold anomalies southward. For the same eastward-increasing basic thermal wind in the Southern Hemisphere, $f_0<0$ and the basic buoyancy gradient reverses; the negative northward eddy buoyancy flux is again poleward. For thermal expansion coefficient ${\alpha }_T>0$ and specific heat capacity at constant pressure $c_p$, the corresponding sensible heat flux in a thermal Boussinesq approximation is ${\rho }_0{c}_p\overline{v^{\prime }{T}^{\prime }}={\rho }_0{c}_p\overline{v^{\prime }{b}^{\prime }}/(g{\alpha }_T)$. The growing amplitude makes this flux grow as $A^2$. The tilt supplies the correlation absent in an untilted disturbance and reduces the background available potential energy.

Work in the standard class of real constant-coefficient linear multistep methods using only first-derivative evaluations. Writewith ${\alpha }_s\ne 0$. The Dahlquist equivalence theorem states that such a method converges on each fixed finite time interval, for sufficiently regular Lipschitz initial value problems and every set of consistent starting values, if and only if it is consistent and zero-stable. Consistency means $\rho (1)=0$ and ${\rho }^{\prime }(1)=\sigma (1)$. Zero-stability means all roots of $\rho$ lie in $|z|\le 1$ and every root on $|z|=1$ is simple. In particular the consistency root at $1$ is simple.

Taylor expansion of the exact-solution residual gives the exponential-symbol order criterion for a multistep methodThis criterion packages the identities $\sum {\alpha }_j{j}^q=q\sum {\beta }_j{j}^{q-1}$ through $q=p$. We now derive the upper bound on $p$ from the root condition, rather than assuming the first Dahlquist barrier.

Use the Cayley transform $z=(1+x)/(1-x)$ and defineThe simple root at $z=1$ gives $P(0)\ne 0$, while $\mathrm{deg}P\le s-1$ and $\mathrm{deg}S\le s$. Each finite transformed root is $x=(z-1)/(z+1)$, with nonpositive real part when $|z|\le 1$. A possible root at $z=-1$ simply lowers the degree of $R$. Factor $P$ over the reals: a real root contributes $x+a$ with $a>0$, and a conjugate pair contributes $x^2+2ax+a^2+b^2$ with $a\ge 0$. Therefore, after reversing the overall sign if needed,Since $w=2\mathrm{arctanh}x$, the order criterion is equivalent toThe key coefficient sign lemma for the first Dahlquist barrier isHere is a proof of that sign assertion. Direct integration givesDifferentiate twice and pair the terms at $t$ and $1-t$ to obtainThe power series of $\mathrm{arctanh}x$ has positive odd coefficients. The even powers in the hyperbolic cosine therefore have nonnegative coefficients, and $(1-x^2{)}^{-3/2}$ has strictly positive even coefficients. Every even coefficient of $F^{\prime \prime }$ is consequently negative. Also $F$ is even and $F(0)=1$, proving the assertion.

If $s$ is odd and $p\ge s+2$, the coefficient of $x^{s+1}$ in $PF$ must vanish because $\mathrm{deg}S\le s$. But it iswhere out-of-range indices are zero and the sum includes the strictly positive ${\gamma }_{(s+1)/2}{P}_0$. This is impossible, so $p\le s+1$. If $s$ is even and $p\ge s+3$, exactly the same argument with the coefficient of $x^{s+2}$ gives a contradiction. Thus $p\le s+2$ in that case. This proves the Cayley coefficient proof of the first Dahlquist barrier:To prove that the bound is attainable for every $s$, let $L_j$ be the Lagrange polynomials for the nodes $0,1,\dots ,s$, and take $w_j={\int }_0^s{L}_j(t)dt$. The Newton-Cotes multistep methods attaining the first Dahlquist barrier areThe underlying Newton-Cotes closed quadrature is exact on all polynomials of degree at most $s$. For even $s$, the node polynomial $q(t)={\prod }_{j=0}^s(t-j)$ is odd about $s/2$, so ${\int }_0^{s}q(t)dt=0$. Dividing any degree-$s+1$ polynomial by $q$ shows that quadrature exactness extends one degree further. Integrating the exact derivative therefore gives method order at least $s+1$ for odd $s$ and at least $s+2$ for even $s$.

Finally $\rho (z)=z^s-1$ has only simple unit-circle roots, so these methods satisfy the root condition. The weights sum to $s={\rho }^{\prime }(1)$, proving consistency. The Dahlquist equivalence theorem gives convergence with consistent starts; the proved upper bound makes the attained formal orders exact; starting values accurate to those orders also give the corresponding global accuracy. Hence the highest convergent order is $2\lfloor (s+2)/2\rfloor$. This result concerns the stated linear first-derivative class, not unrestricted multiderivative or nonlinear formulas.

The hyperbolic cosine has the absolutely convergent expansion $\cosh (1/2)={\sum }_{k\ge 0}1/[4^k(2k)!]$. Suppose its value were a rational $a/b$, with $b>0$. For sufficiently large $N$, $b\mid (2N)!$. Multiplying by $4^N(2N)!$ makes the assumed value an integer, and every term of the partial sum through $k=N$ is also an integer. Their difference is the positive remainderLet $q=[4(2N+1)(2N+2){]}^{-1}$. Every new pair of factorial factors is at least $(2N+1)(2N+2)$, so $0<R_N\le {\sum }_{j\ge 1}{q}^j=q/(1-q)<1$ for $N\ge 1$. No integer lies strictly between zero and one. This contradiction proves $\cosh (1/2)$ is irrational. The argument uses a finite integer partial sum and a controlled tail; it does not infer irrationality merely from an infinite expansion.