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At second order, the displaced-boundary Navier slip boundary condition gives $({\psi }_2{)}_Y-\delta [({\psi }_2{)}_{YY}-({\psi }_2{)}_{XX}]=(1+2\delta ){\sin }^2(X-\tau )$ at $Y=0$. Its mean shear is zero in a bounded half-space flow, so the Navier-slip Taylor swimming sheet dimensionless speed is $U^{\ast }={ϵ}^2(1+2\delta )/2+O({ϵ}^4)$. The sheet travels opposite to the wave. This fixed-slip asymptotic expansion predicts a speed enhancement factor $1+2\delta$.

Two sectorial fundamental solutions of the same linear system differ on their overlap by a constant invertible right factor, since $(Y_j^{-1}{Y}_{j+1}{)}^{\prime }=0$. With a common formal normalization these factors encode the exponentially small ambiguities invisible to the asymptotic expansion. They are called Stokes matrices. Equivalently their inverses act on column vectors of solution coefficients; the convention must be specified when ordering products.

Exponentially small contributions can change their asymptotic representation when a complex parameter crosses a Stokes direction. An all-orders asymptotic expansion in inverse powers alone does not resolve these terms.

Restrict the horizontal-field magnetoconvection dispersion relation to $k=0$, $x=m^2>0$. The steady neutral curve becomes $R={\pi }^{2}Q+(Q+3{\pi }^2)x+x^2+3{\pi }^4+{\pi }^6/x$. Its unique minimum satisfies $x^2(Q+3{\pi }^2+2x)={\pi }^6$, since $R^{\prime \prime }(x)=2+2{\pi }^6/x^3>0$. Expanding at large Chandrasekhar number gives $x\sim {\pi }^3{Q}^{-1/2}$ and the stated asymptotic expansion. Unlike field-aligned rolls, these disturbances bend the field, and their stationary threshold grows with field strength.

// Target: fluid-mechanics.bigb

For the asymptotically free two-loop running coupling $da/d\log \mu =-{\beta }_0{a}^2-{\beta }_1{a}^3$, with ${\beta }_0>0$, let $y=1/a$, $c={\beta }_1/{\beta }_0$, and $L=\log (\mu /\Lambda )$. A choice of the scale $\Lambda$ gives the exact implicit equation $y-c\log ((y+c)/{\beta }_0)={\beta }_{0}L$ on the ultraviolet branch. Its expansion isThe remainder after the first two terms is therefore $O(\log L/L)$, and is not $O(1/L)$ when ${\beta }_1\ne 0$. A fixed rescaling of $\Lambda$ cannot change the coefficient of $\log L/L$. This distinction matters when quoting errors for asymptotic expansions.

For $m^2>0$, the zero-dimensional scalar field theory with Euclidean action $m^2{x}^2/2+\lambda x^4/4!$ has a Gaussian integral normalization $I(0)=\sqrt{2\pi /m^2}$. Its Feynman rules assign $1/m^2$ to each propagator, $-\lambda$ to each quartic vertex, and the reciprocal Feynman-diagram symmetry factor to each vacuum diagram. The integral itself includes disconnected vacuum diagrams; its logarithm selects connected ones. At second order the three possible numbers of edges joining the two vertices are zero, two and four, with Feynman-diagram symmetry factors $128,16,48$ respectively. Their weights add to $35{\lambda }^2/(384m^8)$.

The Wick contractions give the general coefficient $I(0)(-1{)}^v(4v-1)!!/[v!(4!{)}^v{m}^{4v}]$. Successive absolute coefficients have ratio $(4v+3)(4v+1)/[24(v+1)m^4]$, which tends to infinity. Thus this is an asymptotic expansion as $\lambda \downarrow 0$, rather than a convergent power series. For every fixed truncation the Taylor theorem remainder estimate for $e^{-u}$, $u\ge 0$, can be integrated against the Gaussian distribution to justify the corresponding remainder bound.