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A strong stability preserving Runge-Kutta method writes its stages as convex combinations of previously computed states and suitable Forward Euler method steps. It transfers any convex-functional nonincrease property of the forward step, such as a norm bound, under a proportionally scaled time-step restriction.

For example, if $E_k(U)=U+kF(U)$ is nonexpansive in a norm, then $Y=E_k(U)$ and $U_{\mathrm{new}}=(U+E_k(Y))/2$ is also nonexpansive: the triangle inequality bounds the new difference by one half of the initial difference plus one half of the twice-advanced difference. Both are at most the initial difference. A Taylor expansion yields $U_{\mathrm{new}}=U+kF(U)+k^2{F}^{\prime }(U)F(U)/2+O(k^3)$, so this is a second-order Runge-Kutta method. The forward-step hypothesis must hold on the stage states; preservation of one chosen convex bound is not the same as unconditional B-stability.

For a differentiable real-valued function on an open convex set, convexity is equivalent to the displayed inequality for all $x,y$ in its domain. To prove the forward direction, convexity gives $[f(x+t(y-x))-f(x)]/t\le f(y)-f(x)$ for $0<t\le 1$. Take the limit as $t\downarrow 0$ using first-order Taylor expansion.

Conversely, put $z=(1-t)x+ty$ and apply the supporting inequality at $z$ to both $x$ and $y$. Multiply the two inequalities by $1-t$ and $t$ and add them. The gradient terms cancel because $(1-t)(x-z)+t(y-z)=0$, leaving $f(z)\le (1-t)f(x)+tf(y)$. This is the defining convexity inequality. The endpoint values $t=0,1$ hold as equalities.

Set $A=p^2-m^2\ne 0$. The denominator product is exactly $(A+k^2{)}^2-4(p\cdot k{)}^2$. Its reciprocal therefore has the Taylor expansionThe coefficients follow by expanding $1/(1+x)=1-x+O(x^2)$ with $x=2k^2/A-4(p\cdot k{)}^2/A^2+O(k^4)$. The expansion is a local algebraic statement away from denominator zeros; integrating it through Feynman propagator poles requires a regulator and the correct Feynman i-epsilon prescription.

For a displaced wall $y=ϵh_1(x,t)$, write fluid velocity as $\mathbf{u}=ϵ{\mathbf{u}}_1+{ϵ}^2{\mathbf{u}}_2+\cdots$ and prescribed wall velocity as $ϵ{\mathbf{v}}_1+{ϵ}^2{\mathbf{v}}_2+\cdots$. A Taylor expansion about $y=0$ gives ${\mathbf{u}}_1(0)={\mathbf{v}}_1$ and ${\mathbf{u}}_2(0)={\mathbf{v}}_2-h_1{\partial }_y{\mathbf{u}}_1(0)$. The second-order correction converts first-order velocity gradients into a mean boundary flow.

Multiply the two exponential series in $e^{-(x-a{)}^2}=e^{-a^2}{e}^{2ax}{e}^{-x^2}$ to obtain the displayed Taylor expansion. Its first three nonzero terms depend on the real parameter $a$. For $a\ne 0,\pm 1/\sqrt{2}$, they end at degree two. At $a=0$, odd coefficients vanish and the first three terms are $1-x^2+x^4/2$. At $a=\pm 1/\sqrt{2}$, the quadratic coefficient vanishes, leaving $e^{-1/2}(1+2ax-4a{x}^3/3)$ as the first three nonzero terms. Parameter-dependent vanishing coefficients must be checked when a question counts nonzero terms rather than prescribing an expansion order.

Differential gravitational acceleration, multiplied by the affected mass, across a spatially extended body. In the linear approximation the acceleration difference is $T\Delta \mathbf{r}$, where $T$ is the tidal tensor. Stellar tidal forces limit the domain of a circumplanetary orbit; sufficiently strong ones can produce tidal disruption.

To isolate a tidal force, subtract the gravitational acceleration shared by the body as a whole. Let a reference test particle follow $\ddot{\mathbf{R}}=\mathbf{g}(\mathbf{R})$, and let another have position $\mathbf{R}+\mathbf{x}$. In a translating accelerated reference frame with parallel, nonrotating axes,for a source point mass, when $|\mathbf{x}|/R\ll 1$. Here $\mu =GM$ is its standard gravitational parameter and the last term uses Big O notation. This Taylor expansion removes the common inward pull and keeps the spatial variation. The near side has an excess pull toward the source; the far side has a deficit of inward pull, which is an outward relative acceleration. Both sides can stretch apart while both still accelerate toward the source.^[ref]

For a material element of mass $m$, the residual external force in this translating accelerated reference frame is $m[\mathbf{g}(\mathbf{R}+\mathbf{x})-\mathbf{g}(\mathbf{R})]$. Actual deformation also depends on elasticity, pressure and self-gravity. A rigid body approximation suppresses deformation but can still have internal forces that resist the tidal force. The point-mass tidal stretching and squeezing pattern describes freely falling, noninteracting test particles, rather than promising the same shape for every solid, star or ocean. A rotating reference frame introduces centrifugal acceleration and Coriolis acceleration, which must be included separately.

The dimensionless product of surface-wave amplitude and wavenumber, $ka$, measures orbital displacement relative to the wavelength. Small wave steepness permits a Taylor expansion of the velocity sampled by moving parcels. For a deep-water gravity wave, orbital amplitudes and the local steepness decrease exponentially with depth.

A phase object for which the relevant relative phase variations are much smaller than one radian. The Taylor expansion $e^{i\varphi }=1+i\varphi +O({\varphi }^2)$ separates a direct reference term from a small specimen term. A constant overall phase can be removed before making this approximation: phase-contrast microscopy measures spatial variations, not an absolute common phase.

A real voltmeter has finite input electrical resistance $R_d$; a galvanometer also has a finite winding electrical resistance. Away from balance, its electric current changes both intermediate electric potentials. This is voltmeter loading when the detector is a voltmeter. Model the detector as a passive resistor, with no offset or injected electric current. Thévenin's theorem supplies the full response without pretending the connected bridge remains two independent voltage dividers.

With the detector removed, the equivalent source is $V_{\mathrm{oc}}$. Replace the ideal supply by a wire to calculate the source electrical resistance seen between $B,C$: nodes $A,D$ merge, making $R_1,R_2$ parallel and $R_3,R_x$ parallel. ConsequentlySince the attenuation factor is strictly positive for $R_d>0$, the null remains $R_2{R}_3=R_1{R}_x$. A smaller detector electrical resistance reduces the measured electric potential difference off balance but does not shift the ideal null. Its electric current sensitivity instead depends on $1/(R_{\mathrm{th}}+R_d)$; detector gain and resolution also matter in deciding how closely a null can be located.

In the plotted example, $R_x=1.100\mathrm{k\Omega }$ gives $V_{\mathrm{oc}}=-119.048\mathrm{mV}$ and $R_{\mathrm{th}}=1.02381\mathrm{k\Omega }$. A detector with $R_d=1.000\mathrm{k\Omega }$ instead reads $V_d=-58.824\mathrm{mV}$. Using that loaded reading in the unloaded voltage divider formula would underestimate the deviation of $R_x$ from $1\mathrm{k\Omega }$. Balancing the bridge avoids this particular systematic measurement error.

The local derivative makes the sensitivity explicit. For $R_1=R_2=R_3=R$ and $R_x=R+\delta R$, the first-order Taylor expansion isAt the plotted null, the open-circuit slope is $-1.25mV/\Omega$, compared with $-0.625mV/\Omega$ for $R_d=1\mathrm{k\Omega }$. Increasing the supply improves this ideal sensitivity but also increases Joule heating. A finite source electrical resistance requires recomputing $R_{\mathrm{th}}$; the displayed parallel-pair expression assumes an ideal supply.