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The set of complex numbers $s$ for which the displayed integral is finite is the region of absolute convergence of a Laplace transform. It depends only on the real part $\sigma =\mathrm{Re}s$. Once the condition holds for one $\sigma$, it holds for every larger real part. For the exponential function $e^{-2t}$ it is exactly $\mathrm{Re}s>-2$; in general, convergence on the boundary must be checked separately. An analytic continuation outside this set need not equal a convergent defining integral.

For probability distributions $P,Q$, relative entropy satisfies $D_{\mathrm{KL}}(P\Vert Q)\ge 0$, with equality exactly when $P=Q$. Thus truthful prediction uniquely minimizes expected log-loss whenever the comparison is well defined, explaining the strictly proper scoring rule property of the logarithmic scoring rule.

For a short proof with probability density functions $f,g$ and $g>0$, set $r=f/g$. Their integrals both equal one, givingThe strictly convex function $r\log r-r+1$ is minimized only at $r=1$, so equality means $f=g$ almost everywhere. Its value at zero is one by $0\log 0=0$. Where $g=0<f$, relative entropy is infinite; where both vanish, the contribution is zero. These conventions extend the argument beyond positive reference probability density functions.

For a one-dimensional free particle with $m>0$ and $p>0$, take the positive total energy from the relativistic energy-momentum relation:A fixed plane wave quantum phase obeys $kx-\omega t=\text{constant}$. Its phase velocity is thereforeFor a narrow wave packet centered at $p_0$, expand its dispersion relation as $E(p)\simeq E(p_0)+E^{\prime }(p_0)(p-p_0)$. In the Fourier inversion integral building the wave packet, the envelope then depends on $x-E^{\prime }(p_0)t$. Thus its group velocity isThe superluminal phase velocity follows a repeating phase marker; it does not describe a localized signal or superluminal transport of energy. Higher derivatives of the dispersion relation produce wave packet spreading, so the translated-envelope argument is a narrow-band approximation. For $v=0.6c$, the Lorentz factor is $1.25$, $p=0.75mc$ and $E=1.25m{c}^2$: the group velocity is $0.6c$ while the phase velocity is $5c/3$.

The total-energy convention matters. Define $\psi =e^{im{c}^{2}t/\hslash }\Psi$ to remove the common rest energy from a wavefunction $\Psi$ with total-energy time dependence. ThenThis is exactly the free-particle momentum convention used by the Schrödinger equation here. Removing a common time-dependent global phase changes the reported phase velocity, while leaving the de Broglie wavelength, relative spatial quantum phases and group velocity unchanged. Adding $m{c}^2$ again to a total energy would count the rest energy twice.

A vector field singular on an axis need not satisfy the classical divergence theorem on the original closed region. Remove a thin tube $\rho <\epsilon$, apply the theorem to the remaining smooth region, then examine the added cylindrical boundary. If its flux tends to zero and the remaining volume integral converges, the improper boundary flux equals the improper integral of the divergence. For $\mathbf{F}=(\log {\rho }^2,\log {\rho }^2,z/\rho )$ with $\rho =\sqrt{x^2+y^2}$, the cylindrical unit normal has zero $z$ component, so the added flux is $O(\epsilon |\log \epsilon |)$ on a bounded axial interval. The divergence $(2\cos \varphi +2\sin \varphi +1)/\rho$ is locally integrable under the cylindrical coordinates measure $\rho d\rho d\varphi dz$. Hence no finite line-source correction survives, despite the field itself being unbounded.

Let $\Gamma$ be a piecewise smooth contour forming a simple closed curve, positively oriented: travel counterclockwise, with the interior on the left. Suppose $f$ is holomorphic on an open set containing $\Gamma$ and its interior, except at finitely many isolated singularities $a_1,\dots ,a_m$ strictly inside. ThenNo isolated singularity may lie on $\Gamma$: the ordinary contour integral must be defined there. Reversing the direction of the contour reverses the sign of the answer. The theorem applies to essential singularities as well as poles. ^[ref]

Return to $f(z)=1/(z^2+1)$. A counterclockwise circle $|z|=2$ encloses both simple poles, soYet $f$ is nonzero everywhere it is defined. A zero contour integral does not imply a zero function, or even absence of poles: the enclosed residues can cancel. By contrast, the counterclockwise circle $|z-i|=1/2$ encloses only $i$, givingThe picture compares these two selections. Changing the shape of a contour while keeping the same enclosed poles leaves the contour integral unchanged, provided the contour deformation stays where $f$ is holomorphic.

This is the practical appeal of the residue theorem: local Laurent coefficients determine a global contour integral. For a rational function, a few simple pole computations can replace a long parametrized calculation. It also makes real improper integrals accessible by treating the real line as part of a closed contour in the complex plane.

To compute $I={\int }_{-\infty }^{\infty }dx/(1+x^2)$, take $R>1$ and travel along the real line from $-R$ to $R$, then return along the upper semicircle $A_R$, parametrized by $z=R{e}^{it}$ for $0\le t\le \pi$. This contour is positively oriented and encloses the simple pole $i$ only; $-i$ is outside. HenceThe added arc must be controlled before taking a limit. On $A_R$, the reverse triangle inequality gives $|z^2+1|\ge R^2-1$, and the arc length is $\pi R$. The estimation lemma therefore yieldsThe denominator grows quadratically while the arc length grows only linearly. This quantitative comparison, rather than a sketch alone, explains why closing the contour does not alter the limiting real improper integral. It even supplies an error bound: the difference between $\pi$ and the finite real integral is at most $\pi R/(R^2-1)$.

For a numerical check, $R=10$ gives the finite real integral $2\arctan (10)\approx 2.94226$, leaving about $0.19934$ to reach $\pi$. The arc bound is $10\pi /99\approx 0.31733$, comfortably larger than that error. The graph shows both quantities approaching zero. The exact tail here can be checked with the antiderivative $\arctan x$; the residue theorem method remains useful for improper integrals where a convenient real antiderivative is unavailable.

Finally, both one-sided improper integrals converge absolutely: for $|x|\ge 1$, $0<1/(1+x^2)\le 1/x^2$, whose tails are integrable. Thus the symmetric limit computes the ordinary improper integral, not merely a Cauchy principal value, andClosing below instead would enclose the simple pole $-i$. With the same left-to-right traversal of the real line, that contour is clockwise, so its contour integral is $-2\pi i\mathrm{Res}(f,-i)=\pi$ again. This provides a useful sign check: the opposite residue is accompanied by the opposite direction around the enclosed region.

The same strategy extends to many other rational functions: choose a contour, identify its enclosed isolated singularities, compute their residues, and bound the added pieces. The choice of contour and the bound are as much part of the calculation as the residues. ^[ref]

A complete orthonormal basis in a Hilbert space resolves the identity as ${\sum }_j|j⟩⟨j|=I$. In infinite dimension the partial sums converge strongly: their action on each fixed vector converges in norm. Generalized position and momentum eigenstates instead give distributional integral resolutions, such as $\int dx|x⟩⟨x|=I$, with their normalization interpreted through the Dirac delta function.

The ridge-push force in a plate model is the depth integral of the horizontal pressure difference between hot and cold compensated columns. To leading order, the old plate's linear excess density gives $F=g{\int }_0^{a}s\delta \rho (s)ds={\rho }_{m}g\alpha \Delta T{a}^2/6$. Here $s$ is depth below the old seafloor. Retaining the geometrical contribution from its lowered surface adds $g({\rho }_m-{\rho }_w)h^2/2$ within the linear-density model. Force per ridge length has units $\mathrm{N}{\mathrm{m}}^{-1}$, distinct from stress.

A right Riemann sum on a partition $a=x_0<\cdots <x_N=b$ samples each subinterval at its right endpoint. For a continuous function its limit as the mesh tends to zero is the integral, by the Riemann sum convergence theorem. On a uniform partition of width $h$, comparison of endpoint weights in the composite trapezoidal rule gives $R_N-T_N=h(f(b)-f(a))/2$. Thus $N^{-5}{\sum }_{j=1}^N{j}^4$ is a right Riemann sum for ${\int }_0^1{x}^{4}dx$, and its limit is $1/5$.

Expanding the squared velocity $v_G+\omega \times r$ and integrating over mass removes the mixed term because the centre-of-mass first moment is zero. The remaining rotational integral is half the scalar product of angular velocity and angular momentum about the centre of mass. Principal axes diagonalize that rotational quadratic form.

The ideal rocket equation gives the change in signed velocity produced by ejecting rocket propellant. Assume one-dimensional Newtonian mechanics in an inertial frame, no external force, and a fixed positive rocket exhaust speed $u$ directed rearward relative to the instantaneous rocket. The rocket's instantaneous mass $M$ includes its remaining rocket propellant; burning it gives $dM<0$. For initial and final masses $M_{\mathrm{initial}}\ge M_{\mathrm{final}}>0$,Here $\ln$ is the natural logarithm, and its argument is the rocket mass ratio. This is the constant-rocket exhaust speed, zero-pressure-correction version of the ideal rocket equation. ^[ref]

For a toy rocket with $u=2km/s$, $M_{\mathrm{initial}}=1000\mathrm{kg}$ and rocket dry mass $M_{\mathrm{final}}=250\mathrm{kg}$ after all rocket propellant is spent, the rocket mass ratio is $4$ andBurning from $1000$ to $500\mathrm{kg}$ gains $2\ln 2km/s\simeq 1.3863km/s$. Burning from $500$ to $250\mathrm{kg}$ gains exactly the same velocity increment, using only $250\mathrm{kg}$ of rocket propellant instead of $500\mathrm{kg}$. Both burns halve the remaining mass. The second burn accelerates a lighter rocket; equal fractional mass losses give equal velocity gains.

The natural logarithm makes the rocket equation both useful and revealing: multiplying successive rocket mass ratios adds their velocity gains. Conversely, the required rocket mass ratio is the exponential function $M_{\mathrm{initial}}/M_{\mathrm{final}}=e^{\Delta v/u}$. Each extra $u\ln 2$ of ideal velocity gain doubles the required initial mass for fixed final mass. This explains why a larger rocket exhaust speed is valuable and why carrying ever more rocket propellant has diminishing returns. Changing the burn duration alone does not change this ideal endpoint result.

For a short proof outline, follow the rocket together with a tiny parcel about to be expelled. Before ejection, their total momentum is $Mv$. After ejection, the remaining rocket has mass $M+dM$ and velocity $v+dv$, while the parcel has positive mass $-dM$ and, to first order, inertial velocity $v-u$. Momentum conservation givesDiscarding the second-order product $dMdv$ gives $Mdv=-udM$; taking the integral gives $\Delta v=-u{\int }_{M_{\mathrm{initial}}}^{M_{\mathrm{final}}}dM/M$. Evaluating the integral yields the natural logarithm above. This is a variable-mass system calculation: the rocket alone is not a closed system, and its momentum changes as the expelled material carries away momentum. The fixed collection consisting of the rocket and all its expelled material has conserved total momentum when external forces vanish.

Fixed rocket exhaust speed means fixed relative speed at release, not fixed inertial velocity. As the rocket's velocity $v$ changes, newly emitted material has changing inertial velocity $v-u$; previously released parcels coast at their own release velocities in this ideal model. If $v>u$, the exhaust even moves forward in the inertial frame, although it is emitted rearward relative to the rocket. No surrounding medium is needed for this exchange of momentum.

Real flight adds gravity, drag force and possibly a pressure correction to rocket thrust. A varying rocket thrust due solely to a changing mass loss rate still gives the same ideal rocket equation if $u$ is constant; it changes the acceleration and burn duration. A changing rocket exhaust speed instead requires $\Delta v=-\int udM/M$, and external forces require their own contribution to the velocity change. The existing constant mass-loss propulsion with quadratic drag and constant-burn rocket with ground friction illustrate such additional forces; the forward-exhaust rocket equation reverses the ejection direction and therefore the sign of the recoil.

A Runge-Kutta method advances an initial value problem $y^{\prime }=f(t,y)$, $y(t_0)=y_0$, by sampling its vector field at several trial states within one time step. These samples are Runge-Kutta stages; a weighted combination of their slopes produces the next state. The construction improves on the single tangent line used by the Euler method while requiring evaluations of $f$, rather than explicit higher derivatives of the solution. It applies to scalar equations and systems of ordinary differential equations alike. The coefficient choices distinguish methods such as Heun's method, the classical fourth-order Runge-Kutta method and implicit Runge-Kutta methods.

For $s$ stages, write $A=(a_{ij})$ for the stage matrix, $c_i$ for relative time nodes and $b_i$ for output weights. A time step of length $h$ computesHere $k_i$ is a slope, with units of state per time; $h{k}_i$ is a state increment. Another common convention writes $K_i=h{k}_i$, so its stage states are $Y_i=y_n+{\sum }_j{a}_{ij}{K}_j$ and its output is $y_n+{\sum }_i{b}_i{K}_i$. These are equivalent conventions: including $h$ both inside $K_i$ and again in the output would count the time step twice. The textbook formulation uses the increment convention.^[ref]

Usually the Butcher tableau obeys $c_i={\sum }_j{a}_{ij}$, called internal consistency. Stage $i$ then combines slopes over an effective time $c_{i}h$. The second-order conditions with independent Runge-Kutta abscissae explain what changes when time nodes are specified independently. A Runge-Kutta stage is generally an auxiliary approximation, not a separate accepted solution value at its node. Several stages can even have the same time node but different state arguments, as in the classical fourth-order Runge-Kutta method.

The useful idea is to approximate the integral $y(t_n+h)-y(t_n)={\int }_{t_n}^{t_n+h}f(t,y(t))dt$ without knowing the intermediate exact trajectory. Trial states provide places to sample $f$, and Butcher order conditions make the resulting combination agree with the exact Taylor expansion. More stages alone do not guarantee a higher order of a numerical method: their coefficients must satisfy the appropriate conditions. Embedded Runge-Kutta pairs reuse these same samples for a second output and a practical local error estimator.

Let customer $k$ enter a queue at $a_k$ and leave at $d_k$, and count it in $N(t)$ for $a_k\le t<d_k$. On a finite observation interval $[0,T]$, the integral of the queue population is exactly the sum of the lengths of the overlaps of customer visits with that interval, as displayed above; $x_{+}=\max (x,0)$ is the positive part. This deterministic identity includes visits already in progress at time zero and visits still unfinished at time $T$. It needs no probability distribution for arrivals or service times. The area argument is also developed in Richard C. Larson and Amedeo R. Odoni, Urban Operations Research, section 4.4.

For a fully computed example, start empty and observe through $T=6$ minutes. Three customers arrive at $a=(0,1,4)$ minutes, begin service at $b=(0,2,4)$ minutes, and depart at $d=(2,3,5)$ minutes. These times describe a single first come first served server with service times $(2,1,1)$ minutes. Their customer sojourn times are $(2,2,1)$ minutes, and their queueing delays are $(0,1,0)$ minutes. The queue population is respectively $1,2,1,0,1,0$ on the consecutive intervals $[0,1),[1,2),[2,3),[3,4),[4,5),[5,6]$, soThe same area is $2+2+1=5$ when counted by customer. Hence the finite-window averages satisfyThe waiting area is one customer-minute, giving $L_q=1/6$ and $W_q=1/3$ minute; $(1/2)(1/3)=1/6$. The server works for four minutes, so its server utilization is $4/6=2/3$ and $L-L_q=5/6-1/6=2/3$. Repeating this six-minute pattern indefinitely makes all these values long-run averages, without a Poisson process of arrivals.

An observation window that cuts visits short requires the overlap terms, rather than full customer sojourn times. Starting empty, let $A(T)$ count admissions before $T$ and write $w_k=d_k-a_k$. ThenFor the example cut at $T=2.5$ minutes, the area is $1+2+0.5=3.5$ customer-minutes, whereas the full durations of the two admitted customers sum to $4$; the remaining half-minute of the second visit is $R(2.5)=0.5$. A precise sufficient sample path condition for the limiting Little's law is $A(T)/T\to {\lambda }_{\mathrm{eff}}\in (0,\infty )$, $A(T{)}^{-1}{\sum }_{a_k<T}{w}_k\to W<\infty$, and $R(T)/T\to 0$. Dividing the identity by $T$ then gives $L={\lambda }_{\mathrm{eff}}W$. Equating admission and departure rates additionally uses $N(T)/T\to 0$, by the balance $N(T)=A(T)-D(T)$ for an initially empty queue, where $D(T)$ counts all departures.

Scalar diffraction theory approximates light by one complex field satisfying a scalar wave equation. Aperture propagation is represented by an integral of secondary contributions, with vector polarization of waves and detailed electromagnetic edge conditions suppressed. The approximation is useful for many apertures large compared with wavelength, but is not an exact substitute for Maxwell equations at arbitrary apertures.

For $a>0$, this integral is an entire function of $x$: the exponential series converges uniformly on bounded subsets after multiplication by the integrable bound $e^{|x|}{u}^{a-1}$. ThusFor positive real $x$, a change of variables gives $F_a(x)=x^{-a}\gamma (a,x)$, where here $\gamma (a,x)={\int }_0^x{e}^{-t}{t}^{a-1}dt$ denotes the lower incomplete gamma function. The singular-looking factor $x^{-a}$ is removed by the integral over $[0,1]$. Applying integration by parts to the derivative of $u^a{e}^{-xu}$ gives $x{F}_a^{\prime }+a{F}_a=e^{-x}$; differentiating and eliminating $e^{-x}$ proves $x{F}_a^{\prime \prime }+(a+1+x)F_a^{\prime }+a{F}_a=0$.

For $a>0$, let $g(x)=f(ax)$. Whenever the Mellin transform of $f$ is absolutely convergent,The Mellin fundamental strip is unchanged: the corresponding integrals of absolute values differ by the finite positive factor $a^{-\sigma }$. The change of variables $y=ax$ gives $x^{s-1}dx=a^{-s}{y}^{s-1}dy$, proving the rule. For $s=\sigma +i\tau$, the transform's absolute value changes by $a^{-\sigma }$, while the factor $e^{-i\tau \log a}$ changes its phase. On the line $\sigma =0$, wherever absolute convergence holds, the transform's absolute value is preserved.

For the Mellin transform of an interval indicator, replacing $1_{[1,4]}(x)$ by $1_{[1,4]}(2x)=1_{[1/2,2]}(x)$ givesAt the removable singularity $s=0$, both sides have value $\log 4$. Scaling leaves the width measured by the natural logarithm unchanged. Likewise, the exponential function $e^{-ax}$ has Mellin transform $a^{-s}\Gamma (s)$ on $\mathrm{Re}s>0$.

For continuous $G$ and $f$, restrict to the closed supremum-norm ball of radius $R$ in $C([0,1])$. Boundedness on the compact kernel and nonlinearity boxes makes the integral map preserve this ball for sufficiently small positive $\mu$. Uniform continuity of $f$ proves continuity of the map; uniform continuity of $G$ gives an equicontinuous bounded image. Arzelà-Ascoli makes that image relatively compact, and Schauder gives a solution. No global Lipschitz hypothesis or uniqueness conclusion is needed.

The Haar wavelet provides an explicit one-vanishing moment comparison:It cancels a constant polynomial, but not a nonconstant affine function. To obtain two vanishing moments by an equally transparent construction, take three adjacent unit intervals with heights proportional to $1,-2,1$:The L2 norm normalization follows from $1^2+(-2{)}^2+1^2=6$. Direct integration givesThus $\eta$ has exactly two vanishing moments. In fact its coefficient is a finite difference of adjacent local integrals:The weights are those of the second-order central difference. The first moment balances the centres of the three intervals; it is not enough merely to balance their signed areas.

This $\eta$ is a normalized analyzing function, with no orthonormal basis assertion. Indeed, its unit translate has nonzero L2 inner product:Hence these translates already fail orthogonality. The Daubechies wavelet construction imposes additional refinement and orthogonality requirements; two vanishing moments and unit L2 norm by themselves do not supply those requirements.

The second virial coefficient is the leading interaction correction to the ideal gas law: at fixed temperature, $B_2(T)={\lim }_{n\to 0}(Z-1)/n$, where $Z=p/(n{k}_{B}T)$ and $n$ is number density. For identical classical particles in a homogeneous three-dimensional gas, with an isotropic pair potential $u(r)$ and $r$ the centre-to-centre separation,The function $f$ is the Mayer f-function. The factor $4\pi r^{2}dr$ counts the volume of a thin spherical shell; the factor $1/2$ counts each unordered pair once. The result has units of volume. Repulsive regions, $u>0$, contribute positively, while attractive regions, $u<0$, contribute negatively. ^[ref]

This is a striking bridge from a microscopic pair potential to a macroscopic pressure reading. At sufficiently small number density, $B_2>0$ raises pressure relative to an ideal gas at the same temperature and number density, while $B_2<0$ lowers it. A single value of $B_2$, however, cannot determine the entire pair potential: many different functions have the same weighted integral. Measurements over several temperatures supply more information.

The integral must exist. For example, a bounded short-range attraction outside a hard core has no difficulty. If instead $u(r)\sim A/r^s$ as $r\to \infty$, then $1-e^{-u/(k_{B}T)}\sim u/(k_{B}T)$ and the radial tail behaves like $\int r^{2-s}dr$: it converges only for $s>3$. A bare unscreened Coulomb potential therefore falls outside this formula's ordinary short-range setting. This condition on the tail does not by itself establish the stability or convergence of the full virial expansion.

For an exactly integrable toy attraction, choose the following square-well pair potential, with diameter $\sigma >0$, attraction depth $ϵ>0$, and range multiplier $a>1$:Set $T^{\ast }=k_{B}T/ϵ$ and $b_{\mathrm{HS}}=2\pi {\sigma }^3/3$, the second virial coefficient of a hard-sphere gas. Integrating the core and attractive shell separately gives the exact toy resultThe core contributes $b_{\mathrm{HS}}$; the shell contributes $-b_{\mathrm{HS}}(a^3-1)(e^{1/T^{\ast }}-1)$. Heating weakens the relative Boltzmann factor enhancement of attractive configurations, so $B_2$ approaches the positive hard-core value as $T^{\ast }\to \infty$. For the illustrative choice $a=3/2$, the zero occurs atAt this temperature, $Z=1+B_3{n}^2+\cdots$: cancellation removes the linear correction, not every correction. At low $T^{\ast }$, the negative shell contribution becomes large, requiring a correspondingly smaller number density for a useful truncated virial expansion. The square well is a deliberately chosen model, not a fit to the carbon dioxide experiment.

The Seebeck effect is the generation of a thermoelectric electromotive force by a temperature difference. In a thermocouple, two different homogeneous electrical conductors A and B connect a measuring junction at $T_h$ to two reference connections at a common $T_c$. With negligible electric current, the measured terminal voltage is determined by the difference of their Seebeck coefficients. A hot junction alone does not specify the reading: the reference temperature, the second material, and the terminal order matter.

For a motivating experiment, join one end of A to one end of B, insulating the wires elsewhere. Connect their free ends to identical meter leads at reference temperature $T_c$, for example in an ice-water heat bath. A meter with input electrical resistance much larger than the thermocouple's electrical resistance approximates an open circuit, meaning $I\approx 0$. Keep the measuring junction in a separately controlled heat bath at $T_h$ and wait for stable junction temperatures. Both reference connections must share $T_c$; identical leads with identical endpoint temperatures have cancelling thermoelectric contributions, allowing the meter to be elsewhere. ^[ref].

Define the reading as $V_{AB}=V_A-V_B$, with the positive input on the A reference terminal and the negative input on B. Use the Seebeck coefficient convention in which the voltage change along a homogeneous wire at zero electric current is $dV=-S(T)dT$. Following the path from A at $T_c$ to the common junction at $T_h$, then back along B to $T_c$, givesThis formula assumes homogeneous materials whose Seebeck coefficients depend only on the local temperature, and no additional uncompensated lead contributions. Here voltage is the electrical transport voltage associated with electrochemical potential, rather than a direct measurement of an isolated equilibrium contact electric potential.

Start with $T_h=T_c$, then increase $T_h$ while holding $T_c$ fixed and record $V_{AB}$ at each stable setting. In the ideal model the equal-temperature reading is zero. If $S_A-S_B>0$ throughout the interval, A reads positive when $T_h>T_c$. Exchanging meter inputs reverses the reading; exchanging the two junction temperatures reverses the oriented integral. Repeating the run with identical homogeneous materials gives zero because $S_A=S_B$. These controls test the stated model; a nonzero experimental baseline should be measured rather than silently removed by assuming it is zero.

For an explicitly toy linear example, stipulate constant Seebeck coefficients $S_A=25\mu V/K$ and $S_B=-15\mu V/K$. Choose reference temperature $T_c=20{}^{\circ }\mathrm{C}$ and measuring-junction temperature $T_h=70{}^{\circ }\mathrm{C}$. A temperature difference has the same numerical size in kelvins and on the degree Celsius scale, soSwapping the inputs gives $-2.00\mathrm{mV}$. Bringing both junction regions to the same temperature gives zero. With temperature-dependent Seebeck coefficients, shifting both junction temperatures equally while keeping their difference fixed can change the reading. No actual material pair is claimed to have the stipulated coefficients over this interval.

Real thermocouples need a calibrated voltage-versus-temperature relation: the constant-Seebeck coefficient approximation is useful only where the coefficient difference changes little. Wire inhomogeneity can make the result depend on where the temperature gradient lies, and reference connections at unequal or unknown temperatures introduce errors. An electronic reference correction uses a measured reference temperature in place of an ice bath. ^[ref]. Burns and colleagues provide the standardized reference functions for common letter-designated thermocouple types; the toy line above is not a replacement for them. ^[ref].

The historical experiment used a different detector. In his 1821–1823 investigations, Seebeck found that a closed electrical circuit of two dissimilar materials with junctions at different temperatures deflected a compass needle. The modern explanation is that thermoelectric electromotive force drives an electric current, whose magnetic field deflects the needle. The open-circuit experiment above detects voltage while drawing negligible electric current. The striking connection is that a thermal imbalance can become an electrical signal and then a magnetic response. ^[ref].

Self-inductance is inductance associated with a coil's own electric current, rather than magnetic flux supplied by another coil. In the linear model, fixed geometry and magnetic response give constant $L$ in $\Lambda =LI$. With the turn orientations defined in magnetic flux linkage, Faraday's law of induction gives the self-induced electromotive forceThe final equality requires constant $L$; if the geometry changes, use $-d\Lambda /dt$. The minus sign expresses Lenz's law: the induced electromotive force opposes the change in electric current. It does not always oppose the electric current itself. When positive electric current decreases, the induced electromotive force acts in its direction. ^[ref].

For terminal sign bookkeeping, label the ideal inductor's terminals $a,b$, choose positive electric current entering $a$ and leaving $b$, and define $v_L$ as the terminal electric potential difference drop from $a$ to $b$ in the lumped electrical circuit model. This passive convention givesThus a coil with self-inductance $0.5\mathrm{H}$ and increasing electric current $dI/dt=12A/s$ has terminal drop $v_L=6\mathrm{V}$ and induced electromotive force ${\mathcal{E}}_{\mathrm{ind}}=-6\mathrm{V}$. These signs describe the same process with different reference conventions. The RL circuit uses the terminal-drop convention throughout.

A finite terminal electric potential difference cannot make the electric current of an ideal inductor with constant $L>0$ jump. Integrating its differential equation givesIf $v_L$ stays bounded, the integral tends to zero as the interval shrinks, ensuring continuity of electric current. A step in terminal electric potential difference changes $dI/dt$ immediately, while electric current remains continuous. At constant electric current, the self-induced electromotive force is zero, although the inductor's magnetic energy can remain nonzero.