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For real $x,y$, the exponential function definitions give $\sinh (x+iy)=\sinh x\cos y+i\cosh x\sin y$. Taking the squared complex modulus and using ${\cosh }^{2}x-{\sinh }^{2}x=1$ gives the identity. In particular $|\sinh (x+iy)|\ge |\sinh x|$, uniformly in $y$. This controls the vertical sides of rectangular contours in residue theorem arguments with hyperbolic functions.

Ice-ocean albedo feedback is the ocean-surface case of ice-albedo feedback: loss of bright sea ice, often covered by snow, exposes darker water, lowers surface albedo and increases absorbed sunlight. The extra stored ocean heat content can promote further melting and delay refreezing. Reversing the sequence can amplify cooling. This is a positive climate feedback, meaning that it reinforces the initial change; the word positive does not mean beneficial, and it does not guarantee runaway warming.

For unchanged downward solar irradiance $S_{\downarrow }$ on a newly exposed opaque patch, the extra absorbed heat flux isFor illustrative local surface albedos $0.60$ and $0.10$, and $S_{\downarrow }=200\mathrm{W}{\mathrm{m}}^{-2}$, the patch absorbs an additional $100\mathrm{W}{\mathrm{m}}^{-2}$. This is a local sunlight calculation, not a whole-planet radiative forcing: newly exposed area, illumination, clouds and season determine its global contribution. With no incident sunlight this particular heating pathway vanishes. NASA's notes describe the contrast between reflective ice and snow and darker water. ^[ref]

To isolate how a positive climate feedback competes with Planck feedback, retain the assumptions of the zero-dimensional planetary energy balance and prescribe the smooth broadband Bond albedoThe exponential function makes a smooth transition from a bright cold state near $0.6$ to a darker warm state near $0.3$. The midpoint $240\mathrm{K}$ and width $3\mathrm{K}$ are chosen to illustrate the dynamics. They are not a seawater freezing law, an observed global relation, or a calibration of modern Earth. No evolving ice thickness or latent heat is represented.

An equilibrium satisfies $F(T_{\ast })=0$. For a small perturbation $\delta T$, linear stability follows fromSince $C>0$, $F^{\prime }<0$ gives an asymptotically stable equilibrium, while $F^{\prime }>0$ gives an unstable equilibrium. With $x=(T-240\mathrm{K})/(3\mathrm{K})$, the derivative isThus the optical term $-S{\alpha }^{\prime }/4$ is destabilizing, while the thermal radiation term $-4\sigma T^3$ is stabilizing. Their sum, not the optical term alone, decides the linear stability.

For the same $S=1360\mathrm{W}{\mathrm{m}}^{-2}$ and Stefan-Boltzmann constant used above, the bisection method on the intervals $(220,223)$, $(238,241)$ and $(253,255)$, with temperature in kelvins, gives:

Unrounded numerical zeros of a function have $|F(T_{\ast })|<10^{-10}\mathrm{W}{\mathrm{m}}^{-2}$; the table rounds them for readability. In the positive-temperature domain the sign of $F$ alternates between heating, cooling, heating and cooling across these three equilibria. The middle equilibrium separates the basins of attraction of the two stable equilibria. For example, a starting temperature of $235\mathrm{K}$ cools toward the cold state, whereas $245\mathrm{K}$ warms toward the warm state: this is bistability at identical external illumination.

The same model illustrates hysteresis. Its equilibrium curve can be written as $S(T)=4\sigma T^4/[1-\alpha (T)]$. The two turning points, where $4[1-\alpha (T)]+T{\alpha }^{\prime }(T)=0$, are saddle-node bifurcations: approximately $(T,S)=(232.637\mathrm{K},1567.801\mathrm{W}{\mathrm{m}}^{-2})$ and $(245.893\mathrm{K},1250.481\mathrm{W}{\mathrm{m}}^{-2})$. For illumination strictly between these thresholds there are two stable equilibria and one unstable equilibrium. Under sufficiently slow increase of $S$, the cold branch ends at the upper threshold; under sufficiently slow decrease, the warm branch ends at the lower threshold. The ensuing warming and cooling transitions differ, so the state depends on its history. Finite sweep rates can delay the transitions; these thresholds describe this model's static equilibrium curve, not predicted tipping points for modern Earth.

The striking feature is that a smooth albedo law can produce distinct long-time states and abrupt transitions, although no discontinuity was inserted into the optical law. It is useful for understanding why positive climate feedback can amplify a response while stabilizing thermal radiation still limits it. Here $0.3<\alpha (T)<0.6$ bounds the incoming heating, whereas outgoing thermal radiation grows as $T^4$; high enough temperatures therefore cool rather than run away indefinitely.

Historically, M. I. Budyko's 1969 paper coupled changing albedo from expanding ice cover to a latitude-dependent temperature and heat transfer calculation, showing how that coupling amplified the effect of changed sunlight. ^[ref] William D. Sellers's 1969 paper constructed an annual energy-budget model in 10-degree latitude belts, including planetary albedo, atmospheric infrared transmission and atmospheric and oceanic exchange. ^[ref] These primary papers provide historical examples of ice-sensitive energy-budget modeling; the smooth single-temperature example above is a separate illustration, not a reproduction of either paper's equations or predictions.

Infrared is electromagnetic radiation with wavelengths longer than the red end of ordinarily visible light. A useful named convention is the CIE range $780\mathrm{nm}$ to $1\mathrm{mm}$; the approximate visible range $400$–$700\mathrm{nm}$ used in some teaching texts puts the boundary nearer $700\mathrm{nm}$. ^[ref]. Neither number is a sharp physical cutoff of human vision. The CIE divisions IR-A, IR-B and IR-C are respectively $0.78$–$1.4\mu \mathrm{m}$, $1.4$–$3\mu \mathrm{m}$ and $3$–$1000\mu \mathrm{m}$; the boundaries of near-, middle- and far-infrared vary between applications. ^[ref].

Use vacuum wavelength $\lambda$ when comparing these ranges. With frequency $\nu$, speed of light $c$, Planck constant $h$ and photon energy $E$,Thus the CIE infrared interval corresponds approximately to $3.00\times 10^{11}$–$3.84\times 10^{14}\mathrm{Hz}$ and $1.99\times 10^{-22}$–$2.55\times 10^{-19}\mathrm{J}$ per photon, with the smaller energy at the longer wavelength. A $940\mathrm{nm}$ infrared photon has $E\simeq 2.11\times 10^{-19}\mathrm{J}$, whereas a $254\mathrm{nm}$ ultraviolet photon has $E\simeq 7.82\times 10^{-19}\mathrm{J}$. These are calculations from $hc/\lambda$, not measurements of a particular source. A television remote is a familiar approximately $940\mathrm{nm}$ source; its information is carried by variations in the emitted radiant flux. ^[ref].

Detection and visual perception answer different questions. A photodetector can respond to infrared even when an observer perceives no light; a thermometer detects an increase in temperature after photon absorption. Conversely, a small readout can reflect a small spectral responsivity, rather than absence of incident electromagnetic radiation. Images made from infrared readouts use display colors chosen by the imaging system; those colors are not the colors of the incoming infrared.

The important microscopic mechanisms also differ. In infrared spectroscopy, photon absorption often increases the energy of a normal mode in which chemical bonds stretch or bend. The corresponding optical spectrum helps distinguish molecules by their motions. For the usual electric dipole mechanism, the motion must change the electric dipole moment; matching the frequency alone is not sufficient. ^[ref]. Electronic processes can also emit or absorb infrared, so it is not exclusively radiation from molecular motion. ^[ref]; ^[ref].

Infrared is especially useful for observing thermal radiation from objects near everyday temperatures. For an ideal $300\mathrm{K}$ blackbody, Wien's displacement law gives a wavelength peak near $9.66\mu \mathrm{m}$. A thermal camera's readout nevertheless depends on emissivity, reflected surroundings and transmission along the path, as well as temperature. ^[ref]. A source emitting infrared need not be hot: an infrared remote is useful precisely because a photodetector can distinguish its signal from a background. Visible light and ultraviolet can also cause heating when absorbed; “infrared” names a wavelength range, not a unique form of heat.

Herschel's 1800 discovery made the distinction between sight and detection experimentally concrete. His report describes a optical prism spreading sunlight across a white, ruled receiving surface, with three thermometers at different positions near the end of the visible optical spectrum. He observed heating beyond the red edge where no visible color appeared. The result established that sunlight contained invisible radiation less strongly refracted by that optical prism than the visible red; it did not measure a photon energy or establish today's infrared boundaries. ^[ref]; ^[ref].

A documented modern version by Caltech/IPAC uses a glass optical prism, white paper in a shaded box and three blackened liquid-filled thermometers. Initial readings in shade provide each thermometer's baseline; subsequent readings compare blue, yellow and just beyond red after the bulbs have had time to warm. The measured quantity is the rise in temperature, not a “temperature of a color”. The lesson reports the expected largest rise beyond visible red and specifically discusses matching the bulb coatings and thermometer calibration. Clouds and changes in the projected optical spectrum affect the comparison. ^[ref].

For a stronger modern interpretation, an additional shaded thermometer would monitor ambient drift, exchanging bulbs between positions would test differences between thermometers, and blocking the incident beam would distinguish beam-dependent heating from warmed supports or air. These are suggested controls, not claims about Herschel's historical procedure. Thermal conduction through the mounting and convection can change the readout. A broad bulb also samples a range of wavelengths; the optical prism's transmission and optical dispersion, the bulb's absorption and heat loss all affect the position of the largest rise. In particular, the result does not show that an infrared photon carries more energy than a visible photon.

A simple analytical illustration makes those limitations explicit. Let $\theta =T-T_{\mathrm{amb}}$ be the bulb's excess temperature, $C$ its heat capacity, $G$ its effective heat-loss coefficient in $\mathrm{W}{\mathrm{K}}^{-1}$, and $P_{\mathrm{abs}}$ the absorbed power. Assume one uniform bulb temperature, constant ambient temperature, constant $C,G$, and losses linear in $\theta$. Thenfor a beam switched on at $t=0$ with $\theta (0)=0$. After switching the beam off at $t=t_s$, $\theta (t)=\theta (t_s)e^{-(t-t_s)/\tau }$. The exponential function follows by solving this linear differential equation. The steady excess temperature measures $P_{\mathrm{abs}}/G$; it is not a direct measurement of the optical spectrum. With illustrative $C=0.020\mathrm{J}{\mathrm{K}}^{-1}$ and $G=0.0010\mathrm{W}{\mathrm{K}}^{-1}$, $\tau =20\mathrm{s}$. Absorbed powers of $1.0$ and $0.5\mathrm{mW}$ give steady rises of $1.0$ and $0.5\mathrm{K}$, regardless of which wavelengths supplied that absorbed power.

The upper schematic represents the logic of the optical prism experiment; its light ray positions are not a calibrated wavelength scale. The lower curves use the illustrative heat-balance model, with the beam blocked at $60\mathrm{s}$; they are not historical or modern measured data. The beauty of the experiment is that a simple change of detector extends observation beyond what the eye can perceive.

Normalization of the reverse law gives an exact finite identity:The sum notation describes discrete paths; the corresponding path integral has the same change-of-variable argument. Jensen's inequality for the exponential function then gives $e^{-⟨{\Sigma }_F{⟩}_F}\le 1$, hence $⟨{\Sigma }_F{⟩}_F\ge 0$. Equivalently this mean is the Kullback-Leibler divergence between the forward path law and the reverse path law pulled back by $\Theta$. Equality requires those two laws to agree. ^[ref]

Rare negative entropy production contributes disproportionately to this expectation through $e^{-{\Sigma }_F}$. Missing such events in a finite sample can spoil an estimate of the identity. Matching probability support is also essential: reverse paths absent from the forward law cannot be recovered by cancelling forward weights. This Second law of thermodynamics statement concerns an ensemble average; an individual negative value is compatible with it.

An integrating factor is a nonzero multiplier that makes a differential equation directly integrable. For a first-order linear differential equation, it gathers the unknown function and its derivative into one product rule expression. Suppose $a$ and $b$ are continuous functions on an interval containing $t_0$, and consider the initial value problemDefine $A(t)={\int }_{t_0}^{t}a(s)ds$ and $\mu (t)=e^{A(t)}$. The exact solution of a differential equation throughout that interval isThe definite integral fixes $\mu (t_0)=1$, so the initial condition appears without an extra constant to determine. If the equation originally has a coefficient multiplying $y^{\prime }$, first divide by that coefficient on an interval where it is nonzero; a zero of that coefficient needs separate attention. See OpenStax's first-order linear equations chapter for this normalization and the integrating factor procedure.

For a concrete balance, solve $y^{\prime }+2y=3$ with $y(0)=0$. Here $\mu =e^{2t}$, so multiplying the entire ordinary differential equation gives $(e^{2t}y{)}^{\prime }=3e^{2t}$. Accumulating from zero to $t$ givesThe computed values are $y(0.5)\approx 0.94818$, $y(1)\approx 1.29700$ and $y(2)\approx 1.47253$. Its derivative is $3e^{-2t}$, which verifies $y^{\prime }+2y=3$ exactly. Interpreting $3$ as a steady input and $2y$ as proportional removal explains the shape: initially all input increases $y$, then removal catches up. The constant particular solution $y_p=3/2$ balances input and removal, and the homogeneous solution $y_h=-\frac{3}{2}{e}^{-2t}$ is the transient response that corrects it to the initial condition. Because this autonomous differential equation has a time-independent right-hand side, $3/2$ is also an equilibrium point.

The first panel adds the particular solution and homogeneous solution; the second shows the same balance in rates, $3=y^{\prime }+2y$. This links the algebra to a physical accounting rule. Newton's law of cooling has this form with $a=k$ and $b=k{T}_a$, while an electrical circuit can have a varying input. In both cases the integrating factor systematically handles the input rather than requiring a guess for a particular solution.

Variable coefficients work just as directly. For $y^{\prime }+2ty=t$, $y(0)=0$, the integrating factor is $e^{t^2}$, andThus $y(0.5)\approx 0.11060$, $y(1)\approx 0.31606$ and $y(2)\approx 0.49084$. The derivative $y^{\prime }=t{e}^{-t^2}$ verifies $y^{\prime }+2ty=t$, including at zero, where both coefficient and input vanish. No division by $t$ is necessary. The general solution is $y=1/2+C{e}^{-t^2}$: $1/2$ is a constant particular solution, and $C{e}^{-t^2}$ solves the associated homogeneous linear differential equation. Although this special forcing term admits a constant trajectory, this is a nonautonomous differential equation: it depends explicitly on time. Merely setting its right-hand side to zero at one instant would not establish a constant solution of a differential equation; at $t=0$ every state has zero slope.

The direction field shows the selected trajectory leaving its initially horizontal tangent line. For $t\ge 0$, the difference between any two solutions of a differential equation is multiplied by $e^{-t^2}$, rather than $e^{-2t}$ in the constant-rate example. This dependence on elapsed history is clearer if the exact formula is rewritten asFor $t\ge t_0$, each contribution from the input at $s$ is weighted by the accumulated coefficient between $s$ and $t$. The homogeneous solution remembers the initial condition; the particular solution given by the second term accumulates the forcing term with zero initial value. This decomposition differs from choosing a constant particular solution, but gives the same total. Initial-data memory tends to zero when ${\int }_{t_0}^{t}a\to +\infty$; a negative accumulated coefficient instead amplifies it. The reciprocal relation between integrating factor and homogeneous solution is developed in MIT's notes on solutions to linear first-order equations.

Why does the trick work? The fundamental theorem of calculus and chain rule give ${\mu }^{\prime }=a\mu$. Consequently the product rule givesIntegrating this identity and dividing by the positive exponential function $\mu$ produces the stated formula. Any nonzero constant multiple of $\mu$ works equally well; normalization is a convenience. This is the beauty of the method: a sum involving an unknown function becomes one exact derivative, leaving only an integral of known functions. See Oregon State's product-rule derivation. The related integrating factor for a differential one-form extends the idea of making an expression exact beyond this linear setting.

Integration by parts transfers a derivative from one factor of a product to the other, while keeping an endpoint contribution. Its usefulness comes from choosing which factor becomes simpler when differentiated, and which has an accessible antiderivative. The integral is then replaced by a boundary term minus another integral.

For real continuously differentiable functions $u,v$ on $[a,b]$, the precise definite integral formula isIn the shorter notation, $du=u^{\prime }(x)dx$ and $dv=v^{\prime }(x)dx$, giving $\int udv=uv-\int vdu$ for indefinite integrals, with a constant of integration understood. The product rule says $(uv{)}^{\prime }=u^{\prime }v+u{v}^{\prime }$; the fundamental theorem of calculus turns its integral into the endpoint difference. Thus the formula packages a differentiation rule into a tool for integration.^[ref]

For example, choose $u=x$ and $dv=e^{x}dx$. Then $du=dx$ and $v=e^x$, since the exponential function is its own derivative:The boundary term is $e\approx 2.718282$; the remaining integral is $e-1\approx 1.718282$. Their difference is exactly one. The endpoint product alone would overcount, because its derivative includes $u^{\prime }v=e^x$ as well as the desired $u{v}^{\prime }=x{e}^x$.

In the figure, the area under $(x{e}^x{)}^{\prime }=(1+x)e^x$ is the endpoint contribution $e$. Its lower band has area $e-1$, and its upper band has thickness $x{e}^x$ and area one. This decomposition makes the subtraction visible. The endpoint contribution is a difference of product values; it need not equal the area under either factor separately.

The same choice repeatedly lowers the degree of a polynomial when integrating a polynomial times an exponential function, while choosing $u=\ln x$ gives $\int \ln xdx=x\ln x-x+C$ for $x>0$. More broadly, moving a derivative onto a convenient test function is the mechanism behind a weak derivative. Always evaluate the boundary term: it disappears only when the relevant endpoint values justify doing so, often through a boundary condition.

The inverse function theorem answers a practical question: can nearby outputs be used to recover nearby inputs uniquely and smoothly? Its useful surprise is that a nonlinear function can be tested using its linear derivative. If the first-order changes preserve all input directions, then, on a sufficiently small region, the nonlinear map can also be reversed. This is the basis for treating a nonlinear map as a coordinate transformation.

Precisely, let $f:U\to {\mathbb{R}}^n$ be continuously differentiable on an open subset $U$ of Euclidean space, let $a\in U$, and put $b=f(a)$. If $Df(a)$ is an invertible linear map, there are open neighbourhoods $U_0$ of $a$ and $V_0$ of $b$ such that $f:U_0\to V_0$ is a bijection with a continuously differentiable inverse function $g$. Thus this restriction is a diffeomorphism. In coordinates the test is $\det Df(a)\ne 0$, using the Jacobian determinant. The derivative of a local inverse map is obtained by inverting the Jacobian matrix, at the corresponding input. ^[ref]

For a one-dimensional example, $f(x)=x^2$ has $f^{\prime }(1)=2$. Near $x=1$, the inverse function is $g(y)=\sqrt{y}$, and the inverse function rule gives $g^{\prime }(1)=1/2$. But $x^2$ has no inverse function on the whole real line, since $x$ and $-x$ give the same output. The theorem supplies a local restriction, not global injectivity. At zero, the derivative test fails; the nonnegative branch still has a square root inverse, but that inverse has no finite derivative at zero. ^[ref]

Here is a two-dimensional example with an invertible derivative everywhere and infinitely repeated outputs:The exponential function turns $s$ into a positive radius, while $\theta$ is the angle in polar coordinates. Restricting, for example, to $|s|<0.35$ and $|\theta |<0.6$ produces the reversible patch pictured below. On its image, with Cartesian coordinates $(u,v)$ and $u>0$, the inverse function isThe arctangent selects the angle near zero. Without an angle restriction, $f(s,\theta +2\pi k)=f(s,\theta )$ for every integer $k$, so there is no single global inverse function. This map is a local diffeomorphism but is not injective.

The short proof idea is to correct a guessed input using the inverse linear response. With $A=Df(a)$, solving $f(x)=y$ is equivalent to finding a fixed point ofAt $a$, the derivative $D{T}_y(a)=I-A^{-1}Df(a)$ vanishes. Continuity makes $T_y$ a contraction mapping on a sufficiently small closed ball, which it maps into itself when $y$ is sufficiently close to $b$. The contraction mapping theorem then gives existence and uniqueness there. Establishing the smooth dependence on $y$ completes the inverse theorem. This explains why the nonlinearity can be tolerated: its correction has a small local slope.^[ref]

Define the instantaneous spin magnetization $m=\frac{1}{4}{\sum }_i{\sigma }_i$. Global reversal sends $m$ to $-m$ without changing the zero-field energy. Pairing every microstate with its reversed partner proves $⟨m⟩=0$ at every positive temperature, even when almost all probability lies on the two fully aligned patterns. A vanishing mean does not imply weak alignment: the limiting probability distribution is half at $m=+1$ and half at $m=-1$.

The Ising ring magnetization fluctuations measure this distinction through variance. In the $d=0$ class, both patterns have $m^2=1$. Among the $d=2$ patterns, eight have $m^2=1/4$ and four have $m^2=0$; the two $d=4$ patterns also have $m^2=0$. ConsequentlyIt equals $6/17$ at $K=(\ln 2)/4$, approaches $1/4$ for four independent fair Ising spins at $K\to 0$, and approaches $1$ as $K\to \infty$. Unlike the energy variance, the spin magnetization variance remains large in the low-temperature symmetric mixture.

The same thermal fluctuation controls the field response. Allow $h$ to vary before setting it to zero; differentiating the Boltzmann distribution inserts $\beta {\sum }_i{\sigma }_i=4\beta m$, givingThis is a spin-normalized magnetic susceptibility with respect to the energy-valued parameter $h$; conversion to physical magnetization and magnetic field requires the magnetic moment and sample volume factors.

For finite $N$, $Z$ is a finite positive sum of exponential functions. Its free energy $-k_{B}T\log Z$ is a real analytic function for $T>0$, so the smooth heat capacity peak here is not a critical phase transition. Spontaneous magnetization concerns the order of limits: first a thermodynamic limit, then removal of a symmetry-breaking field. The four-spin Ising ring alone establishes neither that limit nor any two-dimensional Ising model result.^[ref]

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].

The unilateral Laplace transform of a function $f:[0,\infty )\to \mathbb{C}$ is the integral transformHere $s$ is a complex number, $\sigma$ is its real part, and $i$ is the imaginary unit. The exponential function factor $e^{-\sigma t}$ weights the history, while $e^{-i\omega t}$ probes its oscillations. The integral starts at zero, so it uses only the history from that starting time onward. This section concerns ordinary functions; their value at the single endpoint does not change the integral. See MIT 18.03SC, definition and elementary examples.

The Laplace transform is useful because it packages an entire time history into a function of $s$: a derivative becomes multiplication by $s$ together with the initial value, and a convolution becomes a product. Thus a linear differential equation can become an algebraic equation that already remembers its initial value problem. The worked example below shows both descriptions of the same response.

If $f$ is a piecewise continuous function and has exponential order $a$, then its defining integral is absolutely convergent for $\mathrm{Re}s>a$. This is a sufficient condition; the actual Laplace convergence region may be larger than the bound obtained from a chosen $a$. ^[ref].

The bound has a short explanation. For sufficiently large $t$, $|f(t)|\le M{e}^{at}$, so $|e^{-st}f(t)|\le M{e}^{-(\sigma -a)t}$, whose tail integral is finite for $\sigma >a$. The finite initial interval is covered by piecewise continuity.

Three calculations make the convergence condition concrete. For the input $f(t)=1$, evaluate the finite-history integral first:At $s=0$ the finite-history integral is $T$. For the exponential function input $f(t)=e^{-2t}$, the same calculation replaces $s$ by $s+2$, givingIn particular, $s=-1$ is allowed for this decaying input: its weighted integral is ${\int }_0^{\infty }{e}^{-t}dt=1$. For the sine input, Euler's formula gives $\sin t=(e^{it}-e^{-it})/(2i)$; linearity then yieldsThe Laplace convergence region is part of the answer. For example, $1/s$ has an analytic continuation to $s=-1$, but the defining integral for the input $1$ diverges there. The sine calculation is also developed in MIT 18.03SC, further elementary transforms.

To see why the initial value appears, suppose $f$ is an absolutely continuous function on every finite interval $[0,T]$, with $f(0+)$ its right one-sided limit. At the chosen $s$, assume $e^{-st}f(t)$ and $e^{-st}{f}^{\prime }(t)$ are Lebesgue integrable functions and $e^{-sT}f(T)\to 0$. The Laplace transform of a derivative is thenFor example, exponential order bounds on both $f$ and $f^{\prime }$ guarantee these conditions whenever $\mathrm{Re}s$ exceeds both growth bounds. The initial value is the lower-end boundary contribution of integration by parts.

Consider the initial value problemWrite $Y=\mathcal{L}\{y\}$. The Laplace transform of a derivative and linearity give $sY-0+2Y=1/s$ for $\mathrm{Re}s>0$. A partial fraction decomposition therefore givesWe recognize the two elementary Laplace transforms already computed. Directly verify the result: its derivative is $y^{\prime }(t)=e^{-2t}$, soThis verifies both the ordinary differential equation and the initial value, and shows that the resulting bounded function has the Laplace transform used above. The response rises from zero toward $1/2$: at that value the input $1$ balances the term $2y$. The general method is described in MIT 18.03SC, solving initial value problems.

The same solution is a convolution. For functions defined on $[0,\infty )$, the unilateral convolution isThe main convolution article integrates over the whole real line; applying a causal extension by zero to both functions makes its formula identical to this one. Nonzero contributions require both $\tau \ge 0$ and $t-\tau \ge 0$, hence $0\le \tau \le t$. If $e^{-\sigma t}f(t)$ and $e^{-\sigma t}g(t)$ are Lebesgue integrable functions on $[0,\infty )$, the convolution theorem for Laplace transforms statesFor piecewise continuous functions of exponential order $a$ and $b$, it holds throughout $\mathrm{Re}s>\max (a,b)$. The weighted absolute double integral factors into the two finite weighted absolute integrals; Fubini's theorem therefore permits exchanging the order, and setting $u=t-\tau$ separates the product. ^[ref].

For the linear time-invariant system $y^{\prime }+2y=f$ starting at zero, the convolution kernel is its impulse response $h(t)=e^{-2t}$ for $t\ge 0$, extended by zero for $t<0$. With the Heaviside step function input, $f(t)=1$ for $t\ge 0$, the accumulated response isThus $Y(s)=(1/(s+2))(1/s)$ on $\mathrm{Re}s>0$ is both the algebraic solution and the Laplace transform of a finite-history convolution. Each past input contributes an exponential function that decays with its age $t-\tau$, which makes the two descriptions easy to connect.

Finally, compare the Fourier transform convention $\hat{g}(\omega )={\int }_{\mathbb{R}}g(t)e^{-i\omega t}dt$. If $f_0$ is the causal extension by zero of $f$, define $g_{\sigma }(t)=e^{-\sigma t}{f}_0(t)$. Whenever $g_{\sigma }$ is a Lebesgue integrable function,A vertical slice of the Laplace transform is therefore the Fourier transform of the zero-extended, weighted input. Positive $\sigma$ damps its positive-time history. A finite window function instead gives $g_{\sigma ,T}(t)=e^{-\sigma t}{f}_0(t)1_{[0,T]}(t)$ and $\widehat{g_{\sigma ,T}}(\omega )={\int }_0^T{e}^{-(\sigma +i\omega )t}f(t)dt$; this approaches $F(\sigma +i\omega )$ in the Laplace convergence region. Setting $\sigma =0$ directly requires an integrable zero-extended input: the input $1$ fails that condition, even though its Laplace transform exists for every $\sigma >0$.

The history of the Laplace transform spans several stages. Schiff's historical summary notes that Laplace's 1812 treatise acknowledged earlier work on integral transforms and describes later development of the modern Laplace transform method. ^[ref].

The leaky integrate-and-fire model combines a passive equation for membrane potential with a firing rule. Between events, inward-positive applied electric current $I(t)$ obeyswhere $C>0$ is membrane capacitance, $g_L>0$ is the leak conductance, $R$ is the corresponding electrical resistance, and $E_L$ is the leak reversal potential. Here the leak conductance is constant. The time constant $\tau$ describes passive relaxation. When $V$ first reaches a threshold $V_{\theta }$ from below, record an action potential event and reset $V$ to $V_r<V_{\theta }$. In the version used here, hold $V=V_r$ for a fixed neuronal refractory period $t_{\mathrm{ref}}$, then resume integration. See 1.3 Integrate-And-Fire Models.

For constant $I$, with $t=0$ at the start of an integration interval and $V(0)=V_{\mathrm{initial}}$, the exact subthreshold solution isThe equilibrium point of this passive equation is $V_{\infty }=E_L+RI$. Subtracting $V_{\infty }$ leaves $d(V-V_{\infty })/dt=-(V-V_{\infty })/\tau$, explaining the exponential function. This solution applies only until a threshold event or a change in $I$. Starting below threshold, constant input produces repeated events if $V_{\infty }>V_{\theta }$; if $V_{\infty }=V_{\theta }$, the threshold is approached asymptotically and never reached at finite time.

For a worked example, chooseThese imply $g_L=10\mathrm{nS}$ and $C=200\mathrm{pF}$. All these parameters are illustrative model choices, not measurements from a particular neuron. With $I=200\mathrm{pA}$, Ohm's law gives $RI=20\mathrm{mV}$ and hence $V_{\infty }=-45\mathrm{mV}$. ThusAt threshold, $e^{-t/\tau }=1/4$, so the first event occurs after $T=20\ln 4\mathrm{ms}\approx 27.7259\mathrm{ms}$. Each subsequent cycle includes the $2\mathrm{ms}$ neuronal refractory period, givingThis is the ideal sustained firing rate; the first-event latency is $T$, not $T+t_{\mathrm{ref}}$.

With $I=100\mathrm{pA}$ instead, $RI=10\mathrm{mV}$ and $V_{\infty }=-55\mathrm{mV}$. From the same initial state, $V(t)=-55\mathrm{mV}-10\mathrm{mV}{e}^{-t/\tau }$ stays below $-50\mathrm{mV}$: no events occur, however long this current is held. The critical sustained current is $(V_{\theta }-E_L)/R=150\mathrm{pA}$; equality gives asymptotic approach rather than finite-time firing.

The leak gives old inputs progressively less influence, while capacitance prevents an ordinary finite electric current step from changing $V$ instantly. This combination makes the leaky integrate-and-fire model useful for exploring input summation, event timing and network activity at low computational cost. Its fixed parameters and reset omit the evolving voltage-gated ion channels of the Hodgkin-Huxley model, spatial integration in dendrites and mechanisms of neural adaptation. Treating a synapse as a prescribed $I(t)$ also omits its voltage-dependent driving force unless a conductance term is added. See 1.4 Limitations of the Leaky Integrate-and-Fire Model and constant-input firing rate of a leaky integrate-and-fire neuron.

Compose the lens-to-quadrant Möbius map with the logarithm having argument in $(0,\pi /2)$ on the first quadrant. Polar coordinates show that the latter maps bijectively to the horizontal strip $0<\mathrm{Im}\zeta <\pi /2$. Its inverse is the exponential function, so the combined inverse is $(e^{\zeta }-(1+i))/(e^{\zeta }+(1+i))$. Both derivatives are nonzero, proving conformality. The logarithm's branch fixes the strip's vertical position; adding $2\pi i$ shifts that position without changing the exponential inverse.

Why does an isopycnic centrifugation band return after displacement? Choose a dilute mathematical model with outward coordinate $z$, constant local centrifugal acceleration $a>0$, fixed displaced volume $V>0$, linear friction coefficient $f>0$, and an imposed liquid density ${\rho }_f(z)={\rho }_0+Gz$ with $G>0$. Treat the species' buoyant density ${\rho }_b$ as constant in this medium. The sedimentation coefficient force balance gives the ordinary differential equationIts exponential function solution is $z(t)=z_{\ast }+(z(0)-z_{\ast })e^{-kt}$. The sign of the velocity reverses across $z_{\ast }$, proving stable equilibrium when $z_{\ast }$ lies inside the liquid. Changing $V/f$ changes the approach time $1/k$, while $z_{\ast }$ depends on buoyant density and the imposed density gradient. A real centrifuge rotor has radially varying centrifugal acceleration; the constant-$a$ approximation describes a short radial interval. Here the density gradient is held fixed, rather than modeling its formation.

For an original dimensionless example, let $u=z/L$ and $\tau =t/T$, where $L$ is column length and $T$ a chosen time scale. Write $q=z_{\ast }/L$ and $\lambda =kT$, so $u(\tau )=q+(u(0)-q)e^{-\lambda \tau }$. Species A and B have the same $q=0.45$ but different friction-dependent rates: $({\lambda }_A,u_A(0))=(0.4,0.1)$ and $({\lambda }_B,u_B(0))=(1.6,0.95)$. A moves outward and B inward, both toward $u=0.45$. Species C has $(q_C,{\lambda }_C,u_C(0))=(0.8,0.8,0.1)$ and approaches a separate band at $u=0.8$. At $\tau =4$, their positions are approximately $0.3793$, $0.4508$ and $0.7715$: a finite-time photograph would still show A short of its limiting position.

An equilibrium band has finite width because diffusion opposes focusing. In the same local model, let $c(z)$ be concentration and let the constant diffusion coefficient be $D>0$. Combining the restoring drift with Fick's laws, zero net flux requiresDividing by $c$ and integrating proves the Gaussian function profile, whose standard deviation is ${\sigma }_z=\sqrt{D/k}$. This local result assumes that the band is narrow relative to its distance from the boundaries and that $D$, $k$ and the density gradient vary negligibly across it. Equal buoyant densities therefore imply coincident centers in this model, not identical widths. A band outside the available density range has no such interior restoring point.

For $0<I_1<I_2<I_3$, the transverse linear stability rates about steady rotation with signed component $\Omega \ne 0$ areAxes one and three have transverse eigenvalues $\pm i{\nu }_1$ and $\pm i{\nu }_3$, producing bounded oscillations. Axis two has transverse eigenvalues $\pm {\lambda }_2$, giving growing and decaying exponential functions. A neutral longitudinal perturbation corresponds to a small change of $\Omega$. Generic intermediate-axis disturbances contain a growing component; a precisely chosen decaying direction is exceptional. For the normalized intermediate-axis model with $|\Omega |=1$, the values are ${\nu }_1=1/\sqrt{3}$, ${\nu }_3=1$ and ${\lambda }_2=1/\sqrt{3}$.

Proof outline: the linearization of the Euler equations for a torque-free rigid body about the third principal axis gives transverse perturbations obeyingand henceCyclic permutation gives the other rates. For nonlinear Lyapunov stability of the extreme rotations, use the torque-free rigid-body invariants: the conserved quadratic forms $L^2-2I_{1}E$ and $2I_{3}E-L^2$ have strictly positive transverse coefficients about axes one and three respectively. They bound transverse angular velocity, while $E$ bounds its longitudinal change near nonzero rotation. Imaginary transverse eigenvalues alone would not establish this nonlinear conclusion.^[ref]

The practical lesson is to choose a stable principal axis when using rotation to maintain an instrument's pointing direction. Both extreme principal moments of inertia are stable in the ideal torque-free model; if internal damping lowers rotational kinetic energy while approximately preserving angular momentum, the minimum energy $E=L^2/(2I_3)$ instead favors the largest principal moment of inertia. A control system must account for torque, changing mass distributions and flexibility before applying this ideal result.

For a positive real base $b\ne 1$, the logarithm is the inverse function of the exponential function $t\mapsto b^t$:It asks which exponent produces a given positive number. Thus ${\log }_{2}8=3$ and ${\log }_2(1/8)=-3$. The base restrictions matter: $b=1$ would produce only one value, while a negative base does not define $b^t$ as a real-valued function for every real exponent.

The useful transformation isFor example, $8\cdot 16=128$ corresponds to $3+4=7$ in base two. Products become sums and ratios become differences. The logarithm does not generally convert a sum into a sum of logarithms. Base conversion uses the natural logarithm: ${\log }_{b}x=\ln x/\ln b$. It also gives $d({\log }_{b}x)/dx=1/(x\ln b)$, positive when $b>1$ and negative when $0<b<1$.

Historically, John Napier's 1614 Mirifici logarithmorum canonis descriptio presented logarithmic tables and their use, documented in the Smithsonian Libraries' digitized original. Tables made multiplication-heavy calculations more manageable by replacing products with addition and a lookup. The 1619 Constructio title page below belongs to a later publication from the same work on logarithms. These historical publications preceded the modern inverse function and integral presentation used here.

For fixed vectors $a_1,\dots ,a_m$ in a finite-dimensional Euclidean space, defineThe gradient and Hessian matrix of $\Phi =\log F$ areTo derive the first identity, differentiate each exponential function and divide the sum of its gradients by $F$. For the second, differentiating a normalized weight gives $\nabla w_i=2w_i(\bar{a}-a_i)$. Thus $J_{\bar{a}}=-2{\mathrm{Cov}}_w(a_i)$, where the weighted covariance matrix is the sum in the displayed Hessian matrix. Differentiating $2(x-\bar{a})$ completes the calculation.

The weighted covariance matrix is positive semidefinite, so ${\nabla }^2\Phi ⪰2I$. Also $\Phi (x)\ge \Vert x-a_1{\Vert }^2\to \infty$ as $\Vert x\Vert \to \infty$. Therefore $F$ and $\Phi$ share a unique global minimizer, characterized by $x=\bar{a}(x)$. The weights place that minimizer in the convex hull of the centres.

For stable numerical evaluation, subtract the largest exponent before computing the normalized weights. This is the log-sum-exp principle and avoids floating-point overflow without changing the gradient. A naive iteration $x_{n+1}=\bar{a}(x_n)$ need not converge: in one dimension, centres $-1,1$ give $\bar{a}(x)=-\tanh (2x)$ with derivative $-2$ at its unique fixed point. A Newton root-finding iteration or a suitable quasi-Newton method instead uses the positive curvature of $\Phi$.

The logistic differential equation is the nonlinear ordinary differential equationAs a mathematical model of population dynamics, it treats abundance $N$ as a continuous nonnegative quantity. The constant $r$ has units of inverse time, while the carrying capacity $K$ has the same units as $N$. The useful idea is to combine growth resembling exponential growth at low abundance with a slowing approach to a finite abundance. The three trajectories below already show an important distinction: starting above the carrying capacity gives a decreasing curve, whereas starting below it gives an increasing one.