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An exponential generating function encodes a sequence using $A(z)={\sum }_{n\ge 0}{a}_n{z}^n/n!$. For labelled counting, $a_n$ is the number of structures on a fixed set of $n$ distinct labels. Labels distinguish otherwise similar objects: changing which person belongs to a group can change the structure. Recover the count by coefficient extraction as $a_n=n![z^n]A(z)$. The raw coefficient is $a_n/n!$, not $a_n$.

The factorial makes multiplication account automatically for distributing labels. To combine an $A$-structure on $k$ labels and a $B$-structure on the remaining $n-k$, choose the first label subset in $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ ways, then choose the two structures. The binomial coefficient counts how the two groups of labels interleave within the full label set. ThusThe normalization works because $1/(k!(n-k)!)=\left(\genfrac{}{}{0ex}{}{n}{k}\right)/n!$. This explains the usefulness of exponential generating functions for labelled counting: an ordinary Cauchy product would omit the choices of labels. See Peter Magyar, Math 482, notes for 2/7, labelled product principle.

For a concrete example, there is one undecorated set on any given collection of labels. Its exponential generating function is the exponential function $E(z)=e^z={\sum }_{n\ge 0}{z}^n/n!$. Combine a red set and a blue set, allowing either to be empty:Every label independently receives one of two colours, giving $2^n$ assignments by the rule of product. With three labels, the counts by number of red labels are $1,3,3,1$, totalling eight. Choosing only the two group sizes would give the four possibilities $(0,3),(1,2),(2,1),(3,0)$; it would forget which labels are red.

Indeed, the ordinary generating function for the constant sequence $a_n=1$ is the geometric series $1/(1-z)$. Its square has coefficient $[z^n](1-z{)}^{-2}=n+1$, counting pairs of nonnegative sizes. Both encodings are correct for their respective questions. The decision is whether identity matters when objects are combined.

For unordered sets of labelled components of positive size, an exponential generating function $B(z)$ becomes $e^{B(z)}$: the term $B(z{)}^m/m!$ forgets the ordering of $m$ components. The existing rooted-tree generating function develops the richer example $T=z{e}^T$. A labelled rooted tree consists of its root of a tree and an unordered set of smaller labelled rooted trees, connecting the same label-distribution mechanism to recursive counting.

For fixed real $x$, the base $1+x/n$ is positive for sufficiently large $n$. Applying Taylor's theorem to the natural logarithm at one gives $n\log (1+x/n)=x+O(n^{-1})$. Continuity of the exponential function then proves the displayed limit. In particular $(1-z/n{)}^n\to e^{-z}$ for fixed $z\ge 0$, a basic source of exponential distribution limits in probability theory.

Applying Markov's inequality to the positive random variable $e^{tX}$ givesHere $M_X(t)=\mathbb{E}[e^{tX}]$ is the moment-generating function, and the useful bound requires it to be finite at the chosen $t$. Optimizing over $t>0$ gives the Chernoff bound method: an exponential function weights large values more strongly before applying the same average-budget argument.

For example, let $N$ have the Poisson distribution with parameter one. Its expected value and variance are both one, and its moment-generating function is $M_N(t)=\exp (e^t-1)$. At threshold four, choosing $t=\log 4$ givesThis $t$ minimizes the exponent $e^t-1-4t$ because its derivative is $e^t-4$, with positive second derivative. The direct Markov's inequality bound is $1/4$, and the two-sided Chebyshev's inequality bound, applied to a deviation of at least three from the mean one, is $1/9$. The exact Poisson distribution tail probability is $1-e^{-1}(1+1+1/2+1/6)\approx 0.0189882$. These compare three valid bounds obtained from progressively different information; none is an exact tail probability formula for general random variables.

From $f_1=O(g_1)$ and $f_2=O(g_2)$ one cannot generally deduce $f_2\circ f_1=O(g_2\circ g_1)$. For $f_1(n)=2n$, $g_1(n)=n$ and $f_2(n)=g_2(n)=2^n$, the two premises hold, but the ratio of the compositions is $2^n$, which is unbounded. A constant scaling of an inner argument can change the growth order after an exponential function is applied.

Under the Picard-Lindelöf theorem's hypotheses on $U\subset \mathbb{R}\times {\mathbb{R}}^d$, extend a solution of a differential equation as far as possible, to a maximal interval $(T_{-},T_{+})$. If $T_{+}<\infty$, the time-state trajectory $(t,y(t))$ must eventually leave every compact subset of $U$. Equivalently, confinement near $T_{+}$ to one compact subset of $U$ permits continuation past $T_{+}$. See Teschl, Section 2.6, Lemma 2.14 and Corollaries 2.15–2.16, pp. 52–53.

The domain restriction matters even for a bounded trajectory. Restrict $y^{\prime }=1$, $y(0)=0$, to $U=\mathbb{R}\times (-1,1)$. Then $y(t)=t$ exists in this domain only for $-1<t<1$; at $t=1/2,3/4,9/10$ its values are $1/2,3/4,9/10$. As $t\uparrow 1$, it approaches the excluded boundary point $(1,1)$ while remaining bounded. Its trajectory cannot fit into any compact subset of $U$ near that endpoint. This is escape from a finite domain, without finite-time blowup in norm. Enlarging the domain to $\mathbb{R}\times \mathbb{R}$ gives the same formula $y(t)=t$ for every real $t$.

Conversely, $y^{\prime }=y$, $y(0)=1$, stays in the compact set $[0,T]\times [1,e^T]$ for every fixed $T>0$. Its growing exponential function is unbounded over the entire future, but has no finite-time blowup. At $T=2$, its largest value on $[0,2]$ is only $e^2\approx 7.38905610$. The relevant question is boundedness before each finite endpoint, rather than one bound valid forever. The tangent function in the diagram under the Picard-Lindelöf theorem illustrates the other possibility: its norm escapes every bound before $T=\pi /2$.

This criterion is useful because an estimate can secure continuation without an explicit formula. In a finite-dimensional vector space, a bounded closed ball is compact; a bounded state trajectory therefore supplies the needed compact set when its time-state closure is inside the domain. That final proviso prevents mistaking a domain boundary for finite-time blowup.

A common sufficient estimate comes from global Lipschitz continuity in the state. Assume $U=I\times {\mathbb{R}}^d$ for an open time interval $I$, with $f$ continuous and a Lipschitz constant $L\ge 0$ valid for all states on each compact time interval under consideration. For $[t_0,T]\subset I$, let $B={\sup }_{t_0\le s\le T}\Vert f(s,0)\Vert$, finite by continuity. Then $\Vert f(t,y)\Vert \le B+L\Vert y\Vert$, and the Gronwall inequality gives, for $t_0\le t\le T$ wherever the solution already exists,For instance, $y^{\prime }=y+1$, $y(0)=0$, has $L=B=1$ and the exact solution $y(t)=e^t-1$, attaining this estimate; at $t=1$ the value is $e-1\approx 1.71828183$. Finite bounds of this kind give continuation throughout $I$, with the analogous estimate backwards in time. When $I=\mathbb{R}$, this gives existence for all real times. If $I$ has a finite endpoint, it gives no continuation beyond that prescribed domain. The underlying linear-growth criterion is ^[ref].

A brief explanation of the continuation mechanism suffices: within the protecting compact set, the continuous function $f$ is bounded, say by $M$. The integral equation yields $\Vert y(t)-y(s)\Vert \le M|t-s|$ near the finite endpoint, so $y(t)$ has a limit $y_{\ast }$. The protecting compact set puts $(T_{+},y_{\ast })$ inside $U$. Restarting the initial value problem there and using local uniqueness joins the trajectories.

Take a nonrelativistic particle of constant mass $m$ with energy $E>0$, and choose the same zero of potential energy on both sides of a rectangular potential barrier:The equality of the exterior potential energies matters: it gives the same exterior wavenumber $k=\sqrt{2mE}/\hslash$ and makes $T=|t{|}^2$. Here $\hslash =h/(2\pi )$ is the reduced Planck constant. This model has a finite jump in potential energy, no delta potential, and no absorbing term.

For $0<E<V_0$, the time-independent Schrödinger equation gives the spatial wavefunctionThe full stationary state is $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$. We set the incident probability amplitude to one and impose no incoming wave from the right; this scattering wavefunction is not a normalizable wavefunction. The tunnelling decay constant $\kappa$ describes exponential decay, rather than an oscillatory wavenumber in the barrier. Both exponential functions are needed on the finite interval: the growing term is not excluded by a condition at infinity.

The wavefunction and its derivative are continuous at each face, giving four linear matching conditions:Their solution supplies the transmission amplitudeA short derivation is to use the two right-face equations to express $A{e}^{\kappa a}$ and $B{e}^{-\kappa a}$ in terms of $t$, then substitute into the left-face equations. Taking the squared modulus gives the rectangular barrier transmission probability.

If two nonnegative finite-valued convex functions are defined on all of a finite-dimensional space, their infimal convolution is finite, convex and continuous everywhere. Convex perturbation duality then gives strong duality and dual attainment at every argument. It does not require or imply attainment of the primal decomposition infimum.

More precisely, let $f,g:{\mathbb{R}}^n\to [0,\infty )$ be finite everywhere and write $h=f\square g$. Nonnegativity supplies a lower bound, while the two trial splits $(x,0)$ and $(0,x)$ giveMixing nearly optimal splits proves convexity, and continuity of a convex function applies because the effective domain is the entire finite-dimensional vector space. These statements concern the optimized value, rather than existence of an optimal split. For the continuity theorem and dual attainment under the equality constraint $u+v=x$, see Bertsekas, Convex Optimization Theory: A Summary, Propositions 1.3.11 and 5.3.3(a). The equality has a feasible point in the whole space for every $x$, and the displayed bound makes its optimal value finite.

A finite value without any optimal split. On $\mathbb{R}$, take the convex functions $f(u)=e^u$ and $g(v)=0$. Both are nonnegative and finite everywhere, butfor every $x$. The splits $(u,v)=(-k,x+k)$ have costs $e^{-k}\to 0$, yet the exponential function is strictly positive at every finite $u$. No split attains the infimum. The value function is even constant and differentiable; regularity of the value function does not prevent optimizing splits from escaping to infinity. For example, $k=1,3,6$ gives costs approximately $0.367879$, $0.049787$ and $0.002479$.

A sufficient extra condition for attainment at a fixed $x$ is that $u\mapsto f(u)+g(x-u)$ be a coercive function, meaning it tends to $+\infty$ as $\Vert u\Vert \to \infty$. Under the finite-valued assumptions it is a continuous function, so a sufficiently large closed ball contains a global minimum by the extreme value theorem. If this cost is also a strictly convex function, the optimal split is unique. The quadratic example above satisfies both conditions; the exponential function example does not satisfy coercivity.

Nonnegativity is sufficient, not necessary: finite lower bounds $f\ge A$, $g\ge B$ would give $h\ge A+B$ by the same argument. Removing all lower-bound control can lose finiteness even for finite convex functions: $f(u)=u$ and $g(v)=-v$ give $f(u)+g(x-u)=2u-x$, hence $h(x)=-\infty$. Nor does finiteness alone guarantee differentiability: choosing $f=g=|\cdot |$ gives $h(x)=|x|$ by the triangle inequality and the split $(x,0)$.

An observation interval of finite duration is not an infinitely sharp frequency selector. Combine the lock-in quadratures intoFor unwanted electric potential difference $s(t)=B\cos ((\omega +\delta )t+\psi )$, where $\delta$ is an angular frequency offset and $\omega +\delta >0$, expansion into complex exponentials gives the exact responseThe sinc function formula follows by integrating the exponential function and factoring $e^{\mathrm{i}\nu T}-1=2\mathrm{i}{e}^{\mathrm{i}\nu T/2}\sin (\nu T/2)$; its value at zero follows from a limit. The first term changes slowly after multiplication; the second oscillates at the sum frequency. It vanishes at exact matching when $T$ is an integer number of reference cycles, each lasting one period of a function, but cannot generally be discarded in a short-window calculation.

For $|\delta |\ll \omega$ and many reference cycles, the first term usually dominates: its modulus is $B|\mathrm{sinc}(\delta T/2)|$ and its complex argument includes $\delta T/2$. Thus $|\delta |T\ll 1$ gives almost no rejection. A longer observation separates closer frequencies, with the first zeros of this term at $|\delta |=2\pi /T$. Those zeros are not a brick-wall cutoff: the sinc function has nonzero lobes beyond them. At $\delta =0$ a synchronized contribution remains however large $T$ becomes.

For a numerical check, use $f_0=1\mathrm{Hz}$, $T=4\mathrm{s}$, $\psi =0$ and an unwanted frequency $f=1.125\mathrm{Hz}$. Then $(f-f_0)T=1/2$ and the exact expression giveswhereas the difference-frequency term alone gives $2/\pi \approx 0.636620$. An interference amplitude $B=0.9\mathrm{V}$ therefore contributes $0.539254\mathrm{V}$ to $|Z_T|$ in this constructed example. A nearby frequency can survive strongly even though the three-times-frequency component in lock-in harmonic rejection cancels exactly.

A practical lock-in amplifier's low-pass filter weights past products differently from this rectangular observation interval. Its time constant and roll-off determine near-reference rejection and settling time; narrowing detection slows the response.^[ref]

Floating-point overflow occurs when the magnitude of a computed result exceeds the largest finite representable floating-point number. Large values of an exponential function are a common source. In a normalized weighted sum, replacing exponentials $e^{r_i}$ by $e^{r_i-r_{\max }}$ before division prevents this particular overflow while leaving the normalized weights unchanged; this is an application of the log-sum-exp construction.

Fourier analysis describes a function by the oscillations that compose it. A graph records values at successive positions or times; a frequency spectrum records how much of each frequency is present, together with its wave phase. The same complicated shape can become a short list of simple components. For periodic functions, the components form a Fourier series; on the whole real line, the Fourier transform uses a continuous range of frequencies, and the Fourier inversion theorem puts them back together.

For a concrete example, consider the trigonometric polynomialWith $t$ measured in seconds, its three Fourier harmonics have frequencies $1$, $3$, and $5$ cycles per second and amplitudes $1$, $1/2$, and $1/4$. Each cosine has zero wave phase. Adding them creates the sharper peaks in the figure. The frequency spectrum makes the recipe immediately visible; changing an amplitude changes the shape without changing the component's frequency. Fourier orthogonality lets us recover each Fourier coefficient by averaging against the corresponding sine or cosine, because the other Fourier harmonics cancel over a full period of a function.

The deeper reason these components are so useful is that they simplify operations. By Euler's formula, the exponential function $e^{i\omega x}$ packages a cosine and a sine into one Fourier mode. A translation or a derivative changes only its multiplier:Thus Fourier modes are simultaneous eigenfunctions of these operations. In Fourier series, this gives a literal diagonalization in the Fourier basis; on the real line, the Fourier transform gives the continuous counterpart, with each frequency handled separately. The Translation property of the Fourier transform and the Fourier transform of a derivative turn shifting and differentiating a whole function into multiplication of its Fourier transform. A convolution likewise becomes multiplication by the convolution theorem. This is beautiful because one description simplifies several apparently different operations at once.

For example, the heat equation $u_t=\kappa u_{xx}$, with thermal diffusivity $\kappa >0$, evolves a spatial Fourier mode asFine ripples have larger angular frequencies and decay faster: the formula explains why heating smooths a jagged temperature profile. For a general initial profile with sufficient regularity, the Fourier transform gives $\hat{u}(\omega ,t)=e^{-\kappa {\omega }^{2}t}\hat{u}(\omega ,0)$, and Fourier inversion recovers the temperature. Similarly, multiplying selected parts of a frequency spectrum suppresses unwanted components. Claude E. Shannon's Communication in the Presence of Noise (1949), Section III, describes ideal filtering as a projection that retains some frequency coordinates and removes others.^[ref]

The physical motivation is already visible in Joseph Fourier's Théorie analytique de la chaleur (1822). Chapter III constructs heat profiles using sums of sines and cosines; Chapter IX, articles 355–359, passes from Fourier series to integrals as the spatial interval becomes unbounded. These heat problems explain the motivation for the discrete and continuous descriptions. Their modern formulations specify which functions admit reconstruction and in what sense the expansions converge.^[ref][ref]

For $a>0$, use the Fourier transform convention $\hat{f}(s)={\int }_{\mathbb{R}}f(x)e^{-isx}dx$. Splitting the integral at zero gives $1/(a+is)+1/(a-is)=2a/(a^2+s^2)$. By the Fourier inversion theorem, the dual identity isThis pair connects an exponential function with the kernel of a Cauchy distribution and produces the Poisson kernel for the upper half-plane by inversion of $e^{-y|s|}$.

For real $a>0$ and $\mathrm{Re}s>0$, the change of variables $u=at$ in the defining improper integral for the Gamma function givesThe improper integral converges absolutely: near zero its absolute integrand behaves like $t^{\mathrm{Re}s-1}$, and at infinity the exponential function overcomes every fixed power.^[ref]

For real $s=1,2,4$ and $a=1$, the integrands are $e^{-t}$, $t{e}^{-t}$, and $t^3{e}^{-t}$, with total areas $\Gamma (1)=1$, $\Gamma (2)=1$, and $\Gamma (4)=6$. The third integrand is taller and has six times the area of the first, although the exponential function makes each tail decay.

For real $s>0$, the kernel $t^{s-1}{e}^{-t}$ becomes the probability density function of a Gamma distribution with shape $s$ and unit rate only after division by its normalizing constant $\Gamma (s)$. In particular, the shape-four probability density function is $t^3{e}^{-t}/6$. More generally, the Gamma integral verifies that $a^s{t}^{s-1}{e}^{-at}/\Gamma (s)$ integrates to one for shape $s>0$ and rate $a>0$.

For the half-integer evaluation, the change of variables $t=u^2$ reduces the Gamma integral to the Gaussian integral:Consequently, for real $u>0$,

For $b>0$ and a nonnegative integer $N$, define the symmetric partial sum and its positive remainder bySuccessive omitted terms have ratio $e^{-\pi b(2n+1)}\le e^{-\pi b(2N+3)}$ for $n\ge N+1$. Bounding them by a geometric series givesConsequently the physical approximation to $S(a)$ has error at most $B_N(a)$, while the dual approximation $a^{-1/2}{S}_N(1/a)$ has error at most $a^{-1/2}{B}_N(1/a)$. The prefactor must be included when setting an absolute error tolerance.

For a truncation error below $10^{-12}$, at $a=0.04$ the physical side first passes this bound with $N=15$: $31$ lattice points give $B_{15}(0.04)<2.172\times 10^{-14}$, whereas $N=14$ gives $B_{14}(0.04)>10^{-12}$. The dual side needs only $k=0$. At $a=25$ the physical side needs only $n=0$, whereas the dual side first passes the bound with $N=14$: $29$ lattice points give $25^{-1/2}{B}_{14}(0.04)<2.146\times 10^{-13}$. These counts measure symmetric lattice terms; using the even function symmetry reduces the distinct exponential function evaluations to $N+1$. Truncation bounds do not include rounding error in a numerical implementation.

The generating function holds for $|h|<1$ and fixed $x$. From Rodrigues' formula, $L_n(x)={\sum }_{j=0}^n\left(\genfrac{}{}{0ex}{}{n}{j}\right)(-x{)}^j/j!$. Sum first over $n\ge j$, using ${\sum }_{n\ge j}\left(\genfrac{}{}{0ex}{}{n}{j}\right)h^n=h^j/(1-h{)}^{j+1}$, and then over $j$ to obtain the exponential function. The absolute-value double series is finite for $|h|<1$, justifying the rearrangement. Alternatively, the weighted inner product and the Laplace transform of a Laguerre polynomial extract the coefficient $h^n$.

The geometric mean replaces a list of nonnegative real numbers by one repeated value that preserves their product of numbers. For $n\ge 1$ inputs,For two positive inputs, $G=\sqrt{x_1{x}_2}$ is their positive square root. The geometric mean of $1$ and $9$ is three: replacing $1\cdot 9$ by $3\cdot 3$ retains the same product. In contrast, their arithmetic mean five retains the sum $1+9=5+5$. The geometric mean is zero if any input is zero; the logarithm formula below requires every input to be positive. ^[ref].

Its usefulness becomes immediate for successive growth factors. Suppose a hypothetical positive quantity starts at $Q_0=1000$, is multiplied by $3/2$, then by $2/3$. It goes $1000\to 1500\to 1000$, so the geometric mean factor is $\sqrt{(3/2)(2/3)}=1$: a constant factor of one for both steps gives the same endpoint. The arithmetic mean factor is $13/12$; repeating it would give $1000(13/12{)}^2\approx 1173.61$, a different endpoint. A mean of factors summarizes multiplicative change; subtracting one afterwards gives the equivalent change per step. Equal-duration steps are necessary to interpret this as change per common time interval.

For positive inputs, logarithms turn multiplication into addition, soThus the geometric mean is an arithmetic mean in logarithmic coordinates, followed by the exponential function. For positive measurements with a common unit, one can use dimensionless ratios to a fixed reference $x_0$: $G=x_0\exp (n^{-1}{\sum }_i\log (x_i/x_0))$. Changing the common unit rescales $G$ by the same factor as the inputs. As a check, doubling one input and halving another retains $G$ while usually changing the arithmetic mean.

The arithmetic-geometric mean inequality gives $G\le A$. Apply that same inequality to the multiplicative inverses $1/x_i$ to obtain $H\le G$, where $H$ is the harmonic mean. Consequentlyfor positive inputs, with equality throughout exactly when all inputs agree. For $30$ and $60$, the three values are $H=40$, $G=30\sqrt{2}\approx 42.4264$, and $A=45$. They preserve, respectively, the sum of inverses, the product, and the sum. No one of these summaries preserves all three for unequal inputs.

For an exact dimensionless Hamiltonian $H$ and trial Hamiltonian $H_0$, the free energies obeywhere $F_0=-\log Z_0$ and the expectation uses the trial Gibbs distribution $Z_0^{-1}{e}^{-H_0}$.

The identity $e^{-F}=e^{-F_0}⟨e^{-(H-H_0)}{⟩}_0$ and the Jensen inequality for the convex exponential function give $\log ⟨e^{-(H-H_0)}{⟩}_0\ge -⟨H-H_0{⟩}_0$. Taking minus the logarithm proves the bound.

Here Heun's method means the explicit endpoint-averaging Runge-Kutta method: take an Euler predictor $Y_2=y_n+hf(t_n,y_n)$, sample $k_2=f(t_n+h,Y_2)$ there, and average it with $k_1=f(t_n,y_n)$ for the output. The second Runge-Kutta stage is evaluated at the predictor, not at the corrected endpoint. In particular it requires no implicit solve as the ODE trapezoidal rule does. Naming conventions vary between books; the formulas specify the method unambiguously.^[ref]

For smooth $f$, a Taylor expansion about $(t_n,y_n)$ gives $k_2=f+h(f_t+f_{y}f)+O(h^2)$, where $f_y$ is the state Jacobian matrix. Consequently the output is $y_n+hf+h^2(f_t+f_{y}f)/2+O(h^3)$, matching the exact solution through degree two by the chain rule. This gives local truncation error $O(h^3)$ and, with the usual bounded smoothness and Lipschitz continuity hypotheses on a fixed finite interval, global error $O(h^2)$. The Euler-Heun embedded pair compares this output with its Euler predictor.

For $y^{\prime }=y$, $y_n=1$ and $h=0.2$, the sampled slopes are $1$ and $1.2$. Transporting the second slope to the starting point would end at $1.24$; the average slope $1.1$ instead ends at $1.22$. The exact exponential function reaches $e^{0.2}\approx 1.22140275816$. The magnified endpoint comparison separates the pair difference $0.02$ from the actual Heun local defect $0.00140275816$.

Here is an explicitly synthetic mathematical model of a warming disturbance and its compensatory cooling. It is a teaching analogy for thermoregulation, not a measured human heat budget. Write $x=(T-T_r)/\Delta T$, where $T$ is a lumped internal temperature, $T_r$ is a fixed reference and $\Delta T>0$ is a chosen temperature scale. Let $d$ be an additional heating rate and $u$ an additional cooling rate, both scaled to the same reference rate. All three are dimensionless variables, while $\tau >0$ is a time constant in the chosen time units. The ordinary differential equation isThe term $-x$ idealizes increased passive heat transfer with increased internal temperature, around the baseline. The term $-u$ opposes warming through proportional control with feedback gain $k$. This local signed model does not treat a negative $u$ as negative sweating: it would represent decreased cooling relative to baseline. Changes of the environment and multiple body compartments are omitted.

For a constant disturbance $d=D$ and initial condition $x(0)=0$, the exponential function solution isSubstitution checks the ordinary differential equation directly. Increasing $k$ reduces the lasting offset and, in this instantaneous model, shortens the response time constant to $\tau /(1+k)$. It does not make $x_{\ast }$ zero for a nonzero disturbance and finite $k$: a persistent cooling command requires a persistent displacement under proportional control. At the equilibrium point, the extra heating is balanced by passive loss $x_{\ast }$ plus active cooling $u_{\ast }$. Zero change of temperature means balanced rates, rather than zero rates, consistent with conservation of energy.

The first diagram uses $D=1$, $\tau =4$, and a pulse from $t=2$ to $t=10$, comparing $k=0$ with $k=3$. The latter approaches $x_{\ast }=1/4$ during the pulse, compared with $1$ without active feedback control. After the pulse ends, each displacement decays as $x(10)e^{-(1+k)(t-10)/\tau }$. These are computed curves, not physiological measurements. The $k=0$ curve still has passive loss; it is not a claim that an unregulated body warms without limit.

Real physiological effectors take time to respond. To isolate that issue, set $\tau =1$ and replace the instantaneous command by another ordinary differential equation:Here the disturbance remains $1$ from $t=0$ onward and the effector time constant is also $1$. The equilibrium point is still $(x_{\ast },u_{\ast })=(1/5,4/5)$, but the characteristic polynomial is $(\lambda +1{)}^2+4$. Its eigenvalues $-1\pm 2i$ give decaying oscillations and asymptotic stability. The exact response isIn particular, $\dot{x}=e^{-t}\cos (2t)$, so the first maximum occurs at $t=\pi /4$, with $x=1/5+(2/5)e^{-\pi /4}\simeq 0.3824$. The response exceeds its eventual displacement while remaining above its baseline. The return has the same negative sign as before; the extra state changes the timing. A finite effector time constant is distinct from the fixed time lag in the delayed negative feedback model.

A homogeneous linear differential equation is a linear differential equation with identically zero forcing term, $L[y]=0$. On a common interval, its solutions of a differential equation form the vector space $\ker L$, the kernel of a linear map. They include the zero function and are closed under addition and scalar multiplication. For any two solutions of a differential equation $u,v$ and constant scalars $\alpha ,\beta$, linearity gives $L[\alpha u+\beta v]=\alpha 0+\beta 0=0$: this is the superposition principle. The scalars must be constant; variable multipliers introduce extra derivatives.

For example, $y^{\prime }+2y=0$ has general solution $y=C{e}^{-2t}$, where $C$ is any constant real number. A short verification uses the product rule: $(e^{2t}y{)}^{\prime }=e^{2t}(y^{\prime }+2y)=0$, so $e^{2t}y$ is constant. For $y_1=e^{-2t}$ and $y_2=2e^{-2t}$, the sum is $3e^{-2t}$ and also solves the same homogeneous linear differential equation. At $t=1$, their values are approximately $0.13534$, $0.27067$ and $0.40601$. Every curve is a multiple of one exponential function; the solution space of a homogeneous linear differential equation has dimension one. Both $t$ and $y$ in this example and the following diagram are dimensionless mathematical variables.

Now keep the same differential operator $L[y]=y^{\prime }+2y$ but prescribe the forcing term $1$. The inhomogeneous linear differential equation $y^{\prime }+2y=1$ has particular solution $y_p=1/2$ and general solution $y=1/2+C{e}^{-2t}$. The initial conditions $y(0)=0,1,2$ select, respectively,At $t=1$, their values are approximately $0.43233$, $0.56767$ and $0.70300$. The midpoint $\frac{1}{2}(u_0+u_2)=u_1$ solves the same inhomogeneous linear differential equation, because $L[\frac{1}{2}(u_0+u_2)]=\frac{1}{2}(1+1)=1$. The sum $s=u_0+u_2=1+e^{-2t}$ instead satisfies $s^{\prime }+2s=2$, so it fails the equation with forcing term $1$. More generally, a linear combination of solutions of a differential equation for the same fixed nonzero forcing term retains that forcing term precisely when its scalar weights add to one. This is the affine space structure, rather than unrestricted superposition within that fixed family.

On the left, all three curves solve the same homogeneous linear differential equation. On the right, $u_0,u_1,u_2$ share the particular solution $1/2$, and the shaded difference $u_2-1/2$ is the homogeneous solution $\frac{3}{2}{e}^{-2t}$. The dashed sum has twice the forcing term, even though its starting value coincides with that of $u_2$. Adding the two initial conditions also adds their inputs; taking their midpoint averages both. This distinction makes the superposition principle useful for assembling responses to different inputs without mistaking that assembly for a solution of a differential equation with the original input. See MIT's superposition notes, including the particular solution one half.

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.