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An arithmetic series adds the sequence terms of an arithmetic progression. Its $n$-th partial sum is $S_n={\sum }_{k=1}^n{a}_k$, with the empty sum convention $S_0=0$. For $n\ge 1$,Thus the total is the number of sequence terms multiplied by the arithmetic mean of the two endpoints. The formula also covers decreasing and constant arithmetic progressions.^[ref]

For the eight-entry example, the sum is $8(7+28)/2=140$, and the arithmetic mean of all eight entries is $17.5$. A decreasing example $11,9,7,5,3,1$ gives $6(11+1)/2=36$. Negative values cause no algebraic difficulty: $3,1,-1,-3,-5$ has sum $5(3-5)/2=-5$.

The triangular number $T_n=1+2+\cdots +n$ is the special case $a_1=d=1$, giving $n(n+1)/2$. More generally, an equal increase in each successive allocation makes the total grow as a quadratic function of the number of allocations. The eight-entry example hasIts first four entries add to $46$, its first eight to $140$, and its first twelve to $282$. The sequence terms lie on a straight line when plotted against their sequence indices, while the partial sums follow a quadratic function.

The capacitor energy is the electrostatic energy stored when electric charge is separated. For a linear capacitor of fixed positive capacitance $C$, plate electric charge magnitude $Q$, and terminal electric potential difference $V=Q/C$, it isThe zero is the uncharged state. Thus the capacitor energy is positive for nonzero $Q$, even though the total electric charge of the two plates is zero. ^[ref].

For the parallel-plate capacitor example, capacitance $C=8.854\times 10^{-11}\mathrm{F}$ and terminal electric potential difference $V=10\mathrm{V}$ giveThe plate electric charge magnitude is $Q=0.8854\mathrm{nC}$. Doubling terminal electric potential difference with capacitance unchanged doubles the plate electric charge magnitude and quadruples the capacitor energy.

The electric charge is a linear function of terminal electric potential difference, while capacitor energy is a quadratic function. The second curve in each plot uses twice the capacitance: at a given terminal electric potential difference, both plate electric charge magnitude and capacitor energy double. These relations let us calculate how much energy an electrical circuit can receive when a capacitor discharges.

A brief charging argument explains the factor $1/2$. If the plate electric charge magnitude has reached $q$, the terminal electric potential difference is $q/C$. Transferring a further amount $dq$ of electric charge quasistatically requires work $dW=(q/C)dq$. With the geometry and permittivity fixed, the integral isUsing $Q=CV$ gives the other two forms. This integral accounts for the gradually rising terminal electric potential difference; it is the work stored in the capacitor's electric field, rather than a claim that all energy drawn from a fixed-voltage battery during charging ends up stored.

The field picture gives the same result. In the vacuum gap of a parallel-plate capacitor, the electric field has energy density $u={ϵ}_0{E}^2/2$. Multiplying by the gap volume $Ad$ gives the electrostatic energyFor the worked example, $u=4.427\times 10^{-4}J/{\mathrm{m}}^3$ and $Ad=10^{-5}{\mathrm{m}}^3$ reproduce $U=4.427\mathrm{nJ}$. The electrostatic energy resides in the electric field throughout the gap, despite overall charge neutrality.

When capacitance changes, specify the constraint. At fixed plate electric charge magnitude $Q$, increasing capacitance decreases capacitor energy $Q^2/(2C)$; doubling capacitance halves it. At fixed terminal electric potential difference $V$, increasing capacitance increases capacitor energy $C{V}^2/2$; doubling capacitance doubles it. In the latter case a connected battery supplies additional electric charge and energy, and there may also be mechanical work. Use the fixed-voltage capacitor energy balance to account for all transfers before inferring a force from changing capacitor energy.

The chain rule calculates the derivative of a function composition by combining local rates. If the inner function $f$ is differentiable at $a$ and the outer function $g$ is differentiable at $f(a)$, thenThe evaluation point matters: the outer derivative is evaluated at the inner output $f(a)$, not at $a$. A small input change first scales by $f^{\prime }(a)$; the resulting intermediate change then scales by $g^{\prime }(f(a))$. In terms of linear approximations, composing those two local descriptions multiplies their scale factors. ^[ref].

For example, $h(x)=\sin (x^2)$ applies a quadratic function and then the sine. With $f(x)=x^2$ and $g(u)=\sin u$, the chain rule givesHere all angles are in radians. Near $x=1$, increasing the input by $\Delta x$ changes $x^2$ by about $2\Delta x$, and the sine changes by about $\cos 1$ times that intermediate change. For $\Delta x=0.01$, the linear approximation predicts an output increase of $0.0108060$; the exact increase is $\sin (1.0201)-\sin 1\approx 0.0106894$.

The same rule handles several nested stages. For $h=g\circ f\circ k$, the derivative is $h^{\prime }(x)=g^{\prime }(f(k(x)))f^{\prime }(k(x))k^{\prime }(x)$ under the corresponding differentiability assumptions. Writing each intermediate output prevents evaluating a derivative at the wrong stage. It also makes the chain rule useful for propagating sensitivities through a mathematical model, where one measured quantity influences another through several transformations.

For a quick proof idea, replace $f$ near $a$ by its linear approximation, then replace $g$ near $f(a)$ by its linear approximation. The composite leading term is their product; differentiability ensures the remaining error is smaller than the original input change. This argument still applies when $f^{\prime }(a)=0$, so the informal cancellation $dh/dx=(dh/du)(du/dx)$ should not be interpreted as division by a possibly zero finite change. The hypotheses are sufficient, but are not necessary for every composite to be differentiable, as the nondifferentiable inner function with a differentiable nondegenerate composite example shows.

The codomain of a function is its specified target set. The image of a function is a subset of the codomain, and equals it precisely for a surjective function. Changing a codomain can change whether the same input-output rule is surjective, as the quadratic function examples under function show.

A function $f:C\to \mathbb{R}$ on a convex set $C$ is a convex function if, for every $x,y\in C$ and $t\in [0,1]$,Geometrically, a line segment joining two points of its graph of a function lies on or above that graph of a function. The convex set condition ensures that the intermediate input stays in the domain of a function. An affine function satisfies the defining inequality with equality.

For the quadratic function $f(x)=x^2$, the inputs $1$ and $3$ have equally weighted arithmetic mean $2$. Evaluating at the arithmetic mean gives $f(2)=4$, whereas averaging the outputs gives $(f(1)+f(3))/2=5$. Averaging first saves one unit in this example. More generally, direct expansion givesThis identity proves convexity and also shows exactly when its defining inequality is strict.

For a differentiable convex function, the tangent line in one dimension, or the corresponding supporting affine function in several dimensions, lies below the entire graph of a function. This is the subgradient inequality. The equivalent geometric statement is that the epigraph, consisting of points on or above the graph of a function, is a convex set. These characterizations are developed in Boyd and Vandenberghe's convex-function slides, sections 3.1–3.4.

Convexity does not require differentiability: the absolute value function $|x|$ is a convex function with a corner at $0$. When a function is twice continuously differentiable on an open interval, it is a convex function exactly when its second derivative is nonnegative throughout that interval. In several dimensions, the corresponding criterion on an open convex set is that its Hessian matrix is positive semidefinite everywhere. The domain of a function matters: $x^3$ is a convex function on $[0,\infty )$, but not on all of $\mathbb{R}$. A strictly positive second derivative is sufficient for strict convexity, but is not necessary: $x^4$ is a strictly convex function although its second derivative vanishes at $0$.

The usefulness of convexity is that local information can certify a global conclusion. Every local minimum of a convex function on a convex set is a global minimum: if a lower value existed elsewhere, the defining inequality would produce lower values arbitrarily close to the proposed local minimum. For a differentiable convex function on an open convex set, a zero gradient therefore certifies a global minimum. Strict convexity gives uniqueness when a global minimum exists; it does not supply existence by itself. These facts explain the role of convex functions in convex optimization.

The covariance measures the average product of two centered deviations. For real square-integrable random variables $X$ and $Y$,Deviations with the same sign contribute positively; deviations with opposite signs contribute negatively. Their magnitudes and probabilities determine the net result. Taking $Y=X$ recovers the variance. The definitions and algebraic properties are presented in Orloff and Bloom, MIT 18.05, Covariance and Correlation, sections 2–2.1.

For a worked example, let $X,Z$ be independent random variables, each equally likely to be $-1$, $0$ or $1$, and put $Y=2X+Z$. All nine pairs $(X,Z)$ have probability $1/9$. The expected values are zero, while linearity of covariance and the vanishing covariance of independent random variables giveThe shared $X$ term makes large $X$ accompany large $Y$ on average. Replacing $2X$ by $-2X$ reverses the covariance to $-4/3$ without changing either variance.

The covariance is useful whenever deviations are added or subtracted. Expanding the square givesFor the worked model, these are $20/3$ and $4/3$. Positive covariance increases variation in the sum while reducing variation in the difference: the same shared fluctuation reinforces one combination and partly cancels in the other. The identity applies beyond independent random variables.

The size of a covariance depends on the measurement scale. For constants $a,b,c,d$, linearity of covariance gives $\mathrm{Cov}(aX+b,cY+d)=ac\mathrm{Cov}(X,Y)$. Adding offsets has no effect; rescaling either measurement rescales the covariance. The Pearson correlation coefficient divides by both standard deviations to describe the standardized linear relationship.

Adding one sequence term changes a partial sum by exactly that entry, so $S_{n+1}-S_n=a_{n+1}$. Taking another finite difference givesThus arithmetic sequence terms produce partial sums with constant second finite differences. Conversely, if these second finite differences are a fixed $d$ and $a_{n+1}=S_{n+1}-S_n$, then those entries form an arithmetic progression. For nonzero $d$, the displayed sum formula is a quadratic function; for $d=0$, it reduces to the linear expression $S_n=n{a}_1$.

An infimal convolution is the cheapest way to split a prescribed total between two costs. Suppose two activities contribute $u$ and $v$ to a required total $x$, costing $f(u)$ and $g(v)$. Their combined cost is optimized over all splits $u+v=x$, rather than charging both activities for the entire total.

For proper extended-real functions $f,g:X\to \mathbb{R}\cup \{+\infty \}$ on a real vector space $X$, defineThe definition does not require convexity. An infinite cost forbids that contribution, as with an indicator functional of a constraint set. If no split has two finite costs, set $h(x)=+\infty$; if the costs of admissible splits are unbounded below, $h(x)=-\infty$. Thus proper inputs need not produce a proper extended-real function. The notation $E\square F$ denotes the same operation. See Parikh and Boyd, Proximal Algorithms, §3.1 for the standard convex analysis definition and its connection to the Moreau envelope.

An infimum is a best lower bound, whereas a minimum is a value realized by some split. Writing $\min$ in place of $\inf$ requires an attaining pair $(u_{\ast },v_{\ast })$. For convex functions, mixing two nearly optimal splits and applying convexity to each cost proves that their infimal convolution is convex whenever it is proper; one must use nearly optimal splits because exact minimizers need not exist. Its convex conjugate is $f^{\ast }+g^{\ast }$ whenever the extended real number arithmetic is well defined; see conjugate of an infimal convolution for the precise identity. This turns the optimization over splits into addition in the conjugate variables.

Two quadratic costs: share the load according to its price. Let $X=\mathbb{R}$ and let the convex quadratic functions beFor fixed $x$, eliminate $v=x-u$. The derivative of the total cost is $au-b(x-u)$, so equal marginal costs giveThe cheaper activity receives more of a positive total: when $a<b$ and $x>0$, $u_{\ast }>v_{\ast }$. Completing the square gives an exact certificate for every competing split:Consequently the infimum is a unique minimum, andThe effective quadratic coefficient is $c=ab/(a+b)$, with $1/c=1/a+1/b$. It is smaller than either $a$ or $b$: allowing both activities reduces the cost compared with assigning everything to either one. If $a=b$, each receives half the total and $c=a/2$. This is a simple quadratic optimization problem whose answer preserves the form of the original costs.

Smoothing a corner by splitting its cost. Take the absolute value function $f(u)=|u|$ and the convex quadratic function $g(v)=v^2/(2t)$, with $t>0$. This infimal convolution is the Moreau envelope of the absolute value function:Here $\mathrm{sgn}$ is the sign function, with $\mathrm{sgn}(0)=0$. The optimal $u_{\ast }$ is soft thresholding. In the central region, the quadratic function carries the whole total; outside it, $v_{\ast }$ stays at $t\mathrm{sgn}(x)$ and the absolute value function carries the excess. At $t=1$ the result is the Huber loss in the normalization $x^2/2$ near zero and $|x|-1/2$ outside. Its derivative is $x$ for $|x|\le 1$ and $\mathrm{sgn}(x)$ outside, so the corner becomes a continuous change of slope. Parikh and Boyd's ^[ref] uses twice this normalization for the Huber loss; the explicit formula here fixes the convention. This is also the one-dimensional case of the existing Moreau envelope of the Euclidean norm.

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 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.

For $N>0$, dividing the derivative by abundance gives the instantaneous per-capita rate:It decreases linearly with $N$: it approaches $r$ at low abundance, is $r/2$ at $N=K/2$, vanishes at the carrying capacity, and is negative above it. This per-capita rate differs from the total rate $N^{\prime }$, which also counts how many individuals contribute. It is an instantaneous rate, rather than a fixed percentage increase over a whole day. The linear per-capita assumption and the resulting qualitative picture are developed in MIT 18.03SC's logistic-model notes.

Writing $f(N)=rN(1-N/K)$, the equilibrium points satisfy $f(N)=0$, so the constant solutions of a differential equation are $N(t)=0$ and $N(t)=K$. The phase line points upward for $0<N<K$ and downward for $N>K$. Also $f^{\prime }(0)=r>0$ and $f^{\prime }(K)=-r<0$: the criterion for fixed point stability for an autonomous differential equation makes zero an unstable equilibrium and $K$ an asymptotically stable equilibrium point. Every positive initial condition approaches $K$ as $t\to \infty$; $N_0=0$ stays exactly zero. Instability of zero means that a small positive population grows away from it, not that a zero population spontaneously appears.

For every $N_0>0$, the unique solution of a differential equation for all $t\ge 0$ isThe denominator is positive on this domain, including when $N_0>K$. For $0<N_0<K$, this is a translated and scaled logistic function; its extension to all real times is a sigmoid function. For $N_0>K$ the coefficient of the exponential function is negative, and the future curve instead decreases toward $K$. No nonconstant positive solution of a differential equation crosses $K$ or reaches it at finite time. A short derivation uses the dimensionless variable $z=K/N-1$: the chain rule gives $z^{\prime }=-rz$, so exponential decay yields $z(t)=(K/N_0-1)e^{-rt}$. The substitution divides by $N$, so the zero equilibrium point must be included separately.

The total growth rate is a downward-opening quadratic function of abundance. Completing the square gives its maximum without solving the differential equation:The balance is beautiful: more individuals contribute to growth, but each contributes less, and the strongest total increase occurs halfway to the carrying capacity. This is a maximum over possible abundances. A trajectory starting above $K/2$ never passes through that abundance in its future.

The chain rule connects this maximum to the shape of the time trajectory through its second derivative:For $0<N_0<K/2$, the trajectory initially has convexity and then concavity, with its one future inflection point atFor $N_0=K/2$, this point lies at the initial time, with concavity for $t>0$. For $K/2<N_0<K$, the inflection point belongs to the past extension, so the future trajectory has concavity throughout. Above $K$, the decreasing trajectory has convexity and no inflection point. The two constant solutions of a differential equation have no change of concavity. Thus an entire logistic function has an S shape, but its observed future segment need not show both bends. See also ^[ref].

For the plotted example with $K=1000$, $r=0.4{\mathrm{day}}^{-1}$ and $N_0=100$, time $t$ measured in days givesThe initial total rate is $N^{\prime }(0)=0.4\cdot 100\cdot 0.9=36$ individuals per day, and the initial per-capita rate is $0.36{\mathrm{day}}^{-1}$. The fastest total increase is $rK/4=100$ individuals per day, reached at $t_{1/2}=\ln 9/0.4\approx 5.49306$ days. At that time the per-capita rate has already fallen to $0.2{\mathrm{day}}^{-1}$.

These parameters are illustrative choices, rather than an empirical fit. The equation specifies deterministic evolution under a phenomenological assumption: it summarizes how net growth depends on abundance without separately modelling every process that affects it. The carrying capacity is a parameter for a specified setting, not a universal constant for all populations. In his 1838 paper "Notice sur la loi que la population suit dans son accroissement", Correspondance mathématique et physique 10, pp. 113–121, Pierre-François Verhulst proposed $p^{\prime }=mp-n{p}^2$, equivalent to this equation with $r=m$ and $K=m/n$. He also noted that different assumed growth-retarding terms could describe the observations similarly while giving very different limiting populations. This is an early illustration of why a good-looking curve alone does not establish a unique long-term prediction.^[ref]

The logistic differential equation describes continuous time. The logistic map instead advances in discrete steps and can exhibit a period-doubling cascade and Devaney chaos; the monotone positive trajectories here cannot. Their connection through the forward Euler method is explained under Euler discretization of logistic growth, where a large step can introduce behaviour absent from the original differential equation.

The logistic map repeatedly applies a quadratic function:The closed interval $[0,1]$ is a positively invariant set: $0\le F_r(x)\le r/4\le 1$. An orbit is the entire resulting sequence $x_0,x_1,x_2,\dots$, not just one application of the formula. For example, with $r=2$ and $x_0=0.2$, the first states are $0.2,0.32,0.4352,0.49160192,\dots$, approaching the fixed point $1/2$.

As a simple model of population dynamics, $x_n$ is a scaled abundance in generation $n$. When abundance is small, $x_{n+1}/x_n=r(1-x_n)$ is close to $r$; increasing abundance reduces this per-capita multiplier. The falling side of the quadratic function represents sufficiently severe crowding reducing the next generation. The scale $x=1$ is not generally the carrying capacity: the positive fixed point is $1-1/r$. The logistic map is a discrete-generation model; the continuous-time logistic growth equation has a different evolution rule and does not produce its period-doubling cascade.

The practical question is how abundance behaves after the initial adjustment: does it settle, alternate, or continue varying? The same formula permits all these possibilities. The following figure shows three orbits, their cobweb plots, and a bifurcation diagram. In a cobweb plot, moving vertically to $y=F_r(x)$ calculates the next state, and moving horizontally to $y=x$ makes that value the input for the next step.

The fixed points solve $F_r(x)=x$:The second fixed point is in $[0,1]$ for $r\ge 1$ and coincides with zero at $r=1$. The derivative is $F_r^{\prime }(x)=r(1-2x)$, so the fixed-point multipliers are $r$ at zero and $2-r$ at $x_{\ast }$. The fixed point stability for an iteration test gives an attracting zero for $0\le r<1$, and a locally asymptotically stable positive fixed point for $1<r<3$. A multiplier of modulus one needs a nonlinear argument; the strict derivative test alone settles neither $r=1$ nor $r=3$.

This derivative test has a direct interpretation. A small error $e_n=x_n-x_{\ast }$ obeysAt $r=2.8$, the leading multiplier is $-0.8$: the error changes sign, explaining the alternating approach, while its magnitude eventually shrinks by approximately a factor $0.8$ per step. At $r=3.2$, the multiplier is $-1.2$, and the positive fixed point repels nearby states. However, a different outcome has asymptotic stability: the logistic map two-cycle attracts nearby orbits.

For $3<r<1+\sqrt{6}$, the attracting period-two orbit has the two valuesAt $r=3.2$ they are approximately $0.513045$ and $0.799455$, with $F_r(x_{-})=x_{+}$ and $F_r(x_{+})=x_{-}$. Their periodic-orbit multiplier is $4+2r-r^2=0.16$, so a small displacement shrinks over two steps even though the fixed point is unstable. The brief algebraic justification is to factor $F_r^2(x)-x$, remove the fixed point factors, and multiply the two derivatives using the chain rule.

At $r=3$, the positive fixed-point multiplier passes through $-1$ and a period-doubling bifurcation creates the period-two orbit. At $r=1+\sqrt{6}\approx 3.44949$, its periodic-orbit multiplier passes through $-1$ and an attracting period-four periodic orbit emerges. At $r=3.5$, the period-four values are approximately $0.382820,0.826941,0.500884,0.874997$ in that order. Further period-doubling bifurcations accumulate near $r\approx 3.56995$. Beyond that value there are also parameter windows with attracting periodic orbits, including a period-three window near $r=3.83$. Neither a crowded numerical plot nor a parameter beyond the accumulation value proves sensitive dependence on initial conditions on the attracting set.

The endpoint $r=4$ supplies an analytically transparent example. The homeomorphism $h(u)={\sin }^2(\pi u/2)$ satisfieswhere $T$ is the full tent map. Thus these maps have topological conjugacy; stretching and folding in the tent map explain sensitive dependence on initial conditions and Devaney chaos for $F_4$ on $[0,1]$. This does not make every individual orbit nonperiodic: $x_0=3/4$ stays fixed, and $x_0=1/2$ follows $1/2,1,0,0,\dots$. Sensitivity means that these exceptional orbits also have arbitrarily close initial states that eventually depart substantially.

Robert M. May's "Simple mathematical models with very complicated dynamics" (1976), Nature 261, pp. 459–467, reviewed how simple deterministic recurrence relations can pass from stable abundances through period-doubling cascades to irregular behaviour. May presented the logistic map as an especially accessible example and discussed the relevance of such models to population dynamics. The paper explicitly describes itself as a review of results found by several researchers, bringing their findings together for a broad scientific audience.^[ref]

For a real symmetric invertible matrix $H$, the full undamped Newton step for the displayed quadratic function sends every starting point to the unique stationary point:This follows because the gradient is affine and its first-order expansion is exact. If $H$ is a positive-definite matrix the result is the unique global minimum; a negative-definite matrix gives a global maximum, and an indefinite matrix $H$ gives a saddle point. Thus exact arrival does not by itself show that a minimization method has minimized anything. A line search, damping, or a positive-definite matrix replacing the Hessian matrix changes this exact one-step assertion.

The Cauchy point minimizes the model along the negative gradient ray inside the trust region. With $q=g^{T}Hg$, writeIf $q\le 0$, this last quadratic function decreases throughout the interval. If $q>0$, its unconstrained minimum occurs at $t=\Vert g{\Vert }^2/q$. ThusThe additive constant $a$ has no effect on the Cauchy point. To obtain the Cauchy decrease bound, first suppose $q>0$. Since $\tau q\le \Vert g{\Vert }^2$,The spectral norm inequality $q\le \Vert H\Vert \Vert g{\Vert }^2$ gives $\tau \ge \min \{\Delta /\Vert g\Vert ,1/\Vert H\Vert \}$. HenceIf $q\le 0$, the decrease is at least $\Delta \Vert g\Vert$, which also implies the displayed bound. When $H=0$, interpret $\Vert g\Vert /\Vert H\Vert =+\infty$. Thus the universal choice is $c=1/2$.

The first term is the bending energy of rod 1, whose preferred bending strain of a rod is zero. The second is the bending energy of rod 2 relative to its preferred intrinsic curvature of an elastic filament $\hat{u}$. The third is its stretching energy, with stretching stiffness $k{B}_2$ and stress-free stretch ratio ${\alpha }^{+}=1$. Each term is an energy per unit reference arc length before integration.

Substitute $u^{+}=u$ and ${\alpha }^{+}-1=-au$. The resulting quadratic function of $u$ has densityThere are no derivatives of $u$ or imposed end constraints in this minimum potential energy principle in elasticity. Sincethe unique pointwise minimizer is also the constant global minimizer:For example, completing the square giveswhich also proves uniqueness, up to equality almost everywhere, without merely assuming a uniform curvature.

With $k$, $a$ and $\hat{u}$ fixed, the two limits areIn the first limit, rod 1 dominates the bending energy and suppresses the bend. In the second, rod 1 offers negligible bending resistance, but following rod 2's preferred bending strain of a rod still shortens rod 2 through the bonded geometry. The competition between its bending energy and stretching energy leaves $u_{\ast }<\hat{u}$ when $\hat{u}>0$ and $k{a}^2>0$. Only when that stretching penalty also vanishes does $u_{\ast }\to \hat{u}$.

These formulae describe an admissible bonded elastic rod provided $1-a{u}_{\ast }>0$. If parameters instead force $a{u}_{\ast }\ge 1$, the unconstrained quadratic minimizer is outside the kinematic regime of the model; the given energy does not itself enforce positive stretch ratio.

In the parametric description, $a$ is a point of the straight line and the nonzero vector $b$ gives its direction. The real parameter runs through every point of that line. For the cross product description, take another cross product with $c$ and use the vector triple product identity:Hence every solution has the representationHere $a\cdot b=0$ and $|b|=|c|$, as required. Conversely,Thus the perpendicularity assumption is sufficient, and the solutions form a straight line parallel to $c$. When $d\ne 0$, the vector $d$ is perpendicular both to $c$ and to every position vector on the line; it is the normal to the plane through the origin containing that line. The point $a=c\times d/|c{|}^2$ is the perpendicular foot from the origin. Its distance from the origin is $|d|/|c|$, since $c\cdot d=0$.

For a general parametric straight line, minimize the squared Euclidean distanceThis strictly convex quadratic function has its unique minimum at $\lambda =b\cdot (y-a)/|b{|}^2$, giving the orthogonal projectionFor the cross product description, $a\cdot c=0$, so this becomes

For the two planes, put $s=m\cdot n$ and $c=m\times n$. Nonparallel unit normals give $|s|<1$ and $|c{|}^2=1-s^2$. Seek the perpendicular foot in their span, $a=um+vn$. Its two dot products impose $u+sv=\mu$ and $su+v=\nu$. Solving gives both requested descriptions of the intersection:andIndeed the vector triple product identity gives $x\times (m\times n)=m(x\cdot n)-n(x\cdot m)$. Conversely, since $m,n$ are linearly independent, equality to $\nu m-\mu n$ forces both plane equations. Thus neither description introduces extra points.

Substitute $y=y_0+y_F$ into the energy functional and integrate the resulting polynomial:The term linear in $F$ cancels. HenceAs required, $E_0$ is independent of the force and $\mathcal{E}$ is a quadratic function of it.

A linear operator $\alpha$ on a real inner-product space is self-adjoint whenfor all $v,w$.

To prove the finite-dimensional spectral theorem, first note that the continuous quadratic function $⟨\alpha v,v⟩$ has a maximum on the unit sphere. The Lagrange multiplier equation at a maximizing vector $v$ gives $\alpha v=\lambda v$, so $\alpha$ has a real unit eigenvector. Its orthogonal complement is invariant becauseInduction on the dimension supplies an orthonormal eigenbasis of that complement and hence of $V$.

For $f\ne 0$,while symmetry and bilinearity are immediate, so the displayed formula defines an inner product on $P_n$. Sinceintegration by parts givesThe boundary term vanishes for polynomials, proving that $\alpha$ is self-adjoint.

On the monomial $x^k$,Thus the matrix of $\alpha$ in the monomial basis is triangular with diagonalThese are therefore its eigenvalues. For $n=2$, corresponding eigenvectors arefor eigenvalues $0,-1,-2$, respectively. They are the first Laguerre polynomials up to normalization, as described by the Laguerre differential operator on polynomials.

Setting $\nu ={\nu }_b=0$, the dispersion relation reduces toAn axisymmetric density mode grows whenThe left-hand side is a concave quadratic function of $k$, with maximum $(\pi G{\Sigma }_0{)}^2/v_s^2$ at $k=\pi G{\Sigma }_0/v_s^2$. Growth is therefore possible precisely when the Toomre stability criterion is violated:

Multiply the equation of motion by $\dot{x}$ and integrate. The conserved mechanical energy isThe initial conditions $x=\ell$, $\dot{x}=0$ give $E=-C/\ell$, soInitially $\ddot{x}=-C/(m{\ell }^2)<0$, so the particle starts moving towards the origin. For $0<x<\ell$, factor its kinetic energy asThe quadratic function $x(\ell -x)$ has maximum ${\ell }^2/4$ at $x=\ell /2$. Consequently kinetic energy is strictly positive throughout $0<x<\ell$ exactly when $k{\ell }^3<8C$. Thus the collision threshold isFor ${\ell }_1<\ell <{\ell }_2$ there is no intervening turning point, and the particle collides with the fixed particle. The collision takes finite time: near zero, $|\dot{x}|\sim \sqrt{2C/(mx)}$, so the remaining time is bounded by a constant times ${\int }_0^{\delta }\sqrt{x}dx$.

For $\ell >{\ell }_2$, the first turning point encountered isThe particle oscillates periodically between $x_T$ and $\ell$, around the larger stable equilibrium, without collision. The zeros of $E-V$ at the two endpoints of this allowed region are simple, so travel to each turning point takes finite time. At the boundary $\ell ={\ell }_2$, $x=\ell /2$ is a double zero of $E-V$ and an unstable equilibrium; the inward trajectory approaches it only as $t\to \infty$.

Assume $C,Q>0$. The expected net profit per item is $\Pi (\alpha )=Q\mathbb{E}[T]-C/{\alpha }^2=Q/\alpha -C/{\alpha }^2$. Put $u=1/\alpha >0$ and complete the square:This quadratic function has a unique global maximum at $u=Q/(2C)$. HenceThe calculation uses the expected operating time, rather than replacing lifetime by an arbitrary fixed time.