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The scaled backward finite difference uses the current and preceding nodal values to approximate a first derivative. If a function has two continuous derivatives, Taylor's theorem gives $D_h^{-}u(x)=u^{\prime }(x)-h{u}^{\prime \prime }(x)/2+o(h)$ as $h\to 0$. Composition with the scaled forward difference operator gives $D_h^{+}{D}_h^{-}u(x)=(u(x+h)-2u(x)+u(x-h))/h^2$, the second-order central difference for the second derivative.

For $d(a,b)=a\log (a/b)+(1-a)\log ((1-a)/(1-b))$, the second derivative with respect to $a$ is $1/[a(1-a)]\ge 4$. Its value and first derivative at $a=b$ vanish, so $d(a,b)\ge 2(a-b{)}^2$. Boundary values follow by continuity or the infinite-divergence convention. Together with the binary partition bound for relative entropy, this gives Pinsker's inequality with the natural-logarithm constant.

A clamped cubic spline is a cubic spline whose endpoint first derivatives are prescribed. A clamped cubic spline interpolant additionally matches prescribed values at its spline knots. Distinct spline knots, arbitrary nodal values and arbitrary endpoint slopes determine one such interpolant. The minimum roughness property of a clamped cubic spline proves uniqueness. Unlike the natural cubic spline, this construction prescribes endpoint slopes rather than zero endpoint second derivatives.

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.

Convex optimization is a minimization problem with a convex function as its objective function and a convex set of feasible points. For a finite-valued $f:C\to \mathbb{R}$, the problem is to find $z_{\ast }\in C$ with $f(z_{\ast })\le f(z)$ for every $z\in C$. The displayed infimum describes its value even when no minimizing point exists. The two convexity requirements concern different objects: the allowed decisions and the scores assigned to them.

A convex set contains the whole line segment between any two of its points:A convex function lies below the interpolation of its endpoint values:Thus mixing two allowed decisions preserves feasibility, and the mixed score is no worse than the same mixture of endpoint scores. The objective function may have flat regions or corners. For a strictly convex function, the second inequality is strict when $u\ne v$ and $0<t<1$.

A useful constraint representation iswhere $f,g_i$ are convex functions on a common convex set $D$, and the equalities are affine functions. Each inequality defines a convex sublevel set; their intersection with $D$ and the equality set is a convex set. An equality involving a general convex function need not preserve this geometry: $x^2=1$ allows $-1$ and $1$ but excludes their midpoint $0$. See Boyd and Vandenberghe, Convex Optimization (2004), §4.2.1 for the standard problem form.

Why local information can certify a global answer. Every local minimum of a convex optimization problem is a global minimum, including minima on the boundary. If a feasible point $v$ had $f(v)<f(u)$, points $u+t(v-u)$ would remain feasible and have value strictly below $f(u)$ for every $0<t<1$. Taking $t$ arbitrarily small contradicts a local minimum at $u$. This argument explains the special power of convexity: a better distant decision would reveal an improving direction immediately nearby.

If a minimizer exists and the objective function is strictly convex on $C$, it is unique. Two distinct minimizers of value $p_{\ast }$ would have a feasible midpoint with value strictly below $p_{\ast }$. Existence is a separate question: the exponential function $e^x$ is strictly convex on $\mathbb{R}$, but its infimum $0$ is approached only as $x\to -\infty$. Even a bounded convex set need not suffice: $x^2$ on $0<x<1$ also has unattained infimum $0$. A continuous function on a nonempty compact set attains its minimum by the extreme value theorem.

Nor does strict convexity imply differentiability. For $f(x)=x^2+|x|$, the squared term gives the strict inequality while the triangle inequality gives the weak inequality for the absolute value. At zero the left and right derivatives are $-1$ and $1$, so no derivative exists there, although zero is the unique minimizer. Such objectives require subgradients or proximal operators rather than an ordinary gradient at every point.

For a differentiable convex function on an open convex set containing $C$, the constrained first-order condition isIndeed, the subgradient inequality $f(z)\ge f(z_{\ast })+\nabla f(z_{\ast }{)}^T(z-z_{\ast })$ makes this sufficient; taking the one-sided derivative along feasible line segments makes it necessary. At an interior minimizer, both directions are feasible and the gradient must vanish. At a boundary minimizer it can be nonzero: it only needs to make every feasible direction nondecreasing. See Boyd and Vandenberghe, §§3.1.3 and 4.2.3.

A fully computed nearest-point problem. Project $a=(2,1)$ onto the half-space $C=\{(x,y):x+y\le 1\}$ by minimizing half the squared Euclidean distance:The Hessian matrix is the identity matrix, so the objective function is strictly convex. The unconstrained minimizer $(2,1)$ violates the inequality. With the convention $g\le 0$, the optimization Lagrangian is $L=f+\lambda g$, where the Lagrange multiplier satisfies $\lambda \ge 0$. Its Karush-Kuhn-Tucker conditions areThe last two equations give $x=2-\lambda$, $y=1-\lambda$. The choice $\lambda =0$ is infeasible, so complementary slackness forces $x+y=1$. Consequently $3-2\lambda =1$, giving $\lambda =1$, $(x_{\ast },y_{\ast })=(1,0)$ and $f(x_{\ast },y_{\ast })=1$. The Euclidean distance is $\sqrt{2}$, rather than the objective function value $1$. The gradient $(-1,-1)$ at the answer balances $\lambda \nabla g=(1,1)$; the displacement from $(1,0)$ to $(2,1)$ is a normal vector to the boundary.

There is an exact certificate independent of numerical search. Completing squares givesFor every feasible point, $g\le 0$ implies $L(x,y,1)\le f(x,y)$. Thus every feasible objective is at least $1$, and $(1,0)$ attains this bound. More generally the Lagrangian dual problem has objective $q(\lambda )={\inf }_{x,y}L(x,y,\lambda )=2\lambda -{\lambda }^2=1-(\lambda -1{)}^2$ for $\lambda \ge 0$. Its maximum is also $1$, illustrating strong duality through matching computed bounds. This makes the example a model for Euclidean projection onto a convex set and the projection step in projected gradient descent.

What the multiplier conditions require. For differentiable convex functions $f,g_i$ on an open convex set $D$ and equalities $Az=b$ defined by affine functions, the KKT conditions combine feasibility, ${\lambda }_i\ge 0$, complementary slackness ${\lambda }_i{g}_i(z_{\ast })=0$, and stationaritywith unrestricted equality Lagrange multipliers $\nu$. They suffice to certify a global minimum without an additional regularity assumption: the optimization Lagrangian is convex, stationarity minimizes it, and feasibility with complementary slackness gives the same certificate as above. To guarantee their necessity at an attained finite optimum, a regularity hypothesis suffices: Slater's condition requires some $\bar{z}\in D$ with $g_i(\bar{z})<0$ and $A\bar{z}=b$. The projection example has the strictly feasible point $(0,0)$. See Boyd and Vandenberghe, §§5.2.3 and 5.5.3. This hypothesis guarantees suitable multipliers, not existence of a primal minimizer.

Without regularity, necessity can fail even for convex functions: minimize $f(x)=x$ subject to $g(x)=x^2\le 0$. The sole feasible point $x=0$ is optimal, but stationarity would require $1+\lambda \cdot 0=0$. No Lagrange multiplier works. There is no strictly feasible point, and the gradient of the constraint vanishes at the answer.

The nonconvex contrast. The polynomial $h(x)=(x^2-1{)}^2+0.2x$ has derivative $h^{\prime }(x)=4x^3-4x+0.2$ and second derivative $h^{\prime \prime }(x)=12x^2-4$. Numerically solving the cubic gives stationary points $-1.024120300$, $0.050125947$, and $0.973994353$. Their objective values are respectively $-0.202440434$, $1.005006281$, and $0.197434153$, with second derivatives $8.585869$, $-3.969849$, and $7.383980$. Hence both outer points are local minima, and the middle point is a local maximum. Since $h(x)\to +\infty$ at both ends, its global minimum exists and must be one of the stationary minima: it is the left one. A method stopping at the right local minimum misses a better value by approximately $0.399874587$.

The practical appeal of convex optimization is the ability to check an answer globally using geometry and matching Lagrangian duality bounds. The projection certificate certifies every feasible alternative at once; the tilted quartic shows why a zero gradient alone cannot do that for general nonlinear optimization.

Historically, an early landmark is Kuhn and Tucker's ^[ref]. Its §3 explicitly states a regularity hypothesis for inequality-constrained Lagrange multipliers; its discussion of saddle points explains how concave functions support global conclusions for maximization. This documents an early formulation of the multiplier approach; it does not establish sole discovery of today's KKT conditions or of convex optimization as a whole.

A covalent bond joins chemical atoms through shared electrons. In the familiar two-centre description, one bond pair contributes to a connection between two atomic nuclei. Sharing means that the electrons belong to the combined arrangement and interact with both atomic nuclei; it does not mean that identifiable dots shuttle along drawn paths. The ^[ref] emphasizes the concentration of electrons between atomic nuclei and the resulting attraction and characteristic separation. The pair picture is especially useful for simple molecules, but is not a universal description of every covalent bond.

A Lewis structure turns this picture into practical bookkeeping. Count the valence electrons, indicate which chemical atoms are connected, and allocate two electrons to each line. A bond pair is shared; a lone pair is assigned to one chemical atom without connecting it to another. A double line represents two bond pairs. Dots and lines record allocations, rather than positions, trajectories, or a measured three-dimensional arrangement. ^[ref]

For molecular hydrogen, ${\mathrm{H}}_2$, each neutral hydrogen chemical atom supplies one valence electron. Thus $1+1=2$ valence electrons make one bond pair, represented by $\mathrm{H}-\mathrm{H}$, with no lone pairs. Each hydrogen counts the same shared pair when checking its two-electron arrangement. The molecule still contains two electrons, not four: counting a pair around both endpoints does not create extra electrons.

For water, ${\mathrm{H}}_2\mathrm{O}$, neutral oxygen supplies six valence electrons and the two hydrogen chemical atoms supply one each: $6+1+1=8$. Two O–H connections consume four electrons as two bond pairs. The remaining $8-4=4$ electrons form two lone pairs on oxygen. Around oxygen, the two bond pairs and two lone pairs count as eight electrons; each hydrogen counts two. The whole water molecule has ten electrons, because two additional oxygen electrons are omitted from this valence electron drawing.

For Ammonia, ${\mathrm{NH}}_3$, the total is $5+3\times 1=8$ valence electrons: five from nitrogen and three from hydrogen. Three N–H connections use six electrons as three bond pairs, leaving two electrons as one lone pair on nitrogen. Thus water and Ammonia have the same total of eight valence electrons but different allocations: two plus two pairs versus three plus one. This small calculation already distinguishes their connectivity without a detailed quantum mechanics calculation.

For molecular oxygen, ${\mathrm{O}}_2$, $6+6=12$ valence electrons give the conventional $\mathrm{O}=\mathrm{O}$ Lewis structure. Its two bond pairs use four electrons, leaving eight as four lone pairs, two on each oxygen. Each oxygen consequently counts eight electrons locally. However, actual molecular oxygen has two unpaired electrons and exhibits paramagnetism: it is attracted into a stronger magnetic field. The all-paired Lewis structure cannot capture this observation. Molecular orbital theory describes the unpaired electrons; the successful counting diagram alone is insufficient. ^[ref]

Why does sharing produce a stable separation? For molecular hydrogen, the electronic arrangement responds as the atomic nuclei approach. Their interactions can lower the energy, but pushing them too close raises it. A curve of potential energy against separation therefore has a local minimum, rather than falling forever. Attraction and repulsion balance there. This is a statement about the combined system, not a claim that stationary electrons sit between otherwise motionless atomic nuclei. ^[ref]

The following Morse potential makes those features calculable. Set $x=r/r_e$, where $r$ is separation and $r_e$ its equilibrium value, and let $D$ be the well depth. With the self-chosen parameter $a=2$, define the dimensionless quantitiesHere $F=-dV/dr$ is the net separating force, obtained from the derivative of the potential energy. Positive $f$ pushes the chemical atoms apart; negative $f$ draws them together. These are illustrative curves with $r_e=D=1$ in chosen units, not measured molecular hydrogen data or fitted parameters.

At $x=1$, $u=-1$ and $f=0$. The second derivative is $u^{\prime \prime }(1)=8>0$, confirming a local minimum. At $x=0.5$, $u=(1-e{)}^2-1\approx 1.9525$; at $x=1.5$, $u=(1-e^{-1}{)}^2-1\approx -0.6004$. As $x$ increases without bound, $u$ approaches zero from below. Thus moving from the well bottom to separated chemical atoms requires an increase of $D$ in this model's potential energy. An actual quantum mechanics dissociation threshold also depends on the initial motion of the atomic nuclei, so the well depth alone is not every measured dissociation energy.

The force panel separates $f$ into a positive term $4e^{-4(x-1)}$ and a negative term $-4e^{-2(x-1)}$. At $x=1$ they are $+4$ and $-4$, cancelling exactly; nearer separations favour the positive term. These are algebraic components of the chosen Morse potential, not separately measured microscopic forces. Their usefulness is to show how a balance arises from the slope of one energy curve.

The beauty of the pair picture is its economy: a few marks summarize connectivity and constrain what a proposed molecule can contain. Its historical anchor is Gilbert N. Lewis's 1916 paper ^[ref], also available as an ^[ref]. Lewis structures remain useful because they answer counting questions quickly; quantum mechanics supplies the richer account when energy, shape, or unpaired electrons matter.

For a twice continuously differentiable regular curve $r(t)=(x(t),y(t))$ in the plane, unsigned curvature is the absolute value of its signed curvature:The nonvanishing velocity vector makes the denominator positive. For a graph of a function $(t,f(t))$, this reduces to $|f^{\prime \prime }(t)|/(1+f^{\prime }(t{)}^2{)}^{3/2}$; omitting the absolute value gives the signed curvature of a plane graph for increasing $t$.

For the parabola $r(t)=(t,t^2)$, the derivatives are $r^{\prime }=(1,2t)$ and $r^{\prime \prime }=(0,2)$. ThereforeThe curvature is greatest at the vertex $t=0$, where $\kappa =2$ and the radius of curvature is $1/2$. At $t=\pm 1/2$ the curvature is $1/\sqrt{2}$, so the parabola bends less sharply despite its larger speed $\sqrt{2}$.

Its unit tangent vector and left unit normal areSince $k>0$, $N_L$ is also its principal normal vector. At $t=0$ the osculating circle consequently has center $(0,0)+(1/2)(0,1)=(0,1/2)$ and equationThe lower branch $y=1/2-\sqrt{1/4-x^2}$ has value $0$, derivative $0$, and second derivative $2$ at $x=0$, matching the parabola $y=x^2$ through second order. Away from the vertex the two curves separate: an osculating circle describes a local bend, not the whole curve.

Vanishing curvature at one point does not imply a straight segment. The graph of a function $y=x^3$ has signed curvature $6x/(1+9x^4{)}^{3/2}$, which vanishes at its inflection point $x=0$ and changes sign across it. By contrast, if curvature vanishes throughout a connected regular interval, $dT/ds=0$ throughout that interval: the constant unit tangent vector forces the image to lie on a straight line.

This initial value problem uses a square root and absolute value to define a continuous function on all of ${\mathbb{R}}^2$. For every finite waiting time $c\ge 0$, a forward solution of a differential equation on $t\ge 0$ isThe constant solution of a differential equation $y_{\infty }(t)=0$ is another choice. For example, at $t=3$, waits $c=0,1,2,3$ give values $9,4,1,0$. These are dimensionless mathematical trajectories; no measured physical waiting time is being supplied.

Verification uses the original ordinary differential equation, including the join. Before departure, both $y_c^{\prime }$ and $2\sqrt{|y_c|}$ vanish. For $t>c$, their common value is $2(t-c)$. For $c>0$, the difference quotient at the join isThus the derivative exists at $c$, equals the rate law there, and is continuous across the join. For $c=0$ the same computation gives the right derivative at the starting endpoint. For $c>0$, the left and right second derivatives at departure are $0$ and $2$: these forward solutions of a differential equation are continuously differentiable functions, without being twice differentiable at departure. The local two-sided statement of the Peano existence theorem is separate from this explicitly forward family.

The rate law tends to zero as $y\to 0$, but its secant slopes from zero are unbounded:No finite local Lipschitz constant exists at zero. At $\epsilon =10^{-2},10^{-4},10^{-6}$ the quotient is $20,200,2000$, even though the rates are shrinking to zero. Continuity controls the rate itself; Lipschitz continuity would control its change relative to the state displacement.

There is also a numerical lesson. The forward Euler method starting exactly at zero gives $Y_{k+1}=Y_k+2\Delta t\sqrt{|Y_k|}=0$ at every step. It reproduces the constant solution of a differential equation for every positive step size, while the moving trajectories remain equally valid. A computation producing one trajectory cannot establish uniqueness. Likewise, the separable differential equation calculation that divides by $\sqrt{|y|}$ only describes the moving pieces: checking the original equation recovers the zero pieces and their joins.

Let $X_i$ be an independent and identically distributed sample from probability density function $f$, take $n\ge 2$ and $b>0$, and write $L_b(x)=b^{-1}L(x/b)$. Assume the kernel for density estimation $L$ has square-integrable second derivative. DefineThis U-statistic obeys $\mathbb{E}{\hat{J}}_{2,b}=\Vert (L_b\ast f{)}^{\prime \prime }{\Vert }_2^2$ whenever differentiation and the displayed integrals are justified. Under the usual convolution regularity this is $\Vert L_b\ast f^{\prime \prime }{\Vert }_2^2$.

The exact relation to the second derivative kernel density estimator isIndeed, expand the square into ordered pairs; its diagonal terms are all $A(0)=\Vert L^{\prime \prime }{\Vert }_2^2$. Both subtracting this diagonal and changing the normalization from $n^2$ to $n(n-1)$ are needed for the displayed U-statistic. This removes the diagonal sampling contribution, while retaining smoothing bias of an estimator relative to $\int (f^{\prime \prime }{)}^2$. The corrected value can be negative in a finite sample even though its target is nonnegative.

Consider the original small Poisson equation exampleIts exact solution is the quadratic polynomial $u(x)=1+2x-x^2$. Use a uniform mesh $x_i=ih$ with $h=1/4$ and three interior unknowns $U_1,U_2,U_3$. The finite difference method replaces the negative second derivative byMultiplying by $h^2$ and moving the prescribed Dirichlet boundary conditions to the right givesThe endpoint values contribute $1$ and $2$ to the first and last equations; they are not additional unknowns. The Thomas algorithm givesand back substitution yieldsAll three values equal $u(x_i)$ exactly: the centered finite difference for the second derivative is exact on a quadratic polynomial. This special nodal exactness does not assert exactness for arbitrary forcing or for the piecewise-linear curve joining the nodes.

For any $n$ interior points with $h=1/(n+1)$, the scaled tridiagonal matrix has $b_i=2$ and $a_i=c_i=-1$. Its pivot elements are $p_i=(i+1)/i$: start at $p_1=2$ and use $p_i=2-1/p_{i-1}$. All are positive. Interior rows have equality in diagonal dominance, so strict diagonal dominance is not the reason this example works. Instead it is a symmetric positive-definite matrix, since for a real vector $v\ne 0$ with $v_0=v_{n+1}=0$,Zero would force every neighbouring value equal and hence all values zero. This connects the boundary constraints directly to the safe pivots for tridiagonal elimination.

An energy has $\lambda$-displacement convexity when the displayed inequality holds along every displacement interpolation between endpoints in its domain. The case $\lambda =0$ is Displacement convexity; $\lambda >0$ gives uniform convexity along these metric geodesics. This is distinct from convexity along ordinary linear mixtures of probability measures. On subintervals, the inequality is equivalent to convexity of $t\mapsto E({\mu }_t)-\lambda W_2({\mu }_0,{\mu }_1{)}^2{t}^2/2$; if the energy is twice differentiable, it is equivalent to the corresponding second derivative lower bound. Without that regularity the lower bound is interpreted using distributions.

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.

The integrated squared second derivative measures a smooth density's curvature in L2. It supplies the leading integrated squared bias constant of a second-order kernel density estimator. For nonzero second moment ${\mu }_2(K)$, the asymptotic mean integrated squared error has bias term $h^4{\mu }_2(K{)}^{2}R(f^{\prime \prime })/4$, so greater curvature calls for a smaller smoothing bandwidth.

The Laplacian takes the divergence of a scalar field's gradient. For a real-valued function $f$ on an open region of Euclidean space, with continuous partial derivatives through order two, its expression in a Cartesian coordinate system isThus the Laplacian, also called the Laplace operator, adds the coordinate second derivatives; equivalently it is the matrix trace of the Hessian matrix. Its output is another scalar field.^[ref]

Three original examples distinguish shape from this sum. The bowl $f(x,y)=x^2+y^2$ has $\Delta f=2+2=4$. The saddle $g(x,y)=x^2-y^2$ instead has $\Delta g=2-2=0$: opposite second derivatives cancel. It is a harmonic function, although its graph of a function is not a plane. For the Gaussian function $h(x,y)=e^{-r^2}$, with $r^2=x^2+y^2$, differentiation givesIts Laplacian is negative inside the circle of radius one, zero on that circle, and positive outside it. The Gaussian function itself is positive throughout: its value and its Laplacian describe different properties.

The sum does not favor a particular orientation. Under a change of Cartesian coordinate system represented by an orthogonal matrix $Q$, the Hessian matrix becomes $Q^{\mathsf{T}}(D^{2}f)Q$. The cyclic property of the trace and $Q{Q}^{\mathsf{T}}=I$ preserve its matrix trace. This makes the Laplacian useful when the behavior depends on spatial differences equally in every direction, rather than on the arbitrary choice of axes.

For a vector field, the Euclidean Laplacian acts separately on components expressed in a fixed Cartesian coordinate system. Components in a position-dependent basis require differentiating that basis too; applying a scalar formula separately to polar components omits those terms.^[ref] The scalar formula in Laplacian in polar coordinates is valid away from $r=0$; smoothness at the origin must be handled in coordinates that remain regular there.

The Laplacian detects whether a value lies below or above nearby values, to second order. For a scalar field whose partial derivatives through order four are continuous functions, take the arithmetic mean of four equally spaced neighbors:Applying Taylor's theorem in each coordinate cancels the first- and third-order terms between opposite neighbors. The two pairs leaveThis is an asymptotic expansion, rather than an equality for arbitrary $h$. Rearranging gives the familiar central finite difference approximationThis finite difference replaces partial derivatives by five sampled values. The arithmetic mean lies above the center for a sufficiently small $h$ if the Laplacian is strictly positive there, and below if it is strictly negative. A zero Laplacian cancels this second-order discrepancy without requiring equality of individual values.

At $(0,0)$, the bowl $x^2+y^2$ has center value zero and four neighbor values $h^2$, so the discrepancy is exactly $h^2$. The saddle $x^2-y^2$ has neighbor values $h^2,h^2,-h^2,-h^2$, whose arithmetic mean is exactly zero. For the Gaussian function $e^{-x^2-y^2}$ the discrepancy is $e^{-h^2}-1=-h^2+O(h^4)$, agreeing with $\Delta f(0,0)=-4$.

A harmonic function need not equal this four-neighbor arithmetic mean for a finite $h$. The polynomial $f(x,y)=x^4-6x^2{y}^2+y^4$ has $\Delta f=0$ everywhere, but $A_{h}f(0,0)=h^4$. Higher-order terms remain even when the second-order terms cancel.

At a local minimum, each coordinate second derivative is at least zero, hence the Laplacian is at least zero. At a local maximum the inequality reverses. The sum alone does not characterize an extremum: cancellation at a saddle point of a scalar function can also produce zero. The local-average interpretation explains why the heat equation smooths differences, and helps motivate the heat equation maximum principle.

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 log-likelihood is the natural logarithm of the likelihood function:For positive finite likelihood, the natural logarithm is a strictly increasing function, so the maximizing parameter values are unchanged. Set $\ell =-\infty$ where $L=0$; the same ordering then includes zero-likelihood boundaries whenever a positive finite maximum exists. For independent and identically distributed random variables, a product becomes a sum,This makes derivatives simpler and avoids forming products of tiny numbers in floating-point arithmetic. These equal-maximizer and additive formulas underlie the worked calculations in MIT 18.05, Reading 10b, section 3.1.

For the ordered sample of eight successes and two failures in likelihood function, on $0<p<1$,The score equation ${\ell }^{\prime }(p)=0$ gives $8(1-p)=2p$, hence $\hat{p}=8/10=0.8$. The negative second derivative proves strict concavity, so this is the unique interior maximum. Both counts are positive, so $L(0)=L(1)=0$, while $L(p)>0$ inside: the same point is the unique global maximum on $[0,1]$. Counting only successes adds $\log 45$ to the log-likelihood, leaving both derivatives and the MLE unchanged. For all-success or all-failure samples, the boundary analysis in maximum-likelihood estimator applies instead of an interior score equation.

The two panels below compare $k=2,5,8$ successes out of ten Bernoulli trials. Their peaks are at $\hat{p}=k/10$. The left panel uses relative height; the right panel uses the log-likelihood difference from each peak. These are the same comparison on two scales:A difference of $-2$ means relative height $e^{-2}$, and a difference of $-10$ means $e^{-10}$; the log-likelihood scale makes small relative heights easier to see. Neither vertical axis is a parameter probability density function. The zero likelihood endpoints have log-likelihood $-\infty$ and are outside the finite vertical range in the right panel.

Multiplying a likelihood function by a positive parameter-independent constant adds a constant to its log-likelihood. Subtracting its peak removes that constant. This is why the ordered-data and count-data curves coincide in this figure after rescaling. When differentiable, the gradient of the log-likelihood is the score function; the negative Hessian matrix is the observed information. A solution of a score equation still needs checking against other stationary points and boundaries: maximizing the log-likelihood does not automatically make it concave.

Let $S$ be a clamped cubic spline interpolant and let $T$ have the same values at all spline knots and the same endpoint slopes. For $T\in C^2[x_0,x_n]$, or more generally $T^{\prime }$ an absolutely continuous function with square-integrable second derivative, the exact displayed identity holds. In particular $S$ uniquely minimizes the second derivative roughness penalty $J$ among these competitors.

To prove the identity, put $E=T-S$. On each knot interval, two applications of integration by parts give $\int S^{\prime \prime }{E}^{\prime \prime }=[S^{\prime \prime }{E}^{\prime }-S^{\prime \prime \prime }E]$, since the cubic piece has zero fourth derivative. Terms containing $E$ vanish at the spline knots; terms containing $E^{\prime }$ cancel at interior spline knots because $S^{\prime \prime }$ and $E^{\prime }$ are continuous. The endpoint terms vanish because $E^{\prime }$ is zero there. Thus $\int S^{\prime \prime }{E}^{\prime \prime }=0$, and expanding $(S^{\prime \prime }+E^{\prime \prime }{)}^2$ proves the identity. Equality makes $E^{\prime \prime }=0$, so $E$ is an affine function; a prescribed endpoint value and slope force $E=0$.

If $T$ also matches the first derivative of $S$ at every spline knot, both $E$ and $E^{\prime }$ vanish at the ends of each knot interval. For a sufficiently differentiable $T$, integration by parts separately on that interval gives the stronger local identity $\int E{T}^{(4)}=\int (E^{\prime \prime }{)}^2\ge 0$.

The minimum roughness property of the natural cubic spline interpolant extends to interpolants with an absolutely continuous first derivative and square-integrable second derivative. For $h=f-g$ vanishing at the knots, integration by parts on each cubic interval gives $\int g^{\prime \prime }{h}^{\prime \prime }=0$: the knot boundary terms cancel by continuity, the exterior terms vanish by natural boundary conditions, and the remaining constant-third-derivative term vanishes because $h$ is zero at the interval endpoints. Expanding the square proves minimality. Equality makes $h$ affine; its zeros at two distinct knots force $h=0$, proving uniqueness. Competitors with infinite roughness cannot improve the finite spline penalty.

For a broad, thin oil lens below a flat ice-water interface, define the effective interfacial energy $\Gamma ={\gamma }_{io}+{\gamma }_{ow}-{\gamma }_{iw}$ from the relevant surface tensions. At fixed oil volume $\mathcal{V}=Ah$, the energy is $E=\Gamma \mathcal{V}/h+\frac{1}{2}({\rho }_w-{\rho }_o)g\mathcal{V}h$. Setting its derivative to zero gives $h^2=2\Gamma /[({\rho }_w-{\rho }_o)g]$ for $\Gamma >0$, and the positive second derivative proves a minimum. A contact angle measured through the oil gives $\Gamma ={\gamma }_{ow}(1-\cos \vartheta )$. Oil-ice tension alone does not determine this effective quantity. Complete wetting or rough underside pockets invalidate this finite, flat-lens approximation. With densities $1025$ and $850\mathrm{kg}{\mathrm{m}}^{-3}$, a lens below $1\mathrm{cm}$ requires $0<\Gamma <0.08584\mathrm{N}{\mathrm{m}}^{-1}$.