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A twice-differentiable real-valued function on an open interval is convex if and only if its second derivative is nonnegative everywhere. If $f^{\prime \prime }\ge 0$, the mean value theorem makes $f^{\prime }$ nondecreasing, which implies the supporting-line and chord inequalities for convexity. Conversely, the slopes of a convex function are ordered; where the second derivative exists, this gives $f^{\prime \prime }\ge 0$. Strict positivity of $f^{\prime \prime }$ is sufficient for strict convexity, but is not necessary: $f(x)=x^4$ is strictly convex despite $f^{\prime \prime }(0)=0$.

The nonnegative functional $J(g)={\int }_a^b(g^{\prime \prime }(x){)}^{2}dx$. Its null space on $C^2[a,b]$ consists of affine functions. The natural cubic spline interpolant uniquely minimizes this penalty among functions matching at least two prescribed knot values. It is a squared second derivative penalty, not the exact geometric curvature energy of a graph.

Let $f$ have continuous second partial derivatives near an interior critical point $x_{\ast }$ of an open subset of ${\mathbb{R}}^n$, with $\nabla f(x_{\ast })=0$. The Hessian matrix gives the following sufficient conditions:

A positive semidefinite or negative semidefinite Hessian matrix with a zero eigenvalue does not settle the classification. The higher-order test for separated extrema gives two quartic examples with exactly the same zero Hessian. These conditions apply to an interior critical point; a boundary optimum or a constrained optimization problem requires attention to the allowed directions. ^[ref]

The proof outline is the Taylor expansionFor a positive-definite matrix, the smallest eigenvalue supplies a positive bound proportional to $\Vert h{\Vert }^2$, which dominates the remainder for small enough $h$. Negate the function for the local maximum case. For opposite-sign eigenvalues, choose the two corresponding eigenvectors: the leading terms have opposite signs along their lines.

There are also necessary conditions. At an interior local minimum, the Hessian matrix is positive semidefinite; at an interior local maximum, it is negative semidefinite. Restriction to each straight line reduces this to the one-variable second derivative condition. Since the Laplacian is the trace of the Hessian, it is nonnegative at such a local minimum and nonpositive at such a local maximum. The sign of the Laplacian alone is insufficient: $x^2-2y^2$ has negative Laplacian but a saddle point of a scalar function at the origin.

Let a nonnegative kernel for density estimation $K$ have integral one, zero first moment and finite second moment ${\mu }_2(K)$. If a probability density function $f$ has bounded second derivative, its kernel density estimator has pointwise bias at most $h^2{\mu }_2(K)\Vert f^{\prime \prime }{\Vert }_{\infty }/2$. Apply the Taylor theorem with Lagrange remainder to $f(x-hu)$ and integrate: the first-order term vanishes and the remainder is bounded by $h^2{u}^2\Vert f^{\prime \prime }{\Vert }_{\infty }/2$. For $K=1_{[-1/2,1/2]}$, ${\mu }_2(K)=1/12$.

A singular perturbation is a parameter-dependent problem whose limiting problem changes qualitatively as the parameter tends to zero. In an ordinary differential equation, a small coefficient of the highest derivative can reduce the order of the limiting differential equation, preventing it from satisfying all the original boundary conditions. Different length or time scales often require different asymptotic expansions.^[ref]

Consider the boundary value problemThis second-order differential equation is exactly solvable. Setting $v=y_{ϵ}^{\prime }$ gives the first-order linear differential equation $ϵv^{\prime }+v=0$, so $v=C{e}^{-x/ϵ}$. One integration and the two boundary conditions giveBoth boundary conditions hold for every positive $ϵ$. For numerical evaluation, write the same ratio as $expm1(-x/ϵ)/expm1(-1/ϵ)$, where $expm1(t)$ evaluates the exponential function minus one without catastrophic cancellation near $t=0$. Keep the denominator for finite $ϵ$: at $ϵ=0.2$ it is approximately $0.993262$, rather than exactly one.

Setting $ϵ=0$ gives $y_0^{\prime }=0$. The outer solution selected by the right boundary condition is therefore $y_0=1$, which fails the left boundary condition. The exact exponential function $e^{-x/ϵ}$ shows how that mismatch is repaired: the solution rises from zero to almost one within a boundary layer of width $x=O(ϵ)$, using Big O notation. For each fixed $x>0$, $y_{ϵ}(x)\to 1$, while $y_{ϵ}(0)=0$ always. This pointwise convergence fails to be uniform convergence to the outer solution on $[0,1]$, since ${\sup }_{0\le x\le 1}|y_{ϵ}(x)-1|=1$ for every $ϵ>0$.

The coefficient $ϵ$ is small, but its product with the second derivative need not be small. Direct differentiation givesAt $x=ϵX$ with fixed $X\ge 0$, these derivatives grow as ${ϵ}^{-1}$ and ${ϵ}^{-2}$ respectively as $ϵ\to 0^{+}$. Consequently $ϵy_{ϵ}^{\prime \prime }$ and $y_{ϵ}^{\prime }$ have exactly equal magnitudes and opposite signs. Even away from the boundary layer, their magnitudes remain equal in this example; both become exponentially small at each fixed positive $x$. The outer expansion neglects an exponentially small transient there, rather than establishing that one derivative term is smaller than the other everywhere.

To predict the width without knowing the exact solution, suppose an order-one change occurs over distance $\delta$. The derivative scales are $y^{\prime }=O({\delta }^{-1})$ and $y^{\prime \prime }=O({\delta }^{-2})$. The dominant balance $ϵ/{\delta }^2\sim 1/\delta$ gives $\delta \sim ϵ$. The resulting inner variable $X=x/ϵ$ makes the boundary layer visible at a fixed scale. This is the useful feature of singular perturbation: an apparently disappearing term identifies the missing structure when one measures its derivatives on the appropriate scale.

Interrogation of a subdivision limit needs patch/topology navigation, refinement, limit evaluation and derivatives where they exist, with certified bounds and geometric error estimates. On regular patches one may use the corresponding spline representation. At extraordinary subdivision vertices, local subdivision matrices determine limit positions and tangents under the scheme's smoothness conditions. A normal vector may exist even when second derivatives and Gaussian curvature do not; finite refined polygons are approximations rather than exact limit queries.

The third derivative is the derivative of the second derivative. On the interior of a cubic polynomial piece it is constant. A cubic spline can have different constants on adjacent pieces, so its third derivative need not be continuous across a spline knot.

Let the input alphabet be $\{0,1,2\}$ and let the output be $Y=X+Z$, where independent noise is uniform on two distinct integers. Translating the output alphabet reduces the noise to $\{0,d\}$, $d\ge 1$. The reachable output alphabet has size $3+\min (d,3)$: it is the union of two three-element intervals shifted by $d$.

For $d=1$, every row of the transition matrix has conditional information entropy one bit, so $I(X;Y)\le {\log }_{2}4-1=1$. Input probabilities $(1/2,0,1/2)$ make the output uniform and attain one bit per use. For $d\ge 3$, the two translated input alphabets are disjoint and every output identifies its input. A uniform input then attains ${\log }_{2}3$ bits per use.

For $d=2$, reflection symmetry and concavity of information entropy permit restricting to input probabilities $(s/2,1-s,s/2)$. Conditional on output parity, outputs whose labels are even integers have probabilities $(1/4,1/2,1/4)$ and outputs whose labels are odd integers have probabilities $(1/2,1/2)$. Consequently $I(X;Y)=h_2(s)+s/2$. Its strictly negative second derivative and stationary equation ${\log }_2((1-s)/s)+1/2=0$ give $s=\sqrt{2}/(1+\sqrt{2})$ and channel capacity ${\log }_2(1+\sqrt{2})$. All three values follow from overlap of translated input alphabets, rather than from output alphabet size alone.

The Thomas algorithm is Gaussian elimination specialized to a tridiagonal matrix: each row couples an unknown only to its immediate neighbours. Eliminating one unknown changes just the next diagonal entry and right-hand side. Provided all pivot elements are nonzero, a forward sweep followed by back substitution solves an $n$-unknown linear system in $O(n)$ time and storage, using Big O notation. Treating the same entries as a full matrix wastes the zeros: classical dense LU factorization and solve costs grow cubically for factorization and quadratically for storage.

A concrete reason to care is the Dirichlet Poisson solution by tridiagonal elimination: a three-point finite difference for the second derivative couples adjacent points on a mesh, producing precisely this tridiagonal matrix. The same structure occurs in one-dimensional Crank-Nicolson method calculations. NPTEL's archived Lecture 13 introduces the Thomas algorithm from that time-stepping problem and labels the procedure “Thomas algorithm (1949)”.^[ref]

The triweight kernel is a compactly supported symmetric probability density with variance $1/9$. Its value and first two derivatives vanish at the support endpoints. Its integrated squared second derivative is $35$, and it minimizes that curvature among twice continuously differentiable densities with zero mean and the same variance, by the triweight density minimizes integrated squared curvature inequality.