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For a holomorphic normal family, limits are holomorphic or identically infinity. A uniform bound at one interior point excludes infinity on its connected domain. Every sequence therefore has a finite locally uniform subsequential limit. The Cauchy integral formula makes the derivatives converge locally uniformly too, so the derivative family is normal. Without the point bound this can fail: $n(1+z^2/2)$ tends uniformly to infinity on the unit disk, whereas its derivatives $nz$ are not normal at zero.

Each component of a finite vector measure is absolutely continuous with respect to its variation measure. The Radon-Nikodym theorem gives a vector density $\sigma$. Taking the variation measure of this representation gives $|\mu |=|\sigma ||\mu |$, hence the unit-length property. For a BV derivative this separates the magnitude of variation from its local direction.

Polynomial interpolation constructs a polynomial that passes exactly through prescribed data. Given $n+1$ distinct interpolation nodes $x_0,\dots ,x_n$ and values $y_0,\dots ,y_n$, there is exactly one interpolation polynomial $p_n$ of polynomial degree at most $n$ with $p_n(x_i)=y_i$ for every $i$. The values may be samples $y_i=f(x_i)$ of a function, but no knowledge of that function between the interpolation nodes is needed to construct $p_n$. The Lagrange polynomial and Newton polynomial are two representations of this same answer.^[ref]

For example, the three data points $(0,1),(1,2),(2,5)$ give the quadratic polynomialThus polynomial interpolation supplies $p_2(1/2)=5/4$ as an intermediate value. Whether this approximates an unknown underlying function well is a separate question about polynomial interpolation error.

Both hypotheses in the uniqueness statement matter. The Lagrange cardinal polynomials prove existence constructively. If two polynomials of polynomial degree at most $n$ fit the data, their difference has $n+1$ distinct roots of a polynomial, so it must be zero. Repeated interpolation nodes with conflicting values have no solution, and repeated equal values provide no additional independent condition; prescribed derivatives instead lead to Hermite interpolation. Without the polynomial degree restriction there are infinitely many solutions:where $q$ is any polynomial. By the factor theorem, this describes every polynomial fitting the same data. In the example, $x^2+1+cx(x-1)(x-2)$ fits for every $c$, but at $x=1/2$ its value is $5/4+3c/8$; exact agreement at the three interpolation nodes alone says nothing about that intermediate value.

The usefulness of polynomial interpolation is that data become an explicit function whose values, derivatives, and integrals can be calculated. Differentiating or integrating the Lagrange cardinal polynomials produces differentiation weights from nodal evaluations or a quadrature rule. Exact passage through observations is the defining constraint. In polynomial regression, linear least squares instead chooses a polynomial degree and minimizes a sum of squared residuals, generally without passing through every observation. Interpolating noisy values reproduces their errors at the interpolation nodes and can amplify them between interpolation nodes; polynomial interpolation does not itself smooth measurement noise. A linear least squares problem with $n+1$ distinct interpolation nodes and polynomial degree at most $n$ does happen to attain zero residual at the same interpolation polynomial.

For $0<k<1$, principal upper-half-plane branches give a conformal map onto the upper half-plane cut along $[0,\tau ]$, where $\tau =2k^k(1-k{)}^{1-k}{e}^{i\pi k}$. Its derivative is $(z-1+2k)(z-1{)}^{k-1}(z+1{)}^{-k}$. The two prevertices $-1,1$ map to the slit base, and $1-2k$ maps to its tip. Boundary traversal goes out along the slit and back, while $g(z)=z+1-2k+O(1/z)$ at infinity. The argument principle gives one preimage of every point off the slit.

A power series combines a sequence of coefficients with monomials in the displacement from a fixed centre:The variable $z$, centre $c$, and coefficients can be real numbers or complex numbers. Its degree-$N$ partial sum $P_N(z)={\sum }_{n=0}^N{a}_n(z-c{)}^n$ is a polynomial, so a convergent power series can turn a complicated function into increasingly accurate finite calculations. Convergence, and the error after stopping, are essential parts of that claim.

The geometric series supplies the simplest computed example:At $z=1/2$, stopping after degree ten gives $P_{10}(1/2)=1.9990234375$, differing from the exact value $2$ by $1/1024=0.0009765625$. At $z=0.99$, the same degree gives only about $10.4662$ instead of $100$. The same formula has very different practical accuracy at different points; the geometric power-series remainder quantifies it.

The structure is useful beyond numerical evaluation. Within its radius of convergence, termwise differentiation of a power series turns a question about a derivative into a question about coefficients. A Taylor series chooses $a_n=f^{(n)}(c)/n!$ when it represents the function locally. A formal power series instead uses the coefficients algebraically, without asserting numerical convergence; the existing generating function examples use that distinction to encode counting problems.

A boundary point of the source half-plane that maps to a polygon vertex. If that corner has angle $\pi \alpha$, the Schwarz-Christoffel mapping has local behavior $F(z)-F(a)\sim C(z-a{)}^{\alpha }$ at its prevertex, so the derivative has exponent $\alpha -1$. Infinity may also be a prevertex, expressed in the local coordinate $1/z$.

For flux $c(|\nabla u|)\nabla u$, the derivative matrix is $c(s)I+c^{\prime }(s)\nabla u\otimes \nabla u/s$ with $s=|\nabla u|>0$. Its tangential and normal coefficients are $c(s)$ and $c(s)+s{c}^{\prime }(s)$. Positivity of the scalar conductance alone therefore does not guarantee forward parabolicity.

For two projectively equivalent connections that are Levi-Civita connections, contraction of their difference gives $(n+1)V_c={\bar{\Gamma }}^a{}_{ac}-{\Gamma }^a{}_{ac}$. The Jacobi formula gives ${\Gamma }^a{}_{ac}={\partial }_c\log \sqrt{|\det g|}$. Subtraction proves the displayed expression. The two determinant densities transform by the same coordinate Jacobian factor, so their ratio is a scalar and its derivative a covector. The formula is a necessary consequence of projective equivalence, not a test asserting that arbitrary metric tensors have identical geodesics.

Choose the illustrative local modelHere $t$ is the Celsius temperature, measured relative to $0{}^{\circ }\mathrm{C}$, so $\alpha t$ is dimensionless. For this chosen affine function, the derivative and inverse relation areCalculated values are $R(0{}^{\circ }\mathrm{C})=100\Omega$, $R(20{}^{\circ }\mathrm{C})=107.70\Omega$, $R(25{}^{\circ }\mathrm{C})=109.625\Omega$, and $R(50{}^{\circ }\mathrm{C})=119.25\Omega$. These are model outputs, not measurements or an assertion that any real Pt100 follows this line exactly. The number $0.00385$ here is the chosen slope parameter; it is not a universal local derivative for all platinum resistance thermometers. Curvature, manufacturing differences, and the fitted calibration interval matter. A wider-range calibration uses a validated response curve rather than extrapolating this line indefinitely.

For a separate synthetic thermometer calibration, stipulate $R_a=100.20\Omega$ at $t_a=0{}^{\circ }\mathrm{C}$ and $R_b=119.60\Omega$ at $t_b=50{}^{\circ }\mathrm{C}$. A two-point affine function has slope $0.388\Omega /{}^{\circ }\mathrm{C}$ and givesThus $R=109.90\Omega$ would indicate $25{}^{\circ }\mathrm{C}$. If an additional synthetic check at reference $25{}^{\circ }\mathrm{C}$ instead supplied $110.10\Omega$, the line would indicate $25.515\dots {}^{\circ }\mathrm{C}$. The $0.20\Omega$ residual exposes disagreement that two fitted points cannot reveal. This separate example changes both offset and slope; it is not a replacement specification for the nominal Pt100 local linear model.

Let $t=T(\mathbf{x})$ be a smooth wavefront and suppose the first jump in time derivatives of displacement field is $[{\partial }_t^k\mathbf{u}]=a^{[k]}\mathbf{n}$, with $k\ge 2$. Substitution of $\mathbf{u}=a^{[k]}\mathbf{n}(t-T{)}_{+}^k/k!+\cdots$ into the elastic equation first gives the eikonal equation $|\nabla T|=1/\alpha$ and longitudinal polarization $\mathbf{n}=\alpha \nabla T$. Projecting the next-order compatibility condition onto $\mathbf{n}$ gives $2\rho \alpha \mathbf{n}\cdot \nabla a^{[k]}+a^{[k]}\nabla \cdot (\rho \alpha \mathbf{n})=0$, hence the displayed conservation law. Coefficients and wavefront must be sufficiently differentiable for this classical transport equation; its local form fails at a caustic.

Put $B_j=[m(x)(x-\bar{x}{)}^j{]}_0^L$, $M_j=\int m(x)(x-\bar{x}{)}^{j}dx$, $A=\int \alpha (t)dt$, and $C=B_0(B_2-M_1)-B_1(B_1-M_0)$. For quadratic bending $h_0=-\alpha (t)(x-\bar{x}{)}^2$, the linear recoil correction in elongated-body theory with a stable post-stroke relaxation gives the displayed angle. It follows by setting time derivatives to zero in the integrated lateral and angular momentum equations. The simpler $2UA$ is exact in the zero-body-mass limit, or when $B_1=M_0$, but is not a generic finite-body-mass identity. For $L=1$, $\bar{x}=1/2$, and $m(x)=m_{\ast }{x}^2$, it becomes $2UA/(1+2M_b/m_{\ast })$. Body deformation inertia affects derivative forcing during the stroke, not this final integral relation when the deformation and its time derivative return to zero.

The squared L2 norm of a probability density function derivative measures roughness and appears in optimal smoothing constants. For $r=2$, its directional derivative along a smooth compactly supported perturbation $a$ is $2\int f^{\prime \prime }{a}^{\prime \prime }$. When integration by parts is valid twice, this equals $2\int f^{(4)}a$. Thus regular root-$n$ estimation requires substantially more regularity than merely estimating a smooth probability density function.

A section of the square of the canonical bundle, locally $q=\varphi (z)d{z}^2$; its coefficient changes by the square of the coordinate derivative. A holomorphic quadratic differential has holomorphic coefficient, and its flat metric of a quadratic differential is $|q{|}^{1/2}$.

Let ${\delta }_t=\sqrt{p_t}-\sqrt{p}=(t/2)g\sqrt{p}+r_t$, with $\Vert r_t{\Vert }_2=o(|t|)$. Since $p_t-p=2\sqrt{p}{\delta }_t+{\delta }_t^2$, the Cauchy-Schwarz inequality gives $\Vert p_t-p-tpg{\Vert }_1\le 2\Vert r_t{\Vert }_2+\Vert {\delta }_t{\Vert }_2^2=o(|t|)$. Thus differentiability in quadratic mean suffices to differentiate bounded density integrals, even when the density itself has no useful pointwise derivative.