For a single-valued harmonic function tending to zero both at the origin and at infinity, match an angular Fourier series across the unit circle. A nonzero mode with continuous interface value $K_n\cos (n\theta )$ has interior field $K_n{r}^n\cos (n\theta )$ and exterior field $K_n{r}^{-n}\cos (n\theta )$. Its exterior-minus-interior radial derivative is $-2n{K}_n\cos (n\theta )$. Thus a prescribed jump coefficient $j_n$ determines $K_n=-j_n/(2n)$. Sine modes obey the same relation. The constant mode must have zero jump under these decay conditions.

For the Laplace equation on $0<y<1$, prescribe bottom data $g$ and zero top data. A horizontal Fourier transform reduces the equation to ${\hat{u}}_{yy}-s^2\hat{u}=0$, giving the displayed multiplier and its limit $1-y$ at $s=0$. Integrable transformed data give a bounded inverse-transform solution. The bounded solution is unique by the maximum principle for harmonic functions, using barriers $\epsilon \cosh (\delta x)\cos (\delta (y-1/2))$ for $0<\delta <\pi$ on expanding rectangles. Without a growth condition, one can add $e^{\pi x}\sin (\pi y)$, so the two boundary traces alone do not determine an arbitrary harmonic function on an infinite strip.

For a regular interior and a decaying exterior harmonic potential with continuous values on the unit sphere, the same coefficient multiplies the two radial branches of each Legendre polynomial mode. Its exterior-minus-interior radial derivative is $-(2n+1)a_n{P}_n$. Thus an axisymmetric prescribed jump $j_n{P}_n(\cos \theta )$ produces coefficient $a_n=-j_n/(2n+1)$. This diagonal matching rule reduces a spherical interface problem to its angular-mode coefficients.

For $0<x<L$, $0<y<H$, choose $k=m\pi /H>0$ and seek a harmonic function with zero values on the horizontal edges, boundary value $\sin (ky)$ at $x=L$, and $\varphi +\alpha {\partial }_n\varphi =0$ at $x=0$, with constant $\alpha \ge 0$. Separation of variables gives $\varphi =[\alpha k\cosh (kx)+\sinh (kx)]\sin (ky)/[\alpha k\cosh (kL)+\sinh (kL)]$. The outward derivative at the left edge is $-{\partial }_x$, giving $F(0)-\alpha F^{\prime }(0)=0$. The denominator is positive and Poisson uniqueness with a nonnegative Robin normal coefficient proves uniqueness in the regular solution class.

A single-valued harmonic function in a planar annulus has radial Fourier modes $1,\log r$ and $r^n,r^{-n}$ multiplying $\cos (n\theta ),\sin (n\theta )$ for integers $n\ge 1$. Substitute $R(r)\Theta (\theta )$ in Laplace's equation: periodicity gives ${\Theta }^{\prime \prime }+n^2\Theta =0$, while $r^2{R}^{\prime \prime }+r{R}^{\prime }-n^{2}R=0$. Regularity at the origin excludes $\log r,r^{-n}$; boundedness at infinity excludes $\log r,r^n$. Coefficients must give locally convergent series, with the boundary regularity required by the particular Dirichlet problem. Mere finiteness at each finite exterior point does not imply boundedness at infinity.

A Generalized Lane-Emden equation is a semilinear partial differential equation whose highest-order term is the Laplacian:For a radial function in $d$ dimensions it becomes $u^{\prime \prime }+(d-1)u^{\prime }/r+u^p=0$. The case $d=3$ gives the classical Lane-Emden equation.

The spinodal is the boundary where a metastable phase loses local stability. In a mean-field free-energy description the relevant curvature vanishes there, separating local minima from unstable configurations.

Fix the spatial mean of a scalar order parameter to $M$ and sum the Boltzmann weights of all configurations with that mean. The resulting constrained partition function $Z_M$ defines this free-energy density, with an additive normalization fixed consistently. For an applied conjugate field $h$, $Z(h)=\int dM{e}^{-{\beta }_{T}V[F_c(M)-hM]}$, up to the chosen measure normalization. The thermodynamic limit gives $f(h)={\inf }_M[F_c(M)-hM]$. A homogeneous Landau free energy is an approximation to this constrained potential; phase-separated configurations can convexify it in the full thermodynamic limit.

An isotropic linear elasticity material with equal positive Lamé parameters has Poisson's ratio $1/4$ and the displayed compressional/shear speed ratio. This simplifying model fixes the relative amplitudes in the tensile-crack radiation pattern; it does not describe every stable isotropic solid.

Application $MN$ applies the lambda term $M$ to the lambda term $N$. It associates to the left: $MNP=(MN)P$. Together with variables and lambda abstraction, it generates all lambda terms.

A pointed complete partial order $D$ is reflexive when there are Scott continuous maps $e:[D\to D]\to D$ and $p:D\to [D\to D]$ with $p\circ e=\mathrm{id}$. Define $a\cdot b=p(a)(b)$ and interpret lambda abstraction using $e$. The inverse law gives beta soundness; $e\circ p=\mathrm{id}$ additionally gives eta soundness.

For a locally compact Hausdorff Abelian group, each continuous unitary character defines an algebra character by $f\mapsto \int f(x)\chi (x)dx$. Conversely, if $\Phi$ is a nonzero algebra character and $\Phi (g)\ne 0$, the ratio $\Phi (T_{x}g)/\Phi (g)$ is independent of $g$, continuous and multiplicative. Translation isometries bound all its integer powers, so its modulus is one. The convolution formula $f\ast g=\int f(x)T_{x}gdx$ recovers $\Phi (f)=\int f\chi$. The Gelfand topology corresponds to the compact-open topology on the Pontryagin dual group; conjugating the characters gives the alternative Fourier sign convention.

A measurable function is essentially bounded if some finite nonnegative real number $M$ bounds its absolute value almost everywhere. The exceptional null set may contain arbitrarily large values. With Lebesgue measure on $[0,1]$, a function equal to zero outside $\{1/n:n\in {\mathbb{N}}_{>0}\}$ and equal to $n$ at $1/n$ is essentially bounded despite being unbounded pointwise: its exceptional countable set is a null set, so $M=0$ works. Its L-infinity norm is zero because it represents the zero equivalence class in Lp space.

Contrast this with $f(x)=1/x$ on $(0,1]$. For every $M\ge 1$, the values exceeding $M$ occupy $(0,1/M)$, a set of positive Lebesgue measure. Hence no finite almost everywhere bound works, and this function is not essentially bounded. Being finite at every input does not imply essential boundedness.

For $B>0$ and $0<\beta <1$, the unreferenced relative potential of the Kuzmin reflection method diverges at large $b$. Referencing each component to the origin before integrating gives the finite potential differenceAt small $b$ the subtraction is integrable because $\beta >0$; at large $b$ the difference is $O(b^{-2})$, integrable because $\beta <1$. Its plane force is ${\partial }_R\psi =-C{R}^{\beta -1}$, where $C=(GB/2)B((1+\beta )/2,1-\beta /2)$. A circular speed $V=K{R}^{\beta /2}$ fixes $C=K^2$, givingThe beta function substitutions for the force and surface density have different exponents. Its central mass is locally finite, but its total mass diverges at large radius; a potential zero at infinity cannot be imposed. The force and density approach those of the Mestel disk as $\beta \downarrow 0$.

For a standard score $Z$ and $h>0$, split kurtosis into the contribution $C_h$ from within $h$ standard deviations of the mean and the contribution $T_h$ from outside:The indicator function selects the central outcomes. The bound follows directly from $Z^4\le h^4$ there. Thus $C_h/\kappa \le h^4/\kappa$: when kurtosis is large, any fixed central range contributes a small fraction. This is a precise reason to emphasize extreme standard scores rather than peak height.^[ref]

For the rare-outlier three-point distribution with $q=1/100$, $C_1=0$ and $T_1=100$. At the other endpoint $q=1$, all probability mass lies at $Z=\pm 1$, so $C_1=\kappa =1$: central dominance is possible when kurtosis is small. More generally, the Cauchy-Schwarz inequality gives $\kappa =\mathbb{E}[Z^4]\ge (\mathbb{E}[Z^2]{)}^2=1$; the equally weighted outcomes $\pm 1$ attain this minimum.

Changing units or shifting the origin preserves kurtosis. For real constants $a\ne 0$ and $b$, the mean of $Y=aX+b$ is $a\mu +b$, its variance is $a^2{\sigma }^2$, and its fourth central moment is $a^4\mathbb{E}[(X-\mu {)}^4]$. The factors $a^4$ cancel in the ratio, even for negative $a$. Hence excess kurtosis is also unchanged. For example, $100X+7$ and $X$ have the same kurtosis, although their means and variances differ. The case $a=0$ produces zero variance and is excluded.

For $0<q\le 1$, put probability mass $1-q$ at zero and $q/2$ at each of $-1$ and $1$. This symmetric probability distribution has mean zero. Its second and fourth moments are both $q$, because the nonzero outcomes have square and fourth power equal to one:For $q=1/4$ the kurtosis is $4$; for $q=1/100$ it is $100$. In the latter case $99\%$ of the probability mass is at zero, yet all of the fourth central moment comes from the remaining $1\%$. The standard score $Z=X/\sqrt{q}$ takes values $0$ and $\pm q^{-1/2}$, with variance one. Each nonzero outcome contributesto kurtosis. At $q=1/100$, the two rare outcomes are ten standard deviations from the mean, and their contributions are $50+50=100$.

Every member has finite support of a probability distribution, so all its moments are finite. Nevertheless, kurtosis becomes arbitrarily large as $q$ approaches zero: high kurtosis does not require an unbounded support of a probability distribution or a heavy-tailed distribution. At $q=0$ the variance vanishes and kurtosis is undefined. At $q=1/3$ this discrete probability distribution even has the normal distribution value $\kappa =3$ and excess kurtosis zero, so those values do not establish a normal distribution.

Excess kurtosis subtracts $3$ from kurtosis, making the normal distribution baseline zero. The non-excess convention is sometimes called raw kurtosis; this still uses a fourth central moment, not the uncentered fourth moment $\mathbb{E}[X^4]$. Software may report either kurtosis or excess kurtosis, so the convention must be stated.^[ref]

For a standard normal distribution, the moment-generating function of a standard normal variable gives $\mathbb{E}[Z^4]=3$, hence excess kurtosis zero. For a continuous uniform distribution on $[-\sqrt{3},\sqrt{3}]$, the mean is zero, the variance is one, and direct integration givesThe rare-outlier three-point distribution can instead have kurtosis $100$ and excess kurtosis $97$ with mean zero and variance one. These are calculated properties of specified probability distributions, rather than a measured statistic.

For finite cell complexes, the cohomological cross product gives a split, but not naturally split, short exact sequence $0\to {⨁}_{p+q=n}{H}^p(X;\mathbb{Z})\otimes H^q(Y;\mathbb{Z})\to H^n(X\times Y;\mathbb{Z})\to {⨁}_{p+q=n+1}{\mathrm{Tor}}_1(H^p(X;\mathbb{Z}),H^q(Y;\mathbb{Z}))\to 0$. The Tor degree is $n+1$ for cohomology, unlike the $n-1$ shift in the homological version. If one factor's cohomology is free, the Tor term vanishes and the resulting ring is the graded tensor product.

If $k$ is a field and $H^{\ast }(F;k)$ has finite total dimension, the cohomology cross product gives the displayed graded-ring isomorphism for arbitrary spaces. Over a field, every chain complex splits into its homology with zero differential and contractible summands. The Eilenberg–Zilber theorem thus replaces the fiber chain complex by a bounded finite-dimensional complex; dualizing its tensor product with $C_{\ast }(X;k)$ gives a finite direct sum of shifted cochain complexes of $X$. Its cohomology is the displayed tensor product. Finiteness prevents the infinite-product obstruction that can occur for two infinite discrete spaces. The standard product formula and its finiteness hypothesis are discussed in pi.math.cornell.edu/~hatcher/AT/ATch3.pdf, Theorem 3.15.