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The support or pivot about which a lever turns.

Historical term for an excess of white blood corpuscles, especially the persistent organ-associated disorders subsequently classified as leukaemia. The old term does not make every raised white count malignant.

An optical lens with different refractive powers in two principal meridians. Correcting regular ocular astigmatism requires both the power difference and its orientation.

A lens thicker near its centre than at its edge. It is converging in the usual case of glass in air, but the refractive index relative to the surroundings matters; shape alone is not a medium-independent power specification.

On a bounded open set, suppose $b\in L^{\infty }$ has weak divergence in $L_{\mathrm{loc}}^1$ with $\mathrm{div}b\le 0$. For $\phi \in C_0^{\infty }$, testing the divergence identity with ${\phi }^2/2$ gives $\int (b\cdot \nabla \phi )\phi =-\frac{1}{2}\int (\mathrm{div}b){\phi }^2\ge 0$. The left side is continuous in the zero-boundary Sobolev space $H_0^1$, by the Poincaré inequality and boundedness of $b$. Density therefore extends its nonnegativity to every $u\in H_0^1$. One need not assume that $(\mathrm{div}b)u^2$ is integrable. It follows that $a(u,v)=\int \nabla u\cdot \nabla v+\int (b\cdot \nabla u)v$ is a coercive bilinear form in the gradient norm, giving unique weak Dirichlet solvability by the Lax-Milgram theorem.

A coercive zero-Dirichlet weak form gives a bounded solution map from $L^2$ into $H_0^1$. Composing with the compact Sobolev embedding into $L^2$ gives the compact inverse $K$. Transposing the bilinear form produces its Hilbert-space adjoint inverse $K^{\ast }$. The Fredholm compatibility condition for a shifted equation is orthogonality of $f$ to the homogeneous adjoint solutions. Indeed $(Kf,w)=(f,w)/\lambda$ when $K^{\ast }w=w/\lambda$, and the zero spectral parameter is excluded at resonance by coercivity.

Suppose a meromorphic function is holomorphic on two concentric annuli about $a$, with only finitely many poles between chosen inner and outer coefficient circles. Let $a_n,b_n$ denote the corresponding Laurent coefficients. Subtract their contour integral formulas and apply the residue theorem to the intervening annular region. It givesThe sum is over the poles crossed, including their multiplicities through the residue. This identity applies to every integer index. Thus changing the annulus can change even the coefficients of nonnegative powers; it is not merely a change in the principal part.

For example, $f(z)=1/\sin z$ has simple poles at $\pm \pi$, each with residue $-1$. Moving a coefficient circle centered at zero from radius below $\pi$ to radius between $\pi$ and $2\pi$ changes its $n$th coefficient by $-{\pi }^{-n-1}-(-\pi {)}^{-n-1}$. Even indices remain unchanged, while odd indices acquire a nonzero correction.

For real $p>0$, the Laplace transform $G$ of the order-zero Bessel function of the first kind satisfies $(1+p^2)G^{\prime }+pG=0$. This follows by transforming $x{J}_0^{\prime \prime }+J_0^{\prime }+x{J}_0=0$, using $\mathcal{L}[xg]=-{\partial }_p\mathcal{L}[g]$ and $J_0(0)=1$, $J_0^{\prime }(0)=0$. Thus $G=C/\sqrt{1+p^2}$; the initial value theorem for Laplace transforms gives $pG(p)\to 1$ and fixes $C=1$.

If two piecewise continuous functions of exponential order on $[0,\infty )$ have the same unilateral Laplace transform for every sufficiently large positive real parameter, they agree at every common point of continuity and agree almost everywhere. For continuous functions, the equality holds everywhere. Values at an isolated discontinuity cannot be recovered from an ordinary integral transform.

A bounded function on $[0,\infty )$ has a Laplace transform satisfying the displayed bound, by integration of $|y(t)e^{-pt}|$. Consequently its Laplace transform cannot have a pole of order two or higher at $p=0$. This can eliminate the arbitrary constant introduced when multiplication by $t$ turns an ordinary differential equation into a first-order differential equation for its Laplace transform.

For example, transforming $t{y}^{\prime \prime }+(t-1)y^{\prime }-y=0$ with $y(0)=1$ gives $p(p+1)Y^{\prime }+(3p+2)Y=2$. An integrating factor $p^2(p+1)$ yieldsThe bound forces $C=0$, selecting $y=e^{-t}$. This selection can be justified without assuming unproved endpoint regularity for every solution: $e^{-t}$ and $t-1$ have nonzero Wronskian $t{e}^{-t}$ on $(0,\infty )$, so they form a fundamental system of solutions. Boundedness removes $t-1$, and the limit at zero fixes the remaining coefficient. The endpoint limit by itself does not select a unique solution.

Use normalized measure on the circle. If a trigonometric polynomial has frequency support $\{\pm n_j\}$ with $n_{j+1}>3n_j$, every zero sum of four signed frequencies pairs into opposite equal frequencies: the largest unpaired term would exceed the sum of the other three. Expanding the fourth moment and bounding its three possible pairings gives $\Vert p{\Vert }_4^4\le 3\Vert p{\Vert }_2^4$. The Holder inequality gives $\Vert p{\Vert }_2\le \Vert p{\Vert }_1^{1/3}\Vert p{\Vert }_4^{2/3}$ and hence the displayed estimate. Therefore an $L^1$ function with only these lacunary frequencies has square-summable coefficients: apply the estimate to its uniformly $L^1$-bounded Fejér sums and then pass to finite coefficient blocks. Finitely many extra modes do not affect the conclusion.

For a symmetric stationary frequency distribution, locked oscillators contribute $r={\int }_{-Kr}^{Kr}\sqrt{1-{\omega }^2/(Kr{)}^2}g(\omega )d\omega$. The drifting oscillators' real contribution is zero by integrating a periodic logarithmic derivative, and imaginary contributions cancel by symmetry. Substituting $\omega =Kr\sin \theta$ and dividing by $r>0$ gives the equation. The incoherent solution must be retained before division. For finite positive continuous $g(0)$ the branch reaches it at $K_c=2/[\pi g(0)]$.

For null coordinates normalized by $UV=(2m-r)e^{r/(2m)}$, the displayed field agrees with exterior Schwarzschild time translation. Its flow rescales $U$ and $V$ inversely, preserving their product, the radius and $dUdV$. It therefore preserves the extended metric across both horizons. Its squared norm is $1-2m/r$: it is timelike in both exterior regions, spacelike in the black-hole and white-hole interiors, and nonzero null on the horizons away from the bifurcation two-sphere. At that sphere it vanishes. The zero vector should not be mistaken for a nonzero null generator. This gives an explicit symmetry that survives where the original time coordinate ceases to be regular.

For cell diffusion, self-produced chemotactic agent and logistic proliferation, Neumann cosine modes reduce linear stability of the uniform state to a two-by-two matrix indexed by $s={\pi }^2(n^2/L_1^2+m^2/L_2^2)$. Its trace is negative and its determinant is $D(s)=\alpha s^2+(1+\alpha \kappa -\beta \kappa \eta )s+\kappa$. Chemotaxis can destabilize a mode precisely when $D(s)<0$, despite stability of the spatially uniform reaction system.