A steady spherical Parker wind with constant isothermal sound speed $c$ and negligible wind self-gravity obeys $4\pi r^2\rho u=\dot{M}$ and $(u-c^2/u)u^{\prime }=2c^2/r-GM/r^2$. A regular sonic point has $u=c$ at $r_s=GM/(2c^2)$. With $x=r/r_s$ and $w=u^2/c^2$, integration gives the displayed equation for the smooth accelerating transonic branch. Its slope at the sonic point is $u^{\prime }(r_s)=c/r_s$. Maintaining constant gas temperature during expansion generally requires heat input, so this model is distinct from an adiabatic wind with $\gamma >1$.

The steady isothermal Parker wind differential equation also has positive solutions describing subsonic flow at every radius. Write their first integral as $w-\ln w=4\ln x+4/x+C$, where $w=u^2/c^2$ and $x=r/r_s$. On the small-$w$ branch at large radius, $w\sim e^{-C}{x}^{-4}$, so $u\sim c{e}^{-C/2}{x}^{-2}$. Constant spherical mass flow rate then gives nonzero limiting mass density and hence finite external pressure. This distinguishes a pressure-confined breeze from an escaping wind into vanishing pressure. Specifying outward motion alone does not select the supersonic asymptotic branch.

Oxalate is the dianion ${\mathrm{C}}_2{\mathrm{O}}_4^{2-}$ formed by loss of two protons from oxalic acid. Its oxygen donors can bind a metal centre as a chelate.

Precipitation of calcium oxalate illustrates the difference between a dissolved molecular ion and a solid salt. Changing the metal counterion or the pH can change solubility and hence whether oxalate remains in chemical solution.

For the Sturm-Liouville problem $-(p{y}_n^{\prime }{)}^{\prime }={\lambda }_{n}w{y}_n$, assume that the boundary term $[p{y}_n^{\prime }{y}_m{]}_a^b$ vanishes. One integration by parts gives the displayed identity. Distinct eigenvalues make the right side zero by orthogonality of Sturm-Liouville eigenfunctions, so the derivatives are orthogonal with weight $p$, while the original functions are orthogonal with weight $w$. This conclusion uses the absence of the $qy$ term: for a general potential the extra integral $-\int q{y}_n{y}_m$ need not vanish. For the Chebyshev polynomial of the second kind of degree $n-1$, the two weights are $w=\sqrt{1-x^2}$ and $p=(1-x^2{)}^{3/2}$.

Let $f$ be a real function with $\Vert f{\Vert }_w^2={\int }_a^{b}f(x{)}^{2}w(x)dx<\infty$, and let $V_m$ be the vector space of real polynomials of degree at most $m$. Its orthogonal projection is the least-squares polynomial in an orthogonal-polynomial basisEach coefficient depends on just one weighted inner product with $f$ and one squared norm. In particular, the coefficients $c_0,\dots ,c_m$ stay the same when the allowed polynomial degree increases. This statement assumes a fixed weight, interval and normalization of the orthogonal polynomials.

To see why the least-squares normal equations decouple, first write a trial polynomial as $q={\sum }_{j=0}^m{a}_j{\varphi }_j$ in any basis of $V_m$. The residual must be orthogonal to each permitted direction, givingThe Gram matrix usually couples the unknown coefficients. With ${\varphi }_j=p_j$, it becomes $G_{ij}=h_i{\delta }_{ij}$, leaving the separate equations $h_i{a}_i=b_i$. For the monomial basis on $[-1,1]$ with $w=1$, by contrast, $G_{ij}$ is zero when $i+j$ is odd and $2/(i+j+1)$ when $i+j$ is even. Distinct even powers, or distinct odd powers, remain coupled.

There is also a proof of the minimum that uses no differentiation. The defining coefficients give $⟨f-q_m,p_n{⟩}_w=0$ for $0\le n\le m$. Thus $f-q_m$ is orthogonal to all of $V_m$. For every $v\in V_m$, the Pythagorean theorem in an inner-product space givesThe second term is nonnegative and vanishes exactly when $v=q_m$, proving that the least-squares solution is unique. This is an integrated squared-error optimum; it does not assert a smallest error at every point or a smallest uniform norm.

For $N>m$, the nested orthogonal projections satisfy the exact residual budgetIndeed, $q_N-q_m$ is a linear combination of mutually orthogonal polynomials, and $f-q_N$ is orthogonal to it. Every newly retained component removes exactly $c_n^2{h}_n$ from the squared error. No infinite expansion or completeness assumption is needed for this finite identity. The Legendre least-squares projections of a cubic make every term explicit.

A finite subgroup of $O(2)$ is cyclic or dihedral. Its orientation-preserving part consists of rotations and is cyclic: choose the smallest positive rotation angle, divide any other angle by it with remainder, and use minimality to force zero remainder. The smallest angle is therefore $2\pi /n$ for some $n$. A trivial rotational subgroup is included by $n=1$.

If there is no orientation-reversing element, this is the whole group. Otherwise choose a reflection $s$. Every orientation-reversing element is $r^{j}s$, where $r$ generates the rotations; these form the second coset. Since $srs=r^{-1}$, the group is a dihedral group of order $2n$. Its involutions are all $n$ reflections and, when $n$ is even, the extra half-turn $r^{n/2}$. Consequently it has $n$ order-two cyclic subgroups for odd $n$, and $n+1$ for even $n$.

A homogeneous first-order differential equation has the form $y^{\prime }=F(y/x)$ on a domain where $x\ne 0$. The substitution $v=y/x$, so $y^{\prime }=v+x{v}^{\prime }$, reduces it toWhere $F(v)-v\ne 0$, this is a separable differential equation, with $\int dv/[F(v)-v]=\ln |x|+C$. Values $v_0$ satisfying $F(v_0)=v_0$ give the additional straight-line solutions $y=v_{0}x$, which must be checked before division. This scaling-based use of “homogeneous” is different from a homogeneous linear differential equation.

For example, $y^{\prime }=y/x+x/y$ gives $F(v)=v+1/v$, so $v{v}^{\prime }=1/x$ and $y^2=x^2(2\ln |x|+C)$. Real explicit branches are $y=\pm x\sqrt{2\ln |x|+C}$ on intervals where the expression under the square root is positive. Zeros of a denominator and endpoints of a branch must still be checked in the original ordinary differential equation.

For a twice differentiable solution, put $p=y^{\prime }$. Differentiation gives $[x+f^{\prime }(p)]p^{\prime }=0$. One family consists of straight lines $y=Cx+f(C)$ with constant slope $C$. The other consists of the envelope of those lines, given parametrically by $x=-f^{\prime }(p)$ and $y=f(p)-p{f}^{\prime }(p)$ where these equations define a differentiable graph with derivative $p$. An envelope is a singular solution not obtained by fixing $C$. If only continuous first derivatives are required, a solution may also join a line to its envelope at their point of tangency, provided the joined function satisfies the original equation. Such joins need not have a continuous second derivative.

On a branch where a differentiable function $y(x)$ has a nonzero derivative, the inverse function theorem allows the ordinary differential equation to be rewritten for $x(y)$. A nonlinear equation in one direction can become a linear ordinary differential equation in the other. For example, $y^{\prime }=y/(x+y+c)$ becomesThe integrating factor $1/y$ gives $(x/y{)}^{\prime }=1/y+c/y^2$, hence $x+c=y(\ln |y|+C)$. This family must retain the original denominator restriction $x+y+c\ne 0$. The separate zero solution $y=0$ on intervals avoiding $x=-c$ is lost by the inversion and must be restored.

For independent survival times $T_1,T_2$ with cumulative distribution functions $F_1,F_2$, the probability both components have failed by time $t$ isThe probability at least one has failed isThe first equality uses independence of random variables; the second applies it to the complementary event that both survive. These are the distributions of the two extreme order statistics.

Marginal distributions alone do not determine these probabilities. Without independence, the Fréchet inequalities give $\max (0,F_1+F_2-1)\le P(\text{both})\le \min (F_1,F_2)$, and inclusion-exclusion gives $P(\text{at least one})=F_1+F_2-P(\text{both})$.

For every positive integer $n$, cancellation gives $(ha{h}^{-1}{)}^n=h{a}^n{h}^{-1}$. Thus the power on one side is the identity element exactly when $a^n=e$, proving preservation of the order of a group element, including infinite order. In particular, $ab$ and $ba$ are conjugate because $a^{-1}(ab)a=ba$, and hence have the same order.

For every positive integer $m$, $(g^{-1}{)}^m=(g^m{)}^{-1}$, hence $g^m=e$ if and only if $(g^{-1}{)}^m=e$. The sets of positive powers returning to the identity element are therefore equal. Taking the least positive exponent therefore gives the same finite order of a group element for $g$ and $g^{-1}$; if the sets are empty, both elements have infinite order. Equivalently, $g$ and $g^{-1}$ generate the same cyclic subgroup.

An optical pulse with constant group delay over its spectrum. At fixed spectral amplitude, its power-weighted temporal variance is minimal.

Reduction of an optical pulse's duration, for example by compensating its spectral phase.

A diagnostic proportional to $\int I(t)I(t-\tau )dt$ for delay $\tau$. It does not uniquely reconstruct a complex optical electric field; extracting pulse duration requires a shape model.

The product of a specified temporal width and spectral frequency width. State whether each width refers to power or field amplitude and whether it is a full width at half maximum or a standard deviation.

Dark-field microscopy excludes directly transmitted illumination from the microscope objective, while admitting light scattered by the specimen. A transparent particle can then appear bright against a dark background even when it has little optical absorption.

One implementation illuminates the specimen with a hollow cone whose direct rays pass outside the objective's acceptance cone. Scattering of light at particles or edges redirects some of this illumination into the objective. Dust and scratches also become bright, so a clean blank specimen is a useful control.^[ref]

Detection is not the same as optical resolution: a very small particle may be visible as an unresolved spot whose width is set by the point spread function.

Transmitted-light imaging with an optical microscope in which the directly transmitted illumination contributes the bright background. An absorbing specimen can reduce that background locally. The ideal lossless phase object calculation in coherent zero-mode model of phase contrast shows why ordinary brightness detection need not reveal every optical variation.

The geometrical optics account of an optical trap adds the momentum flux changes of refracted and reflected light rays. It applies when the bead is large compared with the surrounding medium's wavelength. When the bead is much smaller, an induced electric dipole gives a different approximation. For a homogeneous dielectric sphere of radius $a$, define $\mathbf{p}(t)=\mathrm{Re}[\alpha {\mathbf{E}}_0{e}^{-i\omega t}]$, with ${\mathbf{E}}_0$ the peak complex electric field and $\alpha$ its SI polarizability. In a transparent, nonmagnetic, nondispersive host,where $n_p,n_m$ are particle and host refractive indices, $I$ is cycle-averaged light intensity, ${ϵ}_0$ is the vacuum permittivity and $c$ is the speed of light. The quasistatic electric dipole approximation requires $2\pi n_{m}a/{\lambda }_0\ll 1$, with ${\lambda }_0$ the vacuum wavelength, and a field nearly uniform across the bead.

For real quasistatic $\alpha$, building an induced electric dipole gives instantaneous potential energy $-\alpha E^2/2$. Averaging over one optical cycle givesThus $n_p>n_m$ gives positive polarizability and attraction toward larger light intensity; $n_p<n_m$ reverses this contribution. The factor $1/4$ uses a peak field: substituting a root mean square field without changing that factor would halve the force incorrectly. Dimensionally $\alpha |E_0{|}^2$ is energy, so its gradient is force.

This leading conservative term does not remove the optical scattering force. Even a nonabsorbing induced electric dipole reradiates light: at fixed wavelength and index contrast, its Rayleigh scattering cross-section scales as $a^6$, while $\alpha$ scales as $a^3$. A smaller bead can therefore improve the ratio of gradient to scattering forces while weakening the well relative to $k_{B}T$, where $k_B$ is the Boltzmann constant and $T$ the temperature. Optical force balance and resistance to thermal escape are separate questions. Intermediate bead sizes require a wave-scattering calculation; neither a ray calculation nor this small-dipole expression is uniformly valid there.

The frequency range over which an amplifying medium supplies appreciable optical gain. Its width must be stated with a gain-profile convention.