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The radiative efficiency is $\eta =L/(\dot{M}{c}^2)$. For a zero-torque thin disk it is $1-E_{\mathrm{ISCO}}$, the binding energy per unit rest mass released before matter crosses the innermost stable circular orbit.

Conditionally on $X_2=x_2$,

For independent Gaussian random vectors $Z_i\sim N_p(0,\Sigma )$ with known zero mean and positive-definite covariance matrix $\Sigma$, the candidate maximum-likelihood covariance estimator isWhen $S$ is positive definite, the Gaussian log likelihood, up to a constant, is $-\frac{n}{2}[\log \det \Sigma +\mathrm{tr}({\Sigma }^{-1}S)]$. Writing the precision matrix as $\Omega ={\Sigma }^{-1}$ gives the concave objective $\frac{n}{2}[\log \det \Omega -\mathrm{tr}(\Omega S)]$, whose derivative vanishes uniquely at $\Omega =S^{-1}$. This proves the displayed estimator. It is unbiased because the mean is known: $\mathbb{E}S=\Sigma$.

If the sample vectors do not span the whole $p$-dimensional space, $S$ is singular. Shrinking a candidate covariance eigenvalue to zero along an unobserved direction makes the likelihood unbounded, so there is no maximum within the positive-definite parameter space. In particular, $n<p$ cannot yield a positive-definite maximum-likelihood estimate. For a positive-definite Gaussian population and $n\ge p$, the vectors span the space almost surely. When the mean is instead estimated from the same data, centering changes the rank condition and the unbiased covariance uses denominator $n-1$ rather than $n$.

A thin spherical shell of proper surface mass density $\rho$ held at Schwarzschild radius $r_{\mathrm{st}}$ around mass $M$ contributes, to first order in its mass,to the exterior Schwarzschild mass parameter.

The blowup of ${\mathbb{A}}^2$ at the origin isIts projection to ${\mathbb{A}}^2$ is an isomorphism away from the origin, while the fiber above the origin is the exceptional curve ${\mathbb{P}}^1$.

It is an explicit example of blowup of an algebraic variety.

A finite graph is bipartite exactly when it has a proper colouring with two colours. Equivalently,for its chromatic polynomial.

The complete bipartite graph $K_{r,s}$ has vertex classes of sizes $r$ and $s$ and contains all $rs$ edges between the classes.

If a square bipartite adjacency matrix $B$ has nonzero determinant, some product in its Leibniz expansion is nonzero. Its permutation selects one edge at each vertex and therefore gives a perfect matching.

Coefficient extraction from $(1-x{)}^n(1+x{)}^n=(1-x^2{)}^n$ gives

If square matrices $A$ and $B$ commute, thenThe usual proof works because commutativity allows words with the same numbers of $A$ and $B$ factors to be collected.

Partition the vertices into two sets of comparable size. If $np\to \infty$, the first set contains a triangle with probability tending to one. Conditional on any triangle chosen using only internal edges, the probability that none of its vertices has a neighbour in the second set is at mostHence $G(n,p)$ contains a triangle with an attached leaf with probability tending to one.

If $X$ counts triangles in $G(n,p)$, thenandThus $np\to \infty$ implies $\mathrm{var}X/(\mathbb{E}X{)}^2\to 0$, and the second moment method gives $\mathbb{P}(X>0)\to 1$.

If $X_t$ counts copies of the complete graph $K_t$ in $G(n,p)$, thenConsequently $p{n}^{2/(t-1)}\to 0$ implies $\mathbb{P}(X_t>0)\to 0$ by the first moment method.

If $p<1/4$, choose $p<\delta <1/4$. Binary codes of relative minimum distance above $2\delta$ have positive asymptotic rate by a greedy packing bound, and nearest-codeword decoding fails only when more than $\delta n$ channel errors occur. That probability tends to zero.

At crossover probability $p=1/2$, the two rows of the transition matrix of a binary symmetric channel coincide. Its output is then independent of its input and its capacity is zero.

An error is undetected by codeword-membership checking when the received word is a different valid codeword. For a linear code this occurs exactly when the nonzero channel-error vector is itself a codeword.

A Roche lobe is the region around one member of a circular, synchronously rotating binary star bounded by the critical effective potential surface through the inner Lagrange point. The two lobes meet at that saddle point. The construction includes the gravitational effects of both stars and the centrifugal contribution in the rotating frame; it is not simply a sphere at a fixed distance from one star.

A star that expands to fill this region can lose gas through Roche-lobe overflow. A volume-equivalent radius provides a convenient single size for the nonspherical lobe, allowing comparison with a model stellar radius. Changing the binary mass ratio or orbital separation changes that size, so orbital evolution can initiate transfer even without intrinsic stellar expansion.

An equivalence relation says which set elements should count as the same for a particular purpose. In modular arithmetic, for example, distinct integers with the same remainder are equivalent. The relation gathers interchangeable set elements into equivalence classes; the quotient set then treats each whole equivalence class as one object.

Formally, an equivalence relation on a set $X$ is a binary relation $R\subseteq X\times X$, written $x\sim y$, satisfying three conditions for all $x,y,z\in X$:These conditions make the equivalence classes a partition of a set: each set element belongs to exactly one class. ^[ref]

For a worked example, take the finite set $X=\{0,1,\dots ,8\}$ and restrict integer congruence modulo three to $X$:Reflexivity follows because $x-x=0$. The symmetric relation property follows by changing the sign of a difference, and transitivity follows by adding $x-y$ and $y-z$: divisibility by three is preserved in both steps. Thus $0\sim 3$ and $3\sim 6$ force $0\sim 6$. The three equivalence classes are $\{0,3,6\}$, $\{1,4,7\}$ and $\{2,5,8\}$. Their members alternate along the number line; membership in an equivalence class need not mean being close together.

The transitivity requirement rules out a tempting alternative. On $\{0,1,2\}$, define nearness by the absolute value condition $|x-y|\le 1$. This binary relation is reflexive and symmetric, but $0$ is near $1$ and $1$ is near $2$ while $0$ is not near $2$. Its neighborhoods $\{0,1\}$, $\{0,1,2\}$ and $\{1,2\}$ overlap without being equal. They are not equivalence classes or a partition of a set.

A relation $R$ is transitive when $xRy$ and $yRz$ imply $xRz$.

A relation $R$ is symmetric when $xRy$ implies $yRx$.