A basic variable corresponds to a column of the current simplex basis. Given the nonbasic variables, the invertible basis matrix determines the basic variables from the equality constraints.

// Target: mathematical-optimization.bigb

A nonbasic variable corresponds to a column outside the current simplex basis. In standard nonnegative linear programming, it is set to zero at the basic solution. With explicit finite bounds, it is normally fixed at one of its bounds.

// Target: mathematical-optimization.bigb

For positive smooth tube area $A$, the equation $q_Z-q{q}_{\theta }+q(\log A{)}^{\prime }/2=0$ becomes $Q_{\zeta }-Q{Q}_{\theta }=0$ under the displayed transformation. Indeed, $g^{\prime }/g=-A^{\prime }/(2A)$ cancels geometrical spreading and ${\zeta }^{\prime }=g$ makes the remaining nonlinear and transport coefficients equal. Starting with $q(\theta ,Z_0)=f(\theta )$ gives $Q=f(\xi )$ and $\theta =\xi -f(\xi )\zeta$; characteristic crossing first occurs when $\zeta \sup f^{\prime }=1$, if that supremum is positive.

Fix $N>1$ real population values $y_1,\dots ,y_N$. Define the finite population mean and two normalizations of the finite population variance byThe values are fixed; only the selected subset $A$ is random. For its sample mean ${\bar{y}}_A=n^{-1}{\sum }_{i\in A}{y}_i$, the exact design expectation and variance areThe sample mean is an unbiased estimator of the finite population mean. At $n=N$, every unit is observed and ${\bar{y}}_A=\bar{Y}$ exactly, so the variance is zero. The $N=1$ case is deterministic and is handled directly, since $S^2$ is undefined then.^[ref]

The finite population correction depends on the baseline. With the $N-1$ normalization it is $1-n/N$ multiplying $S^2/n$. With the $N$ normalization it is $(N-n)/(N-1)$ multiplying ${\sigma }_N^2/n$, the exact variance for independent uniform sampling with replacement. The standard error uses the square root of the relevant variance factor. Mixing the two normalizations gives a wrong answer even at $n=1$, when replacement cannot matter. For the six-value example, $\bar{Y}=7$, ${\sum }_i(y_i-7{)}^2=70$, $S^2=14$ and ${\sigma }_N^2=35/3$, reproducing $14/3$ without replacement and $35/6$ with replacement.

A short proof exposes the statistical dependence. Let $X_r$ be the value at position $r$ in uniform sequential sampling without replacement, so each $X_r$ has expected value $\bar{Y}$ and variance ${\sigma }_N^2$. Write $d_i=y_i-\bar{Y}$. Every ordered pair of distinct labels is equally likely, and ${\sum }_i{d}_i=0$, hence for $r\ne s$,Linearity of expectation proves unbiasedness without requiring independence. The variance of a sum includes all $n(n-1)$ ordered covariance terms, givingFor the original population, each pair of distinct draw positions has covariance $-7/3$. Drawing a high value leaves a lower mean among the remaining values, balancing subsequent draws. Equal marginal probability distributions therefore coexist with negative covariance.

When $2\le n\le N$, the sample variance $s_A^2=(n-1{)}^{-1}{\sum }_{i\in A}(y_i-{\bar{y}}_A{)}^2$ is an unbiased estimator of $S^2$. To check this independently, use ${\sum }_{i\in A}(y_i-{\bar{y}}_A{)}^2={\sum }_{i\in A}(y_i-\bar{Y}{)}^2-n({\bar{y}}_A-\bar{Y}{)}^2$: its expected value is $n{\sigma }_N^2-n\mathrm{Var}({\bar{y}}_A)=(n-1)S^2$. Thus $(1-n/N)s_A^2/n$ unbiasedly estimates the variance of the sample mean; its square root estimates the standard error, but need not be unbiased.^[ref]

The diagram compares variance with the expected fraction of distinct population labels observed. Without replacement this fraction is exactly $n/N$; with independent uniform sampling with replacement, a label is missed with probability $(1-1/N{)}^n$, so linearity of expectation gives expected fraction $1-(1-1/N{)}^n$. At six draws, replacement covers only about $66.5\%$ of labels on average and retains sample mean variance $35/18$; sampling all six distinct labels covers the population completely and removes selection uncertainty. This population coverage is distinct from confidence interval coverage probability.

The correction matters when the sampling fraction is appreciable. It calibrates uncertainty about this fixed population under the specified selection mechanism. It does not correct unequal selection weights, missing observations or measurement errors; observing every label also does not eliminate uncertainty about a larger unobserved population. Moreover, an exact variance alone does not guarantee the coverage probability of a proposed confidence interval: a small discrete sampling distribution can make a normal approximation inaccurate.

For a fixed bounded interval with no atoms at its endpoints, the occupation count of fully contained half-open dyadic intervals increases to the true count of a simple locally finite point measure. Refinement cannot reduce occupancy. Finitely many distinct atoms in a bounding interval have positive minimum separation; for fine enough cells each atom lies alone, and every interior atom's cell is contained in the target interval. Thus the counts eventually agree, path by path.

For $S=\frac{1}{2}{\int }_0^T({\dot{x}}^2-{\omega }^2{x}^2)dt$ and fixed endpoint values $a,b$, the Euler-Lagrange equation is $\ddot{x}+{\omega }^{2}x=0$. When $\sin \omega T\ne 0$, solve the endpoint equations for the unique trajectory and use ${\dot{x}}^2-{\omega }^2{x}^2=(x\dot{x}{)}^{\prime }$ on that trajectory to obtain the displayed action. For $\omega \ne 0$, at $\omega T=k\pi$ a solution exists only if $b=(-1{)}^{k}a$; then there is a family of solutions and their action is zero. At $\omega =0$ the nonresonant formula has the limit $(b-a{)}^2/(2T)$.

For a finite simple group of this order, $e=0$ gives a normal Sylow $5$-subgroup, and $e=1$ gives a nontrivial sign homomorphism from the regular action, because an involution swaps fifteen pairs. Thus $e\ge 2$. The Sylow $2$-count divides $15$; it cannot be one or three. If it is five, a Sylow subgroup's normalizer has index five. If it is fifteen, the strengthened Sylow congruence from intersections produces two Sylow subgroups meeting in index two. Their common subgroup is normal in both, so its normalizer contains both and has index three or five. Simplicity excludes index three. The resulting faithful degree-five action embeds $G$ in $A_5$ and forces order $60$, proving the conclusion.

A nontrivial finite group whose only normal subgroups are the trivial group and the whole group. A cyclic group of prime number order gives an abelian example; $A_5$ gives a nonabelian alternating group example. Finite versions of classical groups can produce further examples after taking suitable subgroups and quotients; small parameters require separate checks.

A viscometric flow has the local relative deformation history of steady simple shear flow, up to rigid changes of frame. For an isotropic simple fluid, symmetry leaves one shear stress and two independent normal-stress differences. The corresponding viscometric functions depend on the magnitude of the shear rate, while pressure supplies the isotropic part.

A simple closed curve in a connected oriented topological surface is nonseparating when cutting along it leaves a connected topological surface. In that case an arc joining the two new boundary components closes to a simple closed curve meeting the original once. Their algebraic intersection number of curves on an oriented surface is $\pm 1$. Consequently the original curve represents a nonzero primitive homology class.

On a closed oriented topological surface, a separating simple closed curve bounds a compact subsurface, whose fundamental class exhibits it as a boundary in homology. Thus on a closed oriented topological surface, separation is equivalent to vanishing of its integral first homology class.

The closed-surface condition cannot be omitted. In the open annulus $S^1\times \mathbb{R}$, the core $S^1\times \{0\}$ separates the two ends, yet generates $H_1(S^1\times \mathbb{R};\mathbb{Z})\cong \mathbb{Z}$. Either side is noncompact and supplies no finite singular chain bounding that core.

For an unknown nonzero period, a collision $f(x)=f(y)$ reveals the period $x\oplus y$. Birthday problem sampling finds one using $O(2^{n/2})$ queries with bounded error. A matching lower bound follows by choosing the period and output labels randomly: without a collision, $q$ queries rule out at most $\left(\genfrac{}{}{0ex}{}{q}{2}\right)$ candidate periods. For $q=o(2^{n/2})$, almost all periods remain consistent with the transcript. The exponential separation is in query complexity, compared with Simon's algorithm's linear quantum query count.

For real matrices, orthogonal similarity is the special case $B=Q^{-1}AQ=Q^{T}AQ$ with $Q$ an orthogonal matrix. Then a symmetric matrix remains symmetric, because $B^T=Q^T{A}^{T}Q=B$. The expression $QA{Q}^T$ describes the reverse conversion and has the same property. An orthonormal basis retains the ordinary dot product formula for its coordinate vectors.

An arbitrary change of basis need not preserve entrywise symmetry. In the opening example, $A$ is a symmetric matrix but $B=\left(\begin{array}{cc}2& -3\\ 0& 5\end{array}\right)$ is not. The geometric inner product has not disappeared: in the new coordinates it is represented by the Gram matrixFor $p$ with new coordinate vector $y=(1,1{)}^T$, the squared Euclidean norm is $y^{T}Gy=5$, agreeing with $x^{T}x=2^2+1^2$. The naive coordinate sum $y^{T}y=2$ would use the wrong inner product.

If $A^T=A$, direct multiplication givesThis is self-adjointness expressed in the new inner product, rather than entrywise symmetry relative to the identity matrix. For the example, both products equal $\left(\begin{array}{cc}2& 2\\ 2& 7\end{array}\right)$. Thus arbitrary matrix similarity preserves the linear operator, while its geometric interpretation requires transforming the Gram matrix as well. Matrix congruence supplies the corresponding rule for a bilinear form and should be distinguished from matrix similarity.

When $c=e$ in the signed cyclic three-mode amplitude equations, $\dot{X}=2X(\mu -aX)$ and $\dot{V}=3V(\mu -aX)$ for $V=xyz$. Thus $V/X^{3/2}$ is conserved wherever $X>0$, precluding asymptotically stable isolated equilibria at nonzero radius. This does not exclude attraction to the origin: for $\mu <0$ and $a>0$, $X$ tends to zero. When $\mu /a>0$, on the invariant sphere $X=\mu /a$, each octant contains closed contours of $V$ around the equal-magnitude equilibrium, bounded by heteroclinic arcs between coordinate axes if $c\ne 0$. For $c=0$ the sphere consists entirely of equilibria.

The unit tangent of a simple regular closed plane curve traversed anticlockwise turns once. Consequently its integrated signed curvature is $2\pi$; clockwise orientation gives $-2\pi$. Simplicity and one traversal are essential to this value.

For two real primitive nonprincipal Dirichlet characters with nonprincipal product, the real zeros in this interval are counted with multiplicity. The logarithmic-derivative coefficients of $\zeta L({\chi }_1)L({\chi }_2)L({\chi }_1{\chi }_2)$ are $\Lambda (n)(1+{\chi }_1(n))(1+{\chi }_2(n))\ge 0$. The global partial-fraction expansion of a Dirichlet L-function logarithmic derivative therefore gives $0\le 1/(\sigma -1)-{\sum }_{j=1}^2(\sigma -{\beta }_j{)}^{-1}+C\log (q_1{q}_2)$ if two nearby real zeros exist. Put $\sigma =1+a/\log (q_1{q}_2)$, choose $a$ small, and test $1-{\beta }_j<a/(4\log (q_1{q}_2))$. The right side is negative, a contradiction.

// Target: analytic-number-theory.bigb

For $(a,q)=1$ and $q\le {\log }^{A}N$, apply the Siegel–Walfisz theorem to every reduced residue class with error exponent $A+B+2$. The main terms sum to $N{c}_q(a)/\phi (q)$, where the Ramanujan sum satisfies $c_q(a)=\mu (q)$ for coprime $a,q$. Nonunits with nonzero Von Mangoldt function are powers of primes dividing $q$ and contribute at most $\omega (q)\log N$, which is absorbed by the error. For arbitrary $a$, the correct main term is $N{c}_q(a)/\phi (q)$; the reduced numerator hypothesis cannot be omitted.

A spatial sideband is a Fourier perturbation with wavenumber close to a pattern's carrier wavenumber, such as $k_c+p$ or $k_c-p$. A real physical field also includes the conjugate Fourier component. Linearizing a nonlinear complex amplitude equation can couple the two displaced components, so testing only a single uniform amplitude misses their joint linear stability problem.

For a spatially periodic nonlinear pattern, a linear stability disturbance is written $e^{\lambda t+iq\cdot x}v(x)$ with $v$ periodic on the pattern cell. The conjugated operator $L(q)$ acts on that fixed cell and determines the sideband eigenvalues. Translation invariance supplies zero eigenfunctions at $q=0$, namely spatial derivatives of the base pattern. Their small-$q$ eigenvalues determine long-wavelength phase stability; other parts of the spectrum can give independent instabilities.

Near a saddle-focus equilibrium, linearized coordinates satisfy $\dot{z}={\lambda }_{+}z$, $\dot{r}={\lambda }_{-}r$ and $\dot{\theta }=\omega$. A trajectory entering at a small positive coordinate $z$ takesto reach a fixed outgoing section. Hence $r_{\mathrm{out}}\propto z^{\delta }$ with $\delta =-{\lambda }_{-}/{\lambda }_{+}$, while the accumulated angle is logarithmic in $z$. Composing this passage with a smooth global return gives the leading scalar formwhere $|\Omega |=|\omega |/{\lambda }_{+}$ and the phase convention absorbs constant angular shifts. The map applies only on branches whose trajectories return to the section.^[ref]

Differentiation givesFor $\delta <1$, the envelope becomes unbounded as $z\downarrow 0$, but the derivative also has infinitely many zeros. The map therefore has strongly expanding branches, rather than uniform expansion everywhere.

The full two-dimensional return has Jacobian determinant of order $z^{2\delta -1}$ in regular section coordinates: radial scaling contributes $z^{\delta -1}$ and conversion of polar coordinates contributes $z^{\delta }$. For $\delta >1/2$ this contracts area near the singular limit, but area contraction alone does not prove that a scalar reduction is valid. A one-passage periodic trajectory has period $T_g+T_{\mathrm{loc}}$, explaining the logarithmically diverging time near the homoclinic connection.

For an invertible matrix $A$, the rank-one matrix update $B=A+u{v}^T$ is a singular matrix exactly when $1+v^T{A}^{-1}u=0$. This follows directly from the matrix determinant lemma. At that value, $u\ne 0$ and the nonzero vector $w=A^{-1}u$ satisfiesEvery vector in the kernel is a multiple of $w$, since $Bx=0$ implies $x=-A^{-1}u(v^{T}x)$. The matrix rank therefore drops by exactly one.

For $B(t)=I+tu{v}^T$ with $u=(1,2{)}^T$ and $v=(3,-1{)}^T$, the update becomes a singular matrix at $t=-1$, and $B(t)u=(1+t)u$. At $t=-0.99$, this direction is compressed by a factor $0.01$ and the inverse amplifies it by a factor 100. The actual spectral norm of the matrix inverse, which maximizes amplification over every direction, is about $700.17853409$.

The exact formula can expose a large condition number in the updated linear system even when the original one had a small condition number. It does not guarantee numerical stability merely because its evaluation uses fewer arithmetic operations. The denominator condition certifies existence; the condition number and floating-point behavior address different questions about using the resulting answer.