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An entire function bounded by a polynomial in the radius is itself a polynomial, by Cauchy estimates.

If an entire function satisfies $|h(z)|\le |\mathrm{Re}z{|}^{-1/2}$ away from the imaginary axis, then $h=0$. Cauchy's formula on $|z|=R$ bounds every Taylor coefficient by a constant timeswhich tends to zero as $R\to \infty$.

The image of every nonconstant entire function is dense in $\mathbb{C}$. If a disc about $w$ were omitted, then $1/(f-w)$ would be bounded and entire, so Liouville's theorem would make $f$ constant.

A Riccati equation is a first-order equation quadratic in the unknown function.

Every convergent series of positive rational terms has a subseries whose rational partial sums $p_j/q_j$ satisfyChoose each new term small enough to enforce all previous tail bounds. Separation of reduced fractions first makes $\alpha$ irrational, and the Liouville approximation theorem then rules out every finite algebraic degree.

For distinct reduced fractions $p/q$ and $u/v$,If $q>v$, this is strictly greater than $q^{-2}$.

Let $\alpha$ be a real irrational algebraic number of field extension degree $d\ge 2$, and let $P\in \mathbb{Z}[X]$ be its primitive algebraic minimal polynomial. For any rational $p/q$ with positive integer $q$, irreducibility gives $P(p/q)\ne 0$. Consequently the nonzero integer $q^{d}P(p/q)$ givesOn a fixed closed interval around $\alpha$, the mean value theorem gives $|P(p/q)|\le M|p/q-\alpha |$, where $M=\max |P^{\prime }|>0$ on that interval. Thus rational points in the interval satisfy $|\alpha -p/q|\ge M^{-1}{q}^{-d}$. Outside it, a fixed positive distance from $\alpha$ gives the same type of bound after decreasing the constant. Choosing a smaller constant makes the inequality strict, proving the Liouville approximation theorem.

For nonreal algebraic numbers, the positive distance from the real axis already gives a uniform lower bound against rational approximations. The substantive approximation argument is therefore the real case.

An unlink is a link whose components bound pairwise disjoint embedded disks in $S^3$.

For a complex $m$-by-$n$ matrix $A$, the linear system $Ax=b$ has a solution precisely when $b$ is orthogonal to every vector in $\ker A^{†}$, where $A^{†}$ is the conjugate transpose. For a real matrix, use the transpose $A^T$ instead. Indeed,because the defining adjoint operator identity turns orthogonality to every $Ax$ into $A^{†}y=0$. Taking the orthogonal complement again recovers $\mathrm{im}A$ in finite dimensions. Thus a linear functional in the annihilator of a vector subspace gives the exact obstruction to solvability, rather than just a necessary test.

For $A=\mathrm{diag}(1,0)$, its image of a linear map is the first coordinate axis and $\ker A^{†}$ is the second. Consequently $Ax=(b_1,b_2{)}^T$ is solvable exactly when $b_2=0$.

A linear equation is an equation in which the unknowns occur only to the first power and are combined linearly.

Starting at the same value, forward and backward Euler have opposite leading local errors. Their half-difference therefore estimates either local error:For a first-order method, a local-tolerance controller consequently scales the next step by $(\mathrm{tol}/E_n{)}^{1/2}$, usually with a safety factor.

If a linear system has negative eigenvalues with widely separated magnitudes, an explicit method can be forced to resolve the fastest decaying mode solely for stability. For eigenvalues $-1$ and $-100$, forward Euler requires $h\le 0.02$, whereas backward Euler is stable for every positive step size.

A method is A-stable when its stability domain contains the closed left half-plane.

For a real $2\times 2$ Jacobian matrix $J$, both eigenvalues have strictly negative real parts if and only ifThe characteristic polynomial is ${\lambda }^2-(\mathrm{tr}J)\lambda +\det J$. If its roots are real, positive product makes their signs agree and negative sum makes both negative. If they are a complex conjugate pair, each real part is half the trace. Conversely, two eigenvalues with negative real parts have negative real sum and positive product.

For a continuous-time nonlinear system, the linearization stability theorem turns this strict test into local asymptotic stability. If $\det J<0$, the real eigenvalues have opposite signs and the equilibrium is a saddle. Equality in either condition needs separate analysis. A discrete map instead requires eigenvalues inside the unit disk; the trace-determinant inequalities for negative real parts are not its stability test.

A pulled front is selected by growth and diffusion in its small-amplitude leading edge rather than by nonlinear dynamics behind the front.

This criterion concerns travelling-front solutions of Fisher's equation.

For diffusion coefficient $D$ and leading-edge growth rate $r>0$, a Fisher-KPP-type pulled front has minimum speed $2\sqrt{Dr}$.

The transportation problem minimizes linear shipping cost while matching prescribed row supplies and column demands.

This is the finite discrete optimal transport problem with prescribed supply and demand marginals.

For the primal maximization $Ax\le b$, $x\ge 0$ and dual minimization $A^{T}y\ge c$, $y\ge 0$, a feasible pair has the nonnegative objective-gap decompositionThus equal objective values are equivalent toThese are complementary slackness. Combined with primal and dual feasibility they certify optimality by weak duality. Conversely, primal and dual optima satisfy them by strong duality.

A positive dual multiplier forces its primal constraint to be tight, and a positive primal variable forces its dual constraint to be tight. A tight constraint need not have a positive multiplier: both factors in a complementary product may be zero. This distinguishes ordinary complementary slackness from a stronger strict-complementarity assertion. In the example under linear programming, the third primal inequality is slack and its multiplier is zero.^[ref]

Weak duality bounds every feasible primal objective by every feasible dual objective, with the direction set by the primal convention. For the primal maximization $Ax\le b$, $x\ge 0$, dual feasibility means $y\ge 0$ and $A^{T}y\ge c$. HenceThe first inequality uses $x\ge 0$; the last uses $y\ge 0$. For a minimization primal, feasible dual values instead provide lower bounds. The argument is algebraic and needs no assumption that optima exist or are equal.

If a feasible pair attains equal objective values, each side must already be optimal. If a maximization primal is unbounded above, no feasible finite-valued dual vector can exist. The converse of that last statement is not automatic: dual infeasibility can also occur when the primal is infeasible. Strong duality is the additional theorem asserting equality under suitable conditions.

The dual pairsatisfies weak duality because $b^{T}y\le y^{T}Ax=x^T{A}^{T}y\le c^{T}x$ for every feasible pair.