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Suppose every point of ${\mathbb{F}}_q^n$ lies on an affine line in a vector space meeting a subset $N$ in at least $m$ points, with $1\le m\le q$. A smaller $N$ would admit a nonzero multivariate polynomial of total degree at most $m-1$ vanishing on it, by the dimension of a bounded-total-degree polynomial space. Each selected line then has more roots of a polynomial than the degree of its restriction, so the polynomial vanishes at every point. The Schwartz-Zippel lemma forbids this because $m-1<q$. Thus the displayed bound holds, and is at least $m^n/n!$. This is a point-covering condition; the finite-field Kakeya set condition instead quantifies over directions.

Since $2\not\equiv 1(\mathrm{mod}4)$, the ring of integers of $\mathbb{Q}(\sqrt{2})$ iswith integral basis $1,\sqrt{2}$ and field discriminant eight. Multiplication is ordinary polynomial multiplication followed by $(\sqrt{2}{)}^2=2$:For example, $(1+\sqrt{2}{)}^2=3+2\sqrt{2}$ and $(1+\sqrt{2}{)}^3=7+5\sqrt{2}$. The integer coefficients stay integral, as required for a ring.

Under the two field embeddings, the element $a+b\sqrt{2}$ becomes the point $(a+b\sqrt{2},a-b\sqrt{2})$. The integral basis vectors become $(1,1)$ and $(\sqrt{2},-\sqrt{2})$, giving a Minkowski embedding of a real quadratic field. Their determinant has absolute value $2\sqrt{2}=\sqrt{8}$, consistent with the field discriminant.

If a nonzero real polynomial has $N$ roots in an open real interval, counted with multiplicity of a root, its derivative has at least $N-1$ roots in that interval, also counted with multiplicity, provided $N\ge 1$. Each root of multiplicity $m$ becomes a root of the derivative of multiplicity $m-1$. The Rolle theorem supplies one further root between each pair of distinct consecutive roots. Thus $k$ distinct roots of multiplicities $m_j$ give at least ${\sum }_j(m_j-1)+(k-1)=N-1$ derivative roots.

Rolle's theorem is the equal-endpoint case of the mean value theorem. Let $a<b$ and let the real-valued function $f:[a,b]\to \mathbb{R}$ be continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$. If $f(a)=f(b)$, there exists $c\in (a,b)$ with $f^{\prime }(c)=0$. Geometrically, a tangent line somewhere in the interior is horizontal, like the secant line joining the endpoints. ^[ref]

For the polynomial $f(x)=x(x-2)$ on $[0,2]$, both endpoint values are zero and the derivative is $f^{\prime }(x)=2x-2$. Thus $c=1$, where $f(1)=-1$, and the horizontal tangent line is $y=-1$. The horizontal secant line is $y=0$: equal endpoint values do not force the interior tangent line to pass through those endpoints.

The absolute value function $f(x)=|x|$ on $[-1,1]$ exposes the importance of differentiability. It is continuous and $f(-1)=f(1)=1$, so the secant line has slope zero. However,and the derivative does not exist at zero because the one-sided derivatives disagree. Every existing derivative in the interior is nonzero. The corner fails a hypothesis of both Rolle's theorem and the mean value theorem, so neither conclusion is guaranteed.

For a short proof, the extreme value theorem supplies a greatest and a least value on $[a,b]$. If $f$ is a constant function, every interior point works. Otherwise at least one of these values differs from the common endpoint value, so it is attained at an interior extremum. At an interior local maximum or local minimum, differentiability forces the derivative to be zero: the difference quotients from the two sides have opposite weak signs and the same limit.

This turns repeated values into information about derivatives. Two distinct roots of a polynomial force a root of its derivative between them; iterating this idea gives the Rolle root count with multiplicities. More generally, a differentiable function with nonzero derivative throughout a real interval cannot repeat a value there, hence is injective. The beauty is that knowing only two equal values guarantees an exact local feature somewhere between them.

A root bracket is a closed interval known to contain at least one zero of a function. For a real-valued continuous function on $[a,b]$, a strict change of sign between $f(a)$ and $f(b)$ certifies such an enclosure by the intermediate value theorem. An endpoint with value zero already supplies a zero of a function. For example, $[1,2]$ is a root bracket for the polynomial $f(x)=x^2-2$, since $f(1)=-1$ and $f(2)=2$. A root bracket need not contain a unique zero of a function.

A root of an irreducible polynomial in an extension field defines a base-field embedding of the polynomial quotient by evaluation. Conversely the image of the residue class of the variable determines the embedding and is a root. This counts embeddings through separable minimal polynomials.

A root of a polynomial $f$ is a scalar $a$ satisfying $f(a)=0$ in the chosen field or ring. For $p(X)=(X+1)(X-2{)}^2$ over the real numbers, the distinct roots of a polynomial are $-1$ and $2$. The second has multiplicity of a root two, so there are three roots when counted with multiplicity.

The domain matters. $X^2+1$ has no real polynomial root, but has the two complex number roots $i,-i$. The fundamental theorem of algebra guarantees a polynomial root in the complex numbers for every nonconstant complex polynomial, and complete factorization counts its degree of a polynomial many roots with multiplicity of a root. It does not promise that those roots are real or that they have a simple expression.

The Routh-Hurwitz criterion tests whether all roots of a real polynomial have negative real part, by sign conditions on determinants formed from its coefficients. Applied to a characteristic polynomial, it gives an algebraic test for strict linear asymptotic stability.

For the unit disk with the nonpositive real radius removed, compact sets $K_m=\{r{e}^{i\theta }:1/m\le r\le 1-1/m,|\theta |\le \pi -1/m\}$, $m\ge 3$, exhaust the domain. Each has connected complement: its inner and outer complementary regions join through the missing angular sector. The polynomial Runge theorem gives a polynomial within $1/m$ of a given holomorphic function on $K_m$. Every compact subset lies in some $K_m$, so these polynomials converge uniformly on compact subsets.

For a fixed polynomial $f\in \mathbb{Q}[X]$ with at least two distinct complex roots, integer solutions with $m\ge 2$ and $|y|>1$ have an effectively bounded exponent. Factorization over a splitting field and the fixed differences between distinct roots control common ideal divisors. Outside a fixed set of primes, each factor's valuation is divisible by $m/\gcd (m,e_i)$, where $e_i$ is its root multiplicity. The generalized power-equation estimates from linear forms in logarithms of algebraic numbers, including unit and exceptional-prime factors, give the exponent bound. A single-root polynomial and the values $y=0,\pm 1$ cannot be included: $4^m=(2^m{)}^2$, and $1^m=f(0)$ for $f(X)=X(X-1)+1$, give infinite exponent families.

For a nonzero multivariate polynomial of total degree $d$ over a field, and a finite subset $S$ of that field, at most $d|S{|}^{n-1}$ points of $S^n$ are zeros. For one variable this is the root bound for a polynomial. If $d\ge |S|$, the bound is already trivial. Otherwise, in the inductive proof write the polynomial as degree $k$ in its last variable, with a nonzero leading coefficient of degree at most $d-k$ in the other variables. The leading coefficient vanishes on at most $(d-k)|S{|}^{n-2}$ fibers; outside those fibers there are at most $k$ roots of a polynomial per fiber. Counting the bad fibers with the trivial $|S|$ bound gives the asserted inequality.

The Haar wavelet provides an explicit one-vanishing moment comparison:It cancels a constant polynomial, but not a nonconstant affine function. To obtain two vanishing moments by an equally transparent construction, take three adjacent unit intervals with heights proportional to $1,-2,1$:The L2 norm normalization follows from $1^2+(-2{)}^2+1^2=6$. Direct integration givesThus $\eta$ has exactly two vanishing moments. In fact its coefficient is a finite difference of adjacent local integrals:The weights are those of the second-order central difference. The first moment balances the centres of the three intervals; it is not enough merely to balance their signed areas.

This $\eta$ is a normalized analyzing function, with no orthonormal basis assertion. Indeed, its unit translate has nonzero L2 inner product:Hence these translates already fail orthogonality. The Daubechies wavelet construction imposes additional refinement and orthogonality requirements; two vanishing moments and unit L2 norm by themselves do not supply those requirements.

A linear subspace $W\subseteq X^{\ast }$ is separating if every nonzero $x\in X$ has $T(x)\ne 0$ for some $T\in W$. For finite-dimensional $X$, an evaluation map $X\to F^{\dim W}$ is injective, so the rank-nullity theorem forces $\dim W\ge \dim X=\dim X^{\ast }$ and $W=X^{\ast }$. For the infinite-dimensional polynomial space, the span of its coefficient functionals is a proper separating linear subspace of the dual space, which contains all sequences of coefficients, not just finitely supported ones.

For the Landau free energy $G(\varphi )={\varphi }^6/6+g{\varphi }^4/4+\epsilon {\varphi }^2/2$, positive $g$ gives a continuous phase transition at $\epsilon =0$. For negative $g$, phase coexistence between the disordered global minimum $\varphi =0$ and the ordered global minima occurs on $\epsilon =3g^2/16$, with ${\varphi }^2=-3g/4$. Indeed, on that locus the polynomial is $G(\varphi )={\varphi }^2({\varphi }^2+3g/4{)}^2/6\ge 0$, so these three zeros are global minima. For $g<0$, the disordered and ordered spinodal points are respectively $\epsilon =0$ and $\epsilon =g^2/4$; equality of stationary free energies determines phase coexistence, while loss of a local minimum determines a spinodal point.

For $L(f)=f^{\prime }(0)+\frac{3}{2}f(0)-2f(1)+\frac{1}{2}f(2)$ on $C^3[0,2]$, the Peano kernel isThe Taylor expansion with integral remainder gives $L(f)={\int }_0^{2}K(t)f^{(3)}(t)dt$, because $L$ annihilates polynomials of degree at most two. Since $K\ge 0$ and ${\int }_0^{2}K=1/3$, the sharp bound is $|L(f)|\le \frac{1}{3}\Vert f^{(3)}{\Vert }_{\infty }$. Equality holds for $f(x)=x^3/6$.

A polynomial may fail the Eisenstein polynomial test while a translate $f(T+a)$ passes it. Translation does not change the generated extension: if $f(\alpha )=0$, then $f(T+a)$ vanishes at $\alpha -a$. For $f(T)=T^3-m$, the translate by $m$ is Eisenstein at $3$ whenever $3\nmid m$ and $m\not\equiv \pm 1(\mathrm{mod}9)$.