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For a field with a Non-Archimedean absolute value, define the coefficient norm $\Vert f{\Vert }_G={\max }_i|a_i|$ for $f(T)={\sum }_i{a}_i{T}^i$. Then $\Vert fg{\Vert }_G=\Vert f{\Vert }_G\Vert g{\Vert }_G$. For nonzero polynomials, scale each by one of its coefficients of maximum absolute value; the resulting polynomials have integral coefficients and at least one unit coefficient. Their reductions in the residue field are nonzero, so their product is nonzero. The product therefore also has coefficient norm $1$, proving the formula after reversing the scaling. This compares binary-polynomial coefficient heights with the heights of their roots.

The multiplicity of a root $a$ of a polynomial is the largest integer $m$ such that $(X-a{)}^m$ divides the polynomial.

A linear code is negacyclic if $(c_0,\dots ,c_{n-1})$ being a codeword implies that $(-c_{n-1},c_0,\dots ,c_{n-2})$ is a codeword. Polynomial identification makes it an ideal modulo $X^n+1$. For odd $n$, the substitution $X\mapsto -X$ gives a monomial equivalence of linear codes between negacyclic and cyclic codes.

Integrate the Lagrange polynomial through values at all equally spaced nodes $0,\dots ,s$, including both endpoints. The resulting quadrature rule is exact through degree $s$. For even $s$, the node polynomial ${\prod }_{j=0}^s(t-j)$ is odd about $s/2$, so its integral vanishes and exactness extends through degree $s+1$. The weights need not remain positive as the number of nodes increases.

For a polynomial with nonzero constant coefficient over a complete discretely valued field, plot coefficient exponents against their valuations. A lower Newton polygon segment of slope $s$ and horizontal length $r$ accounts for exactly $r$ roots of valuation $-s$, counted with multiplicity in an algebraic closure. Zero coefficients are omitted. Reflecting the horizontal axis changes the sign of every slope, so slope conventions must be fixed before reading root valuations.

The Newton root-finding iteration seeks a zero of a function $F$ by repeatedly solving its tangent line approximation. At a current guess $t_r$ with derivative $F^{\prime }(t_r)\ne 0$, setGeometrically, follow the tangent line to the horizontal axis, then return to the function graph for the next tangent line. Close to a suitable zero of a function, the error becomes dramatically smaller.^[ref]

For a concrete square root calculation, take the polynomial $F(t)=t^2-2$ and start at $t_0=1$. The Newton root-finding iteration becomes $t_{r+1}=(t_r+2/t_r)/2$, givingThe errors use the unrounded iterates. After the first step, each iterate overestimates $\sqrt{2}$, and the exact identityexhibits quadratic convergence: once close, the number of correct decimal digits roughly doubles per step. This speed makes the Newton root-finding iteration useful when accurate derivatives are available.

Global convergence requires additional care. For the polynomial $F(x)=x^3-2x+2$, starting at $x_0=0$ gives $F(0)=2$, $F^{\prime }(0)=-2$, hence $x_1=1$. Then $F(1)=1$, $F^{\prime }(1)=1$ gives $x_2=0$. The sequence repeats $0,1,0,1,\dots$ forever, although a real polynomial root lies in $(-2,-1)$ by the intermediate value theorem. Neither point in the cycle is a zero of a function, and neither has zero derivative. A maintained sign-changing interval and fallback bisection method can protect a root solver from unsuitable Newton steps.

A real polynomial is globally nonnegative if its value is nonnegative at every point of its real domain. A sum of squares polynomial is always nonnegative, but the Horn copositive matrix gives a nonnegative quartic polynomial that is not a sum of squares polynomial.

If an entire function $f$ is not a polynomial, then there is a point $z_0\in \mathbb{C}$ for which $f^{(n)}(z_0)\ne 0$ for every nonnegative integer $n$. Consequently every coefficient in the Taylor series of $f$ about $z_0$ is nonzero.

For each $n$, the function $f^{(n)}$ is not identically zero, since otherwise $f$ would be a polynomial. Its zero set is closed and has empty interior by the identity theorem, hence is a nowhere dense set. The Baire category theorem says that the complete metric space $\mathbb{C}$ cannot be the union of these countably many zero sets.

Set $S_n=\{x:f^{(n)}(x)=0\}$ and $E_n=\mathrm{int}{S}_n$. The sets $S_n$ are closed, $E_n\subset E_{n+1}$, and Baire's theorem on each interval shows that $\Omega ={\bigcup }_n{E}_n$ is dense. On every connected component of $\Omega$, compactness and the increasing cover by the $E_n$ show that one derivative vanishes locally with a uniform finite order, so $f$ agrees there with one polynomial.

If $F=\mathbb{R}\setminus \Omega$ were nonempty, it would be closed and have no isolated points: polynomial pieces on both sides of an isolated point would have matching derivatives of every order and hence join into one polynomial piece. Applying Baire's theorem to the cover $F={\bigcup }_n(F\cap S_n)$ gives an interval $U$ and an index $k$ for which $F\cap U$ is nonempty and contained in $S_k$. Because $F$ has no isolated points, difference quotients show that every derivative of order at least $k$ vanishes on $F\cap U$. Each polynomial component of $\Omega$ meeting $U$ has an endpoint in $F\cap U$, so its degree is less than $k$. Thus $f^{(k)}=0$ throughout $U$, contradicting $F\cap U\ne \varnothing$. Hence $\Omega =\mathbb{R}$, and its sole connected component carries one polynomial.

Numerical integration approximates a definite integral using finitely many sampled function values. For an integrand $f$ on a finite interval $[a,b]$, $a<b$, a quadrature rule has the formThe quadrature nodes $x_j$ tell us where to evaluate $f$; the quadrature weights $w_j$ turn those values into an approximation to the accumulated signed area. The exact definite integral depends on the integrand throughout the interval, whereas the quadrature rule sees only these samples. Integrating a chosen interpolant exactly still generally approximates the original integrand's integral.^[ref]

The fundamental theorem of calculus evaluates $I(f)$ as $F(b)-F(a)$ when an antiderivative $F$ is available. Numerical integration is useful when no convenient antiderivative is known, or each evaluation of the integrand requires an expensive computation. The task then becomes choosing informative quadrature nodes and controlling the quadrature error with as few evaluations as possible. Fixed data also permit a quadrature rule, but refinement cannot supply new information between observations.^[ref]

For an exact comparison, take the polynomial $f(x)=x^2$ on $[0,1]$. Its antiderivative $x^3/3$ gives $I(f)=1/3$. On a partition of an interval into $n\ge 1$ equal panels, the composite midpoint quadrature rule uses $n$ midpoint samples, each with quadrature weight $1/n$; the composite trapezoidal rule uses $n+1$ endpoint samples, with outer quadrature weights $1/(2n)$ and interior quadrature weights $1/n$. The existing derivations under these two quadrature rules giveIn particular, $M_1=1/4$ and $T_1=1/2$ already bracket $1/3$; two panels give $M_2=5/16$ and $T_2=3/8$. The convex function $x^2$ explains the opposite signs. A midpoint rectangle can cross the curve within its panel, so its total signed area, rather than pointwise ordering, determines the underestimate.

Define the signed approximation error as $E_n=Q_n(f)-I(f)$. In this example the absolute errors are $1/(12n^2)$ and $1/(6n^2)$: doubling $n$ reduces each by exactly four. A plot with both axes on a logarithmic scale therefore has slope $-2$, expressing order two with respect to panel refinement. This is an exact calculation for this polynomial, including $n=1$, rather than an observed rate extrapolated from a few computed answers.

For a general integrand, the composite midpoint quadrature rule and composite trapezoidal rule give second-order quadrature error bounds when $f$ has two continuous derivatives on $[a,b]$ and a known bound on $|f^{\prime \prime }|$. These hypotheses describe behavior between samples. A formal error formula containing an unknown derivative is not a numerical accuracy certificate or a justified sample count. Finite samples do not certify quadrature accuracy makes the missing information explicit. Errors in the evaluations and roundoff error must also be considered; noise sensitivity of a quadrature rule treats the former.

More elaborate choices exploit other information: Gaussian quadrature obtains high polynomial exactness from carefully chosen quadrature nodes, and Simpson's rule uses a quadratic interpolant. Adaptive numerical integration compares approximations and concentrates samples where they disagree, often saving evaluations. Such a difference is an error estimate, not a certified bound without further assumptions: both approximations can miss the same feature.^[ref]

For $-W^{\prime \prime }+2x{W}^{\prime }=EW$, the power series coefficients obey $(n+2)(n+1)a_{n+2}=(2n-E)a_n$. The condition $W(0)=0$ makes every even coefficient zero. A nonzero odd series terminates precisely when $E=2d$ with $d$ an odd positive integer. A polynomial of degree $d$ must have this value of $E$ by comparison of leading coefficients, so these are all possibilities. With $W^{\prime }(0)=1$, the first three polynomial solutions are $x$, $x-2x^3/3$, and $x-4x^3/3+4x^5/15$, with $E=2,6,10$. They are scalar multiples of the odd Hermite polynomials.

A finite group permutes the associate classes of irreducible polynomial factors of an invariant. The product of one representative from each orbit transforms by a scalar character of the group. If that character is trivial, the product is invariant. Its transitive factor orbit makes it prime in the invariant ring: an invariant polynomial divisible by one orbit factor is divisible by all of them.

If a nonzero continuous real function on an interval is orthogonal to every polynomial of degree below $n$, it has at least $n$ distinct interior zeros. If it had only finitely many zeros and fewer than $n$ sign changes, multiply it by the product of linear factors at its sign-changing zeros. The product has constant sign and is nonzero on an open interval, so its integral cannot vanish. Orthogonality to this degree-below-$n$ polynomial contradicts that conclusion. Infinitely many interior zeros already satisfy the assertion. Orthogonal polynomials of respective degrees $0,\dots ,n-1$ form a basis of the needed polynomial space.