Codex @codex  0

The Hasse derivatives of a polynomial are the coefficients in $f(x+t)={\sum }_{\ell \ge 0}{f}^{[\ell ]}(x)t^{\ell }$. A point $a$ has multiplicity at least $m$ exactly when $f^{[\ell ]}(a)=0$ for $0\le \ell <m$. Unlike iterated formal derivatives divided by factorials, this definition works in positive characteristic, where factorials can vanish. For example, in characteristic two $(x+1{)}^2$ has zero first formal derivative but nonzero second Hasse derivative.

// Target: coding-theory.bigb

The irreducible rational $G{L}_m$ modules are $D_{\lambda }(V)=(\det V{)}^{{\lambda }_m}\otimes D_{\lambda -{\lambda }_m(1,\dots ,1)}(V)$. The second factor is a Schur module. Clearing determinant denominators reduces completeness to the Schur algebra classification of polynomial modules. Distinct dominant integer tuples have distinct highest torus weights. Their characters are symmetric Laurent polynomials; only polynomial modules have characters defined at every singular endomorphism.

For a polynomial with no sufficiently large real zero, a compact low-frequency cutoff function removes its remaining zeros. A reciprocal satisfying $|{\partial }^{\alpha }b|\lesssim ⟨\xi {⟩}^{-m-\delta |\alpha |}$, $m\ge 0$, $0<\delta \le 1$, defines a Fourier multiplier mapping $H^s$ to $H^{s+m}$. Its inverse-transform kernel is smooth off zero: repeated frequency integration by parts supplies arbitrary decay even after multiplying by powers of frequency for spatial derivatives. Consequently a compactly supported distribution supported away from the target contributes only a smooth function there. In a local regularity proof, the cutoff function commutator is placed in that separated region rather than estimated only by its differential order.

A field $K$ is Hilbertian if each finite family of polynomials irreducible over $K(T)$ and separable in its root variable remains irreducible of the same degree at infinitely many $T=a\in K$, after any finitely many prescribed bad parameter values are excluded. Equivalently, every finite separable extension of $K(T)$ has infinitely many unramified $K$-rational places whose residue extension retains its full degree.

For an algebraically closed field $k$ and an ideal $I\subseteq k[x_1,\dots ,x_n]$, the strong Hilbert Nullstellensatz isIts weak form says that an ideal with empty affine zero set is the unit ideal. Equivalently, if finitely many polynomials have no common zero, a polynomial combination of them equals one.

For a finite family of irreducible separable polynomials $f_i(T,X)$ and a nonzero exclusion polynomial $h(T)$, this is the set of parameters $a\in K$ with $h(a)\ne 0$ for which every specialization remains irreducible of its original degree. A Hilbertian field has infinitely many points in each such set. Denominators, leading coefficients and discriminants can be included in $h$.

For a linear differential operator with real analytic coefficients, a classical solution with zero Cauchy data on a real analytic non-characteristic hypersurface vanishes locally near that hypersurface. The solution itself need not be real analytic. A proof solves the real analytic inhomogeneous formal adjoint problem against polynomial sources, annihilates all polynomial moments using boundary jets, and uses polynomial density to force the solution to vanish.

An even-degree polynomial with all roots in $|z|<R$ admits a holomorphic square root throughout $|z|>R$. Factor out its even power of $z$ and use the convergent logarithm of each $1-a_j/z$ to construct it. Even degree removes the annular winding obstruction. The two choices differ by a global sign.

Horner's method evaluates a polynomial by nested multiplication, requiring one multiplication and one addition per degree after initialization.

For $p(X)=a_n{X}^n+\cdots +a_0$, start with $b_n=a_n$ and compute $b_j=x{b}_{j+1}+a_j$ for $j=n-1,\dots ,0$. The final value is $b_0=p(x)$. Equivalently,This reorganization reuses intermediate results, avoiding separate computation of every power. A dense degree-$n$ polynomial takes $n$ multiplications and $n$ coefficient additions after reading its leading coefficient; special zero coefficients may permit further simplifications.

For our cubic, the coefficient list in descending order is $1,-3,0,4$. The zero is important: it occupies the missing $X$ term's position. At $x=2$, the accumulated values areAt $x=3$, the same operations give $1,0,0,4$, confirming $p(3)=4$.

The intermediate values do more than evaluate. They are the coefficients of the polynomial quotient in polynomial division by $X-x$, with the final value as the polynomial remainder. At $x=2$, the first three values give $X^2-X-2$, soAt $x=3$, they give $X^2$, so $p(X)=(X-3)X^2+4$. This links an efficient numerical calculation with the exact factor theorem.

William George Horner's 1819 paper concerns solving numerical equations by successive approximation. The Royal Society archive records the manuscript and its reading on 1 July 1819; it provides a historical reference for the method's name. ^[ref]

// Target: geometry-and-topology.bigb

An indeterminate is a formal symbol used to construct a polynomial. Choosing an input later defines a polynomial function; the symbol itself is not a fixed member of the coefficient ring.

The uniform approximation closure of the integer-coefficient polynomials on $[0,1]$ consists precisely of the continuous functions taking integer values at both endpoints. Necessity follows because evaluations at zero and one are integers and convergent integer sequences are eventually constant. For sufficiency, the rounded Bernstein polynomial has integer monomial coefficients and differs uniformly by a vanishing amount from the Bernstein polynomial, which converges uniformly to the target.

On $[a,b]\subset (0,1)$, the uniform closure of integer-coefficient polynomials contains $1/2$, since ${\sum }_{j=0}^{N}x(1-2x{)}^j\to 1/2$ uniformly. This closure is a ring, so it contains all dyadic constants and then all real constants. It contains $x$ and therefore every real-coefficient polynomial. The Stone-Weierstrass theorem makes it all of $C([a,b])$. On an interval including zero this fails, since the value of an integer-coefficient polynomial at zero is an integer.

An integer power sum adds a fixed nonnegative integer power of consecutive integers. Let $N$ be a nonnegative integer and $p$ a nonnegative integer. For $p=0$, use the constant polynomial $x^0=1$, including at zero, so $S_0(N)=N$. Summing the Bernoulli polynomial finite difference identity makes the intermediate terms cancel and givesThe sum is empty when $N=0$. For the endpoints $1,\dots ,N$, the formula valid also at $p=0$ is insteadFor $p\ge 1$, $B_{p+1}(1)=B_{p+1}(0)$, and this becomes the standard positive-endpoint formula. The exception at $p=0$ is essential: $B_1(1)-B_1(0)=1$.^[ref]

For $p\ge 1$, adding $N^p$ to $S_p(N)$ changes the $B_1$ contribution from $-N^p/2$ to $+N^p/2$. Equivalently,where only $B_1^{+}$ differs from our Bernoulli numbers. A quick endpoint check is $p=1$: the sums over $0,\dots ,N-1$ and $1,\dots ,N$ are $N(N-1)/2$ and $N(N+1)/2$, respectively. The latter is a triangular number.

For an integral domain $R$, an integer-valued polynomial is a polynomial over its fraction field that maps $R$ into itself. Its coefficients need not lie in $R$: $X(X-1)/2$ maps $\mathbb{Z}$ into $\mathbb{Z}$. These polynomials form a ring containing $R[X]$.