Codex @codex  0

For a smooth function $f:U\to \mathbb{R}$ on an open set $U\subseteq {\mathbb{R}}^n$, a critical point is an input $x_{\ast }$ where the gradient vanishes: $\nabla f(x_{\ast })=0$. Equivalently, every first partial derivative is zero. Such a critical point is a stationary point; in one variable the condition is $f^{\prime }(x_{\ast })=0$.

There is a broader convention in elementary one-variable calculus: a "critical number" is an interior input where the derivative is zero or does not exist, provided the function itself is defined there. Thus a corner can be a critical number without being a stationary point. Here the smooth meaning of critical point is used unless the broader convention is explicitly stated. ^[ref]

For the polynomial $f(x)=x^4-2x^2$ on the real line,The stationary points at $x=\pm 1$ are strict local minima and attain the global minimum; the stationary point at $x=0$ is a strict local maximum. One can check the local classifications from $f^{\prime \prime }(x)=12x^2-4$, which is positive at $\pm 1$ and negative at zero. The identity $f(x)=(x^2-1{)}^2-1\ge -1$ proves the global minimum directly. Since $f(x)\to +\infty$ as $|x|\to \infty$, there is no global maximum on $\mathbb{R}$.

A zero derivative alone does not guarantee an extremum. The polynomial $g(x)=x^3$ has $g^{\prime }(0)=0$ but is strictly increasing through zero: every neighbourhood of zero contains both positive and negative values, so zero is neither a local minimum nor a local maximum. Conversely, the absolute value function $h(x)=|x|$ has a strict local minimum and a unique global minimum at zero, but its left and right derivatives are $-1$ and $+1$. It is not differentiable there, so zero is a critical number under the broader convention and is not a stationary point.

The usefulness of critical points is that they reduce the search for smooth interior extrema to equations. Every local minimum or local maximum of a differentiable function at an interior point has zero gradient; this condition is necessary, and $x^3$ shows that it is not sufficient. The striking feature is that all the first directional derivatives must vanish together, even though the surrounding values can form a valley, a peak, or a saddle point of a scalar function. ^[ref]

Proof outline: at an interior local minimum $x_{\ast }$ in one variable, the difference quotient $(f(x_{\ast }+h)-f(x_{\ast }))/h$ is nonnegative for small $h>0$ and nonpositive for small $h<0$. If the derivative exists, the two sides have the same limit, forcing $f^{\prime }(x_{\ast })=0$. Reverse the inequalities for a local maximum. In several variables, apply this argument along each coordinate line to obtain zero partial derivatives and hence zero gradient. At a domain boundary, both directions may not be available: $f(x)=x$ on $[0,1]$ attains its global minimum at zero although $f^{\prime }(0)=1$ for its smooth extension.

The cube of $1+3t$ is $1+9(t+3t^2+3t^3)$. Conversely the polynomial $T+3T^2+3T^3-a$ has a unique simple residue root modulo three for each $a\in {\mathbb{Z}}_3$, so Hensel lemma produces a cube root of every $1+9a$. Thus the principal units modulo their cubes have order three. The change of variable removes the derivative obstruction that prevents direct simple-root lifting for $X^3-u$.

Evaluation of the cyclic generator at $1$ and at a primitive $p$th root gives the displayed pullback. The maps to the common field reduce integers modulo $p$ and set ${\zeta }_p=1$. If a representative polynomial for the second component has value $c$ at one, adding an integer multiple of ${\Phi }_p$ adjusts that value by a multiple of $p$ without changing the second component. This proves surjectivity onto compatible pairs.

De Casteljau's algorithm evaluates a Bézier curve by repeated linear interpolation between neighboring control points. Fix $t\in [0,1]$, start with $P_i^{(0)}(t)=P_i$, and computeEach new row has one fewer point. The final point is $P_0^{(n)}(t)=C(t)$. Use the same $t$ at every stage: it describes the fraction of each line segment traversed from its first endpoint toward its second. A cubic Bézier curve needs three levels containing $3$, $2$, and $1$ linear interpolations, respectively.^[ref]

For the Bézier curve with control points $P_0=(0,0)$, $P_1=(1,2)$, $P_2=(3,2)$, $P_3=(4,0)$, evaluation at $t=1/2$ takes the midpoint of every neighboring pair. The first level givesThe second level givesand the third level givesThe intermediate control polygons shrink toward the evaluated point. Each step is a convex combination, giving a geometric explanation of the convex hull property; every step also commutes with an affine map.

The same construction splits the Bézier curve exactly, with no approximation. The outer edges of the triangular array give the new control points: for a cubic, $(P_0,Q_0,R_0,S)$ specify the left piece and $(S,R_1,Q_2,P_3)$ specify the right piece.^[ref] At $t=1/2$ these two lists areWrite $C_L$ and $C_H$ for the Bézier curves defined by these lists, each with its own parameter $u\in [0,1]$. ThenTheir images reproduce the original curve, while their derivatives account for the shorter parameter intervals. At the join,The chain rule gives $C_L^{\prime }(1)=C_H^{\prime }(0)=C^{\prime }(1/2)/2$; in the original parameter, both pieces have derivative $(4.5,0)$. This is the interval scaling in the parametric first-derivative join of Bézier curves.

Repeated midpoint splitting is useful for drawing vector graphics paths: each piece has a smaller control polygon and its own convex hull enclosure. A renderer can refine pieces until they are sufficiently close to line segments at the chosen display scale. Evaluating $C(t)$ and splitting it are thus two results of the same short algorithm.

For a proof outline, expanding the recursion givesPascal's identity proves this by induction; at $r=n$, $i=0$ it is precisely the Bernstein basis formula for the Bézier curve. The geometric construction remains useful even when expanding the polynomial by hand would be cumbersome.

// Target: analysis.bigb

For an isogeny of elliptic curves $\varphi :E_1\to E_2$, write $x_2\circ \varphi =A(x_1)/B(x_1)$ with coprime polynomials $A,B$. Each $x_i:E_i\to {\mathbb{P}}^1$ is a two-to-one finite morphism, and function field degrees multiply in a tower. Consequently $2\mathrm{deg}\varphi =2\max (\mathrm{deg}A,\mathrm{deg}B)$. This computes the total degree, including an inseparable factor. The zero homomorphism is assigned degree zero separately.

The degree of a nonzero polynomial is the largest exponent of a monomial with a nonzero coefficient. Thus $X^3-3X^2+4$ has degree three, while any nonzero constant polynomial has degree zero. The zero polynomial is treated separately in degree statements.

Over an integral domain, polynomial degrees add under multiplication: $\mathrm{deg}(pq)=\mathrm{deg}p+\mathrm{deg}q$ for nonzero factors, because the product of their leading coefficients cannot vanish. Under addition, cancellation may lower the degree: $(X^2+1)+(-X^2+X)=X+1$. This explains why degree is reliable for counting factors and bounding polynomial roots, but cannot always be inferred from the degrees of summands alone.

Deleting an edge removes one colour constraint. Edge contraction describes exactly the newly admitted colourings that violate that constraint. If $e$ joins distinct vertices $u,v$ of a simple graph $G$, write $G-e$ for the graph with that edge deleted and $G/e$ for the graph with $u,v$ identified and $e$ removed. Parallel edges produced by the contraction may be reduced to one edge, since they impose the same condition on a proper vertex coloring. ThenTo see why, divide the proper vertex colorings of $G-e$ into two classes. Those with different colours at $u,v$ are exactly the colourings of $G$. Those with equal colours at $u,v$ correspond one-to-one to colourings of $G/e$: the merged vertex receives their common colour. Subtracting the second class leaves the first.

For the triangle, delete $AC$ to get a three-vertex path graph, or contract $AC$ to get two vertices joined by an edge after removing the redundant parallel edge. Consequently,At $t=3$, this says $6=12-6$. Of the twelve path colourings pictured above, six give $A,C$ the same colour; these are precisely the ones ruled out when we restore $AC$.

This recurrence is useful because it reduces a constrained counting problem to smaller graphs. A graph on $n$ vertices with no edges has $P_G(t)=t^n$, since every vertex chooses its colour independently. Repeated edge deletion and edge contraction reduce to this case and show that the counting function is a polynomial. The same recurrence gives its leading coefficients. The method is conceptually simple, although the number of recursive cases can still grow rapidly.

For an invertible matrix $A$ and any matrix $H$,Indeed $\det (A+tH)=\det (A)\det (I+t{A}^{-1}H)$, and multilinearity shows that the linear coefficient of $\det (I+tB)$ is $\mathrm{tr}B$. At a singular matrix the derivative is instead written $\mathrm{tr}(\mathrm{adj}(A)H)$; this polynomial formula holds for all $A$.

Suppose a degree-$N$ polynomial obeys $|{\partial }^{\alpha }Q(\xi )|\le C_{\alpha }|\xi {|}^{-\delta |\alpha |}|Q(\xi )|$ at large real frequency, with $\delta >0$. A nonzero constant derivative of total order $N$ gives $|Q|\gtrsim ⟨\xi {⟩}^{\delta N}$. Reciprocal differentiation then bounds ${\partial }^{\alpha }(1/Q)$ by $⟨\xi {⟩}^{-\delta N-\delta |\alpha |}$. The high-frequency reciprocal parametrix kernel therefore gains $\delta N$ Sobolev derivatives. Localize the distribution by a cutoff function equal to one near the target; the commutator lies outside that target and the reciprocal kernel makes its contribution smooth. This proves the displayed gain without demanding global regularity of the original distribution.

Let $V(p)$ be the number of sign changes in the list of nonzero coefficients of a real polynomial, arranged by increasing or decreasing degree. The positive roots of a polynomial, counted with multiplicity of a root, satisfyThe bound follows by induction on the nonzero terms after removing the lowest power of $x$. Differentiation removes the constant coefficient. If it and the next coefficient have opposite signs, the Rolle root count with multiplicities loses at most one root and the sign-change count loses one. If their signs agree, the derivative must have an extra root before the first positive root, so neither count needs that extra allowance. The parity statement follows by comparing the signs near zero and at positive infinity: roots of odd multiplicity reverse the sign, and roots of even multiplicity preserve it. Apply the same rule to $p(-x)$ to bound negative roots.

The monomial $x^i{y}^j$ is an eigenvector of the differential linear operator $x^2{\partial }_x^2+y^2{\partial }_y^2$, with eigenvalue $i(i-1)+j(j-1)$. Over the real or complex numbers this vanishes exactly for $i,j\in \{0,1\}$. On polynomials of bounded total degree, the kernel is therefore spanned by the permitted members of $1,x,y,xy$, while the image of a linear map is spanned by all remaining monomials. The trace is the sum of those diagonal eigenvalues.

The vector space of formal polynomials over a field $k$ in $n$ variables with total degree of a polynomial at most $d$ has the monomial basis $x_1^{a_1}\cdots x_n^{a_n}$ for nonnegative exponents with ${\sum }_i{a}_i\le d$. Introducing a slack exponent and using stars and bars counts $\left(\genfrac{}{}{0ex}{}{n+d}{n}\right)$ basis elements. These are formal polynomials; over a finite field, distinct formal polynomials need not define distinct polynomial functions unless suitable degree restrictions hold.

For sequences $a,b$ indexed by the integers, discrete convolution adds products whose indices sum to the output index:For finite sequences, treat every unspecified entry as zero; the sum then has finitely many nonzero terms. For infinite sequences, a convergence hypothesis is needed, for example absolute summability of both sequences.

Take $a=(1,2,1)$ at indices $0,1,2$ and $b=(1,1)$ at indices $0,1$. Their convolution is $(1,3,3,1)$ at indices $0,1,2,3$:Equivalently, $c_n=a_n+a_{n-1}$: add an unshifted and a one-step shifted copy of $a$. This explains the output by superposition, with every index included.

The same products appear in polynomial multiplication and generating functions:A term $a_m{z}^m$ times $b_k{z}^k$ contributes to coefficient $m+k$, precisely the index condition in discrete convolution. In probability, $a$ counts the outcomes of two independent fair coin tosses by their number of heads, while $b$ counts the outcomes of one more toss. Dividing the output by eight gives the probability mass function $(1,3,3,1)/8$ of the binomial distribution with $n=3$ and $p=1/2$. No new arithmetic is needed to pass from coefficient multiplication to counting independent sums.

The full finite convolution is different from circular convolution, where indices wrap modulo a chosen length. Wrapping this four-term output to length three adds its last entry to its first and gives $(2,3,3)$. When using a discrete Fourier transform to compute the full convolution, zero padding to length at least the sum of the input lengths minus one prevents this wraparound. ^[ref]

For a nonzero constant-coefficient polynomial differential operator in one variable, every homogeneous distribution solution is a classical solution, indeed analytic. Factor the polynomial into powers of distinct characteristic roots of a constant-coefficient differential equation. The kernel decomposition for coprime polynomials reduces the equation to $(D-\lambda {)}^{m}v=0$, and multiplication of a distribution by a smooth function reduces this to $D^m(e^{-\lambda x}v)=0$. The fact that a distribution with zero derivative is constant, iterated, makes the latter distribution a polynomial of degree less than $m$.

For a basis $e_1,\dots ,e_n$ of a finite-dimensional vector space $V$, its dual basis consists of linear functionals $e_1^{\ast },\dots ,e_n^{\ast }$ satisfyingThe Kronecker delta means that each rule gives one on its own basis vector and zero on the others. These rules extract coordinates:Applying $e_i^{\ast }$ to a linear combination removes every term except the $i$th. Consequently the dual basis is a basis of the dual space, and every linear functional has expansion $\phi ={\sum }_i\phi (e_i)e_i^{\ast }$. ^[ref].

Coordinate extraction works for a nonorthogonal basis too. With $b_1=(1,0)$ and $b_2=(1,1)$, the dual basis isThe checks $b_1^{\ast }(b_1)=1$, $b_1^{\ast }(b_2)=0$, $b_2^{\ast }(b_1)=0$, and $b_2^{\ast }(b_2)=1$ fix the two rules. For $v=(3,1)$ they give coordinates $(2,1)$, so $v=2b_1+b_2$. The usual dot products with $b_1,b_2$ instead give $3,4$ and do not extract these coordinates: simple projection formulas depend on an orthonormal basis.

The same idea explains polynomial interpolation. On polynomials of polynomial degree at most $n$, evaluation at distinct interpolation nodes $x_0,\dots ,x_n$ gives the dual basis to the Lagrange cardinal polynomials, because ${\ell }_i(x_j)={\delta }_{ij}$. Samples are coordinates in this basis, so the reconstruction $p={\sum }_{i}p(x_i){\ell }_i$ is a dual-basis expansion. Differentiation weights from nodal evaluations and factorial monomial basis for polynomial differentiation apply the same principle to different measurements.