PhyslibAlpha.Mathematics.PartialDerivativeTest

For any type $\alpha$ and elements $a, b, c \in \alpha$, updating the first component (index $0$) of the vector $(a, b)$ with the value $c$ results in the vector $(c, b)$.

Let $a, b$, and $c$ be elements of a type $\alpha$. Updating the second component (index 1) of the vector $(a, b)$ with the value $c$ results in the vector $(a, c)$.

Let $V$ be a real normed vector space and $f: V \to \mathbb{R}$ be a function. For a point $x_0 \in V$, the **Hessian bilinear companion** is the bilinear map $B: V \times V \to \mathbb{R}$ defined by $$B(u, v) = D^2 f(x_0)(u, v) + D^2 f(x_0)(v, u)$$ where $D^2 f(x_0)$ is the second Fréchet derivative of $f$ at $x_0$ viewed as a multilinear map on $V^2$.

Let $V$ be a real normed vector space. For a function $f: V \to \mathbb{R}$ and a point $x_0 \in V$, this definition constructs the quadratic map $Q: V \to \mathbb{R}$ such that for any vector $y \in V$, $Q(y) = D^2 f(x_0)(y, y)$, where $D^2 f(x_0)$ denotes the second iterated Fréchet derivative of $f$ at $x_0$.

Let $\mathbb{k}$ be a complete nontrivially normed field and $V$ be a finite-dimensional normed vector space over $\mathbb{k}$. This definition converts a continuous multilinear map $g : V \times V \to \mathbb{k}$ into its corresponding continuous bilinear map $B: V \to \mathcal{L}(V, \mathbb{k})$, defined such that for any $x, y \in V$, $B(x)(y) = g(x, y)$.

Let $V$ be a real vector space. This definition associates a quadratic map $Q : V \to \mathbb{R}$ with its corresponding multilinear map $B : V \times V \to \mathbb{R}$ (where the domain is represented as the space of functions from $\{0, 1\}$ to $V$). The resulting map is defined such that for any $v_0, v_1 \in V$, $B(v_0, v_1)$ is equal to the polar bilinear form of $Q$ evaluated at $v_0$ and $v_1$.

Let $V$ be a vector space over $\mathbb{R}$. For a quadratic map $Q: V \to \mathbb{R}$, this definition constructs the associated multilinear map $M: V \times V \to \mathbb{R}$ corresponding to half of the polar bilinear form of $Q$. Specifically, for any $v_0, v_1 \in V$, the map is defined as $M(v_0, v_1) = \frac{1}{2} (Q(v_0 + v_1) - Q(v_0) - Q(v_1))$.

Let $V$ be a finite-dimensional real normed vector space. For any quadratic map $Q : V \to \mathbb{R}$, its associated multilinear map $B : V \times V \to \mathbb{R}$ (the polar bilinear form) is continuous.

Let $V$ be a finite-dimensional real normed vector space and $Q: V \to \mathbb{R}$ be a quadratic map. Then the associated half-polar bilinear form $B: V \times V \to \mathbb{R}$, defined by $$B(v_0, v_1) = \frac{1}{2} (Q(v_0 + v_1) - Q(v_0) - Q(v_1)),$$ is continuous.

Let $V$ be a finite-dimensional real normed vector space. For any quadratic map $Q: V \to \mathbb{R}$, this definition constructs the corresponding continuous multilinear map $B: V \times V \to \mathbb{R}$ (represented as a map from the index set $\{0, 1\}$ to $V$). The resulting map is defined by the polar bilinear form of $Q$, such that for any $x, y \in V$, $B(x, y) = Q(x + y) - Q(x) - Q(y)$.

Let $V$ be a finite-dimensional real normed vector space and let $Q: V \to \mathbb{R}$ be a quadratic map. This definition constructs the continuous bilinear map $B: V \times V \to \mathbb{R}$ corresponding to one-half of the polar bilinear form associated with $Q$. For any vectors $v_0, v_1 \in V$, the map is defined as: $$B(v_0, v_1) = \frac{1}{2} (Q(v_0 + v_1) - Q(v_0) - Q(v_1))$$ The continuity of this map is guaranteed by the finite-dimensionality of $V$.

Let $V$ be a finite-dimensional real normed vector space and $Q : V \to \mathbb{R}$ be a quadratic map. For any $x, y \in V$, the continuous multilinear map associated with $Q$ (constructed via the polar identity) evaluated at the pair $(x, y)$ is equal to the polar bilinear form of $Q$, defined as $B(x, y) = Q(x + y) - Q(x) - Q(y)$.

Let $V$ be a finite-dimensional real normed vector space and $Q: V \to \mathbb{R}$ be a quadratic map. Let $B_{1/2} : V \times V \to \mathbb{R}$ be the continuous bilinear map constructed as one-half of the polar bilinear form associated with $Q$. For any vectors $x, y \in V$, the value of $B_{1/2}$ applied to the pair $(x, y)$ is equal to $\frac{1}{2} B(x, y)$, where $B$ is the polar bilinear form of $Q$.

Let $V$ be a finite-dimensional real normed vector space. Suppose $F: V \to \mathbb{R}$ is a positive definite quadratic map. Then the associated continuous bilinear form $B: V \times V \to \mathbb{R}$, defined as one-half of the polar form of $F$ (such that $B(v, v) = F(v)$ for all $v \in V$), is coercive. That is, there exists a constant $c > 0$ such that $B(v, v) \geq c \|v\|^2$ for all $v \in V$.

Let $V$ be a real normed vector space and $f : V \to \mathbb{R}$ be a function. For a point $x_0 \in V$, let $Q$ be the quadratic map associated with the second Fréchet derivative at $x_0$, defined by $Q(y) = D^2 f(x_0)(y, y)$. Then the polar bilinear form $B$ of this quadratic map satisfies $$B(x, y) = D^2 f(x_0)(x, y) + D^2 f(x_0)(y, x)$$ for all $x, y \in V$. This identity expresses the polar bilinear form of the second-derivative quadratic map as the symmetrized Hessian.

Let $V$ be a real normed space and $f: V \to \mathbb{R}$ be a function. If $f$ is $C^\infty$ (infinitely differentiable) at a point $x_0 \in V$, then the second iterated Fréchet derivative $D^2 f(x_0)$ is symmetric. That is, for any vectors $x, y \in V$, the following equality holds: $$D^2 f(x_0)(y, x) = D^2 f(x_0)(x, y)$$ This result is known as Schwarz's theorem (or Clairaut's theorem) regarding the symmetry of second derivatives.

Let $V$ be a finite-dimensional real normed vector space and $f: V \to \mathbb{R}$ be a function. Suppose $f$ is infinitely differentiable ($C^\infty$) at a point $x_0 \in V$. If the quadratic form associated with the second Fréchet derivative of $f$ at $x_0$, defined by $y \mapsto D^2 f(x_0)(y, y)$, is positive definite, then the bilinear map $D^2 f(x_0)$ is coercive. That is, there exists a constant $c > 0$ such that for all $y \in V$: $$D^2 f(x_0)(y, y) \geq c \|y\|^2$$

Let $V$ be a finite-dimensional real normed vector space and $f: V \to \mathbb{R}$ be a function. If the second Fréchet derivative of $f$ at $x_0 \in V$ is positive definite—meaning the quadratic map $Q(y) = D^2 f(x_0)(y, y)$ satisfies $Q(y) > 0$ for all nonzero $y \in V$—then the corresponding bilinear form $B(x, y) = D^2 f(x_0)(x, y)$ is coercive. (A bilinear form $B$ is coercive if there exists a constant $c > 0$ such that $B(y, y) \geq c \|y\|^2$ for all $y \in V$).

Let $V$ be a real inner product space and $f: V \to \mathbb{R}$ be a function. For any points $x_0, x \in V$ and a constant $C \in \mathbb{R}$, suppose the following conditions are satisfied: 1. The second Frechét derivative of $f$ at $x_0$ satisfies the inequality $C \|x - x_0\|^2 \leq D^2 f(x_0)(x - x_0, x - x_0)$. 2. The first Frechét derivative of $f$ at $x_0$ satisfies $Df(x_0)(x) = Df(x_0)(x_0)$ (which, by linearity, implies $Df(x_0)(x - x_0) = 0$). 3. The error of the second-order Taylor expansion is bounded by $$\left\| f(x) - \sum_{i=0}^2 \frac{1}{i!} D^i f(x_0)(x - x_0, \dots, x - x_0) \right\| \leq \frac{C}{2} \|x - x_0\|^2,$$ where $D^i f(x_0)$ denotes the $i$-th Frechét derivative of $f$ at $x_0$. Then, $f(x_0) \leq f(x)$.

Let $V$ be a finite-dimensional real inner product space and $f: V \to \mathbb{R}$ be a function that is $C^\infty$ at $x_0 \in V$. Suppose the following conditions hold: 1. The gradient of $f$ vanishes at $x_0$, i.e., $\nabla f(x_0) = 0$. 2. The function $f$ satisfies a second-order Taylor expansion condition at $x_0$: $$f(x) - \sum_{i=0}^2 \frac{1}{i!} D^i f(x_0)(x - x_0, \dots, x - x_0) = o(\|x - x_0\|^2)$$ as $x \to x_0$, where $D^i f(x_0)$ is the $i$-th Fréchet derivative of $f$ at $x_0$. 3. The second Fréchet derivative at $x_0$ is positive definite, i.e., the quadratic map $v \mapsto D^2 f(x_0)(v, v)$ is positive definite. Then $f$ has a local minimum at $x_0$.

Let $V$ be a normed real vector space and $f: V \to \mathbb{R}$ be a function. Suppose $f$ has a formal power series representation $p$ on a ball of radius $r$ centered at $x_0 \in V$. Then the difference between the function $f(x)$ and its partial sum of degree 2 (the quadratic approximation) is little-o of the square of the distance from $x_0$. That is, $$f(x) - \sum_{i=0}^2 p_i(x - x_0, \dots, x - x_0) = o(\|x - x_0\|^2)$$ as $x \to x_0$, where $p_i$ denotes the $i$-th term of the formal multilinear series $p$.

Let $V$ be a topological space that is a subsingleton (meaning it contains at most one element). For any function $f : V \to \mathbb{R}$ and any point $x_0 \in V$, $f$ has a local minimum at $x_0$.

Let $V$ be a finite-dimensional real inner product space and $f: V \to \mathbb{R}$ be a function. Suppose $x_0 \in V$ is a point such that the gradient of $f$ at $x_0$ vanishes, $\nabla f(x_0) = 0$. If $f$ has a formal power series representation on a ball of radius $r > 0$ centered at $x_0$, and the second Fréchet derivative at $x_0$ (represented by the quadratic map $y \mapsto D^2 f(x_0)(y, y)$) is positive definite, then $f$ has a local minimum at $x_0$.

Let $V$ be a finite-dimensional real inner product space. Suppose $f: V \to \mathbb{R}$ is a function that is analytic at a point $x_0 \in V$. If the gradient of $f$ vanishes at $x_0$ (i.e., $\nabla f(x_0) = 0$) and the second Fréchet derivative of $f$ at $x_0$ is positive definite as a quadratic map (meaning $D^2 f(x_0)(y, y) > 0$ for all non-zero $y \in V$), then $f$ has a local minimum at $x_0$.