Physlib.SpaceAndTime.Space.Derivatives.DerivativeIndex

For natural numbers $d$ and $k$, this set consists of all multi-indices $I$ of dimension $d$ whose total order $|I|$ is at most $k$. Given a multi-index $I = (I_0, I_1, \dots, I_{d-1})$, its order is defined as the sum of its components: $$|I| = \sum_{i=0}^{d-1} I_i$$ The definition captures the collection $\{ I \in \mathbb{N}^d \mid |I| \le k \}$.

For natural numbers $d$ and $k$, the set of multi-indices $I = (I_0, I_1, \dots, I_{d-1})$ of dimension $d$ whose total order $|I| = \sum_{i=0}^{d-1} I_i$ is at most $k$ is a finite set.

For any dimension $d \in \mathbb{N}$ and maximum order $k \in \mathbb{N}$, let $I$ and $J$ be multi-indices in the set $\{ I \in \mathbb{N}^d \mid |I| \le k \}$. It is decidable whether $I = J$, meaning there exists an algorithm to determine the truth value of the equality between these two bounded multi-indices.

For natural numbers $d$ and $k$, the zero element of the set of bounded derivative indices $\text{DerivativeIndex}(d, k) = \{ I \in \mathbb{N}^d \mid |I| \le k \}$ is defined as the zero multi-index $0 = (0, \dots, 0)$. This definition is valid because the total order of the zero multi-index is $|0| = 0$, which satisfies $|0| \le k$ for any $k \in \mathbb{N}$.

For any natural numbers $d$ and $k$, the zero element of the set of bounded derivative indices $\text{DerivativeIndex}(d, k) = \{ I \in \mathbb{N}^d \mid |I| \le k \}$ is equal to the zero multi-index $0$ in the space of all $d$-dimensional multi-indices.

For any natural numbers $d$ and $k$, let $\text{DerivativeIndex}(d, k) = \{ I \in \mathbb{N}^d \mid |I| \le k \}$ be the set of multi-indices with total order at most $k$. For any index $i \in \{0, \dots, d-1\}$, the $i$-th component of the zero multi-index $0 \in \text{DerivativeIndex}(d, k)$ is $0$.