Vamos a suponer que estamos en dimensión $latex n$ y miraremos como quedan los diferentes kernels y sus derivadas. Denotaremos con $latex boldsymbol{q}$ a $latex (q_1,ldots,q_n)$ y $latex q = sqrt{sum_{i=1}^n q_i^2}$ será su módulo.
Para poder hablar de las derivadas parciales en espacios de esta dimensión vamos a introducir la notación multi-índice de Schwartz. Un multi-índice de dimensión $latex n$ es una $latex n$-tupla de enteros no negativos $latex alpha = (alpha_1,ldots,alpha_n)$ con $latex alpha_j in mathbb{Z}_+$ para $latex j= 1ldots n$, de manera que $latex alpha in mathbb{Z}_+^n$. Llamamos módulo de $latex alpha$ a:
$latex |alpha| = sum_{i=1}^n alpha_i$
Además, podemos definir
$latex alpha + beta := (alpha_1 + beta_1, ldots ,alpha_n + beta_n) $
y
$latex q^alpha:=Pi_{i=1}^n q_i^{alpha_i}$.
Finalmente, definimos el operador diferencial para $latex alpha in mathbb{Z}_+^n$ como:
$latex D^alpha:=frac{partial^{|alpha|}}{partial q_1^{alpha_1} partial q_2^{alpha_2} ldots partial q_n^{alpha_n}}$
kernel Gaussiano:
$latex W(boldsymbol{q},h) = frac{sigma}{h^n} e^{-(frac{q}{h})^2}$
$latex frac{partial}{partial q_j}W(boldsymbol{q},h) = D^{e_j}W(boldsymbol{q},h) = frac{sigma}{h^n}frac{-2}{h^2} e^{-(frac{q}{h})^2}q_j$
$latex frac{partial^2}{partial q_j partial q_k}W(boldsymbol{q},h)=D^{e_j+e_k}W= frac{sigma}{h^n}(frac{-2}{h^2})^2 e^{-(frac{q}{h})^2} q^{e_j+e_k}=frac{sigma}{h^n}(frac{-2}{h^2})^2 e^{-(frac{q}{h})^2}q_jq_k$
$latex D^alpha W(boldsymbol{q},h) = frac{sigma}{h^n}(frac{-2}{h^2})^{|alpha|} e^{-(frac{q}{h})^2}q^alpha$
kernel cúbico:
$latex W(boldsymbol{q},h) = frac{sigma}{h^n} begin{cases} frac{1}{4}(2-q)^3-(1-q)^3 mbox{ si } 0 leq q < 1 \ frac{1}{4}(2-q)^3 mbox{ si } 1 leq q < 2 \ 0 mbox{ si } q geq 2 end{cases}$
Recordemos que $latex q = sqrt{ sum_{i=1}^n q_i^2 }$, por lo que $latex frac{partial}{partial q_j}q = frac{2 q_j}{2 sqrt{sum_{i=1}^{n} q_i^2}} = frac{q_j}{q}$. Así pues:
$latex frac{partial}{partial q_j}W(boldsymbol{q},h) = frac{sigma}{h^n}(-1)frac{3}{q} q_j begin{cases} frac{1}{4}(2-q)^2-(1-q)^2 mbox{ si } 0 leq q < 1 \ frac{1}{4}(2-q)^2 mbox{ si } 1 leq q < 2 \ 0 mbox{ si } q geq 2 end{cases}$
$latex frac{partial^2}{partial q_j partial q_k}W(boldsymbol{q},h) = frac{sigma}{h^n}frac{3.2}{q^2} q_j q_k begin{cases} frac{1}{4}(2-q)-(1-q) mbox{ si } 0 leq q < 1 \ frac{1}{4}(2-q) mbox{ si } 1 leq q < 2 \ 0 mbox{ si } q geq 2 end{cases}$
$latex frac{partial^3}{partial q_j partial q_k partial q_l}W(boldsymbol{q},h) = frac{sigma}{h^n}(-1)frac{3.2.1}{q^3} q_j q_k q_l begin{cases} frac{1}{4}-1 mbox{ si } 0 leq q < 1 \ frac{1}{4} mbox{ si } 1 leq q < 2 \ 0 mbox{ si } q geq 2 end{cases}$
Por lo tanto, para $latex 0 leq |alpha| leq 3$ (si $latex |alpha|>3$ entonces $latex W equiv 0$) podemos escribir:
$latex D^alpha W(boldsymbol{q},h) = $
$latex =frac{sigma}{h^n}(-1)^{|alpha|}frac{frac{3!}{(3-|alpha|)!}}{q^{|alpha|}} q^alpha begin{cases} frac{1}{4}(2-q)^{3-|alpha|}-(1-q)^{3-|alpha|} mbox{ si } 0 leq q < 1 \ frac{1}{4}(2-q)^{3-|alpha|} mbox{ si } 1 leq q < 2 \ 0 mbox{ si } q geq 2 end{cases}$
kernel cuártico:
$latex W(boldsymbol{q},h) = frac{sigma}{h^n} begin{cases} (frac{5}{2}-q)^4-5(frac{3}{2}-q)^4+10(frac{1}{2}-q)^4 mbox{ si } 0 leq q < frac{1}{2} \ (frac{5}{2}-q)^4-5(frac{3}{2}-q)^4 mbox{ si } frac{1}{2} leq q < frac{3}{2} \ (frac{5}{2}-q)^4 mbox{ si } frac{3}{2} leq q < frac{5}{2} \ 0 mbox{ si } q geq frac{5}{2} end{cases}$
Por tanto, para $latex |alpha| = 0 ldots 4$:
$latex D^alpha W(boldsymbol{q},h) =$
$latex =frac{sigma}{h^n}(-1)^{|alpha|}frac{frac{4!}{(4-|alpha|)!}}{q^{|alpha|}} q^alpha begin{cases} (frac{5}{2}-q)^{4-|alpha|}-5(frac{3}{2}-q)^{4-|alpha|}+10(frac{1}{2}-q)^{4-|alpha|} mbox{ si } 0 leq q < frac{1}{2} \ (frac{5}{2}-q)^{4-|alpha|}-5(frac{3}{2}-q)^{4-|alpha|} mbox{ si } frac{1}{2} leq q < frac{3}{2} \ (frac{5}{2}-q)^{4-|alpha|} mbox{ si } frac{3}{2} leq q < frac{5}{2} \ 0 mbox{ si } q geq frac{5}{2} end{cases}$
kernel quíntico:
$latex W(boldsymbol{q},h) = frac{sigma}{h^n} begin{cases} (3-q)^5-6(2-q)^5+15(1-q)^5 mbox{ si } 0 leq q < 1 \ (3-q)^5-6(2-q)^5 mbox{ si } 1 leq q < 2 \ (3-q)^5 mbox{ si } 2 leq q < 3 \ 0 mbox{ si } q geq 3 end{cases}$
Por tanto, para $latex |alpha| = 0 ldots 5$:
$latex D^alpha W(boldsymbol{q},h) =$
$latex =frac{sigma}{h^n}(-1)^{|alpha|}frac{frac{5!}{(5-|alpha|)!}}{q^{|alpha|}} q^alpha begin{cases} (3-q)^{5-|alpha|}-6(2-q)^{5-|alpha|}+15(1-q)^{5-|alpha|} mbox{ si } 0 leq q < 1 \ (3-q)^{5-|alpha|}-6(2-q)^{5-|alpha|} mbox{ si } 1 leq q < 2 \ (3-q)^{5-|alpha|} mbox{ si } 2 leq q < 3 \ 0 mbox{ si } q geq 3 end{cases}$