I am somewhat new to using multi-index notation, and I am having trouble wrapping my head around how the chain rule would work. Suppose that we have $y = f(x)$, and that $g = g(y)$. Letting $D^\alpha = D_1 ^ {\alpha_1} \dots D_n^{\alpha_n}$, where $D_i^{\alpha_i} = \frac{\partial^{\alpha_i}}{\partial x_i ^{\alpha_i}}$, what would happen to
$$D^\alpha g(y)$$
This has given me quite a headache, and it is most likely due to the fact that I am not entirely comfortable with this notation yet. I have looked around for the multi-index chain rule, but I have not found anything on it. Any help is appreciated.
Thanks!
Best Answer
there are formulas for that. If $f$ and $g$ are both functions from $\mathbb{R}$ to $\mathbb{R}$, the $n$th derivative is given by the so-called Faa di Bruno formula. If $f$ and $g$ are both functions from $\mathbb{R}^d$ to $\mathbb{R}^d$ there are multivariate versions of this formula. Please have a look in the references:
L. E. Fraenkel, Formulae for high derivatives of composite functions, Math. Proc. Cam. Phil. Soc. 83 (1978), 159-165.
H. Gzyl, Multidimensional extension of Faa di Bruno's formula, J. Math. Anal. Appl. 116 (1986), 450-455.
G. M. Constantine and T. H. Savitis, A multivariate Faa di Bruno formula with applications, Trans. Amer. Math. Soc. 348 (1996), 503-520.
L. H. Encinas and J. M. Masque, A short proof of the generalized Faa di Bruno's formula, Appl. Math. Lett. 16 (2003), 975-979.
Tsoy-Wo Ma, Higher chain formula proved by combinatorics, The Electronic Journal of Combinatorics 16 (2009).
The formula you are looking for is difficult from the combinatorial point of view and not easy to handle. However, in the special case where the $i$th componen function of $f$ depends only on the $i$th coordinate, i.e. $f (x) = (f_1 (x_1) , \ldots , f_d (x_d))$, the formula gets much simpler.
Best wishes, Tauti