Faà di Bruno's Formula/Proof 4

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Theorem

Let $D_x^k u$ denote the $k$th derivative of a function $u$ with respect to $x$.

Then:

$\displaystyle D_x^n w = \sum_{j \mathop = 0}^n D_u^j w \sum_{\substack {\sum_{p \mathop \ge 1} k_p \mathop = j \\ \sum_{p \mathop \ge 1} p k_p \mathop = n \\ \forall p \mathop \ge 1: k_p \mathop \ge 0} } n! \prod_{m \mathop = 1}^n \dfrac {\left({D_x^m u}\right)^{k_m} } {k_m! \left({m!}\right)^{k_m} }$


Proof

$D_x^n$ can be expressed as a determinant:

$D_x^n = \begin{vmatrix} \dbinom {n - 1} 0 u_1 & \dbinom {n - 1} 1 u_2 & \dbinom {n - 1} 2 u_3 & \cdots & \dbinom {n - 1} {n - 2} u_{n - 1} & \dbinom {n - 1} {n - 1} u_n \\ -1 & \dbinom {n - 2} 0 u_1 & \dbinom {n - 2} 1 u_2 & \cdots & \dbinom {n - 2} {n - 3} u_{n - 2} & \dbinom {n - 2} {n - 2} u_{n - 1} \\ 0 & -1 & \dbinom {n - 3} 0 u_1 & \cdots & \dbinom {n - 3} {n - 4} u_{n - 3} & \dbinom {n - 3} {n - 3} u_{n - 2} \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & -1 & \dbinom 0 0 u_1 \end{vmatrix}$

where $u_j := \left({D_x^j u}\right) D_u$.

Both sides of this equation are differential operators which are to be applied to $w$.



Source of Name

This entry was named for Francesco Faà di Bruno.


Sources