Faà di Bruno's Formula/Example/0/Proof
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Theorem
Consider Faà di Bruno's Formula:
Let $D_x^k u$ denote the $k$th derivative of a function $u$ with respect to $x$.
Then:
- $\ds D_x^n w = \sum_{j \mathop = 0}^n D_u^j w \sum_{\substack {\sum_{p \mathop \ge 1} k_p \mathop = j \\ \sum_{p \mathop \ge 1} p k_p \mathop = n \\ \forall p \mathop \ge 1: k_p \mathop \ge 0} } n! \prod_{m \mathop = 1}^n \dfrac {\paren {D_x^m u}^{k_m} } {k_m! \paren {m!}^{k_m} }$
When $n = 0$ we have:
- $D_x^0 w = w$
Proof
In the summation:
- $\ds \sum_{j \mathop = 0}^0 D_u^j w \sum_{\substack {\sum_{p \mathop \ge 0} k_p \mathop = j \\ \sum_{p \mathop \ge 0} p k_p \mathop = 0 \\ \forall p \ge 0: k_p \mathop \ge 0} } 0! \prod_{m \mathop = 0}^1 \dfrac {\paren {D_x^m u}^{k_m} } {k_m! \paren {m!}^{k_m} }$
the only element appearing is for $j = 0$, and the continued product is vacuous.
Thus:
- $\ds \prod_{m \mathop = 0}^1 \dfrac {\paren {D_x^m u}^{k_m} } {k_m! \paren {m!}^{k_m} } = 1$
and we are left with:
\(\ds \sum_{j \mathop = 0}^0 D_u^j w\) | \(=\) | \(\ds D_u^0 w\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds w\) | Definition of Zeroth Derivative |
$\blacksquare$