Faà di Bruno's Formula/Example/1/Proof

Theorem
Consider Faà di Bruno's Formula:

When $n = 1$ we have:

Proof
In the summation:
 * $\displaystyle \sum_{j \mathop = 0}^1 D_u^j w \sum_{\substack {\sum_{p \mathop \ge 1} k_p \mathop = j \\ \sum_{p \mathop \ge 1} p k_p \mathop = 1 \\ \forall p \ge 1: k_p \mathop \ge 0} } 1! \prod_{m \mathop = 1}^1 \dfrac {\paren {D_x^m u}^{k_m} } {k_m! \paren {m!}^{k_m} }$

we need to consider $j = 0$ and $j = 1$.

Let $j = 0$.

Consider the set of $k_p$ such that:
 * $k_1 + k_2 + \cdots = 0$
 * $1 \times k_1 + 2 k_2 + \cdots = 1$
 * $k_1, k_2, \ldots \ge 0$

It is apparent by inspection that no set of $k_p$ can fulfil these conditions.

Therefore when $j = 0$ the summation is vacuous

Let $j = 1$.

Consider the set of $k_p$ such that:


 * $k_1 + k_2 + \cdots = 1$
 * $1 \times k_1 + 2 k_2 + \cdots = 1$
 * $k_1, k_2, \ldots \ge 0$

By inspection, it is seen that these can be satisfied only by:
 * $k_1 = 1$

and all other $k_p = 0$.

When $k_m = 0$:
 * $\dfrac {\paren {D_x^m u}^{k_m} } {k_m! \paren {m!}^{k_m} } = 1$

by definition of zeroth derivative and factorial of $0$.

Thus any contribution to the summation where $k_m = 0$ can be disregarded.

Thus we have: