Faà di Bruno's Formula/Proof 1

Proof
The proof proceeds by induction on $n$.

For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
 * $\ds D_x^n w = \sum_{j \mathop = 0}^n D_u^j w \sum_{\substack {\sum_{p \mathop \ge 1} k_p \mathop = j \\ \sum_{p \mathop \ge 1} p k_p \mathop = n \\ \forall p \mathop \ge 1: k_p \mathop \ge 0} } n! \prod_{m \mathop = 1}^n \dfrac {\paren {D_x^m u}^{k_m} } {k_m! \paren {m!}^{k_m} }$

The case for $n = 0$ is:

which is consistent with the definition of the zeroth derivative.

The case for $n = 1$ is:

which is consistent with Derivative of Composite Function.

Basis for the Induction
$\map P 2$ is the case:

This is consistent with Derivative of Composite Function: Second Derivative.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that, if $\map P r$ is true, where $r \ge 2$, then it logically follows that $\map P {r + 1}$ is true.

So this is the induction hypothesis:
 * $\ds D_x^r w = \sum_{j \mathop = 0}^r D_u^j w \sum_{\substack {\sum_{p \mathop \ge 1} k_p \mathop = j \\ \sum_{p \mathop \ge 1} p k_p \mathop = r \\ \forall p \mathop \ge 1: k_p \mathop \ge 0} } r! \prod_{m \mathop = 1}^r \dfrac {\paren {D_x^m u}^{k_m} } {k_m! \paren {m!}^{k_m} }$

from which it is to be shown that:
 * $\ds D_x^{r + 1} w = \sum_{j \mathop = 0}^{r + 1} D_u^j w \sum_{\substack {\sum_{p \mathop \ge 1} k_p \mathop = j \\ \sum_{p \mathop \ge 1} p k_p \mathop = r \mathop + 1 \\ \forall p \mathop \ge 1: k_p \mathop \ge 0} } \paren {r + 1}! \prod_{m \mathop = 1}^{r + 1} \dfrac {\paren {D_x^m u}^{k_m} } {k_m! \paren {m!}^{k_m} }$

Induction Step
This is the induction step:

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

which shows that any contribution to the summation where $k_m = 0$ can be disregarded.

Let $j = 0$.

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

It is apparent by inspection that, for all $r > 0$, no set of $k_p$ can fulfil these conditions.

Therefore when $j = 0$ the summation is vacuous.

Also note that from:
 * $1 \times k_1 + 2 k_2 + \cdots = r$

it follows that:
 * $k_{r + 1} = k_{r + 2} = \cdots = 0$

Thus, while there are only finitely many $k$'s, their upper limit need not be explicitly considered.

Let $\map c {r, j, k_1, k_2, \ldots}$ be the coefficient of $D_u^j w$ in $D_x^r w$.

We establish some lemmata:

Lemma 1:

Lemma 2:

By differentiating $x$:

The equations:
 * $k_1 + k_2 + \cdots = j$

and:
 * $k_1 + 2 k_2 + \cdots = r$

are preserved by this induction step.

Thus it is possible to factor out:
 * $\dfrac {r!} {k_1! \paren {1!}^{k_1} \cdots k_r! \paren {r!}^{k_r} }$

from each term on the of the equation for $\map c {r + 1, j, k_1, k_2, \ldots}$.

Thus we are left with:
 * $k_1 + 2 k_2 + 3 k_3 + \cdots = r + 1$

So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\ds \forall n \in \Z_{\ge 0}: D_x^n w = \sum_{j \mathop = 0}^n D_u^j w \sum_{\substack {\sum_{p \mathop \ge 1} k_p \mathop = j \\ \sum_{p \mathop \ge 1} p k_p \mathop = n \\ \forall p \mathop \ge 1: k_p \mathop \ge 0} } n! \prod_{m \mathop = 1}^n \dfrac {\paren {D_x^m u}^{k_m} } {k_m! \paren {m!}^{k_m} }$