Faà di Bruno's Formula/Proof 1

Proof
The proof proceeds by induction on $n$.

For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition:
 * $\displaystyle D_x^n w = \sum_{j \mathop = 0}^n D_u^j w \sum_{\substack {k_1 \mathop + k_2 \mathop + \mathop \cdots \mathop + k_n \mathop = j \\ k_1 \mathop + 2 k_2 \mathop + \mathop \cdots \mathop + n k_n \mathop = n \\ k_1, k_2, \ldots, k_n \mathop \ge 0} } n! \prod_{m \mathop = 0}^n \dfrac {\left({D_x^m u}\right)^{k_m} } {k_m! \left({m!}\right)^{k_m} }$

Note that when $k_m = 0$:
 * $\dfrac {\left({D_x^m u}\right)^{k_m} } {k_m! \left({m!}\right)^{k_m} } = 1$

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

Let $c \left({n, j, k_1, k_2, \ldots}\right)$ be the coefficient of $\left({D_x^n u}\right)^{k_n}$.

$P \left({1}\right)$ is the case:

Thus $P \left({1}\right)$ is seen to hold.

Basis for the Induction
$P \left({2}\right)$ is the case:

For $j = 0$ we have:


 * $k_1 + k_2 = 0$
 * $k_1 + 2 k_2 = 2$
 * $k_1, k_2 \ge 0$

By inspection, there are no $k_1$ and $k_2$ which satisfy these three equations.

Therefore when $j = 0$ the summation is vacuous

For $j = 1$ we have:


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

By inspection, it is seen that these can be satisfied only by:
 * $k_1 = 0, k_2 = 1$

For $j = 2$ we have:


 * $k_1 + k_2 = 2$
 * $k_1 + 2 k_2 = 2$
 * $k_1, k_2 \ge 0$

By inspection, it is seen that these can be satisfied only by:
 * $k_1 = 2, k_2 = 0$

Thus:

This is the basis for the induction.

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

So this is the induction hypothesis:
 * $\displaystyle D_x^r w = \sum_{j \mathop = 0}^r D_u^j w \sum_{\substack {k_1 \mathop + k_2 \mathop + \mathop \cdots \mathop + k_r \mathop = j \\ k_1 \mathop + 2 k_2 \mathop + \mathop \cdots \mathop + r k_r \mathop = r \\ k_1, k_2, \ldots, k_r \mathop \ge 0} } r! \prod_{m \mathop = 0}^r \dfrac {\left({D_x^m u}\right)^{k_m} } {k_m! \left({m!}\right)^{k_m} }$

from which it is to be shown that:
 * $\displaystyle D_x^{r + 1} w = \sum_{j \mathop = 0}^{r + 1} D_u^j w \sum_{\substack {k_1 \mathop + k_2 \mathop + \mathop \cdots \mathop + k_{r + 1} \mathop = j \\ k_1 \mathop + 2 k_2 \mathop + \mathop \cdots \mathop + \left({r + 1}\right) k_{r + 1} \mathop = r \mathop + 1 \\ k_1, k_2, \ldots, k_{r + 1} \mathop \ge 0} } \left({r + 1}\right)! \prod_{m \mathop = 0}^{r + 1} \dfrac {\left({D_x^m u}\right)^{k_m} } {k_m! \left({m!}\right)^{k_m} }$

Induction Step
This is the induction step:

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:
 * $\dfrac {r!} {k_1! \left({1!}\right)^{k_1} \cdots k_r! \left({r!}\right)^{k_r} }$

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

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

Note that while there are only finitely many $k$'s, it is convenient to consider an infinite sum because $k_{r + 1} = k_{r + 2} = \cdots = 0$.

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

Therefore:
 * $\displaystyle \forall n \in \Z_{\ge 0}: D_x^n w = \sum_{j \mathop = 0}^n D_u^j w \sum_{\substack {k_1 \mathop + k_2 \mathop + \mathop \cdots \mathop + k_n \mathop = j \\ k_1 \mathop + 2 k_2 \mathop + \mathop \cdots \mathop + n k_n \mathop = n \\ k_1, k_2, \ldots, k_n \mathop \ge 0} } n! \prod_{m \mathop = 0}^n \dfrac {\left({D_x^m u}\right)^{k_m} } {k_m! \left({m!}\right)^{k_m} }$