Faà di Bruno's Formula/Proof 1

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Theorem

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

Then:

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


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:

$D_x^0 w = w$

which is consistent with the definition of the zeroth derivative.


The case for $n = 1$ is:

$D_x^1 w = D_u^1 w D_x^1 u$

which is consistent with Derivative of Composite Function.


Basis for the Induction

$\map P 2$ is the case:

$D_x^2 w = D_u^2 w \paren {D_x^1 u}^2 + D_u^1 w D_x^2 u$


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:

$\map {D_x} {\paren {D_x^m u}^{k_m} } = k_m \paren {D_x^m u}^{k_m - 1} D_x^{m + 1} u$


Lemma 2:

$\ds \map {D_x} {\prod_{m \mathop = 1}^r \paren {\dfrac {\paren {D_x^m u}^{k_m} } {k_m! \paren {m!}^{k_m} } } } = \prod_{m \mathop = 1}^r \paren {\dfrac {\paren {D_x^m u}^{k_m} } {k_m! \paren {m!}^{k_m} } } \sum_{m \mathop = 1}^r k_m \dfrac {D_x^{m + 1} u} {D_x^m u}$



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By differentiating with respect to $x$:

\(\ds \map c {r + 1, j, k_1, k_2, \ldots}\) \(=\) \(\ds \map c {r, j - 1, k_1 - 1, k_2, \ldots}\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \paren {k_1 + 1} \map c {r, j, k_1 + 1, k_2 - 1, k_3, \ldots}\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \paren {k_2 + 1} \map c {r, j, k_1, k_2 + 1, k_3 - 1, k_4, \ldots}\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \cdots\)

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 right hand side 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:

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


Source of Name

This entry was named for Francesco Faà di Bruno.


Sources

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