Faà di Bruno's Formula/Proof 3

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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

We have that:

$\dfrac {D_x^k u} {k!}$ is the coefficient of $z^k$ in $\map u {x + z}$
$\dfrac {D_u^j w} {j!}$ is the coefficient of $y^j$ in $\map w {u + y}$.

Hence the coefficient of $z^n$ in $\map w {\map u {x + z} }$ is:

$\dfrac {D_x^n w} {n!} = \ds \sum_{\substack {\sum_{p \mathop \ge 1} k_p \mathop = j \\ \sum_{p \mathop \ge 1} p k_p \mathop = n \\ \forall p \mathop \ge 1: k_p \mathop \ge 0} } \dfrac {j!} {k_1! \, k_2! \cdots k_n!} \paren {\dfrac {D_x^1 u} {1!} }^{k_1} \paren {\dfrac {D_x^2 u} {2!} }^{k_2} \cdots \paren {\dfrac {D_x^n u} {n!} }^{k_n}$


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Source of Name

This entry was named for Francesco Faà di Bruno.


Sources

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