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:

$\displaystyle D_x^n w = \sum_{j \mathop = 0}^n D_u^j w \sum_{\substack {\sum_{p \mathop \ge 1} k_p \mathop = j \\ \sum_{p \mathop \ge 1} p k_p \mathop = n \\ \forall p \mathop \ge 1: k_p \mathop \ge 0} } n! \prod_{m \mathop = 1}^n \dfrac {\left({D_x^m u}\right)^{k_m} } {k_m! \left({m!}\right)^{k_m} }$


Proof

We have that:

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

Hence the coefficient of $z^n$ in $w \left({u \left({x + z}\right)}\right)$ is:

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



Source of Name

This entry was named for Francesco Faà di Bruno.


Sources