Derivative of Generating Function/General Result/Corollary

Theorem

Let $\map G z$ be the generating function for the sequence $\sequence {a_n}$.

Let $m$ be a positive integer.

Let the coefficient of $z^n$ extracted from $\map G z$ be denoted:

$\sqbrk {z^n} \map G z := a_n$

Then:

$\sqbrk {z^m} \map G z = \dfrac 1 {m!} \map {G^{\paren m} } 0$

where $G^{\paren m}$ denotes the $m$th derivative of $G$.

Proof

\(\ds \dfrac {\d^m} {\d z^m} \map G z\)

\(=\)

\(\ds \sum_{k \mathop \ge 0} \dfrac {\paren {k + m}!} {k!} a_{k + m} z^k\)

Derivative of Generating Function: General Result

\(\ds \)

\(=\)

\(\ds \dfrac {m!} {0!} a_m + \sum_{k \mathop \ge 1} \dfrac {\paren {k + m}!} {k!} a_{k + m} z^k\)

\(\ds \)

\(=\)

\(\ds m! a_m + \sum_{k \mathop \ge 1} \dfrac {\paren {k + m}!} {k!} a_{k + m} z^k\)

Factorial of Zero

\(\ds \leadsto \ \ \)

\(\ds \map {G^{\paren m} } 0\)

\(=\)

\(\ds m! a_m\)

\(\ds \leadsto \ \ \)

\(\ds \dfrac 1 {m!} \map {G^{\paren m} } 0\)

\(=\)

\(\ds a_m\)

$\blacksquare$