Derivative of Generating Function

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Theorem

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

Then:

\(\ds \frac \d {\d z} \map G z\) \(=\) \(\ds \sum_{k \mathop \ge 0} \left({k + 1}\right) a_{k + 1} z^k\)
\(\ds \) \(=\) \(\ds a_1 + 2 a_2 z + 3 a_3 z^3 + \cdots\)


General Result

Let $m$ be a positive integer.

Then:

$\ds \dfrac {\d^m} {\d z^m} \map G z = \sum_{k \mathop \ge 0} \dfrac {\paren {k + m}!} {k!} a_{k + m} z^k$


Corollary

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

\(\ds \frac \d {\d z} \map G z\) \(=\) \(\ds \frac \d {\d z} \paren {\sum_{k \mathop \ge 0} a_k z^k}\) Definition of Generating Function
\(\ds \) \(=\) \(\ds \sum_{k \mathop \ge 0} \paren {\frac \d {\d z} a_k z^k}\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop \ge 0} k a_k z^{k - 1}\) Power Rule for Derivatives
\(\ds \) \(=\) \(\ds \sum_{k \mathop \ge 1} k a_k z^{k - 1}\) as the zeroth term vanishes when $k = 0$
\(\ds \) \(=\) \(\ds \sum_{k \mathop \ge 0} \paren {k + 1} a_{k + 1} z^k\) Translation of Index Variable of Summation

$\blacksquare$


Sources

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