Derivative of Generating Function/General Result

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Theorem

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

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

Proof by induction:


Basis for the Induction

When $n = 0$, we have:

$\dfrac {\d^0} {\d z^0} \map G z = \map G z$

Also:

$\ds \sum_{k \mathop \ge 0} \dfrac {\paren {k + 0}!} {k!} a_{k + 0} z^k = \sum_{k \mathop \ge 0} a_k z^k$

This is the basis for the induction.


Induction Hypothesis

$\ds \forall n \in \N: n \ge 0: \dfrac {\d^n} {\d z^n} \map G z = \sum_{k \mathop \ge 0} \dfrac {\paren {k + n}!} {k!} a_{k + n} z^k$

This is the induction hypothesis.

It is to be demonstrated that:

$\ds \dfrac {\d^{n + 1} } {\d z^{n + 1} } \map G z = \sum_{k \mathop \ge 0} \dfrac {\paren {k + n + 1}!} {k!} a_{k + n + 1} z^k$


Induction Step

This is the induction step:

Consider $m = n + 1$.

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


Hence the result by induction.

$\blacksquare$