Laplace Transform of Generating Function of Sequence

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Theorem

Let $\sequence{a_n}$ be a sequence which has a generating function which is convergent.

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


Let $\map f x$ be the step function:

$\map f x = \ds \sum_{k \mathop \in \Z} a_k \sqbrk{0 \le k \le x}$

where $\sqbrk{0 \le k \le x}$ is Iverson's convention.


Then the Laplace transform of $\map f x$ is given by:

$\laptrans {\map f s} = \dfrac {\map G {e^{-s} }} s$


Proof

We note that:

\(\text {(1)}: \quad\) \(\ds \laptrans {\map f s}\) \(=\) \(\ds \int_0^\infty e^{-s t} \map f t \rd t\) Definition of Laplace Transform

and:

\(\text {(2)}: \quad\) \(\ds \map G z\) \(=\) \(\ds \sum_{n \mathop \ge 0} a_n z^n\) Definition of Generating Function


Then:

\(\ds \int_n^{n + 1} e^{-s t} \map f t \rd t\) \(=\) \(\ds \int_n^{n + 1} \paren{a_0 + a_1 + \cdots + a_n} e^{-s t} \rd t\) Definition of $f \left({x}\right)$
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^n a_k \sqbrk{\dfrac {-1} s e^{-s t} }_n^{n + 1}\) Primitive of Exponential Function
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^n \dfrac {a_k} s \paren{e^{-s n} - e^{-s \paren{n + 1} } }\)
\(\ds \leadsto \ \ \) \(\ds \laptrans {\map f s}\) \(=\) \(\ds \int_0^\infty e^{-s t} \map f t \rd t\) from $(1)$
\(\ds \) \(=\) \(\ds \sum_{n \mathop \ge 0} \sum_{k \mathop = 0}^n \dfrac {a_k} s \paren{e^{-s n} - e^{-s \paren{n + 1} } }\)
\(\ds \) \(=\) \(\ds \dfrac {a_0} s \paren{1 - e^{-s} } + \dfrac {a_0 + a_1} s \paren{e^{-s} - e^{-2 s} } + \dfrac {a_0 + a_1 + a_2} s \paren{e^{-2 s} - e^{-3 s} } + \cdots\)
\(\ds \) \(=\) \(\ds \dfrac 1 s \paren{a_0 + a_1 e^{-s} + a_2 e^{-2 s} + \cdots}\)
\(\ds \) \(=\) \(\ds \dfrac 1 s \sum_{n \mathop \ge 0} a_n e^{-s n}\)
\(\ds \) \(=\) \(\ds \dfrac 1 s \map G {e^{-s} }\) from $(2)$

$\blacksquare$


Sources

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