Laplace Transform of Generating Function of Sequence

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Theorem

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

Let $G \left({z}\right)$ be the generating function for $\left\langle{a_n}\right\rangle$.


Let $f \left({x}\right)$ be the step function:

$f \left({x}\right) = \displaystyle \sum_{k \mathop \in \Z} a_k \left[{0 \le k \le x}\right]$

where $\left[{0 \le k \le x}\right]$ is Iverson's convention.


Then the Laplace transform of $f \left({x}\right)$ is given by:

$\mathcal L \left\{ {f \left({s}\right)}\right\} = \dfrac {G \left({e^{-s} }\right)} s$


Proof

We note that:

\((1):\quad\) \(\displaystyle L \left\{ {f \left({s}\right)}\right\}\) \(=\) \(\displaystyle \int_0^\infty e^{-s t} f \left({t}\right) \rd t\) Definition of Laplace Transform

and:

\((2):\quad\) \(\displaystyle G \left({z}\right)\) \(=\) \(\displaystyle \sum_{n \mathop \ge 0} a_n z^n\) Definition of Generating Function


Then:

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

$\blacksquare$


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