Discrete to Continuous Representations of Functions via Laplace Transforms?

The Laplace transform can be thought of as the continuous analogue of a power series, as in this video.

From this perspective, think of the function $ a : \mathbb{N} \rightarrow \mathbb{R}$ as a function giving the coefficients of the taylor series of a function $f$ in $ f(x) = \sum _{i=0}^\infty a(n)x^n$. In other words the ‘sequence’ $a$ is a discrete representation of the function $f$. What is the continuous analogue? Can we use this to find a continuous representation for the function $f$? We basically want:

$$ f(x) = \sum _{i=0}^\infty a(n)x^n \rightarrow f(x) = \smallint_0^\infty a(t) x^t dt$$

Then applying ‘cosmetic’ adjustments like $ x = e^{\ln(x)}$, $ x^t = e^{t\ln(x)}$ & $ \ln(x) = – s$ we get

$$ f(x) = \sum _{i=0}^\infty a(n)x^n \rightarrow f(x) = \smallint_0^\infty a(t) e^{-st} dt = F(s)$$

So every time you apply the Laplace transform to a function, you can interpret it as saying that we are saying that this function, restricted to lying over $\mathbb{N}$, represents the Taylor series of some other function, & the continuous representation we’ve been given is merely a tool we’re using in finding a continuous representation of that other function via the Laplace transform.

Is there any general relationship between $a(n)$ & $a(t)$? The fact that Laplace transforms can be taken for functions without Taylor expansions would seem to indicate there isn’t, however maybe there is a relationship of Laplace transforms to general sequences?

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