Laplace transform and Power series

Question

As it can be read here, Discrete to Continuous Representations of Functions via Laplace Transforms? the Laplace transform is a continuous analog of a power series in which the discrete parameter n is replaced by the continuous parameter t, and x is replaced by exp(-s).

Therefore, I ask the following: computing a discrete power series or a continuous laplace transform should converge to the same function?

Let's apply it for the simplest case: a(x)=1

• For the discrete power series it converges to 1/1-x (provided that -1 < x < 1)
• For the continuous power series it converges to 1/s (provided that s>0)

Now, this two should be equivalent right? If you substitute s=-ln(x) you get -1/ln(x), which is not the same as 1/1-x.

Why it is not equal?

Answer 1

Give you more a counterexample:

$\mathcal{L}_{x\to s}\biggl\{\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{2n}}{4^n(n!)^2}\biggr\}=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!}{4^n(n!)^2s^{2n+1}}$

The former is suitable for $x\in\mathbb{C}$ but the latter is only suitable for $|s|\geq1$ .

Answer 2

Note that Laplace transform is NOT an interpolation of the discrete case and hence it should NOT give the same result obtained from power series.

The discrete power series $$A(x)=\sum_{n=0}^\infty a(n)x^n$$ is rewritten as $$A(x)=\sum_{n=0}^\infty a(n)x^n \Delta n$$ where $\Delta n = 1$ in this case.

So the question is, does the limit

$$\lim_{q\to \infty}\lim_{r\to \infty}\sum_{n=0}^{r q}x^\frac{n}{q}\frac{1}{q}$$ converge to the expected result of Laplace transform of $\frac{1}{s} = \frac{-1}{ln(x)}$ or not?

I wrote a Mathematica program to answer this question and here are the results:

For $ r=10$ and $q=1$

For $ r=100$ and $q=1$

For $ r=100$ and $q=10$

As you can see, it converges nicely.