Power series and Laplace transform

1. Jul 12, 2014

ricard.py

As it can be read here, http://en.wikipedia.org/wiki/Laplace_transform#Relation_to_power_series
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, computing a discrete power series or a continuous laplace transform should converge to the same function, is it right?

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.

What I am doing wrong?

2. Jul 12, 2014

pasmith

"Replace the discrete parameter n by the continuous paramater t" is not that straightforward.

Doing that literally would require that
$$\int_0^\infty f(t)e^{-st}\,dt = \sum_{n=0}^\infty f(n)e^{-ns}$$ which, as your example shows, is in general false.
What is true is that if you define $f(t) = \sum_{n=0}^\infty a_n\delta(t - n)$ where $\delta$ is the Dirac delta distribution then
$$\int_{0^{-}}^\infty f(t)e^{-st}\,dt = \sum_{n=0}^\infty a_n e^{-ns}.$$ Observe that here $f(t) = 0$ for all non-integer $t$ and that $f(t)$ is not technically defined for integer $t$; it is not the case that $f(n) = a_n$.

3. Jul 12, 2014

HallsofIvy

In general, your statement "computing a discrete power series or a continuous laplace transform should converge to the same function" is incorrect. Interchanging two limit process does NOT always give the same result.