Re: Why exp(-st) in the Laplace Transform?

From: Han de Bruijn (Han.deBruijn_at_DTO.TUDelft.NL)
Date: 11/25/04 

Date: Thu, 25 Nov 2004 12:11:56 +0100

cdj wrote:

> heard something somewhat similar tho - that laplace transforms are the
> continuous generalization of power series. I can't recall exactly how
> the explanation went, it was in an ODE book by a guy named Simmons.

Perhaps you have seen something like this.

Let F(p) = integral exp(-px) f(x) dx (: Laplace Transform)
==> F(0) = integral f(x) dx = area beneath f(x)
And F'(p) = integral (-x) exp(-px) f(x) dx
==> F'(0) = - integral x f(x) dx = - first order moment
                                    / center of gravity / midpoint
And F''(p) = integral x^2 exp(-px) f(x) dx
==> F''(0) = integral x^2 f(x) dx = second order moment
                                   / moment of inertia / variance

And then you have F(p) = F(0) + p.F'(0) + p^2/2.F''(0) + ...

Interpretation: subsequent terms of the series expansion of the Laplace
Transform involve an area, a midpoint, a moment of intertia .. In short:
the most important (physical/global) characteristics of a function come
first, when considered in the Laplace domain.

Han de Bruijn

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