Series form of the Laplace transform


by dydxforsn
Tags: form, laplace, series, transform
dydxforsn
dydxforsn is offline
#1
Nov28-12, 05:26 PM
P: 111
I thought it would be obvious, but I can't find a series representation of the Laplace transform. I'm looking for something analogous to the Fourier series and how it can be used to derive the Fourier transform. I though it would simply be [tex]f(x) = \sum_{s=-\infty}^{\infty}{C_{s} e^{sx}}[/tex]
, but that was under the assumption that the Laplace transform was merely a decomposition of a function into an exponential basis (I just simply noticed the analogies to the Fourier decomposition of a function into its basis imaginary exponential functions.) However, I'm increasingly thinking the interpretation is wrong, there appears to be no mention in any literature of a "Laplace series" other than people sometimes referring to the series of spherical harmonics as the "Laplace series".

I notice that the inverse Laplace transform appears to be a contour integral, and having not taken complex analysis before, I'm further mystified by how I am to interpret the Laplace transform. I had long thought it was a just an exponential decomposition, but I can't justify such a claim because the inverse Laplace transform is not as simple as the inverse Fourier transform where it is completely obvious that the inverse Fourier transform linearly combines basis sine and cosine terms to produce the original function.

Is there a series by which the Laplace transform can be derived? What is the Laplace transform if not a method to represent a function in a particular basis of an infinite dimensional vector space? Maybe the basis representation it produces is something I don't have in mind? Why do conceptual analogies between the Laplace and Fourier transform break down?

Sorry if the subject of this post doesn't seem completely clear. I learned the Laplace transform in ODE before I learned the Fourier transform in PDE. The Fourier transform has a fairly clean infinite dimensional vector space basis interpretation and I've tried to go back and apply it to the Laplace transform thinking it would be an easy analogy, but I am suddenly surprised by the Laplace transform's complexities and am essentially asking for reasons and explanations behind the conceptual differences between Laplace and Fourier transforms coming from the angle of somebody who understands basic Fourier decomposition theory.

Thanks to anyone who tries to bear with my apparent nonsense.. heh
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jasonRF
jasonRF is offline
#2
Nov29-12, 08:28 AM
P: 653
I'm sure a mathematician can shed some light on this, but here is my take as an engineer.
There is absolutely nothing wrong with thinking of the Laplace transform as the projection of a function onto the space of complex exponentials, it is just that the space is continuous instead of discrete. Indeed, my professor talked about it that way to help build our intuitions. Likewise for the Fourier transform, of course - it represents a continuous spectrum while a Fourier series represents a discrete (countable) spectrum. In your infinite dimensional vector space framework that you like, differential operators can have a discrete spectrum (represented by a sum), a continuous spectrum (represented by an integral) or sometimes both (so you get an integral plus a sum). For pdes, continuous spectra are typical for unbounded domains , and discrete spectra are typical for bounded domains.

Note that Engineers (at least electrical engineers) usually think of Fourier transforms as a special case of Laplace transforms, since they usually allow [itex]s[/itex] to be complex but force the Fourier transform variable (usually [itex]\omega[/itex]) to be real. Roughly speaking, letting [itex]s=i\omega[/itex] turns the Laplace transform into a Fourier transform. So conceptually I usually think of them as the same thing; indeed, when using Fourier transforms to solve complicated PDEs, it is often convenient to allow [itex]\omega[/itex] to be complex so the difference between Fourier and Laplace simply becomes a 90 degree rotation in the complex plane.

By the way, so-called Z transform is usually considered the "series" version of the Laplace transform. Near the bottom of the wikipedia page (http://en.wikipedia.org/wiki/Z-transform) you can find a discussion of the connection.

Regarding complex analysis, for now don't worry about it. You only need it to invert the laplace transform; but you can use the fact that given a Laplace transform and the region of convergence the inverse Laplace transform is unique, so you can use tables of computed Laplace tranforms to identify inverses. I do recommend eventually taking complex analysis, though, as I have found it to be a lot of fun as well as a very useful tool (even for inverting the Fourier transform, as well as performing many other integrals). I took it my last semester as an undergrad - it was the first math I ever saw that was "beautiful" and it made me wish i had taken a lot more math classes!
best of luck,

jason
lurflurf
lurflurf is offline
#3
Nov29-12, 08:48 AM
HW Helper
P: 2,149
A Dirichlet series can be seen as a generalization of a power series

[tex]F(e^{-s})=\sum_{n=0}^\infty a_n e^{-\lambda_n s}[/tex]

the Laplace transform is the continuous version of this


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