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# Laplace Transform Via Limits

Posted Jul14-12 at 02:23 PM by John Creighto
Updated Jul14-12 at 06:24 PM by John Creighto

Quite a while back I had a brief interest in wheather the Laplace transform made numerical sense. I was interested in this for its own sake:

Laplace Transform Convergence
Laplace Transform (Numeric Computation?)

and because if we could numericaly understand the Laplace transform it might tell us more about a dynamic system then a fourier transform.

using the fft to calculate the laplace transform

Memory of my failures in obtaining the desired results led me to believe that the Laplace transform was not,"Riemann integratable".

However when chalanged on this in another forum:

andrewk for the laplace transform was:

(1) - $$F(s)= \int_0^{oo} f(t) dt = \lim_{m->oo} \lim_{n->oo} \sum_{k=1}^{n} f(km/n) \frac{m}{n}$$

It looks simple enough that I may have or should have considered it but it is not*entirely*obvious for how the double sum should work. Moreover, Wikipedia discusses some problems with trying to use*Riemann*integration with improper integrals.

Wikipedia:Riemann_integral#Generalizations

Needless to I am able to use a Riemann like result and get the right answer (Whether the approach is mathematically sound is an entirely different question).

To analyze the convergence of the above expression lets look at a particular example:

(2) - $$f(x)=exp(- \alpha t)$$

substituting this into (1) with t=km/n:

(3) - $$L\{f(x)\} = \lim_{m->oo} \lim_{n->oo} \sum_{k=1}^{n} e^{-skm/n} exp(- \alpha km/n) \frac{m}{n}$$

combining terms

(4) - $$L\{f(x)\} = \lim_{m->oo} \lim_{n->oo}\sum_{k-1}^n exp(-(s+ \alpha )km/n) \frac{m}{n}$$

Expressing as a geometric series:

(5) - $$L\{f(x)\}= \lim_{m->oo} \lim_{n->oo}\sum_{k-1}^n \left( e^{-(s+ \alpha )m/n} \right)^k \frac{m}{n}$$

Using the geometric series:

(6) - $$L\{f(x)\}= \lim_{m->oo} \lim_{n->oo} \frac{1- \left( exp(-(s+ \alpha )m/n) \right)^n}{1-exp(-(s+ \alpha )m/n)} \frac{m}{n}$$

Now here is where the Voodoo comes into play. We need to let m/n*approach*zero and m approach infinity. Rather them attempting to do this formally lets just cut to the chase.

When m/n approaches zero we can we can approximate: $$e^{-(s+ \alpha )m/n}$$ as $$1 - (s+ \alpha ) \frac{m}{n}$$

and as m approach infinity $$\left( exp(-(s+ \alpha)) \frac{m}{n} \right)^n$$ approaches zero.

Making these substitution into equation (6) we get:

(7) - $$L\{f(x)\}= \lim_{m->oo} \lim_{n->oo} \frac{1}{1-(1+(s+ \alpha )m/n)} \frac{m}{n}$$

which further simplifies to

(8) - $$L\{f(x)\}= \lim_{m->oo} \lim_{n->oo} \frac{m/n}{(s+ \alpha )m/n)}$$

cancelling m/n from the numerator and denominator one obtains:

(9) - $$L\{f(x)\}= \lim_{m->oo} \lim_{n->oo} \frac{1}{(s+ \alpha )}$$

and now the limits are irrelivant:

(10) - $$L\{f(x)\}= \frac{1}{(s+ \alpha )}$$

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A related writing by me:

Greens Functions, PDEs and Laplace Transforms
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1.  There's no need for voodoo. We just take the n limit first to get a function of m and then take the m limit. $L\{f(x)\}= \lim_{m->oo} \lim_{n->oo} \frac{1- \left( exp(-(s+ \alpha )m/n) \right)^n}{1-exp(-(s+ \alpha )m/n)} \frac{m}{n} = \lim_{m->oo} \lim_{n->oo} \frac{1- \left( exp(-(s+ \alpha )m) \right)}{1-1+(s+ \alpha )m/n+O(n^{-2})} \frac{m}{n}$ $= \lim_{m->oo} \lim_{n->oo}m \frac{1- \left( exp(-(s+ \alpha )m) \right)}{(s+ \alpha )m+O(n^{-1})} = \lim_{m->oo} \lim_{n->oo}\frac{1- \left( exp(-(s+ \alpha )m) \right)}{(s+ \alpha )+O(n^{-1})} = \lim_{m->oo} \frac{1- \left( exp(-(s+ \alpha )m) \right)}{(s+ \alpha )} = \frac{1}{(s+ \alpha )}$ Provided that $(s+ \alpha )>$, which is exactly the same condition as is needed for the proper integral to have a finite value. Posted Jul18-12 at 11:53 PM by andrewkirk