# Earth Cubed

## Laplace Transform Via Limits

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

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

https://earthcubed.wordpress.com/2009/08/30/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:

I rethought a suggested solution and decided that I was perhaps partly incorrect. We can derive the integral from sums & limits bit it converges in a very strange way. The sum converges exactly where the geometric series diverges. All that is left prior to cancellation is a ratio of infinitesimally small quantities.

The summation suggested by 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.

http://en.wikipedia.org/wiki/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. In the above equation if we are trying to integrate an exponentail function of the form

(2) – $f(x)=exp(- \alpha t)$

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

combining terms

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

Expressing as a geometric series:

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

Using the geometric series:

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

Now here is where the Voodoo comes in 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:$latex e^{-(s+alpha)m/n}$ as $1 - (s+alpha) (m/n)$

and as m approach infinity $\left( e^{-(s+alpha)m/n \right}^n$ 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) – $L{f(x)}=lim_{m->oo}lim_{n->oo} \frac{1}{(s+alpha)m/n)} x m/n$

cancelling m/n from the numerator and denominator (

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

and now the limits are irrelivant:

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