Origin of Laplace Transform (alec green)
In the first 15 minutes of this MIT lecture, Arthur Mattuck delivers a clear illustration of what the Laplace transform really is: a continuous analogy of the discrete power series (specifically, the Maclaurin Series).
Below I've merely summarized his explanation.
(1) Power series = discrete summation
We start with this power series:
$ A(x) = \sum_{n=0}^{\infty} a(n)x^{n} \ \mid \ a(n) \in \R \ \ \forall \ n \in \N $
In case you're not familiar with all the above notation, here's the explicit translation starting starting after the summation term, where each quoted term corresponds to each symbol: 'such that' a(n) 'is an element of' 'the set of real numbers' 'for all' n 'which is an element of' 'the set of natural numbers'.
Note that $ a(n) $ is a function here, and just defines the coefficient of each polynomial term in the power series, since a power series is $ = a(0) + a(1)x + a(2)x^{2} + ... + a(n)x^{n} + ... $. However, because the power series is a discrete summation, $ a(n) $ is only guaranteed to be defined if $ n $ is a natural number (non-negative integer), as indicated above. So for example, $ a(23) $ is defined, but not necessarily $ a(-2) $, $ a(.5) $, or $ a(10.001) $.
(2) Integral = continuous summation
Now we'll make the following conversions:
- from discrete function $ a $ to continuous function $ f $
- from discrete dependent variable $ n $ to continuous dependent variable $ t $
to arrive at:
$ F(x)=\int_{0}^{\infty} f(t)x^{t} \ dt \ \mid \ f(t) \in \R \ \ \forall \ t \in (0,\infty) $
The only difference now is that we sum the contributions of $ f(t)x^{t} $ for all real numbers instead of all natural numbers from 0 to infiniti, and we can expect $ f(t) $ to be defined at all those points.
(3) Define variable $ s $ in terms of $ x $
$ F(s)=\int_{0}^{\infty} f(t)e^{-st} \ dt $
