(1:2) Where the Laplace Transform comes from (Arthur Mattuck, MIT)

What people said

Strong rec E.P. · 2013-07 · Mathematics Stack Exchange · on “Fourier series is to Fourier transform what Laurent series is to ...?”

In the same way that there is a discrete $\rightarrow$ continuous transition as one goes from Fourier series to transforms, one can make a continuous analogue to Taylor series in the form of the Laplace transform . This is beautifully explained by Arthur Mattuck in this video [this video] and this one [LINK] , which I heartily recommend, but the gist is this: Start with a Taylor series of the form $\sum_{n=0}^\infty a_nx^n=:A(x)$, and change the discrete variable $n=0,1,2,\ldots$ for a continuous variable $t\in(0,\infty)$, to get $\int_0^\infty a(t) x^t dt$. From this it is only a cosmetic change to define the variable $s=-\ln(x)$, which is positive and real in the interesting range for $x$, $0<x

Strong rec Fahd Siddiqui · 2014-10 · Mathematics Stack Exchange · on “basics needed for studying Laplace Transform”

The best video I have found for Laplace Transform was this: [this video] [LINK] MIT professor explains where it comes from, its analogue: Power Series and so on. It will take 13 minutes of your time but you will understand the concept completely!

Recommends bolbteppa · 2013-09 · Mathematics Stack Exchange · on “Discrete to Continuous Representations of Functions via Laplace Transforms?”

The Laplace transform can be thought of as the continuous analogue of a power series, as in this video [this video] . From this perspective, think of the function $ a : \mathbb{N} \rightarrow \mathbb{R}$ as a function giving the coefficients of the taylor series of a function $f$ in $ f(x) = \sum _{i=0}^\infty a(n)x^n$. In other words the 'sequence' $a$ is a discrete representation of the function $f$. What is the continuous analogue? Can we use this to find a continuous representation for the function $f$? We basically want: $$ f(x) = \sum _{i=0}^\infty a(n)x^n \rightarrow f(x) = \smallint_0^\infty a(t) x^t dt$$ Then applying 'cosmetic' adjustments like $ x = e^{\ln(x)}$, $ x^t = e^{t\ln(x)}$ &

Recommends user169852 · 2016-04 · Mathematics Stack Exchange · on “Why is the Laplace Transformation defined the way it is?”

This [this video] helps to explain mathematically why it is defined as it is. Not sure what, if any, physical significance it has. The Fourier transform (a special case) does have physical significance, for example one can take a Fourier transform using an optical lens [LINK]

Recommends Aravindh Vasu · 2019-12 · Mathematics Stack Exchange · on “How is Laplace transform more efficient?”

I saw that power series analogy( [this video] ), but it wasn't as convincing as this operator origin. I think so because... I'm not able to digest ... that we just "arrived" at the equation out of nowhere.

Cites MGA · 2014-08 · Mathematics Stack Exchange · on “'Deriving' the Laplace Transform from the $z$ Transform: Missing a $\Delta t$”

Textbooks normally give the following 'derivation' (or justification, if you prefer) of the z-Transform from the Laplace Transform. Let $x(t)$ be a signal defined on $t\geq 0$, and write a discretized version of $x(t)$ as: $$ x_d(t)=\sum\limits_{n=0}^{\infty} x(n\Delta t)\delta(t-n\Delta t) $$ where $\delta()$ is the Dirac delta function. Take the Laplace Transform of $x_d(t)$: $$ \mathcal{L}\{x_d(t)\}= \sum\limits_{n=0}^{\infty} x(n\Delta t)\mathcal{L}\{\delta(t-nT)\}=\sum\limits_{n=0}^{\infty} x(n\Delta t) e^{-sn\Delta t} =^{z=e^{s\Delta t}} \sum\limits_{n=0}^{\infty} x(n\Delta t) z^{-n} = X(z) $$ I wish demonstrate to someone who is already familiar with the $z$ Transform that the Laplace