Question

What's a Laplace transform? : Mostly a guide

Answer 1

BabyKainui, awww, cute c;

Answer 2

It helps with discontinuous coefficients, so we have the integral transform \[F(s) = \int\limits_{\alpha }^{\beta} k(s,t) f(t) dt\] where k(s,t) is the kernal of the transform, it's also part of fourier transforms.

Answer 3

Alright so you have power series representations of functions:

\[F(x) = \sum_{n=0}^\infty a_n x^n\]
Let's make some aesthetic changes and write \(a_n = f(n)\)
\[F(x) = \sum_{n=0}^\infty f(n) x^n\]
Now let's decide to include infinitely more terms in here so we have to throw in an infinitesimal weight to make it not explode:
\[F(x) = \sum_0^\infty f(n) x^n dn\]
Alright let's use some algebra to make this simpler to evaluate the integral:
\[x^n = e^{n \ln x}\]
now let's substitute \(-s = \ln x\) just for the sake of making convergence easier to see/control. Purely cosmetic really.
\[F(x) = \int_0^\infty f(n) e^{-ns} dn\]
Oh and also let's change n to t cause we like that better
\[F(x) = \int_0^\infty f(t) e^{-ts} dt\]

So what tho?

Answer 4

o.o

Answer 5

That clueless look, lol

Answer 6

I "sum"d when I shoulda "intergralled" so w/e hopefully that much follows plausibly speaking.

But there are 2 other pretty important points to hit upon. First linearity, it's really just a giant _invertible_ linear map and you can sorta visualize the matrix equaton as this:

\[\left( \int_0^\infty e^{-st} dt \right) f(t) = F(s)\]
Think:
\[A u = v\]
except the vectors f(t) and F(s) have an infinite continuum of vector components instead of a usual vector which has components of only v(1), v(2), and v(3). Go figure.

Alright so there's that and really you really sorta said this, It just ends up being useful for solving differential equations since it turns it into an algebra problem we can sorta more or less play around with easily and then use the invertibility of the linear map to get the answer.

Answer 7

I guess I should write the integral more like you would the matrix equation to give it better off:

\[A u = v\]
in summation notation is:
\[\sum_i A_{ij} u_i = v_j\]
Which is really where this integral form comes from:
\[\int a(x,y) f(x) dx = g(y)\]
Then if it's invertible you will have some other thing, I think you'll probably first see this kind of thing done with Fourier transforms/series if you haven't seen it yet and that inverse transform is how you'll define the Dirac/Kronecker delta AKA identity matrix really of your transformation.

Yeahhhh

Answer 8

Anyways main thing is I wanted to draw some parallels between the Laplace transform to both Power Series and Matrix Multiplication so I hope I accomplished that as in some sorta motivation. But I probably left you with some questions.

Answer 9

honestly, after I asked this question I realized I was too tired to actually understand any answer... I'll probably read it when I feel not brain fried from all those silly midterms this week

Answer 10

Hahaha I thought so but idc I started a blog so like after people ask questions / get lost on stuff here I can fix it up better and then I'll copy paste edit it to throw over there. link in my profile and here: https://orbitaloverlap.wordpress.com/

lolol this is all just a shameless plug for a blog no one will read :X

Answer 11

@Kainui Laplace transform and Fourier transform what is difference ?
iknow laplace use for solve DF

Answer 12

Good question, both are pretty closely related and both are used to solve differential equations by turning them into algebra.

The Fourier transform can be built up from a similar process to this except instead of starting at the power series you start at the Fourier series which is a way of writing periodic functions as a sum of their weighted frequency components... Kinda confusing but I'm a bit rushed for time my family's about to go out and eat pizza haha.

Answer 13

@kainui ty for information

Answer 14

finally read this. Kinda get the intuition/motivation - Thanks!