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Anonymous Mon 13 May 12:19:22 2019 No.10640478 View Reply Original Report

Quoted By: >>10640607 >>10640676 >>10640699 >>10640728 >>10640766 >>10640856 >>10640857 >>10640985

Someone explain me the Laplace transform please

I kind of understand the Fourier transform
I just don't understand what the Laplace transform does.
I understand it helps with differential equations somehow.

My textbook isn't helping either it just jumps straight into it. Have searched online but not very helpful

Am I a brainlet?

>Inb4 enginigger

Anonymous

Anonymous Mon 13 May 2019 14:26:08 No.10640607 Report

Quoted By: >>10640617 >>10640628 >>10640641

>>10640478

It takes a function of a real variable t (often time) to a function of a complex variable s (complex frequency). The transform has many applications in science and engineering.

The Laplace transform is similar to the Fourier transform. While the Fourier transform of a function is a complex function of a real variable (frequency), the Laplace transform of a function is a complex function of a complex variable.

from jewpedia.

what's not clear about it?

Anonymous

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Anonymous Mon 13 May 2019 14:31:56 No.10640617 Report

Quoted By: >>10640633 >>10640647 >>10640862 >>10640933

>>10640607
Not OP, but this is how I envision the explanation going:
>the fourier series breaks the function into sine waves
OK!
>the fourier transform breaks it into sine waves, but does it continuously instead of as a discrete series
OK!
>the laplace transform breaks the function into _______
I also don't know the answer by the way

Anonymous

Anonymous Mon 13 May 2019 14:36:03 No.10640628 Report

Quoted By:

>>10640607
wow this answered nothing

not OP either by the way

Anonymous

Anonymous Mon 13 May 2019 14:36:44 No.10640632 Report

Quoted By:

All you need to know if you're an engineer is that you can build take your function and turn it into something easier to work with, then solve your problem and turn your solution back into the form you want. You get a table of what functions map to what and it's linear so you already know the rules of how to apply it.

Anonymous

Anonymous Mon 13 May 2019 14:37:27 No.10640633 Report

Quoted By: >>10640645 >>10640647

>>10640617
>>the laplace transform breaks the function into _______

fine.

http://tutorial.math.lamar.edu/pdf/Laplace_Table.pdf

as I said before, the laplace transform of a function is a complex function of complex variable.

it's like when people ask what an eigenvalue and an eigenvector is, they actually want "real examples" but there is no "real example", just cases where we can apply the eigenvalue and the eigenvector, they do not exist per se.

Anonymous

Anonymous Mon 13 May 2019 14:39:55 No.10640639 Report

Quoted By:

>While the Fourier transform of a function is a complex function of a real variable (frequency), the Laplace transform of a function is a complex function of a complex variable.
on a practical level just think of it as a version of the Fourier transform that works on complex variables

Anonymous

Anonymous Mon 13 May 2019 14:40:15 No.10640641 Report

Quoted By:

>>10640607
Thanks for explaining fucking nothing.

Anonymous

Anonymous Mon 13 May 2019 14:43:03 No.10640645 Report

Quoted By: >>10640652 >>10640750

>>10640633
Except you can explain eigenvectors pretty well. It's the spaces that linear transformations, in some sense, preserve but only scale. Knowing what stays the same is important because there are numerous examples for where this is a "real" thing.

I mean, look at almost the entire study of quantum mechanics. The only reason it works out is because of this idea that the only linear transformations we care about are those that conform to this idea of "staying the same." It honestly sounds like you're a purist who has no intuition for the mathematics outside of formalism

>t. pure math grad

Anonymous

Anonymous Mon 13 May 2019 14:44:29 No.10640647 Report

Quoted By: >>10640661 >>10640750 >>10640883

>>10640633
>it's like when people ask what an eigenvalue and an eigenvector is, they actually want "real examples" but there is no "real example"
Yes there is. When you rotate in 3d there is an axis along which things don't move. Are you thinking of determinants? Either way the ten-word summary of Fourier in >>10640617 btfos anyhing you've written so far.

Anonymous

Anonymous Mon 13 May 2019 14:45:43 No.10640652 Report

Quoted By: >>10640653

>>10640645

yeah I know, eigenvectors can be applied to everything, states, stresses, etc
I was only making a point about them being an abstract object.

Anonymous

Anonymous Mon 13 May 2019 14:47:09 No.10640653 Report

Quoted By: >>10640659

>>10640652
if all OP cared about was an explanation that only treated laplace transformations as an abstract object then OP wouldn't have made this thread.

Anonymous

Anonymous Mon 13 May 2019 14:50:00 No.10640659 Report

Quoted By: >>10640666 >>10640677

>>10640653

look, I have tried to answer to the following question from the OP

>I just don't understand what the Laplace transform does.

I told him that you can do some special transformations of equations in a form which might be more computationally friendly.

Is it necessary to specify the application of said equations?

Anonymous

Anonymous Mon 13 May 2019 14:51:29 No.10640661 Report

Quoted By: >>10640663 >>10640750

>>10640647
I mean, even then determinants have an actual geometric explanation.

The way I'd explain Laplace transform is something like:
"A Laplace transform always exists while a Fourier transform may not. Basically, it shows that can represent differentiation like multiplication, do the easy algebra, and then transform back. By the end, the effect is the same as had you solved the equation 'honestly.' It turns differentials into basic algebra."

Anonymous

Anonymous Mon 13 May 2019 14:52:31 No.10640663 Report

Quoted By:

>>10640661
>*it shows that we can can represent

Anonymous

Anonymous Mon 13 May 2019 14:53:19 No.10640666 Report

Quoted By: >>10640676

>>10640659
>Is it necessary to specify the application of said equations?
i think it is. OP said they "kind of understand the Fourier transform" and that they don't understand the book just jumping straight into the theory. i would bet that OP's understanding of all the pure theory of the Fourier transform is already on shaky ground.

Anonymous

Anonymous Mon 13 May 2019 14:57:39 No.10640676 Report

Quoted By:

>>10640666

Fair enough.

>>10640478

OP they are a technique which is useful for multiple calculations, probably you will use them in evaluating some improper integrals, other applications might depend upon your field of studies.

Study the theory, the applications will come, if it seems somewhat rough or difficult remember that people took centuries to understand stuff like this.

Anonymous

Anonymous Mon 13 May 2019 14:57:45 No.10640677 Report

Quoted By:

>>10640659
You're a fucking idiot, dude.
How far fucking gone do you need to be to not understand what OP is asking for here? I don't have an answer myself but man this is just pathetic.
Telling someone what something is useful for and how it works is not the same as giving them an intuition for its meaning as a solitary object. OP wants something directly analogous to "The fourier transform gives us a function's frequency profile by measuring how well it coincides with particular frequencies"
I suppose there isnt a great way to say this about Laplace, except for that you're measuring how the function "looks like" certain exponentials.

Anonymous

Anonymous Mon 13 May 2019 15:11:02 No.10640699 Report

Quoted By:

>>10640478
I don't have a good answer for OP but I do have an interesting data-point: even smart and opinionated autists have struggled with the question of motivating Laplace in an elementary way. See section 9 of Rota's article:
>https://web.williams.edu/Mathematics/lg5/Rota.pdf

Anonymous

Anonymous Mon 13 May 2019 15:34:00 No.10640728 Report

Quoted By: >>10640735

>>10640478
Holy shit there's a lot of non-answers here.
The Laplace Transform essentially breaks a complex valued time domain function into simple complex exponential components exp(st), where the value of the frequency domain function for that value of s is more or less the complex amplitude you multiply to the associated exponential (there's an extra factor of 1/(2i*pi) in there).
The main difference with Fourier transforms is that there's no unique way to reconstruct the time domain function with the inverse transform. Instead, you pick a real component of s and integrate along the imaginary component (choosing a real component such that the integral converges). Usually the Laplace transform is also restricted to s with no negative real component so that the forward transform of integrating f(t)exp(-st) tends to converge.
In the special case that, during the inverse transform, you integrate along the line with real component of s equals 0, you are reconstructing the time domain function from sine waves, much like the inverse Fourier transform.

Anonymous

Anonymous Mon 13 May 2019 15:37:24 No.10640735 Report

Quoted By:

>>10640728
Meant to say s with nonnegative* real component

Anonymous

Anonymous Mon 13 May 2019 15:50:55 No.10640750 Report

Quoted By: >>10640765

>>10640645
>Based geometer
>>10640647
>Based geometer
>>10640661
>Based geometer

Does anyone actually have mathematical intuition beyond geometry? It's like you either have geometric intuition, or you jerk off to definitions and fool yourself into thinking you have it

Anonymous

Anonymous Mon 13 May 2019 15:58:25 No.10640765 Report

Quoted By: >>10640767

>>10640750

visualization is a big deal, but not the only way

Anonymous

Anonymous Mon 13 May 2019 15:58:45 No.10640766 Report

Quoted By: >>10640793 >>10640801

>>10640478
Here ya go OP, enjoy

https://www.google.com/url?sa=t&source=web&rct=j&url=https://www.youtube.com/watch%3Fv%3D6MXMDrs6ZmA%26vl%3Den&ved=2ahUKEwjDi6PD75jiAhWHwcQHHVmjA28QwqsBMAF6BAgFEAo&usg=AOvVaw314rR04lT0elBbNjHFSgHL

Anonymous

Anonymous Mon 13 May 2019 16:00:29 No.10640767 Report

Quoted By: >>10640803

>>10640765
Examples?

Anonymous

Anonymous Mon 13 May 2019 16:12:42 No.10640793 Report

Quoted By: >>10640808

op's post made me realize; i just finished an advanced control system course and still have basically no idea what a Laplace transform is

>>10640766
based. why the fuck do we still learn from 2D depictions (whiteboards/overhead/textbooks) when almost everyone that's learning these topics have a phone/laptop.

and why are spirals in everything

Anonymous

Anonymous Mon 13 May 2019 16:17:00 No.10640801 Report

Quoted By:

>>10640766
Based Eugene

Anonymous

Anonymous Mon 13 May 2019 16:17:38 No.10640803 Report

Quoted By:

>>10640767

I can only speak for myself, I'll make you an example from when I was a first year student.

When I first encountered the concept of linear operation, I necessarily needed a "visualization" for an example.
Then I grew comfortable with treating the transformation as an abstract entity, dealing only with the properties, from there I abstracted all the transformations based on their relative definitions and operations.
Then one day I found myself more comfortable at looking at things based on their properties, instead of visualizing them.
I still use visualization on some things, but I've noticed that the better I get at some mathematical things, the less I need to visualize them.

Anonymous

Anonymous Mon 13 May 2019 16:19:18 No.10640808 Report

Quoted By:

>>10640793
>and why are spirals in everything

because of complex logarithms

Anonymous

Anonymous Mon 13 May 2019 16:41:26 No.10640856 Report

Quoted By:

>>10640478
It takes a differential equation into an algebraic equation, which is easier to solve in many situations. Then you take the solution, apply the inverse transform and get the solution to your differential equation

All integral transforms in a way or another do this, it is the same idea of homeomorphisms: map your problem into an easier framework, get solutions there and bring them back through the inverse map

Anonymous

Anonymous Mon 13 May 2019 16:41:43 No.10640857 Report

Quoted By: >>10641049

>>10640478
If you work with taylor series, it simply transforms terms of the form (x^n)/n! into terms of the form (1/t)^(n+1).

Abstractly, it sort of encodes information about half convolutions.

$e^{-xD_{t=0}}[g(t)]= \sum\limits_{n=0}^{\infty}{(-x)^n \over n!}D_{t=0}^{n}[g(t)]=g(-x). \\ If\ \int_0^\infty f(x)e^{-xs}dx = F(s)\ then,\ \int_0^\infty f(x)e^{-xD_{t=0}}dx = F(D_{t=0})\ and\\ \int_0^\infty f(x)e^{-xD_{t=0}}[g(t)]dx = \int_0^\infty f(x)g(-x)dx =F(D_{t=0})[g(t)]. $

Anonymous

Anonymous Mon 13 May 2019 16:46:15 No.10640862 Report

Quoted By:

>>10640617
>I also don't know the answer by the way

What happens when you Fourier transform sin(x) or e^ix? Laplace replaces i? with s = i? - ? and integrate from t>0. This makes it universal for periodic and aperiodic signals/functions. The actual value of the real constant ? is irrelevant as long as the integral converges.

Anonymous

Anonymous Mon 13 May 2019 17:01:05 No.10640883 Report

Quoted By: >>10641055 >>10641418

holy shit engineers are so fucking dumb. look at this whole thread.
also lol at the retard
>>10640647
who thinks every matrix is a rotation

if you are clueless stop posting and thinking you are hot shit jesus christ

Anonymous

Anonymous Mon 13 May 2019 17:37:48 No.10640933 Report

Quoted By:

>>10640617

The Laplace transform breaks functions into exponentially increasing/decreasing sine functions. It is to see which frequencies grows and which frequencies decays with time.

Anonymous

Anonymous Mon 13 May 2019 18:01:20 No.10640985 Report

Quoted By:

>>10640478
It's the continuous version of power series.
In $\sum a_n x^n$ replace the sequence $a_n$ with a function $f(n)$ , where n is now a real number, then rewrite x^n as $e^{n \log(x)}$ , and then replace $log(x)$ with $-s$ $(s>0)$ so that the "sum"(integral) converges.

An example of the discrete Laplace transfrom (power series):
The constant sequence 1, -1/2! , 1/4!, -1/6!, ... corresponds to the power series 1-\frac{x^2}{2!}+\frac{x^4}{4!} - \frac{x^6}{6!} + \cdots which is the function \sin(x).
But it goes the other way around as well, you can recover the sequence from sin(x).
Or, the constant sequence 1,1,... coresponds to 1/(1-x).

An example of the continuous Laplace transform:
The constant function f(t)=1 (continuous version of 1,1...) corresponds to 1/s.

Anonymous

Anonymous Mon 13 May 2019 18:30:40 No.10641049 Report

Quoted By:

>>10640857
>Abstractly, it sort of encodes information about half convolutions.
t. reddit

Anonymous

Anonymous Mon 13 May 2019 18:37:28 No.10641055 Report

Quoted By: >>10641058

>>10640883
Every matrix is a rotation, a stretch, and another rotation. Moron.

Anonymous

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Anonymous Mon 13 May 2019 18:38:44 No.10641058 Report

Quoted By: >>10641461

>>10641055
>who thinks every matrix is a rotation
your reply:
>>Every matrix is a rotation or something else

Anonymous

Anonymous Mon 13 May 2019 20:26:48 No.10641361 Report

Quoted By: >>10641465

https://youtu.be/sZ2qulI6GEk?t=1m5s
$ \sum_{n=0}^{\infty} a_n x^n = A(x) \\ \sum_{n=0}^{\infty} a(n) x^n = A(x) \\ a(n) \sim \! \sim \! \sim \! \! \! > A(x) \\ a(n) = 1 \sim \! \sim \! \sim \! \! \! > \frac{1}{1-x} \: \: \, ,\: \: \: \left | x \right | < 1 \\ a(n) = \frac{1}{n!} \sim \! \sim \! \sim \! \! \! > e^x \\ \text{continuous analog:} \\ \int_{0}^{\infty} a(t) x^t dt = A(x) \: \: \, ,\: \: \: 0 < x < 1 \Rightarrow ln(x) < 0 \\ x^t = \left ( e^{ln(x)} \right )^t \: \: \, ,\: \: \: -s = ln(x) \Rightarrow x = e^{-s} \\ \int_{0}^{\infty} f(t) e^{-st} dt = F(s) $

Anonymous

Anonymous Mon 13 May 2019 20:44:47 No.10641418 Report

Quoted By:

>>10640883
>this isn’t a real example of an eigenvalue because it’s not the only possible real example of an eigenvalue
low IQ

Anonymous

Anonymous Mon 13 May 2019 20:55:41 No.10641451 Report

Quoted By: >>10641462

I'm not really satisfied with any of your answers. They're not straight to the point.

The Laplace transform is simply a very nice functional. A functional changes functions into other functions. This procedure is, for all intents and purposes, invertible on the functions you care about, so you can undo it. This is an important fact that tends to get glossed over in introductory differential equations. I'll call this Property 1.

There is a nice formula for the transform of a derivative that converts it into a polynomial in terms of the transform of the unknown solution, which I'll call Property 2. So if you apply the transform to a differential equation, all the derivatives in the equation disappear and it terns into an equation in terms of the transform of the solution only. So if you can solve the equation for the transform of the solution (just with algebra) then you can hopefully undo the transform (because of Property 1) and you're done.

You can play around with the definition of the Laplace transform using by-parts integration to see why Property 2 is true. Property 1 is a little trickier to show.

There is not some super great intuitive visualization of the Laplace transform. It works because of how the definition works, and the special properties of the exponential function. Don't read too much into it.

Anonymous

Anonymous Mon 13 May 2019 21:00:03 No.10641461 Report

Quoted By:

>>10641058
He was giving an example of "eigenvalue and eigenvector" dipshit. No one claims that every matrix is a rotation.

Anonymous

Anonymous Mon 13 May 2019 21:00:37 No.10641462 Report

Quoted By:

>>10641451
>There is not some super great intuitive visualization of the Laplace transform.

https://www.youtube.com/watch?v=6MXMDrs6ZmA

Obviously it doesn't make the useful properties of the Laplace Transform apparent, but as far as intuitive visualizations go it's damn good.

Anonymous

Anonymous Mon 13 May 2019 21:02:26 No.10641465 Report

Quoted By:

>>10641361
Holy fuck that's wonderful. Thank you so much, laplace transform just went from my shit list to my based transforms list.

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