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If you understand fourier transform well, then perhaps this viewpoint will help. The fourier transform is 'just' a change of basis, with the basis being the sin
by bollu 6y ago
If you understand fourier transform well, then perhaps this viewpoint will help. The fourier transform is 'just' a change of basis, with the basis being the sinusoidal functions. Why these? Well, because if we take a look at the discrete fourier transform, the matrix that changes the basis is both unitary [which is has to be as a change of basis] and vandermonde. So we can think of it as both
a 'change of basis' and as a 'evaluation of polynomial'. This is where most of the power of the fourier transform comes from.
Similarly, the laplace transform is also a change of basis. But the basis it chooses is a very special one --- it's the eigenvectors of the differential operator. Note that
d/dx(e^(ax)) = ae^ax
So e^(ax) literally an eigenvector of `(d/dx)`. And as we all know, going to
the eigenbasis of a given operator/linear transform/matrix makes it easier
to manipulate. The laplace transform is a change of basis that digonalizes
the differential operator. This makes it easy to solve differentials.
- chombier 6y agoExactly what I needed, thanks!
- hnra 6y agoDo you have any good text book where this is included? Seems like something very interesting to read about formally.
- pokoleo 6y agoMath 213 at uwaterloo has publicly available notes in spades: https://www.gabrielwong.net/notes/math213_notes.pdf#page9 https://www.gabrielwong.net/notes/math213_notes.pdf#page9
- novaRom 6y agosearch for "Introduction to Signals & Systems"
- abecedarius 6y agoI'm not sure this is what you're looking for, but the Feynman lectures give the basic idea in an elementary way: https://www.feynmanlectures.caltech.edu/I_23.html https://www.feynmanlectures.caltech.edu/I_23.html > This idea of using exponentials in linear differential equations is almost as great as the invention of logarithms, in which multiplication is replaced by addition. Here differentiation is replaced by multiplication. . . . See how simple it is! Differential equations are immediately converted, by sight, into mere algebraic equations
- spopejoy 6y agoHaven't read it but this currently-free Springer textbook covers DEs, Laplace and Fourier transformations, "Differential Equations and Their Applications": https://link.springer.com/book/10.1007%2F978-1-4612-4360-1 https://link.springer.com/book/10.1007%2F978-1-4612-4360-1
- edouard-harris 6y agoThis is an excellent explanation.
- jjoonathan 6y agoIs the Laplace eigenbasis considered a basis in a relaxed sense that permits non-orthogonality or does the notion of orthogonality itself change to counteract the apparent redundancy?
- Twisol 6y agoBases only need to be linearly independent and spanning; non-orthogonality is already permitted. Not all bases are orthogonal.
- jjoonathan 6y agoRight, thanks, it's been too long.
- dzolob 6y agoOne also can think of transforms as the eigenfunctions of the continuous part of the spectrum of a differential operator. In differential equations (DE) theory, a well posed DE has a structure (the DE itself), a domain and enough independent boundary conditions. Fourier transform will show up for an harmonic oscilator in the whole real line with incoming and outgoing wave boundary conditions, while Laplace will show up when working on semi-infinite interval whit initial conditions and proper convergence at infinity. These are the most common, but not the only transforms one can build. There are also Melin and Hankel transforms, and by playing with the operator, the domain and the boundary conditions, we can construct the adequate transform for each given problem. Spectral theory of DE’s is such a beautiful topic.