Why do laplace and z and fourier transforms work? I can plug

>>8584260
B/c L^2 spaces

>>8584260

It's just like a change of basis in linear algebra. Each basis "vector"/function is orthogonal: <e^(i n*k x), e^(i m*k x)> = 0 if n≠m so you can project onto them. They are also complete ie they span the space: ∑|e^inkx><e^inkx| = 1 so you can perfectly represent any function as them and not just approximate.

>>8584290
Whew lad

>>8584290
decent, now quote a post made by yourself some time in the future

>>8584260
>laplace
https://www.youtube.com/watch?v=sZ2qulI6GEk#t=60

>>8584290
Dude straight up divided by zero...

>>8584290
W E W
E
W

>>8584290
witnessed

File: wtf.png (102KB, 1772x1232px) Image search: [Google]

102KB, 1772x1232px

>>8584320

File: rar.jpg (749KB, 3280x4928px) Image search: [Google]

749KB, 3280x4928px

This may a little bit opaque, but maybe it's of some value.

All questions of "why the function E(x):=e^x" come down to
>because it's the group homomorphism between the two operations in the ring:
E(x+y) = E(x) · E[y]

Consider the collection of all complex numbers of norm 1.
This is e^{iu} for u in [0, 2pi].
It's a representation of U(1), which lives in the category of locally compact abelian groups.

For any group G, the set Hom(G,U(1)) inherits a group structure from U(1). (If you can't see what this means, ask.)
This is called the dual group and its elements are called "characters" X.

For example, consider R with addition. We'll see that this group is actually self-dual. Fix a real number p and then e.g. [math] x \mapsto e^{ipx} [/math] is an element [math] X_p [/math] of Hom(R,U(1)). It's a homomorphism, because [math] e^{ipa} · e^{ipb}= e^{ip(a+b)} [/math] . We can use the multiplication of phases in U(1) to define a group structure * on these X's by
[math]X_p*X_q \equiv (x \mapsto e^{ipx}) · (x\mapsto e^{iqx}) := x \mapsto e^{i(p+q)x} [/math].
We have a X_p for all real numbers p and so, in this case, as promised, we find Hom(R,U(1)) = R.
(That "new" R from constructing the dual is the domain of Fourier space!)

Now since we see Hom(-,U(1)) maps groups to groups, we can consider it as an auto-functor in the category of locally compact abelian groups.
It now turns out that the twice applied functor [math] Hom(Hom(-,U(1)),U(1)) [/math] is naturally isomorphic to the identity functor. Actually: If X in Hom(G,U(1)), then we can map g\in G to [math]g\mapsto (X\mapsto X(g))[/math].

Since locally compact groups have the Haar measure, one can form the space of integrable functions G -> C and carry it along with these functors. This procedure is the Fourier transform and it really works for all groups in this category! E.g. for U(1) itself, i.e. the periodic interval, we find that the dual group is Z and the associated transform is the Fourier series.

File: Screen Shot 2017-01-07 at 00.58.04.png (814KB, 965x663px) Image search: [Google]

814KB, 965x663px

Here's a more workshop elaboration of the above:
https://en.wikipedia.org/wiki/Pontryagin_duality
https://en.wikipedia.org/wiki/Circle_group

----

Now here a take on Laplace transforms that comes from a very different angle:

We're interested in series coefficients [math]a_n[/math] vs. inverse integral transforms [math]a(t)[/math].
We may compare the sum and integral of monomials [math] \sum_{n=0}^\infty z^n [/math] and [math]\int_0^\infty z^t\,{\mathrm d}t[/math].
We find they differ by Euler-MacLaurin sum terms 1/2, -1/12 etc:

[math]\sum_{n=0}^\infty z^n=\dfrac{1}{1-z}[/math]

[math]\int_0^\infty z^t\,{\mathrm d}t=\int_0^\infty {\mathrm e}^{t\log(z)}\,{\mathrm d}t=-\dfrac{1}{\log(z)}=-\dfrac{1}{1-z}-\dfrac{1}{2}+\dfrac{1}{12}(z-1)+{\mathcal O}\left((z-1)^2\right).[/math]

To match the integral computation to the sum, i.e. [math]-\log(z)\leftrightarrow 1-z[/math], we would first have to perform the conformal mapping [math]z\mapsto {\mathrm e}^{z-1}[/math]. Indeed,

[math]\int_0^\infty ({\mathrm e}^{z-1})^t\,{\mathrm d}t=\int_0^\infty {\mathrm e}^{(z-1)t}\,{\mathrm d}t=\dfrac{1}{1-z}[/math].

In other words, the smooth version of summands given by pairing [math]a_n z^n[/math] doesn't so much correspond to the pairing [math]a(t)\,z^t[/math] under the integral, but rather [math]a(t)\,{\mathrm e}^{(z-1)t}[/math].
Make that [math]a(t)\,{\mathrm e}^{-st}[/math] after a shift [math]z\mapsto 1-s[/math], so that [math]a(t)=1[/math] is connected to [math]\dfrac{1}{s}[/math].

If [math]a_0,a_1,a_2,\dots[/math] is a series, the function

[math]G[a](z)=\sum_{n=0}^\infty a_nz^n[/math]

is called it's generating function.
If [math]a(t)[/math] is a function, the corresponding function (shifted by one) is the Laplace transform

[math]L[a](s):=\int_0^\infty a(t)\,{\mathrm e}^{-st}\,{\mathrm d}t[/math].

For those two, the constant series/function [math]a_n=a(t)=1[/math] both lead to [math]\frac{1}{1-z}=\frac{1}{s}[/math].

File: 16d.jpg (10KB, 480x360px) Image search: [Google]

10KB, 480x360px

>>8584451
>>8584481

File: 1454412974732.png (25KB, 655x509px) Image search: [Google]

25KB, 655x509px

>>8584451
What about the Fourier transform for compact non-abelian groups as given via the Peter-Weyl theorem? How can you reconcile this with the locally compact abelian case? Is there a generalization encompassing both, and more definitions of FT?

File: rabbit hole.jpg (160KB, 1280x720px) Image search: [Google]

160KB, 1280x720px

>>8584804
https://en.wikipedia.org/wiki/Gelfand_representation

https://en.wikipedia.org/wiki/Wavelet_transform

https://en.wikipedia.org/wiki/Pontryagin_duality

https://en.wikipedia.org/wiki/Cartier_duality

https://en.wikipedia.org/wiki/Tannaka%E2%80%93Krein_duality

...

*smiley face*

>>8584260
https://www.youtube.com/watch?v=r18Gi8lSkfM

File: D2BEC632-95AF-492E-9B2F-013E0BA9982C-3605-000003F4D4F3D4DB_tmp.png (224KB, 447x489px) Image search: [Google]

224KB, 447x489px

>>8584290

Huly fug

>>8585244
Thanks bro. I always avoided C*-algebras. Looks like I have to overcome my dislike for algebra now.

>>8584260
Just look at this.

https://ncatlab.org/nlab/show/Fourier-Mukai+transform

File: 1483341921992.png (1MB, 1009x960px) Image search: [Google]

1MB, 1009x960px

>>8584290
Wait wat