I'm excited, Gottfried. I have a fair amount of time over the holidays, and then come January 10th, I'm homefree for 2-3 months. And I plan to work a lot on some of the ideas I've been mulling over--specifically cementing a lot of iteration theory using the Mellin transform! I'm busting out a lot of old notes which hand waived the crucial details, and rigorously justify them much more. Look forward to working with you! 
I'm excited to deep dive into the "Matrix method is the same as the integral method."
Regards, James
EDIT! Just a quick proof to show that the \(1,3,5,7,...\) sequence in Gottfried's original post is nothing special.
\[
f(x) \,\,\text{is C^2 in}\,\,(-\pi,\pi)\\
\]
Additionally, assume that:
\[
\int_{-\pi}^\pi |f(x)|^2\,dx < \infty\\
\]
This is true for any iterate of \(\sin\). Now:
\[
f(\pi - x) = f(x)\\
\]
What I think is more important here, is that \(f(x-\pi) = -f(x)\). Which is what happens for every iterate of \(\sin\). So let's work with the identity: \(f(x+\pi) = - f(x)\).
Therefore:
\[
2\pi a_k = \int_{-\pi}^\pi f(x)sin(kx)\,dx = \int_0^\pi \left(f(x)\sin(kx)-(-1)^kf(x)\sin(kx)\right)\,dx
\]
Even values of \(k\) are zero; odd values non-zero. Additionally, since:
\[
\sin^{\circ n}(-x) = -\sin^{\circ n}(x)\\
\]
We know:
\[
\int_{-\pi}^\pi f(x) \cos(kx)\,dx =0\\
\]
This is because this integrand is odd, across a symmetrical domain. Whereby, \(f(-x)\cos(-kx) = -f(x)\cos(kx)\); and since we are integrating about zero over the same domain, the negative domain cancels the positive.
And so, when we call on the traditional Fourier series:
\[
f(x) = f(0) + \sum_{k=1}^\infty c_k \cos(kx) + a_k \sin(kx)\\
\]
We know, ABSOLUTELY, that \(c_k = 0\) and \(a_{2k} = 0\) and \(f(0) = 0\). So that:
\[
f(x) = \sum_{k=0}^\infty a_{2k+1} \sin(2k+1)x\\
\]
Is the default expansion.
So the fact the MSE question pulled out this expansion is very much non-surprising, lol.
We want to use this to show that the integral transform coefficients I generate, are the same as the Bessel function coefficients this MSE question pulled out. I'd bet 50$ they're the same expansion. But the matrix approach looks much more promising. Definitely, the Bessel matrices are quicker and more accurate--but they are the same thing mathematically as my integrals...
Anyway... I'm dead for tonight....
I'm excited to deep dive into the "Matrix method is the same as the integral method."Regards, James
EDIT! Just a quick proof to show that the \(1,3,5,7,...\) sequence in Gottfried's original post is nothing special.
\[
f(x) \,\,\text{is C^2 in}\,\,(-\pi,\pi)\\
\]
Additionally, assume that:
\[
\int_{-\pi}^\pi |f(x)|^2\,dx < \infty\\
\]
This is true for any iterate of \(\sin\). Now:
\[
f(\pi - x) = f(x)\\
\]
What I think is more important here, is that \(f(x-\pi) = -f(x)\). Which is what happens for every iterate of \(\sin\). So let's work with the identity: \(f(x+\pi) = - f(x)\).
Therefore:
\[
2\pi a_k = \int_{-\pi}^\pi f(x)sin(kx)\,dx = \int_0^\pi \left(f(x)\sin(kx)-(-1)^kf(x)\sin(kx)\right)\,dx
\]
Even values of \(k\) are zero; odd values non-zero. Additionally, since:
\[
\sin^{\circ n}(-x) = -\sin^{\circ n}(x)\\
\]
We know:
\[
\int_{-\pi}^\pi f(x) \cos(kx)\,dx =0\\
\]
This is because this integrand is odd, across a symmetrical domain. Whereby, \(f(-x)\cos(-kx) = -f(x)\cos(kx)\); and since we are integrating about zero over the same domain, the negative domain cancels the positive.
And so, when we call on the traditional Fourier series:
\[
f(x) = f(0) + \sum_{k=1}^\infty c_k \cos(kx) + a_k \sin(kx)\\
\]
We know, ABSOLUTELY, that \(c_k = 0\) and \(a_{2k} = 0\) and \(f(0) = 0\). So that:
\[
f(x) = \sum_{k=0}^\infty a_{2k+1} \sin(2k+1)x\\
\]
Is the default expansion.
So the fact the MSE question pulled out this expansion is very much non-surprising, lol.
We want to use this to show that the integral transform coefficients I generate, are the same as the Bessel function coefficients this MSE question pulled out. I'd bet 50$ they're the same expansion. But the matrix approach looks much more promising. Definitely, the Bessel matrices are quicker and more accurate--but they are the same thing mathematically as my integrals...
Anyway... I'm dead for tonight....
