Hey, Gottfried.
This is precisely the Fourier expansion of \(\sin^{\circ 1/2}(x)\). It's well known this function is odd, and is also periodic. Thereby, when we expand the fourier expansion:
\[
\sin^{\circ 1/2}(x) = \sum_{k=1}^\infty a_k \sin(kx)\\
\]
Where:
\[
a_k = \frac{1}{2\pi}\int_{-\pi}^\pi \sin(kx)\sin^{\circ 1/2}(x)\,dx\\
\]
We can also deduce that \(a_{2k} = 0\), because \(\sin^{\circ 1/2}(x)(\pi - x) = \sin^{\circ 1/2}(x+\pi/2)\)/// EDIT: SORRY I SCREWED UP THIS. I MEANT TO WRITE \(\sin^{\circ 1/2}(x)(\pi - x) = \sin^{\circ 1/2}(x)\)--which then eliminates even powers. I apoligize screwed that up a bit, lol. Which follows from basic trigonometric identities. Whereby we have a working:
\[
\sin^{\circ 1/2}(x) = \sum_{k=0}^\infty a_{2k+1} \sin(2k+1)x\\
\]
That being said, I've never seen this connection to the Bessel function. Cool find!!!! It makes sense though because one of the formulae for the bessel function of the first kind is an integral of nested sinusoids over \([0,\pi]\); and expanding these \(\sin^{\circ 1/2}(x)\), will result in something very similar.
I also think, this provides a cool interchange of variables.
Write:
\[
\sin^{\circ 1/2}(x) = \frac{1}{\sqrt{\pi}}\int_0^\infty \frac{\vartheta(t,x)}{\sqrt{t}}\,dt\\
\]
Where:
\[
\vartheta(t,x) = \sum_{n=0}^\infty \sin^{\circ n+1}(x) \frac{(-t)^n}{n!}\\
\]
Then; then we can interchange these integrals, so that:
\[
a_{k}(1/2) = \frac{1}{\sqrt{\pi}} \int_0^\infty \sum_{n=0}^\infty a_{k}(n+1)\frac{(-t)^nt^{-1/2}}{n!} \,dt\\
\]
where:
\[
a_k(s) = \frac{1}{2\pi} \int_{-\pi}^\pi \sin^{\circ s}(x) \sin(kx)\,dx\\
\]
I've been aware of this formula for a long time, but I never saw the relation to Bessel before!
Also, note--I've never been able to prove this expansion of \(\sin^{\circ 1/2}(x)\) is analytic. I only managed to show that it was \(C^{\infty}\) on \((-\pi,\pi)\). But I absolutely suspect it is analytic here. Maybe with closed forms of \(a_k\) we'll have better luck!!!! I'd be really interested to see if this Fourier transform converges analytically, or only \(C^\infty\). The sin iteration tends to be an odd ball out trying to get analycity out of. And showing uniform convergence of Fourier series in the complex plane is always a lot harder than it sounds (I think they've awarded like 4 fields medals for tiny step theorems on the convergence of Fourier series! lol!
)
This is precisely the Fourier expansion of \(\sin^{\circ 1/2}(x)\). It's well known this function is odd, and is also periodic. Thereby, when we expand the fourier expansion:
\[
\sin^{\circ 1/2}(x) = \sum_{k=1}^\infty a_k \sin(kx)\\
\]
Where:
\[
a_k = \frac{1}{2\pi}\int_{-\pi}^\pi \sin(kx)\sin^{\circ 1/2}(x)\,dx\\
\]
We can also deduce that \(a_{2k} = 0\), because \(\sin^{\circ 1/2}(x)(\pi - x) = \sin^{\circ 1/2}(x+\pi/2)\)/// EDIT: SORRY I SCREWED UP THIS. I MEANT TO WRITE \(\sin^{\circ 1/2}(x)(\pi - x) = \sin^{\circ 1/2}(x)\)--which then eliminates even powers. I apoligize screwed that up a bit, lol. Which follows from basic trigonometric identities. Whereby we have a working:
\[
\sin^{\circ 1/2}(x) = \sum_{k=0}^\infty a_{2k+1} \sin(2k+1)x\\
\]
That being said, I've never seen this connection to the Bessel function. Cool find!!!! It makes sense though because one of the formulae for the bessel function of the first kind is an integral of nested sinusoids over \([0,\pi]\); and expanding these \(\sin^{\circ 1/2}(x)\), will result in something very similar.
I also think, this provides a cool interchange of variables.
Write:
\[
\sin^{\circ 1/2}(x) = \frac{1}{\sqrt{\pi}}\int_0^\infty \frac{\vartheta(t,x)}{\sqrt{t}}\,dt\\
\]
Where:
\[
\vartheta(t,x) = \sum_{n=0}^\infty \sin^{\circ n+1}(x) \frac{(-t)^n}{n!}\\
\]
Then; then we can interchange these integrals, so that:
\[
a_{k}(1/2) = \frac{1}{\sqrt{\pi}} \int_0^\infty \sum_{n=0}^\infty a_{k}(n+1)\frac{(-t)^nt^{-1/2}}{n!} \,dt\\
\]
where:
\[
a_k(s) = \frac{1}{2\pi} \int_{-\pi}^\pi \sin^{\circ s}(x) \sin(kx)\,dx\\
\]
I've been aware of this formula for a long time, but I never saw the relation to Bessel before!
Also, note--I've never been able to prove this expansion of \(\sin^{\circ 1/2}(x)\) is analytic. I only managed to show that it was \(C^{\infty}\) on \((-\pi,\pi)\). But I absolutely suspect it is analytic here. Maybe with closed forms of \(a_k\) we'll have better luck!!!! I'd be really interested to see if this Fourier transform converges analytically, or only \(C^\infty\). The sin iteration tends to be an odd ball out trying to get analycity out of. And showing uniform convergence of Fourier series in the complex plane is always a lot harder than it sounds (I think they've awarded like 4 fields medals for tiny step theorems on the convergence of Fourier series! lol!
)
