今回は、ただひたすらに計算していくだけの回です。回転積分した反転公式

\begin{align} 8\pi^3f(z)=\int_0^{2\pi}\bigg\{\mathrm{pv.}\int_0^{2\pi}\dfrac{1+\cos(\theta-\tau)}{\sin(\theta-\tau)}\mathscr{H}f(\theta+\pi,z-re(\tau)+re(\theta))\,d\theta\bigg\}\,d\tau+2\pi\int_0^{2\pi}\psi(r,z-re(\tau))\,d\tau \end{align}

からスタートしましょう。右辺の第 $2$ 項は既に半径 $r$ のフィルタになっているので、第 $1$ 項(以下 $A$ とおきます)だけ注目していきます。

\begin{align} A&=\int_0^{2\pi}\bigg\{\mathrm{pv.}\int_{-\pi}^{\pi}\dfrac{1+\cos\theta}{\sin\theta}\mathscr{H}f(\theta+\tau+\pi,z-re(\tau)+re(\theta+\tau))\,d\theta\bigg\}\,d\tau\\ &=\int_0^{2\pi}\bigg\{\int_0^{\pi}\dfrac{1+\cos\theta}{\sin\theta}\{\mathscr{H}f(\theta+\tau+\pi,z-re(\tau)+re(\theta+\tau))\\ &\quad-\mathscr{H}f(-\theta+\tau+\pi,z-re(\tau)+re(-\theta+\tau))\}\,d\theta\bigg\}\,d\tau\\ &=\int_0^{\pi}\dfrac{1+\cos\theta}{\sin\theta}\bigg\{\int_0^{2\pi}\{\mathscr{H}f(\theta+\tau+\pi,z-re(\tau)+re(\theta+\tau))\\ &\quad-\mathscr{H}f(-\theta+\tau+\pi,z-re(\tau)+re(-\theta+\tau))\}\,d\tau\bigg\}\,d\theta \end{align}

最後の項を $\tau=\tau+\theta-\pi$ に置き換えて

\begin{align} A&=\int_0^{\pi}\dfrac{1+\cos\theta}{\sin\theta}\bigg\{\int_0^{2\pi}\{\mathscr{H}f(\theta+\tau+\pi,z-re(\tau)+re(\theta+\tau))\\ &\quad-\mathscr{H}f(\tau,z-re(\tau+\theta-\pi)+re(\tau-\pi))\}\,d\tau\bigg\}\,d\theta\\ &=\int_0^{\pi}\dfrac{1+\cos\theta}{\sin\theta}\bigg\{\int_0^{2\pi}\{\mathscr{H}f(\theta+\tau+\pi,z-re(\tau)+re(\theta+\tau))\\ &\quad-\mathscr{H}f(\tau,z+re(\tau+\theta)-re(\tau))\}\,d\tau\bigg\}\,d\theta \end{align}

$\mathscr{H}f$ の引数を書き換えます。

\begin{align} -e(\tau)+e(\theta+\tau)&=(-\cos\tau+\cos(\theta+\tau),-\sin\tau+\sin(\theta+\tau))\\ &=(-2\sin(\theta/2)\sin(\tau+\theta/2),2\sin(\theta/2)\cos(\tau+\theta/2))\\ &=2\sin(\theta/2)e(\tau+\theta/2+\pi/2) \end{align}

を利用して

\begin{align} A&=\int_0^{\pi}\dfrac{1+\cos\theta}{\sin\theta}\bigg\{\int_0^{2\pi}\{\mathscr{H}f(\theta+\tau+\pi,z+2r\sin(\theta/2)e(\tau+\theta/2+\pi/2))\\ &\quad-\mathscr{H}f(\tau,z+2r\sin(\theta/2)e(\tau+\theta/2+\pi/2))\}\,d\tau\bigg\}\,d\theta \end{align}

$\tau=\tau-\theta/2-\pi/2$ に置き換えて

\begin{align} A&=\int_0^{\pi}\dfrac{1+\cos\theta}{\sin\theta}\bigg\{\int_0^{2\pi}\{\mathscr{H}f(\theta/2+\tau+\pi/2,z+2r\sin(\theta/2)e(\tau))\\ &\quad-\mathscr{H}f(\tau-\theta/2-\pi/2,z+2r\sin(\theta/2)e(\tau))\}\,d\tau\bigg\}\,d\theta \end{align}

$\mathscr{H}f$ の差分を積分で書くと

\begin{align} A&=-\int_0^{\pi}\dfrac{1+\cos\theta}{\sin\theta}\bigg\{\int_0^{2\pi}\bigg\{\int_{-(\pi-\theta)/2}^{(\pi-\theta)/2}\partial_{\theta}\mathscr{H}f(\varphi+\tau-\pi,z+2r\sin(\theta/2)e(\tau))\,d\varphi\bigg\}\,d\tau\bigg\}\,d\theta \end{align}

$\partial_{\theta}\mathscr{H}f$ を

\begin{align} \dfrac{1}{r}\partial_{\theta}\mathscr{H}f(\theta,z)=(\partial_r-\partial_{e(\theta)})\psi(r,z+re(\theta)) \end{align}

を使って書き換えると

\begin{align} A&=-r\int_0^{\pi}\dfrac{1+\cos\theta}{\sin\theta}\bigg\{\int_0^{2\pi}\bigg\{\int_{-(\pi-\theta)/2}^{(\pi-\theta)/2}(\partial_r+\partial_{e(\varphi+\tau)})\psi(r,z-re(\varphi+\tau)+2r\sin(\theta/2)e(\tau))\,d\varphi\bigg\}\,d\tau\bigg\}\,d\theta \end{align}

この時点でフィルタサイズは半径 $r$ になっていますが、もう少し式を簡単にしていきます。$\tau$ だけ回転させる行列を $R_{\tau}$ として

\begin{align} &\nu_0=\dfrac{-e(\varphi)+2\sin(\theta/2)e(0)}{\sqrt{1-4\sin(\theta/2)\cos\varphi+4\sin^2(\theta/2)}},\\ &\nu=R_{\tau}\nu_0,\ \tilde{\nu}=R_{\tau+\pi/2}\nu_0 \end{align}

とおきましょう。

\begin{align} &\int_0^{2\pi}(\partial_r+\partial_{e(\varphi+\tau)})\psi(r,z-re(\varphi+\tau)+2r\sin(\theta/2)e(\tau))\,d\tau\\ &=\int_0^{2\pi}\partial_r\psi(r,z+r\sqrt{1-4\sin(\theta/2)\cos\varphi+4\sin^2(\theta/2)}R_{\tau}\nu_0)\,d\tau\\ &\quad+\int_0^{2\pi}\langle e(\varphi+\tau),\nu\rangle\partial_{\nu}\psi(r,z+r\sqrt{1-4\sin(\theta/2)\cos\varphi+4\sin^2(\theta/2)}R_{\tau}\nu_0)\,d\tau\\ &\quad+\int_0^{2\pi}\langle e(\varphi+\tau),\tilde{\nu}\rangle\partial_{\tilde{\nu}}\psi(r,z+r\sqrt{1-4\sin(\theta/2)\cos\varphi+4\sin^2(\theta/2)}R_{\tau}\nu_0)\,d\tau\\ &=\int_0^{2\pi}\partial_r\psi(r,z+r\sqrt{1-4\sin(\theta/2)\cos\varphi+4\sin^2(\theta/2)}R_{\tau}\nu_0)\,d\tau\\ &\quad+\int_0^{2\pi}\langle e(\varphi),\nu_0\rangle\partial_{\nu}\psi(r,z+r\sqrt{1-4\sin(\theta/2)\cos\varphi+4\sin^2(\theta/2)}R_{\tau}\nu_0)\,d\tau\\ &\quad+\int_0^{2\pi}\langle e(\varphi),R_{\pi/2}\nu_0\rangle\partial_{\tilde{\nu}}\psi(r,z+r\sqrt{1-4\sin(\theta/2)\cos\varphi+4\sin^2(\theta/2)}R_{\tau}\nu_0)\,d\tau\\ &=\int_0^{2\pi}\partial_r\psi(r,z+r\sqrt{1-4\sin(\theta/2)\cos\varphi+4\sin^2(\theta/2)}R_{\tau}\nu_0)\,d\tau\\ &\quad+\int_0^{2\pi}\dfrac{-1+2\sin(\theta/2)\cos\varphi}{\sqrt{1-4\sin(\theta/2)\cos\varphi+4\sin^2(\theta/2)}}\partial_{\nu}\psi(r,z+r\sqrt{1-4\sin(\theta/2)\cos\varphi+4\sin^2(\theta/2)}R_{\tau}\nu_0)\,d\tau\\ &\quad+\int_0^{2\pi}\dfrac{2\sin(\theta/2)\sin\varphi}{\sqrt{1-4\sin(\theta/2)\cos\varphi+4\sin^2(\theta/2)}}\partial_{\tilde{\nu}}\psi(r,z+r\sqrt{1-4\sin(\theta/2)\cos\varphi+4\sin^2(\theta/2)}R_{\tau}\nu_0)\,d\tau \end{align}

第 $1,2$ 項は $\varphi$ に関して偶関数、第 $3$ 項は奇関数となることに注意して

\begin{align} A&=-2r\int_0^{\pi}\dfrac{1+\cos\theta}{\sin\theta}\bigg\{\int_0^{2\pi}\bigg\{\int_0^{(\pi-\theta)/2}\partial_r\psi(r,z-re(\varphi+\tau)+2r\sin(\theta/2)e(\tau))\\ &\quad +\dfrac{-1+2\sin(\theta/2)\cos\varphi}{\sqrt{1-4\sin(\theta/2)\cos\varphi+4\sin^2(\theta/2)}}\partial_{\nu}\psi(r,z-re(\varphi+\tau)+2r\sin(\theta/2)e(\tau))\,d\varphi\bigg\}\,d\tau\bigg\}\,d\theta \end{align}

$R=2\sin(\theta/2),\rho=\cos\varphi$ と変数変換しましょう。

\begin{align} A&=-4r\int_0^2\dfrac{1}{R}\bigg\{\int_0^{2\pi}\bigg\{\int_{R/2}^1\dfrac{1}{\sqrt{1-\rho^2}}\partial_r\psi(r,z+r\sqrt{1-2R\rho+R^2}R_{\tau}\nu_0)\\ &\quad +\dfrac{1}{\sqrt{1-\rho^2}}\dfrac{-1+R\rho}{\sqrt{1-2R\rho+R^2}}\partial_{\nu}\psi(r,z+r\sqrt{1-2R\rho+R^2}R_{\tau}\nu_0)\,d\rho\bigg\}\,d\tau\bigg\}\,dR\\ &=-4r\int_0^2\dfrac{1}{R}\bigg\{\int_0^{2\pi}\bigg\{\int_{R/2}^1\dfrac{1}{\sqrt{1-\rho^2}}\partial_r\psi(r,z+r\sqrt{1-2R\rho+R^2}e(\tau))\\ &\quad +\dfrac{1}{\sqrt{1-\rho^2}}\dfrac{-1+R\rho}{\sqrt{1-2R\rho+R^2}}\partial_{e(\tau)}\psi(r,z+r\sqrt{1-2R\rho+R^2}e(\tau))\,d\rho\bigg\}\,d\tau\bigg\}\,dR \end{align}

$X=\sqrt{1-2R\rho+R^2}$ と変数変換しましょう。

\begin{align} A&=-4r\int_0^1\bigg\{\int_0^{2\pi}\bigg\{\int_{1-X}^{1+X}\dfrac{2X}{R\sqrt{(2X)^2-(R^2-1-X^2)^2}}\partial_r\psi(r,z+rXe(\tau))\\ &\quad +\dfrac{2X}{R\sqrt{(2X)^2-(R^2-1-X^2)^2}}\dfrac{R^2-1-X^2}{2X}\partial_{e(\tau)}\psi(r,z+rXe(\tau))\,dR\bigg\}\,d\tau\bigg\}\,dX \end{align}

$\sin\phi=(R^2-1-X^2)/(2X)$ 変数変換しましょう。

\begin{align} A&=-2r\int_0^1\bigg\{\int_0^{2\pi}\bigg\{\int_{-\pi/2}^{\pi/2}\dfrac{1}{\sin\phi+(1+X^2)/(2X)}\partial_r\psi(r,z+rXe(\tau))\\ &\quad +\dfrac{\sin\phi}{\sin\phi+(1+X^2)/(2X)}\partial_{e(\tau)}\psi(r,z+rXe(\tau))\,d\phi\bigg\}\,d\tau\bigg\}\,dX \end{align}

$\phi$ に関する積分は実行できそうですね。今回はここまで。次回も引き続きこの式を簡潔にしていきましょう。(written by J.Nishigata)

研究53- 最小な反転公式フィルタの導出