Matt Pitkin edited Expanding_out_alpha_we_get__.tex  over 8 years ago

Commit id: b7bc925ac9629e3563cf4d5e7aadd8456daaff46

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\begin{align}  \beta &= -\frac{b}{3a}\left(\frac{b^2}{9a} - \frac{b^2}{3a} + c\right), \nonumber \\  &= \frac{b}{3a}\left(\frac{2b^2}{9a} - c\right), \nonumber \\  &= \frac{2\pi\dot{f_k}}{\ddot{f_k}}\left(\frac{\dot{f_k}^2}{3 \ddot{f_k}} - (f_k-f)\right) (f_k-f)\right).  \label{eq:beta} \end{align}  Putting this back into equation~\ref{eq:fourier2} gives  \begin{equation}  H_k(f) = y_k\left(\frac{\pi}{3}\ddot{f_k}\right)^{-1/3}e^{i\left(\phi_k-\pi f \Delta t + \beta \right)}\int_{z_1}^{z_2} \exp{\left[i\left(z^3 + \alpha z\right) \right]} {\rm d}z,  \end{equation}  where  \begin{align}  z_1 &= \left[\left(\frac{\pi}{3}\right)^{4/3}\dot{f_k}\ddot{f_k}^{1/3}-\frac{\Delta t}{2}\right]\left(\frac{\pi}{3}\right)^{1/3}, \nonumber \\  z_2 &= \left[\left(\frac{\pi}{3}\right)^{4/3}\dot{f_k}\ddot{f_k}^{1/3}+\frac{\Delta t}{2}\right]\left(\frac{\pi}{3}\right)^{1/3}.  \end{align}