deletions | additions       diff --git a/We_can_define_the_Fourier__.tex b/We_can_define_the_Fourier__.tex  index 0faf8c5..5125c77 100644  --- a/We_can_define_the_Fourier__.tex  +++ b/We_can_define_the_Fourier__.tex 
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The second term in this equation involving $(f+f_k)$ is negligible for our purposes, so only the first term is required.  We want to convert this into a more easily \href{http://math.stackexchange.com/a/1470809/95147}{solvable} form e.g.\  \begin{equation}\label{eq:fourier2}  H_k(f) = y_k' e^{i\phi_k-i\pi f \Delta t} \int_{z_1}^{z_2} \exp{\left[i\left( z^3 + \alpha z + \beta \right) \right]} {\rm d}z.  \end{equation}  Firstly, lets put the cubic in equation~\ref{eq:fourier} in the form  \begin{equation}  M = at^3 + bt^2 + ct + d,  \end{equation}  where $a=(\pi/3)\ddot{f}_k$, $b = \pi \dot{f}_k$, $c = 2\pi(f_k-f)$ and $d=0$ and make the substitution $t = Az+B$. To get the derivative ${\rm d}z$ we have  \begin{equation}  H_k(f) \frac{{\rm d}t}{{\rm d} z}  = y_k' e^{i\phi_k-i\pi f \Delta t} \int_{z_1}^{z_2} \exp{\left[i\left( z^3 + \alpha z + \beta \right) {\rm d}z \right]} A,  \end{equation}  so, ${\rm d}t = A{\rm d}z$ and therefore in equation~\ref{eq:fourier2} we have $y_k' = Ay_k$.  %This integral is not analytic, but integral of cubic trigonometric functions (through e.g.\ use of \href{http://docs.sympy.org/latest/index.html}{sympy}) can be written as  %\begin{equation}