deletions | additions       diff --git a/We_can_define_the_Fourier__.tex b/We_can_define_the_Fourier__.tex  index 0b27990..0faf8c5 100644  --- a/We_can_define_the_Fourier__.tex  +++ b/We_can_define_the_Fourier__.tex 
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\end{align}  The second term in this equation involving $(f+f_k)$ is negligible for our purposes, so only the first term is required.  This integral is not analytic, but integral of cubic trigonometric functions (through We want to convert this into a more easily \href{http://math.stackexchange.com/a/1470809/95147}{solvable} form  e.g.\use of \href{http://docs.sympy.org/latest/index.html}{sympy}) can be written as  \begin{equation}  \int_x^y \cos{(z^3)} {\mathrm{d}}z H_k(f)  = \left( y\, {}_1F_2(1/6; 1/2, 7/6; -y^6/4) - x\, {}_1F_2(1/6; 1/2, 7/6; -x^6/4)\right) 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]}  \end{equation}  and  \begin{equation}  \int_x^y \sin{(z^3)} {\mathrm{d}}z = \frac{1}{4}\left( y^4\, {}_1F_2(2/3; 3/2, 5/3; -y^6/4) - x^4\, {}_1F_2(2/3; 3/2, 5/3; -x^6/4) \right),  \end{equation}  where ${}_1F_2$ is a \href{http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric1F2/06/ShowAll.html}{generalised hypergeometric function}. So, we need to convert our equation into an equivalent form.  To factor equation~\ref{eq:fourier} into a form that we can use with the hypergeometric %This integral is not analytic, but integral of cubic trigonometric  functions we want to have  \begin{equation}  z^3 + D = 2\pi\left((f_k-f)t + \frac{t^2}{2}\dot{f}_k + \frac{t^3}{6}\ddot{f}_k \right).  \end{equation}  If we (through e.g.\  use $z of \href{http://docs.sympy.org/latest/index.html}{sympy}) can be written as  %\begin{equation}  %\int_x^y \cos{(z^3)} {\mathrm{d}}z  = (A+B)t + C$, then $z^3 \left( y\, {}_1F_2(1/6; 1/2, 7/6; -y^6/4) - x\, {}_1F_2(1/6; 1/2, 7/6; -x^6/4)\right)  %\end{equation}  %and  %\begin{equation}  %\int_x^y \sin{(z^3)} {\mathrm{d}}z  = (A+B)^3t^3 + 3C(A+B)^2t^2 + 3C^2(A+B)t + C^3$. \frac{1}{4}\left( y^4\, {}_1F_2(2/3; 3/2, 5/3; -y^6/4) - x^4\, {}_1F_2(2/3; 3/2, 5/3; -x^6/4) \right),  %\end{equation}  %where ${}_1F_2$ is a \href{http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric1F2/06/ShowAll.html}{generalised hypergeometric function}. So, we need to convert our equation into an equivalent form.  Firstly, %To factor equation~\ref{eq:fourier} into a form that we can use with the hypergeometric functions we want to have  %\begin{equation}  %z^3 + D = 2\pi\left((f_k-f)t + \frac{t^2}{2}\dot{f}_k + \frac{t^3}{6}\ddot{f}_k \right).  %\end{equation}  %If we use $z = (A+B)t + C$, then $z^3 = (A+B)^3t^3 + 3C(A+B)^2t^2 + 3C^2(A+B)t + C^3$.  %Firstly,  we can factor it into sine and cosine parts to give \begin{align}  H_k(f) %\begin{align}  %H_k(f)  \approx & y_k e^{i\phi_k-i\pi f \Delta t} \Bigg( \int_{-\Delta t/2}^{\Delta t/2} \cos{\left[2\pi \left((f_k-f)t + \frac{t^2}{2}\dot{f}_k + \frac{t^3}{6}\ddot{f}_k \right) \right]} {\rm d}t \nonumber \\ & %&  + i\int_{-\Delta t/2}^{\Delta t/2} \sin{\left[2\pi \left((f_k-f)t + \frac{t^2}{2}\dot{f}_k + \frac{t^3}{6}\ddot{f}_k \right) \right]} {\rm d}t \Bigg). \end{align} %\end{align}