Matt Pitkin edited We_can_define_the_Fourier__.tex  over 8 years ago

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\end{equation}  so, ${\rm d}t = A{\rm d}z$ and therefore in equation~\ref{eq:fourier2} we have $y_k' = Ay_k$. If we make the substitution into equation~\ref{eq:subst} then we get  \begin{equation}\label{eq:Mexpanded}  M = A^3az^3 + A^2(3Ba + b)z^2 + A(3B^2a + 2Bb + c)z + B(B^2 B(B^2a  + Bb + c) + d. \end{equation}  If we equation the terms in equation~\ref{eq:Mexpanded} with those in exponential term within the integral in equation~\ref{eq:fourier2} we get  \begin{align} 

and  \begin{align}  \alpha &= A(3B^2a + 2Bb + c), \nonumber \\  \beta &= B(B^2 B(B^2a  + Bb + c) + d. \label{eq:secondterms} \end{align}  %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