#' @section Process Haar WV First Derivative:
#' Taking the derivative with respect to \eqn{\omega}{omega} yields:
#' \deqn{\frac{\partial }{{\partial \omega }}\nu _j^2\left( \omega \right) = \frac{{\tau _j^2\omega }}{8}}{d/domega nu[j]^2 (omega) = (tau[j]^2 * omega)/8}
#' \strong{Note:} We are taking the derivative with respect to \eqn{\omega}{omega} and not \eqn{\omega^2}{omega^2} as the \eqn{\omega}{omega}
#' relates to the slope of the process and not the processes variance like RW and WN. As a result, a second derivative exists and is not zero.
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