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\subsection{minimising the Jensen-Shannon distance between $P$ and $dP_+ + dP_-$ and $\rho c d  dU$ and $dP_+ - dP_-$} The separation of the measured $dP$ and $dU$ can be done for any arbitrary value of $c$ usng the water hammer equations and the assumption that the forward and backward waves are additive. Thus for any value of $c$ (assuming that $\rho$ is a constant) we obtain distributions of $dP_+$ and $dP_-$. For noiseless measurements and the 'true' value of $c$, the distribution of the measured $dP$ should equal the sum of the distributions of $dP_+$ and $dP_-$ and the distribution of $\rho c dU$ should equal the sum of the distributions of $dP_+$ and $-dP_-$. For noisy measurements these distributions will never be equal. We argue, however, that the difference sum of the distances  between thedifferences in  these distribution distributions  will be minimum when we use the true value of $c$. This argument follows from looking at the extreme cases. For $c=0$, $dP_+ = dP_- = \frac{1}{2}dP$ and so the so the distance between the distribution of $dP$ and the sum of $dP_+$ and $dP_-$ is zero but the distance between the distribution of $\rho c dU$ and the distribution of $dP_+$ and $-dP_-$ is equal to $H(\rho c dU)^{1/2}$. At the other extreme $c=\infty$, $dP_+ = - dP_- = \frac{1}{2}\rho c dU$. Thus the distance between the distribution of $\rho c dU$ and the sum of $dP_+$ and $-dP_-$ is zero but the distance between the distribution of $dP$ and the distribution of sum of $dP_+$ and $dP_-$ is equal to $H(dP)^{1/2}$.