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I read the book Mechanic of fluids shames and I find this relationship:

$\frac{1+kM_1^2}{1+kM_2^2} =\frac{M_1}{M_2} \left ( \frac{1+\dfrac{(k-1)}{2}M_1^2}{1+\dfrac{(k-1)}{2}M_2^2} \right )^{0.5}$

where $M_1$ is the Mach number of supersonic flow and $M_2$ is the Mach number for subsonic flow.

How can I find $M_2$ as a function of $M_1$, say $M_2 = f(M_1)$?

Sorry for my English.

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    Dear user, I tried to edit your post a little bit to try to make the equatio$n$ easier to read, please check if this is precisely what you want.2012-07-14

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A small amount of work will show that squaring the expression above gives:

$(m_1-m_2)(m_1+m_2)(2km_1^2 m_2^2-k (m_2^2+m_1^2)+m_1^2+m_2^2-2) = 0.$ Two solutions are obvious (and presumably uninteresting): $m_2 = \pm m_1$. The other two are also straightforward (assuming I haven't made a mistake): $m_2 = \pm \sqrt{\frac{1+\frac{k-1}{2} m_1^2}{k m_1^2+\frac{1-k}{2}}}.$ The original equation rules out the negative solutions.