I've got the expression:
$ x = \tanh^{-1}(p) - \sqrt{\frac{2}{3}} \tanh^{-1}\left( \sqrt{\frac{2}{3}} p\right) $
How can I invert this function so I have a function $p(x)$? I thought about using angle-sum formulas for $\tanh$, but this has lead me nowhere. If it helps with anything, I only care about the inverse in the range that $p \in (-1,1)$, where $x$ is real.
Is it even possible to invert this at all?
EDIT: I have tried;
using the angle formula $\tanh(\alpha-\beta)=\frac{\tanh(\alpha)-\tanh(\beta)}{1-\tanh(\alpha)\tanh(\beta)}$, and this leaves me with a function of $p$ to the power of $\sqrt{2/3}$, so this I can't use.
splitting the function apart into logarithms, but I get some $p$'s to the power of $\sqrt{2/3}$ again
putting the above into the power series $\sum_{n=1}^{\infty} \left( 1 - (\frac{2}{3})^n \right) \frac{p^{2n-1}}{2n-1}$, and I've tried to use reversion of series to get the inverse. I can't get this to work either.
Does any one have an alternate approach? I thought perhaps doing some kind of a "curve-fitting" in Maple, but I don't know if it is possible to get an exact answer with a method like this.
some context: I'm really just looking for the solution to the DE $\frac{dp}{dx}=(p^2-1)(2p^2-3)$.