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I have this:

sol = a^2 + b^2 - c^2 + 2 (c^2 - a^2)*JacobiDN[Sqrt[c^2 - a^2]*x, m] ==
u /. {m -> 1}

And I want to find its inverse, i.e. express x in terms of u. How might I do that. I tried using

Solve[sol, u]

And it hangs on the computation...

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  • $\begingroup$ Hmmm, out of interest, what is the inverse function? $\endgroup$ Mar 24, 2023 at 21:00

1 Answer 1

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With m=1 dn is the sech =1/cosh

a^2 + b^2 - c^2 + 2 (-a^2 + c^2) Sech[Sqrt[-a^2 + c^2] x] == u

If your Solve[sol,u] hangs, its strange, because u is already explicit.

Solving for the inverse, you have to solve for x

In[3]:= 
Solve[sol, x]


Out[3]= {{x -> ConditionalExpression[(-ArcCosh[(2 (a^2 - c^2))/( a^2 + b^2 - c^2 - u)] +  2 I \[Pi] ConditionalExpression[1, \[Placeholder]])/    Sqrt[-a^2 + c^2],   ConditionalExpression[1, \[Placeholder]] \[Element] 
     Integers]}, {x ->  ConditionalExpression[( ArcCosh[(2 (a^2 - c^2))/(a^2 + b^2 - c^2 - u)] +  2 I \[Pi] ConditionalExpression[1, \[Placeholder]])/
    Sqrt[-a^2 + c^2],  ConditionalExpression[1, \[Placeholder]] \[Element] Integers]}}
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