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I am to solve for $r(\rho)$ given the function,

 \[Rho]Asymp[r_,b_,q_] := 1/(1 - q) Gamma[1/(1 - q)]/Gamma[(q - 2)/(q - 1)] r Sqrt[1 - (b/r)^(1 - q)]

This can be solved in a straightforward way by converting this into a quadratic equation and use Solve to find for the root corresponding to $r(\rho)$ for every value of $q$. But I have difficulty implementing the InverseFunction to this problem. Can someone help me with this? Also i want to scan $r(\rho)$ for all values of $q<1$, how can I implement this by using a Module? Thank you

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  • $\begingroup$ Recommend that you change your definition to \[Rho]Asymp[r_, b_, q_] := 1/(1 - q) Gamma[1/(1 - q)]/Gamma[(q - 2)/(q - 1)] r Sqrt[ 1 - (b/r)^(1 - q)] // FullSimplify // Evaluate which will be simplified to Sqrt[1 - (b/r)^(1 - q)]*r $\endgroup$ – Bob Hanlon Jan 15 at 15:55
  • $\begingroup$ I did not see that. Thank you for pointing out. Is there an efficient way of getting $r(\rho)$? for all allowed values of $q$? $\endgroup$ – user583893 Jan 16 at 2:35
  • $\begingroup$ Have you tried using FullSimplify on your function? The gamma functions all disappear and I get ρ = r*Sqrt[1-(b/r)^(1-q)]. Doesn't solve your problem though. $\endgroup$ – Roman Jun 14 at 9:13
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Looking at the examples in the documentaion you could do the following:

g = InverseFunction[Function[{r,b,q},\[Rho]Asymp[r,b,q]],1,3]

which gives the inverse of \[Rho]Asymp[r,b,q], a function of 3 arguments, with respect to it's first argument r. Evaluation is then

g[1,2,3]
(*out:*) Sqrt[2]

The function g may now be used as a usual function.

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  • $\begingroup$ What about symbolically? If i only specify, $b$ and $q$ $\endgroup$ – user583893 Jan 15 at 7:43
  • $\begingroup$ @user583893 sadly that doesn't work for this particular function and my suggested solution. $\endgroup$ – gothicVI Jan 15 at 8:12

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