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1

If you really want to parse the output you can do this (including a check for validity ) Module[{m, n}, {n, m} = { First@Cases[# , Sqrt[1 + n_ k^2] :> n, {2}] , 1/#[[1]]} ; If[Simplify[(1 + Sqrt[1 + n k^2 ])/(m k) == #] , {n, m} ]] &@ myQuad[4, 3] This only works because your function seems to always return exactly the same ...


3

Consider the function $f(k) = (1 + \sqrt{1 + n k^2})/m$ (note that this is your expression times $k$). We have $$ f(0) = \frac{2}{m} $$ $$ f''(0) = \frac{n}{m} $$ So the following Mathematica code does the trick: (2 D[f, {k,2}] / f) /. k -> 0 In your case, you would need to define f = (your quantity)*k. Of course, this isn't really a solution by ...



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