Exact Inverse of (gamma function) = ? --------------> for example: Exact Inverse of (?x) = x²
Might we suggest the inverse of Stirling's Approximation? Assuming the isolation of the domain ( x ) of Stirling's Approximation for n! always has a fixed point x_k, for all independently chosen range values ( y ), this isolation is a function f(x) and if function f is composed infinitely for y is a member of the set of real numbers and x is a member of the set of real numbers, then the infinite composition is the inverse of Stirling's Approximation for n!
closest to an exact inverse is
https://reference.wolfram.com/language/ref/InverseGammaRegularized.html
which only gives exact results for some arguments, according to the documentation
thanks a lot. sometimes IGR doesn't match IG As you know. what should I do?
There isn't really anything I guess as it is multivalued as you can see...
--SH
It was just an example to summarize. surely it is invertible, isn't it?
well, it IS invertible. But like x^2 => sqrt(x), you have to decide what to do for multiple values:
should it return +3 or -3? when one asks for x=9. Well this is by default +3 in Mathematica, you can imagine that this 'choice' becomes much more tricky with the Gamma function...
What should InverseGamma[4] return? You have to decide which one to return; I see like 6 possible values in your plot...
thanks a lot. as you know ? is an extension of ! . for x Real ? 1 => x!=y => x=(x!)?¹ = x(y) = ?
I humbly thank you. what is the mathematical formula for your Numerical approach? e.g ?
It will just find the root of the the equation Gamma[x]==y, so no mathematical formula. It will find it exactly but not using 'elementary' mathematical functions.
The starting point of the root-finding (which is what I do), can be neatly done by solving the Stirling approximation:
x == n! ~~ (2 Pi n)^0.5 (n/E)^n = (2 Pi)^0.5 (n/e)^n n^0.5 = (2 Pi)^0.5 (n/e)^n (n/E)^0.5 E^0.5 = (2 Pi E)^0.5 (n/e)^(n+0.5)
which can be solved approximately using PowerLog (LambertW function). This gives a good starting point.
If you want a Taylor series, you can just simply use:
Normal[InverseSeries[Series[Gamma[x], {x, 4.0, 10}]]] Plot[%, {x, 1, 10}]
Where the expansion is around 4, with 10 terms.
Im sorry to have troubled you . Please Kindly find the files attached. is it exact?
no
please help me
?(x) = digamma function = ?'/ ? , ?? = (positive zero of ?) - 1 , ??= ??!
