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Does the Gamma Function have an Inverse? (Is there an "arc-gamma" function?)

Where $\Gamma(x) = y... \Gamma^{-1}(y) = x\ (arc\Gamma(y)=x)$.

I've searched and found something called DiGamma Function, but when I substituted it didn't seem to be an inverse ("arc") but something else. I am not yet developed enough to understand haha...


Edit, I'd like to draw special attention to the comment below by u/G Cab, be sure to check it out, has a very useful answer to this question, but comments often go overlooked.

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    $\begingroup$ You can define an inverse function on any interval on which $\Gamma$ is monotone, say $x\geq1$. $\endgroup$ Sep 15, 2014 at 3:18
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    $\begingroup$ @YuvalFilmus but how do I evaluate it? $\endgroup$ Sep 15, 2014 at 3:29
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    $\begingroup$ That's a current question. There are many ways. You can use binary search, for example, using some good lower and upper bounds for $\Gamma$ which can be inverted explicitly. $\endgroup$ Sep 15, 2014 at 3:31
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    $\begingroup$ @YuvalFilmus Okay, so there is no like known integral or anything, it's all just iterating guesses until you get the number of decimal places you need. $\endgroup$ Sep 15, 2014 at 5:59
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    $\begingroup$ see this post and this paper by M:Uchiyama $\endgroup$
    – G Cab
    Sep 3, 2018 at 23:10

2 Answers 2

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Is there an arc-gamma function ?

Of course there is. Is it also expressible in terms of elementary functions ? No. $($I believe that this is what you were ultimately trying to ask$)$.

Sorry if "arc" is the wrong term.

It is! But hey, life's too short to be sorry. ;-)

I've searched and found something called DiGamma Function.

The digamma, trigamma, and polygamma functions are the derivatives of the $\Gamma$ function, not its inverse.

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    $\begingroup$ Thankyou for your very kind answer :D Okay, so the answer is that there is CURRENTLY no easily-expressible inverse-gamma function? $\endgroup$ Sep 15, 2014 at 5:58
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    $\begingroup$ @AlbertRenshaw: No. It can be determined $($though not by me$)$, using certain elements of abstract algebra $($which I never actually studied, but of whose existence I am aware$)$, whether a function possesses or not an inverse and/or antiderivative expressible as a finite combination of elements of a specific set of given functions. However, if the function is analytical $($and $\Gamma$ certainly is$)$, then the Lagrange inversion theorem can be employed to express that inverse as an infinite series. $\endgroup$
    – Lucian
    Sep 15, 2014 at 6:28
  • $\begingroup$ @Lucian Could you tell me where I could find a wikipedia page/pdf/the name/etc about this abstract algebra proof to show if a function possesses an inverse expressible as a finite combination of elements of a specific set of given functions? $\endgroup$
    – P-S.D
    Jun 11, 2017 at 10:07
  • $\begingroup$ @P-S.D: Unfortunately, I am not able to remember anymore. At a first glance, Lagrange's inversion theorem seems to come pretty close, but, in and of itself, appears rather useless. However, if one were to use it in conjunction either with his reversion theorem, or with Liouville's theorem, in the eventuality that the infinite power series given by the former might be shown to satisfy... (to be continued) $\endgroup$
    – Lucian
    Jun 12, 2017 at 4:24
  • $\begingroup$ @P-S.D: ...certain implicit functional and/or differential equations “attackable” by the latter two, then all these things taken together might ultimately amount to precisely just such a theorem. $\endgroup$
    – Lucian
    Jun 12, 2017 at 4:25
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Despite the inverse of gamma function is not expressible in terms of elementary functions, It seems to be possible for positive real numbers to start an algorithm from reversing Stirling approximation and using newton formula for roots of an equation, in order to converge to the argument of gamma function, given its result.

A friend of mine tested and so far, it seems to work for real numbers. But since the logarithm of a complex value is multivalued, this approximation would not be suitable for complex expression of gamma function.

$\Gamma{(n+1)}=n!$

$n!=\sqrt{2\pi n}(\frac{n}{e})^n$

$\ln{(n!)}=\frac{1}{2}\ln{(2\pi)}+\frac{1}{2}ln{(n)}+n\ln{(n)}-n$

$n_1=n_0-\frac{\frac{1}{2}\ln{(2\pi)}+\frac{1}{2}ln{(n_0)}+n_0\ln{(n_0)}-n_0-\ln{(n!)}}{-\ln{n_0}}$

$n_1=\frac{n_0+\ln{(\Gamma{(n+1)})}-\frac{1}{2}\ln{(2\pi)}}{\ln{n_0}}-\frac{1}{2}$

so applying this algorithm with a guess value for $x_0$ will converge quickly to the result of argument of gamma function, but only for positive real numbers and it remains an approximation especially if $n\leq2$.

Here is a racket algorithm my friend wrote to detect if a value is a factorial :

#lang racket
(require math/special-functions)
(require math/base)
(define (fact n) (gamma (+ 1 n)))
(define (invfact n)
  (define w (log n))
  (define c (/ (log (* 2 pi)) -2))
  (define (invfact-iter x)
    (let ((xn (improve x)))    
      (if (< (abs (- x xn)) 0.001)
          (check (exact-round xn))
          (invfact-iter xn))))
  (define (improve x) (- (/ (+ x w c) (log x)) 0.5))
  (define (check res)
    (if (= n (fact res))
        res
        "not a factorial"))
  (if (< n 3)
      n
      (invfact-iter 10)))

If you try it under racket with for example : (invfact 720) you will get 6 after very few iterations, but the algorithm detects it by rounding, its result being really around 6.00739...

So @Gary's reverse approximation (see in the comments) is good too, even probably better, I didn't check.

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  • $\begingroup$ This is awesome, thank you!! I haven't confirmed anything yet as it's been years since I worked on this problem but at first glance, great work! $\endgroup$ Jun 8, 2021 at 22:56
  • $\begingroup$ You're welcome, I slightly corrected my post regarding the fact the function is not defined for negative real numbers. $\endgroup$ Jun 10, 2021 at 8:41
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    $\begingroup$ @AlbertRenshaw See my answer here: math.stackexchange.com/q/461207 $\endgroup$
    – Gary
    Jun 10, 2021 at 9:36

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