5
$\begingroup$

When is $\Gamma'(x)=0$, given $x>0$?

There is only one value in this range which is $x≈1.461632$, but I can’t seem to get it in exact form.

I’ve tried taking the derivative of the definition of the Gamma function, but that doesn’t seem to get me very far.

Any help is appreciated.

$\endgroup$
4
  • 1
    $\begingroup$ $\psi(x)$ should have the same roots as $\Gamma'(x)$. Check out this page en.wikipedia.org/wiki/…. I don't believe there is a closed-form expression for the real root you want. $\endgroup$
    – Byte _
    Sep 21, 2021 at 15:25
  • $\begingroup$ Hello and welcome to math.stackexchange. To elaborate on the comment by @Byte, your question is equivalent to asking where the function $\psi(x) = \frac{\Gamma'(x)}{\Gamma(x)}$ (Digamma function) has a zero on the positive axis. A lot is known about this function - see the linked Wikipedia article. $\endgroup$
    – Hans Engler
    Sep 21, 2021 at 15:44
  • $\begingroup$ I agree with @Byte_ , you can find the root only numerically. $\endgroup$
    – Quillo
    Sep 21, 2021 at 19:19
  • $\begingroup$ Is there any reason that make you think that there's an explicit formula for that number? $\endgroup$
    – jjagmath
    Sep 22, 2021 at 2:23

1 Answer 1

Reset to default
2
$\begingroup$

You can obtain decent approximations of the root.

We have $$\Gamma'(x)=\Gamma (x)\, \psi (x)$$ which means that we look for the zero of $\psi (x)$ (the digamma function). The solution is "close" to $1.5$.

Developed as a series $$\psi (x)=(2-\gamma -2 \log (2))+\left(\frac{\pi ^2}{2}-4\right) \left(x-\frac{3}{2}\right)+(8-7 \zeta (3))\left(x-\frac{3}{2}\right)^2 +$$ $$\left(\frac{\pi ^4}{6}-16\right) \left(x-\frac{3}{2}\right)^3+ (32-31 \zeta (5))\left(x-\frac{3}{2}\right)^4+\left(\frac{\pi ^6}{15}-64\right) \left(x-\frac{3}{2}\right)^5+O\left(\left(x-\frac{3}{2}\right)^6\right)$$

Using series reversion $$x=\frac{3}{2}+t+\frac{2 (7 \zeta (3)-8)}{\pi ^2-8}t^2+$$ $$\frac{ \left(-2688 \zeta (3)+1176 \zeta (3)^2+768+96 \pi ^2+8 \pi ^4-\pi ^6\right)}{3 \left(\pi ^2-8\right)^2}t^3+O\left(t^4\right)$$ where $t=-\frac{2 (2-\gamma -2 \log (2))}{\pi ^2-8}$.

Using this very truncated series, an approximation of the root is $x=1.461632068$ while the solution given by Newton method is $x=1.461632145$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .