1
$\begingroup$

I'm trying to evaluate the Gamma function in Bessel's equation for $v=1/4$.

What is $\Gamma(m+5/4)$, or when solving, what is $\Gamma(1/4)$?

$\endgroup$
  • $\begingroup$ $$\Gamma\left(m+\dfrac{5}{4}\right)=(m+\dfrac{1}{4})\Gamma\left(m+\dfrac{1}{4}\right)$$ $\endgroup$ – Nosrati Sep 6 '17 at 4:55
  • $\begingroup$ what does the Γ(m+1/4) term equal? $\endgroup$ – watdoing Sep 6 '17 at 4:58
  • $\begingroup$ See general form $x\Gamma(x)=\Gamma(x+1)$ $\endgroup$ – Nosrati Sep 6 '17 at 5:00
  • $\begingroup$ What do you mean by "solve"? $\Gamma(1/4) \approx 3.625609909$. It is a transcendental constant, which can be related to some others (e.g. see MathWorld), but I don't think there's any expression for it that I'd call "simpler" than $\Gamma(1/4)$. $\endgroup$ – Robert Israel Sep 6 '17 at 5:11
2
$\begingroup$

$$\Gamma\left(m+\frac{1}{4}\right)=\Gamma\left(\frac{1}{4}\right)\prod _{k=1}^m\left(k-\frac{3}{4}\right)$$ $\Gamma\left(\frac{1}{4}\right)$ is a remarkable constant. $$\Gamma\left(\frac{1}{4}\right)\simeq 3.62560990822190831...$$ In the book "An Atlas of Functions",J.Spannier, K.B.Oldham, this constant is noted U and called the "ubiquitous constant". It appears throughout many relationships involving special functions. One most notable is : $$\Gamma\left(\frac{1}{4}\right)=2\sqrt{\sqrt{\pi}\text{K}\left(1/2\right)}$$ where K$(x)$ is the complete elliptic integral of the first kind.

An efficient iterative process to evaluate $\Gamma\left(\frac{1}{4}\right)$ is : $$\Gamma\left(\frac{1}{4}\right)=\text{Arithmetic-Geometric Mean of }1 \text{ and }\frac{1}{\sqrt{2}}$$

The arithmetic-geometric mean of two numbers $A_0\:,\:B_0$ is approached by repeating : $A_{k+1}=(A_k+B_k)/2$ and $B_{k+1}=\sqrt{A_kB_k}$.

$\endgroup$
  • 1
    $\begingroup$ What is called the "common mean" in this answer is more conventionally called the "arithmetic-geometric mean", or the AGM. $\endgroup$ – J. M. is a poor mathematician Sep 6 '17 at 6:33
  • $\begingroup$ Yes, "arithmetic-geometric mean" is more popular. I will change the notation in my answer. Thanks you. $\endgroup$ – JJacquelin Sep 6 '17 at 6:48
0
$\begingroup$

Some properties of that value are gathered in the associated wikipedia article.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.