Is it possible to simplify this expression? $$\frac{\displaystyle\Gamma\left(\frac{1}{10}\right)}{\displaystyle\Gamma\left(\frac{2}{15}\right)\ \Gamma\left(\frac{7}{15}\right)}$$ Is there a systematic way to check ratios of Gamma-functions like this for simplification possibility?
Asked
Active
Viewed 5,767 times
60
-
7Thanks to se.math user A. Walker, I recently became aware of the paper Expressions for the values of the Gamma Function, which gives values for $\Gamma(a/b)$ whenever $b$ divides 120. – MJD May 30 '13 at 01:28
-
See a lot of related material at MathOverflow: http://mathoverflow.net/questions/7616 – GEdgar Jun 16 '13 at 12:25
1 Answers
109
Amazingly, this can be greatly simplified. I'll state the result first:
$$\frac{\displaystyle\Gamma\left(\frac{1}{10}\right)}{\displaystyle\Gamma\left(\frac{2}{15}\right)\Gamma\left(\frac{7}{15}\right)} = \frac{\sqrt{5}+1}{3^{1/10} 2^{6/5} \sqrt{\pi}}$$
The result follows first from a version of Gauss's multiplication formula:
$$\displaystyle\Gamma(3 z) = \frac{1}{2 \pi} 3^{3 z-1/2} \Gamma(z) \Gamma\left(z+\frac13\right) \Gamma\left(z+\frac{2}{3}\right)$$
or, with $z=2/15$:
$$\Gamma\left(\frac{2}{15}\right)\Gamma\left(\frac{7}{15}\right) = 2 \pi \,3^{1/10} \frac{\displaystyle\Gamma\left(\frac{2}{5}\right)}{\displaystyle\Gamma\left(\frac{4}{5}\right)}$$
Now use the duplication formula
$$\Gamma(2 z) = \frac{1}{\sqrt{\pi}}\, 2^{2 z-1} \Gamma(z) \Gamma\left(z+\frac12\right)$$
or, with $z=2/5$:
$$\frac{\displaystyle\Gamma\left(\frac{2}{5}\right)}{\displaystyle\Gamma\left(\frac{4}{5}\right)} = \frac{\sqrt{\pi} \, 2^{1/5}}{\displaystyle\Gamma\left(\frac{9}{10}\right)}$$
Putting this all together, we get
$$\frac{\displaystyle\Gamma\left(\frac{1}{10}\right)}{\displaystyle\Gamma\left(\frac{2}{15}\right)\Gamma\left(\frac{7}{15}\right)} = \frac{\displaystyle\Gamma\left(\frac{1}{10}\right) \Gamma\left(\frac{9}{10}\right)}{\sqrt{\pi^3} \, 2^{6/5} \, 3^{1/10}}$$
And now, we may use the reflection formula:
$$\Gamma(z) \Gamma(1-z) = \frac{\pi}{\sin{\pi z}}$$
With $z=1/10$, and noting that
$$\sin{\left(\frac{\pi}{10}\right)} = \frac{\sqrt{5}-1}{4} = \frac{1}{\sqrt{5}+1}$$
the stated result follows. This has been verified numerically in Wolfram|Alpha.
Ron Gordon
- 138,521
-
3Wonderful! Wouldn't it look even better as $$\frac{\sqrt{5}+1}{\sqrt[10]{12288}\sqrt{\pi}}$$?
Maybe not...
– Kieren MacMillan Sep 17 '14 at 21:00 -
-
The last line is FALSE:
$$\sin\left(\frac{\pi}{10}\right) \neq \frac{1}{1 + \sqrt{5}}$$
– Enrico M. Jan 19 '16 at 20:52 -
2@KimPeek: Why...because you say so? I say it is true. I have analysis and numerics on my side. Prove me wrong. – Ron Gordon Jan 19 '16 at 20:53
-
$$\sin\left(\pi/10\right) = 0.587 \ldots$$
$$\frac{1}{1 + \sqrt{5}} = 0.309\ldots $$
– Enrico M. Jan 19 '16 at 20:55 -
-
2
-
uhm... ok I did it with my calculator and I repeated it with wolfram Alpha and they gave me two different results.. I think my calculator has some bug lol! Forgive me! – Enrico M. Jan 19 '16 at 20:58
-
lol this is SOOO strange!! I tried to compute
$$\sqrt{5}$$
with my calculator and it gives me
$$\sqrt{5} = 1.982283324\ldots$$
There is definitely something wrong....
– Enrico M. Jan 19 '16 at 20:59 -
@KimPeek: You need to develop instincts for this stuff rather than rely on a calculator. Know that, when $x$ is small, $\sin{x} \approx x$ in radians. Thus it should be plain that when $x = \pi/10$, $x$ and $\sin{x}$ should not be that far off. – Ron Gordon Jan 19 '16 at 21:00
-
1
-
1Anyway, this is awesome because then we can state that
$$\sin\left(\frac{\pi}{10}\right) = \frac{1}{2\phi_0}$$
Where $\phi_0$ is the golden ratio!! _
– Enrico M. Jan 19 '16 at 21:01 -
2Nice solution. I particular like that software like Mathematica is not able to simplify it directly. The expression can also be written $\frac{\phi }{\sqrt[10]{12}\sqrt{\pi}}$ where $\phi$ is the golden ratio. – Winther Jan 19 '16 at 21:02
-
4@Winther: lots of ways to express, but the important thing is that we can write this horrific-looking beast in terms of things a pocket calculator can produce. – Ron Gordon Jan 19 '16 at 21:03