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Playing with gamma functions by randomly inputting numbers to Wolfram Alpha, I got the following beautiful result

\begin{equation} \frac{\Gamma\left(\frac{3}{10}\right)\Gamma\left(\frac{4}{10}\right)}{\Gamma\left(\frac{2}{10}\right)}=\frac{\sqrt[\large5]{4}\cdot\sqrt{\pi}}{\phi} \end{equation}

where $\phi$ is golden ratio.

Could anyone here please help me to prove it by hand? I mean without using table for the specific values of $\Gamma(x)$ except for $\Gamma\left(\frac{1}{2}\right)$. As usual, preferably with elementary ways (high school methods)? Any help would be greatly appreciated. Thank you.

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    $\begingroup$ Assuming that it is true by adjusting something, I would bet that it follows from combining math.stackexchange.com/questions/374551/… with the duplication formula for the $\Gamma$ function, or from the Rogers-Ramanujan continued fraction. $\endgroup$ – Jack D'Aurizio Sep 2 '14 at 9:52
  • $\begingroup$ Sorry Mr. @FelixMarin, I've made a typo. Please see my edited OP. $\endgroup$ – Anastasiya-Romanova 秀 Sep 2 '14 at 10:19
  • $\begingroup$ @JackD'Aurizio I think we only need the Euler reflection formula and the Gauss multiplication formula $\endgroup$ – Anastasiya-Romanova 秀 Sep 2 '14 at 10:23
  • $\begingroup$ I imagined something like that. That's fine and interesting. $\endgroup$ – Felix Marin Sep 2 '14 at 10:26
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    $\begingroup$ Other similar formulas can be found here $($starting at formula #$63)$ and here. $\endgroup$ – Lucian Sep 2 '14 at 18:30
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We will show that

$$ \frac{\Gamma\left(\frac{3}{10}\right)\Gamma\left(\frac{4}{10}\right)}{\Gamma\left(\frac{2}{10}\right)}=\frac{\sqrt[\large5]{4}\cdot\sqrt{\pi}}{\phi}, $$

where $\phi$ is the golden ratio.

Because of the Gauss's multiplication formula we know that $$\Gamma(2z)=\frac{1}{\sqrt{\pi}}2^{2z-1}\Gamma(z)\Gamma\left(z+\frac{1}{2}\right).$$ Because $\Gamma$ is nowhere zero, we can divide the formula by $\Gamma(z)$, and we get $$\frac{\Gamma(2z)}{\Gamma(z)}=\frac{1}{\sqrt{\pi}}2^{2z-1}\Gamma\left(z+\frac{1}{2}\right).$$

We put $z:=2/10$ into the formula and get $$\color{red}{\frac{\Gamma\left(\frac{4}{10}\right)}{\Gamma\left(\frac{2}{10}\right)}}=\frac{1}{\sqrt{\pi}}2^{\frac{4}{10}-1}\color{blue}{\Gamma\left(\frac{7}{10}\right)}.$$

Now take a look at the Euler's reflection formula. $$\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(z\pi)}.$$

With $z:=3/10$ we get

$$\color{green}{\Gamma\left(\frac{3}{10}\right)}\color{blue}{\Gamma\left(\frac{7}{10}\right)}=\frac{\pi}{\sin\left(\frac{3}{10}\pi\right)}.$$

Putting this all together, we get

$$ \color{red}{\frac{\color{green}{\Gamma\left(\frac{3}{10}\right)}\Gamma\left(\frac{4}{10}\right)}{\Gamma\left(\frac{2}{10}\right)}}=\frac{1}{\sqrt{\pi}}2^{\frac{4}{10}-1}\frac{\pi}{\sin\left(\frac{3}{10}\pi\right)}.$$

Now we need that

$$\sin\left(\frac{3}{10}\pi\right)=\frac{1+\sqrt{5}}{4},$$

and of course we also know that $\pi / \sqrt{\pi} = \sqrt{\pi}$.

Using this we get

$$ \frac{\Gamma\left(\frac{3}{10}\right)\Gamma\left(\frac{4}{10}\right)}{\Gamma\left(\frac{2}{10}\right)}=2^{-\frac{3}{5}} \cdot \sqrt{\pi} \cdot\frac{4}{1+\sqrt{5}} = \frac{\sqrt[\large5]{4}\cdot\sqrt{\pi}}{\phi},$$ and this completes the proof.


One last fun fact, that we can generalize the problem like this

$$\frac{\Gamma\left(\frac{1}{2}-z\right)\Gamma(2z)}{\Gamma(z)}=\frac{2^{2z-1} \cdot \sqrt{\pi}}{\cos(z\pi)},$$

for all $z \notin -\mathbb{N}$ and $z \neq n-1/2, \ n \in \mathbb{Z}$. You can get your result by $z:=2/10.$

And really at last an other related formula is the following

$$\frac{\Gamma\left(\frac{2}{15}\right)\Gamma\left(\frac{7}{15}\right)}{\Gamma\left(\frac{1}{10}\right)}=\frac{\sqrt[10]{3} \cdot \sqrt[5]{2} \cdot \sqrt{\pi}}{\phi}.$$

This is even more interesting, because with the same idea you have to use the multiplication formula twice for $3z$ and also for $2z$ and after that the reflection formula. Because the equation $6z=z+2/3$ only has one solution, and it is $2/15$, that's why $2/15$ has a very important role in the formula, and this problem does not have a generalization like I gave above. I could imagine other ways to generalize, but it would not be nice.

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    $\begingroup$ Wow! Very detail. I like it! +1 $\endgroup$ – Anastasiya-Romanova 秀 Sep 2 '14 at 11:34
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    $\begingroup$ You're welcome. Nice result! $\endgroup$ – user153012 Sep 2 '14 at 11:49
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You need two standard results about the Gamma function: the Euler reflection formula $\Gamma(s)\Gamma(1-s)=\frac{\pi}{\sin\pi s}$ and the Legendre duplication formula $\Gamma(s)\Gamma(s+\frac{1}{2})=2^{1-2s}\sqrt{\pi}\Gamma(2s)$. If you start with the duplication formula with $s=0.2$ and then substitute for $\Gamma(0.7)$ from the Euler formula you get your result (except you also need to know that $\sin 0.7\pi=\frac{1+\sqrt{5}}{4}$.

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  • $\begingroup$ I thought$$\Gamma(x)\Gamma\left(x+\frac{1}{n}\right)\cdots\Gamma\left(x+ \frac{n-1}{n}\right)=n^{\large\frac{1}{2}-nx}(2\pi)^{\large\frac{n-1}{2}}\Gamma(nx)$$is the Gauss multiplication formula. +1 $\endgroup$ – Anastasiya-Romanova 秀 Sep 2 '14 at 10:53
  • $\begingroup$ Something went wrong with your TeX there. But try putting $n=2$ in that formula. $\endgroup$ – almagest Sep 2 '14 at 10:59
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In this answer, we derive Gauss's Multiplication Formula: $$ \prod_{k=0}^{n-1}\Gamma\left(x+\frac kn\right) =\sqrt{n2^{n-1}\pi^{n-1}}\frac{\Gamma(nx)}{n^{nx}}\tag{1} $$ In this answer, we derive Euler's Reflection Formula: $$ \Gamma(x)\Gamma(1-x)=\pi\csc(\pi x)\tag{2} $$ Multiply $(2)$ by $\frac{\Gamma\left(\frac8{10}\right)}{\Gamma\left(\frac8{10}\right)}$ to get $$ \begin{align} \frac{\Gamma\left(\frac3{10}\right)\Gamma\left(\frac4{10}\right)}{\Gamma\left(\frac2{10}\right)} &=\frac{\color{#C00000}{\Gamma\left(\frac3{10}\right)}\Gamma\left(\frac4{10}\right)\color{#C00000}{\Gamma\left(\frac8{10}\right)}}{\color{#00A000}{\Gamma\left(\frac2{10}\right)\Gamma\left(\frac8{10}\right)}}\tag{3a}\\ &=\frac{\color{#C00000}{2\sqrt{\pi}\frac{\Gamma\left(\frac35\right)}{2^{3/5}}}\Gamma\left(\frac25\right)}{\color{#00A000}{\pi\csc\left(\frac\pi5\right)}}\tag{3b}\\ &=\frac{2^{2/5}\sqrt\pi\,\pi\csc\left(\frac{2\pi}5\right)}{\pi\csc\left(\frac\pi5\right)}\tag{3c}\\ &=\frac{2^{2/5}\sqrt\pi}{2\cos\left(\frac\pi5\right)}\tag{3d} \end{align} $$ Explanation:
$\mathrm{(3a)}$: multiply numerator and denominator by $\Gamma\left(\frac8{10}\right)$
$\mathrm{(3b)}$: in red, apply $(1)$ with $x=\frac3{10}$ and $n=2$; in green, apply $(2)$ with $x=\frac15$
$\mathrm{(3c)}$: apply $(2)$ with $x=\frac25$
$\mathrm{(3d)}$: $\sin(2x)=2\sin(x)\cos(x)$

We can use the identity $\cos(5x)=16\cos^5(x)-20\cos^3(x)+5\cos(x)$ to get $$ \begin{align} -1&=16\cos^5\left(\frac\pi5\right)-20\cos^3\left(\frac\pi5\right)+5\cos\left(\frac\pi5\right)\\ 0&=16\cos^5\left(\frac\pi5\right)-20\cos^3\left(\frac\pi5\right)+5\cos\left(\frac\pi5\right)+1\\ &=\left[4\cos^2\left(\frac\pi5\right)-2\cos\left(\frac\pi5\right)-1\right]^2\left[\cos\left(\frac\pi5\right)+1\right]\tag{4} \end{align} $$ which implies, since $\cos\left(\frac\pi5\right)\gt0$, that $$ 2\cos\left(\frac\pi5\right)=\phi\tag{5} $$ Combining $(3)$ and $(5)$ yields $$ \frac{\Gamma\left(\frac3{10}\right)\Gamma\left(\frac4{10}\right)}{\Gamma\left(\frac2{10}\right)} =\frac{2^{2/5}\sqrt\pi}{\phi}\tag{6} $$

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  • $\begingroup$ Too bad Mr. Rob John, you're a bit late. If your answer posted first, I would accept it since it's very detail and well-organized. I can only give you +1 (>‿◠)✌ $\endgroup$ – Anastasiya-Romanova 秀 Sep 2 '14 at 15:16
  • $\begingroup$ @V-Moy: no problem. Everyone is free to vote and accept as they see fit. $\endgroup$ – robjohn Sep 2 '14 at 15:21
  • $\begingroup$ Anyway, I'll post an interesting problem about 5-10 minutes again, you may want to answer it. It's an integral problem $\endgroup$ – Anastasiya-Romanova 秀 Sep 2 '14 at 15:25
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    $\begingroup$ @V-Moy I'll vote +1 for this answer. I guess you can switch your 'accept option' whenever you like. I suggest ( for the future ) you wait a couple of days ( or maybe three ) before you set the green mark and meanwhile you can up vote if you like. Remember that it's a worldwide forum and some people is sleeping while others are over here. Thanks. $\endgroup$ – Felix Marin Sep 4 '14 at 6:08
  • $\begingroup$ OK Mr. @FelixMarin, I'll remember your suggestion (ô‿ô) $\endgroup$ – Anastasiya-Romanova 秀 Sep 4 '14 at 12:55

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