Gosper's identity for the golden ratio: $\frac{2^{2/5}\sqrt{5} \, \Gamma(1/5)^4}{\Gamma(1/10)^2 \,\Gamma(3/10)^2} = \phi$

Question

Towards the end of a talk by Knuth (one of his Christmas talks, maybe the one from 2017), he mentioned in passing the following identity communicated to him by Bill Gosper (without proof, IIRC):

$$\frac{2^{2/5}\sqrt{5} \, \Gamma(1/5)^4}{\Gamma(1/10)^2 \,\Gamma(3/10)^2} = \phi$$

where $\phi = (1 + \sqrt{5})/2 \approx 1.61803398874989$ is the golden ratio, and $\Gamma$ is the gamma function (extension of the factorial function).

Trying it numerically, e.g. with a computer algebra system like Sage, it seems to hold:

How could one prove this?

Answer 1

(As no one posted an answer, summarizing what I learned from the comments.)

Following up on the reference given by user @WhatsUp in the comments on the question, the Wikipedia article on “Particular values of the gamma function”, has in its section on products, as the last item under “Other rational relations include”, the relation

$$\frac{\Gamma\left(\frac{1}{5}\right)^2}{\Gamma\left(\frac{1}{10}\right)\Gamma\left(\frac{3}{10}\right)} = \frac{\sqrt{1+\sqrt{5}}}{2^{\tfrac{7}{10}}\sqrt[4]{5}}$$

which when squared would give

$$\frac{\Gamma(1/5)^4}{\Gamma(1/10)^2\Gamma(3/10)^2} = \frac{1+\sqrt{5}}{2^{7/5}\sqrt{5}}$$

which is what the question asks for.

The reference given on Wikipedia for this relation, namely the paper “Expressions for values of the gamma function” by Raimundas Vidūnas (seems to be published in Kyushu Journal of Mathematics, 2005, Vol 59, pp 267–283; DOI), does not seem to have this relation exactly.

(It's an interesting question how the relation got onto Wikipedia then. Looking at the edit history of the Wikipedia page, it appears that in February 2018 someone added it to the Wikipedia page, citing Knuth's 2017 lecture, as I did in the question here. The edit was immediately undone less than an hour later. Then a week later the user “WorstUsernameEver” restored it as a conjecture, and the next day found that it can be obtained by multiplying together two formulae from Vidūnas's paper. As I was typing below...)

However, the paper contains (section 2, page 3), among others, the expressions:

$$Γ(1/10) = \frac{\sqrt \phi}{\sqrt\pi 2^{7/10}} Γ(1/5)\,Γ(2/5) \tag 1$$

and

$$Γ(3/10) = \frac{\sqrt \pi \phi^⋆}{2^{3/5}\sqrt5} Γ(1/5)\,Γ(2/5)^{−1} \tag 2$$

where $\phi = 5 + \sqrt5$ and $\phi^⋆ = 5 - \sqrt{5}$.

Multiplying these two together gives

$$Γ(1/10)Γ(3/10) = \frac{\sqrt{5+\sqrt{5}}(5-\sqrt{5})}{2^{13/10}\sqrt5}Γ(1/5)^2$$

and so

$$\frac{Γ(1/5)^2}{Γ(1/10)Γ(3/10)} = \frac{2^{13/10}\sqrt5}{\sqrt{5+\sqrt{5}}(5-\sqrt{5})} \stackrel{?}{=} \frac{\sqrt{1+\sqrt{5}}}{2^{7/10}\sqrt[4]{5}}$$

and the algebra seems to work out.

(TODO: Understand the paper enough to prove $(1)$ and $(2)$.)

Answer 2

The identity can be proven by using two classic tools: the Legendre duplication formula $${\Gamma(2z)\over\Gamma(z)} ={\Gamma(z+1/2)\over2^{1-2z}\sqrt\pi},$$ and Euler’s reflection formula $$\Gamma(z)\Gamma(1-z) ={\pi\over\sin(\pi z)}.$$ Let's begin with some housekeeping. Rearranging the original identity, it suffices to prove that $${\Gamma(1/10)\over\Gamma(1/5)}{\Gamma(3/10)\over\Gamma(1/5)} ={2^{1/5}5^{1/4}\over\sqrt\phi},$$ where $\phi=(1+\sqrt5)/2$ is the golden ratio. The first fraction on the left yields immediately to the Legendre duplication formula: $${\Gamma(1/10)\over\Gamma(1/5)} ={2^{4/5}\sqrt\pi\over\Gamma(3/5)}.$$ The second fraction needs a little coaxing — we rewrite it a little, apply the Legendre duplication formula, then Euler’s reflection formula: $$\begin{align*} {\Gamma(3/10)\over\Gamma(1/5)} &={\Gamma(3/10)\over\Gamma(2/5)}{\Gamma(2/5)\over\Gamma(1/5)} \\ &={\Gamma(3/10)\over\Gamma(2/5)}{\Gamma(7/10)\over2^{3/5}\sqrt\pi} \\ &={1\over\Gamma(2/5)2^{3/5}\sqrt\pi}{\pi\over\sin(3\pi/10)}. \end{align*}$$ Since $\sin(3\pi/10)=\phi/2$, we deduce that $${\Gamma(3/10)\over\Gamma(1/5)} ={2^{2/5}\sqrt\pi\over\Gamma(2/5)\phi}.$$ Putting it all together now, with one more application of Euler’s reflection formula: $$\begin{align*} {\Gamma(1/10)\over\Gamma(1/5)}{\Gamma(3/10)\over\Gamma(1/5)} &={2^{4/5}\sqrt\pi\over\Gamma(3/5)}{2^{2/5}\sqrt\pi\over\Gamma(2/5)\phi} \\ &={2^{6/5}\pi\over\phi}{\sin(2\pi/5)\over\pi} \end{align*}$$ Conveniently, we have $\sin(2\pi/5)=5^{1/4}\sqrt\phi/2$, and so we are done.

Appendix: How do you compute weird sine values?

Firstly, one can verify $$\sin(5x)=16\sin^5x-20\sin^3x+5\sin x,$$ by using, say, de Moivre’s formula. Setting $x=\pi/10$ and writing $y=\sin x$, we get $$16y^5-20y^3+5y-1=0.$$ Now the coefficients add to zero, which tells us that $y=1$ is a root, so we may factor out $(y-1)$ to get $$16y^4+16y^3-4y^2-4y+1=0.$$ This turns out to be $$(4y^2+2y-1)^2=0,$$ and so $$y=\sin\Bigl({\pi\over10}\Bigr)={\sqrt5-1\over4}$$ (we reject the negative solution to the quadratic, as $\sin(\pi/10)$ must be positive).

From here, it is easy to obtain the values of sine that we used earlier. We have $$\begin{align*} \sin\Bigl({3\pi\over10}\Bigr) &=3\sin\Bigl({\pi\over10}\Bigr)-4\sin^3\Bigl({\pi\over10}\Bigr) \\ &=3\Bigl({\sqrt5-1\over4}\Bigr)-4\Bigl({\sqrt5-1\over4}\Bigr)^3 \\ &={3\sqrt5\over4}-{3\over4}-{5\sqrt5-15+3\sqrt5-1\over16} \\ &={1\over4}+{\sqrt5\over4} \\ &=\phi/2. \end{align*}$$ Similarly, we have $$\sin\Bigl({2\pi\over5}\Bigr) =\sin\Bigl({\pi\over2}-{\pi\over10}\Bigr) =\sin\Bigl({\pi\over2}\Bigr)\cos\Bigl({\pi\over10}\Bigr) -\cos\Bigl({\pi\over2}\Bigr)\sin\Bigl({\pi\over10}\Bigr) =\cos\Bigl({\pi\over10}\Bigr),$$ and so $$\begin{align*} \sin\Bigl({2\pi\over5}\Bigr) =\cos\Bigl({\pi\over10}\Bigr) &=\sqrt{1-\sin^2\Bigl({\pi\over10}\Bigr)} \\ &=\sqrt{1-\Bigl({\sqrt5-1\over4}\Bigr)^2} \\ &=\sqrt{1-{5-2\sqrt5+1\over16}} \\ &=\sqrt{5+\sqrt5\over8} \\ &={1\over2}\sqrt{\sqrt{5}(1+\sqrt5)\over2} \\ &=5^{1/4}\sqrt\phi/2. \end{align*}$$