Please consider the following infinite product series which I found by pure happenstance:

$$\frac{1+\sqrt{5}}{2}= e^{\pi/6} \prod_{k=1}^\infty \frac{1+e^{-5(2k-1)\pi}}{1+e^{-(2k-1)\pi}}$$

My question: Would this formula fit within, or be obtainable by, the theory of modular units (by Kubert and Lang)? I am entirely unfamiliar with that level of math, but I was told that the theory produces algebraic values for special infinite series products at CM-points.

Of note: The infinite product series (which adjusts the logarithmic spiral at $30$ degrees) has as its repeating term: $(1 + e^{-k\pi})$, with $k$ representing the sequence of all odd integers excluding those which are divisible by five $(1,3,7,9,11,13,17,19,21,23,27,\ldots)$.

Further, the same repeating term with $k$ representing all odd integers adjusts the logarithmic spiral at $7.5$ degrees to the fourth root of $2$.

And further, the same repeating term with $k$ representing only the odd integers that are divisible by $5$, adjusts the logarithmic spiral at $37.5$ degrees to the value of the product of the golden ratio with the fourth root of $2$. In this case, the first term of the series is so small that it is easy to see the close relation (at $e^{5\pi/24}$) with a calculator.

P.S. There doesn't seem to be an official resource for formula from infinite series, so I do not know if this equation has been recognized before or not. I found a good source containing many series formulae at: http://pi.physik.uni-bonn.de/~dieckman/InfProd/InfProd.html
And he was kind enough to place this formula on his list.

Thanks in advance for any comments that can explain the theory of modular units, or CM-points, in a simplified manner for me, or its relation to infinite product series such as this one.

  • 2
    $\begingroup$ Please consider reading this $\endgroup$ – Bman72 Sep 6 '14 at 11:33

It turns out that OP also posted this question on my blog page. And I provide the answer I posted there.

The formula can be expressed as $$\frac{1 + \sqrt{5}}{2} = e^{\pi/6}\prod_{k = 1}^{\infty}\frac{1 + e^{-5(2k - 1)\pi}}{1 + e^{-(2k - 1)\pi}} = \frac{2^{-1/4}e^{5\pi/24}}{2^{-1/4}e^{\pi/24}}\prod_{k = 1}^{\infty}\frac{1 + e^{-5(2k - 1)\pi}}{1 + e^{-(2k - 1)\pi}}$$ If we check the definition of Ramanujan's Class Invariant $$G_{n} = 2^{-1/4}e^{\pi\sqrt{n}/24}\prod_{k = 1}^{\infty}\left(1 + e^{-(2k - 1)\pi\sqrt{n}}\right)$$ then we find that the given formula says that $$\frac{1 + \sqrt{5}}{2} = \frac{G_{25}}{G_{1}}$$ I have already proved in my post that $G_{1} = 1, G_{25} = (1 + \sqrt{5})/2$. This establishes your formula.

I have given extensive details of the theory of Ramanujan's theta functions which is needed to understand the proofs in my blog posts. All of it is based on real-analysis / calculus and does not require the deep and complicated theory of Modular forms and Complex multiplication.

Update: Based on your comments below I think the formula $$e^{\pi/6} = 2^{3/8}\prod_{k = 1}^{\infty}(1 + e^{-2k\pi})$$ is wrong. Let us put $q = e^{-2\pi}$ then we have $$\begin{aligned}\prod_{k = 1}^{\infty}(1 + e^{-2k\pi}) &= \prod_{k = 1}^{\infty}(1 + q^{k})\\ &= \prod_{k = 1}^{\infty}\frac{1 - q^{2k}}{1 - q^{k}}\\ &= \prod_{k = 1}^{\infty}\frac{1}{1 - q^{2k - 1}}\\ &= \prod_{k = 1}^{\infty}\frac{1}{1 - e^{-2(2k - 1)\pi}}\\ &= \dfrac{1}{2^{1/4}e^{-\pi/12}g_{4}}\end{aligned}$$ We have the definition $$g_{n}= 2^{-1/4}e^{\pi\sqrt{n}/24}\prod_{k = 1}^{\infty}\left(1 - e^{-(2k - 1)\pi\sqrt{n}}\right)$$ so that $g_{4} = 1/g_{1} = (2\sqrt{2})^{1/12}$ which gives the product as $2^{-3/8}e^{\pi/12}$. Your formula should consist of $e^{\pi/12}$ and not $e^{\pi/6}$. Your second product $$e^{\pi/4} = \sqrt{2(1 + \sqrt{2})}\prod_{k = 1}^{\infty}\frac{1 + e^{-4k\pi}}{1 + e^{-2(2k - 1)\pi}}$$ (which is correct) can also be calculated in similar fashion and it is linked with Ramanujan class invariants discussed in my blog post. The idea is to put $q = e^{-\pi\sqrt{n}}$ for some suitable value of $n$ and link the products with the invariants $G_{n}$ or $g_{n}$.

  • $\begingroup$ Would Ramanujan's functions produce relations with the logarithmic function if the series progressed with even numbers (2,4,6,8,10...), as in the case with the relation of e^(pi/6) and 2^(3/8) $\endgroup$ – newby Sep 11 '14 at 10:18
  • $\begingroup$ Also, the relation of e^(pi/4) with the value [2*(2^.5 +1)]^.5 uses the series (2,6,10,14,18...) over the series (4,8,12,16,20...). Having read further through your blog, I believe that the first series would be G4, with n=4, so that the odds (1,3,5,7,9...) are all multiplied by 2. However, I am not sure how to achieve the second series which requires 'evens'. $\endgroup$ – newby Sep 11 '14 at 10:30
  • $\begingroup$ The first equation mentioned above is: e^(pi/6) = 2^(3/8) * PI [1 + e^(-2kpi)] with k = 1 to inf. and the second equation mentioned above is: e^(pi/4) = [2*(2^.5 + 1)]^.5 * PI [1 + e^(-4kpi)] / PI [1 + e^(-2(2k-1))] with k = 1 to inf. I am sorry for my poor formatting. Thanks for the help. $\endgroup$ – newby Sep 11 '14 at 10:41
  • $\begingroup$ @newby: you need to learn more about ramanujan theta functions (my blog may help to some extent). The even number series should be even easier to handle. Also you need to use latex for formatting otherwise very difficult to read. $\endgroup$ – Paramanand Singh Sep 11 '14 at 11:49
  • $\begingroup$ @newby: I have updated my post with more information related to your products. $\endgroup$ – Paramanand Singh Sep 11 '14 at 16:59

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