Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $w_k$ be a primitive k th root of unity, where k is a power of 2. In response to a question, Robert Israel gave the solution,

$$\sum_{n=0}^\infty \frac{(-1)^n}{\binom{kn}{kn/2}} = \frac{2^k}{2^k + 1} + \frac{1}{k} \sum_{j=0}^{k-1} \frac{\omega_k^{j+1/2} \arcsin(\omega_k^{j+1/2}/2)}{2 (1- \omega_k^{2j+1}/4)^{3/2}}$$

The sum is real, but one can see the terms are complex. Renzo Sprugnoli found a version for k = {2,4} in real terms as,

$\sum_{n=0}^\infty \frac{(-1)^n}{\binom {2n}n} = \frac{4}{5} - \frac{4\sqrt{5}}{25}\ln\left(\frac{1+\sqrt{5}}{2}\right)$

$\sum_{n=0}^\infty \frac{(-1)^n}{\binom {4n}{2n}} = \frac{16}{17}+\frac{4\sqrt{34}(-2+\sqrt{17})}{289\sqrt{-1+\sqrt{17}}}\arctan\left(\frac{\sqrt{2}}{\sqrt{-1+\sqrt{17}}}\right) -\frac{2\sqrt{34}(2+\sqrt{17})}{289\sqrt{1+\sqrt{17}}} \ln\left(\frac{\sqrt{1+\sqrt{17}}+\sqrt{2}}{\sqrt{1+\sqrt{17}}-\sqrt{2}}\right )$

I found that, given the Dedekind eta function $\eta(z)$ and,

$t_1 = \frac{1+\sqrt{-5}}{2}$

$t_2 = \frac{1+\sqrt{-17}}{2}$

$\zeta_{48} = \exp(\pi \rm i/24)$

then,

$$\frac{1}{2} \left(\frac{\zeta_{48}\eta(t_1)}{\eta(2t_1)}\right)^4 = \frac{1+\sqrt{5}}{2}$$

$$\frac{1}{2} \left(\frac{\zeta_{48}\eta(t_2)}{\eta(2t_2)}\right)^4 = \frac{\sqrt{1+\sqrt{17}}+\sqrt{2}}{\sqrt{1+\sqrt{17}}-\sqrt{2}}$$

Question: Is this coincidence? For the next case k = 8, how do we convert Israel's complex terms into real ones (akin to Sprugnoli's) to check if the argument of the natural log can be similarly expressed by an eta quotient?

share|improve this question

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.