Can anyone confirm the following values of the $\eta $ function to increase the table of the post What is the exact value of $\eta(6i)$? ?

$\eta(9i)$ = $\frac{1} {6} \big(\sqrt{6}\, (2+\sqrt{3})^{1/6} -3 \big)^{1/3} \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$

$\eta(10i) = \frac {1} {2^{11/8} \sqrt{5} \varphi^{1/2}} \frac {5^{1/4}-1} {\sqrt{2}} \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$

where $\varphi$ is golden ratio.

$\eta(12i) = \frac {1} {2^{31/16} 3^{3/8}} (2+\sqrt{3})^{5/48} (\sqrt{2}-3^{1/4})^{3/8} (\sqrt{2}-1)^{1/4} (\sqrt{3}-\sqrt{2})^{1/4} (3^{1/4}-1)^{1/2} \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$.

$\eta(14i)=\frac{1} {2^{11/4} 7^{7/16}} \big(\sqrt{\sqrt{3\sqrt{7}-7}+\sqrt{5-\sqrt{7}}}-\sqrt{\sqrt{27\sqrt{7}-7}-\sqrt{7\sqrt{7}+21}}\big) \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$

$\eta(15i)=\frac{1} {4 \sqrt{10}* 3^{3/8}}(\sqrt{5}-2)^{1/2}(2-\sqrt{3})^{11/12} \big(\frac{\sqrt{4+\sqrt{15}}-15^{1/4}} {2} \big)^{2} (540^{1/4}+60^{1/4}+2) \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$

$\eta(18i)=\frac{1} {2^{91/72} 3} \frac{\big(1-(2. 108^{1/4}-2\sqrt{3}-2)^{1/3}\big)^{1/3}} {\big(( 3.12^{1/4}+108^{1/4}+2\sqrt{3}+4)^{1/3}+2^{1/3}\big)^{1/3}} \big((6\sqrt{3}+18)^{1/3}-3\big)^{1/3} \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$

$\eta(20i)=\frac{1} {2^{29/16}.\sqrt{5}} (\sqrt{2}-1)^{1/2} (5^{1/4}- \sqrt{2})^{1/2} (\sqrt{10}-3)^{1/4} \big(\frac{5^{1/4}-1} {\sqrt{2}}\big)^{3/2} \varphi^{1/4} \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$

where $\varphi$ is golden ratio.

$\eta(21i)=\frac{1} {2^{11/8} 7^{7/16}\sqrt{3}} \frac{z a} {b c^{2} d e^{1/3}} \big(1+2\sqrt{2} \frac{b^{3/2} d^{3/2} e^{1/2}} {a^{3/2} c^{6}} \big)^{1/4}\frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$




$c=\frac{\sqrt{4+\sqrt{7}}+7^{1/4}} {2}$

$d=\frac{\sqrt{7}+\sqrt{3}} {2}$



$\eta(25i)=\frac{1} {40 \varphi^{10}} \big( 1+(4\varphi)^{1/5} \big( (3+\frac{5^{1/4}} { \varphi^{3/2}} )^{1/5} + (3-\frac{5^{1/4}} { \varphi^{3/2}} )^{1/5} )\big)\big). \big(1+\varphi^{3} \big( 1-(4/\varphi)^{1/5} \big( (3+5^{1/4} \varphi^{3/2})^{1/5} + (3-5^{1/4}\varphi^{3/2})^{1/5} \big) \big)^{2}\big)^{2} \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$

where $\varphi$ is golden ratio.

$\eta(27i)=\frac{ ( k-3\sqrt{3} )^{1/9}} {2^{35/36} 3^{95/72} }(\frac{\sqrt{3}+1} {2})^{1/18}\big(k-3^{5/6}(k^{2}-3\sqrt{3}k+9)^{1/3}\big)^{1/3}\frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$



$\eta(30i)=\frac{1} {8.\sqrt{5}.3^{3/8}}\big(\frac{\sqrt{4+\sqrt{15}}-15^{1/4}} {2} \big)^{4} \frac{(540^{1/4}+60^{1/4}+2)} {(2+\sqrt{3})^{19/12}.\varphi^{7/2}}\frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$

where $\varphi$ is golden ratio.

$\eta(32i)=\frac{1} {8.2^{33/128}}\frac{ (2^{1/4}-1)^{1/8} (\sqrt{1+\sqrt{2}} – 2^{5/8} )^{5/4} } { ( \sqrt{2}+1)^{1/32} (2^{1/4}+1+2^{13/16} (\sqrt{2}+1)^{1/4} )^{1/2} } \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$

$\eta(35i)= \frac{\sqrt { \sqrt{7+3 \sqrt{7}} + \sqrt{13+ \sqrt{7}}}} {\sqrt{5} . 2^{11/8}.7^{7/16} \varphi^{2}. \sqrt{b}. c^{3}. d^{2}} \sqrt{1+2 \frac{ a b ^{1/4} d} {c^{7/2} } } \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$

where $\varphi$ is golden ratio


$$c=\frac{\sqrt{4+\sqrt{7}}+7^{1/4}} {2}$$

$$d=\sqrt{\frac{43+15 \sqrt{7}+(8+3\sqrt{7})\sqrt{10 \sqrt{7}}} {8}}+\sqrt{\frac{35+15 \sqrt{7}+(8+3\sqrt{7}) \sqrt{10 \sqrt{7}}} {8}}$$.

$\eta(45i)=\frac{1} {12 \sqrt{5}} (\frac{\sqrt{5}-1} {2})^{5/2} (3+\sqrt{5}+(\sqrt{3}+\sqrt{5}+60^{1/4}) (2+\sqrt{3})^{1/3}\big(\frac{\sqrt{2}(\frac{\sqrt{5}+1} {2})^{2} (2-\sqrt{3})^{1/3} \frac{\sqrt{4+\sqrt{15}}-15^{1/4}} {2}-1} {\sqrt{2}(\frac{\sqrt{5}-1} {2})^{2} (2+\sqrt{3})^{1/3} \frac{\sqrt{4+\sqrt{15}}+15^{1/4}} {2}+1}\big)^{2/3}\frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$

$\eta(63i)=\frac{\sqrt{\sqrt{5-\sqrt{7}}-\sqrt{3\sqrt{7}-7}}} {2^{13/8}.3.7^{7/16}} \big( \frac{ 2 ( \sqrt{3+\sqrt{7}}-252^{1/4}) (\frac{\sqrt{4+\sqrt{7}}+7^{1/4}} {2})^{4}} {\sqrt{3+\sqrt{7}}+252^{1/4}) (\frac{\sqrt{7}+\sqrt{3}} {2}) (\frac{\sqrt{3}+1} {\sqrt{2}})^{2/3}} + \frac{\sqrt{2}( \sqrt{3+\sqrt{7}}+252^{1/4}) (\frac{\sqrt{7}+\sqrt{3}} {2})^{1/2} (\frac{\sqrt{3}+1} {\sqrt{2}})^{1/3}} {( \sqrt{3+\sqrt{7}}-252^{1/4}) (\frac{\sqrt{4+\sqrt{7}}+7^{1/4}} {2})^{2}}-3 \big)^{1/3} \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$

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    $\begingroup$ A genereric method is to set $f_n(z)=\Delta(nz)/\Delta(z)\in \mathbb{C}(X_0(n))$ then $\prod_{ad=n,b\bmod n} (Y-f_n(\frac{az+b}{dn}))=g_n(1/j(z),Y)$ with $g_n\in \mathbb{Z}[X,Y] $ computable from the Fourier expansions. Then from $j(i) = 1728$ you have a polynomial whose $1/f_n(i)$ is a root, from which you can find the minimal polynomial of $f_n(i)^{1/24} = \eta(ni)/\eta(i)$ and since $\mathbb{Q}(i,f_n(i)) \subset \mathbb{Q}(i,j(i),j(ni))/\mathbb{Q}(i)$ is abelian $\mathbb{Q}(f_n(i)^{1/24})/\mathbb{Q}$ is radical and you obtain the expressions you mentioned $\endgroup$ – reuns Feb 13 at 4:40
  • $\begingroup$ My intent is only to have "physically" the radical, written in unit factors, which generates the value of Dedekind's function, as it is aesthetically appealing, beautiful! Anyway thank you so much! $\endgroup$ – giuseppe mancò Feb 13 at 10:43
  • $\begingroup$ How did you arrive at these values in radical form? I think the desired calculations would be formidable to do by hand (unless one is Ramanujan). $\endgroup$ – Paramanand Singh Feb 17 at 8:57
  • $\begingroup$ You're very kind! Thank you $\endgroup$ – giuseppe mancò Feb 17 at 10:06

Using this answer one can easily verify the value of $\eta(9i)$.

We have by definition $$\eta(9i)=e^{-3\pi/4}\prod_{n=1}^{\infty} (1-e^{-18n\pi})$$ And using the answer linked above we can see that $$\eta(9i)=\frac{\sqrt[3]{\sqrt[3]{18+6\sqrt{3}}-3}}{6}\cdot\frac{\Gamma (1/4)}{\pi^{3/4}}$$ One can use a little bit of algebra to verify that $$\sqrt{6}(2+\sqrt{3})^{1/6}=\sqrt[3]{18+6\sqrt{3}}$$ and get the value of $\eta(9i)$ mentioned in your question.

The modular equation given in the linked answer can be used to evaluate $\eta(27i)$ given the values of $\eta(3i),\eta(9i)$ and in general one can get the values of $\eta(3^ni)$ in similar fashion. Using $\eta(2i),\eta(6i)$ one can also verify the value of $\eta(18i)$. You should use the value of $\eta(7i)$ (given in linked question in your post) and $\eta(63i)$ of your post together with Ramanujan's modular equation to get the value of $\eta(21i)$ and add it to your table.

Remaining set of values of eta function in your question require more effort to verify.

  • $\begingroup$ Thanks for the suggestions, now I try to calculate $ \eta (21i) $. I just posted $ \eta (32i) $. $\endgroup$ – giuseppe mancò Feb 17 at 10:31

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