Additional values of Dedekind's $\eta$ function in radical form

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.

$$\eta(11i)=\frac{1}{2^{1/12}*3^{1/4}*11^{11/24}}\Big(4*22^{1/3}-(306 \sqrt{33}-837\sqrt{3}-351\sqrt{11}+1490)^{1/3}-(-306 \sqrt{33}+837\sqrt{3}-351\sqrt{11}+1490)^{1/3}\Big)^{1/4}\frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$$

A modular equation of 11th degree of Dedekind's $\eta$ function.

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

$$\eta(13i)=\frac{1} {2 \sqrt{3} \sqrt{13} } \sqrt{ -5- (15 \sqrt{39}+39 \sqrt{3}-18 \sqrt{13}-91 )^{1/3}+ (15 \sqrt{39}+39 \sqrt{3}+18 \sqrt{13}+91)^{1/3}} \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$$ A modular equation of 13th degree of Dedekind’s $\eta$ function.

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

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

$$\eta(17i)=\frac{1}{4\sqrt{34}} \sqrt{272^{1/8}\big(\sqrt{61-7\sqrt{17}}-\sqrt{5\sqrt{17}+17}\big)-17^{3/4}+3\sqrt{17}-17^{1/4}-1} \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$$

A modular equation of 17th degree of Dedekind’s $\eta$ function. $$\eta(18i)=\frac{1} {2^{91/72} 3} \frac{\big(1-(2. 108^{1/4}-2\sqrt{3}-2)^{1/3}\big)^{1/3}} {\big(( 3.12^{1/4}+108^{1/4}+2\sqrt{3}+4)^{1/3}+2^{1/3}\big)^{1/3}} \big((6\sqrt{3}+18)^{1/3}-3\big)^{1/3} \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$$

$$\eta(18i) =\frac{1} {2^{11/8}*3^{2/3}}\Big(\big(-\frac{5*12^{1/4}}{6}+\frac{7*\sqrt{3}}{9}+\frac{108^{1/4}}{6}+\frac{2}{3}\big)^{1/3}-1\Big)^{1/3} \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$$

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

$$\eta(20i)≈\frac{1} {2^{9/4}.\sqrt{5}} (\frac{\varphi-5^{1/4}}{\varphi^{5/2}}) \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$$

where $$\varphi$$ is golden ratio.

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

$$a=\sqrt{3+\sqrt{7}}-252^{1/8}$$

$$b==\sqrt{3+\sqrt{7}}+252^{1/8}$$

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

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

$$e=2+\sqrt{3}$$

$$z=\sqrt{\sqrt{13+\sqrt{7}}+\sqrt{7+3\sqrt{7}}}$$.

$$\eta(22i)=\frac{(a-b-c)^{1/4}} {2^{35/24}*3^{1/4}*11^{11/24}*\sqrt{G}}*\big(G^{12}-\sqrt{G^{24}-1}\big)^{1/8}\frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$$

$$G=\frac{r(s+t)+2}{3*\sqrt{2}};$$ $$r=(3*\sqrt{11}+11)^{1/3};$$ $$s=(3*\sqrt{11}+3*\sqrt{3}+4)^{1/3};$$ $$t=(3*\sqrt{11}-3*\sqrt{3}+4)^{1/3};$$ $$a=4*22^{1/3};$$ $$b=(306*\sqrt{33}-837*\sqrt{3}-351*\sqrt{11}+1490)^{1/3};$$ $$c=(-306*\sqrt{33}+837*\sqrt{3}-351*\sqrt{11}+1490)^{1/3};$$

$$\eta(24i) =\frac{\sqrt{d}*c^{1/4}*e^{1/8}*f^{1/4}} {2^{75/32}*3^{3/8}*a^{3/8}*b^{1/12}}*\sqrt{b^{1/16}*c^{3/8}-\sqrt{2a*f}*e^{1/4}}\phi^{1/4} \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$$

$$a=\sqrt{2}-1; b=2-\sqrt{3}; c=\sqrt{2}-3^{1/4}; d=2^{1/4}-1; e=\sqrt{3}-\sqrt{2}; f=3^{1/4}-1$$

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

where $$\varphi$$ is golden ratio.

$$\eta(26i)=\frac{\sqrt{-A+B-5}} {\sqrt{39}*2^{11/8}*\sqrt{G}}*\Big(G^{12}-\sqrt{G^{24}-1}\Big)^{1/8}*\frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$$

$$G=\frac{\sqrt{a+b+26}+\sqrt{a+b+14}} {2*\sqrt{3}};$$ $$a=(10673-936*\sqrt{3})^{1/3};$$

$$b=(10673+936*\sqrt{3})^{1/3};$$ $$A=(15*\sqrt{39}+39*\sqrt{3}-18*\sqrt{13}-91)^{1/3};$$ $$B=(15*\sqrt{39}+39*\sqrt{3}+18*\sqrt{13}+91)^{1/3}.$$

$$\eta(27i)=\frac{(\sqrt{3}-1)^{1/6}} {2^{13/12}*3^{95/72}}\frac{\Big(-3^{5/6}\big((48\sqrt{3}-72)^{1/3}+(16\sqrt{3}-16)^{1/3}+3\sqrt{3}-3\big)^{1/3}+(72-24\sqrt{3})^{1/3}+4^{1/3}\Big)^{1/3} } { \big((48\sqrt{3}-72)^{1/3}+(16\sqrt{3}-16)^{1/3}+3\sqrt{3}-3\big)^{1/9} }\frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$$

$$\eta(27i)=\frac{\Big(-3^{5/6}*a^{5/9}+\big(\sqrt{2}-\sqrt{3}*a^{1/6}\big)^{1/3}*\big(2*\sqrt{3}*a^{1/6}+\sqrt{2}\big)\Big)^{1/3}} {3^{23/72}*a^{11/108}}\frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$$

$$a=2-\sqrt{3}$$

$$\eta(28i)=\frac{1}{2^{29/16}*7^{7/16}} \frac{ \sqrt{ 8*\sqrt{m-n}+ ( \sqrt{n+p}-\sqrt{q-r} )^{3}} -2*\sqrt{2}*(m-n)^{1/4} } {a^{1/4}*(m-n)^{1/12}}*\frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$$

$$a=\sqrt{2}-1;m=\sqrt{83-31*\sqrt{7}};n=\sqrt{3*\sqrt{7}-7};p=\sqrt{5-\sqrt{7}};$$

$$q=\sqrt{27*\sqrt{7}-7};r=\sqrt{7*\sqrt{7}+21}.$$

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

where $$\varphi$$ is golden ratio.

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

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

where $$\varphi$$ is golden ratio

$$b=6+\sqrt{35}$$

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

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

$$\eta(36i)=\frac{a^{1/12}}{2^{29/16}*3^{5/6}*b^{1/18}}\Big(\sqrt{a}*\big(\sqrt{2}-\sqrt{3}*b^{1/6}\big)-3^{1/4}*\sqrt{e}*b^{5/16}*c^{11/8}*d^{1/4}\Big)^{1/3}\frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$$

$$a=\sqrt{2}-1; b=2-\sqrt{3}; c=\sqrt{2}-3^{1/4}; d=\sqrt{3}-\sqrt{2}; e=3^{1/4}-1.$$

$$\eta(40i)=\frac{\sqrt{d*e}} {2^{77/32}*\sqrt{5*\varphi}*a^{3/8}} (\sqrt{b}*a^{1/4}*c^{1/4}*\varphi^{3/4}-d^{3/2})*\frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$$

$$a=\sqrt{2}-1; b=5^{1/4}-\sqrt{2}; c=\sqrt{10}-3; d=\frac{5^{1/4}-1}{\sqrt{2}};e=2^{1/4}-1; \varphi=\frac{\sqrt{5}+1}{2}.$$

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

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

$$\eta(81i)=\frac{\Big(-3^{17/18}*a^{5/27}*b^{4/3}+\sqrt{ 2}\big(-3^{5/6}*a^{5/9}+\sqrt{2}*b^{1/3}*(\sqrt{6}*a^{1/6}+1)\big)^{1/3}*(\sqrt{2}*3^{5/6}*a^{5/9}+\sqrt{6}*a^{1/6}*b^{1/3}+b^{1/3})\Big)^{1/3} } {2*3^{49/27}*a^{19/162}*b^{1/9}}\frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$$

$$b=\sqrt{2}-\sqrt{3}*a^{1/6};$$ $$a=2-\sqrt{3}.$$

• 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 – reuns Feb 13 at 4:40
• 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! – giuseppe mancò Feb 13 at 10:43
• 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). – Paramanand Singh Feb 17 at 8:57
• You're very kind! Thank you – 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.
• Thanks for the suggestions, now I try to calculate $\eta (21i)$. I just posted $\eta (32i)$. – giuseppe mancò Feb 17 at 10:31
• @giuseppemancò Why do you care of $\eta(ni)$ for a few particular values of $n$, why not look instead at the general formulas valid for every $n$ ? Do you know that $\Bbb{Q}(i,j(Ni))/\Bbb{Q}(i)$ is an abelian extension and how to find its minimal polynomial ? – reuns Apr 29 at 0:05