Another algebraic connection from a Ramanujan theorem.

66 Views Asked by At

I think I have interpreted a Ramanujan theorem well, but I ask someone for help to confirm it. In the Notebook II of Ramanujan, we read: enter image description here

Berndt (Vol. IV p. 24 Entry 13), algebraically demonstrates the theorem. However, as in , there is also a reading key here.

The reading key is as follows:

$x=\varphi(q)$

$y=\varphi(q^{4})$

$z=(\varphi(q^{2}))^{2}$

and the theorem becomes:

"If $\varphi(q)=\varphi(q^{4})+\sqrt{(\varphi(q^{2}))^{2}-(\varphi(q^{4}))^{2}}$

then

$2 \varphi(q^{4})=\varphi(q)+\sqrt{2(\varphi(q^{2}))^{2}-(\varphi(q))^{2}}$

I think I interpret these equations in the following way;

from the second degree modular equation:

$\frac{(\varphi(q))^{2}} {(\varphi(q^{2}))^{2}}=1+\alpha_{4n}^{1/2}=\frac{2} {1+\beta_{n}^{1/2}}$

and substituting the moduli you get the fourth degree modular equation:

"If $\sqrt{m}=1+\alpha_{16n}^{1/4}$

then

$\frac{2} {\sqrt{m}}=1+\beta_{n}^{1/4}$

where $m$ is the multiplier $\frac{(\varphi(q))^{2}}{(\varphi(q^{4}))^{2}}$

1

There are 1 best solutions below

0
On BEST ANSWER

You are correct in your calculations. The two Ramanujan equations are both equivalent to my $\ \eta$-product identity $\ q_{16,,22,124a}.\ $ The second and fourth degree modular equations seem correct also.