tag:blogger.com,1999:blog-9004452969563620551.post1058945243644426266..comments2019-05-15T12:04:42.529+05:30Comments on Paramanand's Math Notes: Ramanujan's Class InvariantsUnknownnoreply@blogger.comBlogger2125tag:blogger.com,1999:blog-9004452969563620551.post-83898044116844652362014-09-12T09:37:33.400+05:302014-09-12T09:37:33.400+05:30@Anonymous, I have already given the solution for ...@Anonymous,<br />I have already given the solution for the formula $$e^{\pi/6} = 2^{3/8}\prod_{k = 1}^{\infty}(1 + e^{-2k\pi})$$ on MSE(http://math.stackexchange.com/a/923319/72031). Only thing to note is that this formula is wrong. We need to replace $\pi/6$ with $\pi/12$ and here $g_{4}$ is involved and not $G_{4}$ as you think.<br /><br />The second product $$e^{\pi/4} = \sqrt{2(1 + \sqrt{2})}\prod_{k = 1}^{\infty}\frac{1 + e^{-4k\pi}}{1 + e^{-2(2k - 1)\pi}}$$ requires more calculation. We can proceed as follows \begin{aligned}S &amp;= \prod_{k = 1}^{\infty}\frac{1 + e^{-4k\pi}}{1 + e^{-2(2k - 1)\pi}}\\<br />&amp;= \prod_{k = 1}^{\infty}\frac{1}{1 - e^{-4(2k - 1)\pi}}\prod_{k = 1}^{\infty}\frac{1}{1 + e^{-2(2k - 1)\pi}}\\<br />&amp;= \frac{1}{2^{1/4}e^{-\pi/6}g_{16}}\cdot\frac{1}{2^{1/4}e^{-\pi/12}G_{4}}\\<br />&amp;= \frac{e^{\pi/4}}{2^{3/4}g_{4}G_{4}^{2}}\end{aligned} If we calculate the value of $g_{4}, G_{4}$ then we get the desired formula.Paramanand Singhhttps://www.blogger.com/profile/03855838138519730072noreply@blogger.comtag:blogger.com,1999:blog-9004452969563620551.post-91495808323470072762014-09-11T16:21:10.465+05:302014-09-11T16:21:10.465+05:30Thanks so much for the reply on stackexchange. I ...Thanks so much for the reply on stackexchange. I see now that Ramanujan&#39;s equations produce the relation inquired about. I was curious as to whether they could produce relations that require that k, or the exponents, be even integers as well as odds.<br /> <br />As an example:<br /> e^(pi/6) = 2^(3/8) * PI [1 + e^(-2kpi)] with k = 1 to inf. <br /><br />and also:<br /> e^(pi/4) = [2*(2^.5 + 1)]^.5 * PI [1 + e^(-4kpi)] / PI [1 + e^(-2(2k-1))] with k = 1 to inf. <br /><br />This second relation of e^(pi/4) with the value [2*(2^.5 +1)]^.5 uses the series (2,6,10,14,18...) and 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 even integers.<br /><br />I am sorry for my poor formatting. Thanks again. <br />Anonymousnoreply@blogger.com